2

3
INTRODUCTORY
CHEMISTRY

4
INTRODUCTORY
CHEMISTRY

KEVIN REVELL
Murray State University

5
Vice President: Ben Roberts
Editorial Program Director: Brooke Suchomel
Program Manager: Beth Cole
Marketing Manager: Maureen Rachford
Marketing Assistant: Savannah DiMarco
Market Development Manager: Leslie Allen
Development Editor: Erica Champion
Assistant Editor: Allison Greco
Director of Content: Kristen Ford
Content Development Manager: Stacy Benson
Director of Design, Content Management: Diana Blume
Senior Design Manager: Blake Logan
Director, Content Management Enhancement: Tracey Kuehn
Art Manager: Matthew McAdams
Managing Editor: Lisa Kinne
Senior Content Project Manager: Harold Chester
Photo Editor: Cecilia Varas
Photo Researcher: Bruce Carson
Media Project Manager: Daniel Comstock
Senior Workflow Project Manager: Paul Rohloff
Production Supervisor: José Olivera
Composition: Lumina Datamatics, Inc.
Illustrations: Troutt Visual Services
Cover Photo: Masterfile

Library of Congress Control Number: 2017935153

ISBN 978-1-319-08195-9 (print)

ISBN 978-1-319-20175-3 (epub)

© 2018 by W. H. Freeman and Company

All rights reserved

First Printing

W. H. FREEMAN AND COMPANY

One New York Plaza
Suite 4500
New York, NY 10004-1562
www.macmillanlearning.com

6
Brief Contents
Preface

CHAPTER 1 Foundations

CHAPTER 2 Measurement

CHAPTER 3 Atoms

CHAPTER 4 Light and Electronic Structure

CHAPTER 5 Chemical Bonds and Compounds

CHAPTER 6 Chemical Reactions

CHAPTER 7 Mass Stoichiometry

CHAPTER 8 Energy

CHAPTER 9 Covalent Bonding and Molecules

CHAPTER 10 Solids, Liquids, and Gases

CHAPTER 11 Solutions

CHAPTER 12 Acids and Bases

CHAPTER 13 Reaction Rates and Equilibrium

CHAPTER 14 Oxidation-Reduction Reactions

CHAPTER 15 Organic Chemistry and Biomolecules

CHAPTER 16 Nuclear Chemistry

7
Answers to Odd-Numbered Problems

Glossary

Index

8
Contents
Preface

CHAPTER 1
Foundations
Taxol
1.1 Chemistry: Part of Everything You Do
1.2 Describing Matter
Composition and Structure
Pure Substances and Mixtures
States of Matter
Properties and Changes
1.3 Energy and Change
1.4 The Scientific Method
Summary
You Can Do This
Key Terms
Additional Problems

9
CHAPTER 2
Measurement
A Strange Death
2.1 Measurement: A Foundation of Good Science
Scientific Notation: Working with Very Large and Very
Small Numbers
Converting from Scientific Notation to Standard Notation
Converting from Standard Notation to Scientific Notation
Calculations Involving Scientific Notation
Units of Measurement
Describing the Quality of Measurements
Describing Precision: Significant Digits
Determining the Number of Significant Digits in a
Measurement
Working with Exact Numbers
Using Significant Digits in Calculations
2.2 Unit Conversion
Dimensional Analysis
Problems Involving Multiple Conversions
Converting between Volume Units
Converting Units Raised to a Power
2.3 Density: Relating Mass to Volume
Converting between Mass and Volume

10
 Will It Float?
2.4 Measuring Temperature
Summary
A Strange Death
Key Terms
Additional Problems

CHAPTER 3
Atoms
Mercury Contamination from Small-Scale Gold Mining
3.1 Atoms: The Essential Building Blocks
Uncovering the Atom: From Democritus to Dalton
Can We See Atoms?
3.2 The Periodic Table of the Elements
Regions of the Periodic Table
Metals, Nonmetals, and Metalloids
Groups (Families) of Elements
3.3 Uncovering Atomic Structure
The Discovery of Charged Particles
The Discovery of the Nucleus
3.4 Describing Atoms: Identity and Mass
Atomic Number and Mass Number

11
 Average Atomic Mass
3.5 Electrons—A Preview
The Bohr Model and the Quantum Model
The Formation of Ions
Summary
Mercury: Ancient Treasure, Modern Toxin
Key Terms
Additional Problems

CHAPTER 4
Light and Electronic Structure
Edging toward Solar Energy
4.1 The Electromagnetic Spectrum
Wavelength and Frequency
The Energy of a Photon
4.2 Color, Line Spectra, and the Bohr Model
Color and Line Spectra
The Bohr Model
4.3 The Quantum Model and Electron Orbitals
The Uncertainty Principle and the Wave Nature of
Electrons
Energy Levels and Sublevels
4.4 Describing Electron Configurations

12
 Valence Electrons and the Octet Rule
Electron Configurations for Larger Atoms
Electron Configurations for Ions
4.5 Electron Configuration and the Periodic Table
Summary
Solar Cells: Converting Light into Electric Current
Key Terms
Additional Problems

CHAPTER 5
Chemical Bonds and Compounds
An Unexpected Combination: Lithium Carbonate and Bipolar
Disorder
5.1 Lewis Symbols and the Octet Rule
5.2 Ions
Cations: Ions with a Positive Charge
Naming Cations
Anions: Ions with a Negative Charge
Naming Anions
Polyatomic Ions
Naming Polyatomic Ions
A Summary of the Common Ions
5.3 Ionic Bonds and Compounds

13
 Ionic Bonds and Ionic Lattices
Predicting Formulas for Ionic Compounds
Naming Ionic Compounds
5.4 Covalent Bonding
Nonmetal–Nonmetal Bonds
Covalent Compounds
Naming Covalent Compounds
5.5 Distinguishing Ionic and Covalent Compounds
5.6 Aqueous Solutions: How Ionic and Covalent Compounds
Differ
5.7 Acids—An Introduction
Naming Acids
Binary Acids
Oxyacids
Summary
Continuing Cade’s Work
Key Terms
Additional Problems

CHAPTER 6
Chemical Reactions
Lost Cities of the Maya
6.1 Chemical Equations

14
 Balancing Equations
Strategies for Balancing Equations
Equations with Phase Notations
6.2 Classifying Reactions
6.3 Reactions between Metals and Nonmetals
6.4 Combustion Reactions
6.5 Reactions in Aqueous Solution
Representing Dissociation: Molecular and Ionic
Equations
Solubility Rules and Precipitation Reactions
Acid-Base Neutralization Reactions
Acids and Bases
Neutralization Reactions
Summary
The Mayan Lime Cycle
Key Terms
Additional Problems

CHAPTER 7
Mass Stoichiometry
Process Development
7.1 Formula Mass and Percent Composition
Formula Mass

15
 Percent Composition
How Chemists Measure Formula Mass and Percent
Composition
Mass Spectrometry
Elemental Analysis
7.2 Connecting Atomic Mass to Large-Scale Mass: The Mole
Concept
Avogadro’s Number and the Mole
Converting between Grams, Moles, and Particles
7.3 The Mole Concept in Balanced Equations
Stoichiometry Problems
Gram-to-Gram Questions
Strategies for Solving Stoichiometry Problems
Calculations with Limiting Reagents
Finding the Leftovers
7.4 Theoretical and Percent Yield
Summary
Process Development, Continued
Key Terms
Additional Problems

CHAPTER 8
Energy
The Corn Ethanol Debate
16
 8.1 Energy, Work, and Heat
Units of Energy
Heat and Work in Chemical Changes
Endothermic and Exothermic Changes
The Law of Conservation of Energy
8.2 Heat Energy and Temperature
Specific Heat and Heat Capacity
Calorimetry: Measuring Heat Flow
Coffee Cup Calorimetry
Bomb Calorimetry
8.3 Heat Energy and Chemical Reactions
Fuel Value
Reaction Enthalpy
Summary
Gasoline or Ethanol—Which Fuel Is Better?
Key Terms
Additional Problems

CHAPTER 9
Covalent Bonding and Molecules
The Shortest Race

17
 9.1 Covalent Molecules
Representing Covalent Structures
Exceptions to the Octet Rule
Drawing Lewis Structures
9.2 Molecules and Charge
Polyatomic Ions and Formal Charge
Drawing Lewis Structures for Polyatomic Ions
Choosing the Best Lewis Structure
Resonance
9.3 Shapes of Molecules
9.4 Polar Bonds and Molecules
Electronegativity and Polar Covalent Bonds
Molecules with Dipoles
Identifying Molecules with a Net Dipole
How Dipoles Affect Properties—A Preview
Summary
Building a Nanocar
Key Terms
Additional Problems

CHAPTER 10
Solids, Liquids, and Gases
The North Dakota Boom

18
10.1 Interactions between Particles
10.2 Solids and Liquids
Ionic Substances
Metallic Substances
Molecular Substances
Dipole–Dipole Interactions
Hydrogen Bonding
London Dispersion Forces
Summarizing Intermolecular Forces
Covalent Networks and Polymers
10.3 Describing Gases
Pressure
Measuring Pressure
10.4 The Gas Laws
Boyle’s Law
Charles’s Law
Using Charles’s Law to Find Absolute Zero
Solving Volume–Temperature Problems Using Charles’s
Law
The Combined Gas Law
Avogadro’s Law
The Ideal Gas Law
Mixtures of Gases
A Molecular View of the Gas Laws
10.5 Diffusion and Effusion
10.6 Gas Stoichiometry
Summary

19
 Rethinking Gas Storage
Key Terms
Additional Problems

CHAPTER 11
Solutions
The Perfect Cup of Coffee
11.1 Describing Concentration
Concentration by Percent
Percent by Mass and Volume
Mass/Volume Percent
Very Dilute Solutions: ppm and ppb
Molarity
Preparing Solutions of Known Molarity
Preparing Dilute Solutions
Using Square Brackets to Represent Concentration
11.2 Electrolyte Solutions
Electrolyte Concentrations
Colligative Properties
Freezing Point Depression
Boiling Point Elevation
Osmotic Pressure
11.3 Reactions in Solution—A Review and a Preview

20
 Precipitation Reactions
Acid-Base Neutralization Reactions
Metal Displacement Reactions
11.4 Solution Stoichiometry
Gravimetric Analysis
Advanced Stoichiometry Problems
Summary
Amazing Coffee—The Importance of Concentration
Key Terms
Additional Problems

CHAPTER 12
Acids and Bases
Cocaine: Ruin and Recovery
12.1 Introduction to Acids and Bases
The Arrhenius Definition
Polyprotic Acids
The Brønsted-Lowry Definition
12.2 Acid-Base Equilibrium Reactions
Are Conjugate Bases Basic?
12.3 Reactions Involving Acids and Bases
Neutralization Reactions
Reactions of Acids with Metal

21
 Formation of Acids from Nonmetal Oxides
12.4 Acid and Base Concentration

Concentrations of H+ and OH− in Aqueous Solutions

Math Review: Exponential and Scientific Notation
Connecting [OH−] and [H+]
The pH Scale
12.5 Measuring Acid and Base Concentration
Determining pH in the Laboratory
Acid-Base Titrations
Acid-Base Titrations with Different Coefficients
12.6 Buffers and Biological pH
Summary
The War on Drugs, Then and Now
Key Terms
Additional Problems

CHAPTER 13
Reaction Rates and Equilibrium
The Haber-Bosch Process
13.1 Reaction Rates
How Concentration and Temperature Affect Reaction
Rates
How Changes in Energy Affect Reaction Rates

22
 Describing Energy Changes in Chemistry: Reaction
Energy Diagrams
Catalysts
13.2 Equilibrium Reactions
13.3 Equilibrium Expressions
Equilibrium Expressions Involving Solvents
Equilibrium Expressions Involving Solids
Solubility Products
Equilibrium Expressions Involving Gases
13.4 Le Chatelier’s Principle
Equilibrium and Concentration
Equilibrium and Temperature
Equilibrium and Pressure
Summary
Miracles and Monstrosities: The Brutal Ironies of Fritz Haber
Key Terms
Additional Problems

CHAPTER 14
Oxidation-Reduction Reactions
Volta’s Marvel
14.1 Oxidation and Reduction
Oxidation Numbers

23
 14.2 Types of Redox Reactions
Reactions of Metals with Nonmetals
Combustion Reactions
Metal Displacement Reactions
The Activity Series
Reactions of Metals with Acid and Water
14.3 Half-Reactions and Batteries
Half-Reactions
Batteries
14.4 Balancing Redox Equations
14.5 Other Applications of Redox Reactions
Electroplating
Fuel Cells
Summary
Charging Ahead: Batteries Today and Tomorrow
Key Terms
Additional Problems

CHAPTER 15
Organic Chemistry and Biomolecules
Forming New Bonds: The Grubbs Catalyst
15.1 Organic Chemistry and the Carbon Cycle
15.2 Covalent Bonding with Carbon and Other Nonmetals

24
15.3 Drawing Covalent Structures
Condensed Structures
Skeletal Structures
15.4 Major Functional Groups
Hydrocarbon Functional Groups
Alkanes and Cycloalkanes
Alkenes and Alkynes
Aromatic Compounds
Oxygen-Containing Functional Groups
Alcohols and Ethers
Carbonyl Groups
A Summary of Oxygen-Containing Groups
Nitrogen-Containing Functional Groups
15.5 Polymers and Plastics
15.6 Biomolecules—An Introduction
Carbohydrates
Amino Acids and Proteins
Amino Acids
Peptides and Peptide Bonds
DNA
Summary
How Catalysts Work
Key Terms
Additional Problems

25
CHAPTER 16
Nuclear Chemistry
Fukushima
16.1 Nuclear Changes
The Nucleus—A Review
Nuclear Reactions
16.2 Radioactivity
Types of Radioactive Decay
Alpha Decay
Beta Decay
Gamma Decay
Radioactive Decay Series
Half-Life
Health Effects of Radiation Exposure
Measuring Radiation
Common Exposure Levels
Uses of Radioactive Nuclides
Uses in Medicine
Uses in Geology and Archaeology
16.3 Energy Changes in Nuclear Reactions
Mass Defect, Binding Energy, and Einstein’s Famous
Equation
Nuclide Stability
16.4 Nuclear Power: Fission and Fusion

26
 Fission
Uranium Enrichment
Fission Reactor Design
Waste from Nuclear Fission
Fusion
Replicating Fusion on Earth
Are We There Yet?
Summary
Powering the Future
Key Terms
Additional Problems
Answers to Odd-Numbered Problems
Glossary
Index

27
A word from the author

KEVIN REVELL teaches introductory, general, and organic chemistry at Murray

State University and also serves as the assistant dean for the MSU Jones College of
Science, Engineering, and Technology. A passionate educator, his teaching
experience includes high school, community college, small private, state
comprehensive, and state flagship institutions. His work encompasses curriculum,
technology-enhanced pedagogy, assessment, and active-learning design. He has
hosted multiple science education workshops and is the senior editor for
flippedchemistry.com, an online community for college-level instructors
implementing active-learning pedagogies. A synthetic chemist by training, his
research involves the synthesis and evaluation of functional organic materials.
With his wife, Jennifer, Kevin has three children—James, Julianne, and Joshua—
and two grandchildren.

Welcome to Introductory Chemistry!

For many students, introductory chemistry is a general-education

requirement en route to a degree in education, business, or a liberal-arts
field. For others, it is a stepping stone toward a challenging general
chemistry course and a career in health care, agriculture, or even science
and engineering. Some are traditional students, but many others are
nontraditional students balancing jobs, family, and the dream of
completing a college degree. And for many, chemistry can seem elusive,
mystical, and intimidating.

My vision is to make chemistry accessible to these students, not just

through a textbook, but through an integrated learning experience that

28
addresses different learning styles and draws on a variety of pedagogical
techniques to engage and challenge students.

Let’s begin with the text. I’ve tried to write in a friendly, casual style—
using analogies, stories, and images to make the important ideas stick.
I’ve blended this with digital interactives to create an active reading
experience. In some of these interactive figures, you can explore chemical
changes, choosing different substances to see how they react. In others,
you’ll be able to practice key knowledge and skills through simple games.

For instructors, Introductory Chemistry is more than a textbook—it’s a

complete curriculum, suited for traditional, flipped, or blended active-
learning classrooms. I’ve created video lectures for every section, with
corresponding PowerPoint decks that you can modify as you see fit. I’ve
included in-class activities, clicker questions, and speed drills, developed
over many semesters. Think of it as a tool belt, equipping you with the
curriculum to suit the needs of your classroom.

The pages that follow describe many of these features in more detail.
Whether you are teaching a class, taking a class, or just exploring
chemistry for the first time, Introductory Chemistry is designed to help
you achieve your goals. You can do this—let’s get started.

Best wishes,

Introductory Chemistry, First Edition I Kevin Revell, Murray State

University
December 2017 (©2018), Cloth: 978-1-319-08195-9, Loose-leaf: 978-1-
319-13390-0
For your review copy, contact your local Macmillan Learning
representative or visit macmillanlearning.com/requestrevell1e.

29
Acknowledgments
The Introductory Chemistry textbook and curriculum result from the
combined efforts of a fabulous team, without whose abilities and
dedication this project could not have happened. I am grateful to all those
at W.H. Freeman, at Macmillan Learning, and at Roberts and Company
that have brought this effort to fruition.
First, thanks to Ben Roberts, who got this vision off the ground, and
who supported it throughout its development. Thanks also to Erin
Mulligan and Beth Marsh for their indispensable guidance through the
early stages of this project.
I could not imagine this project without the leadership of Beth Cole.
She has fiercely advocated for the vision of this project, and her attention
to a thousand details has kept the project on schedule. I am grateful for her
honest and constructive feedback, for her willingness to listen, and for her
exceptional ability to see the big picture among the details.
Erica Champion has been an amazing developmental editor, who I’ve
enjoyed working with every step of the way. From the outset, I’ve
appreciated her ability not just to see problems, but to propose solutions.
I’ve relied on her perspectives and insights throughout this text. I’m so
appreciative of her tireless work to keep us on schedule and to manage the
back-and-forth of the revision and transmittal process. Her eye for layout
and pattern is superb. And through it all, her good nature and sense of
humor have made the process far more fun than I would ever have
expected.
I’m grateful to Maureen Rachford for her work in articulating and
refining the vision of this project. She has been vital not only to this
project, but also to the Flipped Chemistry community, which would not
exist today were it not for her tenacity in advocating for it.
I’m also grateful to Leslie Allen for her enthusiasm and dedication in
promoting this project, managing reviews, and coordinating the faculty
advisory board. She has spent many hours gathering feedback from
educators across North America. Her insights have fueled this project, and
will continue to do so for years to come.
Several others have been instrumental to the text: Thanks to Chad
Snyder for producing the problem solutions found at the end of this
textbook and in the solutions guide. Thanks to Erin Inks for managing and
tracking our progress on multiple fronts throughout the media-creation

30
process. Thanks to Donna Brodman for coordinating the reviews, and to
Allison Greco for her work in reviewing and editing the text, as well as
coordinating so many details. And thanks to Chris Thillen for her
meticulous copyediting.
I appreciate the support of the senior leadership team at Macmillan
Learning, including Ken Michaels, Susan Winslow, Brooke Suchomel, and
Lindsey Jaroszewicz.
From its conception, this project was designed for digital
environments. I owe many thanks to the media team, who have made this
dream possible. First, thanks to Kristen Ford, who has directed the
workflow and supported this project from the outset.
Among the media team, special thanks are due to the chemists at
Sapling Learning, whose expertise has enriched and refined this project.
Thanks to Kelly Lancaster, who has served the sometimes challenging
position of liaison between an author’s vision and a designer’s reality. She
has tenaciously advocated for approaches that make sense both
pedagogically and practically. Thanks also to Stacy Benson, for his work
in designing and overseeing the Watch Explanation videos that are such an
integral part of this project, and for his work to integrate the text into the
Sapling environment. Thanks to Kris Hiebner, who coordinated the
animations, who built the Sapling questions to align with the text, and
whose proofreading and meticulous accuracy checks have been vital to the
integrity of the text. And thanks to Heather Southerland, who coordinated
the chemical reaction videos that make the Cause and Effect interactives
so engaging.
Many others have played vital roles in the development of the media
elements: Thanks to Angela Piotrowski and Jamie Carberry for their
voicework in the solutions videos, and to Andrew Waldeck for his work in
scripting and animating the solutions. Thanks to Alex Gordon, who
produced the excellent artwork in the animations with the help of
animators Sara Egner, Clarissa Cochran, Tommy Turner, and Cheryl
McCutchan. Thanks to the SAVI (Simulations, Animations, Videos, and
Interactives) team at Macmillan: Damien Brockmann, Jose Gomez, Jerry
Lee, Lindsay McMahon, Hope Miller, Kincade Park, and Jeff Sims.
Thanks also to the user experience team: Alex Britez, Daniel Cole, and
Juliet Dyer.
I appreciate the efforts of those who brought together the artwork for
this project. Thanks to Cecilia Varas and Bruce Carson for their patient
tenacity in obtaining photos for the textbook and ancillaries. Thanks to
Troutt Visual Services for the illustrations for this text, to Blake Logan
who produced the cover and text design, and to Matthew McAdams who

31
served as the art manager.
Many thanks also to the content team: Harold Chester, Diana Blume,
Tracey Kuehn, Paul Rohloff, José Olivera, and the team at Lumina
Datamatics.
Outside of the Macmillan Learning team, there are many others whose
contributions I very much appreciate:
I would like to thank my colleagues, friends, and family whose
research, stories, insights, and talents have contributed to this project. This
includes Lauren Waugh, Adam Kiefer, Isabel Villaseñor, Yareli Jáidar,
Jared Fennell, James Tour, Christian Joachim, Leonhard Grill, Francesca
Moresco, Alexiane Agullo, Omar Yaghi, Jarrod Eubank, Aaron Blanco,
Susan Hurley, Kimberly Revell, Gary and Cindy Deaton, Julie Revell,
Daniel Johnson, Robert Grubbs, and Jack Revell.
I would like to extend my thanks to the many colleagues who reviewed
the textbook. Your thoughtful feedback has shaped and polished this
project. I especially want to think Brandon Tenn (Merced College) for his
enthusiasm, unwavering support, and very thoughtful feedback. I also want
to thank Paul Forster (University of Nevada, Las Vegas), whose very
perceptive observations brought me a good deal of clarity early on. I am
also thankful to the others on the Faculty Advisory Board: Loren
Barnhurst (Southern Adventist University), Kristin Clark (Ventura
College), Kate Hayden (Birmingham-Southern College), Tamari
Narindoshvili (Blinn College), and Alison Soult (University of Kentucky).
Thanks also to Shawn Batchelder, who conducted many of these faculty
interviews.
I am also grateful to my colleagues at Murray State: Thanks to Ricky
Cox, who continues to inspire my teaching. Thanks to Harry Fannin, Steve
Cobb, and Bommanna Loganathan for their unwavering support. And
thanks to Angela Guyton, Carruth Kittrell, and Kim Hall, who make
everything in the office work so smoothly.
I would also like to acknowledge key mentors whose efforts years ago
made this possible, including Bob Chandler (Lake Gibson High School),
Mark Trudell (University of New Orleans), Walt Trahanovsky (Iowa State
University), Bob Herron (Southeastern University), Deborah Hazelbaker
(Southeastern University), and especially Ed Turos (University of South
Florida).
Finally, thanks to my family, who have supported and encouraged me
so much through this process. Thanks especially to Jennifer, for her
constant love, encouragement, and sacrificial support to make this dream
happen.

32
We thank the many reviewers who aided in the development of this
text.
Samuel Melaku Abegaz, Columbus State University
Mireille J. Aleman, Palm Beach Atlantic University
John U. Alexander, Marymount California University
Jeffrey Allison, Austin Community College
Sarah Alvanipour, Houston Community College
Vicki Amszi, Blinn College
Ilija Arar, Cayuga Community College
Premilla Arasasingham, Moorpark College, El Camino College
Luis Avila, Columbia University
Yiyan Bai, Houston Community College
C. Eric Ballard, The University of Tampa
Brian F. Bartlett, Central New Mexico Community College
Vladimir A Benin, University of Dayton
Stacey-Ann Benjamin, Broward College
Marguerite Healy Benko, Ivy Tech Community College
Terrence M. Black, Nassau Community College
Simon Bott, University of Houston
Ivanna Campbell, McLennan Community College
Ken Capps, College of Central Florida
Sevada Chamras, Glendale Community College
Kaiguo Chang, Metropolitan Community College
Kristin Clark, Ventura College
J. De Anda, Long Beach City College
Maria Cecilia de Mesa, Baylor University
Milagros Delgado, Florida International University
Dylan D. Drake, Wilhelm, Metropolitan Community College
Daniel J. Dwyer, University of Montana
Sarah Edwards, Western Kentucky University
Jack F. Eichler, University of California, Riverside
Julie Ellefson, Harper College
Melinda Findlater, Texas Tech
Paul Forster, University of Nevada, Las Vegas
C. Karen Fortune, Houston Community College
Steve Gentemann, Southwestern Illinois college
Jennifer N. Gotcher, Austin Community College
Pierre Goueth, MiraCosta College
Carol Green, St. Charles Community College
Lindsay Groce, Big Bend Community College
Yi Guo, Chandler Gilbert Community College

33
Christopher G. Hamaker, Illinois State University
Zachariah M. Heiden, Washington State University
Prof. Xiche Hu, University of Toledo
Thomas E. Janini, The Ohio State University Agricultural Technical
Institute
Eugenio Jaramillo, Penn State Altoona
Crisjoe Joseph, California State University, Channel Islands
Joshua Kellogg, University of North Carolina Greensboro
Edith Preciosa Kippenhan, University of Toledo
Joy Kobayashi, Ventura College
Sujatha Krishnaswamy, Chandler Gilbert Community College
Allison C. Lamanna, Boston University
Jean-Marie Magnier, Springfield Technical Community College
Arnulfo Mar, University of Texas, Rio Grande Valley
Matt Marlow, Nicholls State University
J. Aaron Matthews, Florida State College at Jacksonville
N. Alpheus Mautjana, Santa Fe College
Troy Milliken, Holmes Community College
Katherine A. Moga, The Ohio State University
Reza Mohseni, East Tennessee State University
Daniel Moriarty, Siena College
Christopher P. Morong, Minnesota State University, Mankato
Tamari Narindoshvili, Blinn College Bryan Campus
Jung Oh, Kansas State University Polytechnic
Bruce Osterby, University of Wisconsin, La Crosse
Franklin Ow, East Los Angeles College
Monica Rabinovich, The University of North Carolina at Charlotte
Bhavna Rawal, Houston Community College
Mario C. Raya, Bristol Community College
Jimmy Reeves, University of North Carolina Wilmington
Anthony Revis, Saginaw Valley State University
Harrison Rommel, Central New Mexico Community College, Santa Fe
Community College
James Ross, East Los Angeles College
Emily Rowland, University of Mississippi
Gerald B. Rowland, University of Mississippi
Niladri Sarker, Rio Salado College
Einhard Schmidt, Santa Monica College
James Selzler, Ventura College
Mary Snow Setzer, University of Alabama in Huntsville
Amanda Charlton-Sevcik, Baylor University

34
Supriya Sihi, Houston Community College
Chad Snyder, Grace College
Brandon Tenn, Merced College
Susan Tansey Thomas, University of Texas at San Antonio
Harold Trimm, SUNY Broome
Lucas J. Tucker, Siena College
Melanie Veige, University of Florida
Elaine Vickers, Southern Utah University
B. Blairanne Williams, Western Kentucky University
William Williams, Hudson Valley Community College
Shaun M. Williams, Lenoir-Rhyne University
Neil M. Wolfman, Boston College
Kevin A. Wood, North Central Texas College
Marilyn Wooten, Trinity University
Tammy Wooten-Boyd, Community College of Philadelphia
Curtis Zaleski, Shippensburg University
Rong Zhang, University of South Florida
Chen Zhou, University of Central Missouri
Tom Zownir, Manchester Community College

35
 To the cloud of witnesses who urge me on:
To my grandparents, for their courage, dedication, perseverance, and
love.
To Mom and Dad, for their constant love and encouragement,
and for the tremendous example they have set.
To Gary and Cindy, and to Mom and Dad Kruger,
for their support and prayers through the years.

To my wife and best friend Jennifer, the love of my life,

looking forward to great things to come.

To my children, Jamie and Megan, Julie, and Joshua; and my

grandchildren,
Grace and Shawn: Run with patience the race set before you.

And to God our Father, who gives life and peace:

I dedicate this book in love and thankfulness.

36
Chapter One
Foundations

Taxol
Along the cool, coastal areas of the Pacific Northwest, a small evergreen tree
grows. Called the Pacific yew, it is a plain-looking tree with soft needles and red
berries (Figure 1.1). In 1962, a botanist named Arthur Barclay came across a stand
of these trees in the forests near Mount St. Helens, in Washington State. Barclay
worked for the U.S. Department of Agriculture, and he was searching for plants
that might provide new medicines.

Figure 1.1 (a) The Pacific yew tree contains a substance with powerful anticancer
properties. (b) Mansukh Wani (left) and Monroe Wall were the chemists who first
identified Taxol®. (c) In this model of Taxol®, black spheres represent carbon. Red

37
spheres represent oxygen, white are hydrogen, and blue is nitrogen. Lines between the
spheres illustrate the way the particles are joined together. (d) Susan Horwitz discovered
how Taxol® works. (e) Today, Taxol® is the leading treatment for many forms of
cancer.

Barclay collected a sample of twigs, needles, and bark from the trees. He sent
this sample and others like it to Mansukh Wani and Monroe Wall, two chemists in
North Carolina. Using a process much like brewing tea, Wani and Wall ground up
the plant matter and then mixed it with hot water and other liquids to pull material
out of the crushed wood and leaves. They next purified and tested these substances
to see how they would affect the growth of tumor cells. One substance, found only
in the bark of the Pacific yew, showed a remarkable ability to destroy the tumor
cells.
In 1971, after nearly a decade of work, Wani and Wall published a paper
describing the isolation, structure, and anticancer properties of this substance. They
called the substance taxol, after the scientific name for the Pacific yew, Taxus
brevifolia. A few years later, Susan Horwitz, a biochemist at Yeshiva University in
New York City, discovered that this substance prevented cancer cells from
reproducing. It did this in a way that had never been seen before.
Horwitz’s findings led other scientists to explore this substance further. By
1984, it had entered clinical trials (that is, testing on human patients). The
substance was given the trade name Taxol® and the common name paclitaxel.
However, the promise of Taxol created a new problem. The Pacific yew was
the only known source of the drug, and harvesting the bark killed the trees. The
head of the National Cancer Institute estimated that keeping up with the demand
for yew bark would require 360,000 trees per year—well beyond the capacity of
the slow-growing Pacific yew to replenish itself. What could be done?
This situation led to dozens of questions: How does the Pacific yew produce
Taxol? Do other trees also produce this substance? Can Taxol be produced
synthetically in a laboratory or factory? Is it possible to produce simpler substances
that mimic Taxol’s anticancer properties? By the early 1990s, research teams
across the world were searching for answers to these urgent questions.
A major breakthrough came when scientists discovered that the English yew, a
more common variety of the yew tree, produces a similar substance in its needles
that can be converted into Taxol in a laboratory. Because this process involves the
renewable needles rather than the bark, farmers can harvest the substance from
English yews without killing the trees.
More recently, researchers have learned to grow yew cells in the laboratory and
to collect the anticancer compound directly from the cells. Today, pharmaceutical
companies use a combination of these techniques to produce Taxol, which is
among the most widely used medicines for the treatment of cancer.
Think for a moment about the many facets of developing Taxol: From
harvesting the first yew bark to using it in hospitals across the world, the Taxol
story involved people whose expertise ranged from biology and medicine to
business and agriculture. It involved government agencies and private companies.

38
It involved conservation and innovation. And at the center of the story is a single
substance, buried in the bark of an obscure little tree. Through careful study, this
simple, naturally occurring substance changed the way we treat cancer and
improved the lives of people all over the world.

Intended Learning Outcomes

After completing this chapter and working the practice problems, you should be
able to:

1.1 Chemistry: Part of Everything You Do

Describe the impact of chemistry on a variety of other fields.

1.2 Describing Matter

Describe the difference between composition and structure.
Differentiate between elements, compounds, homogeneous mixtures, and
heterogeneous mixtures.
Describe the three phases of matter.
Compare and contrast physical and chemical properties and physical and
chemical changes.

1.3 Energy and Change

Define heat energy in terms of the motion of particles.
Describe the relationship between the potential energy of a system and its
potential for change.

1.4 The Scientific Method

Describe the key components of the scientific method.
Explain the differences between a hypothesis, a theory, and a scientific law.

39
1.1 Chemistry: Part of Everything You Do
The substances around us fill our lives with textures, flavors, sights, and
smells. From the aroma of a delicious meal to the color of the sunset, from
the hardness of a diamond to the softness of a favorite shirt, our world is
filled with materials that have widely varying properties. To understand
these properties, we must look closely at the smaller building blocks that
make up the materials. We must explore how these building blocks are
arranged and how they interact with the world around them.
Understanding what substances are made of and how they behave are
the essence of the field of chemistry. Regardless of your background, your
interests, or your career, your day-to-day activities involve chemistry. Are
you interested in working in a health-care field? Chemistry is essential to
understanding how our bodies function and how they interact with their
surroundings. Are you interested in physics, engineering, or construction?
What about computers and technology? The properties of trade materials
—from concrete and steel to semiconductors and solar cells—depend on
the composition and properties of the substances they are made of. Are
you interested in agriculture or the environment? Everything from
exploring rock formations to monitoring air, soil, and water quality
involves chemistry. Chemistry is sometimes referred to as “the central
science,” because it connects with every other field in the sciences (Figure
1.2). Whether your passion lies in art, music, business, or sports, you can
find ways that chemistry overlaps with these disciplines (Figure 1.3).

40
Figure 1.2 Chemistry connects to every other science and to our day-to-day lives.

41
Figure 1.3 Chemistry is all around us. The properties of different substances produce
the feel of a basketball, the sound from a fine violin, the subtle colors on a painting, the
lifesaving properties of a new medicine, and the flavor of great coffee.

In the remainder of this chapter, we’ll lay a foundation for

understanding chemistry. We’ll consider substances that exist all around us
and the building blocks from which they are made. We’ll look at the
changes that occur within these substances. We’ll discuss the idea of
energy and begin to see how energy drives change. Finally, we’ll look at
the method that scientists use as they explore chemistry and the other
sciences.
The chapters that follow build on this foundation. The goals of this
book are to help you gain a broad understanding of the key principles of
chemistry, to help prepare you for your chosen career, and to help you
connect chemical principles to phenomenon you observe all around you.
The world around us is fascinating; let’s explore it!

42
1.2 Describing Matter
The world around us is composed of matter. Matter is anything that has
mass and takes up volume. The water we drink, the air we breathe, the
ground under our feet: All of these are composed of matter. Although we
broadly defined chemistry in the preceding section, let’s begin here with a
slightly more formal definition: Chemistry is the study of matter and its
changes.
Our definition of chemistry covers some very broad questions. For
example, what is a substance made of? How are its components arranged
—that is, what is its structure? How does structure affect the properties of
a substance? What types of changes can a substance undergo? In the
coming chapters, we will explore the answers to these questions.

Composition and Structure

At the beginning of this chapter, we saw how scientists discovered a
substance in the Pacific yew tree that destroyed cancer cells. One of their
first challenges was to answer the question, “What is this substance made
of?” In other words, they wanted to determine the composition and
structure of the substance.
Composition refers to the simple components that make up the
material. Structure refers to both the composition and arrangement of
those simpler substances. For example, in Figure 1.4 we see a deck, a
canoe, and a baseball bat. Each of these has basically the same
composition: They are all made of wood. However, the wood is arranged
in different structures that help each item fulfill its own function.

Figure 1.4 A deck, canoe, and baseball bat share a similar composition, but each has a
different structure.

In chemistry, we ask similar questions on a smaller scale: “What is the

composition of wood?” That is, what is it made of? Or, “What is the
structure of wood?” That is, how are its components arranged? Answering

43
these questions helps us understand why substances behave the way they
do.

Pure Substances and Mixtures

To understand more about composition and structure, we begin with
atoms. Atoms are the fundamental units of matter. Scientists represent
atoms as tiny spheres, or by using one- or two-letter symbols, or by using
both together (Figure 1.5). We will discuss atoms in more detail in the
following chapters; for now, you can think of an atom as a tiny unit of
matter.

Figure 1.5 We represent atoms using a colored sphere, a one- or two-letter symbol, or a
combination of these two.

The simplest form of matter is an element. An element is a substance

that is made of only one type of atom. Common elements include gold,
silver, iron, oxygen, and nitrogen. Each element is composed of a unique
type of atom. For example, gold contains only gold atoms, and silver
contains only silver atoms.
Compounds are substances composed of more than one element,
bound together in fixed ratios. For example, water is made of two parts
hydrogen and one part oxygen. If the ratio of hydrogen to oxygen changes
at all from 2:1, it is no longer water, but some other substance.
Many compounds form groups of atoms called molecules. In
molecules, the atoms bind tightly together and behave as a single unit. For
example, a water molecule contains two hydrogen atoms and one oxygen
atom (Figure 1.6). Some elements also exist as molecules. In hydrogen,
nitrogen, and oxygen, the atoms of the element pair together to form
diatomic (“two-atom”) molecules (Figure 1.7).

44
Figure 1.6 Water is a compound composed of hydrogen and oxygen. In a water
molecule, two hydrogen atoms bind to one oxygen atom. To represent this idea, we can
use elemental symbols connected by lines; or, we can use spheres (balls) to signify each
atom. Chemists normally use red spheres to represent oxygen atoms and white spheres
to represent hydrogen atoms. See Appendix A2 for a list of colors used to represent
different atoms.

Figure 1.7 Elemental oxygen exists as diatomic molecules. A molecule of oxygen

contains two oxygen atoms joined together. This image depicts four oxygen molecules.

Figure 1.8 shows water flowing from a copper pipe. Copper is an

element, and so the only type of atom present in copper is the copper atom.
In contrast, water is a compound. Each molecule of water is composed of
two hydrogen atoms and one oxygen atom. Both the element copper and
the compound water are examples of pure substances: materials that are
composed of only one element or compound.

45
Figure 1.8 Copper is an element (composed of only one type of atom), but water is a
compound made of hydrogen and oxygen atoms.

In contrast to pure substances, mixtures contain more than one

substance—and the substances are not bound in a fixed ratio. For example,
brass and bronze are both mixtures of metals, known as alloys. Brass is a
mixture of the pure substances copper and zinc, while bronze is a mixture
of the pure substances copper and tin (Figure 1.9). In these mixtures, the
ratio of copper to the other metals is not fixed; it may be altered to affect
the properties of the alloy, such as hardness and color.

46
Figure 1.9 Bronze, a mixture of copper and tin, has been in use since nearly the
beginning of recorded history. The ratio of copper to tin may vary, leading to
differences in properties such as color, melting point, and hardness. This image is of an
Akkadian ruler, possibly Sargon (Syria, approximately 23rd century B.C.E.). Inset: An
atomic view of bronze, showing a mixture of copper and tin atoms.

Mixtures whose components are evenly blended throughout are called

homogeneous mixtures. Metal alloys, a cup of coffee, and the air around
us are all homogeneous mixtures. In contrast, heterogeneous mixtures
contain regions with significantly different composition (Figure 1.10).
Sand and water, gravel, and chocolate-chip cookie dough are all
heterogeneous mixtures. They each have regions where one component is
present in far greater proportion. Figure 1.11 summarizes the categories of
pure substances and mixtures.

47
Figure 1.10 (a) A mixture of salt and water is homogeneous. (b) A mixture of sand and
water is heterogeneous.

Figure 1.11 Matter can be classified by the arrangement of atoms. In these images, gold
and violet spheres represent metal atoms, red represents oxygen, blue represents
nitrogen, black represents carbon, white represents hydrogen, and green represents

48
chlorine.

A key difference between compounds and mixtures is that mixtures

can be separated into their individual components without changing the
identity of the substances. For example, chemists commonly use a
technique called filtration to separate heterogeneous mixtures. If you pour
a mixture of sand and water through a filter, the sand remains trapped in
the filter while the water passes through (Figure 1.12). The mixture has
separated, but the substances have not changed. By contrast, compounds
cannot be separated without changing the substances into their elemental
forms.

Figure 1.12 Mixtures can be separated. Chemists use filtration to separate

heterogeneous mixtures like sand and water.

Example 1.1 Elements, Compounds, and Mixtures: An Atomic

View
In the image shown here, the colored spheres represent different kinds of
atoms. Does this figure represent an element, a compound, or a mixture?

49
 In this figure, two species are present: One type is composed of only red
atoms—this is an element. The other type is composed of two different
types of atoms (red and black)—this is a compound. However, since both
species are present, this is a mixture containing both an elemental form and
a compound. •

IT
TRY

1. Describe each substance shown here as an element, compound, or

mixture. Remember that the colored spheres represent different kinds
of atoms.

2. Describe each of the following substances as an element, a compound,

a homogeneous mixture, or a heterogeneous mixture:
a. table salt, a substance made up of sodium atoms and chlorine atoms
bound in a 1:1 ratio
b. a fruit drink in which frozen concentrate is combined with water, then
stirred until it is evenly blended
c. a substance made up only of zinc atoms
d. blue cheese dressing, a creamy white sauce containing small chunks of
cheese

States of Matter

50
Matter typically exists in one of three forms, which we call the three states
of matter, or sometimes the three phases of matter. The three basic states
of matter are solid, liquid, and gas. We distinguish between these states by
properties that are visible on a macroscopic level (visible to the naked
eye):
Solids have both a definite shape and a definite volume.
Liquids have a definite volume, but no definite shape. Liquids
adopt the shape of their container.
Gases do not have a definite shape or a definite volume. Gases fill
any container they occupy.
If we heat a substance, it can transition from solid to liquid (melting)
or from liquid to gas (vaporization). If we cool it, the reverse occurs: The
gas changes to a liquid (condensation), and the liquid changes to a solid
(freezing). We can see its visible properties change in an obvious way as
the substance transforms between these states, but it also changes on the
atomic level.
Let’s use the element aluminum as an example. At room temperature,
aluminum is a solid. Its atoms are close to each other, packed into an
ordered framework (Figure 1.13). If we were able to “zoom in” to see the
atoms, we would notice something else: Even though this framework holds
each atom in place, the atoms are not perfectly still. Rather, they are
vibrating, moving within their spaces in the framework.

51
Figure 1.13 As a sample is heated, it transitions from solid to liquid to gas.

Explore

Figure 1.13

Next, let’s begin to heat the solid aluminum. When we do this, the
atoms vibrate faster. The more heat we add, the faster the atoms move.
Eventually, they begin to break out of the rigid framework and travel
freely past each other. The substance is now in the liquid state. The
particles in a liquid move randomly, but they remain close to each other.

52
 If we continue heating the liquid aluminum, the atoms move faster and
faster until they begin to break out of the liquid phase and enter the gas
phase. Particles in the gas phase move about freely and have very little
interaction with each other.

As a substance is heated, its particles move faster.

If we cool the aluminum, we observe the opposite effect: As the

temperature drops, the fast-moving gas particles slow and begin clustering
together. Particles drop to the bottom of the container, transitioning from
the gas phase to the liquid phase. Similarly, as we cool a liquid, the
particles move more and more slowly until eventually they settle back into
a solid framework.
The states of matter offer an important lesson: The behavior of any
substance is determined by the arrangement of the particles that compose
the substance. As the aluminum in the example transitioned from solid to
liquid to gas, the atoms did not change, but their arrangement did—with a
huge impact on how the substance behaves. Because of this, it is important
to think not just in terms of the macroscopic world, but also to ask, “What
is going on at the atomic level?” The better you can visualize motion and
interaction at the atomic level, the easier it is to understand and predict the
way a substance behaves.

The behavior of any substance is determined by the arrangement of the particles

in the substance.

Properties and Changes

Chemists study the properties of matter and also the ways that matter can

53
change. We commonly describe both properties and changes as being
physical or chemical.
Physical properties are the properties of a substance that we can
measure without changing the identity of the substance. For example, we
can measure color, temperature, mass, volume, shape, hardness, flexibility,
and a host of other factors without changing the composition of that
substance. Similarly, physical changes are changes that occur without
altering the identity of the substance. For example, we can take an iron bar,
melt it, pour it into a mold, and let it cool, solidifying into a new shape.
We have altered the shape and the phase of the iron, but it is still
composed of iron, and so each of these changes is a physical change.
In a physical change, the properties change but the composition stays the
same.

Phase changes are physical changes.

Chemical properties are properties that we cannot measure without

changing the identity of a substance. For example, flammability is a
chemical property. If we want to determine if something is flammable, we
must try to burn it. If it burns, it is no longer the original substance. When
a substance changes into something different, it has undergone a chemical
change. Chemical changes are also called chemical reactions. In some
chemical changes, elements combine to form compounds. For example,
when heated, zinc and sulfur combine to form a new compound, zinc
sulfide (Figure 1.14). In other chemical changes, compounds break apart
to produce elemental substances or rearrange to form new compounds. For
example, methane gas can react with oxygen gas to produce two new
compounds, carbon dioxide and water (Figure 1.15). The key idea is that a
chemical change always involves the formation of a different substance.

54
Figure 1.14 Zinc and sulfur react to form zinc sulfide. This is a chemical change.

Explore

Figure 1.14

55
Figure 1.15 The flame of a gas stove results from a chemical change. Methane
combines with oxygen gas to produce carbon dioxide gas and water.

In a chemical change, a new substance is formed.

Example 1.2 Physical and Chemical Properties

Describe each property as physical or chemical:
a. A paint has a deep blue color.
b. An iron bar heats up quickly if left out in the sun.
c. Zinc metal reacts with hydrogen chloride gas to form zinc chloride
and elemental hydrogen.
To determine whether a property is physical or chemical, we must ask
whether a new substance is formed. If the property involves the formation
of a new substance, it is a chemical property. If not, it is a physical
property.
The color of paint is a physical property. Observing its color does not
change its identity. Similarly, the fact that an iron bar heats up quickly is a
physical property. Although the temperature of the bar may change (a

56
 physical change), the bar is still composed of iron.
However, the reaction of zinc metal with hydrogen chloride gas involves
the formation of two new substances (zinc chloride and elemental
hydrogen). The fact that zinc reacts in this way is a chemical property. The
change that results is a chemical change. •

Example 1.3 Physical and Chemical Changes

Gold bars are produced by pouring molten gold into a mold, as shown here.
As the gold cools, its atoms transition from moving freely past each other
into a well-ordered framework. Is this a physical change or a chemical
change?

When gold transitions from liquid to solid, its arrangement of atoms

changes, but it does not form a new element or compound. Because no new
substance has been formed, this is not a chemical change. It is a physical
change. Phase changes are physical changes. •

IT
TRY

3. Each of these statements describes a property or change. Identify each

property or change as physical or chemical.
a. The element bromine has a deep red color.
b. Magnesium burns brightly in oxygen to form magnesium oxide.
c. A saltwater solution conducts electricity.
d. A copper slug has a mass of 15.2 grams.
e. Over time, iron hinges rust.

57
f. Your body converts glucose and oxygen into carbon dioxide and
water.

58
1.3 Energy and Change
We often describe chemical and physical changes in terms of their energy.
Energy is the ability to do work. We discuss energy in more detail in
chapter 8, but a foundational concept of energy is essential for
understanding chemistry. Broadly, energy takes two forms: kinetic and
potential. Potential energy refers to energy that is stored. Kinetic energy
is the energy of motion. The faster an object is moving, the greater the
kinetic energy it has.

At higher temperatures, particles have more kinetic energy.

Heat energy is a type of kinetic energy. Heat energy involves the

kinetic energy of the particles within a substance. When a substance is
heated, the particles within the substance vibrate or move more and more
quickly. For example, if a stove is hot, the particles on the hot surface
vibrate more rapidly than if the stove is cold. On the macroscopic level,
the stove doesn’t have kinetic energy (it is sitting still). But on the atomic
level, the particles on the surface of the stove are moving rapidly.
Physical and chemical changes involve changes in energy. Often, an
object or substance changes in a way that releases energy. For example, a
bicycle releases energy as it quickly rolls downhill (Figure 1.16). Changes
that release energy often happen spontaneously. On the other hand, when
we push the bicycle up the hill, we move it to a higher energy state through
an input of energy.

59
Figure 1.16 A bike spontaneously moves downhill to a lower energy state.

Energy changes take many forms. For example, a tree grows by

absorbing energy from the sun to convert two simple chemicals, carbon
dioxide and water, into plant material (Figure 1.17). The energy that the
plant harvests from the sun is stored in the plant material. If the tree falls,
we can use it for firewood—releasing the stored potential energy as heat
and converting the plant material back into carbon dioxide and water.

Figure 1.17 Energy is stored and released in many forms.

60
 In chemistry, we often describe substances as either high energy or
stable. When something has high energy (either kinetic or potential), it can
bring about a change. On the other hand, if something is stable, it has less
energy and is therefore less likely to react.

Substances with a large amount of potential energy are more likely to react.

For example, in Zimbabwe there is an extraordinary rock formation

(Figure 1.18). Time and erosion have carved away much of the soil
beneath the rocks, leaving them precariously balanced. Because these
rocks sit high off the ground, they have a good deal of potential energy.
Eventually, these rocks will fall. When they do, they will release this
stored energy. Right now, you probably wouldn’t want to camp underneath
the rocks. But once the rocks do fall, you have nothing more to worry
about—the energy has been released, and the rock formation will be
stable.

Figure 1.18 The Balancing Rocks in Zimbabwe contain potential energy.

Consider a second example: a mousetrap (Figure 1.19). If we put

energy in, we can bend back the spring and set the trap. If we touch the
trigger—whap! The bar snaps with a large force. Energy is stored in the set
trap, but once this energy is released, the trap is stable. We don’t have to
worry about getting our fingers snapped if the spring on the trap is not
coiled. If there’s no potential, there’s no problem.

61
Figure 1.19 Objects tend to move spontaneously from higher to lower potential energy.
Objects with lower potential energy are more stable. This happens with everyday
objects like roller coasters and mousetraps, but it also happens with atoms and
molecules.

Ethylene oxide (Figure 1.19c) is a simple compound that behaves a lot

like a mousetrap. It contains two carbon atoms and an oxygen atom, but
the structure of the compound is strained. If the compound encounters a
water molecule, it snaps open like the mousetrap to form a product that
contains less potential energy. The new product is lower in energy, so it is
less likely to react with other compounds.
In the coming chapters, we talk frequently about the energy or the
stability of matter. As a general rule, systems move toward the lowest
energy state possible. As we just noted, substances that have a low amount
of energy are said to be stable—they do not react as readily as high-energy
substances do.

The potential energy in any substance depends on its structure.

Many chemical changes absorb or release heat energy. An exothermic

change is one that releases heat energy. Wood burning on a campfire is an
example of an exothermic change. Changes that require energy to occur
are endothermic. For example, we must heat water to convert it to steam;
this is an endothermic change.

Example 1.4 Energy and Chemical Changes

When charcoal burns, the charcoal (which is composed of carbon) reacts
with oxygen to produce carbon dioxide and heat, as represented in the

62
 equation shown here. Which has the higher potential energy: carbon plus
oxygen, or carbon dioxide? Which is more stable? Is this reaction
endothermic or exothermic?
Carbon+oxygen→carbon dioxide+heat

This chemical change involves the release of energy. This means the
substance formed by the change (carbon dioxide) has less potential energy
than the substances that were present before the change (carbon and
oxygen). Since carbon dioxide has less potential energy, it is more stable.
And because this change released energy, we refer to it as an exothermic
change. •

IT
TRY

4. How do the particles in hot coffee differ from those in cold coffee?

5. When we heat water, it becomes steam. Is this an endothermic or

exothermic process? Which has the higher energy, liquid water or
steam?

6. Which has more stored potential energy, a new battery or a dead

battery? Which is more stable?

63
1.4 The Scientific Method
At the beginning of this chapter, we described the discovery of Taxol, a
cancer-fighting compound found in the bark of the Pacific yew tree. From
the discovery of the compound to its widespread use today, scientists
followed a systematic method to formulate and test their ideas.
The approach that scientists take to solving problems is called the
scientific method. At its core, the scientific method is a cyclical process of
making observations, formulating new ideas, and then testing those ideas
through experiments (Figure 1.20).

Figure 1.20 The scientific method is a cyclic process of making observations,

formulating new ideas, and then testing those ideas.

When scientists encounter a new or unexpected occurrence, they

propose explanations for what they have observed. Scientists use the term
hypothesis to describe a tentative explanation that has not been tested.
To test a hypothesis, scientists devise experiments. The results of
experiments provide support for or against a hypothesis, and they lead
scientists to embrace, refine, or discard their ideas. The observations that
scientists make from experiments often lead to new hypotheses, which in
turn lead to new experiments.
Through this cyclic process, scientists refine their ideas over time. A
theory is an idea that is supported by experimental evidence. This term
can also have a broader meaning: Sometimes it is used to mean a
paradigm—a way of thinking about a particular topic.

When scientists talk about a theory, they mean an idea that is supported by
evidence, or even a way of thinking about the world around us.

64
 To many people, the term theory means an idea is unproven, but
scientists use this term differently. For example, in the coming chapters we
discuss the atomic theory. The evidence for atoms is indisputable, and
atoms can be indirectly observed through a number of techniques. Rather
than conveying uncertainty, the term atomic theory describes our modern
understanding of atoms. When we explain how different substances
behave, we do so using the atomic theory. Atomic theory is not just a
guess supported by evidence—it’s a way of thinking about the world
around us.
Scientists use the term scientific law to describe observations that are
true in widely varying circumstances. A scientific law does not explain
why something occurs; it simply observes that it is true. Often, a scientific
law describes a mathematical relationship. For example, consider the law
of gravity: This law describes observations and phenomenon that
consistently occur. Scientists use the law of gravity mathematically, to
calculate the weight of a building or the trajectory of a rocket. This law
predicts what will happen, but it does not explain why it happens.
A common misconception is that scientific research progresses from
hypothesis to theory to law. In fact, theories and laws are two very
different ends. Scientific laws provide a concise description of a behavior,
while theories explain how or why things happen.

Scientific laws describe what happens, but not why. Theories describe how
or why something happens.

IT
TRY

7. Classify each of the following as a hypothesis, theory, or law:

a. When leaving your house, you notice that your trash has been strewn
across the driveway. You guess that the neighbor’s dog is responsible.
b. In most coastal areas, high tides occur one to two times each day.
c. After repeated experiments, scientists believe that a class of antibiotics
functions by weakening the cell walls of certain bacteria.

8. Since ancient times, humans have used plants for medicinal purposes.
In 1962, Arthur Barclay suspected that the Pacific yew might contain
substances that could fight disease. At the time, was this a hypothesis,
a theory, or a scientific law? What types of experiments were
conducted to test this idea?

65
Summary
This chapter introduced several key ideas that are foundational for the
study of chemistry. Because chemistry involves the world all around you,
knowledge of chemistry will serve you well throughout your career,
whether you work in health care, manufacturing, agriculture, or some other
field.
The study of chemistry begins with matter. We define matter as
anything that has mass and takes up volume. All matter is composed of
tiny units called atoms. The arrangement of atoms within a substance
determines its properties.
Elements are substances that are made of only one type of atom.
Compounds are substances made of more than one type of atom, bound
together in fixed ratios. In many compounds, the atoms form discrete
groups called molecules. Mixtures contain more than one element or
compound. The ratios of components in mixtures may vary.
We commonly describe matter by its physical or chemical properties.
We can measure physical properties without changing the identity of a
substance. In contrast, chemical properties are properties of a substance
that we cannot measure without changing the identity of a substance.
Chemical properties describe the ways in which substances change to form
new substances.
Physical and chemical changes involve changes in energy. Energy is
the ability to do work. Energy can exist as kinetic energy (including heat
energy) or as potential energy. As a general rule, spontaneous changes
involve moving from high-energy to low-energy states. Objects or
substances that do not have a large amount of stored energy are said to be
stable.
When scientists encounter questions, they use a systematic approach
called the scientific method to formulate and test their ideas. The scientific
method is a cyclical process of making observations, formulating new
ideas, and testing ideas through experiments. Scientists often propose
untested ideas called hypotheses. An idea that is supported by
experimental evidence is called a theory. Theories often explain how or
why something works. A scientific law is something that is observed to be
true in a variety of circumstances. A scientific law predicts what will
happen but does not explain why it happens.

66
You Can Do This

“It is possible to move a mountain by carrying away small stones.”

In my early 30s, with a wife, a full-time job, and three small children, I decided
to go back to school. I was more than a little reluctant—my previous university
experience had left a bad taste in my mouth, and I dreaded going back.
Nonetheless, I knew I had to complete the degree to accomplish my own career
goals and to provide for my family’s future.
Over the next couple of years, I determined to make a little progress each day. I
taped a picture of a bulldog over my desk to remind me of my goal. Bulldogs
aren’t the prettiest of animals, or the smartest. But they grab on tight, and they
don’t let go. They win by gaining an inch at a time.
As you begin your own chemistry course, I want to encourage you: You can do
this. It’s impossible to learn chemistry in one evening, but you can learn a little bit
each day. If you will commit to read, to study, and to do practice problems, you
can do this. If you will find others who can hold you accountable, and if you will
seek help when you need it, you can do this. Start carrying away stones; soon
you’ll see your mountain begin to move.

67
 Key Terms
1.1 Chemistry: Part of Everything You Do
chemistry The study of matter and its changes.

1.2 Describing Matter

matter Anything that has mass and takes up volume.
composition The components that make up a material.
structure The arrangement of simple units within a substance. In chemistry,
structure refers to both the composition and arrangement of simple units within
a substance.
atoms The fundamental units of matter.
element A substance made of only one type of atom.
compounds Pure substances composed of more than one element in a fixed
ratio.
molecules Groups of atoms that are held tightly together.
pure substances Substances composed of only one element or only one
compound.
mixtures Substances containing more than one substance.
homogeneous mixtures Mixtures in which the components are evenly
blended throughout.
heterogeneous mixtures Mixtures in which the components are not evenly
blended throughout.
states of matter The classification of matter as a solid, liquid, or gas (also
called the phases of matter).
solid A state of matter having a definite shape and a definite volume. The
particles in a solid are held in fixed positions.
liquid A state of matter having definite volume but no definite shape. The
particles in a liquid are close together but move freely past each other.
gas A state of matter that does not have a definite shape or a definite volume.
The particles in a gas move freely with very little interactions.
melting A transition from the solid phase to the liquid phase.
vaporization A transition from the liquid phase to the gas phase.
condensation A transition from the gas phase to the liquid phase.
freezing A transition from the liquid phase to the solid phase.

68
physical properties The properties of a substance that can be measured
without changing the identity of the substance.
physical changes Changes that do not alter the identity of the substance.
chemical properties Properties of a substance that cannot be measured
without changing the identity of a substance.
chemical changes Changes that produce new substances; also called chemical
reactions.
chemical reactions Changes that produce new substances; also called
chemical changes.

1.3 Energy and Change

energy The ability to do work.
potential energy Energy that is stored.
kinetic energy The energy of motion.
heat energy A type of kinetic energy, involving the movement of particles
within a substance.
exothermic change A change that releases heat energy.
endothermic change A change that absorbs energy.

1.4 The Scientific Method

scientific method A cyclical process of making observations, formulating new
ideas, and then testing those ideas through experiments.
hypothesis A tentative explanation that has not been tested.
theory An idea that has been tested and refined; also a way of thinking about a
particular topic.
scientific law A statement that describes observations that are true in widely
varying circumstances. Scientific laws often describe mathematical
relationships. However, they do not explain why something occurs; they only
observe that it occurs.

69
 Additional Problems

1.1 Chemistry: Part of Everything You Do

9. What are your career goals? What substances are unique or integral to your
field of study? In which aspects of your chosen field might knowledge of
chemistry help you?

10. The chart shown here includes a number of career pathways and suggests
some of the chemistry topics associated with each one. Fill in the chart
with some examples where an overlap between each field and chemistry
might occur. Feel free to add more lines for other careers that interest you.

Field Overlap with Chemistry

Human or Veterinary Human and animal physiology, drug design, blood
Medicine and tissue analysis, etc.
Monitoring of air and water quality, plant and animal
Ecology and biology
Environmental Science
Engineering
Forensics
Computer Science
Farming
Geology
Business and
Management
Psychology
Manufacturing

1.2 Describing Matter

11. Determine whether or not the following are made of matter:

a. air
b. a football
c. sunlight
d. water

70
12. Determine whether or not the following are made of matter:
a. sweet tea
b. happiness
c. sound
d. sand

13. What is the difference between composition and structure?

14. What is the difference between an element and a compound?

15. Describe each of the following as an element, a compound, a homogeneous

mixture, or a heterogeneous mixture:
a. Earth’s atmosphere, which contains about 78% nitrogen, 21% oxygen,
and small amounts of other gases
b. fluorine gas, a substance that contains only fluorine atoms
c. carbon monoxide, a substance containing carbon and oxygen atoms in a
fixed 1:1 ratio

16. Describe each of these as an element, a compound, a homogeneous

mixture, or a heterogeneous mixture:
a. an alloy of gold and tin
b. phosphorus trichloride, a substance containing phosphorus and chlorine
atoms in a 1:3 ratio
c. titanium metal, a substance containing only titanium atoms

17. In the graphics shown here, the colored spheres represent different atoms.
Indicate if each image represents an element, compound, homogeneous
mixture, or heterogeneous mixture.

18. An atomic-level representation of saltwater is shown here. Is saltwater best

described as an element, compound, homogeneous mixture, or
heterogeneous mixture?

71
19. How does particle movement differ in a solid, a liquid, and a gas?

20. In the image shown here, colored spheres represent different atoms that are
moving freely around a container. Does this image represent an element, a
compound, or a mixture? Does it represent a solid, liquid, or gas?

21. Identify each of the following as a solid, liquid, or gas:

a. a substance that has a definite shape and a definite volume
b. a substance that fills its container and takes on the shape of the
container
c. a substance in which the particles remain in fixed positions

22. Identify each of the following as a solid, liquid, or gas:

a. a substance made up of particles that are close together but able to move
freely past each other
b. a substance made up of particles that are far apart and have almost no
interaction with each other
c. a substance with a definite volume but no definite shape

23. Describe each property or change as physical or chemical:

a. The paint on a new truck has a shiny red color.
b. The metal wire conducts electricity.
c. Magnesium metal reacts with hydrogen chloride gas to form magnesium
chloride and hydrogen gas.
d. When cooked on a stove, eggs form a fluffy yellow solid.

72
 e. Water is heated and it becomes steam.

24. Describe each change as physical or chemical:

a. Sugar dissolves in water.
b. Charcoal burns, producing carbon dioxide and ash.
c. After you eat, the acids in your stomach break complex molecules into
simpler pieces.
d. A chocolate bar is melted down and poured over pretzels.

25. In the introduction to this chapter you read about Taxol, a compound that
is commonly used in the treatment of cancer. Describe each of the
following properties of Taxol as chemical or physical:
a. Taxol is a white solid.
b. When Taxol enters a patient’s liver, it is broken down into simpler
compounds.
c. Taxol melts at about 216 °C.

26. Aspirin is a substance that is commonly used as a pain reliever. Describe

each of the following properties of aspirin as chemical or physical:
a. Aspirin is a solid at room temperature but melts at 136 °C.
b. Aspirin reacts with the compound sodium hydroxide to form two new
compounds.
c. When a person takes an aspirin, a substance in the stomach reacts with
the aspirin, breaking it into two simpler compounds.

27. Consider the transition from solid to gas shown here. Is this a physical
change or a chemical change?

28. A mixture of nitrogen gas and hydrogen gas (a) undergoes a change to
form a new compound, ammonia (b). Is this a physical change or a
chemical change?

73
29. Seawater is a mixture of salt and water. Across the world, people isolate
salt by adding seawater to shallow ponds, called salt pans (Figure 1.21).
As the water evaporates, salt crystals form. Does this separation use a
physical change or a chemical change?

Figure 1.21 This worker is isolating salt by evaporating the water from a saltwater
mixture.

30. When the two leads of a battery are connected to a saltwater solution, it
causes the water to be broken into the elements hydrogen and oxygen
(Figure 1.22). Is this a physical change or a chemical change?

Figure 1.22 An electric current produces bubbles of hydrogen and oxygen gas as it splits

74
 water into its elemental components.

1.3 Energy and Change

31. When heat energy is transferred to an object, what happens to the atoms in
the object?

32. How do the water molecules in ice differ from the water molecules in
liquid water? Which molecules move more quickly?

33. Which item in each pair has more potential energy? Which is more
energetically stable?
a. a rock at the top of a hill or a rock at the bottom of a hill
b. wood or ashes
c. a tightly wound spring or a spring that is relaxed

34. Indicate whether each event results in a higher-energy (less stable) or a

lower-energy (more stable) product:
a. Heat is added to water, converting it to steam.
b. A firecracker explodes, leaving scattered fragments of ash and paper.
c. A candle burns, producing carbon dioxide gas and water.
d. Sunlight provides energy to convert carbon dioxide and water into
woody material, causing a tree to grow.

35. Container (a) contains hydrogen and oxygen. When these elements react,
they form water, as shown in container (b).

a. Is the material in container (a) a pure element, a compound, or a

mixture?

75
 b. Is the material in container (b) a pure element, a compound, or a
mixture?
c. Is the change that takes place between (a) and (b) a physical change or a
chemical change?
d. This change releases a large amount of energy. Based on this reaction,
is the arrangement of atoms more stable in container (a) or container
(b)?

36. Container (a) contains boron and chlorine. When these elements react, they
form boron trichloride, shown in container (b). The reaction releases heat
energy.

a. Is the material in container (a) a pure element, a pure compound, or a

mixture?
b. Is the material in container (b) a pure element, a pure compound, or a
mixture?
c. Is the change that takes place between (a) and (b) a physical change or a
chemical change?
d. This change releases a large amount of energy. Based on this reaction,
is the arrangement of atoms more stable in container (a) or container
(b)?

37. In photosynthesis, plants absorb energy from the sun, carbon dioxide, and
water. As a result the plants grow. Is this reaction endothermic or
exothermic?

38. Electric energy can be used to break water into its elemental forms,
hydrogen and oxygen. We can write this reaction as:
Energy+water→hydrogen+oxygen

76
 a. Based on this reaction, which is more stable—water, or a combination
of hydrogen and oxygen?
b. Is this change endothermic or exothermic?

1.4 The Scientific Method

39. What are the key differences between a hypothesis and a theory?

40. What are the key differences between a theory and a scientific law?

41. After teaching middle school science for several years, you notice that
your students seem to be most engaged between 9 a.m. and 11 a.m. You
develop a hypothesis that students perform better in critical subject areas
during this time period. How might you test this hypothesis? How could
you make sure your results were reliable?

42. While gardening, you notice that plants on the north end of your garden are
growing much better than those on the south end. You suspect that those
on the south end may not be getting enough sunlight. How could you test
this hypothesis?

77
Chapter Two
Measurement

A Strange Death
In the summer of 2011, the body of a 47-year-old male was wheeled into the West
Virginia Office of the Chief Medical Examiner. The coroner determined that the
cause of death was cardiac arrest, but there was more to the story.
Family members said the man had been acting oddly. While suffering from
insomnia and paranoia and awake for several days, he had become increasingly
aggressive. His aggression reached a crescendo when, after destroying furniture
and walls inside his home, he ran outside yelling, began kicking trash cans, and
assaulted a neighbor. By the time police officers arrived, the man was incoherent
and hallucinating. Within a short time, he was dead.
Investigators interviewed his family and combed through the remains of his
home. Surprisingly, they found no sign of conventional drug use: no pipes, no
needles, no pills. They did, however, find a clue in his car: a trace of residual
powder inside a small plastic container with a strange black label. It wasn’t cocaine
or heroin or methamphetamine—it was something new. What was it? Was the
powder harmless, or was it connected to his death?
During the man’s autopsy, the coroner collected urine, blood, and tissue
samples. James Kraner and Lauren Waugh, toxicologists in the Medical
Examiner’s Office (Figure 2.1), received these samples. In the following weeks,
they began to investigate the man’s mysterious death.

78
Figure 2.1 (a) The Chief Medical Examiner’s Office was the site of the autopsy. (b) A
mysterious container was found in the vehicle of the deceased. (c) A scientist prepares a
blood sample for analysis. (d) Dr. Waugh is photographed here with one of the
instruments used in the analysis.

Kraner and Waugh began their search by testing for the presence of popular
street drugs in the victim’s blood and urine. But their tests showed that none of
these drugs were present. They suspected that the strange powder was involved,
but how could they prove it?
Using an array of instruments and measurement techniques, the toxicologists
identified the substance as MDPV (methylenedioxypyrovalerone), a synthetic drug
similar to methamphetamine.
Next they tested the victim’s blood and urine, and they found that both
contained high levels of MDPV. From these findings, the toxicologists connected
the victim’s erratic behavior and sudden death to an overdose of MDPV.
At the time, MDPV was legal. Smoke shops and online stores sold it as “bath
salts”—a clever ruse to conceal a devastating drug. The toxicologists’ discovery
marked the first time in West Virginia that MDPV had been identified as the cause

79
of a death. In the following months, a string of MDPV-related deaths occurred
across the country. States quickly moved to ban the sale of MDPV and related
drugs.
Kraner and Waugh’s success in solving this problem hinged on two factors: the
quality of their methods and the quality of their measurements. Measurement is
critical to the scientific method. It forms the bedrock of chemistry, forensics, and
all other sciences. In the pages that follow, we’ll explore how scientists make
measurements and communicate their findings to others. At the end of the chapter,
we’ll circle back to this case and examine some of the measurements and
techniques that helped to provide the answer.
Whether your career takes you into criminal justice, health sciences, or some
other area, the concepts in this chapter will serve you well beyond this course.
Knowing how to measure carefully, assess the limits of your information, and think
about numerical data and possible outcomes will strengthen your understanding of
the world and expand your ability to succeed in any field.

Intended Learning Outcomes

After completing this chapter and working the practice problems, you should be
able to:

2.1 Measurement: A Foundation of Good Science

Convert between standard and scientific notation, and solve multiplication
and division problems involving scientific notation.
Describe the quality of measurements using the terms accuracy and
precision.
Identify significant digits in a measured number, report measurements to an
appropriate number of significant digits, and apply the rules for significant
digits to simple calculations.

2.2 Unit Conversion

Perform unit conversions using the factor-label method.

2.3 Density: Relating Mass to Volume

Relate the density, mass, and volume of a substance.

2.4 Measuring Temperature

Convert between Celsius, Fahrenheit, and Kelvin temperature scales.

80
2.1 Measurement: A Foundation of Good Science
Measurement is a foundation of good science, but the value of good
measurement extends beyond science. Almost anything you choose to do
—from graphic design to fantasy football, from cooking to construction—
involves measurement. In studying chemistry (or any science), it is
essential to be able to make good measurements, to understand the limits
of measurements, and to communicate clearly what you have found.

Builders sometimes say, “Measure twice, cut once.” It is important to make sure
your measurement is accurate; otherwise, you may waste valuable time and
materials.

Scientific Notation: Working with Very Large and Very Small

Numbers
One of the challenges of science is dealing with measurements that are
very large or very small. For example, the distance from the Earth to the
Sun is approximately 149,600,000,000 meters. On the other hand, in the
forensic analysis described at the beginning of this chapter, the
toxicologists found that each cubic centimeter of blood contained
0.00000109 grams of MDPV. That’s small, but not nearly as small as the
mass of a hydrogen atom, which is 0.000000000000000000000001672
grams. How can we work with such unruly numbers?
Scientific notation is a way to show very large and very small
numbers in a concise format. Scientific notation expresses numbers as the
product of two values, called the coefficient and the multiplier. In the
coefficient, we write a measured value with only one digit before the
decimal point. The multiplier is 10 raised to an exponent (10x). The
multiplier shifts the decimal point to the left or right to give the value in
standard form. For example, we use scientific notation to express the
number 5,100 like this:

81
 Look at the values of the powers of ten in Table 2.1. Notice that each
time the exponent is increased by one, the value becomes ten times greater.
Stated differently, each unit of the exponent moves the decimal point one
place.

TABLE 2.1 Powers of Ten

Exponent Value

103 1,000.

102 100.

101 10.

100 1.

10–1 0.1

10–2 0.01

10–3 0.001

Converting from Scientific Notation to Standard Notation

When converting from scientific notation to standard notation, we move
the decimal the number of units indicated by the exponent in the
multiplier. With positive exponents, the decimal point moves to the right:

In scientific notation, the exponent tells how many spaces the decimal point
must be moved.

With negative exponents, we move the decimal the opposite direction:

Remember that positive exponents represent large numbers, and

82
negative exponents represent small numbers. Table 2.2 details the trends
for positive and negative exponents.

TABLE 2.2 Multipliers in Scientific Notation

Large Numbers

Example Decimal point moves

5.1 × 101 = 51. 1 space

5.1 × 102 = 510. 2 spaces

5.1 × 103 = 5,100. 3 spaces

5.1 × 104 = 51,000. 4 spaces

Small Numbers

Example Decimal point moves

5.1 × 10−1 = 0.51 −1 space

5.1 × 10−2 = 0.051 −2 spaces

5.1 × 10−3 = 0.0051 −3 spaces

5.1 × 10−4 = 0.00051 −4 spaces

Large numbers have a positive exponent in scientific notation. Small

numbers have a negative exponent.

Converting from Standard Notation to Scientific Notation

To write a number in scientific notation, we move the decimal point in the
coefficient to the right or the left until only one nonzero digit remains in
front of the decimal point. The number of digits we move the decimal
point in the coefficient is the exponent in the 10x multiplier. Again, we use
negative exponents for small values and positive exponents for large
values.
For example, we write 0.0000234 in scientific notation as follows:

Calculations Involving Scientific Notation

83
When multiplying numbers using scientific notation, we multiply the
coefficients and add the exponents in the multipliers:

When dividing numbers using scientific notation, we divide the

coefficients and subtract the exponent in the denominator from the
exponent in the numerator:

Example 2.1 Multiplication with Scientific Notation

Multiply 3.95 × 10−22 and 4.00 × 1024. Express your answer in proper
scientific notation.
When we multiply the coefficients and add the exponents, we get an answer
of 15.8 ×102. This is mathematically correct, but it is not proper scientific
notation because there is more than one digit before the decimal point in the
coefficient. To convert to proper notation, we need to move the decimal
over one space and then add one unit to the exponent. This gives us a final
answer of 1.58 × 103.

If you use a calculator to solve problems, here’s a word of caution:

Different calculators input and represent scientific notation differently. I

84
 encourage you to practice these techniques on your calculator until you are
confident you can do them correctly. •

IT
TRY

1. Write each number in scientific notation.

a. 130,000,000,000
b. 4,700,000
c. 0.005
d. 4.1

2. Write each number in standard notation.

a. 5.31 × 105
b. 1.213 × 103
c. 4.091 × 10−4

3. Solve each calculation. See if you can solve them by hand, and then
solve them with a calculator. Make sure your answers for both
methods agree!

a. (2.1×109)×(3.0×105)

b. (3.0×105)/(1.5×102)

c. (2.0×10−3)(6.0×102)

d. (4.2×10−3)/(2.1×10−4)

Units of Measurement
I love to make enchiladas (Figure 2.2). When preparing this meal, I follow
a recipe that requires two pounds of roasted chili peppers, ¼ cup of lime
juice, and 1 teaspoon of salt (as well as other ingredients). After
assembling the enchiladas, I bake them at a temperature of 350 °F for 25
minutes.

85
Figure 2.2 Enchilada night in the Revell household is a great tradition. For the meal to
come out right, the ingredients must be measured correctly.

Every component of the meal—ingredients, cooking temperature, and

cooking time—has a unit associated with the measurement, such as
pounds, cups, teaspoons, degrees, and minutes. Units of measurement are
quantities with accepted values that can be communicated between people.
When I follow this recipe, I trust that my ¼ cup measure is the same size
as the ¼ cup measure of the person who gave me the recipe. Without units,
and without an accepted standard for what units mean, the measurements
are worthless.
There are many different units. For example, if you wanted to measure
the length of your kitchen table, you could use inches, meters, yards,
leagues, miles, or cubits. Which is most appropriate?
In the United States, people commonly use English units, such as
inches and feet (length), gallons (volume), and pounds (weight). Most of
the world uses a different system, called the metric system. Table 2.3 lists
the relationships between some of these measurements.

Practice
Metric System
It is important to know the common metric prefixes. Test your knowledge
with this interactive quiz.

TABLE 2.3 Relationships between Some Common English and

Metric Units
Measurement Metric Unit English Unit Relationship

Length meter (m) foot (ft) 1 m = 3.281 ft

kilometer (km) mile (mi) 1 km = 0.621 mi
Mass or weight kilogram (kg) pound (lb) 1 kg = 2.205 lb

Volume liter (L) gallon (gal) 1 liter = 0.264 gal

86
 For easier communication, the international scientific community
developed an accepted set of fundamental units, sometimes called SI units
(the abbreviation SI comes from the French for “International System”).
Table 2.4 presents these units, which are based on the metric system.
From these fundamental units, we can derive a host of other units. For
example, a liter (L) is a derived unit of volume corresponding to 0.001
cubic meters (m3). We use meters per second (m/s) to describe velocity, or
kilograms per liter (kg/L) to describe density. (We discuss these units in
more detail in the sections that follow.)

TABLE 2.4 Fundamental (SI) Units of Measurement

Measurement Unit

Mass kilogram (kg)

Length meter (m)

Time second (s)

Temperature kelvin (K)

Light intensity candela (cd)

Electric current ampere (A)

Amount mole (mol)

The metric system uses a series of prefixes to describe larger or smaller

amounts. Table 2.5 summarizes the key metric prefixes. For example,
many wireless networks transmit data at 160 million bits (b) every second.
Rather than writing this quantity of information as 160,000,000 bits, it is
easier to use the metric prefix, and express this as 160 megabits
(abbreviated as 160 Mb). As with scientific notation, the metric prefixes
allow us to express very large or small numbers in a concise way.

TABLE 2.5 Common Metric Prefixes

Prefix Symbol Meaning

tera- T 1012 1,000,000,000,000

giga- G 109 1,000,000,000

mega- M 106 1,000,000

kilo- k 103 1,000

deci- d 10−1

110

87
 centi- c 10−2

1100
milli- m 10−3

11,000

micro- μ 10−6

11,000,000

nano- n 10−9

11,000,000,000

pico- p 10−12

11,000,000,000,000

Working with Small Units

From Table 2.5, we see that

1 mg=11,000g.

This means that 1,000 mg=1 g.

Even for common measurements such as mass and temperature,

scientists don’t always use the fundamental units in Table 2.4. For
example, the official SI unit of mass is the kilogram. But chemists often
work with far smaller amounts than this in the laboratory, so they routinely
deal with grams or with milligrams.

IT
TRY

4. Refer to Tables 2.3, 2.4, and 2.5 to write the appropriate abbreviations
for each unit.
a. meters
b. milliliters
c. kilometers
d. microamperes
e. picoseconds
f. megacandelas
g. millimoles
h. nanograms

88
 5. Refer to Table 2.5 to answer the following questions:
a. How many bytes are in a kilobyte?
b. How many microseconds are in a second?
c. How many mg are in a g?
d. How many MA are in an A?

Describing the Quality of Measurements

Let’s go ahead and measure something. Figure 2.3 shows a ball bearing, a
small steel ball used in many machine parts. What is its diameter? (I
encourage you to stop, look carefully, and commit to a value before
reading the next paragraph.)

Figure 2.3 What is the diameter of this ball? How sure are you about this?

When I pose this question to a classroom full of people, I typically get

a range of measurements. Often, the range looks something like this:

Measurement Percentage of
Students
2.6 cm 25%
2.7 cm 50%
2.8 cm 25%

Each of these responses is reasonable and is a correct reading. If you

looked at the ball in Figure 2.3 carefully, you probably considered more
than one of these responses. You know that the ball is between 2.5 and 3
cm. Most likely, it is between 2.6 and 2.8 cm. But how can you know for
sure?
In fact, you can’t. Uncertainty is a part of science, and it is a part of
measurements. You are certain that the ball is two-point-something
centimeters, but you can’t be sure of the last decimal point. This is

89
perfectly normal. In science, it is acceptable to estimate the last number in
a measurement. When other scientists read this measurement, they
understand that the last measured digit is an estimated value.
The last decimal place in a measurement is an estimated value.

But what if we need to measure something more exactly? Estimating

the diameter of a ball to plus or minus a couple tenths of a centimeter is
fine in general; but if we are doing more meticulous work, like designing a
part for a high-performance engine, that sort of measurement is not
sufficient. In this case, we need a tool like a caliper (Figure 2.4), which
measures much more precisely than a tape measure or a ruler.

Figure 2.4 This caliper is much more precise than a ruler. It is used to measure to the
nearest 0.01 mm.

Scientists use two terms to describe the quality of measurements:

accuracy and precision. Accuracy refers to how reliable measurements
are. That is, does a measurement (or average of measurements) reflect the
true value? If it does, it is accurate. On the other hand, precision refers to
how finely a measurement is made, or how close groups of measurements
are to each other.
It is possible to be accurate without being precise. Accurate but not
precise describes our measurement of the ball in Figure 2.3. It’s also
possible to be precise without being accurate. For example, a laboratory
balance can measure the mass of a sample to ±0.001 gram. However, if
that balance is not calibrated correctly, numbers read from the balance may
be inaccurate even if they are reported to several decimal places. A classic
way of visualizing precision and accuracy is with a series of targets, as
shown in Figure 2.5.

90
Figure 2.5 Precision means the results are very close together. Accuracy means the
average of the results is correct.

Chemists measure mass using balances like the one shown here. They routinely
test the balances by measuring objects of known mass. If the balance is not
reading correctly, they adjust it so the reading is accurate. This is known as
calibrating the balance.

IT
TRY

6. Your laboratory is conducting tests on the nutritional energy value in

different types of candies. At the beginning of this process, you ask a
new colleague to record the mass of a set of eight individually
wrapped peppermints. Your colleague handles the candies very
carefully and records the mass of each candy as shown in the table.

91
 Candy Mass
1 4.3231 g
2 4.3577 g
3 4.2902 g
4 4.3341 g
5 4.3209 g
6 4.3076 g
7 4.3725 g
8 4.3326 g

When you compare the measured masses with those from previous
experiments, the average mass is about 0.1 g greater. When you
consult with your colleague, you find that she did not remove the
wrappers before measuring the masses. Describe the value of these
measurements using the terms precise and accurate.

Describing Precision: Significant Digits

Figure 2.6 shows two different pieces of glassware we use to measure the
volume of liquids: a beaker and a graduated cylinder. In this figure, we can
see that the volume in the beaker is between 40 and 50 mL, and closer to
50. Based on this observation, we might assert that the volume is 48 mL.
We are not certain about the last digit, but it is a reasonable estimate. The
graduated cylinder contains the same amount of water as the beaker, but it
is marked more finely so we are able to make a more precise measurement.
We can estimate the volume of the graduated cylinder to be between 48
and 49 mL, but closer to 48. For example, we might estimate it to be 48.1
mL.

92
Figure 2.6 We can use a beaker (a) or a graduated cylinder (b) to measure volume. The
graduated cylinder gives us a more precise measurement than the beaker.

From these examples, we can infer a general rule. When we use a tool
with clearly marked units to measure a quantity such as length, volume, or
temperature, we can estimate one digit between the marked values. The
volume markings on the beaker in Figure 2.6 show to the tens place, and
so we are able to estimate to the ones place. In the graduated cylinder in
Figure 2.6, the volume is marked to the ones place, so we are able to
estimate to the tenth place.

When measuring, estimate one digit between the marked values.

Significant digits, which are also sometimes called significant figures,

are the digits contained in a measured value. The number of significant
digits indicates how precisely a measurement is made. In our previous
example, we measured the volume of liquid in the beaker (48 mL) to two
significant digits. We measured the volume of liquid in the graduated
cylinder (48.1 mL) to three significant digits. The more precise a
measurement is, the more significant digits it contains.

Significant digits indicate the precision of a measurement.

Example 2.2 Measuring Volume in a Graduated Cylinder

93
 What is the volume of water in the graduated cylinder? Report your answer
to the correct number of significant digits.

In narrow glass cylinders, water adheres to the glass surface, resulting in a

slightly curved surface called a meniscus. The correct way to read a
graduated cylinder is to report the volume at the bottom of the meniscus.
Notice that the bottom of the meniscus lies between the 23-mL and the 24-
mL marks. We estimate the volume to the nearest tenth of a milliliter, one
decimal place beyond what is marked on the cylinder. By careful
inspection, we can estimate the meniscus to be 23.3 mL—although 23.2 mL
or 23.4 mL are also acceptable measurements. •

IT
TRY

7. What is the temperature reading on the thermometer? Make sure your

answer includes the correct number of significant digits.

94
 8. The image shown here is of a liquid sample in a graduated cylinder.
What is the volume of the sample, reported to the appropriate number
of significant digits?

Determining the Number of Significant Digits in a Measurement

Sometimes it is important to recognize how precisely a measurement is
made. There are several rules for recognizing the number of significant
digits in a measurement:
1. All nonzero digits are significant, and all zeros between nonzero
digits are significant. For example, both of the following measured
numbers contain five significant digits:

95
 2. If a decimal point is present, zeros to the right of the last nonzero
digit are significant. A number containing zeros after the decimal point
signifies that the number has been measured that precisely.

3. Zeros to the left of the nonzero numbers are never significant. Zeros
to the left of a number and to the left of the decimal point are meaningless.
Zeros to the left of the measured values are useful for values less than one,
because they show where the decimal point lies; however, they are not part
of the measurement and are not significant. For example, if we measure
the width of an object to be 4.5 millimeters, we say we measured to two
significant digits. What if we then convert this measurement to meters? A
width of 4.5 millimeters is equal to 0.0045 meters. Did our measurement
suddenly become more precise? No—we changed the units, but the
number of significant digits did not change.

4. If there is no decimal point present, zeros to the right of the last

nonzero digit may or may not be significant. If a number contains zeros to
the right of the last nonzero digit, but does not contain a decimal point, it is
hard to know how precisely the number is measured. For example, the cost
of a program may be reported as $10,000,000 per year. Does that mean it
costs exactly $10 million dollars? Probably not. But how closely has it
been rounded? Is the cost between 9 million and 11 million? Or is the cost
plus or minus $1,000, meaning it is between $9,999,000 and $10,001,000?
It’s not clear.

96
 For numbers that end in zeros but do not contain a decimal point, there
are two ways we can specify how precisely we’ve measured. The first
approach is to specify the uncertainty right after the number. For example,
we might know that a large object has a mass of 10,000 kg, measured to
the nearest 100 kg. We write this as 10,000 ± 100 kg. Alternatively, we
can write the value in scientific notation as 1.00 × 104kg. By doing so, we
specify the number of zeros that we are fairly certain about (in this case,
two zeros).
Working with Exact Numbers
Exact numbers are values for which there is no uncertainty. When dealing
with exact numbers, we do not need to be concerned about significant
digits. For example, Figure 2.7 shows a group of seven pennies. We don’t
debate over whether there are 7.0 pennies or 7.1 pennies; clearly, there are
exactly seven pennies. Counted values are exact numbers.

Figure 2.7 Counted numbers are exact numbers; significant digits do not apply. There
are seven pennies in the photo.

A second type of exact number is a number that comes from a

definition. For example, there are 12 items in a dozen. There are 1,000
milligrams in a gram. These are exact relationships, and so we don’t worry
about significant digits for this type of number.

Counted values and defined relationships (such as metric prefixes) are exact
numbers.

97
 Let’s consider a second example to help clarify this point. Consider the
plate of tacos shown in Figure 2.8. If we want to count the number of
tacos, we can do so exactly: There are three tacos. On the other hand, to
determine the mass of the guacamole that was added to the tacos, we
would make a measurement, then report this value to as many significant
digits as we could measure.

Figure 2.8 The number of tacos is exact, but the amount of guacamole on each taco is
measured.

IT
TRY

9. How many significant digits are in each of the following measured

units?
a. a distance of 14.3 kilometers
b. a toxin concentration of 0.0079 milligrams per liter
c. a mass of 4.300 kilograms
d. a time of 000024.3 seconds

10. Identify each of the following values as measured or exact:

a. There are 96.1 nutritional Calories in a jelly donut.
b. The burn marks from a fire cover an area of 24.1 square meters.
c. There are 13 floats at a homecoming parade.
d. There are 1 × 109 bytes of data in a gigabyte.
e. A bag contains 1 pound of coffee.

Using Significant Digits in Calculations

In the previous section, we discussed the measurement of a ball. Let’s say
that we measured the diameter of the ball to be 2.7 cm, and we were
relatively certain to ±0.1 cm. That is, we think the diameter is 2.7 cm, but

98
it could be slightly less (2.6 cm) or slightly more (2.8 cm).
Now suppose we need to calculate the circumference of the ball. The
circumference of a sphere is pi (π) times the diameter. If we enter 2.7* π
into a calculator, we’ll get something like the following answer:

2.7 cm * π=8.48230016 cm

So your calculator reports the circumference to eight decimal places. Calculators

say lots of things. If you type in 0.7734 and turn it upside down, a calculator
will even say hello. It is your responsibility to decide how precisely you really
know the circumference and not to simply accept what your calculator says.

In this example, the calculator reports the circumference of the ball

measured to eight decimal places. Do we really know the circumference
this precisely, if we only measured the diameter to one decimal place? No!
So, how precisely do we know the circumference?
To answer that, let’s recalculate the circumference using the low and
high values of 2.6 and 2.8 cm, and see how it affects the value of the
circumference:

Notice that when we started the calculation, we were not sure about the
second significant digit. After the calculation, we’re still unsure about the
second significant digit. Based on this, we round the answer to the second

99
significant figure, or 8.5 cm.

If the last dropped digit is 5 or greater, we round up. If the last dropped digit is 4
or less, we round down.

Based on problems like this, scientists have developed two common

rules for reporting calculations with significant digits:
Rule 1: When multiplying or dividing, report the same number of digits
as are in the least precise starting measurement.
Rule 2: When adding or subtracting, round to the last decimal place of
the least precise starting measurement.
The following examples demonstrate how these rules are applied in
different calculations.

Example 2.3 Significant Digits: Multiplying and Dividing

In nutrition, the energy available from a food is often measured in
Calories. In one study, researchers find that a 55.023-gram candy bar
contains 283.1 Calories. How many Calories are present in each gram of
this candy bar? Report your answer with the correct amount of significant
digits.
To solve this problem, we divide 283.1 Calories by 55.023 grams. The
calculator reports a value of 5.145121131. We round this value to have the
same number of significant digits as the starting quantity with the fewest
significant digits. Although we know the mass to five significant digits
(55.023), we know the energy to only four digits (283.1), so we only report
four significant digits. Based on this observation, we round the answer to
5.145 Calories/gram. •

Example 2.4 Significant Digits: Multiplying by an Exact

Number
According to the United States Mint, a quarter is made from a mixture of

100
copper and nickel, and it has a mass of 5.670 grams. What is the mass of
eight quarters?
In this case, we are multiplying a measured quantity (5.670 grams) by an
integer (8). Because the integer value is an exact number, it is not included
in the significant digit consideration. Therefore, the fewest measured
significant digits is four. We multiply 5.670 × 8, then round to four
significant digits to arrive at the correct answer of 45.36 grams. •

Example 2.5 Significant Digits when Adding and Subtracting

A chemist collects three samples of a liquid. The first sample has a volume
of 3.62 L, the second has a volume of 0.255 L, and the third has a volume of
21.2 L. What is the total volume, reported to the correct number of
significant digits?
When we add these three volumes together, we obtain a value of 25.075 L.
However, because our least precise value only goes to the tenth place, we
round the number to that value, giving a final answer of 25.1 L. •

Example 2.6 Significant Digits with Multiple Calculations

Nitrate is a common component of fertilizers that can contaminate water
supplies. A chemist tested three samples of water and found they contained
nitrate in the amounts shown. Together, the three samples had a volume of
1.514 L. What was the total mass of nitrate in the three samples? What was
the average mass of nitrate per liter of water? Answer each question to
significant digits.

Sample Mass of Nitrate

A 12.83 mg
B 11.2 mg

101
 C 14.391 mg

Adding the three masses gives us a value of 38.421 mg. However, since we
don’t know the mass of sample B more precisely than a tenth of a
milligram, it would be incorrect to report the total mass to three decimal
places. Therefore, we round our answer to the tenth place and report the
mass as 38.4 mg. We are confident of this answer to three significant digits.

To find the average mass of nitrate per liter of water, we divide the total
mass by the volume. However, to minimize rounding errors, we wait until
the end of the calculation to round to significant digits. That is, we use the
full, unrounded value from the first measurement (38.421 mg), rounding
only after our final calculation. We are confident of the mass to three
significant digits, and the volume to four significant digits. Because this is a
division problem, we keep the least of these values, and therefore round our
answer to three significant digits. •

IT
TRY

11. In 2002, Paula Radcliffe broke a world record when she ran the
Chicago marathon in 137.30 minutes. A marathon is 26.2 miles.
What was her average speed in minutes per mile? Make sure your
answer has the correct number of significant digits.

12. Four runners run an 800-m race. Their respective times are 98.12 s,
98.64 s, 101.33 s, and 104.04 s.

102
 a. Find the sum of their times, reported to significant digits.
b. Find the average of their times, reported to significant digits.

13. A toxicologist measures the concentration of an illegal drug in a

urine sample. She measures five times and obtains the following
values, in milligrams per liter: 12.1, 12.9, 11.5, 13.1, and 12.5.
Calculate the average concentration. Make sure your answer has the
correct number of significant digits.

103
2.2 Unit Conversion
When working with measured quantities, it is often necessary to convert
between units. If you are traveling internationally, you may have to
convert from one currency to another (Figure 2.9). If you are measuring a
distance between cities, you might need to relate kilometers to miles. If
you are working in a lab, it is often necessary to convert between units of
mass or volume.

Figure 2.9 This board displays exchange rates between currencies.

Dimensional Analysis
Sometimes, you can do simple unit conversions in your head. However, as
we delve deeper into chemistry and deal with increasingly complex units,
it is important to have a systematic approach to unit conversions that
allows us to keep careful track of units and to check our work. One
common way to approach unit conversion problems is a technique called
dimensional analysis, also known as the factor-label method.
The cornerstone of the factor-label method is this: When performing a
mathematical operation involving measured quantities, whatever we do to
the number, we also do to the units. For example, to calculate the area of a
square, we multiply the length and the width (Figure 2.10). If the length
and width measurements are in feet, then the area is in units of feet times
feet, or feet squared (ft2).

104
Figure 2.10 When calculating the area of a square, we multiply both the numbers and
the units.

How can we use this concept to convert between units? To understand

how the factor-label method works, consider an example. Suppose we
have a copper pellet with a mass of 231 milligrams. We need to express
this mass in grams. How do we do this?
First, we need to know the relationship between milligrams and grams.
From Table 2.5, we see that 1 milligram = 1/1,000th of a gram. Stated
differently, there are 1,000 milligrams in 1 gram:

1,000 mg=1 g
Because these two quantities are equal, we can write them as two fractions,
both with a value of one:

These fractions are examples of conversion factors. Conversion

factors are fractions that contain equivalent amounts of different units in
the numerator and in the denominator. To convert from one unit to
another, we multiply by one of these two conversion factors. Because a
conversion factor is equal to one, we are not changing the value of the
quantity; we are only changing the units. When we multiply quantities
having units, we cancel any units that appear in both the numerator and the
denominator.
Use conversion factors to change from one unit to another.

To convert between units, we choose the conversion factor that

cancels the old units and retains the new units. In this example, we want to
get rid of milligrams and end up with an answer in grams. We therefore

105
use the conversion factor with grams in the numerator:

This gives us a final answer of 0.231 grams.

What if we select the wrong conversion factor? If we multiply by

1,000 mg1 g, we get units of mg2g, as the following

incorrect calculation shows:

Verifying the units is a good way to check our work. If we work the
problem correctly, the units come out right. If the units do not make sense,
we probably set up the problem incorrectly.

Example 2.7 Converting between Units

Heat energy is commonly measured in calories or in joules (1 calorie =
4.184 joules). A particular chemical reaction releases 150.0 calories of
heat energy. What is this energy in joules?
As in the previous example, we begin with the amount that we want to
convert (in calories). To convert this value from calories to joules, we must
multiply it by a conversion factor. Because we want to cancel out the
calories, we write this unit in the denominator and write joules in the
numerator:

We are able to cancel out the calories, leaving us with a final answer of
627.6 joules. •

106
 Example 2.8 Converting between Units
A brick has a mass of 815.2 grams. What is this mass in kilograms?
We begin with the amount in grams. To convert this value from grams to
kilograms, we must multiply it by a conversion factor. We know from
Table 2.5 that 1,000 g = 1 kg. Because we want to cancel out the grams, we
write this unit in the denominator and write kilograms in the numerator:

This gives us a final answer of 0.8152 kg. •

IT
TRY

14. For each of the following, indicate the correct conversion factor to
use, and find the value in the new unit:
a. Convert 125.31 mg to g.
b. Nanoparticles are tiny capsules that are a promising new way to
introduce medicine into the body. These particles are much too small
to see with the naked eye. One such capsule has a diameter of 6.0 ×
10−8 meters. What is this diameter in nanometers?
c. The average home in Louisiana uses about 15,000 kilowatt hours
(kWh) per year. What is this energy consumption in kilojoules (kJ)?

(1 kWh=3,600 kJ.)

Problems Involving Multiple Conversions

To solve many problems, we need to make more than one unit conversion.
When this occurs, we can place the unit conversions adjacent to each
other. Consider the following examples:

Practice
Unit Conversions
The way to get good at unit conversions is to practice. Try this activity to
test your skills.

107
Example 2.9 Problems Requiring Multiple Conversions
How many micrometers are in 0.0129 inches? (2.54 centimeters [cm] = 1
inch and 10,000 micrometers [μm] = 1 cm.)
In this example, we need to make two conversions. We first convert from
inches to centimeters, then from centimeters to micrometers. For the first
conversion, we write 2.54 cm over 1 inch, as shown below. This allows us
to cancel out the inches, leaving units of centimeters. In the second step, we
write 10,000 micrometers over 1 centimeter, so the centimeters cancel out,
leaving micrometers.

In this problem, the initial measured value (0.0129 inches) contained only
three significant digits. As a result, we round the answer to three digits as
well. •

A conversion factor contains two equivalent amounts. The old unit cancels
out, leaving the new unit.

Example 2.10 Converting Units in the Denominator

In air, sound waves travel at 768 miles per hour. How fast is this in meters
per second (m/s)? (1 mile = 1,609.3 meters.)
Let’s begin by converting miles to meters. Because we need meters in the
numerator, we write a conversion factor that has meters in the numerator
and miles in the denominator. This allows us to cancel out the miles.
In the second conversion step, we use the relationship between seconds and
hours (3,600 s = 1 h). Because we need to have seconds in the denominator
of the final expression, we place hours in the numerator of the conversion
factor. Rounded to significant digits, this gives us an answer of 343 m/s.

108
 This type of unit conversion problem pops up in many of the following
chapters, and I encourage you to make sure you can do them efficiently.
Before moving on, try to master the problems in the following question
box. •

IT
TRY

15. Amoxicillin is a powerful antibiotic. For mild infections, a doctor

may prescribe 0.750 grams/day of amoxicillin. What is this dosage in
milligrams per hour?

16. Upon graduating with a good GPA and work experience, you are
pleased to receive two job offers. Company A offers a salary of
$42,000/year. Company B offers an hourly pay of $25.00/hour.
Assuming that you will work 40 hours/week for 50 weeks per year,
use the factor-label method to calculate your annual income at
Company B. Which offer is more lucrative?

Converting between Volume Units

Chemists commonly make measurements involving volume (that is, the
space something occupies). To measure volume, we use units of length
raised to the third power, such as cm3 (cubic centimeters) or m3 (cubic
meters); see Figure 2.11. The units we choose depend on the size of the
quantity we are measuring. For example, if we are measuring the volume
of a shipping container, we probably want to express this as cubic meters.
To describe the volume of a soft drink or the amount of liquid in a vaccine
injection, we would choose a smaller unit (Figure 2.12).

109
Figure 2.11 The volume of a cube is equal to its length times width times height.

Figure 2.12 We commonly encounter very large and very small volumes. The best units
to use depend on how much we are measuring.

110
 The liter (L) is a common derived unit of volume. A liter is defined as
1 cubic decimeter (1 dm3), which is slightly smaller than a box of tissues
(Figure 2.13).

Figure 2.13 One cubic decimeter, also called a liter, is approximately the volume of a
tissue box.

1 liter (L) = 1 dm3

Another common unit of measurement is the cubic centimeter (cm3). A

cubic centimeter is the same unit of volume as a milliliter. A cubic
centimeter is about the same size as a standard game die (Figure 2.14).

Figure 2.14 One cubic centimeter is approximately the volume of a standard game die.
This volume is also referred to as a milliliter.

1 milliliter (mL) = 1 cm3

Converting Units Raised to a Power

111
We must be careful when converting between units that are raised to a
power, like those used to measure area or volume. To illustrate this point,
let’s ask a question: How many cubic decimeters are in one cubic meter?
We know that 1 meter is equal to 10 decimeters:

1 m=10 dm
To find the relationship between cubic meters and cubic decimeters, we
must cube both sides. When we do this, we must cube both the numbers
and the units

(1 m)3=(10 dm)31 m3=1,000 dm3

Although 1 meter contains 10 decimeters, 1 cubic meter contains 1,000
cubic decimeters. The following examples illustrate this idea further.

A cubic meter is about the size of a washing machine. A cubic decimeter (or
liter) is about the size of a tissue box. One m3 contains 1,000 dm3.

Example 2.11 Conversions with Cubic Units

How many cubic centimeters (cm3) are in 1 cubic meter (m3)?
From the common metric unit conversions in Table 2.5, we know that

100 cm=1 m

112
To convert to cm3 and m3, we cube both sides of the equation:

(100 cm)3=(1 m)3

We cube the numbers, and we also cube the units. This means that

1,000,000 cm3=1 m3
So although there are just 100 centimeters in 1 meter, there are 1 million
cubic centimeters in a cubic meter! •

Example 2.12 Conversions with Volume Units

A liquid has a volume of 15.3 cubic centimeters (cm3). What is this volume
in liters?
To make this conversion, we first identify our unit relationships. We know
that 1 cubic centimeter is equal to 1 milliliter (mL), and there are 1,000 mL
in 1 L. To solve this problem, we begin with 15.3 cm3, then multiply by
two conversion factors. The first cancels out cm3 (leaving us with mL). The
second cancels out mL, leaving us with liters. •

Example 2.13 Conversions with Volume Units

A hospital IV (Figure 2.15) is set to drip at a rate of 125 mL/h. How many
liters of fluid is this patient receiving per day?

113
 Figure 2.15 This intravenous (IV) drip is set to deliver fluid at 125 mL/hour.

In this problem, we need to convert both our volume and time units. We
begin with the drip rate in mL/h, then use two conversion factors. The first
converts volume from mL to L, and the second converts the time unit from
hours to days. This gives us a final answer of 3.00 L/d:

Notice that the answer has three significant digits. The two unit conversions
are both exact numbers, so they do not limit the number of significant digits
in the calculations. Because the only measured quantity (the drip rate of
125 mL/h) has three significant digits, we can keep this many digits in our
final answer. •

IT
TRY

17. A raindrop has a volume of about 0.05 cm3. Express this volume in
the following units:
a. milliliters
b. liters
c. cubic meters

18. On average, a volume of 84,760 ft3 of water flows over Niagara Falls
each second. At this rate, how many m3 of water flow over the falls

114
in an hour? (1 meter = 3.281 feet.)

115
2.3 Density: Relating Mass to Volume
Materials are usually measured by mass or by volume. It is often necessary
to convert between the mass and volume of a substance. To do this, we use
a property of the material called the density. Density is defined as the mass
of a substance per unit volume. That is:

density=massvolume
Or, written in abbreviated form:

d=mv
Density is an important physical property. For example, aluminum and
titanium metal are highly valued for uses from automobiles to laptop
computers to prosthetic limbs (Figure 2.16). They are strong, and because
they have low densities, an object made from one of these metals weighs
less than the same object made from other metals, like iron or lead. Table
2.6 summarizes the densities of several common materials.

Figure 2.16 Prosthetic limbs, such as the one worn by U.S. Army Sergeant and 100-m
gold medalist Jerrod Fields, must be very strong but also very lightweight.

TABLE 2.6 Densities of Common Materials

116
 TABLE 2.6 Densities of Common Materials
Material Density (g/cm3)

Aluminum 2.70

Titanium 4.51

Iron 7.87

Copper 8.96
Lead 11.34

Gold 19.31

Water* 1.00
Seawater* 1.02

Air* 0.001

* At 25 °C and standard atmospheric pressure

To measure the density of a sample, we must first determine the mass

and volume. For example, if we want to measure the density of a solution
of sodium chloride, we could measure the mass of the solution on a
balance, and measure the volume of the solution using a graduated
cylinder (Figure 2.17).

Figure 2.17 A balance measures mass. A graduated cylinder measures the volume of a
liquid. By measuring both quantities, we can determine the density of a liquid.

117
 Explore

Figure 2.17

If the mass is 11.29 grams, and the volume is 10.4 milliliters, we

calculate the density this way:
d=mv=11.29 g10.4 mL =1.09 g/mL

Notice that as before, we’ve rounded the answer to the correct number of
significant digits.

Converting between Mass and Volume

One of the most important uses of density is in unit conversions. If we
know the density of a sample, we can relate its mass and volume. This
process is illustrated in the following examples.

Example 2.14 Conversions Using Density I

A sample of iron metal has a volume of 8.83 cm3. If the density of iron is
7.87 g/cm3, what is the mass of the iron sample?
By rearranging the density equation, we see that the mass is equal to

118
 density times volume. Substituting in the values of the density and the
volume, we make the following calculation:
m=dV=(7.87gcm3)(8.83 cm3)=69.5 g

The mass of the iron sample is 69.5 grams. In this calculation, the units of
volume (cm3) canceled out, leaving grams as the unit of mass. It is
important to make sure that the units of volume are the same in the density
and volume terms. If the units do not cancel out correctly, the answer will
not be correct. •

Example 2.15 Conversions Using Density II

Mercury is a liquid with a density of 13.534 g/mL. What is the volume of a
1.213-kg sample of mercury?
To solve this problem, we first need to make sure that our mass units are
the same. Because the density of mercury is given in units of g/mL, we
need to convert the mass of the sample from 1.213 kg to 1,213 g. This will
allow us to cancel out the units in the next step.
We next rearrange the density equation to show that volume is equal to
mass over density. We then substitute in the values of the mass and the
density, and solve. The grams cancel out, leaving us with units of
milliliters, as follows:
V=md=1,213 g13,534 g/mL=89.63 mL

Rounding to four significant digits, we get an answer of 89.63 milliliters. •

We use density to convert between mass and volume.

Will It Float?

119
The density of an object determines whether it floats or sinks in water.
Pure water has a density of 1 g/cm3. Objects that are less dense than water
(d < 1 g/cm3) float, while objects more dense than water (d > 1 g/cm3)
sink (Figure 2.18).

Figure 2.18 In order to float, an object’s density must be less than that of water.

The density of water is 1 g/mL.

Example 2.16 Will It Float?

A toy submarine has a volume of 14.3 mL and a mass of 15.2 grams. Will
the toy float or sink?
To solve this problem, we first find the density:
d=mV=15.2 g14.3 mL=1.06 g/mL

The density of this toy is greater than the density of water (1 g/mL), so the
toy will sink. •

IT
TRY

19. An unknown liquid has a volume of 15.0 mL and a mass of 11.3 g.

What is the density of this liquid? Based on this information, what
mass would you expect for a 2.50 L sample of this liquid?

20. You recently began working for a rock quarry that produces gravel.

120
 The company has purchased a truck with a capacity of 10.0 m3, and
you have been asked to determine how many kilograms of gravel this
truck will carry. To answer this question, you measure 1.0 L of
gravel and find that it has a mass of 1.8 kg. Based on this result, what
is the density of gravel in kg/L? And in kg/m3? What is the mass of a
10.0-m3 load of this gravel? (1,000 L = 1 m3.)

21. A cylindrical steel tank with a diameter of 20.0 cm, a length of 150.0
cm, and an overall mass of 150.0 kg is accidentally dropped into a
lake. What is the cylinder’s density in g/cm3? Will it sink or float?

121
2.4 Measuring Temperature
In chemistry, we often need to measure and report temperature. There are
three common temperature scales. In the United States, the Fahrenheit
scale is most commonly used. Most of the world uses the Celsius scale.
On the Celsius scale, the freezing point of water is 0 °C, and the boiling
point is 100 °C. Notice in Figure 2.19 that the difference between the
freezing point and boiling point of water is 180 °F, but only 100 °C. This
means that a degree Fahrenheit is a smaller unit than a Celsius degree. To
convert between Celsius and Fahrenheit, we use the following
relationships:

Figure 2.19 Degrees in the Celsius and Kelvin scales are the same size, but a degree
Fahrenheit is a smaller unit.

°F=95 °C+32°C=59 (°F-32)

A third temperature scale is the Kelvin scale. Scientists use this scale
to describe events at very low temperatures and also to predict the way
gases behave. The Celsius and Kelvin scales have the following
relationship:

K=°C+273.15
A unit on the Kelvin scale is the same as a degree Celsius. The difference

122
between the two scales is the zero point. A temperature of 0 °C
corresponds to the freezing point of water, while a temperature of 0 K
corresponds to absolute zero (the lowest possible temperature, at which
particles have zero kinetic energy).

When reporting the temperature in Fahrenheit or Celsius, the word degrees

follows the number. For example, “32 degrees Fahrenheit equals 0 degrees
Celsius.” In contrast, with the Kelvin scale, the word degrees is not included, so
it is correct to say, “Zero degrees Celsius is equal to 273.15 Kelvin.”

IT
TRY

22. While on a business trip to Brussels, Belgium, you hear a weather

forecast predicting a high temperature of 15 °C. What is this
temperature in degrees Fahrenheit? Should you take a jacket when
you leave your hotel?

123
Summary
In this chapter, we examined the key concepts of measurement and
method, which are both critical to the scientific process.
When conducting measurements, it is important to use units of
measurement that can be clearly communicated to others. The international
scientific community has adopted a set of fundamental units, called SI
units, that are based on the metric system. There are seven fundamental
units of measurement; other units are derived from these fundamental
units. With any measurement, it is important to be clear what units are
used.
We can describe measurements in terms of their accuracy (how close
they are to the true value) or their precision (how finely we are able to
make a measurement). The appropriate number of significant digits
communicates the precision of a measurement.
Dimensional analysis, also called the factor-label method, is a
standard, broadly effective way of converting between units. When
converting between units, we use a fractional relationship containing
equivalent amounts (but different units) in the numerator and the
denominator. In these problems, we cancel units in the same way we
cancel numerical values.
Often, we need to convert between measurements of mass and volume
of a substance. To do this, we use the density of the substance. Density is
defined as the mass of a substance divided by its volume. In addition to its
usefulness as a conversion factor, density is an important physical property
that we use for many applications.
When measuring temperature, we use the Celsius, Kelvin, or
Fahrenheit scales. The Celsius scale is the most commonly used scale for
scientific measurements, although the Kelvin scale is important for
calculations involving gases or very low temperatures.

124
A Strange Death

At the beginning of this chapter, we described the case of a man who had died as a
result of consuming MDPV, a synthetic drug similar to methamphetamine. This
was the first death in West Virginia to be linked to MDPV. Following the findings
in this case, however, investigators attributed the abuse of this substance to a series
of other deaths. The case provides an interesting example of the role of
measurement and scientific method in solving problems. Let’s briefly review how
the scientists approached this problem, and examine their findings in more detail.
The toxicologists assigned to the case, James Kraner and Lauren Waugh, first
determined whether other common drugs were involved. They used routine tests
for detecting the presence of popular street drugs, such as cocaine and
methamphetamine, in the victim’s blood and urine. One of the keys to effective
science is to understand and apply what is already known. Sometimes a day in the
library can save a year in the lab.
These routine tests indicated that none of these more common drugs were
present at the time of death. Having ruled out the other drugs, the toxicologists
hypothesized that the substance in the container might have caused the erratic
behavior. To identify the substance, they used an instrument called a mass
spectrometer, which enables scientists to measure the masses of individual
molecules (we will describe this instrument in more detail in Chapter 7). By
combining this measurement with other tests, Kraner and Waugh were able to
positively identify the substance as MDPV.
After identifying the unknown compound, they needed to determine how much
(if any) MDPV was in the body when the person died. To do this, Dr. Waugh
prepared a set of samples of blood serum (a liquid component of blood) that
contained different known amounts of the drug (Figure 2.20). She also tested
samples that had no drug present, to make sure the test would give a negative
response in the absence of the drug.

125
Figure 2.20 Forensic chemists measure the amount of drugs such as MDPV in blood
samples such as those shown here.

After all of this preparatory work, Dr. Waugh was finally ready to test the
victim’s blood and urine for MDPV levels. By comparing the results of these tests
with known solutions, she determined that the blood of the deceased contained
1.09 milligrams of MDPV per liter of blood (that is, 1.09 mg/L). The MDPV
concentration in the urine was 60.3 mg/L.
Analysis of blood and urine samples is a common way to test for illegal drugs
or other toxic substances. In combination with preparation, carefully planned
methods, and meticulous measurement, tests like this can yield a wide range of
forensic and health information.

126
 Key Terms
2.1 Measurement: A Foundation of Good Science
scientific notation A way to show very large and very small numbers in a
concise format. Scientific notation expresses numbers as the product of two
values, called the coefficient and the multiplier.
units of measurement Quantities with accepted values.
liter (L) A common unit of volume, defined as 1 cubic decimeter (1 dm3).
accuracy A measure of how reliable measurements are—that is, how closely
they reflect the true value.
precision A measure of how finely a measurement is made, or how close a
group of measurements are to each other. Precision is often denoted by
significant digits.
significant digits The digits contained in a measured value. The number of
significant digits indicates how precisely a measurement is made. Also called
significant figures.
exact numbers Numbers for which there is no uncertainty. Counted integers
and defined relationships (such as metric prefixes) are exact numbers.

2.2 Unit Conversion

conversion factors Fractions that are used to convert from one unit to another.
A conversion factor contains equivalent amounts of different units in the
numerator and the denominator.

2.3 Density: Relating Mass to Volume

density A physical property of a substance, defined as the mass per unit
volume.

2.4 Measuring Temperature

Fahrenheit scale (°F) A temperature scale commonly used in the United
States. On the Fahrenheit scale, water freezes at 32 °F and boils at 212 °F.
Celsius scale (°C) A temperature scale commonly used throughout the world.
On the Celsius scale, water freezes at 0 °C and boils at 100 °C. Sometimes
called the Centigrade scale.
Kelvin scale (K) The official SI scale for measuring temperature. The Kelvin
scale sets absolute zero at 0 K, the freezing point of water at 273.15 K, and the
boiling point of water at 373.15 K.

127
 Additional Problems

2.1 Measurement: A Foundation of Good Science

23. Write the following values in scientific notation:

a. 1,500,000 km
b. 0.000398 g
c. 1,200,000,000 J
d. 0.02019 s

24. Write the following values in scientific notation:

a. 13,201 kg
b. 0.00000593 g
c. 1,400,000,000 km
d. 0.00322 s

25. Convert the following values to standard notation:

a. 5.192 × 10−4 kg
b. 1.23 × 103 J
c. 4.2 × 108 °C
d. 6.02 × 1023 atoms

26. Convert the following values to standard notation:

a. 8.13 × 10−3 ng
b. 3.4 × 107 lb
c. 4.2 × 105 molecules
d. 1.301 × 102 m

27. Each of these numbers is written in exponential notation, but not in proper
scientific notation. Write each number correctly. (Hint: The coefficients do
not have one nonzero digit in front of the decimal point. Shift the decimal
point to correct this, and adjust the exponents to correspond to the change.)
a. 52.1 × 109 min
b. 0.83 × 109 g

128
 c. 435 × 10−9 m

28. Each of these numbers is written in exponential notation, but not in proper
scientific notation. Write each number correctly. (Hint: The coefficients do
not have one nonzero digit in front of the decimal point. Shift the decimal
point to correct this, and adjust the exponents to correspond to the change.)
a. 533.1 × 10−9 m
b. 0.083 × 103 J
c. 435 × 10−16 kg

29. Complete each calculation:

a. (4.4 × 10−3)(1.2 × 10−5)
b. (1.4 × 10−3) ÷ (1.2 × 10−5)
c. (5.4 × 105)(2.2 × 103)
d. (8.132 × 10−2)/(4.19 × 103)

30. Complete each calculation:

a. (1.4 × 106)(9.1 × 10−5)
b. (1.4 × 10−3)2
c. (5.4 × 105)/(3.7 × 10−5)
d. (2.1 × 104) ÷ 3

31. Using the units and prefixes in Tables 2.3 and 2.4, determine the name of
the unit represented by each of these abbreviations:
a. A
b. K
c. ms
d. mg
e. kA

32. Using the units and prefixes in Tables 2.3 and 2.4, determine the name of
the unit represented by each abbreviation:
a. µA
b. nm
c. kg
d. mm

129
 e. µs

33. Refer to Table 2.5 to answer the following:

a. How many grams are in a megagram?
b. How many calories are in a kilocalorie?
c. How many ms are in 1 s?
d. How many A are in 1 GA?

34. Refer to Table 2.5 to answer the following:

a. How many nanoseconds are in a second?
b. How many watts are in a gigawatt?
c. How many µg are in a g?
d. How many J are in a kJ?

35. Which is a more precise measurement of distance: a number that is

rounded to the nearest gigameter or a number that is rounded to the nearest
nanometer?

36. Which is a more precise measurement of mass: a number that is rounded to

the nearest milligram or a number that is rounded to the nearest
microgram?

37. In the laboratory, you measure out a compound to be used in a chemical

reaction using two different balances. The first balance reports the mass as
10.23 grams. The second balance reports the mass as 11.1925 grams. Can
you tell which balance is more accurate? Can you tell which measurement
is more precise?

38. In a forensic laboratory, a toxicologist uses an analytical balance (an

instrument for measuring small masses) to determine that a skin sample
has a mass of 0.30 milligrams. Nearby, at the local salvage yard, a truck
pulls onto a scale, and the operator determines that the scrap metal in the
truck has a mass of 4,813 kilograms. Which measurement has more
significant digits? Which measurement is more precise?

39. Consider this image of a test tube and ruler. Should the length of the test
tube be reported as 5 cm, as 5.1 cm, or as 5.12 cm? Why?

130
40. This graduated cylinder shows a volume that is more than 5 mL but less
than 6 mL. How precisely can you estimate this volume? Report the
volume to the correct number of significant digits.

41. Identify the number of significant digits in the following measured

numbers:
a. 0.0005123 g
b. 32.01 lb
c. 43.300 m
d. 4.0 GPA

42. Identify the number of significant digits in the following measured values:
a. 2.304 km
b. 0.002000 g
c. 12.0 mA
d. 3.023 × 104m

43. A builder measures the length of a wall for a new home and finds it is
20.31 meters. How many significant digits are in this measurement?

44. The mass of a wood shard recovered from a fire is 0.0397 g. How many

131
 significant digits are in this measurement?

45. The following values are written in standard notation. Convert each value
to scientific notation, using the correct number of significant digits.
a. 5,200,000 ng, measured to four significant digits
b. 0.0003920 L, measured to four significant digits
c. $1,800,000, measured to six significant digits

46. The following values are written in standard notation. Convert each value
to scientific notation, using the correct number of significant digits.
a. 4,260,000,000 bytes of data, measured to four significant digits
b. 0.000320 L, measured to three significant digits
c. 52,000,000,000 km, measured to six significant digits

47. What is the difference between a measured number and an exact number?

48. A dairy farmer reads a study that says lactating cows (cows that are
producing milk) consume an average of 27 pounds of dry hay each day.
The farmer has 44 cows and projects hay costs based on these numbers.
Which of these values is measured and which is exact?

49. Identify each value as measured or exact:

a. There are 40 quarters in a $10.00 quarter roll.
b. The distance from your house to your workplace is 6.2 miles.
c. There are 105 nutritional calories in a medium banana.
d. An octagon has 8 sides.

50. Identify each value as measured or exact:

a. A cyclist rides 82 kilometers in one day.
b. The cyclist’s bicycle has 2 wheels.
c. The month of July contains 31 days.
d. There are 1,000 mg in 1 gram.

51. Carry out each calculation, rounding to the appropriate number of

significant digits.
a. the area of a property: 103.2 ft × 59.3 ft
b. the perimeter of a triangle: 11.502 cm + 18.00 cm + 1.92312 cm
c. the density of a block of iron: 25.3261 grams ÷ 3.6 cm3

132
52. Carry out each calculation, rounding to the appropriate number of
significant digits:
a. the total time spent running during a week: 25.24 minutes + 28.1
minutes + 56 minutes
b. the circumference of a circle: C = 2πr, where r = 1.32 inches and the
number 2 is an exact number
c. the wavelength of blue light (in meters), found by dividing the speed of
light by the frequency: (3.00 × 108 m/s) ÷ (6.911 × 1014 s−1)

53. You and a friend win $500.00 and divide it evenly. How much does each
of you get? Make sure your answer has the correct number of significant
digits.

54. Your sales team wins a bonus of $900.00, to be divided evenly. If there are
seven people on your sales team, how much will each of you get? Make
sure your answer has the correct number of significant digits.

55. To measure the perimeter of a square, you measure the length of one side
to be 12.6 cm. The perimeter is equal to the length of one side times the
number of sides: 12.6 cm/side × 4 sides. Report the perimeter of the square
with the correct number of significant digits.

56. You are considering a new job with a long commute. To measure the
distance, you drive from your home to the worksite and back. Your car’s
odometer reports the round-trip distance as 53.8 km. If you made exactly
20 trips to this new job in a month, how many miles would you commute?
Report your answer to the correct number of significant digits.

57. A quality-control process measures the mass of new screws coming off a
press. A sample of ten screws is collected, and the following masses are
recorded: 10.31 g, 10.33 g, 10.27 g, 10.31 g, 10.50 g, 10.31 g, 10.28 g,
10.39 g, 10.41 g, and 10.36 g. Calculate the total mass and the average
mass of these screws. Make sure your answer has the correct number of
significant digits.

58. While working with a wildlife biologist, you have the task of sampling the
average size of bluegills (Figure 2.21) in a nearby river. In one day you
catch eight fish with the following length measurements: 10.2 cm, 15.1
cm, 15.3 cm, 19.5 cm, 12.1 cm, 14.3 cm, 23.4 cm, and 16.1 cm. What is
the average length of the fish you caught? Make sure your answer has the
correct number of significant digits.

133
 Figure 2.21 Bluegills, like this one, are members of the sunfish family common in U.S.
lakes east of the Rocky Mountains.

2.2 Unit Conversion

59. Perform the following metric conversions using the factor-label method:
a. How many ng are in 0.0213 g?
b. Convert 17,397.4 m to km.
c. Express 0.000310 L in μL.

60. Perform the following metric conversions using the factor-label method:
a. Convert 1.5 × 1014 A to GA.
b. Convert 0.0320 cm to μm.
c. Convert 25.4 ng to g.

61. Complete the following metric conversions:

a. Convert 23.21 µL to L.
b. Convert 50,000 g to kg.
c. Convert 5.40 × 10−7 m to nm.

62. Complete the following metric conversions:

a. Convert 4.3 L to mL.
b. Convert 4,320 mg to g.
c. Convert 1.53 × 10−7 g to ng.

134
63. The unit of electrical resistance is the ohm (Ω). A circuit has a resistance
of 5.4 × 105 Ω. What is this amount in megaohms (MΩ)?

64. The explosive power of a nuclear weapon is usually expressed as the

number of tons of TNT that would be required to produce the same
explosion. The atomic bomb used in Hiroshima, Japan, in 1945 had a
power of approximately 1.5 × 104 tons of TNT. In 1961, the Soviet Union
detonated a test bomb that measured a power of 5.0 × 1010 tons of TNT.
Express the strength of these two weapons in kilotons and in megatons.

65. Refer to Table 2.7 to perform the following conversions:

a. 15.2 inches to centimeters
b. 47.23 kilocalories to joules
c. 1.2 tons to kilograms
d. 3.55 miles to kilometers

TABLE 2.7 Common Conversions

Length

1 inch = 2.54 cm

1 mile = 1.609 km

Volume

1 gallon = 3.785 liters (L)

Mass/Weight (at Earth’s gravity)

1 kg = 2.20 pounds (lb)

1 ton = 2,000 lb

Energy

1 calorie (cal) = 4.184 joules (J)

66. Refer to Table 2.7 to answer the following conversion questions:

a. How many milliliters are in 2.0 gallons?
b. How many calories are in 8.14 × 106 joules?
c. How many meters are in 4.23 × 103 miles?
d. How many kilocalories are in 3.2 × 104 joules?

135
67. While trying to remove a bolt with a socket wrench, you find that the bolt
is sized in units of millimeters and is slightly smaller than a ½-inch socket.
What size metric socket should you use? (This will be an integer value.)

68. Allen wrenches are small, L-shaped wrenches used to assemble bicycles,
furniture, and other common items. Allen wrenches come in both metric
and English sizes. A 6-mm wrench is almost, but not exactly, the same size
as a ¼-inch wrench. Which is smaller?

69. A plane flies at a speed of 540 miles per hour. What is this speed in meters
per second?

70. A leaky faucet drips water at a rate of 35 mL per hour. Express this rate in
terms of liters per day.

71. A small car has a fuel efficiency of 34.3 miles per gallon. What is this fuel
efficiency in kilometers per liter?

72. A large pickup truck has a fuel efficiency of 16.2 miles per gallon. Express
this fuel efficiency in kilometers per liter.

73. Rocephin® is an antibiotic sometimes used to treat skin infections. In

children, the maximum recommended daily dose of this medicine is 75
mg/kg of body mass. Express this dosage in g/lb, using the conversion
(2.20 kg = 1 lb).

74. Adderall® is used to treat attention-deficit hyperactivity disorder (ADHD).

In children, a dosage of 10 mg per day is fairly common. At this rate, how
many grams of Adderall does a child take over the course of a year?

75. Large amounts of the herbicide glyphosphate have been linked to kidney
problems and reproductive difficulties. The U.S. Environmental Protection
Agency limits the levels of glyphosphate in drinking water to a maximum
of 0.7 mg/L. What is this value in g/m3?

76. Dioxin is a highly toxic compound that has been shown to increase the risk
of cancer in humans. Dioxin is sometimes produced from waste
incineration. The U.S. Environmental Protection Agency limits the levels
of dioxin in drinking water to a maximum of 30 ng per liter. If a 1,500-mL
sample of water contains 1.2 × 10−8 g of dioxin, is it above the maximum

136
 level?

77. How many square meters are in a square kilometer? (Hint: It’s not 1,000.
Watch your units.)

78. How many square feet are in a square yard?

79. The length, width, and height of a box are each 1 yard. What is the volume
of this box in cubic feet (ft3)?

80. How many dm3 are in 1 m3? How many cm3 are in 1 m3?

81. How many liters are in 1 m3? How many milliliters are in 1 m3?

82. How many cm3 are in 1 milliliter? How many cm3 are in 1 liter?

2.3 Density: Relating Mass to Volume

83. Cobalt has a density of 8.90 g/cm3 at room temperature. What is this
density in kg/m3?

84. Aluminum has a density of 2.70 g/cm3 at room temperature. What is this
density in kg/L?

85. A sample of an unknown solid has a mass of 15.23 g and a volume of 23.7
mL. What is the density of this material? Report your answer to the correct
number of significant digits.

86. A sample of an unknown liquid has a mass of 20.365 g and a volume of

28.1 mL. What is the density of this material? Report your answer to the
correct number of significant digits.

87. To determine the density of a liquid, you measure out a volume of 9.3 mL.
The mass of the liquid is 9.649 g. Report your answer to the correct
number of significant digits.

88. A small metal cube is 1.34 cm on each side and has a mass of 18.936
grams. What is the density of this cube?

137
89. Indicate whether each of the following would float or sink in water:
a. a solid having a mass of 42.3 grams and a volume of 40.1 cm3
b. a solid having a mass of 42.3 grams and a volume of 44.1 cm3

90. Indicate whether each of the following would float or sink in water:
a. a rectangular block of length 5.0 cm, width 4.0 cm, and height 2.0 cm,
with a mass of 0.583 kg
b. a diving tank with a volume of 80 cubic feet filled with air; the filled
tank has a weight of 21 pounds

Questions 91–94 refer to Table 2.8.

TABLE 2.8 Densities of Common Metals

Material Density (g/cm3)

Aluminum 2.70

Titanium 4.51

Iron 7.87

Copper 8.96

Lead 11.34

Gold 19.31

91. Calculate each of the following:

a. the volume of a bar of pure gold with a mass of 5.00 grams
b. the volume of an iron statue with a mass of 1.8 kg

92. Calculate each of the following:

a. the mass of a copper pipe in which the copper has a volume of 70.5 cm3
b. the mass of a titanium rod having a volume of 0.81 cm3

93. You have three blocks—one made of iron, one made of aluminum, and one
made of lead. Each block is 5.0 cm × 4.0 cm × 1.0 cm. What is the mass of
each block? Make sure your answer has the correct number of significant
digits.

94. To measure the density of a small statue, you determine the mass of the
statue to be 2.219 kg. Next, you pour 400 milliliters of water into a
graduated cylinder (measured to the nearest mL). You then gently lower

138
 the statue into the water and observe that the volume rises to 682
milliliters. Calculate the density of the statue. Based on Table 2.8, what
element might the statue be made of?

2.4 Measuring Temperature

95. What is absolute zero? What are the values of absolute zero on the
Fahrenheit, Celsius, and Kelvin temperature scales?

96. The average temperature of the human body is 98.6 °F. What is this
temperature in °C and K?

97. Which is a larger temperature change: a change of 1.3 °C or a change of

2.5 °F?

98. Which is larger, 1 degree Celsius or 2 degrees Fahrenheit?

99. Room temperature is typically around 72 °F. Convert this temperature to

degrees Celsius and to kelvins.

100. Convert 400 K to degrees Celsius and degrees Fahrenheit.

Challenge Questions

101. A family needs to replace the flooring in their house. The local hardware
store has two options that they like: carpet, which sells for $5.80 per
square yard, or hardwood, which sells for $1.40 per square foot. Which
option is less expensive? If they need to replace 600 ft2 of flooring, how
much will each option cost?

102. For your new candle-making business, you need to purchase a large
amount of a unique scented wax. You plan to charge $9.95 per large
candle. The wax you need is available from a U.S. supplier for $24.00/lb
and also from a German supplier for €9.20/kg. If the current exchange
rate is $1 = €0.76, and 1 kg = 2.20 lb, which supplier is giving the better
price?

103. You are on the management team of a company that is considering

purchasing a tanker truck. The truck you want has a volume of 30,000
liters and an empty mass/weight of 15,800 kg. However, the route you
wish to travel has a bridge with a weight limit of 25 tons. If you purchase

139
the truck and fill it to capacity with a liquid whose density is 0.80 kg/L,
will the truck be too heavy to cross the bridge? (1 kg = 2.20 lb)

140
Chapter Three
Atoms

Mercury Contamination from Small-Scale Gold

Mining
Every summer Adam Kiefer packs a tent, minimal personal supplies, and several
suitcases of testing equipment. He then boards a plane bound for a developing
country such as Mozambique, Ecuador, or Guyana. For several weeks he, along
with colleague Kevin Drace and their team, travels through areas rich with gold.
But for Kiefer—a chemistry professor at Mercer University—and his team of
students, gold mining is not the objective; it’s the problem. Mining practices in
these regions produce large amounts of toxic mercury vapors that contaminate the
soil, air, and water, causing severe health problems for the people who live nearby.
Kiefer and his team work with the miners and residents, measuring mercury levels,
teaching safer techniques, and working to clean up some of the most highly
polluted regions in the world (Figure 3.1).

141
Figure 3.1 (a) Community mines like this one are common in Mozambique. (b) A
miner grinds ore into fine pieces. (c) This worker is sifting the ore from the lighter soil.
Mercury is added during this phase. (d) The mixture of mercury and gold that forms this
tiny pellet is called an amalgam. (e) A miner heats the amalgam on hot coals to
evaporate the mercury. (f) Kiefer collects soil around a mining site.

Why does gold mining release toxic mercury? To answer that question, we
need to know a bit more about the mining process in places like Mozambique.
Despite the abundance of gold, millions of people in the country are desperately
poor. Many residents make their living as artisanal miners in small-scale gold
mines. Each day, the miners descend into community mines to collect coarse gold-
containing ore. They carry the ore to a grinding facility, where they mill the hard
rock down to fine particles. Miners then place the particles in large bowls, add
water, and stir. This causes the lighter soil to rise to the top, where it can be drained
off. The heavier ore remains in the bottom of the bowl. After removing the soil, the
miners stir in elemental mercury. The gold particles dissolve in the mercury,
forming small, shiny pellets containing a mixture of the two elements, called an
amalgam. To separate the gold from the mercury, the miners place the solid

142
amalgam onto a smoldering log. The heat from the fire evaporates the mercury,
leaving behind tiny pellets of purified gold. The miners sell the gold to local
dealers, who come around each day. For their efforts, miners earn just enough to
survive.
Mercury is highly toxic, affecting the central nervous system. Artisanal miners
in Mozambique suffer the effects of mercury poisoning: tremors, anxiety, memory
loss, and eventually insanity. These health problems extend beyond the miners to
the larger community. Small-scale gold-mining operations release more mercury
into the environment than any other human activity, producing widespread effects.
Each year, Kiefer and his team work to make measurements and raise
awareness in the surrounding communities. As he puts it, “These miners don’t
pollute the environment because they don’t care about their community; they do it
because they need to feed their families. By using science and technology, we
teach miners about the hazards of elemental mercury and how to protect
themselves, their families, and the environment.”
Gold and mercury have many similarities but also key differences. Both
elements are metals. They are both shiny, melt at fairly low temperatures, and
conduct electricity. However, gold is a solid at room temperature while mercury is
a liquid. And although many people have died fighting over gold, gold itself does
not pose the health hazards that mercury does.
Why is this? Why are these elements similar? What causes their differences?
The answers to these questions are complex. However, as we saw in Chapter 1, the
properties of any substance are connected to its structure. In this chapter, we’ll
begin to unravel the story of the elements—how they were discovered and how
their structures are alike and different. We’ll lay the groundwork that will allow us
to answer the question, “Why does this element behave the way it does?” These are
important concepts; let’s begin to dig.

Intended Learning Outcomes

After completing this chapter and working the practice problems, you should be
able to:

3.1 Atoms: The Essential Building Blocks

Describe the development of atomic theory and its key observations about
atoms.
Apply the law of conservation of mass to solve mass problems related to
chemical reactions.
Describe chemical changes using atomic theory.

3.2 The Periodic Table of the Elements

Describe the organization of the periodic table.

143
3.3 Uncovering Atomic Structure
Describe the behavior of charged particles.
Describe how the discovery of the battery and Rutherford’s gold foil
experiment shaped our understanding of atomic structure.
Describe the relative mass and charge of protons, neutrons, and electrons and
their arrangement within an atom.

3.4 Describing Atoms: Identity and Mass

Relate the number of protons to atomic number and the sum of nuclear
particles to mass number.
Describe the nuclear structure of isotopes and calculate average atomic mass
from a distribution of isotopes and relative abundances.
Differentiate between mass number and average atomic mass.

3.5 Electrons—A Preview

Contrast the description of electrons in the Bohr model and the quantum
mechanical model.
Identify the overall charge of an atom or ion based on the number of protons
and electrons.

144
3.1 Atoms: The Essential Building Blocks
Right now, somewhere, a breeze is blowing across turquoise-blue water. A
boat creaks as it rocks in the gentle waves. Somewhere else, snow is
falling. Somewhere a person is sitting down with a cup of coffee and
reading the news on her tablet. Somewhere a masterpiece is being painted.
The world around us—everything we can see, touch, taste, hear, or
smell—arises from matter. And all matter is composed of a basic set of
elements. At present, there are 118 known elements, but only about 98 of
those are naturally occurring. The stunning diversity of the world around
us arises from a very small set of building blocks.
At the heart of these elements is the atom. The atom is the fundamental
unit of matter. Each of the elements is composed of a different type of
atom, which gives each element its unique properties. As atoms combine
into compounds and mixtures, materials with new and different behaviors
are produced (Figure 3.2).

Figure 3.2 The world around us is made of atoms. The type and arrangement of atoms
give each material its unique properties.

Uncovering the Atom: From Democritus to Dalton

The word atom derives from the Greek word atomos, meaning
“indivisible.” The Greek philosopher Democritus (Figure 3.3), who lived
around 400 B.C.E., argued that if you cut a substance in half, what results is
two pieces of the same substance. He said that if you cut something in half
over and over, you would eventually reach a point where you could no
longer divide it and still have the same substance. At that point it was
atomos, “indivisible.”

145
Figure 3.3 This bust of Democritus is housed in the Naples National Museum in Italy.

The idea of the atom remained dormant until about the 1700s. Then,
building on the technological innovations of the Renaissance, scientists
began exploring chemical behavior more systematically and unlocking
clues about atomic structure.
In the late 1700s, Antoine Lavoisier cataloged a large number of
chemical reactions. He observed that when a chemical change takes place,
the total mass before and after the reaction is the same. Based on this
research, he formulated the law of conservation of mass, which states that
in chemical reactions, matter is neither created nor destroyed.
For example, hydrogen and oxygen react explosively to form water
(Figure 3.4). Suppose we combined 4 grams of hydrogen with 32 grams of
oxygen in a sealed container and then caused the two elements to react.
The water produced would have a mass of 36 grams. That is, we have the
same amount of mass at the end of the reaction as we had at the beginning.

146
Figure 3.4 The law of conservation of mass tells us that matter is not created or
destroyed in a reaction. The mass before the reaction is the same as the mass afterward.

Explore

Figure 3.4

We can use the law of conservation of mass to predict the amount of

substances that will be consumed or produced in chemical reactions. The
following example illustrates this principle.

Example 3.1 Using the Law of Conservation of Mass

The bright light of a signal flare is caused by a chemical change in which
magnesium combines with oxygen to produce a compound known as

147
 magnesium oxide. If 48.6 grams of magnesium combines with 32.0 grams
of oxygen, how many grams of magnesium oxide will form?
We can represent this change as follows:
magnesium48.6 g+oxygen32.0 g→changemagnesium oxide

The law of conservation of mass states that the total amount of matter
before and after a chemical change is the same. Together, the masses of
magnesium and oxygen are 80.6 grams. Based on this law, 80.6 grams of
magnesium oxide will form. •

IT
TRY

1. In an acetylene torch, acetylene gas reacts with oxygen to produce

carbon dioxide and water (and a lot of heat). If 26 kg of acetylene is
burned, 88 kg of carbon dioxide and 18 kg of water are produced.
How many kilograms of oxygen gas are consumed in this reaction?

In 1808, John Dalton, an English schoolteacher and weather enthusiast,

published a paper that articulated a framework for the modern atomic
theory (Figure 3.5). Dalton’s theory described several concepts that laid
the foundation for our understanding of chemistry:

148
Figure 3.5 This depiction of atoms and compounds was published by John Dalton in
1808.

Elements are made of tiny, indivisible particles called atoms.

The atoms of each element are unique.
Atoms can join together in whole-number ratios to form
compounds.
Atoms are unchanged in chemical reactions.
To appreciate the impact of Dalton’s ideas, let’s consider an example.
When charcoal burns (Figure 3.6), the charcoal—which is essentially the
element carbon—reacts with oxygen in the air to form a new compound,
carbon dioxide. From Dalton we know that carbon atoms are distinct from
oxygen atoms. When they combine, they do so in a whole-number ratio.
One carbon atom combines with two oxygen atoms to produce one

149
molecule of carbon dioxide. The atoms themselves are not changed, but
the properties of carbon dioxide are different from the properties of carbon
and oxygen.

Figure 3.6 Carbon (charcoal) reacts with oxygen to form carbon dioxide.

Atoms are not created or destroyed in chemical reactions.

From this brief history, here are three fundamental ideas that you
should be sure to understand:

1. All matter is composed of atoms.

2. The atoms of each element have unique characteristics and properties.
3. In chemical reactions, atoms are not changed, but combine in whole-
number ratios to form compounds.

The study of chemistry springs from these important concepts.

Can We See Atoms?

Today the ideas that Dalton developed are essential to our understanding
of chemistry. But given the importance of atoms, is it possible to actually
see them? Historically, the answer has been no. Atoms are too small to see
with the naked eye or even with a microscope, and so we’ve always had to
gather information about atoms indirectly. However, in recent decades,
scientists have developed several techniques that make it possible to
visualize atomic structure. A dramatic example took place in the early
1990s, when scientists at IBM labs used a technique called scanning
tunneling microscopy to not only visualize atoms on a metal surface, but
even to move them to form the letters IBM (Figure 3.7). Recently, IBM
scientists took this technique one step further, producing a simple stop-
motion movie made from individual atoms.

150
Figure 3.7 IBM visualized and manipulated atoms using a technique called scanning
tunneling microscopy.

Explore

Figure 3.7

A more common technique for visualizing atoms is X-ray

crystallography. In this technique, a solid is bombarded with a form of
energy called X-rays. Based on the patterns the X-rays form as they pass
through the material, scientists are able to visualize the arrangement of
atoms in the solid. For example, Figure 3.8 shows the structure of aspirin,
a common painkiller. Because the properties of any substance depend on
how its atoms are arranged, the ability to “see” atomic arrangements
through X-ray crystallography helps scientists study diseases, develop new
medicines, and design materials with unique new properties.

151
Figure 3.8 Using X-ray crystallography, scientists can construct images showing how
atoms combine to form molecules. This image shows the structure of aspirin, a common
pain killer.

Explore

Figure 3.8

IT
TRY

2. One of the three chemical reactions shown here is drawn incorrectly,

because it does not follow the law of conservation of mass. Which one
is incorrect? Explain your answer.

152
3. Consider the following statement: “Phosphorus oxide is formed when
1 phosphorus atom combines with 2.5 oxygen atoms.” How does this
statement conflict with the concepts of atomic theory? What is a better
way to express the idea?

153
3.2 The Periodic Table of the Elements
By the late 1860s approximately 60 elements were known, and more were
being discovered each year. A Russian scientist named Dmitri Mendeleev
organized the known elements into a table based on their atomic masses.
He also arranged elements with similar properties, such as chlorine,
bromine, and iodine, into columns. By doing so he was able to predict the
existence of several elements, such as germanium and gallium, that had
not yet been discovered. Mendeleev’s table became the basis for the
modern periodic table of the elements (Figure 3.9), which organizes all
known elements based on their properties.

Figure 3.9 The modern periodic table contains 118 known elements.

Elements in the same column of the periodic table have similar properties.

The term periodic means “cyclical,” and it indicates how the table is
organized. You’ve probably used a different periodic table—a calendar
(Figure 3.10). On a calendar, each horizontal row represents an entire
week and covers the whole spectrum of activities that are part of your

154
normal routine. For example, on Sundays I usually attend church and
enjoy a quiet afternoon with my family. On Monday my work week
begins. I have certain duties that must be done each day: a lab on Tuesday,
a meeting on Thursday, and so on. Friday night is family night, and on
Saturdays I do yard work and hang out with my kids. From Sunday to
Saturday, I’ve covered an entire spectrum of activity. The next Sunday, the
cycle starts again. Although no week is exactly the same, certain days tend
to be similar.

Figure 3.10 A calendar is organized like the periodic table. A horizontal row (red
arrow) represents a full cycle of days, and a column (blue arrow) groups days that are
similar.

The periodic table is organized in much the same way. As we move

from left to right across a horizontal row of the periodic table, we cover
the entire spectrum of chemical properties. The rows in the periodic table
are called periods.
On a calendar, particular days (such as Monday) are arranged into
columns. These days have similar characteristics. Likewise, atoms that are
in the same column have similar properties. The columns of the periodic
table are called groups, or families.

Co versus CO
Each of the atomic symbols has either a capitalized one-letter symbol or a two-
letter symbol whose second letter is lowercase. Be careful with capitalization!
For example, Co refers to the elemental metal cobalt, but CO is a compound
composed of carbon and oxygen—the toxic gas carbon monoxide.

The periodic table uses an accepted set of one- and two-letter

abbreviations for different elements. Chemists routinely use these
chemical symbols to describe molecules, compounds, and chemical

155
reactions. To understand these chemical descriptions, it is essential for you
to know the symbols and names of the most common or important
elements. The elements and their symbols are listed in Table 3.1. Most of
the symbols are based on the English words; but some—such as iron (Fe),
gold (Au), silver (Ag), sodium (Na), and potassium (K)—are derived from
the Latin words for these elements.

TABLE 3.1 The Elements

Atomic Number Atom Symbol Discovery Year

1 Hydrogen H 1776
2 Helium He 1895

3 Lithium Li 1817

4 Beryllium Be 1797

5 Boron B 1808

6 Carbon C ancient

7 Nitrogen N 1772

8 Oxygen O 1774

9 Fluorine F 1886
10 Neon Ne 1898

11 Sodium Na 1807

12 Magnesium Mg 1755

13 Aluminum Al 1825

14 Silicon Si 1824
15 Phosphorus P 1669

16 Sulfur S ancient
17 Chlorine Cl 1774

18 Argon Ar 1894
19 Potassium K 1807

20 Calcium Ca 1808
21 Scandium Sc 1879

22 Titanium Ti 1791
23 Vanadium V 1830
24 Chromium Cr 1797

25 Manganese Mn 1774
26 Iron Fe ancient

156
27 Cobalt Co 1735

28 Nickel Ni 1751
29 Copper Cu ancient
30 Zinc Zn ancient

31 Gallium Ga 1875

32 Germanium Ge 1886

33 Arsenic As ancient

34 Selenium Se 1817
35 Bromine Br 1826

36 Krypton Kr 1898

37 Rubidium Rb 1861

38 Strontium Sr 1790

39 Yttrium Y 1794

40 Zirconium Zr 1789

41 Niobium Nb 1801
42 Molybdenum Mo 1781

43 Technetium Tc 1937

44 Ruthenium Ru 1844

45 Rhodium Rh 1803

46 Palladium Pd 1803

47 Silver Ag ancient

48 Cadmium Cd 1817
49 Indium In 1863

50 Tin Sn ancient
51 Antimony Sb ancient

52 Tellurium Te 1783
53 Iodine I 1811

54 Xenon Xe 1898
55 Cesium Cs 1860
56 Barium Ba 1808

57 Lanthanum La 1839
58 Cerium Ce 1803

59 Praseodymium Pr 1885
60 Neodymium Nd 1885

157
61 Promethium Pm 1945

62 Samarium Sm 1879

63 Europium Eu 1901
64 Gadolinium Gd 1880

65 Terbium Tb 1843

66 Dysprosium Dy 1886

67 Holmium Ho 1867
68 Erbium Er 1842

69 Thulium Tm 1879
70 Ytterbium Yb 1878

71 Lutetium Lu 1907

72 Hafnium Hf 1923

73 Tantalum Ta 1802

74 Tungsten W 1783

75 Rhenium Re 1925

76 Osmium Os 1803

77 Iridium Ir 1803

78 Platinum Pt 1735

79 Gold Au ancient
80 Mercury Hg ancient

81 Thallium Tl 1861

82 Lead Pb ancient
83 Bismuth Bi ancient
84 Polonium Po 1898

85 Astatine At 1940
86 Radon Rn 1900

87 Francium Fr 1939
88 Radium Ra 1898

89 Actinium Ac 1899
90 Thorium Th 1829

91 Protactinium Pa 1913
92 Uranium U 1789
93 Neptunium Np 1940

158
 94 Plutonium Pu 1940

95 Americium Am 1944
96 Curium Cm 1944

97 Berkelium Bk 1949

98 Californium Cf 1950

99 Einsteinium Es 1952
100 Fermium Fm 1952

101 Mendelevium Md 1955

102 Nobelium No 1958

103 Lawrencium Lr 1961

104 Rutherfordium Rf 1964

105 Dubnium Db 1967

106 Seaborgium Sg 1974

107 Bohrium Bh 1981

108 Hassium Hs 1984

109 Meitnerium Mt 1982

110 Darmstadtium Ds 1987

111 Roentgenium Rg 1994

112 Copernicum Cn 1996

113 Nihonium Nh 2003

114 Flerovium Fl 1998

115 Moscovium Mc 2003

116 Livermorium Lv 1999
117 Tennessine Ts 2010

118 Oganesson Og 2005

Regions of the Periodic Table

The blocks of elements on the left and right sides of the periodic table are
the main-group elements (Figure 3.11). We can predict many properties
of these elements based on their location on the periodic table.

159
Figure 3.11 The rows of the periodic table are numbered 1 through 7. Two numbering
systems identify the columns in the periodic table. An older system (shown in red)
numbers the main-group columns as 1A–8A. The modern system numbers all 18
columns that include the main-group and transition elements.

The left- and right-hand blocks of the periodic table contain the main-group
elements.

The elements in the middle of the table are called the transition
elements. At the bottom of the periodic table are two additional rows,
called the inner transition elements. Although these elements actually
belong in the last two rows of the table, they are typically shown below the
table to make it easier to read.
The periodic table contains seven periods (horizontal rows), which we
number from the top of the table. There are two ways to identify the
columns in the periodic table. In the modern system, the numbers 1–18
designate the columns that contain the main-group and transition elements.
An older but very useful system numbers the main-group columns as 1A–
8A. We will use both numbering systems in the chapters to come.

Metals, Nonmetals, and Metalloids

The elements on the left-hand side of the periodic table are metals. Metals
are usually solid at room temperature. They can be molded into different
shapes, and they conduct heat and electricity. The transition elements in
columns 3–12 of the periodic table are often called the transition metals.

160
These tend to be harder and less reactive than the metals of columns 1–2.
The transition metals include common building materials like iron and
copper as well as precious metals such as gold and silver (Figure 3.12).

Figure 3.12 The transition metals include (a) iron; (b) copper; (c) gold; and (d) silver.

The inner transition elements are also metals. The first row of these
elements is the lanthanide series. This row contains heavier, naturally
occurring metals called the rare earth metals. The second row is called the
actinide series. The first few elements in this series (up to uranium) are
naturally occurring. The elements that occur after uranium are mostly

161
human-made (we’ll discuss this in Chapter 16), although scientists have
identified trace amounts of a few of these elements in nature.
The nonmetals are found on the upper right side of the periodic table.
The physical properties of the nonmetals vary widely. For example, carbon
and sulfur are both solids; bromine is a liquid; and nitrogen, oxygen, and
fluorine are all gases. Nonmetals form a rich variety of compounds ranging
from simple gases to plastics to biomolecules. These elements form the
atomic basis for life (Figure 3.13).

Figure 3.13 Nonmetals such as carbon, nitrogen, oxygen, and hydrogen are the atomic
building blocks for plant and animal life.

The left side of the periodic table contains the metals. The upper right side
contains the nonmetals.

Between the metals and nonmetals, there is a stairstep pattern of

elements called the metalloids. Metalloids are semiconductors—they
conduct electricity, but not as efficiently as metals do. The metalloids,
especially silicon, are essential components of modern electronics (Figure
3.14).

Figure 3.14 The metalloid element silicon is a critical component in computer

processors.

Groups (Families) of Elements

162
Elements in the same column on the periodic table tend to have similar
properties. Let’s briefly examine the properties of several important
families of main-group elements.
Alkali metals lie on the far left of the periodic table, in column 1A.
These elements, such as lithium, sodium, and potassium, are very soft
metals that react violently with air or even with moisture (Figure 3.15).

Figure 3.15 Alkali metals such as sodium are soft metals that react violently with water.

Explore

Figure 3.15

Just to the right of the alkali metals, in column 2A, are the alkaline
earth metals (Figure 3.16). Magnesium and calcium are the most
common elements in this group. These elements are also reactive, but less
violently so than the alkali metals. They react slowly with water but burn

163
brightly when combined with oxygen.

Figure 3.16 The alkaline earth metals include beryllium, magnesium, and calcium.

The elements in column 7A are the halogens (Figure 3.17). These

elements all exist as diatomic (“two-atom”) molecules in their elemental
form. The halogens react quickly with metals and other nonmetals to form
many different compounds. Substances as diverse as bleach, Teflon®, fire
retardants, antiseptics, and table salt include halogens in their composition
(Figure 3.18).

164
Figure 3.17 The halogens are reactive nonmetals.

Figure 3.18 Halogens are used in a wide variety of compounds and applications.

Finally, the elements in column 8A are the noble gases (Figure 3.19).
This family is much different from the other elements. The noble gases are
very stable and generally do not react with other elements to form
compounds. There are no known compounds involving helium or neon and
only a handful involving the heavier atoms in this family. These elements
are all gases at room temperature.

165
Figure 3.19 The noble gases do not generally react with other elements.

The noble gases are in the rightmost column of the periodic table.

Example 3.2 Navigating the Periodic Table

Identify each of the following elements as a metal, a nonmetal, or a
metalloid:
a. a transition element
b. the lightest element in the halogen family
c. the element in row 4, column 14
d. an element in group 8A
To solve these problems, we can refer to the periodic table in Figure 3.9.
(a) The transition elements all have metallic properties. They are
commonly called the transition metals.
(b) The lightest element in the halogen family is fluorine (F). This
element is a nonmetal.
(c) The element in row 4, column 14 is germanium (Ge). This element is
a metalloid.
(d) The elements in group 8A are the noble gases. These are nonmetals.

166
 •

IT
TRY

4. Identify the elements located in each position on the periodic table:

a. row 3, column 14
b. row 2, column 4A
c. row 6, column 9

5. Based on its location in the periodic table, which of these elements

would behave most similarly to calcium?
a. potassium
b. magnesium
c. scandium
d. plutonium

6. Using the periodic table in Figure 3.9, identify each element as a

main-group metal, a transition metal, an inner-transition metal, a
metalloid, or a nonmetal.
a. sodium
b. sulfur
c. copper
d. samarium
e. silicon
f. argon
g. oxygen
h. gold

167
3.3 Uncovering Atomic Structure
By the early 1800s, John Dalton had articulated the key ideas of atomic
theory: Each element is composed of a unique type of atom; atoms
combine in whole-number ratios to form compounds; and atoms are not
created or destroyed in chemical reactions. Dalton’s theory was an
enormous step toward understanding our world, but not every aspect of his
model was correct. Dalton stated that atoms could not be broken into
smaller pieces, but later discoveries showed Dalton was incorrect. In fact,
atoms are composed of even smaller components, called subatomic
particles. The arrangement of these subatomic particles determines the
identity and behavior of atoms.
An important property of atoms and subatomic particles is their mass.
Because atoms are so small, we use a unit of mass called the atomic mass
unit (u or amu). One atomic mass unit is equal to 1.66 × 10–27 kg. An
atom of the lightest element, hydrogen, has a mass of 1.0 u.

1 u = 1.66 × 10–27 kg

Another characteristic property of subatomic particles is their charge.

Charge affects how particles interact with each other. Particles may have a
positive charge, a negative charge, or no charge. Particles with opposite
charges attract each other; those with like charges repel each other. We see
the effects of charged particles through the familiar phenomenon called
electricity or electrical energy, a form of energy that involves the motion
of charged particles.

The Discovery of Charged Particles

In 1800 the Italian scientist Alessandro Volta built the first battery (more
formally called an electrochemical cell)—a device capable of moving
electrically charged particles (Figure 3.20). Volta built his device by
placing alternating plates of zinc and copper on either side of pieces of
cardboard soaked in sulfuric acid. When he connected the two sides with a
metal wire, the result was an electrical current, the flow of charged
particles from one side of the battery to another.

168
Figure 3.20 (a) A battery, like the car battery shown here, produces a flow of charged
particles called an electric current. (b) Alessandro Volta built the first battery using
plates of zinc and copper, separated by cardboard soaked in acid. (c) A battery like
Volta’s can be made at home using aluminum foil, zinc washers, copper pennies, and
paper or cardboard that has been soaked in saltwater or vinegar.

Explore

Figure 3.20

Volta’s battery was a critical milestone in the development of science.

Using the controlled electrical energy of the battery, scientists soon
discovered they could use electrical energy to separate some compounds,
such as water, into their elements (Figure 3.21). In addition, scientists
began to identify charged particles and to measure how charged particles
interact with each other (Figure 3.22)

169
Figure 3.21 The electrical energy from a battery can split water into the elements
hydrogen and oxygen.

Figure 3.22 Scientists like J. J. Thomson studied charged particles by passing them
between two charged plates and then measuring where they hit a target on the other
side. This technique became the basis for early television screens.

170
 Explore

Figure 3.22

By the end of the nineteenth century, scientists had discovered that

elements could produce both positive and negative charges. In 1897 J. J.
Thomson, an English scientist, discovered the electron—a tiny, negatively
charged particle that was 2,000 times smaller than the lightest atom.
Thomson showed that electricity involves the flow of electrons.
Electricity involves the flow of electrons.

Using Volta’s battery, scientists discovered many new elements in a few short
years. Much of the information on Mendeleev’s periodic table was a direct
result of Volta’s discovery.

171
The Discovery of the Nucleus
Thanks to the work of Thomson and others, scientists in the early 1900s
knew that atoms contain charged particles. But they still did not have a
clear picture of how these particles are arranged. Thomson knew that
atoms contain small, negatively charged electrons. To balance the negative
charge, he assumed that atoms must also contain positive charges. He
hypothesized that the small, negative electrons were spread throughout a
positive atomic substance the way blueberries are scattered in a muffin
(Figure 3.23). This model came to be called the plum pudding model, after
a dessert (similar to blueberry muffins) that was popular at the time.

172
Figure 3.23 The “plum pudding model” envisioned atoms as negative electrons spread
throughout a positively charged material, like blueberries in a muffin.

However, like good muffins, this model did not last long. Within just a
few years, Ernest Rutherford, one of Thomson’s former students, showed
that the model was incorrect (Figure 3.24).

Figure 3.24 In the gold foil experiment, Rutherford and his students fired alpha
particles into a gold film. While most of the alpha particles passed straight through the
film, a few were deflected back toward the particle source.

In 1909 Ernest Rutherford and his students were studying the behavior
of positively charged particles called alpha particles. They were interested
in the pattern these heavy particles made as they passed through a thin
gold film. To study this pattern, they surrounded the gold film with a
material that glowed when particles struck it. Rutherford expected the
alpha particles to pass through the film and hit behind it. However, to his
surprise, a small number of the particles reflected off the film and back
toward the alpha particle source.
This surprising result led Rutherford to conclude that the plum pudding

173
model was not correct. Instead, he postulated that most of the atom was
empty space with a very dense nucleus at the center. Because of this
empty space, most alpha particles passed through the atoms in the film.
However, the particles that hit the nucleus deflected back toward the
source (Figure 3.25).

Figure 3.25 Most of the particles passed right through the film (black arrows), but some
were deflected (red arrows). This observation led Rutherford to conclude that the atom
was mostly empty space with a tiny, dense nucleus at the center.

We now know that Rutherford was correct. Atoms are mostly empty
space. We often draw atoms as a nucleus surrounded by orbiting electrons;
however, these pictures are never drawn to scale. Although the nucleus
contains almost all of the mass of the atom, it is incredibly dense, packing
the mass into a very tiny volume. The volume of the nucleus compared to
the volume of the atom has been compared to the size of an insect inside a
football stadium.
The nucleus contains two particles, protons and neutrons. Protons
have a mass of about 1 atomic mass unit (u) and a charge of +1. Neutrons
also have a mass of about 1 u, but they have no charge (Table 3.2).

TABLE 3.2 Subatomic Particles

Particle Mass (u) Charge Location

Proton 1.0073 +1 Nucleus

Neutron 1.0087 – Nucleus

Electron 0.0005 –1 Electron cloud

The rest of the atom is occupied by the tiny, negatively charged

electrons. Electrons have a charge of −1. The space around the nucleus is
called the electron cloud (Figure 3.26). The electrons are much lighter

174
than the nuclear particles. Their mass is only about 1/2000th of the mass of
the proton or neutron.

Figure 3.26 Protons and neutrons make up the nucleus of an atom, such as the helium
atom shown here. The electrons occupy the space around the nucleus.

The nucleus contains nearly all the mass of the atom.

As an example, consider the helium atom in Figure 3.26. A helium

nucleus contains four particles—two protons and two neutrons. Around the
nucleus, two electrons occupy the electron cloud. To be electrically
neutral, the number of protons and electrons in an atom must be the same.

175
3.4 Describing Atoms: Identity and Mass

Atomic Number and Mass Number

So what makes a hydrogen atom a hydrogen atom? In the previous section,
we saw that atoms contain three subatomic particles—protons, neutrons,
and electrons. But which of these particles determines the identity of the
atom?
The answer to this question is that the number of protons determines
the identity of the atom. As we will see, hydrogen atoms can have different
numbers of neutrons. Similarly, hydrogen can gain or lose electrons. What
defines hydrogen is the number of protons in its nucleus.
Because the number of protons determines the identity of an atom, we
call this number the atomic number. The atomic number is the integer
value normally seen above the atomic symbol on the periodic table
(Figure 3.27). Because neutral atoms contain the same number of protons
and electrons, the atomic number also tells us the number of electrons in a
neutral atom.

Figure 3.27 The atomic number is found above the symbol on the periodic table. For
this element, platinum, the atomic number is 78.

The number of protons defines the atom.

In addition to the atomic number, we also commonly describe atoms

by their mass number. The mass number is the sum of the number of
protons and the number of neutrons in an atom. Note that both the atomic
number and the mass number are integers. Typically, the periodic table
does not show the mass number.
Sometimes, atoms have the same atomic numbers, but different mass
numbers. For example, the hydrogen atom has three forms (Figure 3.28).
By definition, every hydrogen atom has one proton. Most hydrogen atoms

176
contain no neutrons, so they have a mass number of one. However, some
hydrogen atoms have one neutron and therefore a mass number of two. A
tiny percentage of hydrogen atoms even have two neutrons, and therefore a
mass number of three. Atoms that have the same atomic number but
different mass numbers are called isotopes. The three isotopes of hydrogen
are referred to as protium, deuterium, and tritium.

Figure 3.28 Hydrogen exists as three isotopes. The lightest of these, protium, is the
most common.

The atomic number is the number of protons in an atom. The mass number
is the number of protons plus the number of neutrons.

If we want to show the atomic number and mass number of an atom,

we typically write these numbers alongside the chemical symbols. We
place atomic numbers at the lower left-hand side of the atomic symbol and
mass numbers at the upper left-hand side of the atomic symbol. For
example, chlorine has two common isotopes. The first isotope has 17
protons and 18 neutrons; the second isotope has 17 protons and 20
neutrons. The two symbols may therefore be written like this:

Nuclear power plants use the element uranium as fuel. Uranium exists as two
common isotopes: 235U and 238U.

177
Example 3.3 Writing Atomic Symbols with Atomic Number and
Mass Number
Uranium has two common isotopes: The lighter isotope has a mass number
of 235; the heavier isotope has a mass number of 238. Write atomic
symbols for both of these isotopes, including the atomic number and mass
number.
To write these symbols, we must identify the symbol and atomic number.
The symbol for uranium is U, and the atomic number is 92. We write these
symbols with the mass number in the upper left and the atomic number in
the lower left:

U92235 and U92238

When writing chemical symbols, the atomic number and the chemical
symbol are redundant: Atomic number 92 is always uranium, and uranium
is always atomic number 92. While it is sometimes helpful to write both the
atomic and mass numbers, scientists often write just the mass number and
symbol. For example, using just the mass number and symbol, we can write
the two isotopes of uranium as 235U (typically read as “U-235”) and as
238U (read as “U-238”).
•

Example 3.4 Determining the Nuclear Structure from the

Atomic Number and Mass Number
One isotope of cadmium has a mass number of 114. How many protons and

178
 neutrons are in a nucleus of 114Cd?
To solve this problem, we need to identify the atomic number. Using a
periodic table, we determine that cadmium is atomic number 48. This
means that cadmium has 48 protons. To find the number of neutrons, we
subtract the atomic number from the mass number:
Mass number−Atomic number=Number of neutrons114−48=66

A 114Cd atom contains 48 protons and 66 neutrons. Because the number of

protons and electrons are equal in a neutral atom, we also know that this
atom contains 48 electrons. •

IT
TRY

7. Write the chemical symbol for each of these atoms, including the mass
number and the atomic number.
a. an atom of gold, containing 79 protons and 118 neutrons
b. an atom of mercury, containing 80 protons and 122 neutrons

8. The most common isotope of silicon has a mass number of 28. How
many neutrons are in this isotope?

9. Find the number of protons and neutrons in each of the following:

a. an atom with atomic number 15 and mass number 31
b. an isotope of copper that has 34 neutrons
c. an isotope of carbon with an equal number of protons and
neutrons

Average Atomic Mass

In the previous section, we noted that the mass number is not usually
included on the periodic table. Because the periodic table is a quick-
reference guide, it contains the information that we most often need.
Although the mass number is important, chemists more often use another
value, called the average atomic mass.

179
 The average atomic mass is the weighted average of the different
isotopes. To understand what a weighted average means, consider a simple
example involving poker chips (Figure 3.29). Suppose we have a mixture
of chips whose values are either $1 or $2. If we have equal numbers of
each chip, then our average is $1.5—right in the middle of the two values.
However, if we have more of the $2 chips, then our average is closer to $2
than to $1. Notice that the values of the chips are integer values, but the
average value is a decimal value somewhere between the two.

Figure 3.29 A weighted average considers how many units of each value are present.

Most periodic tables show each element’s average atomic mass, not its mass
number.

For a small number of chips, we can calculate the average simply by

adding up the values of each chip and dividing by the total number of
chips. However, if the number of chips is very large, this approach is not
practical. An easier way of doing this calculation is to use the percentage
of each type of chip present.
For example, imagine we have a dump truck full of poker chips. There
are too many chips to count, but we know that 10% of the chips are $1
chips, and 90% of the chips are $2 chips. To find the average value of the
chips, we do the following calculation:
Weighted average value=(fraction A×value A)+(fraction B×value B)

where fraction A and fraction B are the percentages of chips that are type
A and B, expressed as a decimal:

180
 Average value of chips=(0.10×$1)+(0.90×$2)=$1.9

We use this same approach to calculate the average mass of isotopes. For
example, carbon exists primarily as two isotopes: 12C has a mass of
12.000 u, and 98.93% of all naturally occurring carbon is carbon-12. 13C
has a mass of 13.0034 u, and 1.07% of all naturally occurring carbon is
carbon-13. What is the average atomic mass for carbon?
Before working this question mathematically, stop and think about it:
Out of 100 carbons, 99 have a mass of 12 u, and only 1 has a mass of 13 u.
Thus the average mass is much closer to 12 than to 13. Now let’s calculate
the answer. We find the average atomic mass by multiplying the mass of
each isotope by its percent abundance (expressed as a decimal) and adding
these values together:

The average atomic masses are normally included on the periodic

table, just beneath the atomic symbol. We don’t often have to calculate the
average atomic mass, but it is important to understand what this value
means, how it is different from the mass number, and how it is commonly
used.

Remember, u stands for “atomic mass units.”

Example 3.5 Calculating the Average Mass of an Element

The element antimony (Sb) exists mainly as two isotopes. Of naturally
occurring antimony, 57.21% is 121Sb, which has a mass of 120.9 u. The
remaining 42.79% is 123Sb, whose mass is 122.9 u. What is the average
atomic mass for this element?
We find the average atomic mass by multiplying the mass of each isotope
by its percent abundance, expressed as a decimal, and then adding these
values together. •

Average mass of antimony=(120.9 u)(0.5721)+(122.9 u)(0.4279)=121.8 u

181
 IT
TRY

10. You have a bag containing poker chips. In the bag, 25% of the chips
have a value of $1, and 75% of the chips have a value of $3. What is
the average value of the chips? If the bag contains 80 chips, what is
the total value of the chips in the bag?

11. Indium has two isotopes, 113In and 115In. The average atomic mass
of indium is 114.8 u. Which isotope of indium is more common?

12. Silicon exists as a mixture of three isotopes, shown in the table

below. What is the average atomic mass for silicon? Check your
answer by comparing it to the average atomic mass on the periodic
table (you might find slight differences in rounding).

Isotope Mass (u) Abundance

28Si 27.9769 92.2297
29Si 28.9765 4.6832
30Si 29.9738 3.0872

182
3.5 Electrons—A Preview
Rutherford’s gold foil experiment demonstrated that the nucleus,
composed of protons and neutrons, occupies a very tiny, very dense space
at the center of the atom. The remaining volume of the atom is occupied
by electrons. As we’ll see in the chapters ahead, the arrangement of
electrons around the nucleus determines how atoms combine to form
compounds. In Chapter 4, we will explore electronic structure within a
single atom. In Chapter 5, we will see how atoms combine to form
compounds by gaining, losing, or sharing electrons. For now, let’s
complete our survey of atomic structure with a broad overview of electron
behavior and then preview some of the concepts that lie ahead.

The Bohr Model and the Quantum Model

In the early twentieth century, Ernest Rutherford and Niels Bohr
introduced a new model for atomic structure, commonly called the Bohr
model. This model treated the atom like a tiny solar system, with the
nucleus at the center, and the electrons orbiting the nucleus, much like the
different planets orbit the Sun (Figure 3.30).

Figure 3.30 In the Bohr model, electrons orbit the nucleus much like planets orbit the
Sun in fixed orbits.

The Bohr model was a significant advance over the plum pudding
model of atomic structure, and it enabled chemists to explain the properties
of some elements. However, the Bohr model does not give a complete
picture of electron behavior. In the early 1900s, scientists began to realize
that electrons cannot be described simply as negatively charged particles.

183
Sometimes their behavior can be understood only by describing the
electrons as waves moving at different energies. This description of
electrons is called the quantum model. We will explore this model in
more detail in the next chapter.
Table 3.3 summarizes the different models of atomic structure. Notice
that each new model built on the concepts that came before it. Dalton’s
original atomic theory described the behavior of atoms and compounds,
but it treated atoms as indivisible particles. A century later, J. J.
Thomson’s plum pudding model was a first attempt at describing the
structure of the atom. After Rutherford’s gold foil experiment showed this
model was incorrect, the Bohr model emerged. The Bohr model correctly
described the dense, positively charged nucleus with orbiting electrons, but
it treated electron motion too simply. Although the Bohr model could
explain some common trends, it left many questions unanswered. Finally,
the modern quantum model emerged. This model explains a wide variety
of physical and chemical properties and has led to many technological
advances. Technologies ranging from solar panels to supercomputers rely
on our understanding of the arrangement and behavior of electrons.

TABLE 3.3 Development of Models of Atomic Structure

Model Year Key Ideas

Dalton’s
1808 Atoms are indivisible particles.
atomic theory

Plum pudding Atoms are solid, with negative electrons spread throughout a
1904
model positively charged matrix.

Bohr model 1913 Electrons orbit the nucleus like planets orbit the Sun.

Quantum Electrons behave both as particles and as waves. They occupy

1920s
model an electron cloud around the nucleus.

The Formation of Ions

Atoms frequently gain or lose electrons. To understand why this is so,

184
consider the structure of the atom: Unlike the protons and neutrons, which
pack tightly into the nucleus, the electrons occupy the outer volume of the
atom. As a result, atoms sometimes lose electrons, or they pick up
electrons from neighboring atoms.
When this happens, the number of protons and electrons is no longer
the same. As a result, the atom gains a net positive or negative charge.
Atoms or groups of atoms that contain an overall charge are called ions.
For example, a lithium atom often loses one electron. When this happens,
it becomes a positively charged lithium ion (Figure 3.31):

Figure 3.31 Some atoms lose or gain electrons to form ions.

Atoms can gain or lose electrons to form ions.

Protons Electrons Overall Charge

Lithium atom 3 3 0
Lithium ion 3 2 +1

On the other hand, fluorine atoms tend to gain one electron. When they do,
they become negatively charged fluoride ions:

Protons Electrons Overall Charge

Fluorine atom 9 9 0
Fluoride ion 9 10 −1

Two atoms are sitting in a bar. Suddenly, one atom says, “Ouch! I just lost an
electron!”
The other atom looks over dubiously. “Are you sure?”
“Yes,” the first replies, “I’m positive!”

185
 Example 3.6 Finding the Charge on an Ion
A calcium ion has 20 protons but only 18 electrons. What is the charge on
the ion?
Each proton has a charge of +1, and each electron has a charge of –1. This
calcium ion has a total charge of (+20 – 18), or +2. •

Example 3.7 Relating Ion Charge to Subatomic Particles

Sulfur is atomic number 16. Sulfur atoms commonly form sulfide ions,
which have a charge of –2. How many electrons are in the electron cloud of
a sulfide ion?
The atomic number gives us the number of protons in the nucleus and also
the number of electrons in the cloud of a neutral atom. Because the ion has
a charge of –2, we know that it has two extra electrons present. Based on
this, the sulfide ion has 18 electrons present. •

16 electrons (neutral atom)+2 additional electrons=18 electrons total

IT
TRY

13. Determine the net charge on the following atoms and ions:
a. an atom with 5 protons, 6 neutrons, and 5 electrons
b. an ion with 16 protons, 16 neutrons, and 18 electrons
c. a potassium ion with 18 electrons
d. an ion with atomic number 13, but only 10 electrons

186
14. Magnesium has atomic number 12. The magnesium ion has a charge
of +2. How many electrons are in this ion?

Explore
Atom Builder
Try this interactive to see how protons, neutrons, and electrons combine in
atoms and ions.

187
Summary
Although the Greek philosopher Democritus postulated the existence of
the atom in ancient times, it was not until the eighteenth century that
scientists began to explore chemical reactivity more systematically. In the
late 1700s, Antoine Lavoisier expressed the law of conservation of mass,
which states that matter is not created or destroyed in chemical changes. In
the early 1800s, John Dalton developed the foundation for the modern
atomic theory. Key facets of this theory are that all matter is made of
atoms; that atoms of different elements are unique; that atoms combine in
whole-number ratios to form compounds; and that atoms are not created or
destroyed in chemical reactions.
The invention of the battery in 1800 spawned the discovery of many
new elements. This wealth of new knowledge enabled Dmitri Mendeleev
to develop the periodic table. The periodic table is organized based on
atomic number and also on chemical reactivity. A row on the periodic
table represents a full cycle of chemical behavior, while elements in a
single column share physical and chemical traits.
In the nineteenth and early twentieth centuries, technical innovations
enabled scientists to probe the structure of the atom. J. J. Thomson first
described the electron. Ernest Rutherford, building on the work of
Thomson and others, discovered that atoms are almost entirely empty
space and have a very dense nucleus at the center.
The nucleus is composed of two particles, positively charged protons
and uncharged neutrons. The atomic number is the number of protons in
the nucleus, as well as the number of electrons in an electrically neutral
atom. The number of protons determines the identity of the atom. Atoms
with the same number of protons but different numbers of neutrons are
isotopes. The mass number of any isotope is the sum of the protons and
neutrons in the nucleus. The mass that is usually displayed on periodic
tables is the average atomic mass. This is a weighted average of the masses
of the different isotopes of that element.
The discovery of the nucleus led to the development of the Bohr model
of the atom, which describes the atom as a very dense nucleus with
electrons orbiting around it. The Bohr model gave way to the quantum
mechanical model, which treats electrons as waves rather than simply as
orbiting particles.
In an atom the negatively charged electrons occupy the region around
the nucleus, called the electron cloud. Although the electron cloud

188
accounts for most of the atom’s volume, electrons have a much smaller
mass than the particles in the nucleus, so that nearly all the mass of the
atom arises from the protons and neutrons.
At the beginning of this chapter, we described the environmental
damage and health issues caused by the use of mercury in small-scale gold
mining. In the process, we posed a question: What is the reason for the
differences between gold and mercury? Based on what we’ve learned so
far, we know that gold and mercury have different numbers of protons,
neutrons, and electrons in their atomic structures. These differences in
structure cause gold and mercury to exhibit different physical and
chemical properties. One is a solid, the other a liquid. One is yellow in
color, the other is silver. One is safe to handle, the other is toxic. Small
distinctions make a big difference.

189
Mercury: Ancient Treasure, Modern Toxin

At the beginning of this chapter, we saw how the use of mercury in small-scale
gold mining has led to disastrous effects for many miners and the surrounding
communities. As we wrap up our discussion of atoms and elements, let’s look at
this intriguing element in a little more detail.
From ancient times, mercury has fascinated people. The only naturally
occurring metal that is liquid at room temperature, mercury’s features are
mesmerizing (Figure 3.32). It shines like silver, yet flows effortlessly. It is dense
enough that lead floats on it.

Figure 3.32 Mercury is the only metal that is a liquid at room temperature.

Mercury has played a bit role in the quest for immortality. In the third century
B.C.E., the first emperor of unified China, Qin Shi Huang, ordered the building of
an enormous burial mound. An ancient Chinese historian wrote that construction of
the site employed 700,000 workers and featured mercury replicas of all the great
rivers in China. In fact, some accounts state that Emperor Qin died after drinking a
mixture of powdered jade and mercury that was formulated to give him eternal life.
In lesser amounts, mercury has also been found in both Mayan and Egyptian
tombs.
We now know that mercury can hasten the journey to the afterlife. Mercury is a
powerful toxin that affects the central nervous system. Exposure to mercury can
lead to tremors, mood changes, dementia, and ultimately death. As mentioned at

190
the beginning of this chapter, mercury poisoning acutely affects the gold-mining
communities in developing nations such as Mozambique, Ecuador, and Guyana.
The level of mercury in seafood is an environmental concern. Although
seafood is generally healthy, aquatic plants and animals can absorb mercury
released into waterways. Fish have no mechanism for excreting mercury. Over
time large fish, such as tuna and shark, may build up significant amounts of
mercury in their tissue. In 1958 and again in 1965, outbreaks of illnesses and
deaths in Japan were caused by eating fish contaminated by the discharge of
mercury into the water supply.
For years, my family has fished for shark in the waterways around Tampa,
Florida (Figure 3.33). Served with cheesy grits, cornbread, and sweet tea, shark is
delicious. While modest consumption of shark is acceptable for most people, the
Florida Department of Health recommends that people limit their consumption and
avoid eating larger fish, which have accumulated higher mercury levels.

Figure 3.33 Grandpa Revell in his kayak, catching a small shark, which makes for a
great dinner the next day.

191
 Key Terms
3.1 Atoms: The Essential Building Blocks
law of conservation of mass In chemical reactions, matter is neither created
nor destroyed.
atomic theory A theory describing matter in terms of fundamental units called
atoms.

3.2 The Periodic Table of the Elements

periodic table of the elements A chart that organizes all the known elements
based on their masses and properties.
period A horizontal row on the periodic table; a period encompasses a range
of behavior from metallic to nonmetallic.
group A vertical column on the periodic table; also called a family. Elements
within a group exhibit similar behaviors.
main-group elements The elements in columns 1–2 and 13–18 (or 1A–8A) of
the periodic table.
metals The elements on the left-hand side of the periodic table; these elements
can be molded into different shapes, and they conduct heat and electricity.
transition metals The metals in columns 3–12 of the periodic table; these
metals are harder and less reactive than those in columns 1–2.
nonmetals Elements on the upper-right-hand side of the periodic table; these
elements have widely varying properties and form many different compounds.
metalloids Elements whose properties lie between those of metals and
nonmetals.
alkali metals Metal elements in column 1 (or 1A) of the periodic table; these
metals are very reactive.
alkaline earth metals Metal elements in column 2 (or 2A) of the periodic
table; these metals are very reactive.
halogens Nonmetal elements in column 17 (or 7A) of the periodic table; these
elements form many different types of compounds.
noble gases The nonmetal elements in column 18 (or 8A) of the periodic table;
these elements usually do not form compounds.

3.3 Uncovering Atomic Structure

subatomic particles The particles from which atoms are composed. The three

192
major subatomic particles are protons, neutrons, and electrons.
atomic mass unit (u or amu) A unit of mass equal to 1.66 × 10–27 kg.
charge A characteristic property of subatomic particles that affects how
particles interact with each other.
electron A negatively charged subatomic particle; the electrons occupy the
space around the nucleus.
nucleus The tiny, dense center of an atom; the nucleus contains protons and
neutrons.
proton A positively charged subatomic particle that resides in the nucleus of
the atom.
neutron A subatomic particle having no charge that resides in the nucleus of
the atom.
electron cloud The space around the nucleus; the electron cloud accounts for
nearly the entire volume of the atom.

3.4 Describing Atoms: Identity and Mass

atomic number The number of protons in an atom; also the number of
electrons in a neutral atom.
mass number The sum of protons and neutrons in an atom.
isotopes Atoms that have the same atomic number, but different mass
numbers.

3.5 Electrons —A Preview

Bohr model An early model of atomic structure that treated the atom like a
tiny solar system, with the nucleus at the center, and the electrons orbiting the
nucleus.
quantum model The modern description of electronic behavior that treats
electrons both as particles and as waves.
ion An atom or group of atoms with an overall charge.

193
 Additional Problems

3.1 Atoms: The Essential Building Blocks

15. What is the fundamental unit of matter?

16. How many naturally occurring elements are there?

17. Which scientist is credited with each of these accomplishments?

a. first proposing the idea of atoms
b. developing the law of conservation of mass
c. developing the modern atomic theory

18. List four key features of Dalton’s atomic theory.

19. Alka-Seltzer® is a common treatment for upset stomach. When the tablets
are mixed with water, a chemical change produces gas bubbles (Figure
3.34). To test Lavoisier’s law, a chemist freezes a small amount of water in
a plastic bottle. She then adds an Alka-Seltzer tablet and seals the bottle
tightly. She measures the mass of the bottle. She then sets the bottle on the
counter. As the ice melts, it begins to react with the tablet, producing
bubbles inside the bottle. After all of the ice has melted and the tablet has
dissolved, she measures the mass of the bottle again. Assuming the bottle
does not burst, would you expect the mass afterward to be more than, less
than, or the same as the mass before? How do you explain this?

Figure 3.34 Alka-Seltzer® is a common treatment for upset stomach. When the tablets
are mixed with water, a chemical change produces gas bubbles.

194
20. On a crisp October night, you build a campfire using four heavy logs. The
next morning, you notice that the charred remains of the logs are much
lighter than the logs were the night before. Does this contradict the law of
conservation of mass? What happened to the atoms in the logs?

21. Sodium hydroxide reacts with hydrogen chloride to form two new
compounds, sodium chloride and water. When 200 grams of sodium
hydroxide are combined in a sealed flask with 100 grams of hydrogen
chloride, a reaction takes place. How many grams of material will be
present in the flask after the reaction takes place?

22. When natural gas (methane) burns, it reacts with oxygen in the air to form
two new compounds, carbon dioxide and water. If 100 grams of methane
and 100 grams of air are combined in a sealed container, and the mixture is
allowed to react, what mass will be present in the container after the
reaction takes place?

23. When it burns, 24.3 grams of magnesium can react completely with 16.0
grams of oxygen gas to form a new compound, magnesium oxide. No
other compounds are formed. What mass of magnesium oxide is formed in
this reaction?

24. Zinc metal reacts with hydrochloric acid to form zinc chloride and
hydrogen gas, as shown in this equation:
Zinc + hydrochloric acid → zinc chloride + hydrogen gas
If 65.4 grams of zinc reacts in this way, it will consume 72.9 grams of
hydrochloric acid and produce 2.0 grams of hydrogen gas. Based on this
information, how many grams of zinc chloride will be produced?

25. Strontium and chlorine combine in only one ratio: one strontium atom for
every two chlorine atoms. Based on this information, indicate whether the
following combinations are possible:
a. 10 strontium atoms and 20 chlorine atoms
b. 10 strontium atoms and 30 chlorine atoms
c. 140 billion strontium atoms and 220 billion chlorine atoms
d. 3.8 × 1021 strontium atoms and 7.6 × 1021 chlorine atoms

26. Aluminum and bromine atoms combine in only one ratio: one aluminum
atom for every three bromine atoms. Based on this information, indicate

195
 whether the following combinations are possible:
a. 3 aluminum atoms and 9 bromine atoms
b. 4.2 × 104 aluminum atoms and 8.4 × 104 bromine atoms
c. 50,000 aluminum atoms and 1.5 × 105 bromine atoms
d. 9 dozen aluminum atoms and 27 dozen bromine atoms

27. Each of these panels depicts a chemical reaction. Which of them does not
follow the law of conservation of mass? Explain your answer.

28. Each of these panels depicts a chemical reaction. Which of them does not
follow the law of conservation of mass? Explain your answer.

3.2 The Periodic Table of the Elements

29. Write the name of the element that corresponds with each of the following
symbols:
a. Cs
b. Ti
c. K

196
 d. B
e. Xe

30. Write the name of the element that corresponds with each of the following
symbols:
a. Np
b. O
c. Fe
d. S
e. Th

31. Name the element that corresponds with each of the following symbols:
a. W
b. B
c. U
d. S
e. C

32. Name the element that corresponds with each of the following symbols:
a. O
b. Ba
c. Mg
d. W
e. Pb

33. Write the atomic symbol for each of the following elements:
a. fluorine
b. germanium
c. tin
d. magnesium

34. Write the atomic symbol for each of the following elements:
a. potassium
b. argon
c. ruthenium
d. lithium

35. Write the atomic symbol for each of the following elements:

197
 a. iodine
b. copper
c. silver
d. hydrogen

36. Write the atomic symbol for each of the following elements:
a. iron
b. nickel
c. gold
d. titanium

37. On the periodic table, how are atoms with similar chemical properties
grouped?

38. As you move across a horizontal row of the periodic table, where do you
find the metals, metalloids, and nonmetals?

39. Use the periodic table to identify each element as a main-group metal, a
transition metal, an inner-transition metal, a metalloid, or a nonmetal:
a. boron
b. gallium
c. chromium
d. barium
e. uranium
f. oxygen
g. tin
h. iron

40. Use the periodic table to identify each element as a main-group metal, a
transition metal, an inner-transition metal, a metalloid, or a nonmetal:
a. carbon
b. lithium
c. vanadium
d. calcium
e. plutonium
f. platinum
g. einsteinium
h. argon

198
41. What are the main-group elements?

42. What series of elements contains most of the rare-earth elements?

43. Identify each of these elements as a metal, metalloid, or nonmetal:

a. platinum
b. oxygen
c. lithium
d. uranium

44. Identify each of these elements as a metal, metalloid, or nonmetal:

a. carbon
b. mercury
c. californium
d. tellurium

45. Potassium, rubidium, and cesium all belong to which family of elements?

46. Calcium, strontium, and barium all belong to which family of elements?

47. Identify the family of elements to which each of these elements belongs:
a. chlorine
b. krypton
c. rubidium
d. strontium

48. Identify the family of elements to which each of these elements belongs:
a. francium
b. fluorine
c. helium
d. calcium

49. Based on its location on the periodic table, which of these elements would
behave most like magnesium?
a. sodium
b. beryllium
c. aluminum

199
 d. argon

50. Based on its location on the periodic table, which of these elements would
behave most like oxygen?
a. nitrogen
b. phosphorus
c. sulfur
d. fluorine

51. Based on its location on the periodic table, which two of these elements
would not behave like chlorine?
a. fluorine
b. bromine
c. argon
d. sulfur

52. Based on its location on the periodic table, which two of these elements
would not behave like strontium?
a. rubidium
b. calcium
c. yttrium
d. barium

53. Identify each element based on its location on the periodic table:
a. row 4, column 5
b. row 3, group 7A
c. row 2, column 18

54. Identify each element based on its location on the periodic table:
a. row 1, column 1
b. row 4, group 12
c. row 2, group 3A

3.3 Uncovering Atomic Structure

55. Charge is a fundamental property of subatomic particles. What types of

charges are possible in a subatomic particle?

200
56. What types of charges attract each other? What types of charges repel each
other?

57. Match each of the scientists in the first column with the discovery in the
second column:

(1) Democritus a. the periodic table

(2) Antoine Lavoisier b. first theorized about the existence of atoms
(3) John Dalton c. the law of conservation of mass
(4) Dmitri Mendeleev d. the modern atomic theory

58. Match each of the scientists in the first column with the discovery on the
right:

(1) Alessandro a. discovered the electron

Volta
(2) J. J. Thomson b. model in which electrons orbit the nucleus like planets
orbit the sun
(3) Ernest c. idea that the atom has a dense nucleus
Rutherford
(4) Niels Bohr d. invented the battery

59. When Rutherford and his students conducted the gold foil experiment,
what did they expect to see? How did their actual observations differ from
their hypothesis?

60. What conclusions was Rutherford able to draw from his gold foil
experiment?

61. Which has more volume, the nucleus or the electron cloud? Which has
more mass?

62. How does the density of the nucleus compare to the density of the rest of
the atom?

63. John Dalton’s original atomic theory contained the following key ideas.
Which part(s) of these ideas was/were incorrect?
a. Elements are made of tiny, indivisible particles called atoms.

201
 b. The atoms of each element are unique.
c. Atoms can join together in whole-number ratios to form compounds.
d. Atoms are unchanged in chemical reactions.

64. What findings led J. J. Thomson to articulate the plum pudding model
rather than considering atoms to be indivisible particles as John Dalton
did?

65. Which two subatomic particles each have a mass of about one atomic mass
unit? Which two particles are located in the nucleus? Which two particles
have an overall charge?

66. Of the three major subatomic particles (protons, neutrons, and electrons),
which is the smallest? Which has no charge? Which particle occupies the
space outside the nucleus?

3.4 Describing Atoms: Identity and Mass

67. Using the periodic table, identify the atomic number for each of these
elements:
a. Ti
b. S
c. Te
d. Li

68. Using the periodic table, identify the atomic number for each of these
elements:
a. Rf
b. Ar
c. He
d. Es

69. Using the periodic table, find the symbols for the elements that correspond
to each atomic number:
a. atomic number 13
b. atomic number 86
c. atomic number 99

70. Using the periodic table, find the symbols for the elements that correspond

202
 to each atomic number:
a. atomic number 29
b. atomic number 5
c. atomic number 16

71. Give the atomic number and the mass number for each of the following:
a. an atom with 14 protons and 16 neutrons
b. an atom with 27 protons and 32 neutrons
c. an atom with 20 protons and 26 neutrons

72. Give the atomic number and the mass number for each of the following:
a. an atom with 35 protons and 44 neutrons
b. an atom with 35 protons and 46 neutrons
c. an atom with 82 protons and 124 neutrons

73. Write the atomic symbol, including the atomic number and the mass
number, for each of the following:
a. a potassium atom with 20 neutrons
b. an argon atom with 22 neutrons
c. a fermium atom with 157 neutrons

74. Write the atomic symbol, including the atomic number and the mass
number, for each of the following:
a. a silicon atom with 14 neutrons
b. a gallium atom with 39 neutrons
c. a hafnium atom with 106 neutrons

75. The most common isotope of chromium has a mass number of 52. How
many neutrons are in an atom of 52Cr?

76. The most common isotope of nickel has a mass number of 58. How many
neutrons are in an atom of 58Ni?

77. Using the periodic table, complete this table:

Atom Symbol Protons Neutrons Atomic Number Mass Number

Hydrogen H 1 0
Sulfur 16

203
 52 128
He 2

51 40

78. Using the periodic table, complete this table:

Atom Symbol Protons Neutrons Atomic Number Mass Number

Fr 87 223
Tungsten 110
U 238
86 136
57 44

79. You are trying to find enough change in your room to buy a candy bar.
After a thorough search, you come up with 8 pennies, 5 nickels, 4 dimes,
and 3 quarters. What is the total value of the coins you found? What is the
average value of the coins you found?

80. You are given a bag of coins that contains 70% dimes and 30% nickels.
What is the average value of the coins in the bag?

81. Gallium exists as two isotopes. The first has 31 protons and 38 neutrons.
The second has 31 protons and 40 neutrons. Write an atomic symbol for
each of these isotopes, including the atomic number and mass number.

82. Indium exists as two isotopes. The first has 49 protons and 64 neutrons.
The second has 49 protons and 66 neutrons. Write an atomic symbol for
each of these isotopes, including the atomic number and mass number.

83. Silver exists as two isotopes, 107Ag and 109Ag. How many protons and
neutrons are in the nucleus of each isotope?

84. Iridium exists as two isotopes, 191Ir and 193Ir. How many protons and
neutrons are in the nucleus of each isotope?

204
85. Bromine exists as a mixture of two isotopes. Of bromine atoms, 50.7% are
79Br, and 49.3% are 81Br. Use the mass numbers and the relative
abundance to estimate the average atomic mass of bromine.

86. Boron exists as a mixture of two isotopes. Of boron atoms, 19.80% are
10B, and 80.20% are 11B. Use the mass numbers and the relative
abundance to estimate the average atomic mass of boron.

87. Lead exists as a mixture of four major isotopes, shown in the table.
Calculate the average atomic mass for lead.

Isotope Mass (u) % Abundance

204Pb 203.9730 1.40
206Pb 205.9745 24.10
207Pb 206.9759 22.10
208Pb 207.9766 52.40

88. Zinc exists as a mixture of five major isotopes, shown in the table.
Calculate the average atomic mass for zinc.

Isotope Mass (u) % Abundance

64Zn 63.9291 48.63
66Zn 65.9260 27.90
67Zn 66.9271 4.10
68Zn 67.9248 18.75
70Zn 69.9253 0.62

89. What is the difference between the mass number and the average atomic
mass?

90. Why does the periodic table commonly show the average atomic mass
rather than the mass number?

3.5 Electrons–A Preview

91. How did the Bohr model differ from the plum pudding model of atomic
structure?

205
92. How does the quantum mechanical model differ from the Bohr model of
atomic structure?

93. How many electrons are in each of the following neutral atoms?
a. hydrogen
b. helium
c. carbon
d. selenium

94. How many electrons are in each of the following neutral atoms?
a. copper
b. gold
c. mercury
d. xenon

95. Find the number of protons and electrons in each of the following neutral
atoms:
a. Li
b. Ge
c. Bi
d. Ba

96. Find the number of protons and electrons in each of the following neutral
atoms:
a. Y
b. N
c. O
d. Cf

97. A beryllium ion has 4 protons but only 2 electrons. What is the charge on
this ion?

98. An oxide ion has 8 protons and 10 electrons. What is the charge on this
ion?

99. Find the net charge on each of the following:

a. an ion with 7 protons, 7 neutrons, and 10 electrons
b. an ion with 38 protons, 50 neutrons, and 36 electrons

206
 c. an atom with 35 protons, 44 neutrons, and 35 electrons

100. Find the net charge on each of the following:

a. a phosphorus ion with 18 electrons
b. an argon atom with 18 electrons
c. a potassium ion with 18 electrons
d. a calcium ion with 18 electrons

101. How many electrons are in a neutral silver atom? How many electrons are
in a silver ion with a charge of +1?

102. How many electrons are in a neutral zinc atom? How many electrons are
in a zinc ion with a charge of +2?

103. Titanium commonly forms two ions. The first ion has a charge of +2. The
second ion has a charge of +4. In total, how many electrons are present in
each of these ions?

104. Iron commonly forms two ions. The first ion has a charge of +2. The
second ion has a charge of +3. In total, how many electrons are present in
each of these ions?

105. Under extreme conditions, gold will lose one electron to form a +1 ion.
Mercury often loses two electrons to form a +2 ion. Find the number of
protons and electrons in each of these ions.

106. Sodium atoms easily lose one electron to form a +1 ion. Magnesium
atoms easily lose two electrons to form a +2 ion. How many electrons are
in each of these ions?

107. Fluorine, chlorine, and bromine all gain one electron to form ions with a
−1 charge. Find the number of protons and electrons in each of these
ions.

108. Lithium, potassium, and rubidium all lose one electron to form ions with a
+1 charge. Find the number of protons and electrons in each of these
ions.

207
Chapter Four
Light and Electronic Structure

Edging toward Solar Energy

What if the power company paid you for electricity?

Some believe this seemingly far-fetched dream is edging closer to reality, thanks to
recent advances in solar energy technology and to the vision of innovators like
Elon Musk. We’ll get to your power bill and Elon Musk in just a moment. First, a
little history: In 1839 Edmond Becquerel, a French scientist, discovered that
shining light on certain materials produced an electric current. Over a century later,
in 1954, Bell Labs produced the first solar cell, a device that could convert sunlight
into electricity.
In the years since, many have hailed solar energy as the power source of the
future: clean, renewable energy that would finally replace fossil fuels. Scientists
and engineers dreamed of creating homes that were energetically self-sufficient,
disconnected from the power grid.
Despite these dreams, large-scale solar power has been agonizingly slow to
develop. One problem is location: Solar power is much more promising for sun-
drenched areas like Southern California or Central Africa than for northern areas
like Minnesota or Finland. Then there’s the issue of cost: Even though the price of
solar cells has dropped dramatically in the past decades (Figure 4.1), solar power
is still more expensive than coal or natural gas.

208
Figure 4.1 (a) Solar cells now provide a fraction of the electrical energy for many
homes. (b) The price of solar cells has dropped dramatically in recent years, but solar
energy still costs more than other energy sources. (c) Elon Musk is the founder and/or
CEO of several innovative companies, including Tesla Motors. (d) In 2015, Musk
unveiled his new home battery, based on the battery used in the Tesla electric cars. (e)
In 2016, Musk unveiled a new “solar roof” in which the shingles are solar cells that
provide power to the home.

But perhaps the most challenging issue has been energy storage. Because our
homes are not exposed to sunlight all the time, solar power requires either a way of
storing the energy for later use or a backup power supply.
Enter Elon Musk. He is the founder of SpaceX, a leader in private space
exploration, and the CEO of Tesla Motors, which produces high-performance
electric cars. He also helped create Solar City, a company that produces and
installs home solar energy systems.
In April 2015, Musk introduced a remarkable new product: a rechargeable
battery system capable of powering a small home. Based on the Tesla automotive
technology, these long-term battery systems can connect to solar panels. During
the day, solar panels charge the battery; at night the battery helps power the home,
reducing monthly power bills. Musk and others envision a future in which
homeowners can even sell excess power back to the power company.

209
 The Tesla battery system is an impressive achievement and an important step in
the development of solar energy. And though it’s not likely to completely replace
fossil fuels in our lifetimes, solar power is an important part of our energy future.
So, how does solar power work? To begin answering that question, we must
first understand the relationship between light and the structure of the atom. In this
chapter, we’ll see that both the absorption and production of light are intimately
connected to the configuration of electrons around the nucleus. And light is just the
beginning: As we delve deeper, we’ll see how the properties of the elements relate
to their electronic structure and learn how to determine a great deal of information
with a quick glance at the periodic table. Let’s take a look.

Intended Learning Outcomes

After completing this chapter and working the practice problems, you should be
able to:

4.1 The Electromagnetic Spectrum

Qualitatively and quantitatively describe the relationships between the
wavelength, frequency, and energy of electromagnetic radiation.

4.2 Color, Line Spectra, and the Bohr Model

Describe line spectra, the Bohr model, and how the two are related.
Describe the absorption or emission of light as a function of electron
transitions.

4.3 The Quantum Model and Electron Orbitals

Describe Heisenberg’s uncertainty principle and the wave nature of
electrons.
Identify the number of orbitals and the maximum electron capacity of the s,
p, d, and f sublevels.
Correlate each primary energy level with the available sublevels.

4.4 Describing Electron Configurations

Write electron configurations for atoms and ions, using either full notation or
noble gas shorthand.
Identify the inner, outer, and valence electrons in an atom or ion.
Apply the octet rule to explain the exceptional stability of the noble gases.

4.5 Electron Configuration and the Periodic Table

Use the periodic table to quickly identify the highest occupied energy level
and sublevel of an element.

210
Use the periodic table to identify the number of valence electrons for main-
group elements.

211
4.1 The Electromagnetic Spectrum
Our world is immersed in light and soaked in color. From the camouflage
of a leopard to the glow of a sunset, from the placid green of the forest to
the attention-grabbing text of a billboard, the perception of light and color
plays a vital role in areas ranging from survival to fashion to
communication (Figure 4.2).

Figure 4.2 Light and color are closely connected to atomic structure.

The production of light is closely connected to atomic structure and

especially to the configuration of electrons around the nucleus. In fact,
understanding electronic structure is a key to understanding how light is
produced.
So, what is light? And where does it come from? In the most basic
terms, light is a type of electromagnetic radiation. Electromagnetic
radiation is a form of energy that travels in waves that are produced when
charged particles move or vibrate relative to each other. These energy
waves exist in small increments called photons. We can think of a photon
as a packet of light (Figure 4.3).

Figure 4.3 Light travels in packets of energy called photons.

Electromagnetic radiation ranges from very low-energy waves (TV and

radio waves) to very high-energy waves (X-rays and gamma rays). This
broad continuum of electromagnetic energy is referred to as the

212
electromagnetic spectrum (Figure 4.4).

Figure 4.4 The electromagnetic spectrum covers a wide range of energies. Visible light
is a narrow sliver of the entire spectrum.

In the middle of this spectrum is a small range of radiation that our

eyes can detect—we perceive it as visible light. This narrow range of
radiation is the visible spectrum. If we look more closely at the visible
spectrum (Figure 4.5), we can see that it is made up of the colors of the
rainbow: red, orange, yellow, green, blue, and violet.

Figure 4.5 The visible spectrum ranges from about 700 nm to about 370 nm. The colors
of the spectrum are red (R), orange (O), yellow (Y), green (G), blue (B), and violet (V).

Wavelength and Frequency

We use several parameters to describe electromagnetic waves.
Wavelength, symbolized by the Greek letter lambda (λ), is the distance
from a point on one wave to the same point on the next wave (Figure 4.6).
Typically, we measure wavelength in meters or nanometers (Recall that 1
nm = 10–9 m).

213
Figure 4.6 Wavelength is the distance from a point on one wave cycle to the same point
on the next cycle.

Explore

Figure 4.6

The color of light is related to its wavelength. On the low-energy end

of the visible spectrum, red light has a wavelength of about 700 nm.
Wavelengths longer than this fall into the infrared (IR) region. On the
high-energy end of the visible spectrum, violet light has a wavelength of
about 400 to 370 nm. Wavelengths shorter than this fall into the ultraviolet
(UV) region.
Frequency, symbolized by the Greek letter nu (ν), is the number of
waves that pass through a point in one second. A frequency of one wave
cycle per second is a hertz. The units corresponding to hertz are written as
Hz, as 1/s, or sometimes as s–1. For example, if a wave has a frequency of
10,000 hertz, we can write this in three different ways: as 10,000 Hz, as
10,000/s, or as 10,000 s–1. Each of these notations indicates 10,000 cycles
per second.

1/s=1 s−1=1 Hz

214
 Wavelength and the frequency are inversely related to each other. This
means that as the wavelength decreases, the frequency increases, and vice
versa. Mathematically, we describe this principle using the following
relationship, where λ is the wavelength and ν is the frequency:

C=λv
Before we define c, let’s look at the units involved in this expression:
The wavelength, λ, is often measured in meters. The frequency, ν, is the
number of cycles per second. If we multiply m × 1/s, we get m/s, which is
a unit of speed (Figure 4.7). Because we are describing light waves, c is
the speed of light. In a vacuum, the speed of light has a constant value of
3.00 × 108 m/s. Example 4.1 illustrates how we use this relationship to
convert between wavelength and frequency.

Figure 4.7 This equation gives the relationship between speed, wavelength, and
frequency.

1 nm=10−9 m

Example 4.1 Relating the Wavelength, Frequency, and Speed of

a Light Wave
A beam of green light has a wavelength of 500 nm. What is the frequency of
this light?
To solve this problem, we rearrange the equation above to solve for
frequency:

v=Cλ
Because c is given in units of meters per second, we also need to express

215
 the wavelength in meters so the units will cancel. One nanometer is equal to
10–9 m, so we can say that

500 nm=500×10−9 m
We then insert the values into the equation above:
v=Cλ=3.00×108 ms500×10−9 m=6.00×1014 s−1=6.00×1014 Hz

Notice that the meters cancel out, leaving us with units of 1/s. We can also
write this as s–1, or as Hz. This is the standard unit for frequency. •

IT
TRY

1. A beam of red light has a wavelength of 720 nm. What is the

frequency of this light?

2. A beam of blue light has a frequency of 6.41 × 1014 Hz. What is the
wavelength of this light in meters? In nanometers?

The Energy of a Photon

We often describe light in terms of its energy. The energy of a photon of
light depends on its frequency and wavelength. For example, compare the
two light waves shown in Figure 4.8. Notice that the red wave has a
longer wavelength than the blue one does. This means that the red wave
oscillates up and down more slowly, resulting in fewer cycles per second
(that is, a lower frequency). On the other hand, the blue wave oscillates
with a higher frequency. The energy of a wave depends on how quickly
the wave oscillates. The red wave (with its longer wavelength) has lower
energy while the blue wave (with its shorter wavelength) has higher
energy.

216
Figure 4.8 Blue waves have a shorter wavelength but higher frequency than red waves.
The blue wave has higher energy.

Explore

Figure 4.8

We can relate the energy of a single photon to the frequency using this
relationship:

E=hv
where E is the energy (measured in joules), ν is the frequency, and h,
which is referred to as Planck’s constant, has a value of 6.63 × 10–34 J·s.
To relate energy to wavelength, we can also write this expression as

E=hc/λ
Example 4.2 illustrates how we use these relationships to convert
between frequency, wavelength, and energy.

Example 4.2 Relating the Wavelength, Frequency, and Energy

217
 of Light
A photon has a frequency of 7.50 × 1014 Hz. What is the wavelength of this
light? What color is this light? What is the energy of the photon?
To find the wavelength, we rearrange the equation c = λν and substitute the
given values for c and ν:
λ=Cv=3.00×108 ms7.50×1014 1s=4.00×10−7 m=400 nm

By comparing this answer to the visible spectrum in Figure 4.5, we can see
that light with a wavelength of 400 nm is violet (or purple). Finally, we can
solve for the energy of the wave. •

E=hv=(6.63×10-34 J⋅s)×(7.50×1014 s-1)=4.97×10-19 J

Planck’s constant is named for Max Planck (1858–1947), a German physicist

who was instrumental in developing quantum theory (described in Section 4.3).
For his work, Planck was awarded the 1918 Nobel Prize in Physics.

IT
TRY

218
3. In the original Star Wars movie, Luke Skywalker was given a blue
light saber. After losing his hand (and saber) in The Empire Strikes
Back, he returned in the third movie with a green light saber. Which
light saber emitted light waves with higher energy, the blue or the
green?

4. What is the energy in joules of a beam of light whose frequency is

6.98 × 1014 Hz?

5. What is the wavelength of a photon with an energy of 3.75 × 10–19 J?

Based on the visible spectrum shown in Figure 4.5, approximately
what color of light is this?

219
4.2 Color, Line Spectra, and the Bohr Model

Color and Line Spectra

Our study of light and electronic structure begins with a technology that is
over a thousand years old: fireworks (Figure 4.9). The explosive powder
in fireworks often contains metals that produce distinctive colors as the
mixture burns.

Figure 4.9 Colored fireworks arise from different metals in the explosive mixtures.

We can easily observe this behavior in the laboratory by conducting an

experiment called a flame test. In this experiment, a wire is dipped in a
solution containing metal ions (recall that ions are charged particles). The
ions of each element give off a characteristic color when heated (Figure
4.10). For example, when a wire is dipped in a solution containing
calcium, the flame burns bright orange. When dipped in a solution of
copper, the flame burns bright green.

220
Figure 4.10 Different elements give off characteristic colors when heated in a flame.

Explore

Figure 4.10

We see a similar effect in gas lamps, such as those used in neon signs.
These lamps produce light by passing an electric current through a tube
filled with a gas such as neon, helium, argon, or krypton (Figure 4.11).
Like the metals in a flame test, each gas in a lamp produces a characteristic
color.

Figure 4.11 Gas lamps produce characteristic colors depending on which element is
present. Each of these lamps contains a different noble gas.

We can use a glass prism to analyze the light from these lamps.When
light passes through a prism, it separates into its constituent colors. Many
sources of white light produce all of the visible colors. If we pass this light

221
through a prism, we see the complete rainbow of colors (Figure 4.12). But
if we do the same thing with light from a gas lamp, we see a fascinating
result (Figure 4.13): Rather than producing a continuous spectrum of
colors, gas lamps produce colors of only certain energies. These bands of
color are called spectral lines. For example, the light from a hydrogen
lamp contains only four lines of color: red, light blue, deep blue, and
violet. These distinctive patterns are unique to each element and are called
line spectra.

Figure 4.12 When white light passes through a prism, it produces all the colors of the
rainbow.

Figure 4.13 When light from a gas lamp passes through a prism, it separates into
distinct lines of color that are called spectral lines. In this figure the lamp is filled with
hydrogen, and the spectrum is unique to that element.

222
 Explore

Figure 4.13

Scientists often use line spectra as “fingerprints” to identify elements.

For example, when scientists analyze light from the Sun, they find lines
that correspond to the spectral lines of hydrogen and helium. From this,
they know that the Sun and other stars are composed largely of these two
elements.

Light can be separated into its constituent colors by using either a prism or a
diffraction grating—a surface containing very close parallel lines. You can see
this effect on the surface of a CD or DVD. Modern instruments use diffraction
gratings rather than prisms to separate light.

The Bohr Model

By the early twentieth century, scientists knew that atoms contain a dense

223
nucleus surrounded by electrons. They also knew that high-energy light
could knock electrons off of some atoms—a phenomenon known as the
photoelectric effect. This effect showed that light energy was somehow
connected to electron structure. But how did it relate to the line spectra?
In 1913, the Danish scientist Niels Bohr unveiled a theory (now called
the Bohr model) that connected line spectra with electron structure. Bohr
proposed that electrons orbit the nucleus in much the same way that
planets orbit the Sun (Figure 4.14). He suggested that orbits closer to the
nucleus are lower in energy than those that are farther away, and he
posited that only certain orbits, or energy levels, are “allowed.”

Figure 4.14 The Bohr model proposed that electrons orbit the nucleus the way planets
orbit the Sun and that the electrons could jump from one energy level to another.

Bohr also proposed that electrons can jump from one energy level to
another. He theorized that when an electron absorbs light, it jumps to a
higher energy level. When it drops to a lower energy level, it releases that
energy as light. When an atom’s electrons are in the lowest possible levels,
the atom is said to be in the ground state. If electrons jump to higher
levels, the atom is in the excited state.

When atoms absorb energy, electrons jump to excited states.

Atoms release energy when electrons relax back down.

The Bohr model neatly explained the line spectra of hydrogen (Figure
4.15). In its ground state, hydrogen’s one electron occupies the lowest
possible level—energy level 1. However, this electron can absorb energy
(in the form of heat, light, or electrical energy). When this happens, the
electron jumps from level 1 (the ground state) to a higher energy level (an
excited state). Eventually, the electron relaxes from the excited state back

224
down to lower energy states, and ultimately to the ground state. As it does
so, it releases energy as electromagnetic radiation.

Figure 4.15 According to the Bohr model, electrons orbit the nucleus in certain allowed
energy levels. When an electron absorbs energy, it jumps to a higher energy level.
When the electron comes back down, it releases that energy—often as visible light. (a)
An electron absorbs energy and jumps up to level 5. (b) A drop from level 5 to level 2
releases indigo (dark blue) light. (c) A drop from level 3 to level 2 releases red light.

Explore

225
 Figure 4.15

In a hydrogen atom, four transitions produce visible light, resulting in

four spectral lines (Table 4.1). Notice that the smallest transition produces
the lowest-energy light (red), while the biggest transition produces the
highest-energy light (purple). Other transitions also take place, but they
produce radiation that is either too high in energy (ultraviolet) or too low
in energy (infrared) for our eyes to detect.

TABLE 4.1 Transitions in the Hydrogen Line Spectrum

Transition Color Produced

3→2 Red

4→2 Light blue

5→2 Indigo (deep blue)

6→2 Purple (violet)

Imagine standing on a staircase. You can jump from one step to another, but you
can’t hover between steps—you will always land on one of the steps. Similarly,
electrons occupy only specific energy “steps.” Electrons can absorb energy to
move to a higher level (step) or release energy to move to a lower level.

Bohr’s ideas about light and electron energy levels were a critical
advance. They provided a foundation for understanding the interplay of
light and electrons that occurs all around us. Is your shirt colored? It is
absorbing some colors, but reflecting others. The absorption of light
energy causes electrons to jump to higher levels. Do you have a light on as
you’re reading this? The electrical current is continually exciting the

226
electrons to higher energy levels. As they relax back down, this energy is
released as visible light. Here are a few other examples related to this
concept:

From bonfires to smartphones, light is produced when electrons are excited to

higher levels and then relax back down.

1. Why does the Sun give off light? The Sun is powered by a type of
reaction called nuclear fusion, which releases an incredible amount
of energy (we’ll discuss nuclear fusion in Chapter 16). This energy
continually excites electrons (and other charged particles), which
release this energy as electromagnetic radiation. The Sun radiates
electromagnetic energy across the spectrum, not just visible light.
2. Why does a fire give off light? When a substance like wood burns,
the carbon and hydrogen in the substance react with oxygen in the
air. This reaction releases heat energy as well as gaseous products
such as carbon dioxide and water vapor. The electrons within these
compounds are excited to higher levels. As these gaseous products
exit the reaction, their electrons release this energy as visible light
that we see as flames.

Fluorescence
Not every electronic transition involves visible light. Sometimes, a substance
absorbs invisible UV energy and then releases it in lower-energy steps as visible
light. This phenomenon is known as fluorescence. Ultraviolet lamps (also called
black lights) cause many substances to fluoresce. For example, bark scorpions
are nearly impossible to spot with the naked eye, but they contain substances
that fluoresce and show up easily under a black light.

227
3. Do you give off light? People don’t give off visible light; however,
as your body produces heat, you release this energy in the form of
infrared radiation. Specialized cameras can convert infrared energy
into visible images, making it possible to “see” someone in the dark
(Figure 4.16). This technique is called infrared imaging or thermal
imaging.

Figure 4.16 (a) This image shows the infrared energy given off by two factory
workers. (b) In the classic action movie Predator, Arnold Schwarzenegger
covered himself in mud to mask the infrared radiation produced by his body
heat.

228
4.3 The Quantum Model and Electron Orbitals
The Bohr model proved useful for describing the chemical properties of
the main-group elements, and the concept of electron energy levels was a
key advance that elegantly explained the line spectrum of hydrogen.
However, Bohr’s model was unable to explain the properties of the
transition elements or the complex line spectra of elements larger than
hydrogen.
In the 1920s and 1930s, a series of discoveries led scientists to
abandon much of the Bohr model in favor of a more nuanced description
of electron behavior, called the quantum model, which describes
electrons as both particles and waves. These early discoveries marked the
dawn of quantum mechanics, a field of study that deals with the unique
and surprising behavior of subatomic particles. In this section we will
examine some key ideas from quantum mechanics, and we’ll use the
quantum model to describe electron behavior more completely.

The Uncertainty Principle and the Wave Nature of Electrons

In 1927 the German scientist Werner Heisenberg introduced a startling
idea called the uncertainty principle. This principle deals with the mass,
velocity, and location of subatomic particles. A central idea of this
principle is that it is impossible to know the exact velocity and location of a
particle. Now, this doesn’t make much sense to us in our everyday world
—but as we deal with tiny, fast-moving particles, uncertainty becomes
more significant.
We never know the exact location of electrons.

As a crude analogy to help think about this concept, consider an

electric fan (Figure 4.17). When the fan is off, we know exactly where the
blades are located. However, if the fan is turned on, the blades move so
quickly that we no longer know exactly where they are—we just know
they are moving in an area that occupies a circle. We don’t stick a finger in
that circle, because we know the blades occupy that area.

229
Figure 4.17 As the blades on a fan move faster and faster, we can no longer pinpoint
their location. Instead we describe them by the shape of the region they occupy.

Similarly, in quantum mechanics, we never talk about the exact

position of an electron. According to Heisenberg, this position is
impossible for us to know. Instead, we talk about the most probable
locations of the electrons or the energies that the electrons possess.
A second principle of quantum mechanics is equally surprising: When
dealing with tiny particles (such as electrons), we can’t describe them
simply as particles, because many of their behaviors more closely
resemble energy waves. For example, the line spectra of the elements
clearly show that electrons exist only at certain energies. If electrons are
simply negatively charged particles, we can’t explain why this is so. But
what if electrons behave like waves? Many types of waves, such as those
produced by a guitar string (Figure 4.18), oscillate at specific energies. By
treating electrons as energy waves, quantum mechanics can explain why
electrons exist only at specific energy levels (Figure 4.19) and can even
predict the energy changes that produce line spectra.

Figure 4.18 Each string of a guitar vibrates with a specific wavelength and frequency,
producing a unique note.

230
Figure 4.19 Describing electrons as waves helps explain why they exist only at specific
energy levels.

Together, these two ideas—the uncertainty principle and the wave

nature of electrons—are the foundation for our modern understanding of
electron structure. In the quantum model, we don’t try to pinpoint the
location of electrons. Rather, we talk about the energies of electron waves
and the regions around the nucleus where the electrons are most likely to
be found.

Electrons behave like waves that have different energies.

The story of quantum mechanics is fascinating and full of surprises. It

is a wild, strange, counterintuitive world at the subatomic level. And
though the math behind quantum mechanics is beyond the scope of this
book, the complex equations lead to some straightforward rules governing
the configuration of electrons within an atom. In the pages that follow,
we’ll explore these rules and see how they explain the behavior of
elements across the periodic table.

Energy Levels and Sublevels

The quantum model describes how electrons are arranged within an atom.
To describe these arrangements, let’s begin with four basic rules:
1. Electrons occupy different energy levels. In quantum mechanics, the
energy level is identified by a whole number called the principal
quantum number (Figure 4.20). The lowest energy level (level 1)
lies closest to the nucleus. Higher energy levels (2, 3, 4, etc.) lie
farther from the nucleus. Each energy level can hold a maximum
number of electrons (Table 4.2). Higher energy levels can hold
more electrons than lower energy levels.

231
 Figure 4.20 Electrons occupy different energy levels, which are represented by
the principal quantum number, n. Energy level 1 is the lowest in energy, and it is
closest to the nucleus.

TABLE 4.2 Electrons in Each Energy Level

Level Possible Electrons

1 2

2 8

3 18

4 32

2. Each energy level contains one or more sublevels. There are four
common energy sublevels, which are designated by the letters s, p,
d, and f. Each sublevel can hold a set number of electrons, as
summarized in Table 4.3.

TABLE 4.3 Energy Sublevels

Sublevel Number of Orbitals Electron Capacity

s 1 2

p 3 6

d 5 10

f 7 14

3. Each sublevel contains one or more orbitals.An orbital is the region

where the electrons are most likely to be found. Each sublevel
contains a different number of these orbitals. An s sublevel contains

232
 1 orbital; a p contains 3, a d contains 5, and an f contains 7.
4. Each orbital can hold up to two electrons. Electrons have a tiny
magnetic field, called spin. When electrons pair together in orbitals,
their spins orient in opposite directions. An orbital is filled if it
contains two paired electrons.

Energy levels and sublevels are sometimes referred to as energy shells and
subshells, respectively.

Now that we have the fundamental rules of electron arrangements, let’s

look at each of the energy levels and sublevels.
Energy level 1 contains only the s sublevel. The s sublevel is the
simplest of the sublevels. The s sublevel has only one orbital, shaped like a
sphere (Figure 4.21).

Figure 4.21 The s sublevel can hold two electrons. The shading represents the region
the electron will most likely occupy.

Energy level 2 contains both s and p sublevels. The 2s sublevel is

spherical in shape, but it lies farther from the nucleus than the 1s sublevel
(Figure 4.22). The p sublevel contains three different orbitals. Each
orbital has a shape similar to an infinity symbol (Figure 4.23). The three
orbitals can be thought of as orienting along the x, y, and z axes. Because
each orbital can hold up to two electrons, a p sublevel can contain a total
of six electrons.

233
Figure 4.22 This image shows a cutaway view of the 1s and 2s sublevels. Both
sublevels have a spherical shape, but the 2s sublevel is farther from the nucleus.

Figure 4.23 The p sublevel is composed of three orbitals (shown here in blue, red, and
green), occupying the regions around the x, y, and z axes.

The difference between orbits and orbitals can be confusing. The Bohr model
described electrons as particles orbiting the nucleus, like planets orbit the Sun.
The quantum model states that we can never actually know the location of the
electron. The term orbital describes the region around the atom where the
electron is most likely to be.

234
 If you place two bar magnets side by side, they will pair up with their poles
facing in opposite directions. Similarly, when two electrons occupy the same
orbital, their magnetic fields (that is, their spins) orient in opposite directions.
We often show this as two adjacent, half-headed arrows—one pointing up and
one pointing down.

Energy level 3 contains not only s and p sublevels but also the d
sublevel.The d sublevel contains five orbitals, and therefore it can hold a
maximum of 10 electrons. The shapes of the five d orbitals are shown in
Figure 4.24.

235
Figure 4.24 The d sublevel contains five orbitals. These diagrams represent the five
regions occupied by electrons in these orbitals.

The shapes of the orbitals indicate the location the electron most likely
occupies.

In energy level 4, the f sublevel is added (Figure 4.25). The f sublevel

contains seven orbitals and therefore can hold a maximum of 14 electrons.
The orbitals of the f sublevel have complex geometries. You should know
the shapes of the s and p orbitals, but you do not have to know the shapes
of the d or the f orbitals. However, it is important to remember the number
of orbitals and the number of electrons that can fit within each energy
sublevel.

Figure 4.25 The f sublevel contains seven orbitals. These diagrams represent the seven
regions occupied by electrons in these orbitals.

Notice that the energy sublevels follow a predictable pattern: Each new
level adds a new sublevel, and each new sublevel has two more orbitals
(four more electrons) than the one before. Table 4.4 is a summary of the
sublevels for energy levels 1 through 4.

TABLE 4.4 Energy Levels, Sublevels, and Electron Capacity

Energy Level 1 2 3 4

f (14 e–)

Sublevels d (10 e–) d (10 e–)

p (6 e–) p (6 e–) p (6 e–)

s (2 e–) s (2 e–) s (2 e–) s (2 e–)

236
 Electron Capacity 2 8 18 32

Note: the symbol e– means electrons.

Above energy level 4, the trend continues: Level 5 has five sublevels,
level 6 has six sublevels, and so on. However, even the largest elements on
the periodic table fit all of their electrons within the s, p, d, and f sublevels,
so we don’t need to worry about any sublevels beyond these four.
Figure 4.26 shows the relative energy differences between the
different energy levels and sublevels. Notice that within an energy level,
the s orbital is the lowest in energy, followed by the p, d, and f orbitals.
Also notice that as the energy levels get higher, they group together more
closely. As the energy levels branch out into more sublevels, they actually
overlap each other. For example, energy level 4s is actually lower than
energy level 3d. As we’ll see in the next section, this pattern significantly
affects the electron configurations and element behaviors.

Figure 4.26 This diagram represents the energy differences among the energy
sublevels. Each horizontal line represents an orbital that can hold two electrons. Notice
that the higher energy levels overlap each other.

237
4.4 Describing Electron Configurations
To understand how electrons fill the different energy levels and sublevels,
let’s look at the ground-state (lowest energy) electron configurations of a
series of atoms. As we do so, remember that atomic number, located on the
periodic table, tells us the number of electrons present in any neutral atom.
Hydrogen—one electron. The single electron in hydrogen occupies the
lowest energy level and sublevel possible. This electron occupies energy
level 1, sublevel s. Figure 4.27 includes a depiction of the hydrogen atom.

Figure 4.27 Electron configurations may be represented by a written electron

configuration (at the top of each box), by an energy diagram (lower left in each box), or
as a sketch showing the shape of the occupied orbitals (lower right).

Although graphical depictions can help us understand the configuration

of an atom, we sometimes need to express electronic structures without
drawing elaborate pictures. To do this, we write electron configurations,
which show the number of electrons in each occupied energy level and
sublevel. For example, we write the electron configuration of hydrogen as

Hydrogen: 1s1

238
 In this notation, the 1s means that energy level 1, sublevel s, is
occupied. The superscript 1 indicates that one electron resides in this
orbital.
Figure 4.27 also shows the order of electron filling using an energy
diagram. In this diagram, the orbitals are represented by blank lines that
may be populated with electrons. The electrons are shown as half-headed
arrows.
Helium—two electrons. Both electrons go in the lowest level and
sublevel (1s), filling energy level 1. We write the electron configuration of
helium as 1s2. Notice in the energy diagram of helium (see Figure 4.27)
that when we represent two electrons in one orbital, we draw one with the
half-arrowhead up (↾) and the other with the half-arrowhead down (⇃). This
notation signifies that the spins (that is, the magnetic fields of the
electrons) are oriented in opposite directions.
Lithium—three electrons. The first two electrons fill the first energy
level. The third electron occupies the next-lowest sublevel, which is 2s.
We write the electron configuration for lithium as 1s22s1, to indicate that
there are two electrons in level 1s and one electron in level 2s.
Beryllium—four electrons. The electron configuration for beryllium is
1s22s2.
Boron—five electrons. After filling sublevels 1s and 2s, the next-
lowest energy sublevel is 2p. The electron configuration for boron is
1s22s22p1.
Carbon—six electrons. The first four electrons fill sublevels 1s and 2s.
The two remaining electrons occupy level 2p. However, this configuration
introduces a new question: Because three identical p orbitals are available,
do the two electrons pair up in a single orbital, or does each electron singly
occupy its own orbital? In this case, the electrons occupy their own
orbitals rather than pairing up (Figure 4.28). This idea is referred to as
Hund’s rule: If empty orbitals of the same energy are available, electrons
will singly occupy orbitals rather than pairing together.

239
Figure 4.28 If empty orbitals of the same energy are available, electrons will singly
occupy orbitals rather than pairing together. This is referred to as Hund’s rule.

Imagine that you are a manager for a dormitory. Each room can hold two
people, but the dorm is not full. If you gave the residents a choice of having
their own room or sharing a room, what would they choose? If there is no
difference in rent, most people would choose their own room. Similarly, if
empty orbitals are available, each electron singly occupies its own orbital before
pairing up to share orbital space.

The electron configurations of the remaining elements in row 2 of the

periodic table are shown in Figure 4.29. Notice that as we move across the
periodic table in row 2, the second energy level fills with electrons. Neon,
the last element in row 2, has a completely filled energy level.

240
Figure 4.29 Electron configurations for the elements in row 2 of the periodic table are
shown here.

Row 3 of the periodic table contains the elements whose highest-

energy electrons occupy level 3. Example 4.3 illustrates the electron
configuration for a row 3 element.

Example 4.3 Writing Electron Configurations

Silicon is a vital component of electronic devices. What is the electron
configuration of silicon? Refer to the periodic table and Figure 4.26 as
needed.
To answer this question, we first need to identify the number of electrons in
a silicon atom. From the periodic table, we can see that silicon is atomic
number 14: This means there are 14 electrons in a neutral silicon atom.
We begin by placing two electrons in the lowest energy level (1s). Sublevel
2s can hold two more electrons. Sublevel 2p can hold six. The next-lowest
level is 3s, which can hold two electrons. The remaining two electrons go
in the next-lowest level, which is 3p. Based on this information, we can
write the electron configuration for silicon as 1s22s22p63s23p2. The
superscript numbers indicate the number of electrons in each level. •

241
 The atomic number tells us both the number of protons and the number of
electrons in an atom.

The superscripts show the number of electrons in each sublevel. These

numbers should add up to equal the number of electrons in the atom.

IT
TRY

6. Without referring to Figure 4.29, write the electron configurations for

each atom.
He N O F Ne Na

Valence Electrons and the Octet Rule

Electron configurations are important because they help determine how an
atom behaves. As we’ll see in the chapters to come, chemical changes
involve the gain, loss, or sharing of electrons. These changes take place at
the outermost (highest energy) levels and sublevels. Understanding the
electron configuration enables us to explain and predict an element’s
chemical properties.

Valence electrons occupy s and p sublevels.

The highest-occupied electron energy level is the valence level

(sometimes called the valence shell). As a general rule, atoms hold up to
eight electrons in their valence (highest occupied) level. To see why this is
so, consider the simplified representation of the electron energy levels in
Figure 4.30. Notice that the level above 2p is 3s. The level above 3p is 4s,
and the level above 4p is 5s. If an atom has its s and p sublevels filled, the
next electron goes into a higher energy level.

242
Figure 4.30 Because of the way sublevels overlap, we always move to the next energy
level once the p sublevel is filled.

For example, argon has 18 electrons. Therefore, we write its electron

configuration as 1s22s22p63s23p6. An argon atom has eight electrons in its
valence level (energy level 3). The next atom on the periodic table,
potassium, has an electron configuration of 1s22s22p63s23p64s1.
Potassium has one electron in its valence level (energy level 4). Once the s
and p orbitals of the valence level are filled, additional electrons go in the
next-lowest energy level.
The elements with filled valence levels are the noble gases, which are
located in the far-right column of the periodic table. Helium, with two
electrons, has energy level 1 completely filled. The other noble gases
(neon, argon, krypton, xenon, and radon) all have eight electrons in their
valence level, which completely fill the s and p sublevels (Table 4.5).
These elements are stable and unreactive—they generally do not combine
with other atoms to form compounds. We describe this stability using the
octet rule, which states that an atom is stabilized by having its highest-
occupied (valence) energy level filled. If an atom has eight electrons (an
octet) in its valence shell, the s and p sublevels are completely filled.

TABLE 4.5 Electron Configurations for Noble Gases

Element Electron Configuration Valence Electrons

He 1s2 2

Ne

243
 1s22s22p6 8

Ar 1s22s22p63s23p6 8

Kr 1s22s22p63s23p64s23d104p6 8

As you practice writing electron configurations for larger atoms, you will
probably want to use the chart in Figure 4.30. However, you don’t need to
memorize it: As you’ll see later in this chapter, the periodic table contains all the
information required—you just have to know how to look for it.

Electron Configurations for Larger Atoms

Let’s look at the electron configurations for some larger atoms. For
example, the following three atoms lie in row 3 of the periodic table:
Sodium (11 electrons): 1s22s22p63s1
Phosphorus (15 electrons): 1s22s22p63s23p3
Chlorine (17 electrons): 1s22s22p63s23p5
Notice that in each of these configurations, the first two levels
(sometimes called the inner electrons, shown in red) do not change. If we
add electrons, the change affects only the valence level. Because the inner
electrons do not change, we often represent them in a simpler form, called
noble gas notation. For example, 1s22s22p6 is the configuration of neon.
Rather than writing this configuration out, we represent it by enclosing the
symbol for neon in square brackets: [Ne]. Using this shorthand notation
makes the electron configurations simpler to write:
Sodium (11 electrons): [Ne]3s1
Phosphorus (15 electrons): [Ne]3s23p3
Chlorine (17 electrons): [Ne]3s23p5
In this notation, we always use the noble gases, which have complete
octets in the inner levels.
Besides convenience, using noble gas notation has another advantage.
It allows us to focus on outer electrons (those beyond the largest filled
noble gas configuration). The outer electrons include the valence level and
partially filled d and f sublevels (Figure 4.31). Chemical bonds involve
changes in the outer electrons. We will explore these changes in Chapter 5.

244
Figure 4.31 Chemical bonds involve the outer electrons (those beyond the largest filled
noble gas configuration). These include both valence electrons and d and f orbital
electrons.

Chemical bonds involve outer electrons.

Example 4.4 Identifying Valence Electrons

How many valence electrons are present in oxygen? How many are present
in sulfur?
Oxygen has the electron configuration 1s22s22p4. Its valence level (level 2)
contains a total of six electrons.
Sulfur has the electron configuration 1s22s22p63s23p4. Its valence level
(level 3) also contains a total of six electrons.
Notice that both elements contain the same number of valence electrons,
and they are located in the same column of the periodic table. As we’ll see
in the coming section, the columns of the periodic table contain elements
with similar valence configurations. •

Example 4.5 Identifying Inner, Outer, and Valence Electrons

Write the electron configuration for selenium using noble gas shorthand.
Identify the inner electrons, the outer electrons, and the valence electrons.
The full electron configuration of selenium is 1s22s22p63s23p64s23d104p4.
Because the first 18 electrons (1s22s22p63s23p6) correspond to the electron
configuration of argon, we can rewrite this using noble gas shorthand:
Selenium: [Ar]4s23d104p4
The electrons within the [Ar] configuration are the inner electrons. The
remaining electrons in the 4s, 3d, and 4p sublevels are the outer electrons.

245
 Selenium has a total of 16 outer electrons.
The valence electrons are only those in the highest energy level—that is,
energy level 4. Selenium has six valence electrons. •

IT
TRY

7. Write electron configurations for each atom, using noble gas notation.
P I Fe Sr

8. Identify the inner, outer, and valence electrons in the electron

configurations of each of the atoms in Question 7.

Electron Configurations for Ions

In Chapter 3, we said that atoms are able to gain or lose electrons to form
ions—that is, particles with an overall positive or negative charge. Ions are
a vital part of chemistry, so it is important to understand and be able to
describe the electron configuration of ions as well as neutral atoms. Let’s
begin with two examples.

The octet rule: Eight electrons in the valence shell is a very stable
configuration.

Example 4.6 Writing Electron Configurations for Ions

What is the electron configuration of a sodium atom? What is the electron
configuration of a sodium ion with a charge of +1?
The sodium ion has a charge of +1, so it must have one electron fewer than
neutral sodium. The electron that is removed is from the highest occupied
(valence) energy shell (3s). Notice that once this electron has been
removed, the electron configuration of the sodium ion is identical to the
very stable electron configuration of neon, as shown here. •

Full Noble Gas

Species Symbol
Configuration Shorthand

246
 Sodium atom Na 1s22s22p63s1 [Ne]3s1
Sodium ion (+1 Na+ 1s22s22p6 [Ne]
charge)

Example 4.7 Writing Electron Configurations for Ions

What is the electron configuration of an oxide ion, which is an oxygen ion
with a charge of –2?
Because the oxide ion (that is, a negatively charged oxygen atom) has a
charge of −2, it must have two more electrons than neutral oxygen. The
electrons that are gained by the atom fill the lowest available energy shell.
Notice that when this happens, the electron configuration of the oxygen ion
is identical to the very stable electron configuration of neon. •

Species Symbol Full Configuration Noble Gas Shorthand

Oxygen atom O 1s22s22p4 [He]2s22p4
Oxide ion (–2 charge) O2– 1s22s22p6 [Ne]

In Example 4.6, the sodium atom lost an electron and ended up with
the same electron configuration as neon. In Example 4.7, the oxygen atom
gained two electrons and also ended up with the same electron
configuration as neon. Na+, O2–, and Ne are isoelectronic, meaning they
have the same electron configurations.
The gain, loss, and sharing of electrons is central to the understanding
of chemistry. The idea that atoms will gain or lose electrons and either fill
or empty their highest-occupied electron energy levels is a recurring
theme. You will see it again and again as we study patterns of bonding and
reactivity in future chapters.

IT
TRY

9. Write electron configurations for these ions.

Li+ Cl– S2–

247
10. In Section 4.4, Na+, O2–, and Ne are described as isoelectronic. Which
of the following atoms or ions are also isoelectronic with those ions?

He F F– Mg2+ P

248
4.5 Electron Configuration and the Periodic Table
When Mendeleev organized the periodic table of the elements in the late
1800s, he based it on the chemical and physical properties he observed. He
organized elements with similar behaviors into columns. Earlier in this
chapter, we said that electron configuration is critically important to the
properties of atoms. So how do electron configurations relate to the
periodic table?
To answer that question, let’s take a look at three atoms that behave
very similarly: lithium, sodium, and potassium. All react quickly—even
violently—with water, and they all tend to form ions with a +1 charge.
Here are the electron configurations of these three atoms:
Lithium (3 electrons): [He]2s1
Sodium (11 electrons): [Ne]3s1
Potassium (19 electrons): [Ar]4s1
Do you see a pattern? In their highest-occupied shell, these atoms all
have an s1 configuration. These elements exhibit similar behaviors because
their electron configurations are similar. Now think about where we find
these elements on the periodic table. They are located in a single column,
on the far left-hand side of the table (Figure 4.32).

Figure 4.32 Lithium, sodium, and potassium have similar properties because they have
similar electronic structure. They are grouped in a column on the periodic table.

249
 Let’s look at another example: fluorine, chlorine, and bromine all form
ions with a charge of −1. Here are their electron configurations:
Fluorine (9 electrons): [He]2s22p5
Chlorine (17 electrons): [Ne]3s23p5
Bromine (35 electrons): [Ar]4s23d104p5
What are the similarities this time? In their valence shell, they all have
a configuration of s2p5. (Bromine has the 3d level filled between the s and
p sublevels, but this does not affect the overall trend.) How are these
elements organized on the periodic table? As in the previous example, they
are located in a single column; but these elements are on the right-hand
side of the periodic table (Figure 4.33).

Figure 4.33 Fluorine, chlorine, and bromine have similar properties. They all have a p5
electron configuration.

The periodic table is organized like a calendar. A row on a calendar shows the
whole range of days—Sunday through Saturday. But the days in a calendar tend
to be similar—for example, you might have class every Monday—and days that
are similar are grouped into columns.

250
 Now let’s take a close look at the periodic table in Figure 4.34. The
organization of the periodic table is based on element masses and
properties, but it is also based on electronic structure. For example,
elements in the first row (H and He) have electrons only in energy level 1,
elements in the second row (Li through Ne) have electrons in energy level
2, and so forth. For any atom, we can determine its highest-energy electron
shell from the row it occupies on the periodic table.

Figure 4.34 The periodic table is organized based on electronic structure.

251
 Explore
Figure 4.34

Family Similarities
Families (groups) of elements share similar electron configurations:
Alkali metals: s1
Alkaline earth metals: s2
Halogens: s2p5
Noble gases: s2p6

Columns of the periodic table correspond to the sublevel electron

configuration. Every atom in the first column has an electron configuration
of s1, every atom in the second column has an electron configuration of s2,
and so on. Once you understand the periodic table, you can quickly
determine the electron configuration of the highest-occupied energy levels.

Elements within a group have similar valence configurations.

252
 Helium has its valence energy level filled, and it behaves like a noble gas. On
the periodic table, helium is placed above the p6 noble gases even though its
electron configuration is s2.

Understanding periodic table organization also makes it unnecessary to

memorize the filling sequence of the sublevels. With a glance at the
periodic table, you should be able to tell the sequence in which the
sublevels are filled. For example, look at the diagram of energy levels in
Figure 4.35. What energy sublevels come after 5s? Moving left to right
across the table, you can see that the next sublevels are 4d, then 5p, then
6s.

Figure 4.35 We can use the periodic table to determine the energy sublevel filling
sequence.

Exceptions to the Rule

A few exceptions to the filling sequence rules are not shown on the periodic
table in Figure 4.34. For example, copper (Cu) is expected to have an electron
configuration of [Ar]4s23d9, but its actual configuration is [Ar]4s13d10. And
because the 4f and 5f orbitals are so close in energy to the 5d and 6d orbitals,
respectively, there are multiple exceptions to the filling sequences in the f
blocks. Do not worry about remembering these exceptions; instead, focus on

253
 understanding the general trends.

As you become more comfortable with how the periodic table is

organized, you should be able to quickly identify the valence configuration
of an atom. For example, what is the valence configuration of sulfur? The
strategy for solving this question is shown in Figure 4.36: The row
indicates the energy level (3), and the column shows the sublevel and
filling (s2p4).

Figure 4.36 The periodic table allows us to quickly identify the valence electron
configuration of an atom based on its location.

The four blocks of the periodic table, s, p, d, and f, are based on the highest
energy sublevel that an element’s electrons occupy.

In Chapter 3, we identified the main-group elements using the

numbering system 1A–8A (Figure 4.37). These numbers also give us a

254
quick reference to identify the number of valence electrons in a main-
group element. For example, carbon, silicon, and germanium are all
located in column 4A, and each of these elements has four valence
electrons.

Figure 4.37 The main-group numbers 1A–8A tell us the number of valence electrons in
each main-group element.

Example 4.8 Identifying the Highest-Shell Electron

Configuration from the Periodic Table
Without writing the full electron configuration, identify the electron
configuration for the highest-energy occupied sublevel of each of the
following atoms: potassium, germanium, and iron.
Based on the periodic table (Figure 4.34), potassium (K) is in the fourth
period and in the group with the outermost electron configuration, s1.
Therefore, potassium’s highest-sublevel configuration is 4s1.
Similarly, germanium (Ge) lies in the fourth period, in the p2 column.
Therefore, germanium’s highest sublevel configuration is 4p2.
Iron is slightly different. The d block fills after the s block of the level
higher, so the d block in period 4 actually corresponds to energy level 3
(see Figure 4.35). Based on this information, we can say that iron has a
highest-energy electron configuration of 3d6. •

255
 Example 4.9 Identifying Electron Configurations from the
Periodic Table
Using the periodic table as a guide, write the electron configuration for
strontium. Use noble gas shorthand, and try to find the electron
configuration without counting all the electrons.
On the periodic table, notice that strontium (atomic number 38) comes two
elements after the noble gas krypton (atomic number 36). The inner
electrons have the [Kr] configuration. Strontium falls in row 5, column 2 of
the s block, so its outer configuration is 5s2. Based on this information, we
write its electron configuration as [Kr]5s2. •

IT
TRY

11. Oxygen has a valence-level electron configuration of s2p4. What are

the other atoms that have an s2p4 valence-level configuration?

12. Write correct electron configurations for the following atoms. Try to
do this using the periodic table as a reference, without adding up the
electrons.
Mg Si Zr Sb

256
Summary
In this chapter we explored electronic structure in detail. Beginning with
the electromagnetic spectrum, we saw that the absorption and emission of
light (or other electromagnetic radiation) provides critical clues about
electronic structure. In the early 1900s, Niels Bohr theorized that (1)
electrons orbit the nucleus in much the same way that planets orbit the
Sun; (2) electrons transition from one orbit to another by absorbing or
releasing energy; and (3) line spectra result from these transitions.
Although the planetary model has been discarded, the idea that line spectra
result from electron transitions from one energy level to another is critical
to understanding electron structure.
The Bohr model was replaced by the quantum model, which describes
electronic structure in terms of the energies and most probable locations of
electrons. Electrons do not occupy planet-like orbits (as the Bohr model
suggested) but occur in more complex, wave-like patterns called orbitals.
Each orbital can hold up to two electrons. These orbitals belong to energy
levels and sublevels.
Energy level 1 has only one sublevel, but each additional energy level
includes a new sublevel. The sublevels are s, p, d, and f, and they hold 2, 6,
10, and 14 electrons, respectively. Because chemical reactions involve the
gain, loss, or sharing of electrons, it is important to understand and be able
to express the electron configurations of atoms. Electron configurations in
the highest-occupied energy level are especially important. These electrons
are referred to as the valence electrons.
The periodic table is organized by electronic structure as well as by
atomic number and chemical reactivity. We can quickly determine the
electronic structure of elements on the periodic table based on the row
(primary energy level) and the column (sublevel and electron count) where
they lie. Columns on the periodic table contain elements with similar
electronic structures and therefore similar patterns of behavior.

257
Solar Cells: Converting Light into Electric Current

At the beginning of this chapter, we introduced the concept of solar energy. We

saw that light and electron structure are closely connected. We also observed that
when a material absorbs light, electrons are excited to higher energy levels. But
how does this produce electricity?
Most solar cells use a design called a p–n junction to convert sunlight into
electrical current. In this design, two slightly different semiconductors (called p-
type and n-type) are placed next to each other. The metalloid element silicon is the
main component of both types of semiconductors; the difference is in which other
elements are present. An n-type semiconductor contains tiny amounts of an
element with an s2p3 valence configuration, such as arsenic. These elements have
high-energy electrons that are easily removed from the atom. (The term n-type
refers to the negative charges [electrons] that are available in the solid.) On the
other hand, p-type semiconductors contain tiny amounts of an element with an s2p1
valence configuration, such as gallium. These elements have empty orbitals in their
highest energy level, which can hold additional electrons.
When a p-type material is placed next to an n-type material, some of the
electrons from the n-type move to the p-type, creating a barrier of positive and
negative charges (Figure 4.38).

Figure 4.38 Most semiconductors use silicon doped with elements like gallium and
arsenic. When the sunlight hits the cells, electrons are excited to a higher energy level,
which allows them to move from one atom to another. A line of charge along the p–n
junction pushes electrons, creating an electric current.

258
 When a photon of light hits the semiconductor, it excites an electron to a higher
energy level. In a semiconductor, excited electrons can move from atom to atom.
When this happens, the charges along the p–n junction push electrons away from
the negative charge and toward the positive charge. This push produces a
movement of electrons—an electric current.

259
 Key Terms
4.1 The Electromagnetic Spectrum
electromagnetic radiation A form of energy produced when charged particles
move or vibrate relative to each other; electromagnetic radiation exists as
waves.
photon A small increment or packet of electromagnetic energy (often visible
light).
electromagnetic spectrum All forms of electromagnetic energy, ranging from
low-energy waves (TV and radio) to visible light to high-energy waves such as
gamma rays.
visible spectrum The narrow range of electromagnetic energy that we
perceive as light.
wavelength (λ) The distance from a point on one wave to the same point on
the next wave.
frequency (ν) The number of waves that pass through a point in one second;
typically measured in hertz.
hertz (Hz)A unit of frequency equal to one wave cycle per second.

4.2 Color, Line Spectra, and the Bohr Model

line spectrum A pattern of light energies called spectral lines; they are formed
when gas-phase elements release energy. Each element has a characteristic line
spectrum.
Bohr model An early model of atomic structure that treated the atom like a
tiny solar system, with the nucleus at the center, and the electrons orbiting the
nucleus.

4.3 The Quantum Model and Electron Orbitals

quantum model The modern description of electronic behavior that treats
electrons as both particles and waves.
uncertainty principle The idea that it is impossible to know the exact velocity
and location of a particle; this principle becomes important when studying
electrons.
principal quantum number An integer number that identifies the energy
level an electron occupies.
sublevel A set of electron orbitals that occurs in an electron energy level; the
four main sublevels are s, p, d, and f.

260
orbital A region where electrons are most likely to be found; each orbital can
hold up to two electrons.
s sublevel A sublevel that contains one orbital and can hold up to two
electrons; the s sublevel is present in every energy level.
p sublevel A sublevel that contains three orbitals and can hold up to six
electrons; the p sublevel is present in energy levels 2 and higher.
d sublevel A sublevel that contains five orbitals and can hold up to 10
electrons; the d sublevel is present in energy levels 3 and higher.
f sublevel A sublevel that contains seven orbitals and can hold up to 14
electrons; the f sublevel is present in energy levels 4 and higher.

4.4 Describing Electron Configurations

electron configuration The number of electrons in each energy level and
sublevel.
valence level The highest-occupied electron energy level in an atom.
octet rule The principle that an atom is stabilized by having its highest-
occupied (valence) energy level filled.
isoelectronic Describes atoms or ions that have the same electron
configuration.

261
 Additional Problems

4.1 The Electromagnetic Spectrum

13. Identify which region of the electromagnetic spectrum has higher energy in
each of the following pairs:
a. infrared or ultraviolet
b. red light or green light
c. microwave or radio

14. Identify which region of the electromagnetic spectrum has higher energy in
each of the following pairs:
a. green light or microwave
b. gamma rays or ultraviolet
c. X-rays or blue light

15. In each pair, identify the member that has lower energy:
a. a high-frequency wave or a low-frequency wave
b. a photon of yellow light or a photon of infrared light

16. In each pair, identify the member that has lower energy:
a. a wave with a longer wavelength or one with a shorter wavelength
b. a photon of blue light or a photon of ultraviolet light

17. Which of the following has a lower frequency than green light?
a. red light
b. light with a wavelength of 400 nm
c. light with a wavelength of 1.0 × 10–6 m

18. Which of the following has higher energy than blue light?
a. ultraviolet light
b. light with a frequency higher than blue light
c. light with a wavelength longer than blue light

19. In each pair, which one has the longest wavelength?

a. a radio wave or a microwave

262
 b. a microwave or visible light
c. infrared light or ultraviolet light

20. In each pair, which one has the longest wavelength?

a. visible light or gamma rays
b. microwaves or X-rays
c. visible light or infrared light

21. Determine the wavelength of each wave. Assume each wave is traveling at
3.00 × 108 m/s.
a. a wave with a frequency of 750,000 s–1
b. a wave with a frequency of 20 Hz
c. a wave with a frequency of 2.5 × 105 Hz

22. Find the frequency for each wave. Assume each wave is traveling at 3.0 ×
108 m/s.
a. a photon of green light with a wavelength of 5.5 × 10–7 m
b. a photon of red light with a wavelength of 700 nm
c. a photon of light with a wavelength of 5.2 cm

23. The highest intensity of photons from the Sun have a wavelength of about
5.00 × 10–7 m. What color is this light? Refer to Figure 4.5.

24. While traveling through an intersection, you encounter a traffic light with a
frequency of 4.30 × 1014 Hz. What color is this light? Should you stop,
keep going through the intersection, or proceed with caution? Refer to
Figure 4.5.

25. A beam of light has a frequency of 4.17 × 1014 Hz. What is the energy per
photon of this light?

26. A beam of light has a frequency of 6.29 × 1014 Hz. What is the energy per
photon of this light?

27. A medical X-ray machine has a wavelength of 2.10 × 10–11 m. What is the
frequency (in Hz) and energy (in J) of this wave?

263
28. A microwave has a wavelength of 1.20 cm. What is the frequency (in Hz)
and energy (in J) of this wave?

29. Carbon dioxide readily absorbs radiation with an energy of 4.67 × 10–20 J.
What is the wavelength and frequency of this radiation? Does this
radiation fall in the ultraviolet, visible, or infrared range?

30. Determine the wavelength and frequency for each of the following:
a. a photon of green light with an energy of 3.32 × 10–19 J
b. a photon of ultraviolet radiation with an energy of 3.49 × 10–18 J

4.2 Color, Line Spectra, and the Bohr Model

31. What are line spectra? How are line spectra observed?

32. How did the Bohr model describe electron motion? How did this model
explain line spectra?

33. When an atom absorbs energy, what happens to its electrons?

34. When an atom releases light energy, what happens to its electrons?

35. When an electric current passes through a tube filled with neon, the tube
begins to glow. How are electrons involved in producing this light?

36. Molten iron glows with a reddish-orange color. How is heat energy
converted into light energy in this process?

37. For each of these processes, indicate whether light is absorbed or emitted:
a. An electron moves from level 2 to level 4.
b. An electron moves from level 6 to level 3.
c. An electron jumps from level 3 to 6.
d. An electron relaxes from level 7 to level 2.

38. For each of these processes, indicate whether light is absorbed or emitted:
a. An electron moves from level 1 to level 5.
b. An electron moves from level 4 to level 2.
c. An electron is excited to a higher energy level.
d. An electron relaxes to a lower energy level.

264
39. A hydrogen atom absorbs a photon of ultraviolet light, exciting an electron
from energy level 1 to energy level 5. The excited electron drops back
down in three steps, as shown. Using the table below, identify the energy
released in each transition as ultraviolet, visible, or infrared. The first
transition has been done as an example.

Transition Wavelength Region

5→4 4,057 nm infrared
4→2 487 nm
2→1 122 nm

40. A hydrogen atom absorbs a photon of ultraviolet light, exciting an electron

from energy level 1 to energy level 7. The excited electron drops back
down in three steps, as shown. Using the table below, identify the energy
released in each transition as ultraviolet, visible, or infrared.

265
 Transition Wavelength Region
7→5 4,668 nm infrared
5→2 434 nm
2→1 122 nm

4.3 The Quantum Model and Electron Orbitals

41. Why do we describe electrons in terms of orbitals or energy levels instead

of describing the exact location of the electron?

42. How is an orbital in the quantum model different from the orbits described
in the Bohr model?

43. How many electrons can occupy a single orbital?

44. When an orbital contains two electrons, how do the electron spins align?

45. For each of these energy levels, list all sublevels that are present:
a. level 1
b. level 2
c. level 3
d. level 4

46. In total, how many electrons can fit in energy level 4?

266
47. How many electrons can fit in a p sublevel?

48. How many electrons can fit in a d sublevel?

49. What is the difference between an orbital and a sublevel?

50. Sketch the shape of an s orbital and a p orbital.

51. How many electrons can occupy an atom’s lowest energy level? How
many can occupy its second-lowest energy level?

52. Complete this table to show the number of orbitals and maximum number
of electrons in each energy sublevel.

Sublevel Number of Orbitals Number of Electrons

s 2

5 10

53. Indicate whether each of these level/sublevel combinations are possible:

a. level 1, sublevel s
b. level 4, sublevel s
c. level 1, sublevel p

54. Indicate whether each of these level/sublevel combinations are possible:

a. level 2, sublevel d
b. level 3, sublevel f
c. level 3, sublevel d

4.4 Describing Electron Configurations

55. Complete the orbital energy diagrams, using arrows to represent the
electrons. The first one (nitrogen) is filled out as an example.

267
56. Complete the orbital energy diagrams, using arrows to represent the
electrons:

57. Write electron configurations for each of these atoms:

a. H
b. Be
c. C
d. Ne

58. Write electron configurations for each of these atoms:

a. N
b. Co
c. Ru
d. I

59. What is the valence level? Which two sublevels are involved with an
atom’s valence?

60. Why is energy level 3 considered “filled” with only eight electrons?

268
61. What is the octet rule? What family of elements illustrates the octet rule?

62. If an atom fulfills the octet rule, which sublevels are filled?

63. Rewrite the following electron configurations using noble gas shorthand:
a. 1s22s1
b. 1s22s22p63s2
c. 1s22s22p63s23p5

64. Rewrite the following electron configurations using noble gas shorthand:
a. 1s22s22p3
b. 1s22s22p63s23p64s23d104p65s24d105p66s1
c. 1s22s22p63s23p64s23d104p3

65. Use the noble gas shorthand to write electron configurations for the
following halogens: F, Cl, Br, and I.

66. Use the noble gas shorthand to write electron configurations for the
following alkaline earth metals: Be, Mg, Ca, and Sr.

67. Write the electron configuration for magnesium. How many inner-shell
and valence electrons does magnesium have?

68. Write the electron configuration for phosphorus. How many inner-shell
and valence electrons does phosphorus have?

69. Identify the inner electrons, outer electrons, and valence electrons in each
electron configuration:
a. nitrogen: 1s22s22p3
b. potassium: 1s22s22p63s23p64s1
c. gallium: [Ar]4s23d104p1
d. tungsten: [Xe]6s24f143d4

70. Identify the inner electrons, outer electrons, and valence electrons in each
electron configuration:
a. lithium: 1s22s1

269
 b. chlorine: 1s22s22p63s23p5
c. promethium: [Xe] 6s24f5
d. iodine: [Kr]5s24d105p5

71. Write electron configurations for each of these ions:

a. O2–
b. K+
c. Br–
d. N3–

72. Write electron configurations for each of these ions:

a. Na+
b. Mg2+
c. F–
d. S2–

73. Each of these atoms can gain or lose electrons to become isoelectronic
with neon. Write the number of electrons each atom would need to gain or
lose, and indicate the charge that would result:
a. O
b. F
c. Na
d. Mg

74. Each of these atoms can gain or lose electrons to become isoelectronic
with argon. Write the number of electrons each atom would need to gain or
lose, and indicate the charge that would result:
a. S
b. Cl
c. K
d. Ca

75. List three ions that are isoelectronic with helium.

76. List three ions that are isoelectronic with krypton.

270
77. Fluorine tends to gain one electron while sodium metal tends to lose one
electron. Why is this so? What rule can be used to describe this behavior?

78. Potassium metal tends to lose one electron while calcium metal tends to
lose two electrons. Why is this so? What rule can be used to describe this
behavior?

79. Based on the octet rule, which of these atoms or ions have the stability of a
filled valence level?
a. K
b. K+
c. S–
d. S2–

80. Based on the octet rule, which of these atoms or ions have the stability of a
filled valence level?
a. Ca2+
b. Al
c. Xe
d. I–

4.5 Electron Configuration and the Periodic Table

81. Why do we sometimes refer to the two left-hand columns of the periodic
table as the s block?

82. Why do we sometimes refer to the transition metals as the d-block

elements?

83. Refer to the periodic table to determine which of these atoms have their
valence electrons in energy level 3. (Note: You do not need to write the
electron configurations to answer this question).
a. Li
b. Si
c. P
d. Se
e. He
f. Ar

271
84. Refer to the periodic table to determine which of these atoms have their
valence electrons in energy level 4. (Note: You do not need to write the
electron configurations to answer this question).
a. K
b. Na
c. Y
d. Mn
e. Cl
f. Br

85. What is the inner electron configuration of elements in row 2?

86. What is the inner electron configuration of elements in row 4?

87. The elements in group 2A have how many valence electrons?

88. The elements in group 6A have how many valence electrons?

89. Without writing electron configurations, identify the number of valence

electrons in each of these main-group elements:
a. calcium
b. aluminum
c. bromine
d. neon

90. Without writing electron configurations, identify the number of valence

electrons in each of these main-group elements:
a. carbon
b. selenium
c. xenon
d. rubidium

91. Refer to the periodic table to identify the number of valence electrons
(outer-shell s and p electrons) present in each of these atoms:
a. Mg
b. N
c. P
d. As

272
 e. Cl
f. Ca

92. Refer to the periodic table to identify the number of valence electrons
(outer-shell s and p electrons) present in each of these atoms:
a. Be
b. Sr
c. Ne
d. Br
e. As
f. Si

93. Using the periodic table as a reference, answer these questions:

a. What atom has an electron configuration of [Ar]4s1?
b. What atom has an electron configuration of [He]2s22p3?
c. What atom has an electron configuration of [Kr]5s24d8?

94. Using the periodic table as a reference, answer these questions:

a. What atom has an electron configuration of [He]2s1?
b. What atom has an electron configuration of [Ne]2s22p5?
c. What atom has an electron configuration of [Ar]4s23d6?

95. Using the periodic table as a reference, answer these questions:

a. What subshell is filled after 4s?
b. What subshell is filled after 4d?
c. What subshell is filled after 3p?

96. Using the periodic table as a reference, answer these questions:

a. What subshell is filled after 3s?
b. What subshell is filled after 3d?
c. What subshell is filled after 6p?

97. Using the periodic table, find the electron configuration of the highest-
filled sublevel for each of these elements. Try to do this without writing
the full electron configuration. The first answer is given as an example.
a. germanium (answer: 4p2)
b. tellurium

273
 c. technetium
d. boron

98. Using the periodic table, find the electron configuration of the highest-
filled sublevel for each of these elements. Try to do this without writing
the full electron configuration. The first answer is given as an example.
a. iron (answer: 3d6)
b. astatine
c. gold
d. barium

99. Identify the group of elements with these characteristics:

a. This group has a valence electron configuration of s2.
b. This group of elements does not react to form compounds.
c. This group of elements has two valence electrons.

100. Identify the group of elements with these characteristics:

a. This group of elements has a valence electron configuration of s2p5.
b. This group of elements has a completely filled valence.
c. This group of elements has one valence electron.

101. Oxygen, sulfur, and selenium belong to a family of elements that is

sometimes called the chalcogens. What is the valence electron
configuration of this group?

102. What is the valence electron configuration of the group of elements that
includes boron, aluminum, and gallium?

103. What is the outermost electron configuration for the elements in group 12
of the periodic table?

104. What is the outermost electron configuration for the elements in group 6A
of the periodic table?

274
Chapter Five
Chemical Bonds and Compounds

An Unexpected Combination: Lithium Carbonate and

Bipolar Disorder
Sometimes, combinations produce surprising results. Consider the story of lithium
carbonate–a simple compound used worldwide to treat bipolar disorder–and the
combination of bad luck, good luck, and keen observation that led to its discovery.
In February 1942, during the heart of World War II, Japanese forces attacked
the Allied stronghold at Singapore. After a week of fighting, the Allies surrendered
(Figure 5.1). Among those captured was John Cade, a young psychiatrist serving
in the Australian Army Medical Corps. He spent the next three years in a prisoner-
of-war camp.

275
Figure 5.1 (a) Allied soldiers captured by the Japanese, February 1942. (b) John Cade
discovered that lithium carbonate could treat bipolar disorder. (c) Mogens Schou carried
on Cade’s work, extensively studying and promoting the effects of lithium carbonate.
(d) Lithium carbonate is made from lithium, carbon, and oxygen atoms. (e) Today
millions of people take lithium carbonate for the treatment of bipolar disorder.

While imprisoned, Cade observed a number of prisoners who suffered from

bipolar disorder: They fluctuated between wildly aggressive behavior (called the
manic phase) and deep depression. Cade began to suspect that a toxic chemical
caused the prisoners’ erratic behavior and that their moods stabilized after the toxin
was expelled through their urine.
After his release at the end of the war, Cade returned to Australia and resumed
his career in psychiatry. On the side, he began exploring the ideas he had
developed in captivity. He collected urine samples from bipolar patients and
injected the urine into guinea pigs. Interestingly, the guinea pigs treated with urine
from bipolar patients died faster than those treated with urine from healthy people.
Cade delved deeper. He suspected that a compound called uric acid might be the
mysterious toxin. He began to study the effects of pure uric acid and related
compounds on the guinea pigs. He found that one such compound, lithium urate,
reduced the toxic effects of the other compounds present. Intrigued by this result,
he decided to test a simpler lithium-containing compound: lithium carbonate.
When he injected guinea pigs with pure lithium carbonate, the animals became
sedate.
Ultimately Cade’s ideas about toxins in the urine were discarded, but the
effects he observed from lithium carbonate opened a new door. Cade wondered if
lithium carbonate would also sedate patients suffering from the manic phase of
bipolar disorder. To see if it was safe, he first tested it on himself. Finding no long-
term effects, he treated the manic patients in his ward with lithium carbonate. This
human testing was remarkably successful, and in 1949 he published his results. In
the decades that followed, another psychiatrist, Mogens Schou, extensively studied
the effects of lithium carbonate. Today, lithium carbonate remains one of the most
common and least expensive treatments for bipolar disorder.
Like most science—and most other human activities—this story is messy.
Cade’s ideas were conceived in the harshest of circumstances. His initial ideas
were incorrect. And by today’s standards, Cade’s experiments seem reckless. But
his careful observations, both in the prison camp and the laboratory, led him to
insights that changed the way we treat mental illness. Out of all the messy pieces, a
beautiful discovery emerged.
In this chapter, we’ll begin to study how chemical bonds bring atoms together
to form compounds. Just as a complete story can be much different from the pieces
that comprise it, compounds behave much differently from the elements they are
composed of. As atoms combine to form compounds, new and intriguing
properties emerge–properties that create new materials, new medicines, and new
opportunities.

276
 Intended Learning Outcomes
After completing this chapter and working the practice problems, you should be
able to:

5.1 Lewis Symbols and the Octet Rule

Use the periodic table to identify the number of valence electrons in an atom.
Represent valence electrons using Lewis dot symbols.

5.2 Ions
Describe and predict the formation of main-group ions using the octet rule.
Identify common monatomic and polyatomic ions by name, symbol or
formula, and charge.

5.3 Ionic Bonds and Compounds

Predict ionic formulas based on the charges of the cation and anion.
Broadly describe the arrangement of ions in an ionic solid.
Convert between the name and formula for an ionic compound.

5.4 Covalent Bonding

Describe how nonmetals fulfill the octet rule through covalent bonds.
Describe how covalent compounds lead to the formation of molecules.
Describe the difference between a molecular formula and an empirical
formula.
Name binary covalent compounds.

5.5 Distinguishing Ionic and Covalent Compounds

Distinguish ionic and covalent compounds based on their molecular
formulas.

5.6 Aqueous Solutions: How Ionic and Covalent Compounds Differ

Contrast the behavior of ionic compounds and covalent compounds in
aqueous solutions.

5.7 Acids–An Introduction

Describe the ionization of acids in aqueous solution.
Name binary acids and oxyacids.

277
5.1 Lewis Symbols and the Octet Rule
In Chapter 4, we saw that families of atoms such as the alkali metals, the
halogens, and the noble gases exhibit similar behaviors because they have
similar electronic configurations. In this chapter, we will explore how
these electronic configurations lead to the formation of chemical bonds.
Chemical bonding involves changes in an atom’s outer or valence
electrons. Recall that valence electrons are the electrons in the highest-
occupied energy level of an atom. Because of the sublevel filling
sequence, the valence level involves only the s and p sublevels. Since two
electrons can fit in an s sublevel, and six electrons can fit in the p sublevel,
up to eight electrons can occupy the valence level. For main-group
elements, we can quickly determine the number of valence electrons from
the periodic table: The column (group) number for the main groups is also
the number of valence electrons (Figure 5.2). For example, nitrogen (N) is
in group 5A, so it has five valence electrons. Neon (Ne) is in group 8A, so
it has eight valence electrons.

Figure 5.2 The main-group numbers (1A–8A) also indicate the number of electrons in
each atom’s valence level.

The valence level holds up to eight electrons.

To visualize chemical bonding, it is often helpful to draw Lewis dot

symbols. These symbols represent the number of valence electrons in an
atom as dots drawn around the atomic symbol. Here are the Lewis symbols
for each of the row 2 elements:

In Chapter 4 we introduced the octet rule, which states that an atom is

stabilized by having its valence energy level filled. For elements in row 2

278
and below, eight electrons are required to fill the valence level. The octet
rule explains why the noble gases are so stable, and it also allows us to
predict how main-group elements form chemical bonds. These elements
fulfill the octet rule by gaining or losing electrons to form ions, or by
sharing electrons between two atoms. We will explore these behaviors in
the sections that follow.
Main-group elements can fulfill the octet rule by gaining, losing, or sharing
electrons.

279
5.2 Ions

Cations: Ions with a Positive Charge

Main-group metals fulfill the octet rule by losing electrons to form
positively charged ions, called cations (pronounced cat-eye-uns). For
example, consider sodium metal: Sodium has an electron configuration of
1s22s22p63s1. Because 1s22s22p6 is the same configuration as neon, we
often write it as [Ne]3s1. To fill its valence level (level 3), sodium would
have to gain seven electrons—an unlikely occurrence. However, by losing
just one electron, sodium becomes electronically identical to neon—a very
stable arrangement that fulfills the octet rule. As a result, sodium easily
loses one electron to form Na+, a common ion (Figure 5.3). All of the
alkali metals (metals in group 1A of the periodic table) lose one electron to
form +1 ions.

Figure 5.3 Sodium has one valence electron. By losing its outermost electron, sodium
becomes electronically identical to the noble gas neon and fulfills the octet rule.

Alkali metals form +1 ions.

Magnesium has an electron configuration of [Ne]3s2. Just as sodium

lost one electron to reach the [Ne] electron configuration, magnesium can
reach this electron configuration by losing two electrons (Figure 5.4).
Because it loses two electrons, it has a charge of +2, which is written as
Mg2+. Each of the alkaline earth metals (group 2A) loses two electrons to
form +2 ions (Figure 5.5).

Na+, Mg2+, and Ne are all isoelectronic–meaning they have the same electron

280
 configuration.

Alkaline earth metals form +2 ions.

Figure 5.4 Sodium and magnesium both lose their valence electrons to become
isoelectronic with neon.

Figure 5.5 The group 1A elements (hydrogen and the alkali metals) form +1 ions.
Group 2A elements (the alkaline earth metals) form +2 ions.

Transition metal ions may have multiple charges.

The transition metals (elements in the d block of the periodic table)

also tend to lose electrons to form positively charged ions. But unlike
main-group metals, transition metal ions do not follow a simple pattern.
Transition metals typically form ions having a charge between +1 and +4,
and some transition metals form multiple charged ions. Metals in the lower
part of the p block also behave this way (Figure 5.6).

281
Figure 5.6 Many common transition and p-block metals form ions with multiple
charges.

Naming Cations
In general, metal cations are given the same name as the neutral metal. For
example, the cation produced from sodium metal is simply called the
sodium ion.
As mentioned earlier, some transition metals can have more than one
charge. For example, iron commonly forms both +2 and +3 ions.
Historically, these two ions were named as ferrous and ferric ions,
respectively. Similarly, copper commonly forms both +1 (cuprous) and +2
(cupric) ions. Although you will encounter these names occasionally, the
modern style of naming these ions puts the charge in Roman numerals in
parentheses immediately after the atom name. For example, the ferrous ion
(Fe2+) is named as iron(II), which is read as “iron-two”; the ferric ion
(Fe3+) is named as iron(III), read as “iron-three” (Table 5.1).

TABLE 5.1 Naming Ions with More than One Charge

Atom Ion Older Name Modern Name

Iron Fe2+ Ferrous Iron(II)

Fe3+ Ferric Iron(III)

Copper Cu+ Cuprous Copper(I)

Cu2+ Cupric Copper(II)

If an atom can form more than one cation, use Roman numerals after the
atom name to specify the charge.

282
 Example 5.1 Naming Cations
Name each of the following ions: Ag+, Pb2+, and Pb4+
From Figure 5.6, we see that silver (Ag) forms only one ion. Therefore, we
refer to Ag+ as a silver ion. However, lead (Pb) forms two different ions.
To distinguish them, we refer to Pb2+ as a lead(II) ion and Pb4+ as a
lead(IV) ion. •

IT
TRY

1. Provide names for each of these cations:

Ca2+ Cr2+ Cr3+ Al3+

Anions: Ions with a Negative Charge

The nonmetals lie on the right-hand side of the periodic table. Unlike
metals, the valence shells of most nonmetals are nearly full. To fulfill the
octet rule, most nonmetals gain electrons to form negatively charged ions,
called anions (pronounced an-eye-uns).
For example, fluorine has an electron configuration of 1s22s22p5. By
gaining one electron, fluorine can achieve an electron configuration of
1s22s22p6, the same electron configuration as neon. This configuration
fulfills the octet rule and provides tremendous stability. As a result,
fluorine tends to aggressively “grab” an electron, forming a very stable ion
with a charge of –1 (Figure 5.7). The other halogens (chlorine, bromine,
iodine) also form –1 ions.

283
Figure 5.7 Fluorine and oxygen both gain electrons to become isoelectronic with neon.

Halogens form –1 ions.

Oxygen, sulfur, and the atoms below them on the periodic table
comprise a family called the chalcogens (group 6A). Each of these
elements is two electrons short of a noble gas configuration. For example,
oxygen has an electron configuration of 1s22s22p4. It needs two electrons
to fill its outer valence level and so it tends to gain two electrons, resulting
in a charge of –2. Sulfur also forms a stable ion with a charge of –2.

Chalcogens form –2 ions.

What about group 5A elements, such as nitrogen and phosphorus?

Consistent with the pattern just described, these atoms gain three electrons
to fill their valence level, and so form ions with a charge of –3 (Figure
5.8).

Figure 5.8 The halogens form –1 ions. Oxygen and sulfur form –2 ions. Nitrogen and
phosphorus form –3 ions. The noble gases (shaded violet) have complete valence shells
and do not form ions.

284
Nonmetals gain electrons to form anions.

Naming Anions
When an atom gains electrons, we name the resulting anion by changing
the end of the atom name to –ide. For example, chlorine atoms form
chloride ions, oxygen atoms form oxide ions, and sulfur atoms form
sulfide ions. A list of common anions is given in Table 5.2.

TABLE 5.2 Common Anions

Atom Anion Symbol Anion Name

Nitrogen N3– Nitride

Phosphorus P3– Phosphide

Oxygen O2– Oxide

Sulfur S2– Sulfide

Fluorine F– Fluoride

Chlorine Cl– Chloride

Bromine Br– Bromide

Iodine I– Iodide

Example 5.2 Naming Ions and Predicting Charges

Predict the ions that would be formed from an atom of calcium and from an
atom of sulfur. Name each ion.
Calcium belongs to the alkaline earth metal family. It has an electron
configuration of [Ar]4s2. Calcium loses its two valence electrons, resulting
in a charge of +2. Cations are given the same name as the parent atom, so
we refer to Ca2+ as the calcium ion.
Sulfur is a nonmetal with an electron configuration of [Ne]3s23p4. To fill
its valence shell, sulfur gains two electrons, giving the ion a charge of –2.
We refer to S2– as the sulfide ion. •

Practice

285
 Monatomic Ions
How quickly can you recognize ion charges? Try this interactive quiz to test
your skills.

IT
TRY

2. Use the periodic table to predict whether each atom would gain or lose
electrons, and write the charge on the ion formed:

Cl Br O Be K

Sports drinks contain ions that are commonly lost during exercise. They include
sodium, potassium, chloride, and phosphate.

Polyatomic Ions
Polyatomic ions are groups of atoms that have an overall charge. Many of
these ions, such as acetate and phosphate, are essential to life and common
in many different materials and applications. Formulas and names for the
most common polyatomic ions are given in Table 5.3. Notice that this
table contains only one common polyatomic cation (ammonium). All
others are anions.

TABLE 5.3 Common Polyatomic Ions

Formula Name

NH4+ Ammonium

NO3– Nitrate

286
CO32– Carbonate

HCO3– Bicarbonate (also called hydrogen carbonate)

NO2– Nitrite

PO43– Phosphate

HPO42– Hydrogen phosphate

C2H3O2– Acetate

OH– Hydroxide

CN– Cyanide

O22– Peroxide

SO42– Sulfate

SO32– Sulfite

HSO4– Bisulfate (also called hydrogen sulfate)

ClO4– Perchlorate

ClO3– Chlorate

ClO2– Chlorite

ClO– Hypochlorite

CrO42– Chromate

Cr2O72– Dichromate

MnO4– Permanganate

Naming Polyatomic Ions

Although Table 5.3 contains many ion names, there are patterns that will
help you keep these names organized. Notice that most of the polyatomic
ions contain oxygen–these are called oxyanions. We name oxyanions by
adding the suffix –ate to the root of the element. For example, the
oxyanion from carbon (CO32–) is carbonate, and the oxyanion formed
from phosphorus (PO43–) is phosphate.

“–ate is great, and –ite is lite”

More oxygen atoms: –ate
Fewer oxygen atoms: –ite

287
 Some elements form more than one oxyanion. In these cases we use
the suffix –ate to indicate the ion with more oxygen atoms present, and the
suffix –ite to indicate the ion with fewer oxygen atoms present. For
example, there are two common nitrogen oxyanions:

NO3− nitrateNO2− nitrite

Chlorine forms four oxyanions. In this case, we use the prefix per–
(meaning “more than”) to indicate the largest number of oxygen atoms,
and the prefix hypo– (meaning “below”) to indicate the least number of
oxygen atoms:
CIO4− perchlorate¯CIO3− chlorate¯ClO2− chlorite¯ClO− hypochlorite¯

A Summary of the Common Ions

As you continue studying chemistry, you will find it essential to know the
structure, formula, and charge of common monatomic and polyatomic
ions. Figure 5.9 summarizes the most common ions. You should be very
familiar with these ions, because you will use them regularly throughout
this course.

288
Figure 5.9 It is important to know the names, formulas, and charges for these common
ions.

Practice
Common Ions
Try this interactive to practice identifying important common ions.

Example 5.3 Gathering Information from Ion Names

The four ions named below are less common and are not listed in Table 5.3.
Which of these are polyatomic? Identify each one as a cation or an anion.

289
 a. bromate
b. bromite
c. palladium(II)
d. selenide
From the suffixes –ate and –ite, we know that both bromate and bromite are
oxyanions of the element bromine. Further, we know that bromate contains
more oxygen atoms than bromite. The actual formula for bromate is BrO3–,
and the formula for bromite is BrO2–.
Palladium is a transition metal, so it forms a cation. The (II) indicates that
this ion is Pd2+.
Finally, the ending –ide indicates that selenide is a monatomic anion
formed from the element selenium. Selenium lies just below sulfur on the
periodic table, so we predict the charge of this ion to be –2. •

IT
TRY

3. Write the symbol and the charge for each ion listed. Refer to the
periodic table as needed.

calcium nitrate scandium(III) telluride

4. Name each of these ions:

CS+ Fe2+ SO42− As3−

290
5.3 Ionic Bonds and Compounds

Ionic Bonds and Ionic Lattices

Opposite charges attract each other. When positive and negative ions come
near each other, they stick tightly together. The force of attraction between
oppositely charged ions is called an ionic bond. A compound composed of
oppositely charged ions is an ionic compound. Because metals form
cations and nonmetals form anions, the compounds formed between metals
and nonmetals are ionic compounds.

Metal cations and nonmetal anions form ionic bonds.

Ionic compounds contain many cations and anions, joined through

ionic bonds. To understand how these ions fit together, let’s consider the
structure of a compound composed of sodium cations (Na+) and chloride
anions (Cl–). A single Na+ ion and Cl– ion adhere to each other in an ionic
bond. But what happens if additional ions are present? They pack together
in a structure of alternating positive and negative charges that stretch out in
three dimensions (Figure 5.10). This array of positive and negative ions is
called an ionic lattice.

Figure 5.10 (a) Ions pack together in a framework of alternating positive and negative
charges. (b) This packing results in a three-dimensional framework called an ionic
lattice.

291
 To represent the composition of compounds like this one, we use a
chemical formula that indicates the type and amount of each element
present. We represent ionic compounds using a specific type of chemical
formula, called an empirical formula. An empirical formula gives the
smallest whole-number ratio of atoms in a compound. Subscripts written
after each atom indicate the number of that atom present. If no subscript is
written, we understand the number of atoms to be one. In this instance, the
formula is written simply as NaCl.

An empirical formula gives the smallest whole-number ratio of atoms in a

compound.

The empirical formula gives the smallest number of ions necessary to

form a compound. This number of ions is called the formula unit. For
example, the empirical formula for sodium chloride is NaCl; a formula
unit of sodium chloride contains one sodium ion and one chloride ion.
When writing empirical formulas for ionic compounds, we write the
symbol or formula for the cation, followed by the anion. In the next
section, we’ll look at several more examples of empirical formulas and
formula units.

Predicting Formulas for Ionic Compounds

Some ionic compounds contain cations and anions with different charges.
For example, consider the solid composed of potassium and sulfide ions.
The potassium cation has a charge of +1 while the sulfide anion has a
charge of –2. To form a neutral solid, the positive charges must equal the
negative charges. To balance the charges, the solid must contain two
potassium ions for every one sulfide ion:

In an ionic compound, the total charge must equal zero.

We therefore write the empirical formula for this compound as K2S. Put
another way, a formula unit of potassium sulfide contains two potassium
ions and one sulfide ion.
We can also predict the formulas for ionic solids containing
polyatomic ions. For example, consider the ionic compound produced

292
from calcium (Ca2+) and nitrate (NO3–) ions: For the positive and negative
charges to balance, there must be two nitrate ions for every one calcium
ion (Figure 5.11). We could write this formula as CaN2O6. But it is better
to write the formula as Ca(NO3)2 because this shows that two nitrates are
attached, rather than some other arrangement of nitrogen and oxygen. If
we have more than one polyatomic ion in the formula, we write that ion
inside parentheses to show that the entire unit is repeating.

Figure 5.11 To balance the charges, this compound requires two nitrate ions for every
one calcium ion.

Practice
Balancing Charges
To write an ionic compound formula correctly, you must balance the
charges on the ions. Try this interactive to practice this skill.

Example 5.4 Writing Formulas for Ionic Compounds

A compound is composed of two ions, aluminum and sulfate. What is the
formula for this compound?
We know the aluminum ion has a charge of +3, and the sulfate ion has a
charge of –2 (see Figure 5.9). For the charges of these ions to balance, we
must have two aluminum ions for every three sulfate ions, as shown in
Figure 5.12. Therefore, we write the formula for this compound as
Al2(SO4)3. As before, we put the repeating polyatomic ion in parentheses.
•

293
Figure 5.12 Three sulfate ions are required to balance the charge on two aluminum
ions.

Example 5.5 Writing Formulas for Ionic Compounds

What is the formula for a compound composed of iron(III) and bromide
ions?
Recall that iron is a transition metal, and it can have more than one possible
charge. The name iron(III) indicates that this ion is Fe3+. The bromide ion
is Br–. For the charges to balance, there must be three bromide ions for
every one iron(III) ion. Therefore, we write this formula as FeBr3. •

IT
TRY

5. Predict the empirical formulas for compounds formed from these ions:
a. magnesium and chloride
b. potassium and phosphate
c. lead(II) and oxide
d. ammonium and carbonate

Naming Ionic Compounds

To name an ionic compound, we give the cation name followed by the

294
anion name. For example, NaCl is sodium chloride, and MgCl2 is
magnesium chloride. Because we know that a magnesium ion has a +2
charge and a chloride ion has a –1 charge, there is no need to indicate the
ratio of cations to anions. Given the name of the cation and anion present,
we can determine the empirical formula.
For transition metals with more than one possible charge, it is
important to include the charge of the ion in parentheses with the name.
For example, copper and chloride ions form two different compounds,
CuCl and CuCl2. In CuCl, the copper ion must have a charge of +1 to
balance the charge from the chloride ion. In CuCl2, the copper ion must
have a charge of +2. Therefore, we name these compounds as follows:
CuCl copper(I) chlorideCuCl2 copper(II) chloride

Compounds containing polyatomic ions are named in the same way as

those containing monatomic ions. For example, the ionic compound
MgSO4 consists of a monatomic cation (magnesium) and a polyatomic
anion (sulfate). Therefore, the name of this compound is magnesium
sulfate.
Table 5.4 summarizes the names, formulas, and uses of several
common ionic compounds. It is important to be able to convert between
the name and empirical formulas for ionic compounds. This process is
further illustrated in the examples that follow.

TABLE 5.4 Common Ionic Compounds

Compound Formula Application

Sodium chloride NaCl Table salt

Sodium fluoride NaF Fluoride treatment

Sodium bicarbonate NaHCO3 Baking soda

Calcium oxide CaO Cement mix

Lithium carbonate Li2CO3 Treatment of bipolar disorder

Ammonium nitrate NH4NO3 Fertilizer

Example 5.6 Naming Ionic Compounds

295
 Name the following compound: Fe(NO2)2.
The keys to solving this problem are to identify the ions present and to
know their charges. The anion in this formula is nitrite, which has a charge
of –1. Because two NO2– ions are present, the charge on the iron (Fe)
cation must be +2. Fe2+ is named as iron(II), and so the total compound is
iron(II) nitrite. •

Example 5.7 Writing the Formula for an Ionic Compound

Write the empirical formula for ammonium sulfide.
We know that ammonium is NH4+ and that sulfide is S2–. For the charges
to balance, there must be two ammonium ions for each sulfide ion. To show
this, we put the ammonium formula in parentheses with a two on the
outside. Listing the cation first, we write the formula for this compound as
(NH4)2S. •

IT
TRY

6. Name each of these compounds:

a. RbCl
b. CuBr2
c. ZnCO3
d. K2SO4

7. Write the empirical formula for each compound named:

a. zinc sulfide
b. iron(III) oxide
c. ammonium phosphate

8. Titanium is a transition metal that can have multiple ionic charges.

The titanium compound TiO2 is commonly used as an additive in
paints. In this compound, what is the charge on the cation? What is the
name of this compound?

296
297
5.4 Covalent Bonding

Nonmetal–Nonmetal Bonds
In the previous section, we saw how ionic bonds form between metal
cations and nonmetal anions. When two nonmetal atoms come together, a
different type of bond occurs, called a covalent bond. In a covalent bond,
two electrons are shared between two atoms.
Covalent bonds form between nonmetal atoms.

For example, consider the bond that forms between two hydrogen
atoms. Each hydrogen atom has one proton and one electron. To form a
covalent bond, the two electrons “pair up” in the space between the two
nuclei (Figure 5.13).
The force of attraction between the nuclei and the two electrons holds
the atoms together. We represent these shared electrons by drawing a dash
between the symbols of the two atoms:

Figure 5.13 Two atoms form a covalent bond by sharing a pair of electrons.

Remember that the first energy level holds only two electrons. By
forming a covalent bond, each hydrogen atom completes its valence level.

Covalent compounds form molecules.

When two hydrogen atoms combine, they form a molecule. In earlier

chapters, we defined molecules as groups of atoms that bind together and
behave as a unit. The bonds that hold molecules together are covalent
bonds. In its elemental form, hydrogen is a gas composed entirely of these
two-atom molecules (Figure 5.14).

298
Figure 5.14 This balloon contains elemental hydrogen gas. The gas is composed of
molecules containing two atoms each.

As a second example, let’s look at the bonding that occurs between

two fluorine atoms. Each fluorine atom contains seven valence electrons,
and therefore needs only one more electron to complete its valence level.
Two fluorine atoms can form a single covalent bond. By forming this
bond, the atoms fill their valence shell with eight electrons and satisfy the
octet rule.

As in the hydrogen example, we use a dash to represent two shared

electrons. We call this type of drawing a Lewis structure. Lewis
structures depict the arrangement of valence electrons within a molecule or
polyatomic ion. We will explore Lewis structures further in Chapter 9,
when we take a more detailed look at the bonding and properties of
molecules.

A dash between two chemical symbols indicates a covalent bond.

Atoms sometimes share two or even three pairs of electrons in covalent bonds.

299
 We represent double covalent bonds using two dashes between the atoms, and
triple covalent bonds using three dashes. We will discuss covalent bonding in
more detail in Chapter 9.

Hydrogen and fluorine are two of seven elements that exist as diatomic
(“two-atom”) molecules in their elemental forms. The others are nitrogen,
oxygen, and the rest of the halogens (Figure 5.15).

Figure 5.15 Seven elements exist as diatomic molecules.

Covalent Compounds
Covalent compounds form when elements combine through covalent
bonds, forming discrete molecules. Water is an example of a covalent
compound. In a water molecule, an oxygen atom covalently bonds to two
hydrogen atoms:

The valence level of each hydrogen atom is filled with two electrons.
What about the oxygen atom? It has four unshared electrons and two
covalent bonds. Between the unshared and the shared electrons, the
oxygen atom has eight electrons in its valence level and fulfills the octet
rule.

300
 Covalent compounds fulfill the octet rule by sharing electrons.

To describe covalent compounds, we often use molecular formulas.

This type of chemical formula gives the actual number of atoms in the
molecule rather than the simplest whole-number ratio.
For example, consider hydrazine, a fuel used for rocket thrusters
(Figure 5.16). A hydrazine molecule contains two nitrogen atoms and four
hydrogen atoms. The empirical formula for this compound is the smallest
whole-number ratio, or NH2. However, chemists usually prefer to write
this compound using the molecular formula: N2H4.

Figure 5.16 We usually describe covalent molecules, such as the rocket fuel hydrazine,
by their molecular formula rather than their empirical formula.

Covalent bonds often lead to complex structures. Consider the

molecule octane, a component of gasoline (Figure 5.17): One molecule of
octane contains 25 different covalent bonds. Larger compounds may
contain hundreds or thousands of covalent bonds. Because of this complex
bonding, elements can often combine in many different ratios.

301
Figure 5.17 A molecule of octane is composed of hydrogen and carbon atoms, held
together by covalent bonds.

Each unique bonding arrangement produces a different compound. For

example, Table 5.5 lists several of the compounds that form between
phosphorus and oxygen.

TABLE 5.5 Covalent Compounds Containing Phosphorus and

Oxygen
Compound Name Formula

Phosphorus monoxide PO

Diphosphorus trioxide P2O3

Diphosphorus tetroxide P2O4

Tetraphosphorus decoxide P4O10

302
 Lithium carbonate is an ionic compound used to treat bipolar disorder. In
polyatomic ions like carbonate (CO32–), the atoms are held together with
covalent bonds.

Example 5.8 Interpreting Lewis Structures

The Lewis structure for a molecule of ammonia (NH3) is shown below. In
this structure, how many electrons does the nitrogen atom share through
covalent bonds? How many of the valence nitrogen electrons are not
shared? Does this nitrogen atom have a complete octet?

Each dash represents two shared electrons. We see from the structure that
nitrogen forms three covalent bonds to hydrogen. Because each dash
represents two electrons, we can say nitrogen has six shared electrons. The
two dots above the nitrogen represent nonbonded (unshared) electrons.
Combining the six shared and two unshared electrons, the nitrogen atom
has eight electrons in its valence shell—a complete octet. •

IT
TRY

9. The Lewis structure for the compound HCl is shown below. How
many bonded and nonbonded electrons are in the valence of the
chlorine atom? Does this atom fulfill the octet rule?

10. Consider the Lewis structure shown below. What is the molecular
formula for this compound? What is the empirical formula?

303
Naming Covalent Compounds
Covalent compounds containing only two elements are called binary
covalent compounds. These compounds are named in a manner that is
similar to ionic compounds. The element that is lower and farther to the
left on the periodic table is named first, and the full element name is used.
The element that is nearer to the upper right on the periodic table is named
as though it were an anion.

When naming covalent compounds, use prefixes to indicate how many

atoms are present.

However, there is one complicating factor: Because covalent

compounds can form in many different ratios, covalent compounds use a
series of prefixes (Table 5.6) to indicate the number of atoms present. A
prefix is assigned to both the first and second part of the name. If the
molecule contains only one atom of the first element, the prefix mono– is
not used.

TABLE 5.6 Prefixes for Naming Covalent Compounds

Atoms Prefix

1 mono–

2 di–

3 tri–
4 tetra–

5 penta–
6 hexa–

7 hepta–
8 octa–

9 nona–
10 deca–

Pent– or Penta–
If the root name of the atom begins with a vowel, we remove the –a from the
end of the prefix to make it easier to pronounce. For example, PCI5 is
phosphorus pentachloride, but P2O5 is diphosphorus pentoxide.

304
 For example, phosphorus and chlorine commonly form two
compounds that have the formulas PCl3 and PCl5. How do we name these
compounds? Phosphorus is to the left of chlorine on the periodic table
(Figure 5.18), so we name phosphorus first and then name chlorine as the
anion (chloride). Using the prefixes in Table 5.6, we refer to PCl3 as
phosphorus trichloride, and PCl5 as phosphorus pentachloride.

Figure 5.18 When naming a covalent compound, the atom that lies farthest left on the
periodic table comes first.

Example 5.9 Naming Covalent Compounds

Nitrogen and oxygen form two covalent compounds, NO2 and N2O4. Name
each of these compounds.
Because nitrogen is to the left of oxygen on the periodic table, we name
nitrogen first (nitrogen) and then oxygen as the anion (oxide). The first
compound, NO2, is called nitrogen dioxide. The second compound, N2O4,
is called dinitrogen tetroxide. Notice that we use the prefix on the first
name only if more than one atom is present. •

IT
TRY

11. Write the names of these covalent compounds:

N2O3 SO2 CF4 P4O9

305
306
5.5 Distinguishing Ionic and Covalent Compounds
Let’s briefly review what we’ve covered so far: To fulfill the octet rule,
atoms can either gain or lose electrons to form ions or they can share
electrons through covalent bonds.
Covalent compounds form between nonmetal atoms. These compounds
form distinct units called molecules. We generally describe covalent
compounds using molecular formulas that indicate the exact number of
each atom contained in one molecule.
Ionic compounds form between oppositely charged ions. We describe
an ionic compound by its formula unit or its empirical formula—that is,
the simplest whole-number ratio of cations to anions in the compound. We
avoid using the term molecule to describe these compounds, because an
ionic solid has no molecular unit.

Metal + Nonmetal: Ionic bond

Nonmetal + Nonmetal: Covalent bond

As we’ll see in the sections and chapters ahead, the differences in ionic
and covalent bonding lead to many unique physical properties (Figure
5.19). Because of this, it is important to be able to distinguish between
ionic and covalent compounds. The key to doing this is to identify the
elements present. Is the compound composed entirely of nonmetals? If so,
it is a covalent compound. Is it composed of a metal and a nonmetal? This
indicates that the compound is ionic. Does it contain any of the common
polyatomic ions described in Table 5.3? Again, this suggests it is ionic.
Example 5.10 illustrates how we can differentiate between covalent and
ionic compounds.

307
Figure 5.19 Limestone is composed of calcium carbonate, an ionic compound. Olive oil
is composed of covalent molecules containing carbon, hydrogen, and oxygen. The
properties of any compound are determined by the types of elements and bonds that are
present.

Example 5.10 Identifying and Naming Covalent and Ionic

Compounds
Identify each of these compounds as covalent or ionic. Provide an
appropriate name for each compound.
a. MgBr2
b. FeCl3
c. SF6
Ionic compounds form between a metal and a nonmetal. Covalent
compounds form between two nonmetals. To solve this problem, the first
step is to identify each element as a metal or a nonmetal; the next step is to
decide whether the compound is covalent or ionic. Note where these
elements fall on the periodic table:

In the first example, MgBr2, magnesium is a metal and bromine is a

nonmetal—so this is an ionic compound. We therefore name the compound
simply by naming the cation first and then the anion. This compound is
magnesium bromide.
In the second example, FeCl3, iron is a metal and chlorine is a nonmetal.
Again, this is an ionic compound. Remember that iron forms cations with
more than one charge, so we must specify the charge in parentheses.
Because the cation is bound to three chloride ions, this ion is Fe3+, or
iron(III). This compound is iron(III) chloride.
In the third example, SF6, both sulfur and fluoride are nonmetals. Therefore
this is a covalent compound, and we must use prefixes to indicate the
number of each atom present. Because sulfur is to the left of chlorine on the

308
 periodic table, it is named first. This compound is sulfur hexafluoride. •

IT
TRY

12. Identify each of these compounds as ionic or covalent, and write its
name:

LiCl ICl BCl3 Al2O3

309
5.6 Aqueous Solutions: How Ionic and Covalent
Compounds Differ
One of the most important differences between ionic and covalent
compounds is how they behave when combined with water. To understand
this critical difference, let’s begin with some fundamental ideas: When a
substance such as salt or sugar mixes with water, it disperses through the
liquid, forming a homogeneous mixture called a solution (Figure 5.20).
(If the liquid is water, we call it an aqueous solution.) When this happens,
we say that the solid has dissolved. Compounds that dissolve in water are
said to be soluble in water; those that do not are said to be insoluble.

Figure 5.20 Ocean water contains many dissolved compounds. It is an aqueous

solution.

Pure water is a poor conductor of electricity (Figure 5.21). However, if

ionic compounds are dissolved in water, the resulting solutions conduct
electricity much more efficiently. Because of this property, we refer to
aqueous ionic solutions as electrolyte solutions, and we call the ionic
compounds electrolytes.

310
Figure 5.21 (a) Pure water is a poor conductor of electricity. (b) Nonionic compounds,
such as sugar, may dissolve in water, but they do not increase the solution’s ability to
conduct electricity. (c) Ionic compounds, like salt, dissociate into ions; the resulting
solution conducts electricity.

Explore
Figure 5.21

Devices that test for water purity often test how well the water conducts
electricity. If a water sample conducts electricity well, we know that ionic
compounds are present.

When ionic compounds dissolve in water, the positive and negative

ions are pulled away from each other and surrounded by water ions
(Figure 5.22). This process of pulling apart the ions in an ionic solid is
called dissociation. The dissolved ions help carry electric current through
the aqueous solutions.

311
Figure 5.22 When an ionic solid like salt dissolves in water, the water molecules pull
the ions away from the solid and into solution.

Explore
Figure 5.22

As a general rule, covalent compounds do not form ions in water.

Because of this, aqueous solutions containing only covalent compounds
are not electrolytic.

312
5.7 Acids—An Introduction
Most covalent compounds do not form ions when dissolved in water, but
this rule has one important exception: Acids are covalent compounds that
produce H+ ions in aqueous solution. Most acids contain a covalent bond
between hydrogen and a species that can form a stable anion. When
dissolved in water, this bond breaks to produce a hydrogen cation and a
corresponding anion.

Acids produce H+ ions in water.

For example, HCl and HNO3 are both acidic molecules. When
dissolved in water, these compounds ionize (form two ions):
HCl ionizes to form H+ and Cl– in aqueous solution.
HNO3 ionizes to form H+ and NO3– in aqueous solution.
We will explore the behavior of acids in Chapters 6 and 12. For now, it
is important that you be able to identify and name common acids. The
most common acids are listed in Table 5.7. When writing the formulas for
acids, we typically write the formula with H first, as though it were the
cation, followed by the anion.

TABLE 5.7 Common Acids

Formula Name

HF Hydrofluoric acid

HCl Hydrochloric acid

HBr Hydrobromic acid

HI Hydroiodic acid
H2CO3 Carbonic acid

HNO3 Nitric acid

HNO2 Nitrous acid

H2SO4 Sulfuric acid

H3PO4 Phosphoric acid

HC2H3O2 Acetic acid

313
 Acids are corrosive, meaning they destroy many substances, including metal
surfaces. They can also cause severe burns to the skin and should be handled
with care.

Naming Acids
Binary Acids

Binary acids consist of H+ and a single nonmetal element. The most

common of these acids are those formed from the halogens: HF, HCl, HBr,
and HI. These acids are named by combining the prefix hydro–, the root
name of the halogen, and the suffix –ic acid:
HF hydrofluoric acid
HCl hydrochloric acid
HBr hydrobromic acid
HI hydroiodic acid

Oxyacids

Oxyacids are compounds that dissociate to form H+ and an oxyanion.

There are two rules for naming acids that dissociate to form oxyanions:
1. If the anion ends in –ate, name the acid by changing the suffix to –ic acid.
For example:
NO3 – nitrate ion HNO3 nitric acid
CO3 2– carbonate ion H2CO3 carbonic acid

Derivatives of the sulfur and phosphorus oxyanions deviate slightly from

this rule:
SO42– sulfate ion H2SO4 sulfuric acid

314
 PO4 3– phosphate ion H3PO4 phosphoric acid

2. If the anion ends in –ite, name the acid by changing the suffix to –ous acid.
NO2– nitrite ion HNO2 nitrous acid

Example 5.11 Naming Acids of Oxyanions

Name each of these acids, using the guidelines described earlier:
a. HClO4
b. H2CrO4

In water, HClO4 ionizes to form H+ and ClO4– ions. Because ClO4– is the
perchlorate ion (see Table 5.3), HClO4 is named perchloric acid. Similarly,
CrO42– is the chromate ion, so H2CrO4 is chromic acid. •

IT
TRY

13. Name these acids:

a. HF
b. HClO
c. HC2H3O2

14. Write a formula for the acidic, ionic, and covalent compounds shown
here.
a. chlorous acid
b. zinc chlorate
c. boron trichloride

315
Summary
Chemical bonding involves the gain, loss, or sharing of valence electrons.
A key factor in chemical bonding is the octet rule, which states that atoms
are stabilized by the presence of eight electrons in their valence shells.
Atoms that fulfill the octet rule have completely filled s and p sublevels in
their valence shell.
To fulfill the octet rule, many atoms gain or lose electrons, forming
ions. Metals tend to lose electrons to form positive ions (cations) while
nonmetals tend to gain electrons to form negative ions (anions).
Polyatomic ions are groups of atoms that contain an overall charge.
Ionic compounds are a combination of positive ions (cations) and
negative ions (anions). In any ionic compound, the total charge must be
equal to zero. When naming an ionic compound, we give the name of the
cation first, followed by the name of the anion. Ionic compounds bind
together in lattices of alternating charges. We describe an ionic compound
by its empirical formula, which is the lowest whole-number ratio of atoms
in that compound.
In covalent bonds, electrons are shared between two nonmetal atoms.
Covalent compounds form discrete units called molecules. We typically
describe a covalent solid by its molecular formula, which gives the number
of each type of atom present in the molecule. When naming covalent
compounds, we use prefixes to indicate the number of each type of atom
present.
Ionic and covalent compounds behave differently in water. When ionic
compounds dissolve in water, they dissociate into their component cations
and anions. Dissolved ions enhance water’s ability to conduct electricity.
Because of this trait, ionic compounds are sometimes referred to as
electrolytes. In contrast, most covalent compounds remain intact when
dissolved in water.
Acids are covalent compounds that ionize in water to produce H+ ions
and a corresponding anion. The names of acids derive from the names of
the anions they produce in solution.

316
Continuing Cade’s Work

Alot has changed since John Cade began using lithium carbonate to treat bipolar
disorder. Today we have a much better (though far from complete) understanding
of how ions and compounds affect our brain’s function. For example, scientists
now know that the covalent compound dopamine (Figure 5.23) plays a critical role
in the working of the brain. Dopamine conveys signals between nerve cells, and it
affects brain functions such as mood, memory, and motor control. Parkinson’s
disease (a degenerative disorder affecting muscle control) arises from a drop in
dopamine levels. Other medical and cognitive issues, including drug addiction,
perception of pain, appetite, and sexual gratification, all involve dopamine levels.

Figure 5.23 Dopamine is connected to mood, memory, and motor control.

To perform its function, dopamine binds to cells in the central nervous system
at special locations on the cell surface called receptor sites. When dopamine docks
to a receptor site, it activates the site in much the same way that a key activates a
lock. Like a key, the molecule’s size and shape (along with other features) are
critically important to its function. The shape of a molecule depends on the
electronic structure of its atoms and on the covalent bonds that hold the atoms
together. We’ll explore the shape of molecules in much more detail in Chapter 9.
Medicinal chemists often search for molecules that can mimic the function of
biological molecules like dopamine. They explore how slight changes in molecular
structure (and therefore in molecule size and shape) affect the molecule’s ability to
bind to a receptor site. For molecules in the brain, these small differences in
structure create profound differences in function.
For example, look at the three molecules in Figure 5.24. Do you notice their
similarity to dopamine? The first molecule is adrenaline, a hormone that stimulates
the nervous system. The second is ephedrine, a commercial decongestant and
appetite suppressant. The third is methamphetamine, a devastatingly addictive,
mood-altering drug. Like dopamine, each of these molecules affects brain function.
But their small differences in size and shape affect how they bind to receptors,

317
causing different responses in mood and behavior.

Figure 5.24 These compounds are similar to dopamine, and they also affect brain
function. The green shading highlights their structural similarities.

318
 Key Terms
5.1 Lewis Symbols and the Octet Rule
Lewis dot symbol A method of representing the valence structure of an atom
or ion that involves using dots around the atomic symbol to indicate valence
electrons.

5.2 Ions
cation A positively charged ion.
anion A negatively charged ion.
polyatomic ion A group of bonded atoms with an overall charge.
oxyanion A polyatomic ion containing oxygen.

5.3 Ionic Bonds and Compounds

ionic bond A force of attraction between oppositely charged ions.
ionic compound A compound composed of oppositely charged ions.
ionic lattice A tightly packed array of alternating positive and negative
charges; the characteristic arrangement of ions in an ionic solid.
chemical formula A representation of the type and amount of each element
present in a compound.
empirical formula A chemical formula that gives the smallest whole-number
ratio of atoms in a compound.
formula unit In ionic compounds, the smallest number of ions necessary to
form a compound; the combination of atoms described by an empirical
formula.

5.4 Covalent Bonding

covalent bond A bond in which two electrons are shared between atoms;
covalent bonds typically form between nonmetals.
Lewis structure A depiction of the arrangement of valence electrons in a
molecule or polyatomic ion, in which the Lewis symbols for atoms are shown
connected by dashes representing covalent bonds.
covalent compounds Compounds formed by covalent bonds; these
compounds form discrete groups of atoms called molecules.
molecular formula A formula that gives the actual number of atoms in the
molecule.

319
5.6 Aqueous Solutions: How Ionic and Covalent Compounds Differ
solution A homogeneous mixture; for example, a solid mixed in a liquid.
soluble Having the ability to be dissolved in a liquid.
electrolyte solution An aqueous solution containing dissociated ions; this type
of solution conducts electricity more effectively than pure water.
dissociation The process by which ions are pulled apart from a solid lattice
when an ionic compound dissolves in water.

5.7 Acids—An Introduction

acid A covalent compound that produces H+ ions in aqueous solution.
oxyacid A covalent compound that dissociates in aqueous solution to form H+
and an oxyanion.

320
 Additional Problems

5.1 Lewis Structures and the Octet Rule

15. Using the periodic table, predict the number of valence electrons in each of
these atoms:
Li C Si Kr Se

16. Using the periodic table, predict the number of valence electrons in each of
these atoms:
Be Mg Ca Ge I

17. Write Lewis dot symbols to show the valence structures of each of these
atoms:
Na N H As Sb

18. Write Lewis dot symbols to show the valence structures of each of these
atoms:
Be F Ar Cs S

19. Write the electron configuration for the following atoms. Indicate which
electrons are the valence electrons.
Mg N P I

20. Write the electron configuration for the following atoms. Indicate which
electrons are the valence electrons.
Be S Ge Br

21. Indicate whether each of these species fulfills the octet rule:
a. a sodium atom
b. a Na+ ion
c. a fluorine nucleus with 9 electrons
d. a fluorine nucleus with 10 electrons

321
22. Indicate whether each of these species fulfills the octet rule:
a. a magnesium nucleus surrounded by 10 electrons
b. a phosphorus atom
c. an argon atom

5.2 Ions

23. What family of elements forms only +1 ions?

24. What family of elements forms only +2 ions?

25. Potassium has an electronic structure of [Ar]4s1. What is the electronic

structure of the potassium ion (K+)?

26. Calcium has an electronic structure of [Ar]4s2. What is the electronic

structure of the calcium ion (Ca2+)?

27. Write the electronic structure for each of these atoms and ions:
a. a lithium atom
b. a lithium ion, Li+
c. a sodium atom
d. a sodium ion, Na+

28. Write the electronic structure for each of these atoms and ions:
a. a magnesium atom
b. a magnesium ion, Mg2+
c. a beryllium atom
d. a beryllium ion, Be2+

29. Using the periodic table as a reference, predict the charge for each of these
ions:
a. a beryllium ion
b. a strontium ion
c. a sodium ion
d. a cesium ion

30. Using the periodic table as a reference, predict the charge for each of these

322
 ions:
a. a potassium ion
b. a barium ion
c. a calcium ion
d. a lithium ion

31. What two charges are most common for a copper ion?

32. What two charges are most common for an iron ion?

33. Name each of the following cations:

a. Na+
b. Mg2+
c. Cr2+
d. Cr3+

34. Name each of the following cations:

a. K+
b. Ca2+
c. Co2+
d. Co3+

35. Name each of the following cations:

a. Fe2+
b. Fe3+
c. Rb+
d. Ba2+

36. Name each of the following cations:

a. Sn2+
b. Sn4+
c. Ag+
d. Be2+

323
37. Using the periodic table as a reference, write the symbol and charge for
each cation:
a. strontium
b. zinc
c. copper(II)
d. manganese(III)

38. Using the periodic table as a reference, write the symbol and charge for
each cation:
a. aluminum
b. lead(II)
c. lead(IV)
d. magnesium

39. What family of elements forms only –1 ions? What family of elements
typically forms –2 ions?

40. Unlike the other nonmetals, the noble gases do not form stable ions. Why
is this so?

41. The electronic structure of fluorine is [He]2s22p5. What is the electronic

structure of the fluoride ion (F–)?

42. The electronic structure of oxygen is [He]2s22p4. What is the electronic

structure of the oxide ion (O2–)?

43. Write the electronic structure for each of these atoms and ions:
a. a chlorine atom
b. a chloride ion, Cl–
c. a bromine atom
d. a bromide ion, Br–

44. Write the electronic structure for each of these atoms and ions:
a. a nitrogen atom
b. a nitride ion, N3–
c. a sulfur atom
d. a sulfide ion, S2–

324
45. Indicate whether each atom would gain or lose electrons to fulfill the octet
rule:
a. Na
b. S
c. Mg
d. Br

46. Indicate whether each atom would gain or lose electrons to fulfill the octet
rule:
a. Ba
b. O
c. K
d. F

47. Determine whether the following would gain or lose electrons to fulfill the
octet rule:
a. a calcium atom
b. an atom in the halogen family
c. an atom with an electron configuration of [Ar]4s23d104p4
d. an atom with an electron configuration of [Xe]6s2

48. Determine whether the following would gain or lose electrons to fulfill the
octet rule:
a. an alkaline earth metal
b. an oxygen atom
c. an atom with an electron configuration of [Ar]4s23d104p5
d. an atom with an electron configuration of [Kr]5s1

49. Using the periodic table as a reference, write the symbol and charge for
each of these ions:
a. fluoride
b. iodide
c. oxide
d. selenide

50. Using the periodic table as a reference, write the symbol and charge for
each of these ions:

325
 a. chloride
b. bromide
c. sulfide
d. phosphide

51. Name each of the following anions:

a. F–
b. S2–
c. O2–
d. I–

52. Name each of the following anions:

a. Cl–
b. Br–
c. P3–
d. Te2–

53. Using the periodic table as a reference, predict the charge of each of these
ions:
a. beryllium ion
b. oxide ion
c. chloride ion

54. Using the periodic table as a reference, predict the charge of each of these
ions:
a. bromide ion
b. sodium ion
c. barium ion

55. What charges would you expect on each of these ions?

a. a halogen ion
b. an alkali metal ion
c. an ion formed from a neutral atom with electron configuration
[Ne]3s23p5

56. What charges would you expect on each of these ions?

326
 a. an ion formed from a calcium atom
b. an alkaline earth metal ion
c. an ion formed from a neutral atom with electron configuration [Ne]3s1

57. Write the name and the charge of the ion formed from each of these atoms:
a. K
b. Rb
c. Cl
d. Br

58. Write the name and the charge of the ion formed from each of these atoms:
a. Mg
b. Ca
c. O
d. S

59. Write the symbol and charge for each of these ions:
a. a fluoride ion
b. a strontium ion
c. a beryllium ion
d. a phosphide ion

60. Write the symbol and charge for each of these ions:
a. a ferrous ion
b. a copper(II) ion
c. a nitride ion
d. a rubidium ion

61. Identify each of these anions as monatomic or polyatomic:

a. nitride
b. nitrate
c. sulfite
d. sulfide

62. Identify each of these anions as monatomic or polyatomic:

a. bromate
b. bromite

327
 c. bromide
d. perbromate

63. What two suffixes commonly indicate oxyanions?

64. Four common oxyanions are formed from bromine: BrO4–, BrO3–, BrO2–,
and BrO–. Name each of these ions.

65. Write the formula and charge for each of these polyatomic ions:
a. ammonium
b. carbonate
c. hydroxide
d. acetate

66. Write the formula and charge for each of these polyatomic ions:
a. nitrate
b. nitrite
c. sulfate
d. bicarbonate

67. Write the formula and charge for each of these polyatomic ions:
a. chlorate
b. sulfite
c. hypochlorite
d. permanganate

68. Write the formula and charge for each of these polyatomic ions:
a. cyanide
b. peroxide
c. dichromate
d. bisulfate

69. Write the symbol or formula and charge for each of these ions:
a. tin(IV)
b. cupric ion
c. fluoride
d. sulfate

328
70. Write the symbol or formula and charge for each of these ions:
a. lead(II)
b. aluminum
c. bromide
d. chlorate

71. Write the symbol or formula and charge for each of these ions:
a. zinc
b. chromate
c. sulfite
d. phosphide

72. Write the symbol or formula and charge for each of these ions:
a. iodide
b. hydrogen phosphate
c. iodate
d. chromium(II)

5.3 Ionic Bonds and Compounds

73. Predict the empirical formulas for compounds formed from these ions:
a. lithium and chloride
b. calcium and bromide
c. oxide and calcium
d. iron(II) and phosphide

74. Predict the empirical formulas for compounds formed from these ions:
a. sodium and fluoride
b. chromium(III) and chloride
c. silver and sulfide
d. lithium and nitrite

75. Write the empirical formula for each of these compounds:

a. aluminum chloride
b. iron(II) sulfide
c. calcium sulfate
d. aluminum oxide

329
76. Write the empirical formula for each of these compounds:
a. iron(III) nitrate
b. copper(II) nitrate
c. ammonium phosphate
d. ammonium phosphide

77. Write the empirical formula for each of these compounds:

a. chromium(III) acetate
b. zinc chlorate
c. silver nitrate
d. lead(II) carbonate

78. Write the empirical formula for each of these compounds:

a. tin(IV) chloride
b. ammonium chlorite
c. lithium bicarbonate
d. cobalt(III) hydroxide

79. Write the empirical formula for each of these compounds:

a. chromium(III) hypochlorite
b. potassium permanganate
c. sodium cyanide
d. lead(II) perchlorate

80. Write the empirical formula for each of these compounds:

a. tin(IV) chloride
b. ammonium chlorate
c. lithium bisulfate
d. sodium hydrogen phosphate

81. In these compounds, determine the charge on the transition metal cation:
a. SnCl2
b. SnCl4
c. Pb(NO3)2
d. FeCO3

82. In these compounds, determine the charge on the transition metal cation:

330
 a. Ag3PO4
b. Cu(C2H3O2)2
c. InBr3
d. Cr2(SO4)3

83. Name each of these ionic compounds:

a. NaBr
b. K2O
c. FeBr3
d. CuS

84. Name each of these ionic compounds:

a. KOH
b. Cu(C2H3O2)2
c. K2CrO4
d. NH4Cl

85. Name each of these ionic compounds:

a. FeCO3
b. Al(NO2)3
c. Ba(NO3)2
d. (NH4)2SO4

86. Name each of these ionic compounds:

a. Na2Cr2O7
b. AgOH
c. ZnCO3
d. Cr2(SO4)3

5.4 Covalent Bonding

87. When two nonmetals bond together, why do they form a covalent bond
rather than an ionic bond?

88. What seven elements exist as diatomic molecules in their elemental forms?

331
89. How many electrons are shared in a covalent bond? When drawing
structures, how do we typically represent covalent bonds?

90. The Lewis structure of an H2S molecule is shown. In this structure, how
many electrons does the sulfur atom share through covalent bonds? How
many of the valence electrons are not shared? Does this sulfur atom have a
complete octet?

91. This figure shows a group of water molecules. How many water molecules
are in this image? How many covalent bonds are present in each molecule?

92. This figure shows a group of CH4 molecules. How many CH4 molecules
are in this image? How many covalent bonds are present in each molecule?

93. Acetic acid, shown here, is the main component of vinegar. Give the
molecular formula and the empirical formula for this compound.

332
94. The structure of oxalic acid is shown here. Give the molecular formula and
the empirical formula for this compound.

95. Compounds such as Freon® 112 (referred to as chlorofluorocarbons, or

CFCs) were used as refrigerants for many years, but they were phased out
because of their harmful effects on Earth’s atmosphere. Write the
molecular and empirical formulas for Freon 112.

96. Propane is a natural gas that is widely used as a heating fuel. Give the
molecular and empirical formulas for propane.

97. Why is it necessary to use prefixes when naming binary covalent

compounds?

98. When naming binary covalent compounds, when is a prefix not used?

99. Name each of these covalent compounds:

a. SCl2
b. NF3
c. N2O4
d. P4O10

333
100. Name each of these covalent compounds:
a. SO3
b. CCl4
c. N2F4
d. S2Cl2

5.5 Distinguishing Ionic and Covalent Compounds

101. When looking at a binary compound (one that has just two elements), how
can we tell if it is ionic or covalent?

102. Why is it acceptable to write the formula of acetylene as C2H2 but not
acceptable to write the formula of magnesium oxide as Mg2O2?

103. Determine whether these compounds contain ionic bonds or covalent

bonds:
a. NaBr
b. PCl3
c. MnF2

104. Determine whether these compounds contain ionic bonds or covalent

bonds:
a. CO2
b. N2
c. KCl

105. Indicate whether these compounds would form an ionic lattice or discrete
molecules:
a. KCl
b. CCl4
c. P4O10
d. Na2S

106. Indicate whether these compounds would form an ionic lattice or discrete
molecules:
a. CO2
b. MgF2

334
 c. Ca(NO3)2
d. Na3PO4

107. Indicate whether each of these compounds is ionic or covalent. Correctly

name each compound.
a. NaBr
b. PBr3
c. MgBr2
d. SBr2

108. Indicate whether each of these compounds is ionic or covalent. Correctly

name each compound.
a. SO3
b. ZnO
c. CO
d. Fe2O3

109. Indicate whether each of these compounds is ionic or covalent. Correctly

name each compound.
a. SiCl4
b. AlCl3
c. BBr3
d. Na2SO3

110. Indicate whether each of these compounds is ionic or covalent. Correctly

name each compound.
a. MgSO4
b. SO3
c. NaHCO3
d. CO2

5.6 Aqueous Solutions: How Ionic and Covalent Compounds Differ

111. What does the term electrolyte mean? What types of compounds are
likely to be electrolytes?

112. By itself, water is a poor conductor of electricity. What must be present

335
 for water to conduct electricity efficiently?

113. When sodium sulfate is dissolved in water, it dissociates. What ions are
present in an aqueous solution of sodium sulfate?

114. Ethylene glycol (C2H6O2) is a covalent compound. Describe what

happens to the C2H6O2 molecules as this compound dissolves in water.

115. Which of these compounds are likely to dissociate in an aqueous solution?

How can you tell?
a. KCl
b. CaBr2
c. CO2
d. C2H6O

116. Which of these compounds are likely to dissociate in an aqueous solution?

How can you tell?
a. SO3
b. Zn(ClO4)2
c. CS2
d. MgF2

5.7 Acids—An Introduction

117. What are acids? How do acids differ from most covalent compounds?

118. Place these compounds in the following Venn diagram: NaCl, CCl4, and
HCl.

119. Name the following acids:

a. HCl

336
 b. HBr
c. HI

120. H2S is an acidic, foul-smelling gas. It is often called hydrogen sulfide.

Name this compound using the rules for naming binary acids.

121. Name the following acids:

a. HNO3
b. HNO2
c. HClO4
d. HClO2

122. Name the following acids:

a. H2SO4
b. H2SO3
c. HClO3
d. HClO

123. The formate ion is a biologically important ion with the formula CHO2–.
Based on this information, what is the name of the acid having the
formula HCHO2?

124. The selenate ion has the formula SeO42–. Based on this information, what
is the name of the acid having the formula H2SeO4?

125. Classify each of these compounds as an ionic compound, a covalent

compound, or an acid. Name each compound.
a. NaNO2
b. N2O4
c. HNO2
d. KNO2

126. Classify each of these compounds as an ionic compound, a covalent

compound, or an acid. Name each compound.
a. K2SO4
b. SO2

337
c. H2SO4
d. NaHSO4

338
Chapter Six
Chemical Reactions

Lost Cities of the Maya

For thousands of years, the Mayan civilization dominated the Yucatán region in
southeast Mexico, Guatemala, and Belize. Massive cities like Tikal and Palenque
glistened with temples, palaces, and stadiums. Their culture flourished, with
remarkable achievements in architecture and mathematics. Mayan sculpture and
writing, which used intricate figures called hieroglyphs, left a stunning record of
their history, culture, and beliefs (Figure 6.1).

Figure 6.1 (a) This image shows the palace ruins at Palenque. (b) These plaster
sculptures are from the Mayan city Ek Balam. (c) Mayan writing used hieroglyphs like

339
these, which were often made from molded plaster. (d) Isabel Villaseñor overlooks the
Temple of the Foliated Cross, Palenque. (e) Villaseñor and coworkers worked inside the
crypt of the Mayan King Tikal, located inside the temple. The sarcophagus lid in front
of her is carved limestone.

But something happened. Around 900 C.E., the cities of the inner Yucatán
began to decline. Their populations quickly diminished. By the time the Spanish
arrived in the early 1500s, most of the Maya had migrated north and east to the
coastal areas. The great cities were abandoned, lost to the jungle.
What caused the decline?
For years, archaeological conservationist Isabel Villaseñor studied the ruins of
Mayan cities like Palenque. Her work focused on a specific feature—the white
plaster covering the great temples and palaces. In many places, ancient artisans
even molded this plaster to create intricate artwork and hieroglyphs.
Over the past several decades, Dr. Villaseñor and other archaeologists have
learned how the Mayans produced plaster for their buildings. The story involves
chemical reactions that are still used today to produce plaster and cement.
The Maya began with limestone, a type of rock composed of calcium carbonate
(CaCO3), which is abundant in the Yucatán Peninsula. They crushed the limestone
into small pieces and then heated it over huge fires. Above about 900 °C, calcium
carbonate decomposes to form two new compounds—carbon dioxide (CO2) and
calcium oxide (CaO). The powdery calcium oxide was then mixed with water to
produce a new compound, called slaked lime. Combined with volcanic ash, this
material formed a soft putty that was easily molded. As the lime mixture slowly
dried, a third chemical reaction took place, producing a hard and beautiful white
surface.
But the plaster came at a price: Producing lime created huge demands on the
local resources. The Maya used freshly cut trees to generate the hot, slow-burning
fires needed to convert limestone to calcium oxide. Quarrying the limestone,
cutting down the trees, and maintaining the fires required immense human effort—
but the environmental cost may have been greater still.
The story of Mayan plaster is a story of chemical changes. In this chapter, we
will explore common chemical changes. We’ll see how these changes are
expressed as chemical equations and how atomic structure determines the way
different elements and compounds react. We’ll explore patterns of chemical
change and use them to predict the outcomes of common reactions. And, at
chapter’s end, we will return to Palenque, to revisit the reactions that produced the
great plastered surfaces, and examine clues that suggest how cities like Palenque
were transformed from thriving cultural centers into abandoned ruins hidden for
centuries in the Yucatán jungles.

Intended Learning Outcomes

After completing this chapter and working the practice problems, you should be
able to:

340
6.1 Chemical Equations
Write chemical equations to express the identity and ratio of species in a
chemical change.
Use a balanced equation to describe the ratio in which atoms or compounds
react.
Correctly balance an equation.

6.2 Classifying Reactions

Classify synthesis, decomposition, single-displacement, and double-
displacement reactions.

6.3 Reactions between Metals and Nonmetals

Predict the products formed from the reaction of metals and nonmetals.
Identify the species that are oxidized and reduced in a metal-nonmetal
combination reaction.

6.4 Combustion Reactions

Predict the products formed from the combustion of metals and from the
combustion of hydrocarbons.

6.5 Reactions in Aqueous Solution

Apply the solubility rules to determine whether common ionic compounds
are water soluble and predict the products of precipitation reactions.
Predict the products of acid-base neutralization reactions.
Describe precipitation and neutralization reactions using molecular, complete
ionic, and net ionic equations.

341
6.1 Chemical Equations
Scientists use chemical equations to describe chemical changes. For
example, we just saw how the Mayans heated limestone (calcium
carbonate) to convert it to lime (calcium oxide) and carbon dioxide gas.
We could represent this change with a chemical equation:

CaCO3→CaO + CO2
In this reaction, CaCO3 is the reactant (sometimes called the reagent or
the starting material). The reactants are shown on the left side of the
arrow. The products that form in the reaction (CaO and CO2) are shown
on the right side of the arrow.
Chemical equations also convey the ratios of reactants and products.
For example, look carefully at the molecular representation of the reaction
in Figure 6.2. In this reaction, two molecules of hydrogen gas react with
one molecule of oxygen gas to produce two molecules of water. Notice
that this reaction follows the law of conservation of mass: Atoms are not
created or destroyed in this chemical change, and the total mass of the
starting materials is equal to the mass of the products (Figure 6.3).

342
Figure 6.2 (Top) When ignited with a flame, a balloon filled with hydrogen reacts in a
dramatic fireball. (Bottom) The reaction involves the combination of hydrogen with
oxygen, as shown. Each oxygen molecule combines with two hydrogen molecules to
produce two new water molecules.

Explore

Figure 6.2

Figure 6.3 The reaction of hydrogen with oxygen to form water follows the
conservation of mass. The total mass of the reactants is equal to the total mass of the
products.

We describe the reaction of hydrogen and oxygen using this chemical

equation:

2 H2+ O2→2 H2O

Recall from Chapter 5 that the subscripts (for example, H2, O2, H2O)

343
show the number of atoms in each molecule or formula unit. The numbers
that precede the chemical formulas are called coefficients. These numbers
show the ratio in which compounds react or are formed (Figure 6.4). The
coefficients are written to show the smallest whole-number ratio in which
molecules react. If a reactant or product does not have a coefficient, we
assume that coefficient to be one.

Figure 6.4 It is important to understand the difference between coefficients and

subscripts.

Chemical equations describe what happens at the atomic level, but they
also provide a ratio for larger reactions. For example, what if 2 million
molecules of hydrogen reacted with 1 million molecules of oxygen? Based
on the ratio in the equation, 2 million molecules of water would form.
If we have correctly described the ratios of reactants and products in a
chemical reaction, the equation is balanced. In a balanced equation, the
number and type of each atom are the same in the reactants as in the
products. A properly balanced equation shows the smallest whole-number
ratio of reactants to products.

A balanced equation contains the same number and type of atom on each
side of the arrow.

Example 6.1 Writing Chemical Equations

When methane gas (CH4) burns, it reacts with molecular oxygen to form
carbon dioxide and water, as represented in the image below. Write a
balanced equation to describe this reaction.

344
In this chemical reaction, the reactants are CH4 and O2. The products for
this reaction are H2O and CO2. Notice that in this picture, the number of
carbon, hydrogen, and oxygen atoms is the same in both the starting
materials and the products. Two CH4 molecules react with four O2
molecules to produce two CO2 molecules and four H2O molecules. We
therefore write this as
2 CH4+4 O2→2 CO2+4 H2O

While this is a correct representation, it is not the smallest whole-number

ratio. After dividing each coefficient by two, we get the properly balanced
equation. •

CH4+2 O2→CO2+2 H2O

Example 6.2 Predicting Ratios from Balanced Chemical

Equations
When heated, CH2N2 undergoes an explosive chemical reaction to produce
two new substances, N2 and C2H4, as shown in the balanced equation
below. If 20 molecules of CH2N2 react in this way, how many molecules of
C2H4 form? How many molecules of N2 and C2H4 form if 2 million
molecules of C2H4 react?

2 CH2N2→2 N2+C2H4
The balanced equation tells us that 2 molecules of CH2N2 produce 2
molecules of N2 and 1 molecule of C2H4. Following this ratio, 20
molecules of CH2N2 react to produce 20 molecules of N2 and 10 molecules
of C2H4. Similarly, 2 million molecules of CH2N2 react to produce 2

345
 million molecules of N2 and 1 million molecules of C2H4. •

IT
TRY

1. Write a balanced equation to describe the reaction of nitrogen and

hydrogen shown here:

2. Based on the balanced equation in Question 1, how many molecules of

NH3 could be made from the reaction of 3,000 molecules of hydrogen
gas with 1,000 molecules of nitrogen gas?

Balancing Equations
Sometimes, when describing a reaction, we begin with the identities of the
starting materials and products without knowing the ratio in which the
reaction takes place. For example, iron can react with oxygen to form
iron(III) oxide—the reddish-brown compound we know as rust. To
describe this process, we could write the equation as

Fe + O2→Fe2O3
However, notice that something is funny with this equation: We start with
two oxygen atoms on the left-hand side, but we have three oxygen atoms
on the right-hand side. Did we conjure up another oxygen atom? Of course
not! Although we have identified the compounds correctly, we have not
accounted for the ratio in which the atoms react.

346
 The Maya used iron oxide to produce the red-brown colors in their artwork.

To more accurately describe this reaction, we must balance the

equation to account for each atom in the starting materials and products.
To do this, we add coefficients to the elements and compounds in the
equation until the number and type of atom on each side are the same. This
process is shown in Figure 6.5.

Figure 6.5 Balancing equations is a step-by-step process.

347
 Explore
Figure 6.5

Notice that when balancing this equation, we never touched the

subscripts. The subscripts show the chemical identity of each molecule or
formula unit present. When we balance equations, we are simply adjusting
the ratio of starting materials and products—not altering the identity of the
substances. Examples 6.3 and 6.4 describe how to balance two more
equations.

Balance equations by adjusting the coefficients, not the subscripts.

Example 6.3 Balancing an Equation

Balance the equation below to show the proper ratio for the reaction of
potassium metal with copper(II) chloride to produce potassium chloride
and copper metal:

348
 K + CuCl2→Cu + KCl not balanced

Let’s begin by balancing the chlorine atoms. To do this, we put the

coefficient 2 in front of KCl. This balances the chlorine atoms, but not the
potassium atoms. However, we can balance the potassium atoms by putting
a 2 in front of the K. •

2 K + CuCl2→Cu + 2 KClbalanced

Example 6.4 Balancing an Equation

Balance the following equation:
Al2O3+C + Cl2→AlCl3+COnot balanced

Notice that in this equation, aluminum and oxygen always appear as part of
a compound while carbon and chlorine appear as elemental forms. This
means that we can add coefficients to the carbon or chlorine without
disrupting the balance of any of the other compounds. In general, it is
easier to balance elemental forms last. So in this example, we would
balance aluminum and oxygen first because they appear only in
compounds:
Al2O3+ C + Cl2→ 2 AlCl3 + 3 COC, Cl not balanced

Now that aluminum and oxygen are balanced, we can balance the elemental
forms. Adding coefficients to the carbon and chlorine gives us the balanced
equation:
Al2O3 + 3 C + 3 Cl2→2 AlCl3 + 3 CObalanced

The stepwise strategy for solving this equation is shown in Figure 6.6. •

349
Figure 6.6 When balancing an equation, it is helpful to balance the elemental forms
last.

350
 Explore
Figure 6.6

It is helpful to balance elemental forms last.

IT
TRY

3. Balance each of these equations:

Mg + O2→MgO HCl +
Co→CoCl2+H2

Br2+Al→AlBr3
C4H8+O2→CO2+H2O

351
Strategies for Balancing Equations
We can use a few other techniques to balance complicated equations more
easily. In some equations, it is possible to balance polyatomic ions rather
than atoms. In other equations, using a fractional coefficient may be
helpful as an intermediate step. Examples 6.5 and 6.6 illustrate these
techniques.

Example 6.5 Balancing Equations with Polyatomic Ions

Balance the following equation:
Ni(NO3)2+NaOH→Ni(OH)2+NaNO3not balanced

We could try to balance this equation by making a table of all the elements
present, as we did before. However, notice that this equation contains two
common polyatomic ions—nitrate (NO3–) and hydroxide (OH–). Although
it is possible to balance the atoms individually, it is much easier to simply
balance the nitrate and hydroxide ions. We have two hydroxide ions on the
right, so we add the coefficient 2 in front of NaOH to balance them on the
left. Similarly, we place the coefficient 2 in front of NaNO3, to balance the
nitrate ions on the left and right:
Ni(NO3)2+ 2 NaOH→Ni(OH)2+ 2 NaNO3balanced

This approach works as long as the ions in the reaction do not change. We
will see several examples of this type of reaction in the sections that follow.
•

Example 6.6 Balancing with Fractional Coefficients

Balance the following equation:
C2H6+O2→CO2+H2Onot balanced

Because oxygen appears in elemental form in this equation, we should

balance it last. We therefore begin by adding coefficients to the CO2 and
the H2O to balance the carbon and hydrogen:

352
 C2H6+O2→2 CO2 +3 H2Ooxygen not balanced

Now we need to balance the oxygen atoms. Notice that on the right-hand
side of the equation, seven oxygen atoms are present. For the equation to
balance, we need seven oxygen atoms on the left-hand side. But oxygen
atoms come in groups of two. So how do we get seven?
The easiest way to solve this problem is to recognize that we need three and
one-half oxygen molecules to give us seven oxygen atoms. We write 3½ as
the improper fraction 7/2, as follows:
C2H6+72O2→2 CO2+ 3 H2O balanced, but not whole numbers

Now the equation is balanced, but it is not properly written as the lowest
whole-number ratio. To correct this, we multiply each coefficient by two,
resulting in the lowest whole-number balanced equation:
2 C2H6+ 7 O2→4 CO2+6 H2Obalanced

Finally, the key to being able to balance equations consistently is practice.

Be sure to try each of the problems below, and do the additional problems
at the end of the chapter. •

Oxygen atoms are like peanut butter cups—they come in packs of two. It’s
important to remember this when writing and balancing equations.

IT
TRY

4. Balance each of these equations, using the techniques described in this

353
 section:

Pb(ClO3)2+Nal→NaClO3+PbI2Al(NO2)3+Ca→Al +
Ca(NO2)2C4H10+O2→CO2+H2O

Equations with Phase Notations

For many chemical processes, it is important to know not only the identity
of the components but also the phase or state of the components. Because
of this, phase notations are sometimes used in chemical equations (Table
6.1). If we wish to indicate that a compound is a solid, we write (s) after
the chemical formula. Similarly, we represent liquids by the symbol (l) and
gases by the symbol (g).

TABLE 6.1 Phase Symbols

Symbol Meaning

(s) Solid
(l) Liquid

(g) Gas
(aq) Aqueous solution (dissolved in water)

Many reactions involve substances that are dissolved in water. A

substance that is dissolved in water is said to be in an aqueous solution.
We represent this by using the symbol (aq) in the balanced equation. For
example, consider the change that occurs when you pour a soft drink
(Figure 6.7): The dissolved carbonic acid (H2CO3) reacts to produce two
new compounds, water and carbon dioxide gas. We can write this equation
using phase notation as follows:
H2CO3 (aq)→H2O (l) + CO2 (g)

354
Figure 6.7 The bubbles that form when a soft drink is poured arise from the reaction of
carbonic acid to produce carbon dioxide and water.

We sometimes use equations with phase symbols to represent physical

changes, such as changes of state. For example, we could use the
following equation to show water freezing:

H2O (l)→H2O (s)

In the sections that follow, we will use the phase symbols extensively
to describe reactions.

355
6.2 Classifying Reactions
As you begin to study chemical reactions, you’ll encounter a large number
of examples. You may find it overwhelming—I certainly did when I
started learning chemistry, and sometimes I still do. So how can you begin
to make sense of it all? One of the keys is to organize reactions into similar
types. This makes it easier to recognize patterns in chemical behavior and
to predict similar reactions that may take place.
Let’s illustrate this idea with an example from the grocery. To make
shopping easier, the store organizes food by groups: meats, fresh
vegetables, breads. It also arranges food by origin: Italian, Mexican, Thai.
These simple organizational systems help us remember, understand, and
communicate with one another.
In this section and those that follow, we’ll explore broad patterns of
chemical behavior. We’ll classify reactions by the types of products, or by
the way the reaction takes place. No single classification system is
universal, but they all serve one purpose—to organize and simplify the
world around us. As we begin to explore chemical reactions, let’s consider
four broad types of reactions: decomposition, synthesis, single
displacement, and double displacement (Figure 6.8).

Figure 6.8 Many reactions fall under the four categories shown here.

In decomposition reactions, a single reactant forms two or more

products. As an example of this type of reaction, let’s return to the Mayan
production of lime plaster. The first step in this process was the
decomposition of calcium carbonate into two simpler compounds, calcium
oxide and carbon dioxide:

356
 CaCO3 (s)→CaO (s)+CO2 (g)

Although it is an ancient technology, this decomposition reaction is

still vitally important today. Calcium oxide is the key ingredient in cement
and mortar, and manufacturers use modern kilns to convert limestone
(calcium carbonate) into calcium oxide (Figure 6.9).

Figure 6.9 Lime kilns like this one convert limestone into calcium oxide.

Synthesis (or combination) reactions occur when two reactants join

together to form a single product. The second step in the production of
lime plaster is an example of a synthesis reaction: In this reaction, calcium
oxide combines with water to produce a single new compound, calcium
hydroxide:
CaO (s)+H2O (l)→Ca(OH)2 (s)

Another example of a synthesis reaction is the combination of hydrogen

gas and bromine gas to form hydrogen bromide:

H2(g) + Br2(g)→2 HBr (g)

We will look extensively at synthesis reactions in Section 6.3.
In a single-displacement reaction, one element replaces another

357
element in a compound. For example, when zinc is combined with
aqueous copper(II) sulfate, elemental zinc displaces copper to form a new
compound. This process converts copper into its elemental form:
Zn (s) + CuSO4 (aq) →ZnSO4 (aq) + Cu (s)

Similarly, the compound HCl reacts with elemental tin to produce a

new compound, tin(II) chloride, and elemental hydrogen:
Sn (s) + 2 HCl (aq) →SnCl2 (aq) +H2 (g)

In a double-displacement reaction, two compounds rearrange to form

two new compounds. These reactions involve a “swap” of cation-anion
pairs. For example, the reaction of KCl with AgNO3 in aqueous solution
produces two new compounds, KNO3 and AgCl. This reaction can be
thought of as simply a swap of anions:

A simple way to think about single- and double-displacement reactions

is to consider the cations and anions as dance partners (Figure 6.10). In a
single displacement, an element “cuts in” on the dance—producing a new
element and a new compound (new dance partners). In a double
displacement, the two dancing couples switch partners, producing two new
compounds.

358
Figure 6.10 Single- and double-displacement reactions are similar to changes in dance
partners.

Example 6.7 Classifying a Chemical Reaction

An aqueous solution of hydrochloric acid reacts with solid zinc metal to
produce hydrogen gas and a new compound, zinc chloride, that dissolves in
water. Write a balanced equation to show this reaction, then classify it as a
synthesis, decomposition, single-displacement, or double-displacement
reaction. Finally, add phase symbols to describe the phase of each reactant
and product.
Our two reactants are hydrochloric acid and elemental zinc. The two
products are zinc chloride and hydrogen gas. From Chapter 5, we know that
the formula for hydrochloric acid is HCl. The formula for zinc chloride is
ZnCl2, and hydrogen gas is H2. We begin with this unbalanced equation:
HCl+Zn→ZnCl2+H2not balanced

Next, we balance this equation. In this case, addition of the coefficient 2 in

front of HCl is all that is needed:
2 HCl + Zn→ZnCl2+H2balanced

Notice that this reaction contains one element and one compound in the

359
 reactants, and a different element and compound in the products. This is an
example of a single-displacement reaction.
From the question, we know that HCl and ZnCl2 are both dissolved in
water, zinc is a solid, and H2 is a gas. In the final step, we represent this
using phase symbols. •
2 HCl (aq) + Zn (s)→ZnCl2 (aq) +H2 (s)balanced, with phase symbols

IT
TRY

5. Classify the following changes as decomposition, synthesis, single-

displacement, or double-displacement reactions:

Pb(NO3)2+2 KI→2 KNO3+PbI24 Al +3 O2→2 Al2O32 H2O2→2 H2O +

O2Fe + SnCl2→Sn + FeCl2

6. When solid lead(II) sulfite is heated, it forms two new compounds,

solid lead(II) oxide and sulfur dioxide gas. What type of reaction is
this? Write a balanced equation for this reaction, including phase
symbols.

360
6.3 Reactions between Metals and Nonmetals
Now that we have a method for classifying reactions, let’s look at some
ways that specific substances react. We’ll begin with one of the most
common types of synthesis reactions—the combination of a metal and a
nonmetal.
Metals and nonmetals react to form ionic compounds. Many of these
reactions are intense, producing bright flames and tremendous heat
(Figure 6.11).

Figure 6.11 Metals and nonmetals react to form ionic compounds. (a) The bright light
of a signal flare results from the reaction of metallic magnesium with oxygen gas. (b)
Sodium and chlorine gas react violently to produce sodium chloride. (c) Rust forms by
the slow reaction of iron with oxygen gas.

361
 Explore

Figure 6.11

Reacting with O2
Most metals react with oxygen to form new compounds. In these reactions, the
metal loses electrons. The term oxidized arises from these reactions, but its
meaning is broader. In any reaction where an atom loses electrons, we say the
atom is oxidized.

In these reactions, the metal loses electrons to form a positively

charged ion (cation) while the nonmetal gains electrons to form a
negatively charged ion (anion). The result is an ionic compound:
Metal+Nonmetal→Ionic compound

The reactions of metals and nonmetals are examples of an oxidation-

reduction reaction. When an atom is oxidized, it loses electrons. When
an atom is reduced, it gains electrons. We will explore oxidation-
reduction reactions in more detail in Chapter 14.
Let’s look at some examples of this type of reaction. We’ll begin with
the reaction of elemental calcium with sulfur to form calcium sulfide. The
calcium loses two electrons to form Ca2+ while the sulfur gains two
electrons to form the sulfide ion, S2–. We can illustrate this reaction using
Lewis symbols, or we can represent it using a chemical equation, as shown
here:

362
 Ca+S→CaS
Consider another example: Magnesium metal reacts with chlorine gas
to produce magnesium chloride (Figure 6.12).

Figure 6.12 Magnesium reacts violently with chlorine gas.

The metal atom (magnesium) loses two electrons while each chlorine
atom gains one electron. In this reaction, a covalent bond breaks as the
ions form:

Mg+Cl2→MgCl2
So, how can we predict the compounds that will form in metal-
nonmetal reactions? Recall from Chapter 5 that metals and nonmetals form
specific, stable ions. By knowing the charges of the common ions, we can
often predict the ionic compounds that result from these reactions. This
process is illustrated in the following two examples.

Many metal-nonmetal reactions produce common ions.

Example 6.8 Predicting the Products when Metals and

Nonmetals React

363
 What compound is formed when aluminum metal reacts with chlorine gas?
Write a balanced equation for this reaction.
We know that aluminum forms a +3 ion while a chloride ion has a charge
of −1. Therefore, the compound formed must have the formula AlCl3. We
can write an unbalanced equation as shown:
Al + Cl2 →AlCl3not balanced

To balance the equation, we add coefficients to make sure the number and
type of atom on each side of the equation are the same, as shown. •

2 Al + 3 Cl2 →2 AlCl3balanced

Example 6.9 Predicting the Products when Metals and

Nonmetals React
When tin metal reacts with bromine, it is oxidized to the tin(IV) ion while
bromine is reduced to form bromide ions. Write a balanced equation for
this reaction.
The question indicates that the tin(IV) ion is formed. Therefore, the neutral
compound produced will have the formula SnBr4. We can write an
unbalanced equation as shown:
Sn + Br2 →SnBr4not balanced

To balance the equation, we only need to add a coefficient in front of the

bromine. This gives us the final balanced equation. •

Sn + 2 Br2 →SnBr4balanced

IT
TRY

7. If zinc metal reacts with bromine gas, which species is oxidized?

Which species is reduced?

364
8. Predict the products from these reactions:
a. the reaction of potassium and sulfur
b. the reaction of silver and oxygen

365
6.4 Combustion Reactions
Many common reactions involve elemental oxygen. Oxygen is a major
component of Earth’s atmosphere, and it reacts with nearly every other
element and with many compounds. Both metals and nonmetals react with
oxygen. For example, consider the way that oxygen reacts with tin (a
metal) and also with carbon and with sulfur (both nonmetals):

Sn + O2 →SnO2C + O2 →CO2S + O2 →SO2

The first product, tin(IV) oxide, is an ionic compound. The other two
products, carbon dioxide and sulfur dioxide, are molecular compounds
with covalent bonds. And yet all three elements react with oxygen in
similar ways. Each of these reactions releases a large amount of heat and
forms an oxide compound. These rapid and exothermic reactions with
oxygen are called combustion reactions.
Some of the most important combustion reactions involve compounds
composed of hydrogen and carbon, called hydrocarbons. These reactions
are important because fossil fuels, such as coal, oil, and natural gas, are
composed primarily of hydrocarbons (Table 6.2). The combustion
reactions of fossil fuels produce most of the energy used worldwide
(Figure 6.13).

TABLE 6.2 Common Hydrocarbons

Formula Name Use

CH4 Methane Natural gas

C2H2 Acetylene Torches for cutting and welding

C2H4 Ethylene Manufacture of plastic

C3H8 Propane Natural gas component; used for heating and power

C4H10 Butane Lighter fluid

C6H6 Benzene Solvent; precursor for many pharmaceutical compounds

C8H18 Octane Component of gasoline

366
Figure 6.13 The combustion of coal and natural gas produces most of our electricity,
and the combustion of gasoline powers most of our transportation.

When hydrocarbons react with oxygen, they form two main products,
carbon dioxide and water. For example, methane (CH4) is the main
component of natural gas (Figure 6.14).

Figure 6.14 Gas stoves use the combustion of methane gas to produce heat.

The combustion reaction of methane follows this balanced equation:

CH4+2 O2 →CO2+2 H2O

Similarly, octane, C8H18, is a key component of gasoline. This
compound also reacts with oxygen to produce carbon dioxide and water:
2 C8H18+25 O2 →16 CO2+18 H2O

Nonmetal elements often react with oxygen to form many different

367
compounds. For example, the combustion of elemental sulfur produces
several different oxides, including sulfur dioxide (SO2) and sulfur trioxide
(SO3). These compounds, commonly referred to as sulfur oxides (SOx), are
toxic gases that contribute to environmental issues such as acid rain.
Because coal contains a small amount of sulfur (in addition to hydrocarbon
compounds), the combustion of coal leads to the release of sulfur oxides
into the atmosphere (Figure 6.15).

Figure 6.15 When oxygen reacts with sulfur, it produces a set of compounds known as
sulfur oxides. Because coal contains small amounts of sulfur, sulfur oxides are produced
when coal is burned for fuel.

The combustion of hydrocarbons produces carbon dioxide and water.

Example 6.10 Predicting the Products of Combustion Reactions

Write a balanced equation for the combustion of propane gas, a common
fuel used for home heating, cooking, and so on. The formula for propane is
C3H8.
Hydrocarbons react with oxygen to produce carbon dioxide and water. We
can write an unbalanced equation for this reaction first:

C3H8+O2 →CO2+H2O
We then balance the equation to give us the final answer:
C3H8+5 O2 →3 CO2+4 H2O

For another example of balancing a combustion reaction, see Example 6.6

368
 earlier in this chapter. •

IT
TRY

9. Write a balanced equation that describes the combustion of

magnesium.

10. Write a balanced equation that describes the combustion of C5H12.

369
6.5 Reactions in Aqueous Solution

Representing Dissociation: Molecular and Ionic Equations

In Chapter 5 we described the behavior of different types of compounds
when dissolved in water (Section 5.6). Recall that when an ionic
compound dissolves in water, it dissociates into cations and anions. That
is, the ions that make up the compound are pulled apart and surrounded by
water molecules (Figure 6.16). This behavior enables different ions to
react in different ways.

Figure 6.16 When an ionic compound dissolves in water, the cations and anions
dissociate.

Ionic compounds dissociate when they dissolve in water.

When describing the chemical processes that take place in solution, we

use two different equation styles. In a molecular equation, we write ions
together as compounds. In an ionic equation, we show dissociated ions as
separate species. For example, potassium bromide dissociates in water. We
can write this as a molecular equation (shown in blue) or as an ionic
equation (shown in red):
Molecular equation:KBr (s)→KBr (aq)Ionic equation:KBr (s)→K+(aq)+Br− (aq)

Similarly, we could show the dissociation of magnesium nitrate in

370
solution. In this example, the formula unit Mg(NO3)2 dissociates to form
one magnesium ion and two nitrate ions:
Molecular equation:Mg(NO3)2(s)→Mg(NO3)2(aq)Ionic
equation:MgNO3(s)→Mg2+(aq)+2 NO3-(aq)

Notice that the ionic equation shows only the dissociated (aqueous) ions
separately; we show solid ionic compounds as a complete unit.

Practice
Ion charges

Do you remember the names, symbols, and charges for the common ions
covered in Chapter 5? If not, you should go back and relearn the list. You
will use them extensively in the sections ahead. Try the Ion charges
interactive game to test your knowledge!

Example 6.11 Writing an Ionic Equation

Write an ionic equation showing how solid potassium sulfate dissociates
when dissolved in water. Include phase symbols.
Potassium sulfate has the formula K2SO4. When dissolved in water, this
compound dissociates into the component potassium and sulfate ions. We
write this as follows:
K2SO4 (s)→2 K+ (aq)+SO42− (aq)

In the reactant, we show the two potassium ions with a subscript because
they are part of a compound. However, when the compound dissociates, the
potassium ions also separate. Because the two aqueous ions are no longer

371
 together, they are shown with a coefficient in the products. •

IT
TRY

11. Write ionic equations to show the following two changes:

a. Solid ammonium bicarbonate dissolves in water.
b. Solid aluminum nitrate dissolves in water.

Solubility Rules and Precipitation Reactions

Not all ionic solids dissolve in water. Many ionic compounds are
insoluble. What determines whether a compound is water soluble?
To answer this question, let’s think about the energy changes that
occur when an ionic solid dissociates in water (Figure 6.17). In an ionic
lattice, ions are stabilized by the oppositely charged ions that surround
them. When they dissociate, the ions are stabilized by water molecules.
Which situation is more energetically favorable? The answer to this
question determines whether a compound will dissolve.

Figure 6.17 Sodium bromide dissolves in water because the energy gain from the
interaction between the ions and water is greater than the energy gain between the ions
in the lattice.

The solubility of ionic compounds depends on many factors, including

the charge of the ions, the size of the ions, and the way they pack together.
The more strongly the ions bind to each other, the less likely they are to
dissolve in water. For example, the greater the charge on an ion, the more

372
tightly it sticks to oppositely charged ions. As a result, compounds
composed of ions with a charge of two or three tend to be insoluble in
water. Iron(III) oxide (Fe2O3), lead(II) sulfide (PbS), and barium
carbonate (BaCO3) are all insoluble in water.

Ions with a charge of +1 or −1 are usually soluble.

The way ions fit together can be hard to predict, and sometimes ionic
compounds are surprisingly stable. For example, ionic compounds
containing the halogen ions (chloride, bromide, and iodide) are nearly
always water soluble—unless they are bonded to silver (Ag+) or lead(II)
(Pb2+).

Have you ever thought about leaving one job to take another? There are many
factors to consider. How does the pay compare? Are the hours better? Is the
work more enjoyable or rewarding? In the same way, solubility is determined by
competing factors. Which is more stable—the ionic lattice, or the dissociated
ions surrounded by water molecules? The answer to this question varies for
different compounds.

The solubilities of common ionic compounds are summarized in

Figure 6.18 and in Table 6.3. Note some of the key trends:
1. Compounds containing alkali metals (Li+, Na+, K+), ammonium
(NH4+), or the large oxyanions nitrate (NO3–), chlorate (ClO3–),
perchlorate (ClO4–), or acetate (C2H3O2–), are always soluble.
2. Compounds containing chloride (Cl–), bromide (Br–), and iodide
(I–) are soluble, except when they are bonded to silver (Ag+) or
lead(II) (Pb2+).
3. Compounds containing sulfate (SO42–) are usually soluble, except
when they are bonded to Ba2+, Ca2+, Pb2+, or Ag+.
4. Most other ionic compounds are insoluble.

373
Figure 6.18 A graphical approach to the solubility rules is presented here. Notice that
compounds containing the alkali metals, NH4+, NO3–, ClO3–, ClO4–, and C2H3O2– are
always soluble.

TABLE 6.3 Solubility Rules

Compounds Containing These Ions Are Always Soluble

Alkali metals Li+, Na+, K+, Rb+

Ammonium NH4+

Large −1 oxyanions NO3−, ClO3−, ClO4−, C2H3O2−

Compounds Containing These Ions Are Usually Soluble

Halides F−, Cl−, Br−, I−

(except Pb2+, Ag+)

Sulfate SO42−
(except Ba2+, Ca2+, Pb2+, Ag+)

Not Soluble

Most other ions

These guidelines can help to determine whether compounds are water

soluble. However, many compounds are slightly soluble—meaning that a
small number of ions will dissolve, but most of the ionic solid will not. In
some applications, you may need to know the solubilities more precisely
than is given here. Solubility also depends on temperature: Many ionic
compounds that are insoluble at room temperature become quite soluble at
higher temperatures.

374
 Magnesium hydroxide is a common treatment for acid indigestion. Because
magnesium hydroxide is insoluble in water, the mixture has a white, milky
appearance, leading to the name milk of magnesia.

Example 6.12 Predicting the Solubility of an Ionic Compound

Using the information in Figure 6.18 and Table 6.3, determine whether
these compounds are soluble or insoluble in water:

a. Na3PO4 b. AlCl3 c. CaCO3

Let’s begin with sodium phosphate, Na3PO4. Notice that Na+ is on the list
of ions that are always soluble. Even though phosphate is not typically a
soluble ion, this compound is soluble because it contains sodium.
The situation with aluminum chloride is similar: Chlorides are nearly
always soluble, and so AlCl3 is water soluble.
Finally, let’s look at calcium carbonate, CaCO3. Neither of these ions is
listed as soluble or as usually soluble. This compound is insoluble. •

IT
TRY

12. Use Table 6.3 to determine if these compounds are soluble or

insoluble in water:

a. NaC2H3O2 b. FeCl3 c. CaSO4 d. AlPO4

13. Which of these compounds is more likely to be water soluble: cesium

fluoride (CsF) or barium chromate (BaCrO4)? Why?

375
 The different solubilities of ionic compounds cause some dramatic
reactions. For example, aqueous lead(II) nitrate and aqueous sodium
iodide are both colorless solutions. However, if we combine these
solutions, a yellow solid immediately forms (Figure 6.19). The yellow
solid is lead(II) iodide.

Figure 6.19 Lead(II) iodide is insoluble in water. When a solution containing dissolved
Pb2+ is combined with a solution containing dissolved I–, PbI2 immediately forms as a
yellow solid.

Explore

Figure 6.19

The other product from this reaction, sodium nitrate, remains in

solution. We describe this reaction using the following equation:

376
 Pb(NO3)2(aq)+2 Nal (aq)→2 NaNO3(aq)+PbI2(s)

This type of reaction, in which two aqueous solutions combine to produce

a solid (insoluble) product, is called a precipitation reaction. The solid
product formed in the reaction is the precipitate. In the equation above,
the precipitate is indicated by the (s) notation after the chemical formula.
Compounds that remain in solution are shown using the (aq) notation.
Precipitation reactions are examples of double-displacement reactions.
Precipitation reactions are double-displacement reactions.

It is helpful to think of this reaction in terms of ionic equations. Before

the solutions are mixed, each contains dissolved ions:
Lead(II) nitrate solution: Pb2+ (aq) + 2 NO3− (aq)Sodium iodide solution: Na+(aq)
+ I− (aq)

When these solutions are combined, all four ions are present in
solution. However, as the lead and iodide ions encounter each other
(Figure 6.20), they bind together and drop out of solution as a solid
(precipitate).

Figure 6.20 The driving force for a precipitation reaction is the formation of the solid.
If a cation and anion precipitate out of solution, the other two ions are left in solution
together.

377
 Sometimes precipitates form distinctive solids, but at other times they appear as
“cloudy” mixtures.

We can write this as an ionic equation. Because all of the ions are
shown, this is sometimes called a complete ionic equation:
Pb2+ (aq) + 2 NO3− (aq)+2 Na+ (aq)+2 l− (aq)→2 Na+(aq)+2 NO3−(aq)+Pbl2 (s)

The driving force for a precipitation reaction is the formation of the solid.
Notice in the ionic equation that the nitrate and the sodium ions don’t
actually react: They are present in the starting materials and are unchanged
in the products. The sodium and nitrate ions are paired together only
because the other ions (Pb2+ and I+, shown in red in the equation above)
have dropped out of the solution. The sodium and nitrate ions are
spectator ions because they are not changed in the reaction.
When ions combine to form an insoluble compound, a precipitation reaction
occurs.

Sometimes it is helpful to show precipitation reactions as a net ionic

equation. In this type of equation, we omit the spectator ions and include
only the ions directly involved in the precipitation (Figure 6.21). For the
reaction above, the net ionic equation is

Pb2+ (aq)+2 l− (aq)→Pbl2 (s)

378
Figure 6.21 We use three types of equations to describe aqueous reactions.

The solubility rules are the key to predicting precipitation reactions. If

a cation and an anion can combine to form an insoluble product, a
precipitation reaction will occur. Figure 6.22 contains several additional
examples of precipitation reactions. Can you identify the products of these
reactions?

We use the solubility rules to predict precipitation reactions.

Figure 6.22 Each of the reactions shown produces a precipitate. Can you identify the
solid formed in each example?

379
Explore
Figure 6.22

Example 6.13 Predicting Precipitation Reactions

Write a balanced equation to show the reaction of aqueous silver acetate
with aqueous barium chloride. Include phase symbols.
To solve this problem, we first need to identify the ions that are present:
Silver acetate solution:Ag+(aq)+C2H3O2−(aq)Barium chloride solution:Ba2+
(aq)+2 Cl− (aq)

According to the solubility chart (Table 6.3), the compound formed from
silver and chloride ions is insoluble. Therefore, one of the products will be
AgCl (s). We can write this as a balanced, complete ionic equation:

380
2 Ag+ (aq)+2 C2H3O2− (aq) + Ba2+ (aq) + 2 Cl− (aq)→Ba2+(aq)+2 C2H3O2−
(aq)+2 AgCl (s)

By removing the spectator ions, we can also represent this as a net ionic
equation:
Ag+ (aq) +Cl− (aq)→AgCl (s)

Or as a balanced molecular equation:

2 AgC2H3O2(aq)+BaCl2(aq)→Ba(C2H3O2)2(aq)+2 AgCl (s)

If we tried this reaction in the lab, silver chloride would form as a white
precipitate (Figure 6.23). •

Figure 6.23 Silver and chloride ions combine to form a white precipitate.

Example 6.14 Predicting and Balancing Precipitation Reactions

Write a balanced molecular equation to show the precipitation reaction of
calcium chloride with sodium phosphate. Include phase symbols.
To solve this problem, we first write out the empirical formulas for the two
starting materials:

381
 CaCl2+Na3PO4
These ionic compounds are both water soluble, so they exist in solution as
dissociated ions. Based on Table 6.3, two of these ions, calcium and
phosphate, form an insoluble solid. The formula for this compound is
Ca3(PO4)2:

The other two ions, sodium and chloride, remain in solution. We can
describe this reaction with an unbalanced equation, showing the precipitate
as a solid:
CaCl2(aq)+Na3PO4(aq)→Ca3(PO4)2(s)+NaCl (aq)unbalanced

Finally, we balance this equation by balancing the ions on each side:

3 CaCl2(aq) + 2 Na3PO4(aq)→Ca3(PO4)2(s) + 6 NaCl (aq)balanced

Another approach to solving this problem is to recognize that precipitation

reactions are double-displacement reactions. For a precipitation reaction to
occur, the cation-anion pairs must swap. We must keep in mind the charges
on the ions and write empirical formulas that result in an overall charge of
zero. In this instance, the anions swap positions, producing calcium
phosphate and sodium chloride:

Referring again to Table 6.3, we see that one of these two products, calcium
phosphate, forms an insoluble solid.
While this approach may make the problem simpler, remember that a
precipitation reaction cannot occur unless one of the two product
compounds is insoluble. If neither product is insoluble, all of the ions
remain dissolved in water, and no reaction takes place. •

382
 IT
TRY

14. Write a complete ionic equation showing the reaction of aqueous

barium nitrate with aqueous sodium phosphate to form barium
phosphate and sodium nitrate. Include phase symbols.

15. Refer to Table 6.3 to identify the precipitate formed in the following
reactions. Balance the reactions.

KOH (aq)+Pb(NO3)2(aq)→FeCl2(aq)+Li2S
(aq)→MgBr2(aq)+K3PO4(aq)→

16. In which of these combinations would no precipitation reaction

occur?
a. Ammonium iodide is combined with silver acetate.
b. Potassium chloride is combined with sodium hydroxide.
c. Zinc sulfate is combined with calcium chloride.

Acid-Base Neutralization Reactions

Acids and Bases
In Chapter 5 we introduced the very important class of compounds called
acids. Recall that acids are covalent compounds that ionize to produce H+
ions and a stable anion when dissolved in water. A list of common acids is
given in Table 6.4.

TABLE 6.4 Common Acids

Formula Name

HF Hydrofluoric acid

HCl Hydrochloric acid

HBr Hydrobromic acid

383
HI Hydroiodic acid
H2CO3 Carbonic acid

HNO3 Nitric acid

HNO2 Nitrous acid

H2SO4 Sulfuric acid

H3PO4 Phosphoric acid

HC2H3O2 Acetic acid

We often use ionic equations to show how acids ionize in water. For
example, aqueous hydrochloric acid (HCl) and aqueous nitric acid (HNO3)
ionize as follows:
HCl (aq)→H+(aq)+Cl−(aq)HNO3(aq)→H+(aq)+NO3−(aq)

Bases are compounds that produce hydroxide (OH−) ions in aqueous

solution. Many common bases are soluble metal hydroxides (Table 6.5).
In water, these compounds dissociate to give metal cations and hydroxide
anions. For example, sodium hydroxide dissociates in water to produce
sodium and hydroxide ions:
NaOH (s)→Na+(aq)+OH−(aq)

TABLE 6.5 Common Hydroxide Bases

Formula Name

LiOH Lithium hydroxide

NaOH Sodium hydroxide

KOH Potassium hydroxide

Ba(OH)2 Barium hydroxide

In aqueous solution, acids produce H+ ions and bases produce OH− ions.

Neutralization Reactions
Acids and bases undergo neutralization reactions. In these reactions, the

384
H+ from the acid combines with the OH– from the base to form water. We
can show this as a net ionic equation:

H+(aq)+OH−(aq)→H2O (l)
For example, hydrochloric acid reacts with sodium hydroxide. We can
write this as a molecular equation or as a complete ionic equation:
HCl (aq)+NaOH (aq)→H2O (l)+NaCl (aq)H+ (aq)+Cl− (aq)+Na+(aq)+OH−
(aq)→H2O (l) + Na+(aq)+Cl− (aq)

The spectator ions, Na+ and Cl–, combine to form an ionic compound.
Similarly, nitric acid reacts with lithium hydroxide:
HNO3 (aq)+LiOH (aq)→H2O (l)+LiNO3 (aq)

Again, the H+ from the acid reacts with the OH− from the base to form
water. The remaining ions end up together as LiNO3. The ionic compound
that is formed from the spectator ions is often called a salt.
Chemists sometimes refer to ionic compounds as salts.

In general, the following expression describes the neutralization of an

acid with a hydroxide base:

Like precipitation reactions, acid-base neutralizations are double-

displacement reactions: The H+ and OH− combine to form water while the
other two ions combine to form the salt. The driving force of the reaction
is the formation of water; the salt forms from the other two ions.

The formation of water is the driving force for a neutralization reaction.

Example 6.15 Predicting the Products from Acid-Base

Neutralization Reactions
Write a balanced equation to show the reaction of sulfuric acid with

385
 potassium hydroxide. Include phase symbols.
We begin by writing an unbalanced equation to describe this reaction. One
product is water, and the other product is the ionic compound produced by
the spectator ions:
H2SO4+NaOH→H2O +Na2SO4unbalanced

A simple way to balance neutralization reactions is to balance the number

of H+ and OH− ions that combine. Because H2SO4 produces two H+ ions,
there must be two OH− ions to neutralize them, and two water molecules
will be produced. Adding phase symbols gives us the balanced form shown
here. •

H2SO4(aq)+2 NaOH (aq)→ 2 H2O (l)+Na2SO4(aq)balanced

Acid-base neutralizations are double-displacement reactions. The H+ and OH−

combine to form water (HOH), and the two spectator ions combine to form a
salt:

HCl + NaOH →HOH+ NaCl

IT
TRY

17. Predict the products for the following neutralization reactions. Make
sure the equations are balanced.

HBr+NaOH→HClO4(aq)+Ba(OH)2(aq)→H3PO4(aq)+KOH (aq)→

18. Write a complete ionic equation and a net ionic equation for the
following neutralization reaction:

386
LiOH (aq)+HNO3(aq)→LiNO3(aq)+H2O (l)

387
Summary
Scientists describe chemical changes, or reactions, using chemical
equations. Chemical equations show the substances present before the
reaction (called the reactants) on the left side of the arrow, and the
substances present after the reaction (called the products) on the right side
of the arrow.
Atoms are not created or destroyed in chemical reactions. To reflect
this, we write balanced equations. An equation is balanced if it contains
the same number and type of atom in the reactants and in the products.
Balancing equations often involves trial and error and requires practice to
do it efficiently. Sometimes chemical equations also show the phase of the
reactants and products in parentheses after each element or compound.
There are many types of chemical reactions and different ways of
organizing reactions. Many reactions can be classified as decomposition,
synthesis, single-displacement, and double-displacement reactions. These
classifications are based on the number and type of reactant and product
formed.
We also classify reactions by patterns of chemical behavior. In this
chapter, we examined four broad classes of reactions: metal-nonmetal
reactions, combustion reactions, precipitation reactions, and acid-base
neutralization reactions (Table 6.6).

TABLE 6.6 Patterns of Chemical Behavior

Reaction Type Classification Starting Materials Products

Metal-Nonmetal Combination Metal + Nonmetal Ionic compound

Combustion Element + Oxygen Oxide compound
Hydrocarbon + Oxygen CO2 + H2O

Precipitation reaction Double Two soluble Insoluble compound

displacement compounds (precipitate)

Acid-Base Double Acid + base Water + salt

neutralization displacement

Oxidation-reduction reactions involve the gain and loss of electrons.

An atom or ion that loses electrons is oxidized; a species that gains
electrons is reduced. Metals and nonmetals combine in these reactions to
produce ionic compounds.
Many elements react with oxygen to produce oxide compounds.

388
Similarly, hydrocarbons (compounds composed of hydrogen and carbon)
react with oxygen to form carbon dioxide and water. Collectively,
reactions with oxygen are called combustion reactions.
When an ionic compound dissolves in water, the ions separate in a
process called dissociation. Not all ionic compounds dissolve in water.
Solubility rules provide broad guidelines for determining which
compounds do and do not dissolve in water.
Precipitation reactions occur when two aqueous solutions combine to
produce an insoluble product. The insoluble product drops out of solution
(precipitates), leaving the other ions—called spectator ions—remaining in
solution. We commonly describe precipitation reactions using ionic
equations. Complete ionic equations show all ions present while net ionic
equations show only those directly involved in the central reaction.
Acids are covalent compounds that ionize in water to produce
hydrogen cations (H+) and a corresponding anion. Bases are compounds
that produce hydroxide anions (OH−) when dissolved in water. The most
common bases are metal hydroxides. When acids and bases combine, they
undergo a neutralization reaction in which the H+ from the acid reacts with
the OH− from the base to form water. Like precipitation reactions, acid-
base reactions are examples of double-displacement reactions.

389
The Mayan Lime Cycle

We began this chapter with the great Mayan ruins of Palenque. Now that we have a
broader understanding of chemical reactions, let’s look at the specific process the
Mayans used to make plaster for their temples and palaces (Figure 6.24):

Figure 6.24 Through the lime cycle, natural limestone (CaCO3) is converted into CaO,
then into Ca(OH)2. Ca(OH)2 can be molded or shaped before it cures, reforming the
hard CaCO3.

The first step in lime production is heating. Above about 900 °C, calcium
carbonate decomposes to produce calcium oxide and carbon dioxide:

CaCO3(s)→CaO (s)+CO2(g)
Calcium oxide is a fine white powder, commonly called quicklime. The calcium
oxide is then mixed with water to produce calcium hydroxide, also called slaked
lime, in a synthesis reaction:
CaO (s)+H2O (l)→Ca(OH)2(s)

390
The slaked lime is often mixed with some other material, such as volcanic ash, that
strengthens the overall structure. As this lime plaster dries, it reacts with carbon
dioxide in the air to return to calcium carbonate:
Ca(OH)2(s)+CO2(g)→CaCO3(s)+H2O (l)

This three-step process is called the lime cycle. But what does this have to do with
the decline of Mayan cities?
Evidence from the hieroglyphs at Palenque suggests that a major building
project was completed just before the population began to decline. To produce
enough lime for this project, the Maya needed massive amounts of firewood. To
obtain the firewood, they cleared huge swaths of the Yucatán jungles. Many
archaeologists believe that this deforestation harmed the water supplies and
promoted widespread drought. This drought may have led to the decline in
population and ultimately to the abandonment of cities. Years later, the jungles
slowly grew back and overtook the ancient structures.
Today, archaeological conservators like Yareli Jáidar work to preserve
archaeological and artistic treasures (Figure 6.25). She often uses a modern
material called nanolime. Nanolime is composed of very tiny particles of calcium
hydroxide, often just 100 nanometers across (from which the name originates).
When conservationists coat a plaster structure with nanolime, the tiny lime
particles penetrate the painting surface to fill pores and cracks. As the lime dries, it
fuses with the ancient lime, sealing the structures against further damage.

Figure 6.25 (a) Dr. Yareli Jáidar applies a coat of nanolime to a Mayan structure. (b)
This image shows a jaguar painting that has degraded over time. (c) This is the same
painting after restoration.

391
 Key Terms
6.1 Chemical Equations
chemical equation A symbolic representation of a chemical change. Such
equations consist of reactants and products separated by an arrow.
reactant The starting material in a chemical change, shown on the left-hand
side of a chemical equation.
product The compounds produced in a chemical change, shown on the right-
hand side of a chemical equation.
subscript In a chemical formula, subscript numbers show the number of each
atom or ion present.
coefficient In a chemical formula, the numbers written before each reactant or
product to indicate the ratios in which components of the reaction are
consumed or produced.
balanced equation A chemical equation in which the number and type of
atom are the same for the reactants and the products.

6.2 Classifying Reactions

decomposition A reaction in which a single reactant forms two or more
products.
synthesis A reaction in which two reactants join together to form a single
product; also called a combination reaction.
single displacement A reaction in which one element replaces another
element in a compound.
double displacement A reaction in which two compounds swap cation-anion
pairs to form two new compounds.

6.3 Reactions between Metals and Nonmetals

oxidation-reduction reaction A chemical change in which one species loses
electrons (oxidation) while another gains electrons (reduction).
oxidation The loss of electrons.
reduction The gain of electrons.

6.4 Combustion Reactions

combustion A reaction in which oxygen gas combines with elements or
compounds to produce oxide compounds.

392
6.5 Reactions in Aqueous Solution
molecular equation A chemical equation in which all species are written as
neutral compounds.
ionic equation A chemical equation that shows dissociated ions as separate
species.
precipitation reaction A type of chemical change in which two aqueous
solutions combine to produce an insoluble product.
precipitate A solid product formed from the combination of two solutions.
complete ionic equation An equation that shows all ions present in a solution.
spectator ion An ion that is present in a solution but not directly involved in a
chemical change.
net ionic equation An equation in which the only ions shown are those
directly involved in the chemical change, and spectator ions are omitted.
base A compound that produces hydroxide ions in aqueous solutions. In later
chapters, bases will be further defined as proton (H+) acceptors.
neutralization reaction A reaction in which an acid and a base combine to
produce water and an ionic compound (called a salt).

393
 Additional Problems

6.1 Chemical Equations

19. Write a balanced equation to represent the reaction shown. Write the
coefficients as the lowest whole-number ratio.

20. Write a balanced equation to represent the reaction shown. Write the
coefficients as the lowest whole-number ratio.

21. Write a balanced equation to represent the reaction shown.

22. Write a balanced equation to represent the reaction shown. Write the
coefficients as the lowest whole-number ratio.

394
23. Describe each of the following changes using chemical equations:
a. One molecule of hydrogen reacts with one molecule of bromine to
produce two molecules of hydrogen bromide.
b. Two atoms of sodium and one molecule of fluorine react to produce
two molecules of sodium fluoride.

24. Describe each of the following changes using chemical equations:

a. One unit of calcium hydroxide reacts to form one unit of calcium oxide
and one molecule of water.
b. Four iron atoms react with three oxygen molecules to form two units of
iron(III) oxide.

25. In this reaction, how many units of C form when 30 units of A react?

2 A+B→C

26. In this reaction, how many units of B are required to form 40 units of D?

A+2 B→C+D

27. The balanced equation here shows the reaction of hydrogen gas with
chlorine gas. Based on this equation,

H2+Cl2 →2 HCl
a. How many molecules of HCl form from 30 molecules of H2?
b. How many molecules of Cl2 are required to produce 12 molecules of
HCl?

28. The balanced equation here shows the reaction of calcium with oxygen to
produce calcium oxide. Based on this equation,

395
 2 Ca+O2 →2 CaO
a. How many molecules of oxygen react with 18 calcium atoms?
b. If 30 calcium atoms react with 15 oxygen molecules, how many units of
calcium oxide form?

29. The reaction of aluminum with chlorine gas is shown here. Based on this
equation,

2 Al+3 Cl2 →2 AlCl3

a. How many molecules of chlorine gas are needed to react with 10
aluminum atoms?
b. How many units of AlCl3 can be produced from 10 aluminum atoms?

30. Hydrobromic acid (HBr) can react with barium hydroxide as shown here.
Based on this equation,
2 HBr+Ba(OH)2 →BaBr2+2 H2O

a. If 5 units of barium hydroxide react in this way, how many water

molecules form?
b. How many HBr molecules are needed to produce 40 units of BaBr2?

31. Balance each of the following equations:

a. PCl3 + F2 → PF3 + Cl2
b. SO2 + O2 → SO3
c. B + F2 → BF3

32. Balance each of the following equations:

a. Zn + HCl → ZnCl2 + H2
b. Li + CuCl2 → LiCl + Cu
c. C3H4O + H2 → C3H8O

33. Balance the following equations:

a. PCl3 + H2O → H3PO3 + HCl
b. O2 + H2S → SO2 + H2O
c. HCl + MnO2 → MnCl2 + 2H2O + Cl2

396
34. Balance the following equations:
a. TiCl4 + H2O → TiO2 + HCl
b. VCl3 → VCl2 + VCl4
c. Cr2O3 + Al → Al2O3 + Cr

35. Balance the following equations by balancing the ions:

a. Hg(NO3)2 + KCl → KNO3 + HgCl2
b. MgBr2 + NaOH → NaBr + Mg(OH)2
c. AgC2H3O2 + BaCl2 → AgCl + Ba(C2H3O2)2

36. Balance the following equations by balancing the ions:

a. Cr(NO3)2 + KOH → KNO3 + Cr(OH)2
b. (NH4)2CO3 + Ba(C2H3O2)2 → NH4C2H3O2 + BaCO3
c. AgC2H3O2 + BaCl2 → AgCl + Ba(C2H3O2)2

37. Balance the following equations. For these equations, it may be helpful to
write a fractional coefficient for the diatomic element before converting to
whole numbers.
a. N2 + H2 → NH3
b. C2H6 + O2 → CO2 + H2O
c. HCl + Al → AlCl3 + H2

38. Balance the following equations. For these equations, it may be helpful to
write a fractional coefficient for the diatomic element before converting to
whole numbers.
a. P + Cl2 → PCl5
b. IrF5 + H2 → IrF4 + HF
c. C5H10 + O2 → CO2 + H2O

39. Balance the following equations:

a. Na + O2 → Na2O
b. Pb(NO3)2 + KCl → PbCl2 + KNO3
c. C2H2 + O2 → CO2 + H2O

40. Balance the following equations:

a. NaOCl + HCl → Cl2 + NaCl + H2O

397
 b. AgC2H3O2 + MgCl2 → AgCl + Mg(C2H3O2)2
c. N2H6 + O2 → NO2 + H2O

41. Write a balanced equation to describe each of these chemical changes.

Include phase symbols.
a. Zinc metal reacts with aqueous copper(II) chloride to produce copper
metal and aqueous zinc chloride.
b. Propane gas, C3H8, reacts with molecular oxygen to produce carbon
dioxide gas and water vapor.
c. Solid iron reacts with chlorine gas to produce solid iron(II) chloride.

42. Write a balanced equation to describe each of these chemical changes.

Include phase symbols.
a. When calcium metal is heated in water, aqueous calcium hydroxide and
hydrogen gas form.
b. When solid magnesium burns (reacts with oxygen gas), it forms solid
magnesium oxide.
c. Aqueous silver nitrate reacts with aqueous sodium chloride to produce
solid silver chloride and aqueous sodium nitrate.

43. Provide the name and the phase of the reactants for each reaction:
a. 2 HCl (aq) + Na2CO3 (s) → 2 NaCl (aq) + CO2 (g) + H2O (l)
b. Fe (s) + 2 HNO3 (aq) → Fe(NO3)2 (aq) + H2 (g)
c. ZnCl2 (aq) + Pb(ClO4)2 (aq) → PbCl2 (s) + Zn(ClO4)2 (aq)

44. Provide the name and phase of the products for each reaction:
a. P (s) + Cl2 (g) → PCl3 (g)
b. Na2SO4 (aq) + BaBr2 (aq) → BaSO4 (s) + 2 NaBr (aq)
c. Cr (s) + 2 HCl (aq) → CrCl2 (aq) + H2 (g)

6.2 Classifying Reactions

45. Classify each of these chemical reactions as a synthesis, decomposition,

single-displacement, or double-displacement reaction:
a. 3 NH3 → 3 N2 + 3 H2
b. CaO + H2O → Ca(OH)2
c. AgNO3 (aq) + KBr (aq) → KNO3 (aq) + AgBr (s)

398
46. Classify each of these chemical reactions as a synthesis, decomposition,
single-displacement, or double-displacement reaction:
a. Sn + 2 Br2 → SnBr4
b. 2 HBr + Sn → SnBr2 + H2
c. HBr + NaOH → NaBr + H2O

47. Classify each of these chemical reactions as a synthesis, decomposition,

single-displacement, or double-displacement reaction:
a. Ca + O2 → 2 CaO
b. Co (s) + 2 HCl (aq) → CoCl2 (aq) + H2 (g)
c. PbNO3 (aq) + 2 KBr (aq) → 2 KNO3 (aq) + PbBr2 (s)

48. Classify each of these chemical reactions as a synthesis, decomposition,

single-displacement, or double-displacement reaction:
a. K2SO4 (aq) + Pb(C2H3O2)2 (aq) → 2 KC2H3O2 (aq) + PbSO4 (s)
b. 2 Zn + O2 → 2 ZnO
c. PCl5 → PCl3 + Cl2

6.3 Reactions between Metals and Nonmetals

49. What types of compounds form when metals and nonmetals react?

50. What is oxidation? Do metals or nonmetals oxidize more easily?

51. In these reactions, identify the element that is oxidized and the element that
is reduced:
a. 2 Mg + O2 → 2 MgO
b. Fe + S → FeS
c. Br2 + Ca → CaBr2

52. In these reactions, identify the element that is oxidized and the element that
is reduced:
a. 2 Cu + O2 → 2 CuO
b. S + Hg → HgS
c. Sn + 2 Cl2 → SnCl4

53. Predict the products from these reactions, and balance the equations:

399
 a. Be + Cl2 →
b. K + Cl2 →
c. Co (s) + Cl2 (g) →
d. Cu (s) + O2 (g) →

54. Predict the products from these reactions, and balance the equations:
a. Ca + O2 →
b. Sr + Br2 →
c. Pb (s) + O2 (g) →
d. Cr (s) + O2 (g) →

55. Predict the products from these combustion reactions, and balance the
equations:
a. Mg + O2 →
b. C2H4 + O2 →
c. C4H10 (g) + O2 (g) →

56. Predict the products from these combustion reactions, and balance the
equations:
a. Zn + O2 →
b. C5H12 + O2 →
c. C8H18 (l) + O2 (g) →

6.4 Combustion Reactions

57. What element is always involved in combustion reactions?

58. Each of the following elements reacts with oxygen in combustion

reactions. Identify whether the products of these reactions would be ionic
compounds or covalent compounds.
a. magnesium
b. calcium
c. carbon
d. nitrogen

59. What are hydrocarbons? What products form from the combustion of
hydrocarbons?

400
60. Fossil fuels like coal contain primarily hydrocarbons, but they also contain
a small amount of other elements, like sulfur. What products result from
the combustion of hydrocarbons? What products result from the
combustion of sulfur?

61. Write a balanced equation for each of these reactions:

a. combustion of C2H2
b. combustion of C4H8
c. combustion of C9H18

62. Write a balanced equation for each of these reactions:

a. combustion of CH4
b. combustion of C3H8
c. combustion of C5H10

6.5 Reactions in Aqueous Solution

63. The following molecular equations each show an ionic compound

dissolving in water. Show the ionization process by rewriting each one as
an ionic equation:
a. KCl (s) → KCl (aq)
b. Li2SO4 (s) → Li2SO4 (aq)
c. (NH4)2CO3 (s) → (NH4)2CO3 (aq)

64. The following molecular equations each show an ionic compound

dissolving in water. Show the ionization process by rewriting each one as
an ionic equation:
a. CaBr2 (aq) → CaBr2 (aq)
b. LiNO3 (s) → LiNO3 (aq)
c. Cu(ClO4)2 (s) → Cu(ClO4)2 (aq)

65. Write ionic equations to show the dissociation of each solid as it is

dissolved in water:
a. sodium bicarbonate
b. aluminum chloride
c. chromium(III) chlorite

401
66. Write ionic equations to show the dissociation of each solid as it is
dissolved in water:
a. ammonium cyanide
b. tin(IV) bromide
c. cobalt(II) chlorate

67. Using the general solubility rules in Section 6.5, state whether each of
these compounds is likely to be soluble or insoluble:
a. an ionic compound containing the ammonium ion
b. an ionic compound containing a +2 cation and a –3 anion
c. an ionic compound containing a nitrate ion

68. Using the general solubility rules in Section 6.5, state whether each of
these compounds is likely to be soluble or insoluble:
a. an ionic compound composed of an alkaline earth metal and the oxide
ion
b. an ionic compound containing a transition metal ion and an acetate ion
c. an ionic compound containing a silver ion and a halogen ion

69. Using the general solubility rules in Section 6.5, state whether each of
these compounds is likely to be soluble or insoluble:
a. an ionic compound composed of ammonium and a chlorate ion
b. an ionic compound containing a transition metal ion and a phosphate
ion
c. an ionic compound composed of a transition metal cation and a
perchlorate anion

70. Using the general solubility rules in Section 6.5, state whether each of
these compounds is likely to be soluble or insoluble:
a. an ionic compound composed of an alkaline earth metal and the sulfide
ion
b. an ionic compound containing an aluminum cation and a halogen anion
c. an ionic compound containing a lead(II) cation and a halogen anion

71. Refer to the general solubility rules or Table 6.3 to determine whether each
of these compounds is soluble or insoluble in water:
a. LiOH
b. K2S
c. CaSO4

402
 d. zinc sulfate
e. copper(II) nitrate
f. Fe(C2H3O2)3

72. Refer to the general solubility rules or Table 6.3 to determine whether each
of these compounds is soluble or insoluble in water:
a. NH4OH
b. BaCl2
c. FePO4
d. iron(II) chloride
e. lead(II) chloride
f. aluminum sulfide

73. In general, which would you expect to be more water soluble: a salt with a
+1 cation and a –2 anion, or a salt with a +2 cation and a –2 anion?

74. Which would you expect to be more soluble, potassium nitrite (KNO2) or
aluminum oxide (Al2O3)? Why?

75. These equations represent precipitation reactions. Rewrite them as

complete ionic equations.
a. AgNO3 (aq) + KCl (aq) → KNO3 (aq) + AgCl (s)
b. Ba(ClO4)2 (aq) + K2SO4 (aq) → BaSO4 (s) + 2 KClO4 (aq)

76. These equations represent precipitation reactions. Rewrite them as

complete ionic equations.
a. MgBr2 (aq) + Pb(ClO3)2 (aq) → PbBr2 (s) + Mg(ClO3)2 (aq)
b. K3PO4 (aq) + AlCl3 (aq) → 3 KCl (aq) + AlPO4 (s)

77. Identify the spectator ions in this ionic equation:

Ca2+ (aq) + 2 NO3− (aq) + 2 Cs+ (aq) + CO32– (aq) → CaCO3 (s) + 2 Cs+
(aq) + 2 NO3− (aq)

78. Identify the spectator ions in this ionic equation:

2 NH4+ (aq) + S2– (aq) + Zn2+ (aq) + 2 ClO3− (aq) → 2 NH4+ (aq) + 2 ClO3−
(aq) + ZnS (s)

403
79. Identify the spectator ions in this equation, and rewrite it as a net ionic
equation:
Na+ (aq) + Br– (aq) + Ag+ (aq) + NO3−(aq) → AgBr (s) + Na+ (aq) + NO3−
(aq)

80. Identify the spectator ions in this equation, and rewrite it as a net ionic
equation:
Fe2+ (aq) + 2 NO3– (aq) + 2 K+ (aq) + 2 OH− (aq) → Fe(OH)2 (s) + 2 K+ (aq)
+ 2 NO3− (aq)

81. Aqueous solutions of potassium hydroxide and iron chloride react in a

precipitation reaction to form insoluble iron(II) hydroxide, as shown here:
2 KOH (aq) + FeCl2 (aq) → 2 KCl (aq) + Fe(OH)2 (s)

a. Rewrite this equation as an ionic equation.

b. What are the spectator ions in this equation?
c. Remove the spectator ions, and rewrite this as a net ionic equation.
d. What is the driving force for this reaction?

82. Aqueous solutions of silver nitrate and potassium bromide react in a

precipitation reaction to form insoluble silver bromide, as shown here:
AgNO3 (aq) + KBr (aq) → KNO3 (aq) + AgBr (s)
a. Rewrite this equation as an ionic equation.
b. What are the spectator ions in this equation?
c. Remove the spectator ions, and rewrite this as a net ionic equation.
d. What is the driving force for this reaction?

83. Each of the following reactions results in one water-soluble product and
one precipitate. Complete and balance each reaction, and show phases to
indicate whether the products are aqueous or solid.
a. KCl (aq) + Pb(NO3)2 (aq) →
b. KOH (aq) + FeCl3 (aq) →
c. BaCl2 (aq) + K3PO4 (aq) →

404
84. Each of the following reactions results in one water-soluble product and
one precipitate. Complete and balance each reaction, and show phases to
indicate whether the products are aqueous or solid.
a. K2SO4 (aq) + FeBr3 (aq) →
b. ZnCl2 (aq) + AgC2H3O2 (aq) →
c. Pb(C2H3O2)2 (aq) + MgBr2 (aq) →

85. Write balanced equations for the following precipitation reactions,

including phase symbols to indicate the insoluble product. If no precipitate
forms, indicate “No reaction.”
a. Iron(III) acetate reacts with barium sulfide.
b. Lithium phosphate reacts with copper(II) sulfate.
c. Calcium iodide reacts with ammonium acetate.
d. Iron(II) sulfate reacts with barium hydroxide.

86. Write balanced equations for the following precipitation reactions,

including phase symbols to indicate the insoluble product. If no precipitate
forms, indicate “No reaction.”
a. Sodium chloride reacts with silver nitrate.
b. Lead(II) nitrate reacts with sodium iodide.
c. Ammonium phosphate reacts with aluminum chloride.
d. Barium hydroxide reacts with iron(III) chloride.

87. Write molecular equations, complete ionic equations, and net ionic
equations to describe the following precipitation reactions. Include phase
symbols.
a. reaction of aqueous lithium iodide with aqueous silver nitrate
b. reaction of aqueous lead(II) acetate with aqueous zinc sulfate
c. reaction of aqueous calcium iodide with aqueous lead(II) nitrate

88. Write molecular equations, complete ionic equations, and net ionic
equations to describe the following precipitation reactions. Include phase
symbols.
a. reaction of aqueous lead(II) nitrate with aqueous sodium bromide
b. reaction of aqueous iron(II) perchlorate with aqueous sodium carbonate
c. reaction of aqueous aluminum sulfate with aqueous lead(III) nitrate

89. What is the difference between an acid and a base?

405
90. What are the two common products of acid-base neutralization reactions?

91. Write ionic equations to show the dissociation of each of these acids:
a. HCl
b. HNO3

92. Write ionic equations to show the dissociation of each of these acids:
a. HBr
b. HNO2

93. Balance these neutralization reactions:

a. H2SO4 + NaOH → Na2SO4 + H2O
b. HNO3 + Al(OH)3 → Al(NO3)3 + H2O
c. H2SO4 + Cr(OH)3 → Cr2(SO4)3 + H2O

94. Balance these neutralization reactions:

a. HNO3 + Ba(OH)2 → Ba(NO3)2 + H2O
b. HF + Ba(OH)2 → BaF2 + H2O
c. NaOH + H3PO4 → Na3PO4 + H2O

95. Show the products from these neutralization reactions. Balance the
equations if necessary.
a. HCl + KOH →
b. H2SO4 + 2 LiOH →
c. Ba(OH)2 (aq) + 2 HNO2 (aq) →

96. Show the products from these neutralization reactions. Balance the
equations if necessary.
a. HBr + NaOH →
b. H3PO4 (aq) + 3 KOH (aq) →
c. Ba(OH)2 + H2SO4 →

97. Write a balanced, complete ionic equation showing the reaction of sodium
hydroxide with hydroiodic acid.

98. Write a balanced, complete ionic equation showing the neutralization of

sulfuric acid with potassium hydroxide.

406
99. Write a net ionic equation for a double-displacement reaction in which
lead(II) iodide is a precipitate.

100. Write a net ionic equation for the neutralization of any acid with a
hydroxide base.

Cumulative Questions

101. Classify the reactions below as synthesis, decomposition, single

displacement, double displacement, precipitation, acid-base
neutralization, oxidation-reduction, or combustion. Some reactions can be
classified with more than one of these terms.
a. Hg (l) + S (s) → HgS (s)
b. Fe (s) + Cu(C2H3O2)2 (aq) → Cu (s) + Fe(C2H3O2)2 (aq)
c. Pb(NO3)2 (aq) + 2 KI (aq) → PbI2 (s) + 2 KNO3 (aq)
d. 2 K (s) + 2 H2O (l) → 2 KOH (aq) + H2 (g)
e. H2SO4 (aq) + Ba(OH)2 (aq) → 2 H2O (l) + BaSO4 (s)
f. C8H16 (l) + 12 O2 (g) → 8 CO2 (g) + 16 H2O (g)

102. Classify the reactions below as synthesis, decomposition, single

displacement, double displacement, precipitation, acid-base
neutralization, oxidation-reduction, or combustion. Some reactions can be
classified with more than one of these terms.
a. Fe (s) + 2 AgC2H3O2 (aq) → Fe(C2H3O2)2 (aq) + 2 Ag (s)
b. AgNO3 (aq) + KCl (aq) → AgCl (s) + KNO3 (aq)
c. HBr (aq) + KOH (aq) → KBr (aq) + H2O (l)
d. 2 C2H2 (g) + 5 O2 (g) → 4 CO2 (g) + 2 H2O (g)
e. Fe (s) + 2 HCl (aq) → FeCl2 (aq) + H2 (g)
f. 2 K (s) + Br2 (g) → 2 KBr (s)

103. The reactions here draw from all the reaction types introduced in this
chapter. Predict the products, and balance each equation.
a. AgC2H3O2 (aq) + NaCl (aq) →
b. Na (s) + Cl2 (g) →
c. HCl (aq) + Ba(OH)2 (aq) →
d. C10H20 (l) + O2 (g) →

407
104. The reactions here draw from all the reaction types introduced in this
chapter. Predict the products, and balance each equation.
a. Co (s) + Br2 (l) →
b. MgBr2 (aq) + K3PO4 (aq) →
c. Mg (s) + O2 (g) →
d. LiOH (aq) + H3PO4 (aq) →

408
Chapter Seven
Mass Stoichiometry

Process Development
“Are we ready?”

Jared Fennell, Ph.D., looks around the room one more time. Seated at an array of
computer monitors, several technicians nod in agreement. Next to them, a team of
chemists and engineers watches anxiously. At the center of the immaculate room
sits a massive stainless steel container. In fact, only the top of the container is in
the room—most of it lies beneath the concrete floor. Pipes, hoses, and sensors feed
into the top of the container.
Jared takes a deep breath. “Okay,” he says, “Let’s go.” The team has spent
months working for this moment. Every detail of this performance has been
discussed, written out, and rehearsed. Everyone knows their roles.
One of the operators enters a command on the control panel. Slowly, the
container begins to fill. Over the next 48 hours, the team will monitor a huge
chemical reaction. Everything has to go according to plan. The safety of the team,
the success of the project, and a million-dollar investment depend on it.
Jared has an extraordinarily challenging job. He is a process chemist at Eli
Lilly, an Indiana-based pharmaceutical company that produces medicines for
diabetes, cancer, brain function, and other conditions. Process chemists develop
techniques for manufacturing materials on a very large scale. For example, Jared
and his team were recently tasked with manufacturing a new medicine. Drug
discovery chemists had produced about 25 grams of the compound. Jared’s job:
Scale the reaction up to 100,000 grams.
The scale of the reaction is not the only challenge. Because the team produces
medicine for human consumption, the manufacturing process must work the same
way every time. Every detail must be carefully considered: How much of each
reactant should be added? What is the best temperature for the reaction? How are
the reagents mixed together? For how long does the reaction go? Do these
reactions produce other substances? How can they be removed? Every parameter is
tested and retested, until Jared and his team are certain that every batch they
produce will be exactly the same (Figure 7.1).

409
Figure 7.1 (a) Using massive reaction vessels like these, (b) pharmaceutical companies
like Eli Lilly produce large amounts of vital medicines. (c) Jared Fennell is a process
chemist at Eli Lilly. (d) These process chemists are in the laboratory, preparing for a
large-scale reaction. (e) An operator tends to a reaction vessel before the start of a
manufacturing reaction.

In manufacturing, chemistry begins with a simple question: How much? For

example, how much product is needed? How much of each starting material is
required? In fact, the importance of this question extends well beyond
manufacturing—it applies to every aspect of chemistry. When you breathe, how
much oxygen does your body require? If you burn a gallon of gasoline while
driving your car, how much carbon dioxide have you produced? If you are cooking
a big pancake breakfast, how much mix do you need to feed 20 people? These
questions apply to very different situations, but each of them relates to the
fundamental chemical question, “How much?”
In this chapter, we’re going to tackle this question. We’ll learn how the masses
of atoms and compounds relate to the world around us and see how we can use

410
these relationships to understand the chemistry that takes place in the laboratory,
the factory, or the kitchen.

Intended Learning Outcomes

After completing this chapter and working the practice problems, you should be
able to:

7.1 Formula Mass and Percent Composition

Calculate the formula mass of a compound.
Calculate the percent composition (by mass) of elements in a compound.
Broadly describe how chemists measure formula mass and percent
composition.

7.2 Connecting Atomic Mass to Large-Scale Mass: The Mole Concept

Use the mole concept to relate masses on the atomic scale to masses on the
laboratory scale.
Convert between grams, moles, and atoms or molecules.

7.3 The Mole Concept in Balanced Equations

Apply the mole concept to solve stoichiometry problems, relating the
amounts of reagents and products in a chemical change.
Identify the limiting and excess reagents in chemical reactions.

7.4 Theoretical and Percent Yield

Differentiate between theoretical, actual, and percent yields.
Describe chemical and physical occurrences that can lead to an actual yield
that is less than the theoretical yield.
Using the theoretical and actual yields, correctly calculate the percent yield
for a chemical reaction.

411
7.1 Formula Mass and Percent Composition

Formula Mass
The masses of elements and compounds play a vital role in chemistry.
Chemists use mass to identify unknown compounds and to understand and
predict chemical reactions. As we saw in Chapter 3, the periodic table
shows the average mass for atoms of each element. When working with
compounds, we use these atomic masses to calculate the mass of a single
molecule or formula unit. This is called the formula mass. For example,
what is the formula mass of a water molecule? We know that a molecule
of water, H2O, contains two hydrogen atoms and one oxygen atom. Using
the masses from the periodic table (Figure 7.2), we calculate the mass of
water as follows:
Formula mass of H2O=2(1.01 u)+1(16.00 u) = 18.02 u

Figure 7.2 We use the average atomic masses from the periodic table to calculate
formula mass.

We found the formula mass by adding up the number and type of each
atom, using the masses given in the periodic table. Chemists sometimes
refer to the formula mass as the formula weight, or as the molecular mass

412
or molecular weight.

Example 7.1 Calculating the Formula Mass of a Compound

Potassium carbonate, K2CO3, is a common water-softening agent. What is
the formula mass of this compound?
To solve this problem, we add the atomic masses of each atom in the
formula unit. •

Formula mass of K2CO3= 2(39.10 u)+1(12.01 u)+3(16.00 u)= 138.21 u

Recall that we measure the mass of atoms in atomic mass units, abbreviated
as u or sometimes as amu.

IT
TRY

1. Find the formula mass for each of these compounds:

a. H2CO
b. Na2CO3
c. lithium nitrate
d. iron(II) bromide

Percent Composition
Chemists sometimes describe compounds by the percentage (by mass) of
each element in a compound. This is called the percent composition of a
compound. We calculate percent composition by dividing the mass of one
element in a compound by the total mass of a compound and then
converting to percent:
Percent composition of one element=mass of one elementmass of entire
compound×100%

Examples 7.2 and 7.3 illustrate this idea.

413
Example 7.2 Finding the Percent Composition of a Compound
Octane, a component of gasoline, has the molecular formula C8H18. What
is the percent composition of carbon and hydrogen in octane?
To solve this problem, we find the mass of the carbon and the hydrogen
atoms in the formula:
Mass carbon = mass (C8)= 8(12.01 u)= 96.08 uMass hydrogen = mass (H18)=
18(1.01 u)= 18.18 u

Next, we find the total formula mass of the compound:

Formula mass of C8H18= 8(12.01 u)+18(1.01 u)= 114.26 u

The percent by mass of each element is the mass of that element divided by
the total formula mass, expressed as a percentage. •

Percent carbon =mass carbontotal formula mass×100%=96.08 u114.26

u×100%=84.09%Percent hydrogen=mass hydrogentotal formula
mass×100%=18.18 u114.26 u×100%=15.91%

Example 7.3 Using Percent Composition

Silicon dioxide (SiO2) is the main component of glass. This compound
contains 46.75% silicon and 53.25% oxygen by mass. What mass of silicon
is in a 2.32-kilogram sample of SiO2?
To find the mass of silicon in the 2.32-kilogram sample, we multiply the
mass of the total sample by the percentage of silicon in the sample,
expressed as a decimal. •

Mass Si= total mass × percent silicon=2.32 kg×0.4675 = 1.08 kg

414
 IT
TRY

2. Acetaminophen is a common painkiller. This compound has a

molecular formula of C8H9NO2. What is the percentage by mass of
carbon, hydrogen, nitrogen, and oxygen in this compound?

3. Table sugar is a covalent compound with the formula C12H22O11.

What is the percentage by mass of carbon in table sugar? What mass
of carbon is present in 2.00 kg of table sugar?

Each year a team of scientists from the University of Alabama at Birmingham

collects and studies sponges from the waters around Antarctica. Some
compounds discovered in these sponges block the growth of cancer cells.
Chemists use the formula mass and percent composition to help determine the
structure of these compounds.

How Chemists Measure Formula Mass and Percent Composition

In Example 7.2, we began with the formula for a compound and then
calculated its formula mass and percent composition. However, chemists
often have the opposite problem: They encounter a new compound—
perhaps a natural product with exciting medicinal properties—and must
determine its structure and composition. Both formula mass and percent

415
composition are important in identifying unknown compounds. In this
section, we’ll briefly look at how chemists measure these important
properties.
Mass Spectrometry
Chemists measure the masses of compounds using a technique called mass
spectrometry. A mass spectrometer contains three vital components: an
ionizing chamber, an electric field, and a detector (Figure 7.3). The
ionizing chamber converts neutral molecules into ions (often by removing
an electron). The ions are then accelerated into an electric field, which
causes the pathways of the charged particles to curve. Lighter particles
curve more sharply than heavier particles. After exiting the electric field,
the particles strike a detector. The spectrometer calculates the mass of the
compound based on the size of the electric field, the charge on the particle,
and the exact point where the particle hits the detector.

Figure 7.3 In a mass spectrometer, a molecule passes through the ionizing chamber and
then into an electric field. The path of the charged molecule curves as it moves through
the electric field. The spectrometer calculates the mass of the molecule based on how
sharply the molecule’s path changes as it passes through the electric field.

416
 Explore

Figure 7.3

Mass spectrometry is routinely used in chemical analysis (Figure 7.4).

For example, forensic chemists use mass spectrometry to monitor for
hazardous or illegal substances. Figure 7.5 shows the mass spectrum of a
sample taken from a dollar bill. Cocaine (C17H22NO4) has a formula mass
of 304.2 u—the signal at this mass indicates the presence of cocaine on the
surface of the bill.

Figure 7.4 A forensic chemist uses a mass spectrometer to analyze a mixture.

417
Figure 7.5 This mass spectrum was taken from the surface of a dollar bill. The peak at
304.2 indicates the presence of cocaine. (Data courtesy of JEOL USA.)

Elemental Analysis
Chemists measure percent composition through a technique called
elemental analysis. This technique uses combustion reactions to convert
compounds into simpler products (carbon dioxide, water, etc.). The mass
of each product indicates the percentage of each element in a sample
(Figure 7.6). Using a combination of elemental analysis and mass
spectrometry, chemists can often determine both the percent composition
and molecular formula of an unknown compound.

Figure 7.6 Elemental analysis uses combustion to separate and measure the percentage
of different elements in a compound. This image shows (a) a chemist preparing the
sample and (b) the sample being slid into the combustion chamber.

418
7.2 Connecting Atomic Mass to Large-Scale Mass: The
Mole Concept

Avogadro’s Number and the Mole

So far, we’ve expressed the masses of atoms and molecules using atomic
mass units. While this measure is useful for calculating formula mass or
percent composition, it presents a problem: Atomic mass units are
incredibly small (1 u = 1.66 × 10–24 grams), far too small to be useful in
laboratory measurements. So how can we relate these masses to larger
amounts—such as grams, kilograms, or tons—that can be measured in the
lab or the factory (Figure 7.7)?

Figure 7.7 Manufacturers use large-scale batch reactors like the ones shown here. It is
important to relate reactions between a few atoms to reactions of this scale.

To answer this question, we’re going to use a quantity called a mole

(often abbreviated as mol). A mole is a number of items, similar to the
term dozen. A dozen simply means a quantity of 12, regardless of the item.
If you have 12 planets, you have a dozen planets. If you have 12
toothpicks, you have a dozen toothpicks. If you have 12 donuts, you have a
dozen donuts. It doesn’t matter what you’re measuring; the term dozen
refers to the quantity 12.
Like the term dozen, a mole describes the number of units present. A
mole is 6.02 × 1023 units of anything. If you have 6.02 × 1023 donuts, you
have a mole of donuts. If you have 6.02 × 1023 carbon atoms, you have a
mole of carbon atoms. If you have 6.02 × 1023 oxygen molecules, you
have a mole of oxygen molecules. We refer to the number of particles in a
mole, 6.02 × 1023, as Avogadro’s number.

419
One mole = 6.02 × 1023 units.

The power of the mole concept is that it allows us to express the

masses of atoms and compounds in grams. Here’s how it works: One atom
of carbon has a mass of 12.01 atomic mass units. One mole of carbon has a
mass of 12.01 grams. We can’t measure out one atom of carbon in a lab,
but we can measure out 12.01 grams (Figure 7.8). So, we describe the
atomic mass in one of two ways: We can say that the mass of carbon is
12.01 u (atomic scale), or we can say that the mass of carbon is 12.01
grams per mole (laboratory scale). When a mass is expressed in grams per
mole, it is referred to as the molar mass.

Figure 7.8 The mole concept connects atoms to grams.

We describe mass in atomic mass units (u) or in grams per mole (g/mol).

Let’s consider one more example: The mass of one molecule of CO2 is
44.01 u, and the mass of one mole of CO2 is 44.01 grams. So, the molar
mass of carbon dioxide is 44.01 g/mol. Using moles connects the atomic
world to a scale we can measure in the lab. Figure 7.9 shows one mole of
several different elements.

420
Figure 7.9 This image shows one mole of the elements sulfur, bromine, carbon,
mercury, zinc, copper, and cobalt.

Use the molar mass to convert between grams and moles.

Converting between Grams, Moles, and Particles

In chemistry we frequently need to convert between grams and moles. To
do this we use the molar mass as the conversion factor. For example, how
many moles of sodium chloride are present in a 305-gram sample? Based
on the formula mass of NaCl, we know that 58.44 grams of NaCl = 1 mole
of NaCl. We use this as our conversion factor:

We often need to convert from moles to grams. For example, to

prepare a solution that contains 1.20 moles of NaCl, how many grams of
NaCl do we need? In this example, we again use the molar mass as the
conversion factor:

Notice that when we write the conversion factor, the unit we wish to
cancel (grams or moles) goes on the bottom, and the unit we wish to keep
goes on the top. This concept is very important, and you must be able to do
these calculations comfortably. Examples 7.4 and 7.5 further illustrate this
idea.

421
 Example 7.4 Converting from Grams to Moles
How many moles of aluminum oxide, Al2O3, are present in a 57.3-gram
sample of this compound?
We first calculate the formula mass of Al2O3 and find that 101.96 grams of
Al2O3 = 1 mole of Al2O3. We use this relationship as a conversion factor.
We place the unit we wish to keep (moles) on top and the unit we wish to
cancel (grams) on the bottom. Rounding to the correct number of
significant digits, we get a final answer of 0.562 moles of Al2O3. •

Example 7.5 Converting from Moles to Grams

A sample contains 0.281 moles of potassium sulfate, K2SO4. What is the
mass of potassium sulfate in this sample?
As in Example 7.4, we use the molar mass to convert between grams and
moles. We first determine that the molar mass of K2SO4 is 174.26
grams/mole. As before, we use this mass as a conversion factor. Because
we want to move from moles to grams, we write the conversion factor with
grams on top and moles on the bottom. Rounding to the correct number of
significant figures gives us a final answer of 49.0 g of K2SO4. •

IT
TRY

4. Convert the following amounts from grams to moles:

a. 22.7 g of silver
b. 43.6 g of sodium nitrate
c. 974.3 g of boron trifluoride

5. Find the mass of each of the following:

422
 a. 6.20 moles of silicon
b. 5.3 moles of molecular bromine (Br2)
c. 1.83 moles of ammonium carbonate

Because we know the number of particles in a mole, we can also

convert between moles and particles, such as atoms or molecules. To do
this we use Avogadro’s number as the conversion factor. For example,
how many atoms are in 4.20 moles of gold? To solve this problem, we
multiply the moles of gold by Avogadro’s number:

Use Avogadro’s number to convert between moles and particles.

The mole concept allows us to directly relate the quantity of atoms and
the mass in grams. For example, what is the mass in grams of 2.53 × 1023
iron atoms? To solve this problem, we begin with atoms, then convert to
moles, and finally convert to grams:

Notice in these problems that we use Avogadro’s number to convert

between moles and atoms, and we use the molar mass to convert between
moles and grams. To relate grams to atoms, we must first convert to moles
(Figure 7.10).

Figure 7.10 To convert between grams and atoms, you must go through moles.

Example 7.6 Converting from Grams to Molecules

A chemist has a sample containing 42,300 grams of sulfur trioxide, SO3.
How many molecules of SO3 are present in this sample?
To solve this problem, we must first convert from grams to moles (as in the

423
 previous examples) and then convert to molecules. To convert from grams
to moles, we use the molar mass of SO3 (80.06 g/mol) as the conversion
factor. We then use Avogadro’s number to convert from moles to
molecules. Rounding to three significant figures, we find that this mass
corresponds to 3.18 × 1026 molecules of SO3. •

IT
TRY

6. A sample contains 0.0623 moles of titanium. How many titanium

atoms are present in this sample? What is the mass in grams of this
sample?

7. Find the mass of each of the following:

a. 1.83 × 1024 iron atoms
b. 3.21 × 1021 CH4 molecules

8. How many molecules of CS2 are in a 15.83-gram sample?

424
I live in the corn country of west Kentucky. Planes flying from our small
regional airport go only to Chicago O’Hare. I can go anywhere I want, but I
must go through Chicago. Like O’Hare, moles are a travel hub in many
calculations. If you want to connect one substance to another, you have to travel
through moles.

425
7.3 The Mole Concept in Balanced Equations
In Chapter 6 we discussed the explosive reaction between hydrogen and
oxygen:

2 H2+O2 →2 H2O

A balanced equation gives both an atomic ratio and a mole ratio.

We can read this equation in one of two ways. First, we can interpret it in a
molecular sense: Two molecules of hydrogen react with one molecule of
oxygen. But we can also interpret this equation as a ratio of moles: Two
moles of hydrogen react with one mole of oxygen (Figure 7.11).

Figure 7.11 A balanced equation represents both the ratio of molecules and the ratio of
moles that are involved in a reaction.

426
 Explore

Figure 7.11

In 1937 the airship Hindenburg caught fire and exploded over New Jersey,
killing 36 people. The airship was filled with hydrogen gas, which reacted
explosively with the oxygen gas in the air.

Stoichiometry Problems
Because a balanced equation relates the moles of substances in a reaction,
we can use it to connect the amount of one reactant or product to the
amounts of every other species in the equation. For example, Figure 7.12
poses a series of questions about a reaction. See if you can answer each
question; then look at the explanations that follow.

427
Figure 7.12 See if you can answer the questions in this figure. The answer to each
question is given in the text below.

Explore

Figure 7.12

Here are the solutions to each of the four questions:

1. How many moles of iron react? To answer this question, we need to
convert from grams of iron to moles of iron. As before, we use the molar
mass of iron (55.85 g/mole) as a conversion factor:

428
 2. How many moles of HCl are needed to react with the iron? Based
on the balanced equation, two moles of HCl are required for every one
mole of Fe. Because we have 4.21 moles of Fe, this means we need twice
that many moles of HCl, or 8.42 moles. More formally, we could solve
this problem by unit conversions:

Notice how we did this: We created a conversion factor (2 moles

HCl/1 mole Fe) from the balanced equation. When relating the moles of
one substance to the moles of another, we use the coefficients in the
balanced equation as a conversion factor.

Use the coefficients in the balanced equation to convert between moles of

compounds.

3. How many moles of iron(II) chloride form in this reaction? In the

balanced equation, there is one mole of FeCl2 per one mole of Fe.
Therefore the moles of FeCl2 produced are the same as the moles of iron
that react (4.21 moles). Again, we can do this more formally by writing a
conversion factor from the balanced equation (1 mole FeCl2/1 mole Fe):

4. How many grams of iron(II) chloride form in this reaction? In this

final question, we have to convert from moles of FeCl2 to grams of FeCl2.
Anytime we convert between grams and moles, we use the molar mass of
the substance in question. Based on the masses on the periodic table, the
molar mass of FeCl2 is 126.75 grams/mole. Because we know there are
4.21 moles of FeCl2 (from question 3), we can set up the final problem
like this:

This type of problem is called a stoichiometry problem. These problems

use the amount of one substance to predict the amount of another
substance that is consumed or produced, according to the balanced
equation. Once we found the moles of Fe (in question 1), we were able to
relate this value to both the moles of HCl and the moles and grams of

429
FeCl2. Given the amount in grams or moles of one substance in the
balanced equation, we can find the moles or grams of any other substance
in the equation. Examples 7.7 and 7.8 further illustrate this idea.

We can relate the amount of one substance to any other substance in a

balanced equation.

Example 7.7 Relating Moles Using the Balanced Equation

When magnesium burns, it combines with oxygen to form a new compound,
MgO. The balanced equation for this reaction is shown below. If this
reaction consumes 3.0 moles of oxygen, how many moles of MgO will
form? How many grams of MgO will form?

2 Mg (s)+O2(g)→2 MgO (s)

There are two ways to approach the first part of this problem. One way is
by visual inspection: From the balanced equation, we see that 1 mole of
oxygen gas produces 2 moles of magnesium oxide. Based on this ratio, 3.0
moles of oxygen gas will produce 6.0 moles of MgO.
For simple ratios, this sort of visual inspection is very helpful. But we can
also solve the problem by using the balanced equation to create a
conversion factor. We begin with 3.0 moles of oxygen gas and then
multiply by a conversion factor, showing two moles of MgO for every one
mole of oxygen gas:

Once we have found the moles of MgO, we use the molar mass of MgO
(40.31 g/mol) to convert from moles to grams. The calculator gives an
answer of 241.9 grams; but we can keep only two significant digits, so we
round the answer to 240 grams. •

Example 7.8 Relating Grams and Moles Using the Balanced

430
 Equation
Sodium metal reacts violently with water, as shown in the balanced
equation below. How many moles of H2 gas are produced by the reaction
of 11.0 grams of sodium with water?
2 Na (s)+2 H2O (l)→2 NaOH (aq)+H2(g)

We are given the grams of sodium and asked to relate this amount to the
moles of H2. To solve this problem, we must first convert the grams of
sodium to moles of sodium and then relate the moles of sodium to the
moles of hydrogen. Graphically, we can express our strategy this way:
Grams Na⇒Moles Na⇒Moles H2

To convert grams of sodium to moles of sodium, we use the molar mass of

sodium: 22.99 g/mol. In order to convert moles of sodium to moles of H2,
we use the balanced equation to relate these two species (1 mole H2/2
moles Na). We could write this process in two individual steps:

For simplicity, we can also write the two steps side by side. •

IT
TRY

9. Lithium metal reacts with nitrogen gas according to the balanced

equation shown here. Based on this equation, how many moles of N2
are required to react with 12.0 moles of lithium? How many moles of
Li3N are produced in this reaction? How many grams of Li3N are
produced?

6 Li (s)+N2(g)→2 Li3N (s)

431
10. Sulfur dioxide can react with oxygen gas as shown in this balanced
equation:

2 SO2 (g)+O2 (g)→2 SO3 (g)

How many moles of SO2 are required to produce 40 moles of SO3?

How many moles of O2 are required?

Gram-to-Gram Questions
One common type of stoichiometry problem is the gram-to-gram problem.
In this type of problem, we are given the mass of one reagent or product
(usually in grams) and asked to find the mass of another reagent or product
(also in grams). Since the balanced equation relates amounts by moles, we
must convert to moles to solve this problem. For any question relating the
grams of substance A to the grams of substance B, the strategy is to
convert from grams of A to moles of A, then to moles of B, then finally to
grams of B. This approach is shown graphically in Figure 7.13. Example
7.9 describes how we used this strategy to solve a problem of this type.

Figure 7.13 We use this strategy for solving gram-to-gram stoichiometry problems.

Example 7.9 Gram-to-Gram Conversion

When heated with a Bunsen burner, MgCO3 decomposes to MgO and CO2,
as shown in the equation. If 5.24 grams of MgCO3 are heated in this
manner, how many grams of MgO can be produced?
MgCO3(s)→MgO (s)+CO2(g)

To solve this problem, we must first convert from grams of MgCO3 to

moles of MgCO3. Next, we use the balanced equation to relate the moles of
MgCO3 to the moles of MgO. Finally, we convert moles of MgO into

432
 grams. We can do one conversion at a time, or we can string these
conversions together, like this. •

IT
TRY

11. Consider the reaction of copper with bromine to produce copper(I)

bromide:

2 Cu (s)+Br2(l)→2 CuBr (s)

How many grams of copper(I) bromide are produced from the reaction
of 24.6 grams of copper in this reaction? How many grams of bromine
are required to react with this amount of copper?

Strategies for Solving Stoichiometry Problems

For many people, stoichiometry problems can seem overwhelming. Don’t
worry—you can do this. The key to solving these problems is to recognize
the simple patterns; then follow the patterns every time. Notice that these
problems rely on just three types of conversions, each with its own
conversion factor.

Conversion Type Conversion Factor

1. Grams and moles of one substance Molar mass
2. Moles and particles of one Avogadro’s number
substance
3. Moles of two different substances Mole ratio from the balanced
equation

Once we understand the conversions, we need to identify where we are

starting from and where we are going. Stoichiometry problems provide the
amount of one compound (A), and ask us to relate this to the amount of
another compound (B). These problems can be solved by using the “mole

433
map” shown in Figure 7.14. Like any map, this guide helps us locate
where we are, where we are going, and how to get there. For example, how
do we get from grams of compound A to particles of compound B?
Looking at the mole map, we can see this requires three steps: Beginning
with grams of A, we must convert to moles of A, then to moles of B, and
finally to particles of B. Examples 7.10 and 7.11 illustrate this type of
problem.

Figure 7.14 This “mole map” shows the strategy for solving any stoichiometry problem
involving grams, moles, and atoms or molecules.

If you’re driving along I-95 in southern Connecticut, how do you get

from Norwalk to New Haven? According to the map, you have to drive
through Bridgeport. Figure 7.14 gives you a similar map for solving
stoichiometry problems. You can start anywhere and end anywhere—
just follow the map. For most stoichiometry problems, the map says you
have to go through moles.

Practice
Stoichiometry conversions
The way to get good at stoichiometry conversions is to practice. This
interactive exercise will help you improve your skills.

434
Example 7.10 Strategies for Stoichiometry Problems
Zinc metal reacts with aqueous copper(II) chloride, as shown in this
equation. If 3.0 × 1021 atoms of zinc react, how many grams of ZnCl2 will
form? Show the sequence of conversions necessary; then calculate the
numerical answer.
Zn (s)+CuCl2(aq)→ZnCl2(aq)+Cu (s)

We can look at the map in Figure 7.13 to solve this problem. We know the
number of atoms of Zn, and we are asked to find the mass of ZnCl2. By
replacing A in the map with Zn and B with ZnCl2, we see that we can do
the calculation by taking this route:
atoms Zn⇒moles Zn⇒moles ZnCl2 ⇒grams ZnCl2

Now we set up the conversions to move from atoms of Zn to grams of

ZnCl2:

After rounding to the correct number of significant digits, we find that this
reaction can produce 0.68 g of ZnCl2. •

Example 7.11 Strategies for Stoichiometry Problems

Aluminum metal reacts with nitric acid (HNO3) as in the reaction below. If
this reaction consumes 5.20 moles of aluminum, how many grams of H2
will form?
2 Al (s)+6 HNO3(aq)→2 Al(NO3)3(aq)+3 H2(g)

This example gives us the number of moles of Al and asks for the number
of grams of H2. Following the map in Figure 7.13, we can write a strategy
for this problem:
moles Al⇒moles H2 ⇒grams H2

435
 Now that the route is mapped out, we can do the conversions. •

IT
TRY

12. Iron reacts with oxygen gas as shown in this balanced equation. How
many moles of oxygen are needed to react with 82.1 grams of iron?

4 Fe (s)+3 O2(g)→2 Fe2O3(s)

13. When heated, calcium carbonate decomposes to form calcium oxide

and carbon dioxide, as shown in this equation. How many moles of
CaCO3 are required to produce 71.5 grams of CaO?

CaCO3(s)→CaO (s)+CO2(g)

Calculations with Limiting Reagents

Before going any further, let’s stop and make a sandwich. We can write a
“recipe” for a sandwich that looks something like this:
2 slices of bread+1 slice turkey+1 slice cheese→1 sandwich

Now imagine that you have 18 slices of turkey, 15 slices of cheese, and 80
slices of bread. Using this recipe, how many turkey-and-cheese
sandwiches could you make?
If you answered 15 sandwiches, you are correct. You have plenty of
bread—enough to make 40 sandwiches. You have enough turkey to make
18 sandwiches. But you only have enough cheese to make 15 sandwiches.
Because the cheese runs out first, it limits the number of sandwiches that
can be made.

436
 This situation is common in chemistry as well as in cooking. For
example, consider the combustion of natural gas, as shown in this reaction:
CH4(g)+2 O2 (g)→CO2(g)+2 H2O (g)

What happens if there is more O2 than CH4 available to react? The

reaction will take place until all of the CH4 is consumed. However, there
will still be O2 left over (Figure 7.15). Because an excess of oxygen is
available, O2 is called the excess reagent. On the other hand, the amount
of CH4 present limits the amount of CO2 and water that can be produced
in this reaction. The compound that is consumed first is called the limiting
reagent. The limiting reagent determines the amount of product that can
form in a reaction.

Figure 7.15 The limiting reagent (CH4) is completely consumed, but the excess reagent
(O2) is not.

Often, just like in the sandwich example, we are given amounts of

different starting materials. To find the limiting reagent, we calculate the
amount of product that each starting material can produce. The starting
material that can produce the least amount of product is the limiting
reagent. Examples 7.12 and 7.13 illustrate this important idea.

The limiting reagent is the one that can form the least amount of product.

Example 7.12 Finding the Limiting Reagent

Potassium reacts violently with chlorine gas to produce potassium chloride,
as in the equation below. If 1.2 moles of potassium are combined with 15

437
moles of chlorine gas, how many moles of potassium chloride can form?
Which reagent is the limiting reagent?

2 K (s)+Cl2(g)→2 KCl (s)

To solve this problem, we determine how much potassium chloride can
form from the available amount of potassium and the available amount of
chlorine:

The potassium limits the amount of KCl that can be produced—it is the
limiting reagent. Even though there is enough chlorine to produce a large
amount of KCl, there is only enough potassium to produce 1.2 moles. •

Example 7.13 Finding the Limiting Reagent

Uranium reacts with fluorine gas according to the equation shown. If 30
moles of uranium combine with 75 moles of F2, how many moles of UF6
will form?

U+3 F2 →UF6
To find the limiting reagent, we calculate the amount of UF6 that could be
produced from the uranium and the amount that could be produced from the
fluorine:

Notice that even though fewer moles of uranium are present, fluorine is the
limiting reagent. Because the reaction consumes three fluorine molecules
for each uranium atom, the fluorine is consumed before the uranium. •

438
 IT
TRY

14. Phosphorus reacts with chlorine gas according to this equation:

2 P (s)+3 Cl2 (g)→2 PCl3 (l)

If 25.0 moles of phosphorus are combined with 50.0 moles of chlorine

gas, what is the limiting reagent? How many moles of PCl3 can be
produced?

Practice
Limiting reagents
This interactive exercise will help you improve your skills.

Finding the Leftovers

In the previous section we used a sandwich recipe as an analogy for a
chemical reaction:
2 slices of bread+1 slice turkey+1 slice cheese→1 sandwich

Based on this equation, we said that if we have 18 slices of turkey, 15

slices of cheese, and 80 slices of bread, we can make 15 sandwiches. The
cheese is the limiting reagent. But here’s another question: If we make all
of the sandwiches, how many slices of turkey and of bread will be left
over? I encourage you to take a moment and try to figure this out before
moving forward.

I love sandwiches.

439
 To solve this problem, we first find the amount of each reagent
(ingredient) that we used. Following the ratios in the balanced equation,
we could summarize our sandwich making as follows:
Used: 30 slices of bread, 15 slices of turkey, 15 slices of cheese
Produced: 15 sandwiches
To find the leftovers, we subtract what we used from our initial amounts:
Bread: 80 slices – 30 slices = 50 slices left over
Turkey: 18 slices – 15 slices = 3 slices left over
Cheese: 15 slices – 15 slices = 0 slices left over (limiting reagent)
Chemical reactions work the same way: If we subtract the amount of a
starting material used in the reaction from the amount available, we can
determine the amounts of starting material that are left over. Limiting
reagents are completely consumed in chemical reactions, but some of the
excess reagent will always be left over. Example 7.14 illustrates this idea.

Example 7.14 Finding the Amounts Left Over

In an acid-base neutralization reaction, a chemist combines 15 moles of
HCl with 20 moles of NaOH. How many moles of H2O form in this
reaction? How many moles of the excess reagent are left over?
To solve this problem, we first write a balanced equation for the reaction:
HCl (aq)+NaOH (aq)→NaCl (aq)+H2O (l)

We have enough NaOH to produce 20 moles of water, but we only have

enough HCl to produce 15 moles of water. Therefore HCl is the limiting
reagent, and the reaction produces 15 moles of water.
How many moles of NaOH are left over? We started with 20 moles of
NaOH and consumed 15 in the reaction. Therefore 5 moles are left over.
Sometimes it helps to make a table showing the moles of each species
present at the beginning of the reaction, the amounts consumed or
produced, and then the amounts of each species at the end of the reaction. •

440
 IT
TRY

15. A reinforced steel cylinder is filled with 20.0 moles of hydrogen and
12.0 moles of oxygen. The mixture is detonated inside the cylinder,
causing this reaction to take place:

2 H2(g)+O2(g)→2 H2O (g)

What is the limiting reagent for this reaction? Calculate the number of
moles of hydrogen, oxygen, and water present in the cylinder after the
reaction takes place.

441
7.4 Theoretical and Percent Yield
Many chemists, like the process chemist described at the beginning of this
chapter, conduct chemical reactions to create valuable new materials.
When carrying out a reaction of this type, we often calculate the
theoretical yield of the products—that is, the amount of product that can
form, based on the amount of starting materials and the balanced equation.

Stoichiometry problems predict the theoretical yield of a reaction.

In practice, however, chemists often obtain less than the theoretical

yield. The amount of a product that a chemist actually isolates from an
experiment is called the actual yield. We report the efficiency of a
reaction as the percent yield, which we calculate as follows:
Percent yield=actual yieldtheoretical yield×100%

Percent yields are often much lower than 100%. This happens for
several reasons: Sometimes part of the material adheres to the container
walls. (If you’ve ever made a cake, you know that it’s impossible to get all
of the batter into the pan—the same holds true in chemical reactions.)
Sometimes a competing reaction forms unwanted side products in a
reaction. Other times, product is lost during the purification process. The
percent yield is a measure of how efficiently a reaction takes place.

Due to the scale and cost of chemical manufacturing, industrial chemists work
hard to optimize the percent yields of reactions. A change in yields of even 1%
can drastically reduce costs and increase profits.

For example, suppose a chemist runs a reaction in which the theoretical

yield is 240 grams. However, after the completion of the reaction, he is
only able to isolate 180 grams. In this case, his percent yield for the
reaction is 75%:

Example 7.15 illustrates the relationships between theoretical, actual, and

percent yields.

442
 Example 7.15 Finding the Theoretical and Percent Yield
Sulfur hexafluoride, SF6, is widely used in the power industry. It is
produced through the following reaction:

S (s)+3 F2 (g)→SF6 (g)

A manufacturer reacts 120.0 kilograms of sulfur with excess fluorine gas.
What mass of SF6 is theoretically possible for this conversion? After the
reaction is complete, the manufacturer isolates 480.2 kg of SF6. What was
the percent yield for this process?
The theoretical yield is the amount of product possible, based on the
balanced equation and the amount of starting material used. This is a
stoichiometry problem. As in the earlier examples, we must go through
moles to relate the mass of the starting materials and products:

The theoretical yield for this reaction is 546.5 kg. The actual yield is 480.2
kg. From this, we calculate the percent yield. •

IT
TRY

16. A chemist carried out the following reaction, beginning with 5.25
grams of MgCO3. After the reaction was complete, she obtained 2.37
grams of MgO.
MgCO3(s)→MgO (s)+CO2(g)

a. What was the theoretical yield of magnesium oxide (MgO) in

this reaction?
b. What was the percent yield?

443
444
Summary
Much of chemistry is concerned with the question, “How much?” When
conducting a chemical reaction, chemists need to know how much starting
material to use and how much product to expect. In this chapter we’ve
explored how the atomic masses relate to the masses we can measure in
the laboratory or the factory.
We describe compounds by their formula mass. The formula mass is
the sum of the masses of the atoms in a molecule or formula unit.
Scientists often determine formula mass experimentally by using mass
spectrometry.
We can also describe compounds by their percent composition.
Scientists determine percent composition using a technique called
elemental analysis.
The mole concept allows us to connect the masses of atoms and
molecules to masses that we can measure on the laboratory scale.
Avogadro’s number, 6.02 × 1023, is the number of particles in a mole. The
mass of one mole of a substance in grams has the same numerical value as
the mass of one atom or molecule of that substance in atomic mass units.
That is, the mass of any chemical substance can be described in atomic
mass units or in grams per mole. The measurement of mass in grams per
mole is called the molar mass.
The mole concept allows us to relate large amounts using chemical
equations. By converting to moles, it is possible to relate the amount of
any material involved in a chemical reaction to the amounts of all other
materials consumed or produced in the reaction.
Frequently, we are not able to mix chemicals in a perfect ratio; one
reagent will be in excess. The limiting reagent determines the amount of
product that can form in a reaction.
Chemists commonly describe reactions in terms of their theoretical and
percent yields. The theoretical yield is the amount of material that would
be produced if the reaction went perfectly. However, in practice, material
can be lost because of competing reactions, incomplete transfers between
containers, or other issues. The percent yield is an indicator of how closely
the actual yield aligns with the theoretical yield. When working in a
laboratory or industrial setting, the percent yield is very important.

445
Process Development, Continued

At the beginning of this chapter, we introduced Jared Fennell, a process chemist at

the pharmaceutical company Eli Lilly, who optimizes chemical reactions for large-
scale manufacturing. Process chemistry is much different from the earlier stages of
pharmaceutical research. In the early phases of a drug development project,
scientists try to identify compounds with useful medicinal properties. Chemists
prepare and test a large number of compounds to understand how different
structures behave in living systems. They don’t worry much about percent yield;
they need only a small amount of each compound, and good yields are often less
important than getting the compounds quickly. However, as compounds move
from laboratory testing to animal testing and ultimately to human trials, more and
more compound is needed, and the purity of the compound becomes paramount.
This is where process chemists come in.
In 2014 the Eli Lilly team published results that offer an insight into the
challenges that process chemists face. Their project objective was to prepare a
promising new medicine. One of the final steps involved a reaction that combined
two molecules by forming a new bond to a nitrogen atom (Figure 7.16). Chemists
in the early and intermediate stages had produced 25 grams of the compound; the
process team had to produce 100,000 grams.

Figure 7.16 A simplified representation of the bond-forming reaction. Shaded areas

represent the rest of the molecules.

To make this reaction work, small-scale chemists used a palladium compound

as a catalyst (a catalyst helps a reaction take place more quickly but is not used up
in the reaction). Palladium is very expensive and must be separated from the
products after the reaction is complete. In addition, the catalyst slowly reacted with
oxygen. This caused two problems: First, it produced potentially harmful side
products. Second, trace amounts of oxygen in the mixture could cause the catalyst
to fail, resulting in very low yields and costing the company hundreds of thousands
of dollars.
The process team set to work on the problem. They carefully tested the time,
temperature, amounts of starting materials, amounts of catalyst, and mixing
techniques used in the reaction. They identified the side products. They even
developed a unique method to monitor the oxygen levels in the reaction, ensuring
that they stayed below the harmful threshold.

446
 Finally, after months of testing and planning, the time came for a large-scale
run. This pivotal experiment required 150 kilograms of starting material, valued at
nearly $1 million. Everything had to be right. The entire team assembled. Every
step of the process had been rehearsed. The reaction started. Over the next 48
hours, the team monitored each step, watching for unexpected changes. At the end
of the arduous process, they were rewarded with success. They isolated the pure
product at 87% yield—a very good yield for a challenging reaction.

447
 Key Terms
7.1 Formula Mass and Percent Composition
formula mass The mass of a molecule or formula unit.
percent composition The percentage (by mass) of each element in a
compound.
mass spectrometry A technique used to measure the mass of a molecule.
elemental analysis A technique used to determine the percent composition of
a substance.

7.2 Connecting Atomic Mass to Large-Scale Mass: The Mole Concept

mole A quantity consisting of 6.02 × 1023 units.
Avogadro’s number The number of particles in a mole; 6.02 × 1023.
molar mass The formula mass of an element or compound, expressed in
grams per mole.

7.3 The Mole Concept in Balanced Equations

stoichiometry problem A problem that relates the amount of one reagent or
product to another in a chemical reaction, using a balanced equation.
excess reagent In a chemical reaction, a reagent that is present in larger
stoichiometric quantities than the other reagents; an excess reagent is not
completely consumed.
limiting reagent In a chemical reaction, a reagent that is completely consumed
and limits the amount of product that can form.

7.4 Theoretical and Percent Yield

theoretical yield The amount of product that can form in a chemical reaction,
based on the balanced equation and the amount of starting materials present.
actual yield The amount that a chemist actually recovers from an experiment.
percent yield A measure of the efficiency of a reaction; the actual yield
divided by the theoretical yield, expressed as a percentage.

448
 Additional Problems

7.1 Formula Mass and Percent Composition

17. Calculate the formula mass for each of these compounds:

a. BaSO4
b. K2S
c. C6H12O6
d. FeCl3

18. Calculate the formula mass for each of these compounds:

a. NaNO2
b. PCl3
c. UF6
d. Mg3(PO4)2

19. What is the percent by mass of oxygen in each of these compounds?

a. glucose, C6H12O6
b. ethanol, C2H6O
c. amoxicillin, C16H19N3O5S

20. What is the percent by mass of nitrogen in each of these compounds?

a. urea, CH4N2O
b. nitric acid, HNO3
c. ammonium perchlorate, NH4ClO4

21. What is the percent by mass of carbon, hydrogen, and oxygen in aspirin,
C9H8O4?

22. What is the percent by mass of carbon, hydrogen, nitrogen, and oxygen in
acetaminophen, C8H9NO2?

23. What is the percent by mass of silver in silver nitrate? What mass of silver
is present in a 100.0-gram sample of this compound?

449
24. What is the percent by mass of iron in iron(III) oxide? What mass of iron is
present in a 314.5-gram sample of this compound?

25. What technique is commonly used to measure the formula mass of a

compound?

26. What technique is commonly used to measure the percent composition of a

compound?

27. The mass spectrum of an unknown explosive (shown below) indicates that
the formula mass of the unknown substance is 227 u. Which of these
molecular formulas could correspond to this mass?
a. nitroglycerin, C3H5N3O9
b. methyl nitrate, CH3NO3
c. ethylene glycol dinitrate, C2H4N2O6
d. trinitrotoluene, C7H5N3O6

28. The mass spectrum of an illegal drug is shown below. The peak showing a
mass of 369 u corresponds to which of the following compounds?
a. cocaine, C17H21NO4
b. methamphetamine, C10H15N
c. tetrahydrocannabinol, C21H30O2
d. heroin, C21H23NO5

450
29. A forensics lab analyzes a small amount of a substance containing carbon,
hydrogen, and nitrogen. The compound is suspected to be either heroin, a
highly addictive narcotic with the formula C21H23NO5, or morphine—a
related compound with the formula C17H19NO3. After conducting an
elemental analysis, the lab finds the sample to contain 71.6% carbon and
6.7% hydrogen by mass. Which conclusion does this evidence support?

30. A forensics lab analyzes a small amount of a substance containing carbon,

hydrogen, and nitrogen. The compound is suspected to be either
methamphetamine, a highly addictive narcotic with the formula C10H15N,
or methylenedioxypyrovalerone (MDPV)—a related compound with the
formula C16H21NO3. After conducting an elemental analysis, the lab finds
the sample to contain 69.8% carbon and 7.7% hydrogen by mass. Which
conclusion does this evidence support?

7.2 Connecting Atomic Mass to Large-Scale Mass: The Mole Concept

31. Calculate the number of grams in one mole of each of the following:
a. propane, C3H8
b. calcium chloride, CaCl2
c. ethylene glycol, C2H6O
d. sucrose, C12H22O11

32. Calculate the number of grams in one mole of each of the following:
a. butane, C4H10
b. magnesium bromide, MgBr2

451
 c. urea, CH4N2O
d. potassium bicarbonate, KHCO3

33. Chemists express the mass of elements and compounds either in atomic
mass units (u) or in grams/mole. Which approach is appropriate in each of
these situations?
a. You need to measure out 14.2 moles of a compound.
b. You need to report the mass of a single molecule, as measured on a
mass spectrometer.

34. Chemists express the mass of elements and compounds either in atomic
mass units (u) or in grams/mole. Which approach is appropriate in each of
these situations?
a. You need to calculate the amount of reactant to add to a 200-liter
industrial reaction vessel.
b. You need to calculate the mass of a single protein molecule.

35. Calculate the number of grams in each of the following:

a. 15.2 moles of KCl
b. 0.319 moles of MgSO4
c. 41.3 moles of neon

36. Calculate the number of grams in each of the following:

a. 1.10 moles of iron(II) nitrate
b. 0.201 moles of sodium phosphate
c. 1.132 × 10–3 moles of PtCl4

37. Calculate the number of grams in each of the following:

a. 1.50 moles of gold
b. 24.3 moles of BF3
c. 0.131 moles of helium

38. Calculate the number of grams in each of the following:

a. 1.29 moles of cobalt(II) chloride
b. 11.1 moles of potassium carbonate
c. 4.922 × 10–2 moles of CuCl2

452
39. Calculate the number of moles in each of the following:
a. 182.5 grams of Mg
b. 29.3 grams of Cl2
c. 304.1 grams of Pb

40. Calculate the number of moles in each of the following:

a. 32.3 grams of NaCl
b. 11.6 grams of CO2
c. 508.3 grams of C2H2

41. Calculate the number of moles in each of the following:

a. 12.6 g of Fe
b. 22.4 g of LiF
c. 1.9 mg of Br2

42. Calculate the number of moles in each of the following:

a. 123.5 g of KClO
b. 17.3 g of H2
c. 11.6 mg of Ca(ClO4)2

43. Convert each of the following to moles and to atoms:

a. 23.4 grams of sodium metal
b. 18.2 grams of phosphorus
c. 192.3 grams of carbon

44. Convert each of the following to moles and to atoms:

a. 18.7 grams of calcium metal
b. 51.9 grams of tellurium, Te
c. 178.4 grams of uranium, U

45. Convert each of the following to moles and to molecules:

a. 11.3 grams of H2O
b. 77.3 grams of PCl3
c. 129.4 kilograms of CO2

46. Convert each of the following to moles and to molecules:

453
 a. 55.2 grams of O2
b. 68.2 grams of CH4
c. 31.2 kilograms of SO2

47. Calculate the solution to each of these problems:

a. How many molecules of HCl are present in 2.0 moles of HCl?
b. How many gold atoms are present in 0.12 moles of gold?
c. How many atoms are present in 140.2 grams of zinc metal?

48. Calculate the solution to each of these problems:

a. How many molecules of HBr are present in 3.0 moles of HBr?
b. How many atoms are present in 0.032 moles of silver?
c. How many atoms are present in 109.4 g of helium?

49. How many moles are in a 15.3-gram sample of elemental copper? How
many atoms of copper are present in this sample?

50. How many moles are in a 201.5-gram sample of chlorine gas, Cl2? How
many molecules of Cl2 are present in this sample?

51. A sample contains 205.2 moles of water. How many molecules of water
are present in this sample? What is the mass of this sample in grams?

52. A room contains 19.4 moles of oxygen gas. How many O2 molecules are
present in this room? What is the mass of this amount of oxygen in grams?

53. What is the mass in grams of 3.192 × 1022 atoms of zinc?

54. What is the mass in grams of 1.191 × 1024 atoms of vanadium?

55. What is the mass in grams of 1.372 × 1023 molecules of hydrogen gas?

56. What is the mass in grams of 1.314 × 1024 molecules of sulfur dioxide?

57. Sucrose (table sugar) has a molar mass of 342.30 g/mole. How many
sucrose molecules are in a 5-pound bag of sugar? (1 pound = 453.6
grams.)

454
58. Methanol (CH4O) has a molar mass of 32.04 g/mole. It has a density of
0.791 g/mL. How many molecules are in a 500-mL bottle of methanol?

7.3 The Mole Concept in Balanced Equations

59. Lead and oxygen gas combine to form lead(IV) oxide, as shown here. If 20
moles of Pb react with oxygen in this way, how many moles of O2 are
consumed? How many moles of PbO2 are produced?

Pb (s)+O2(g)→PbO2(s)

60. Lithium and bromine combine to form lithium bromide, as shown here. If
30 moles of Li react with bromine in this way, how many moles of Br2 are
consumed? How many moles of LiBr are produced?

2 Li (s)+Br2(l)→2 LiBr (s)

61. Zinc metal reacts with hydrochloric acid, as shown in this balanced
equation. If 3.0 moles of Zn reacts in this way, how many moles of HCl
are consumed? How many moles of ZnCl2 and H2 are produced?
Zn (s)+2 HCl (aq)→ZnCl2(aq)+H2(g)

62. Tin metal reacts with silver nitrate solution, as shown in the reaction here.
If 5.0 moles of tin reacts in this way, how many moles of AgNO3 are
consumed? How many moles of Sn(NO3)2 and Ag are formed?
Sn (s)+2 AgNO3(aq)→Sn(NO3)2(aq)+2 Ag (s)

63. Potassium metal reacts with water according to this balanced equation:
2 K (s)+2 H2O (l)→2 KOH (aq)+H2(g)

a. If one mole of potassium reacts in this manner, how many moles of

water are consumed?
b. If one mole of potassium reacts in this manner, how many moles of H2

455
 are produced?
c. How many moles of potassium are required to produce 14.0 moles of
H2?
d. How many moles of KOH are produced if 3,014.2 moles of H2O are
consumed?

64. Elemental boron reacts with fluorine gas according to this balanced
equation:

2 B (s)+3 F2(g)→2 BF3(g)

a. How many moles of BF3 are produced if 10 moles of boron are
consumed?
b. How many moles of F2 are required to react with 10 moles of boron?
c. How many moles of F2 are required to produce 60 moles of BF3?
d. How many moles of B are required to react with 18 moles of F2?

65. Zinc reacts with hydrochloric acid to form zinc chloride and hydrogen gas,
as shown in this equation:
Zn (s)+2 HCl (aq)→ZnCl2(aq)+H2(g)

If 14.2 g of zinc reacts in this way,

a. How many moles of zinc are consumed?
b. How many moles of HCl are consumed?
c. How many moles of ZnCl2 can be produced?
d. How many grams of ZnCl2 can be produced?

66. In aqueous solution, sodium iodide reacts with lead(II) nitrate to give solid
lead(II) iodide and aqueous sodium nitrate, as shown in this equation. In
one experiment, it is found that 8.34 grams of PbI2 were produced.
2 Nal (aq)+Pb(NO3)2(aq)→PbI2(s)+2 NaNO3(aq)

Based on this reaction,

a. How many moles of PbI2 were produced?
b. How many moles of NaNO3 were produced?
c. How many moles of NaI were consumed?

456
 d. How many grams of NaI were consumed?

67. Copper(I) bromide reacts with magnesium metal according to this

equation:
2 CuBr (aq)+Mg (s)→2 Cu (s)+MgBr2(aq)

If 0.253 moles of magnesium react in this way,

a. How many grams of CuBr will be consumed?
b. How many grams of Cu will be produced?
c. How many grams of MgBr2 will be produced?

68. Silver nitrate reacts with calcium chloride according to this equation:
2 AgNO3(aq)+CaCl2(aq)→Ca(NO3)2(aq)+2 AgCl (s)

If 0.403 moles of AgCl are produced in this reaction,

a. How many grams of Ca(NO3)2 are produced?
b. How many grams of AgNO3 are consumed?
c. How many grams of CaCl2 are consumed?

69. The reaction of iron metal with copper(II) bromide is shown in this
balanced equation. How many grams of iron(II) bromide result from the
reaction of 23.6 grams of iron metal?
CuBr2(aq)+Fe (s)→Cu (s)+FeBr2(aq)

70. The reaction of sulfuric acid with potassium hydroxide is shown in this
balanced equation. If 19.07 grams of KOH reacts with excess sulfuric acid,
how many grams of water are produced?
H2SO4(aq)+2 KOH (aq)→2 H2O (aq)+K2SO4(aq)

71. Nitric acid reacts with ammonia to produce ammonium nitrate, as shown
here. In this reaction, how many grams of NH3 are required to produce
1,000 grams of NH4NO3?

457
 HNO3(aq)+NH3(aq)→NH4NO3(aq)

72. Water reacts with sulfur trioxide to produce sulfuric acid, as shown here.
In this reaction, how many grams of SO3 are required to produce 500
grams of H2SO4?
H2O (l)+SO3(g)→H2SO4(aq)

73. Calcium reacts with oxygen gas, as shown in this balanced equation. How
many grams of calcium oxide can be produced from 15.0 grams of oxygen
gas?

2 Ca (s)+O2(g)→2 CaO (s)

74. At high temperatures, nitrogen gas and hydrogen gas react to form
ammonia, as shown in this balanced equation. How many grams of
hydrogen gas are required to react with 200.0 grams of nitrogen gas?

N2(g)+3 H2(g)→2 NH3(g)

75. When heated, calcium carbonate decomposes to form calcium oxide and
carbon dioxide, as shown in this equation. If 150.0 grams of CaCO3 reacts
in this way, how many grams of CaO and CO2 are produced?
CaCO3(s)→CaO (s)+CO2(g)

76. Hydrogen peroxide, H2O2, decomposes to form water and oxygen gas, as
shown in this reaction. If 10.0 grams of H2O2 reacts in this way, how
many grams of H2O and O2 are produced?
2 H2O2(aq)→2 H2O (l)+O2(g)

77. Potassium hydroxide and iron(II) chloride react as shown here. If 3.1

458
 moles of KOH are consumed in this reaction, how many grams of
Fe(OH)2 are produced?
2 KOH (aq)+FeCl2(aq)→2 KCl (aq)+Fe(OH)2(s)

78. Hydrofluoric acid neutralizes magnesium hydroxide as shown here. In this

reaction, how many moles of Mg(OH)2 are needed to react with 0.4 grams
of HF?
2 HF (aq)+Mg(OH)2(aq)→MgF2(aq)+2 H2O (l)

79. Metallic tin reacts with chlorine gas as shown here:

Sn (s)+2 Cl2(g)→SnCl4(s)
If 0.253 moles of tin react in this way,
a. How many grams of Cl2 are consumed?
b. How many molecules of Cl2 are consumed?
c. How many grams of SnCl4 are produced?

80. This balanced equation shows the combustion of propane gas (C3H8):
C3H8(g)+5 O2(g)→3 CO2(g)+4 H2O (g)

If 100.0 g of C3H8 reacts in this way,

a. How many moles of O2 are consumed?
b. How many grams of CO2 are produced?
c. How many molecules of H2O are produced?

81. The reaction of sulfur dioxide with oxygen gas is represented in this
balanced equation and the molecular depiction below. In this depiction,
which reagent is the excess reagent? Which is the limiting reagent?

2 SO2+O2 →2 SO3

459
82. The reaction of boron with chlorine gas is represented in this balanced
equation and the molecular depiction below. In this depiction, which
reagent is the excess reagent? Which is the limiting reagent?

2 B+3 Cl2 →2 BCl3

83. Consider the following generic reaction:

3 A+B→C+D
If this reaction is carried out using 8.0 moles of A and 3.0 moles of B,
a. What is the limiting reagent in the reaction?
b. How many moles of C are formed in this reaction?

84. Consider the following generic reaction:

A+2 B→C+D
If this reaction is carried out using 3.0 moles of A and 5.0 moles of B,
a. What is the limiting reagent in the reaction?
b. How many moles of C are formed in this reaction?

460
85. Calcium oxide reacts with water to form calcium hydroxide, as shown in
this equation:
CaO (s)+H2O (l)→Ca (OH)2 (s)

If a bag of calcium oxide is dropped into a lake, which reagent likely will
be the limiting reagent?

86. Ethanol is a common component in lighter fluid, which is used to ignite

charcoal in a barbeque grill. Here is the reaction for the combustion of
ethanol:
2 C2H6O (l)+6 O2(g)→4 CO2(g)+6 H2O (g)

If you apply the lighter fluid to charcoal in an open grill and ignite it,
which reagent likely will be the limiting reagent? Why is this so?

87. Iron and sulfur react according to this equation:

Fe (s)+S (s)→FeS (s)

If 0.10 moles of iron are combined with 0.13 moles of sulfur, what is the
limiting reagent? How many moles of iron(II) sulfide can form in this
reaction?

88. Copper and bromine react according to this equation:

Cu (s)+Br2 (l)→CuBr2 (s)

If 0.39 moles of copper are combined with 0.25 moles of bromine, what is
the limiting reagent? How many moles of copper(II) bromide can form in
this reaction?

89. Potassium reacts aggressively with water, as shown in this equation. If

0.20 grams of potassium are added to 15.0 grams of water, what is the
limiting reagent? What mass of KOH can be produced in this reaction?
2 K (s)+2 H2O (l)→2 KOH (aq)+H2 (g)

461
90. Phosgene (COCl2) reacts with ammonia (NH3), as shown in this reaction.
If 1.2 grams of COCl2 are added to 10.5 grams of NH3, what is the
limiting reagent? What mass of CH4N2O can be produced in this reaction?
COCl2 (g)+2 NH3 (g)→CH4N2O (g)+2 HCl (g)

91. Sodium metal reacts with elemental iodine to form sodium iodide, as
shown here:

2 Na (s)+I2 (s)→2 NaI (s)

If 10.0 moles of sodium are combined with 6.0 moles of I2,
a. What is the limiting reagent?
b. How much of the excess reagent is used in the reaction?
c. How much of the excess reagent is left over?

92. Aluminum metal reacts with elemental chlorine to form aluminum

chloride, as shown here:
2 Al (s)+3 Cl2 (g)→2 AlCl3 (s)

If 12.0 moles of aluminum are combined with 15.0 moles of Cl2,

a. What is the limiting reagent?
b. How much of the excess reagent is used in the reaction?
c. How much of the excess reagent is left over?

93. Thionyl chloride (SOCl2) reacts violently with water, as shown here:
H2O (I)+SOCl2 (g)→SO2 (g)+2 HCl (aq)

If 20.0 g of water is combined with 50.0 g of SOCl2,

a. What is the limiting reagent in this reaction?
b. What mass of excess reagent is used in the reaction?
c. What mass of excess reagent is left over?

94. Phosphorus trichloride reacts with water, as shown here:

PCl3 (l)+3 H2O (l)→H3PO3(aq)+3 HCl (aq)

462
 If 60.0 g of water is combined with 137.3 g of PCl3,
a. What is the limiting reagent in this reaction?
b. What mass of excess reagent is used in the reaction?
c. What mass of excess reagent is left over?

95. Silicon (28 g, 1.0 mole) and oxygen gas (128 g, 4.0 moles) are reacted in a
sealed container. At the end of the reaction, how much silicon, oxygen,
and silicon dioxide are present in the container? To answer the question,
complete this table:

→ SiO2
Si + O2
Starting Moles 1.0 mol 4.0 mol 0 mol
Change –1.0 mol
Ending Moles
Ending Grams

96. Methane gas (16 g, 1.0 mole) and oxygen gas (96 g, 3.0 moles) are reacted
in a sealed container. At the end of the reaction, how much methane,
oxygen, carbon dioxide, and water are present in the container? To answer
the question, complete this table:

CH4 + 2 O2 → CO2 + 2 H2O

Starting 1.0 mol 3.0 mol 0 mol 0 mol

Moles
Change –1.0 mol
Ending
Moles
Ending
Grams

97. This reaction shows the neutralization of phosphoric acid (H3PO4) with
potassium hydroxide (KOH):
H3PO4(aq)+3 KOH (aq)→K3PO4(aq)+3 H2O (l)

463
 If 96.3 grams of H3PO4 are combined with 150.0 g of KOH,
a. What is the limiting reagent in this reaction?
b. What mass of water can be produced in this reaction?
c. What mass of excess reagent will be left over in this reaction?

98. This equation shows the reaction of ethylene with molecular bromine:
C2H4(g)+Br2(l)→C2H4Br2(l)

If 32.0 grams of C2H4 are combined with 172.0 grams of Br2,

a. What is the limiting reagent in this reaction?
b. What mass of C2H4Br2 can be produced in this reaction?
c. What mass of excess reagent will be left over in this reaction?

7.4 Theoretical and Percent Yield

99. What are two of the most common causes for a percent yield that is below
100%?

100. Is it possible for a reaction to occur with a yield that is greater than 100%?
Why or why not?

101. Based on the amounts of starting materials used, a chemist calculates a

possible yield of 210.3 grams in a reaction. However, after isolating her
purified product, she finds that she has only 194.1 g of product. What is
her percent yield for this reaction?

102. Based on the amounts of starting materials used, a chemist calculates a

possible yield of 51.3 grams in a reaction. However, after isolating her
purified product, she finds that she has only 45.1 g of product. What is
her percent yield for this reaction?

103. Consider this generic reaction:

A+2 B→C+D
If this reaction is carried out using 1.5 moles of A and 4.0 moles of B,
a. What is the limiting reagent in this reaction?

464
 b. What is the theoretical yield (in moles) of compound D?
c. If 1.2 moles of compound D were isolated, what would the percent
yield of D be for this reaction?

104. Consider this generic reaction:

A+2 B→C+2 D
If this reaction is carried out using 2.5 moles of A and 4.0 moles of B,
a. What is the limiting reagent in this reaction?
b. What is the theoretical yield (in moles) of compound C?
c. If 1.6 moles of compound C is isolated, what is the percent yield of C
for this reaction?

105. A chemist carries out this reaction, starting with 14.2 g of C6H6 and an
excess of Br2:
C6H6(l)+Br2(l)→C6H5Br (g)+HBr (g)

a. What is the theoretical yield for the reaction?

b. If he isolated 16.3 grams of C6H5Br, what is his percent yield for this
reaction?

106. A chemist carries out this reaction, beginning with 2.10 grams of NaBr
and 3.25 grams of Pb(NO3)2:
2 NaBr (aq)+Pb(NO3)2(aq) →PbBr2(s)+2 NaNO3(aq)

a. Which starting material is the limiting reagent?

b. What is the theoretical yield of PbBr2 in this reaction?
c. If the chemist actually isolates 3.5 grams of PbBr2, what is the percent
yield of the reaction?

107. A chemist carries out the following reaction, beginning with 4.20 g of
MgCl2 and 3.43 g of Na3PO4. After the reaction is complete, she is able
to isolate 2.14 grams of Mg3(PO4)2 as a white solid.
3 MgCl2(aq)+2 Na3PO4(aq)→Mg3(PO4)2(s)+6 NaCl (aq)

a. Which starting material is the limiting reagent?

465
 b. What is the theoretical yield of Mg3(PO4)2 in this reaction?
c. What was her percent yield of Mg3(PO4)2 in this reaction?

108. A chemist dissolves 0.401 g of AgNO3 in a beaker of water and 0.253 g

of MgCl2 in a second beaker of water. She then combines the two
solutions, which together form AgCl as a white precipitate. She isolates
the AgCl and finds its mass to be 0.292 g. Here is the balanced equation
for this reaction:
2 AgNO3(aq)+MgCl2(aq)→2 AgCl (s)+Mg(NO3)2(aq)

a. Which starting material is the limiting reagent?

b. What is the theoretical yield of AgCl in this reaction?
c. What was the percent yield of AgCl in this reaction?

Challenge Questions

109. You are conducting a set of experiments to see if it is possible to

determine the type of wood used in a campfire by analyzing the ashes. In
the laboratory, you weigh out 500 grams of hickory wood chips. You
then burn the wood in a simulated campfire. After collecting the ashes,
you find that only 37 grams of ashes are present. Does this finding
conflict with the law of conservation of mass? How can you explain your
findings?

110. You are conducting a series of experiments on mass changes in different

chemical reactions. In each reaction you take two compounds dissolved
in water in separate tubes (labeled A and B), measure the mass of both
tubes before the reaction, and then combine the contents of the tubes.
You observe any qualitative changes, and you also measure the mass of
the two tubes after the reaction. Your observations are as follows:

Test Tube A Tube B Observation Combined Combined

Aqueous Aqueous Masses Masses
Solution Solution Before After
Reaction Reaction
1 Sodium Potassium Nothing 20.07 g 20.06 g
chloride bromide observed
2 Lead(II) Sodium Yellow 17.84 g 17.81 g
nitrate iodide precipitate

466
 3 Silver Sodium White 20.44 g 20.42 g
nitrate chloride precipitate
4 Sodium Hydrochloric Bubbles 18.51 g 16.92 g
carbonate acid formed

In tests 1, 2, and 3, the law of conservation of mass seems to be followed.

However, in test 4, you see a drop in mass. Based on your observations,
can you explain why the mass in this reaction appears to have decreased?

111. Ethyl acetate (C4H8O2) is a sweet-smelling liquid commonly found in

nail polish removers. Industrially, this compound is produced from acetic
acid HC2H3O2 and ethanol (C2H6O) through this reaction:
HC2H3O2(l)+C2H6O (l)→C4H8O2(l)+H2O (l)

a. Using the table provided, calculate both the mass and volume of
ethanol required to completely react with 400.0 L of acetic acid.
b. What is the theoretical yield of ethyl acetate in this reaction? Report
your answer in grams, kilograms, and liters.

Compound Molar Mass Density

Acetic acid, HC2H3O2 60.06 g/mole 1.049 g/mL

Ethanol, C2H6O 46.08 g/mole 0.789 g/mL

Ethyl acetate, C4H8O2 88.12 g/mole 0.902 g/mL

112. Coal is a leading source of energy for the world. While coal is primarily
composed of carbon and hydrogen, a small amount of sulfur is also
present. This leads to the production of pollutants such as sulfur dioxide
(SO2) through the following reaction:

S (s)+O2(g)→SO2(g)
A sample of coal was found to contain 1.35% sulfur by mass. If 1,250 kg
of this coal were burned, what mass of sulfur dioxide could theoretically be
formed?

467
Chapter Eight
Energy

The Corn Ethanol Debate

Is it wise to use corn to power our cars?

Over the past decade this question has become more and more pressing, and more
and more heated. At the heart of the issue is a single compound—ethanol—and the
conflicting interests that surround it.
Throughout recorded history, people have produced ethanol (also called ethyl
alcohol or simply alcohol) from grains such as wheat and corn. While most people
think of alcohol in terms of beverages, this colorless liquid burns easily and can be
used as fuel. Henry Ford’s original Model T could run on gasoline or on ethanol
(Figure 8.1).

468
Figure 8.1 (a) The oil shortage of 1973 sparked renewed interest in using ethanol as
fuel. (b) Ethanol is a colorless liquid that burns easily. (c) For millennia, people have
produced ethanol from grains, especially for use as alcoholic beverages. The process
involves a chemical reaction (fermentation), often followed by purification using a
technique called distillation. This image shows a simple still used to make
“moonshine”—a mixture containing mainly ethanol and water. (d) Industrial distilleries
like this one produce large amounts of ethanol from corn. (e) Henry Ford’s
revolutionary Model T ran on either gasoline or ethanol. (f) Some newer vehicles also
accommodate gasoline or ethanol fuels.

For years gasoline was far cheaper than ethanol, and it became the fuel of
choice for automobiles. But in the 1970s things began to shift. Gasoline is
produced from crude oil, and many experts feared the supply of oil would dwindle.
More urgently, an alliance of oil-producing countries called OPEC controlled the
prices and supply of oil to the rest of the world. In a show of force in October
1973, OPEC blocked shipments of oil to the United States and several of its allies.
Crippling gasoline shortages plagued the U.S. until the blockade ended in early
1974.
These factors produced a surge of interest in renewable fuels such as solar and
wind energy and biofuels made directly from plant matter. In 1978 the U.S.
government began to provide subsidies (monetary incentives) to encourage farmers
to produce corn for ethanol. Slowly, companies began producing gasoline blends
that contained a small percentage of ethanol.
In 2005 and 2007, the U.S. Congress passed two laws that created a Renewable
Fuel Standard (RFS). These laws forced oil companies to blend renewable fuels
(especially ethanol) with gasoline. The new regulations dramatically increased the
amount of corn used for ethanol production. Today about 40% of all corn grown in
the United States is used to produce ethanol for fuel.
Not everyone is happy about these changes. Due to increased demand, the price
of corn doubled between 2005 and 2008. And because farmers rely on corn to feed
their livestock, the prices of meat and dairy products have increased sharply.
Advocates of the renewable fuel standard say that it reduces dependence on
foreign sources of oil and provides jobs for rural communities. Critics of the policy
argue that it creates an undue financial burden on people across the nation while
pouring money into a few corn-producing states. Some insist that ethanol is a
clean-burning fuel that improves air quality. Others counter that ethanol production
actually damages the environment. Who is right?
The answers to these issues are complex, so let’s begin with some simpler
questions: How well does ethanol actually work? Does a gallon of ethanol produce
the same amount of energy as a gallon of gasoline? Do you get better gas mileage
with pure gasoline or with an ethanol blend? To answer these questions, we must
examine the energy changes that accompany the combustion of ethanol or gasoline.
In this chapter, we will explore the relationships between energy and chemical
reactions. We’ll see how scientists measure the energy stored in fuels like gasoline
and ethanol. We’ll see how to predict energy changes using the rules of

469
stoichiometry developed in Chapter 7. Finally, we’ll return to the ethanol/gasoline
debate as we answer the question, “Which is better?”

Intended Learning Outcomes

After completing this chapter and working the practice problems, you should be
able to:

8.1 Energy, Work, and Heat

Describe the relationships between heat, work, and total energy change.
Describe the exchange of energy between the system and surroundings that
accompanies a physical or chemical change.

8.2 Heat Energy and Temperature

Explain the difference between heat and temperature.
Use calorimetry measurements to determine energy changes.
Apply the heat capacity or specific heat of a system to solve problems
relating to heat energy and temperature.

8.3 Heat Energy and Chemical Reactions

Use fuel values and reaction enthalpies to calculate the heat absorbed or
released in a chemical reaction.

470
8.1 Energy, Work, and Heat
Most chemical and physical changes are accompanied by changes in
energy and temperature. A campfire releases energy as it burns. When an
ice cube melts, the water around it becomes colder. As our bodies break
down food, we absorb energy stored in the chemical bonds (Figure 8.2).
To fully describe these changes, we must also describe the changes in
energy. The study of energy and temperature changes is called
thermodynamics. Thermochemistry is the part of this field that deals
specifically with chemical changes.

Figure 8.2 Winter campers like this one rely on heat energy from the burning wood to
warm themselves, melt snow into drinkable water, and cook their food. Most physical
and chemical changes also involve changes in heat and temperature.

We first discussed energy in Chapter 1. Recall that energy is the

ability to do work—that is, the ability to bring about a change. Energy can
take several forms, including potential energy (stored energy) and kinetic
energy (the energy of motion).

Heat energy involves the kinetic energy of particles within a substance.

Changes in energy take place in two forms: heat and work. When a
substance absorbs heat energy (often simply called heat), the particles
within the substance vibrate or move faster. Heating a substance increases
the kinetic energy of the particles that compose it. Work is the transfer of
energy from one form to another. Work often involves a force that moves
an object.

471
 It’s helpful to think of energy, heat, and work using the analogy of
wealth and spending (Figure 8.3). What does it mean to have wealth? It
means the ability to spend money. What is spending? Spending transfers
money from one person to another. For example, if I have some money, I
have the ability to spend it on something—say, a hot dog and a Coke. If
I’m hungry, I may transfer my money to the hot dog guy. Now the guy
who sold me the hot dog has the wealth (but I have lunch).

Figure 8.3 In much the same way that spending is the transfer of wealth, heat and work
are the transfer of energy. We describe heat and work using the same units we use to
describe energy.

Energy changes take place through heat or work.

In the same way that spending transfers wealth from one person to
another, heat and work transfer energy from one object or substance to
another.
To describe wealth, we use some form of money, such as dollars. If we
want to describe spending, it makes sense to describe it in the same way—
using dollars. Similarly, we describe heat and work in the same units that
we use to describe energy.

Units of Energy
There are several common units for energy (Table 8.1). The standard unit
of energy is the joule (J). Other common units of measurement include the
British thermal unit (BTU), the kilowatt hour (kWh), and the calorie (c);
see Figure 8.4. In nutritional circles, a kilocalorie is often referred to as a
Calorie, where the letter C is sometimes capitalized. Most scientists prefer
to avoid ambiguity, so they use the term kilocalorie rather than the capital
C nutritional calorie.

A joule is defined as the energy required to accelerate a mass of 1 kg at a rate of

1 m/s2 over a distance of 1 m.

472
Figure 8.4 Energy measurements use many different units. (a) A company advertises
the energy its space heaters release in BTUs. (b) Your electric meter often measures
energy in kilowatt hours. (c) Food products often report energy content in Calories
(meaning kilocalories).

TABLE 8.1 Common Units of Energy

1 joule = 1 kg·m2/s2

1 calorie = 4.184 J
1,000 calories = 1 kcal = 1 Calorie
1 British thermal unit (BTU) = 1,055 J
1 kilowatt hour (kWh) = 3.6 × 106 J

Example 8.1 Converting between Energy Units

A candy bar contains 212.0 kilocalories of stored chemical energy. What is
this energy in joules and in kilojoules?
We first convert from kilocalories to calories; then we multiply by a
conversion factor relating calories and joules, using the relationship given
in Table 8.1. This gives us a final answer of 887,000 J or 887.0 kJ. •

212.0 kcal×1,000 cal1 kcal×4.184 J1 cal=887,000 J=887.0 kJ

473
 IT
TRY

1. One gram of gasoline releases 47.3 kJ of potential energy when it is

burned. Convert this value to calories.

Heat and Work in Chemical Changes

The energy changes that accompany a physical or chemical change take
place in the forms of heat and work. For example, consider a piston in a
gasoline engine (Figure 8.5). As the gasoline burns, it releases energy.
This energy causes the gas in the piston to expand, which pushes the
cylinder up and turns the engine. The energy released has performed work
on the piston.

Figure 8.5 As the fuel mixture in a piston explodes, it produces heat and work.

474
 Explore

Figure 8.5

The two forms of energy transfer are heat and work.

Scientists use the capital Greek letter delta (Δ) to indicate a change. E means
energy, and ΔE means a change in energy.

However, not all of the energy transfers to the piston as work. Part of
the energy escapes to the surroundings as heat. The total energy released
from the gasoline (ΔE) is the sum of the heat (q) and the work (w):

ΔE=q+w
So, here is a question: What if we welded the piston shut and then
ignited the gasoline? (Don’t try this at home.) Assuming the piston
chamber doesn’t explode, all of the energy from the reaction would be
released as heat. Because the piston does not move, none of the energy is
converted into motion—that is, no work is done. The amounts of heat and
work can change, but the total amount of energy released will be the same.

475
 An engine like this one converts the potential energy in gasoline into work and
heat. Car designers try to make engines run as efficiently as possible—that is,
they try to maximize the amount of energy transferred as work and minimize the
amount of energy lost as heat.

Example 8.2 Measuring the Total Energy Change

A small sample of propane burns, producing carbon dioxide and water
vapor. As the hot gas mixture expands, it releases 20.0 kJ of heat and does
31.0 kJ of work pushing against a piston. What is the total amount of
energy released in this reaction?
The change released 20.0 kJ of energy in the form of heat and 31.0 kJ in the
form of work. The total amount of energy released is the sum of the heat
and the work, or 51.0 kJ. •

Energy released=q+w=20.0 kJ+31.0 kJ=51.0 kJ

Endothermic and Exothermic Changes

Let’s examine the reaction inside the piston from a chemical perspective.
When gasoline reacts with oxygen, it produces carbon dioxide and water.
It also releases a large amount of energy. We describe this reaction as an
exothermic change—meaning it releases heat energy. In fact, we
sometimes include “energy” with the reaction products:
2 C8H18 (l)+25 O2 (g)→16 CO2 (g)+18 H2O (g)+energy

Conversely, an endothermic change absorbs energy. When describing

476
an endothermic change, we sometimes show energy as a reactant. For
example, when an ice cube melts, it absorbs energy to convert the solid ice
to liquid. We can express this as an equation:

Energy+H2O (s)→H2O (l)

System: The part of the universe being studied.

When describing energy changes, we must be very careful to define the

system that is changing. In thermochemistry, the system is the part of the
universe being studied. The term surroundings refers to the rest of the
universe. So, when we say that burning fuel is an exothermic process, we
mean that energy is released from the system (the fuel) to the
surroundings. In the example of the melting ice cube, the ice is the system
that absorbs heat from the surroundings.
So, if someone puts an ice pack on an injured ankle, is the flow of heat
endothermic or exothermic? To answer that question, we must first define
the system and the surroundings. We know that heat energy flows from the
ankle into the ice, so the change is exothermic for the ankle, but it is
endothermic for the ice (Figure 8.6).

Figure 8.6 An ice pack is often the first treatment for a sprained ankle. The heat energy
flows out of the ankle and into the ice. This change is exothermic from the perspective
of the ankle, but endothermic from the perspective of the ice.

Endothermic: +q
Exothermic: −q

When describing energy changes, we must be careful to show the

direction of energy flow from the perspective of the system we are
studying. In an exothermic reaction, the system loses heat energy, and so
the energy change for that system is negative. On the other hand, if a
system gains heat (an endothermic reaction), the sign of the energy change
is positive.

477
 Similarly, if a system does work on its surroundings (for example,
when the gas pushes the piston up), the sign of w for that system is
negative. If the surroundings do work on the system (that is, increase the
energy of the system), then the sign of w is positive (Figure 8.7).

Figure 8.7 The signs of q and w depend on the system being studied. (a) As gasoline
explodes, it releases heat to the surroundings and pushes back the air around it. If
gasoline is the system, both q and w are negative. (b) Heat travels from the campfire to
the coffeepot. The heat change, q, is positive for the water in the pot but negative for the
wood in the fire.

Example 8.3 Expressing Energy Changes in Chemical Equations

When ammonium chloride dissolves, it absorbs heat from the surroundings.
Write a chemical expression for this change, including energy where
appropriate.
In this case, the system (the ammonium chloride) absorbs heat from its

478
 surroundings as it dissolves. This is an endothermic change. Because heat
energy is required for the reaction to take place, we can write “energy” as a
reactant. •

Energy+NH4Cl (s)→NH4Cl (aq)

IT
TRY

2. Describe each of these changes as endothermic or exothermic, using

the frame of reference given:
a. Water boils, converting to steam. (system: water)
b. An acid and base react, releasing heat. (system: acid and base
molecules)
c. A log burns in a fireplace. (system: the log)

3. Freezing is an exothermic process. When a liquid solidifies, it releases

heat to its surroundings. Write an equation for the freezing of water,
and include energy in the chemical equation where appropriate.

The Law of Conservation of Energy

In each example we’ve discussed so far, energy moved between a system
and its surroundings. This transfer of energy leads us to a key idea,
commonly called the law of conservation of energy:

The law of conservation of energy is also called the first law of

thermodynamics.

Law of conservation of energy: Energy cannot be created or destroyed.

In a chemical or physical change, energy passes from one system to
another, but the total energy of the universe remains constant. The energy
that flows out of the system flows into the surroundings, and vice versa.
Mathematically we can say that the energy change of the system
(ΔEsystem) is equal to the energy change of the surroundings
(ΔEsurroundings) but is opposite in sign:

479
 ΔEsystem=−ΔEsurroundings

As an analogy, think of the classic board game Monopoly® (Figure 8.8):

At the beginning of the game, you pass out play money to all players. As
the game progresses, the players buy and sell property. If you land on
someone else’s property, you pay them rent. Your wealth goes down, and
the other player’s wealth goes up by the same amount. But the total
amount of money in the game has not changed. At the end of the game, the
game box contains the same amount of money as when you started.

Figure 8.8 In the board game Monopoly®, players may gain or lose money; but the total
money in the game does not change. Similarly, chemical systems may gain or lose
energy, but this does not change the total energy of the universe.

It’s the same thing with the conservation of energy: Energy can be
passed back and forth between the system and its surroundings, but the
total amount of energy remains constant.

In Chapter 16, we’ll explore nuclear reactions—and we’ll see that the law of
conservation of energy is not a complete description of the universe. But for
nonnuclear chemical and physical changes, this law holds true.

480
 Example 8.4 Using the Law of Conservation of Energy
A chemical reaction releases 200 J of heat energy to its surroundings.
Write this change of energy for the system (the chemical reaction) and for
the surroundings.
Because the system loses energy, we write this change as −200 J. Based on
the law of conservation of energy, the energy that the system loses is gained
by the surroundings. Therefore the change in energy for the surroundings is
+200 J. Notice that these values are equal in magnitude but opposite in
sign. •

IT
TRY

4. As a sample of butane gas burns, it releases 55.0 J of heat and

performs 23.0 J of work on its surroundings. What is the total change
in energy for the chemical system? What is the total change in energy
for the surroundings? Include the correct signs in your answers.

481
8.2 Heat Energy and Temperature
We commonly use the words heat and temperature to describe the world
around us. As we explore the effects of energy changes, it’s important to
understand the difference between these terms. Heat refers to the total
kinetic energy transferred from one substance or object to another.
Temperature is a measure of the average kinetic energy of the particles in
a substance. For example, suppose a spoon of boiling water and a pot of
boiling water (Figure 8.9) both have the same temperature—in other
words, the same average kinetic energy. Which one can we use to cook a
meal? The pot, of course: because it holds a larger amount of water, there
is a larger amount of heat energy available to cook the food.

Figure 8.9 The water in the spoon and the water in the pot are the same temperature,
but because the pot has more water, it has a larger amount of heat energy available to
cook the meal.

Specific Heat and Heat Capacity

The precise relationship between heat and temperature is unique to each
substance. For example, if we add 1,000 calories of heat energy to one
kilogram of water, the temperature increases by just one degree Celsius.
However, if we add the same amount of heat energy to one kilogram of
iron, the temperature increases nearly ten times as much. While the
reasons for this are not obvious, the relationships between heat,

482
temperature, and the composition of different materials are well known
and very important.
The relationship between heat energy and temperature depends on the
substance.

Scientists describe the relationship between heat, mass, and

temperature using the term specific heat (often called the specific heat
capacity). The specific heat is the amount of heat required to raise the
temperature of one gram of material by one degree Celsius. Written
mathematically, this relationship can be expressed as
Specific heat=heat(mass)×(change in temperature)

or

s=qmΔT
or, in rearranged form,

q=msΔT
where q is the amount of heat, s is the specific heat, m is the mass, and ΔT
is the change in temperature. The specific heats of several common
substances are given in Table 8.2. Example 8.5 is based on this equation.

The specific heat capacity relates heat, mass, and temperature change.

One calorie (4.184 J) is the amount of heat required to raise one gram of water
by one degree Celsius. That is, the specific heat of water = 1 cal/g · °C, or 4.184
J/g · °C.

TABLE 8.2 Specific Heats for Several Materials

Material Specific Heat (J/g · °C)

Gas Air (dry) 1.01

Liquid Water (liquid) 4.184

Ethanol 2.597

483
 Oil (petroleum) 1.74

Gasoline 2.2

Solid Glass (quartz) 0.70

Concrete 0.880

Ice 2.10

Sand 0.799

Aluminum 0.897

Chromium 0.449

Gold 0.129

Iron 0.449

Lead 0.130

Nickel 0.444

Zinc 0.388

Steel 0.50

Compare the specific heat of water to the other materials in Table 8.2. Do you
notice how much larger the value is for water? This subtle fact is essential to
human life: Although our bodies require a very narrow temperature range (about
80–110 °F) to stay alive, we are able to survive in much colder or somewhat
warmer settings. This is partly because the specific heat of water helps to
regulate our body temperatures, making us less susceptible to short-term
temperature extremes.

Example 8.5 Using Specific Heat in Calculations

How many kilojoules of heat are required to raise the temperature of 120.0
grams of water by 5.0 °C?

484
 Using the relationship above, we can answer this question. After rounding
to two significant digits, we obtain an answer of 2,500 J, or 2.5 kJ. •

q=msΔT=(120.0 g)(4.184Jg⋅°C) (5.0 °C)=2,500 J=2.5 kJ

Sometimes scientists and engineers need to determine heat–

temperature relationships for objects containing many substances or for
objects whose mass is difficult to measure. In these situations a simpler
measure is used, called the heat capacity. Heat capacity (represented by
the letter C) is defined as the amount of heat required to raise the
temperature of an object, regardless of its mass.
Heat capacity=heatchange in temperature

or

C=qΔT

Heat capacity calculations do not involve mass.

Heat capacity is a useful tool in many situations. For example, suppose

a team of workers at a chemical plant needs to raise the temperature of a
steel reactor filled with aqueous sodium hydroxide. Calculating the
specific heats of all the different components of the reactor is too
complicated. Instead, they use the heat capacity of the whole unit (the
reactor vessel and the solution) to predict the temperature changes.
Example 8.6 illustrates a calculation of this type.

Example 8.6 Using Heat Capacity in Calculations

When filled with water, a large reaction vessel in a chemical plant has a
heat capacity of 5.41 × 105 kJ/°C. How many kJ of heat are required to
heat this entire vessel from 25.0 °C to 48.2 °C?

485
 From the initial and final temperature values, we can determine the change
in temperature, ΔT, for this reaction:
ΔT=Tfinal−Tinitial=48.2 °C−25.0 °C=23.2 °C

Rearranging the equation for heat capacity and substituting the numerical
values, we find this change requires 1.26 × 107 kJ of heat. •
q=CΔT=(5.41×105 kJ°C)(23.2 °C)=1.26×107 kJ

IT
TRY

5. If 400 joules of heat flow into a 100-gram block of chromium, how

much will the temperature of the block rise? The specific heat of
chromium is 0.449 J/g · °C.

6. When filled with water, a reaction vessel in a manufacturing plant has

a heat capacity of 28,000 kJ/°C. If 1.41 × 105 kJ of heat are added, by
how much will the temperature of the vessel change?

Calorimetry: Measuring Heat Flow

Scientists and engineers often need to measure how much heat is
transferred from one source to another. Experiments that measure heat
flow are called calorimetry experiments. The two most common
calorimetry techniques are coffee cup calorimetry and bomb calorimetry.
Coffee Cup Calorimetry
Coffee cup calorimetry is exactly what it sounds like. In this method,
water is placed in an insulated cup (a Styrofoam coffee cup works well)
that is covered with a cork and a thermometer (Figure 8.10). The system
to be studied is placed inside the coffee cup, and the reaction or change is
allowed to take place. Because Styrofoam is a very good insulator, nearly
all of the heat involved in the change is absorbed or released by the water.

486
By measuring the temperature change for the water, we can determine the
amount of heat that the system gained or lost.

Figure 8.10 A simple coffee cup calorimeter.

For example, if we wanted to measure the specific heat of a block of

iron, we would heat the block, and then place it in the calorimeter. When
the hot iron is placed in the cool water, heat is transferred from the iron to
the water. According to the law of conservation of energy, the total energy
inside the cup is conserved. This means that the metal loses the same
amount of heat that the water gains. That is:

qwater=−qmetal
In this equation, the minus sign shows that the metal is losing energy. This
type of experiment is shown in Figure 8.11. Because q = msΔT, we can
say that

487
Figure 8.11 The specific heat of a metal can be measured using coffee cup calorimetry.
When a hot metal sample is placed in water, the water temperature rises. Using the
mass, specific heat, and temperature change of water, we can determine the amount of
heat absorbed by the water. This quantity is equal in magnitude to the amount of heat
released by the metal.

Explore

Figure 8.11

mwswΔTw=−mmsmΔTm
where the subscript “m” refers to the metal, and the subscript “w” refers to
the water. We can then rearrange this equation to find the specific heat for
the metal.

488
 sm=−mwswΔTwmmΔTm
Example 8.7 illustrates this type of problem.

The heat lost by the metal is gained by the water.

Example 8.7 Finding the Specific Heat Using Coffee Cup

Calorimetry
To find the specific heat of nickel, a chemist heats a 26.0-g sample to 100.0
°C, and then places it into a coffee cup calorimeter containing 52.1 g of
water at an initial temperature of 20.0 °C. After some time, both the metal
and water reach an equal temperature of 24.0 °C. What is the specific heat
of the metal?
To solve this problem, we first need to find the change in temperature of
the metal and the water:
ΔTmetal=Tfinal−Tinitial=24.0 °C−100.0 °C=−76.0 °CΔTwater=Tfinal
−Tinitial=24.0 °C−20.0 °C=4.0 °C

Next we use the equation for finding specific heat. The specific heat of
water (sw) is equal to 4.184 J/g·°C. Using the rearranged form of the
equation, we can find the specific heat of the metal. •

Sm=−mwswΔTwmmΔTm=−(52.1 g)(4.184Jg⋅°C)(4.0 °C)(26.0 g)(−76.0

°C)sm=0.44 J/g ⋅ °C

IT
TRY

489
 7. To find the specific heat of an unknown metal, a chemist heats a 50.0-
g block of the metal to 100.0 °C. She then places it in a coffee cup
calorimeter containing 50.0 g of water at a temperature of 24.8 °C.
After the metal is placed in the water, both the metal and water reach
an equal temperature of 51.2 °C. What is the specific heat of the
material?

Coffee cup calorimetry is used to measure the gain or loss of heat

energy for many physical and chemical changes. For example, ammonium
chloride has an unusual property: It absorbs heat when it dissolves in
water. This compound is used in “instant cold packs” because it takes heat
from its surroundings. We can write the endothermic reaction like this:
Heat+NH4Cl (s)→NH4Cl (aq)

How much does the temperature drop when we dissolve ammonium

chloride in water? Of course, this depends on how much NH4Cl and how
much water we use. But we can measure the amount of heat absorbed by
NH4Cl by doing a coffee cup calorimetry experiment: Fill a coffee cup
with a known amount of water. Add a known amount of NH4Cl, quickly
cover the calorimeter, and measure the change in temperature that results.
The amount of heat lost from the water (the aqueous solution) will be
equal to the amount of heat gained by the solid:

qsolid=−qaq
Because we know that q = msΔT, this equation becomes

qsolid=−maqsaqΔTaq
Example 8.8 illustrates this type of calculation.

Example 8.8 Finding the Heat of a Reaction Using Coffee Cup

Calorimetry
A 10.4-gram sample of NH4Cl was combined with 100.0 grams of water in
a coffee cup calorimeter, causing the water temperature to decrease by
6.20 °C. Based on this result, how much heat energy was required to

490
 dissolve the sample of NH4Cl? Calculate the heat of solution for NH4Cl in
kJ/mol.
In this example, ammonium chloride absorbs heat energy as it dissolves.
The amount of heat absorbed by the solid is equal to the amount of heat lost
from the aqueous solution. We can solve this using the specific heat
equation:

qsolid=−maqsaqΔTaq
The mass of the resulting solution (maq) includes both the water and the
dissolved ammonium chloride (110.4 g). Because it is mostly water, the
solution will have a specific heat nearly identical to that of water (4.184
J/g·°C). To solve, we substitute these values into the equation:
qsolid=−(110.4 g)(4.184Jg⋅°C)(−6.20 °C)=2,860 J=2.86 kJ

Finally, we are asked to calculate the heat of solution in kJ/mol. We begin

by converting 10.4 grams of NH4Cl from grams to moles:
10.4 g NH4Cl×1 mole NH4Cl53.49 g NH4Cl=0.194 moles NH4Cl

To calculate the heat of the solution in kJ/mol, we divide the heat absorbed
by the number of moles of NH4Cl. •

Heat of solution=2.86 kJ0.194 moles NH4Cl=14.7 kJ/mol

IT
TRY

8. A chemist fills a coffee cup calorimeter with 50.0 g of water. He then

adds a sample of KOH (5.21 g) and stirs the mixture until all of the
KOH dissolves. As he mixes the KOH, the temperature rises from
24.0 °C to 46.4 °C. Based on this result, how much heat energy does
the KOH release? Calculate the heat of solution for KOH in kcal/mol.

491
Bomb Calorimetry
Bomb calorimetry is not what it sounds like. Bomb calorimetry measures
the heats of reaction for fuels and other high-energy chemical changes. In
bomb calorimetry, the test substance is placed in a heavy steel container.
Outside this container is a second container that is filled with water and
equipped with a thermometer (Figure 8.12). A pair of ignition wires
detonates the sample. The gases produced cannot expand, so all of the
energy is released as heat into the calorimeter. Using the heat capacity
(joules/°C) for the calorimeter, the amount of energy produced in the
reaction can be determined.

Figure 8.12 A bomb calorimeter contains a sealed sample chamber that is surrounded
by a water bath. When the sample detonates, heat flows from the sample chamber into
the water bath.

492
 Explore

Figure 8.12

In bomb calorimetry, a reaction takes place inside a sealed container.

We can use bomb calorimetry to answer the important questions we

asked at the beginning of this chapter: How well does ethanol work as a
fuel? Does a gallon of ethanol provide the same energy as a gallon of
gasoline?
To test this question, we would place a small sample of ethanol and
excess oxygen inside the bomb calorimeter. We would then ignite the
sample inside the container and measure how much the temperature of the
calorimeter increased. Using this temperature change, we could determine
the joules of energy per gram of ethanol. We could repeat the experiment
with gasoline and then compare the results of the two experiments. Look at
the data in Example 8.9 and in question 10, and see how the results
compare.

In coffee cup calorimetry, the reaction takes place at a constant pressure. In

bomb calorimetry, the reaction takes place at a constant volume.

Example 8.9 Finding the Heat of a Reaction Using Bomb

Calorimetry
A chemist places a 20.0-g sample of ethanol inside a bomb calorimeter with
a known heat capacity of 28.72 kJ/°C. When the ethanol ignites, the
temperature of the calorimeter rises from 22.04 °C to 42.74 °C. How much
heat did the ethanol release? Calculate the energy released in kilojoules
per gram of ethanol.

493
 From the initial and final temperatures, we see that the calorimeter
undergoes a temperature change, ΔT, of 20.70 °C. Using the heat capacity
of the bomb calorimeter, we can determine the heat released from this
reaction:
q=CΔT=(28.72 kJ°C)(20.70 °C)=594.5 kJ

We can also express this result in kilojoules per gram of ethanol:

heat releasedgrams fuel=594.5 kJ20.0 g=29.7 kJ/g

The practice problem that follows repeats this problem for gasoline. I
encourage you to try this practice problem and then compare the results for
the two fuels. •

IT
TRY

9. When a chemist repeated the experiment in Example 8.9 using a 20.0-

g sample of gasoline, the temperature rose from 22.0 °C to 54.4 °C.
Calculate the energy released by gasoline in kilojoules per gram. How
does this compare to the energy released by ethanol in Example 8.9?

494
8.3 Heat Energy and Chemical Reactions
Calorimetry provides scientists and engineers with valuable information
about the amount of energy that is absorbed or produced in chemical
reactions. Scientists and engineers rely on this type of information for
many applications, from understanding how reactions take place to
designing efficient automobile engines to creating safe manufacturing
environments.
The energy absorbed or released in a reaction is an extensive property
—that is, it relies on how much matter is involved (Figure 8.13). Because
of this, we express energy changes as a function of the amount of
substances involved in the reaction.

Figure 8.13 The energy released from a chemical reaction depends on the amount of
matter that reacts. A small campfire releases much less heat than a massive forest fire.

Fuel Value
Earlier, we looked at how chemists use bomb calorimetry to measure the
energy stored in fuels such as gasoline and ethanol. The fuel value of a
substance is the amount of energy that can be produced by its combustion.
We often express the fuel value in terms of energy per unit mass (for
example, kilojoules/gram). Table 8.3 gives the fuel value of several
common fuels. The higher the fuel value, the more energy it releases when
it burns.

TABLE 8.3 Fuel Values for Common Combustion Fuels

Fuel Fuel Value (kJ/g)

Methane 55.5

Natural gas 54.0

495
 Propane 50.3

Butane 49.5

Gasoline 46.5

Anthracite coal 34.6

Ethanol 29.7

Wood (oak) 18.9

Data from CRC Handbook of Chemistry and Physics, 92nd ed. (Boca Raton, FL: CRC Press,
2011).

Example 8.10 Using Fuel Values

Based on the data in Table 8.3, how much energy (in kJ) would be released
by the combustion of 2.50 kilograms of coal? How much energy would be
released by the combustion of 2.50 kilograms of oak?
To answer these questions, we first convert the mass of each species from
kilograms to grams. Then we use the fuel value to convert from grams to
kilojoules of energy. Rounding to three significant digits, we obtain values
of 86,500 kJ of energy from coal and 47,300 kJ of energy from oak.•

2.50 kg coal×1,000 g1 kg×34.6 kJg coal=86,500 kJ energy from coal2.50 kg

oak×1,000 g1 kg×18.9 kJg oak=47,300 kJ energy from oak

IT
TRY

10. Based on Table 8.3, which releases more energy, the combustion of
1.0 kg of natural gas or the combustion of 3.0 kg of anthracite coal?

496
 Let’s return to the analogy of wealth and spending for a moment. Right now,
what is your total net worth? That is, what is the value of all of your
possessions? It’s hard to answer this question exactly. But if I asked how much
you spent on lunch today, you probably could tell me much more precisely. It’s
hard to measure total wealth, but easy to measure a change in wealth.
The same is true with energy changes. We rarely know the total energy of a
system, with all the different forms of potential energy. However, we can easily
measure energy changes that take place. These energy changes drive the
physical and chemical properties that we can observe in the world around us.

Reaction Enthalpy
One of the most important measures of energy change is the reaction
enthalpy, represented by the symbol ΔHrxn. The reaction enthalpy is the
amount of heat energy that is absorbed or released in a chemical reaction
at constant pressure.
Chemists often write ΔHrxn along with a chemical reaction. For
example, consider the combustion of ethanol:
C2H6O+3 O2 (g)→2 CO2 (g)+3 H2O (g) ΔHrxn=−1,368 kJ

Notice that the sign of ΔHrxn is negative. This means that heat is
released from the system (the reaction). This is an exothermic reaction.

ΔHrxn is negative for an exothermic reaction.

The reaction enthalpy relates the heat change to the number of moles
of reactants and products. The equation above means that each mole of
C2H6O releases 1,368 kJ of heat when it burns. The ΔHrxn is related to the
number of moles of each reactant or product by the coefficients in the
balanced equation. For this equation, we can write the following
conversion factors:

497
ΔHrxn is the heat energy change per number of moles in the balanced
equation.

The enthalpy of a reaction allows us to relate the heat changes in a

reaction to the amount of a substance that reacts. Examples 8.11 and 8.12
illustrate this important calculation.

Example 8.11 Predicting Changes in Heat Energy Using the

Enthalpy of a Reaction
The enthalpy of reaction for the combustion of ethanol (C2H6O) is – 1,368
kJ/mol, as in the equation below. How much heat will be released by the
combustion of 789.0 g of ethanol?
C2H6O (l)+3 O2 (g)→2 CO2 (g)+3 H2O (g) ΔHrxn=−1,368 kJ

In this problem, we’re given the number of grams of ethanol. The ΔHrxn
relates the energy per mole of ethanol. To solve this problem, we must first
convert to moles of ethanol and then to kilojoules of energy:
Grams C2H6O⇒Moles C2H6O⇒Kilojoules of energy

To convert from grams to moles, we use the molar mass of C2H6O (46.08
g/mol). We then use the conversion factor that relates ΔHrxn to moles of
C2H6O:

The negative value in this answer means that this amount of heat is released
to the surroundings as this reaction occurs. •

Example 8.12 Predicting Changes in Heat Energy Using the

Enthalpy of a Reaction

498
 Many manufacturers produce hydrogen gas from methane gas, as in the
reaction below. This reaction is endothermic, with ΔHrxn = 206.1 kJ. How
much heat energy is required to produce 1.00 kg of hydrogen gas?
CH4 (g)+H2O (g)→CO (g)+3 H2 (g) ΔHrxn=206.1 kJ

To solve this problem, we must convert from the mass of hydrogen to

moles of hydrogen and then finally to kilojoules of energy:
Kilograms H2⇒Grams H2⇒Moles H2⇒Kilojoules of energy

As before, we use the molar mass of H2 to convert from grams to moles

and then use the conversion factor from the balanced equation to convert
from moles to kilojoules.

Notice that the final conversion factor related the energy in kilojoules to 3
moles of H2. When writing a conversion factor involving ΔHrxn, the
coefficients from the balanced equation indicate the number of moles
present. •

Physical changes, such as a phase change or the formation of an

aqueous solution, also involve enthalpy changes (ΔH). For example, these
two equations describe the enthalpy changes for melting and freezing
water:
Melting: H2O (s)→H2O (l) ΔH=44.0 kJFreezing: H2O (l)→H2O
(s) ΔH=−44.0 kJ

499
 Notice that the energy absorbed when ice melts is the opposite of the
energy released when ice freezes.

Similarly, we can describe the ΔH for the formation of a solution:

KOH (s)→KOH (aq) ΔH=−57.7 kJ NH4Cl (s)→NH4Cl (aq)ΔH=14.0 kJ

Example 8.13 illustrates the use of enthalpies to describe physical

changes.

Example 8.13 Using Enthalpy to Describe a Physical Change

Dissolving sodium nitrate in water is an endothermic process. What is the
enthalpy change when 200.0 grams of sodium nitrate dissolves in water?
NaNO3 (s)→NaNO3 (aq) ΔH=21.7 kJ

As before, we solve this problem by converting from grams to moles and

then using the ΔH as a conversion factor. •

500
 IT
TRY

11. Acetylene (C2H2) reacts with hydrogen gas in the combination

reaction shown here. How much heat energy is released if 501.6
grams of H2 react with excess acetylene?
C2H2 (g)+2 H2 (g)→C2H6 (g) ΔHrxn=−311.5 kJ

12. A company needs 5.02 × 103 kJ of heat energy to power a

manufacturing process. The enthalpy of reaction for the combustion
of propane is 2,202 kJ/mol. How many kilograms of propane are
needed to produce this amount of heat energy?
C3H8 (s)+5 O2 (g)→3 CO2 (g)+4 H2O (g) ΔHrxn=−2,202 kJ

501
Summary
In this chapter, we’ve explored the relationships between energy and
change. When a physical or chemical change takes place, energy passes
from one location or system to another. Energy transfer takes place in the
form of work or of heat.
When describing energy changes, it is important to define the system
under investigation. In an endothermic reaction, a system absorbs heat
energy from its surroundings. In an exothermic reaction, a system releases
energy to its surroundings. The law of conservation of energy states that
energy is not created or destroyed but moves from one form to another.
Heat and temperature are closely related. Heat energy refers to the total
kinetic energy transferred from one substance or object to another.
Temperature is a measure of the average kinetic energy of the molecules
in a substance. The temperature change brought about by the gain or loss
of heat energy is unique to each substance. For any substance, the specific
heat (also called the specific heat capacity) is the amount of heat required
to raise one gram of a substance by one degree Celsius. A similar term is
the heat capacity—this is the amount of heat required to raise the
temperature of any object by one degree Celsius. Heat capacity is useful
for larger, complex objects where temperature changes are important.
Scientists measure heat changes using a technique called calorimetry.
Two common forms of this technique are coffee cup calorimetry and bomb
calorimetry. These studies typically involve an insulated container that
holds water and the system being studied. As the physical or chemical
change takes place, the temperature of the water changes. Using the
specific heat of water and the heat capacity of the insulated container,
scientists can measure the heat absorbed or released by the system.
The amount of energy absorbed or released in a chemical reaction is an
extensive property, meaning it depends on the amount of the substances
involved. Two common measurements of reaction energy are fuel value
and reaction enthalpy. Fuel value gives the amount of energy released by
different fuels in combustion reactions. Reaction enthalpy, ΔHrxn, gives
the relationship between the heat change in a constant-pressure reaction
and the amount of each substance (in moles) involved in the reaction.
Enthalpy changes are also useful for describing physical changes, such as
phase changes and the formation of solutions.

502
Gasoline or Ethanol— Which Fuel Is Better?

At the beginning of this chapter, we considered a question: Which makes a better

fuel, ethanol or gasoline? Using the concepts in this chapter, we can measure and
compare the energy available from the combustion of a gallon of ethanol and from
a gallon of gasoline.
Gasoline is a mixture of compounds made up primarily of carbon and
hydrogen. A major component of gasoline is octane (C8H18). Using calorimetry
experiments, scientists have carefully measured the reaction enthalpies for the
combustion of ethanol and that of octane. Let’s compare the ΔHrxn for the
combustion of one mole of these two compounds:
Ethanol: C2H6O+72O2→2 CO2+3 H2O ΔHrxn=−1,368 kJOctane:
C8H18+252O2→8 CO2+9 H2O ΔHrxn=−5,470 kJ

Notice that we used fractional coefficients for oxygen. While this is not the
conventional format for a balanced equation, we write it this way so the ΔH for
each reaction is the energy released by one mole of fuel.
If we divide each ΔHrxn by the molar mass and take the absolute value, we can
obtain the fuel value of these compounds in kilojoules per gram:
Fuel value of ethanol (kJ/g)=|−1,367 kJ/mol46.08 g/mol|=29.69 kJ/gFuel value of
octane (kJ/g)=|−5,470 kJ/mol114.26 g/mol|=47.87 kJ/g

Finally, we can use the densities of ethanol and octane to convert this result
from kJ/gram to kJ/gallon:
Fuel value of ethanol=29.69 kJg×0.789 gmL×3,785 mLgallon=88,700
kJ/gallonFuel value of octane=47.87 kJg×0.703 gmL×3,785 mLgallon=127,000
kJ/gallon

503
 This means that a gallon of octane contains 43% more chemical potential
energy than a gallon of ethanol. Because of this, vehicles that use ethanol require
more gallons of fuel than those that use gasoline. The U.S. government website
fueleconomy.gov reports that E15 mixtures (gasoline containing about 15%
ethanol) get 4–5% fewer miles per gallon than fuel containing 100% gasoline.

504
 Key Terms
8.1 Energy, Work, and Heat
thermodynamics The scientific field that deals with energy and temperature
changes.
energy The ability to do work.
work The transfer of energy from one form to another.
potential energy Energy that is stored.
kinetic energy The energy of motion; the faster an object is moving, the
greater kinetic energy it has.
heat energy A form of kinetic energy involving the kinetic energy of the
particles within a substance; heat energy specifically refers to the total kinetic
energy transferred from one substance to another.
joule (J) The standard unit of energy; 1 J = 1 kg·m2/s2.
exothermic change A physical or chemical change that releases energy to the
surroundings.
endothermic change A physical or chemical change that absorbs energy from
the surroundings.
system In thermodynamics, the part of the universe being studied.
surroundings In thermodynamics, everything that exists around the system
being studied.
law of conservation of energy In a chemical or physical change, the total
energy of the universe remains constant.

8.2 Heat Energy and Temperature

temperature A measure of the average kinetic energy of the molecules in a
substance.
specific heat The amount of heat required to raise the temperature of one gram
of a substance by one degree Celsius; sometimes called the specific heat
capacity.
heat capacity The amount of heat required to raise the temperature of a given
object.
calorimetry An experimental technique used to measure heat changes.
coffee cup calorimetry A technique for measuring heat changes that uses an
insulated container (such as a Styrofoam coffee cup) to measure heat changes.
bomb calorimetry A technique for measuring heat changes using a sealed

505
container; commonly used to measure high-energy reactions.

8.3 Heat Energy and Chemical Reactions

fuel value The amount of heat energy that can be released by a combustion
reaction of a certain substance.
reaction enthalpy (ΔHrxn) The amount of heat energy that is absorbed or
released in a chemical reaction at constant pressure.

506
 Additional Problems

8.1 energy, Work, and heat

13. How are energy, work, and heat related?

14. What is the difference between potential energy and kinetic energy?

15. Each of these objects contains potential energy. How may the energy be
released?
a. an anvil positioned at the top of a cliff
b. a plate of spaghetti
c. a tightly pulled bowstring
d. a stick of dynamite

16. Each of these things contains potential energy. How may the energy be
released?
a. a combustible gas such as acetylene
b. a high-calorie lunch
c. an apple hanging from a branch
d. water in a dam-formed lake

17. Refer to Table 8.1 to complete these energy conversions:

a. Convert 12.8 kilowatt hours (kWh) to joules (J).
b. Convert 259.3 kilocalories to joules.
c. Convert 300 kWh to British thermal units (BTUs).

18. Refer to Table 8.1 to complete these energy conversions:

a. Convert 14.8 kilojoules to kilocalories.
b. Convert 25,000 BTU to kWh.
c. Convert 28.3 kWh to J.

19. A recent-model refrigerator is estimated to use 722 kWh each year. What
is this energy consumption in BTUs? What is it in kJ?

20. In 2012, the average U.S. home used 10,837 kWh. What is this energy
consumption in BTUs? What is it in kJ?

507
21. An average banana has about 105 kilocalories of energy content. What is
this value in joules?

22. A jelly donut has a food value of about 925 kilojoules. What is this value
in kilocalories?

23. What are the two ways in which energy transfers take place?

24. How are heat energy and kinetic energy related?

25. Consider the following chemical reaction. Is this change endothermic or

exothermic? Which has the higher potential energy, the starting materials
or the products?
CH4(g)+2 O2 (g)→CO2 (g)+2 H2O (g)+heat energy

26. Consider the following chemical reaction. Is this change endothermic or

exothermic? Which has the higher potential energy, the starting materials
or the products?
heat energy+Cu(OH)2 (s)→CuO (s)+H2O (g)

27. Identify each of these reactions as endothermic or exothermic:

a. heat + H2O (l) → H2O (g)
b. H2O (g) → H2O (l) + heat
c. K2CO3 (s) → K2CO3 (aq) + heat
d. heat + NH4Cl (s) → NH4Cl (aq)

28. Identify each of these reactions as endothermic or exothermic:

a. C3H8 + 5 O2 → 3 CO2 + 4 H2O + energy
b. energy + Zn(OH)2 → ZnO + H2O
c. KOH (s) → KOH (aq) + heat
d. NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l) + heat

29. Write a balanced equation for each of these reactions. Include heat with the
starting materials or products as appropriate.

508
 a. Liquid water freezes, forming ice (exothermic).
b. Magnesium and oxygen combine to form magnesium oxide
(exothermic).
c. Calcium hydroxide forms calcium oxide plus water (endothermic).

30. Write a balanced equation for each of these reactions. Include heat with the
starting materials or products as appropriate.
a. Carbon dioxide changes from a solid to a gas (endothermic).
b. Solid carbon reacts with oxygen gas to form carbon dioxide gas
(exothermic).
c. Potassium (solid) reacts with chlorine gas to form solid potassium
chloride (exothermic).

31. Describe each of these changes as endothermic or exothermic, using the

frame of reference given:
a. The water in a lake releases heat to the air, forming ice. (system: water)
b. The wax in a candle burns. (system: candle)
c. You warm your hands by a fire. (system: your hands)

32. Describe each of these changes as endothermic or exothermic, using the

frame of reference given:
a. A grenade detonates, creating a fireball. (system: the grenade)
b. A stove heats water to boiling. (system: the stove)
c. A stove heats water to boiling. (system: the water)

33. When many ionic compounds dissolve, they release heat. For example,
when calcium sulfate mixes with water, the temperature of the water may
rise by 10 °C. If calcium sulfate is the system, what makes up the
surroundings? Is this reaction endothermic or exothermic?

34. Ammonium chloride is commonly used in instant ice packs. When this
compound dissolves, the temperature of the water quickly drops. If
ammonium chloride is the system, what makes up the surroundings? Is this
reaction endothermic or exothermic?

35. On a cold December day in Indiana, a boy foolishly responds to a dare and
touches a flagpole with his tongue. The cold flagpole quickly freezes the
moisture on his tongue, which then sticks to the pole. Describe this change
as endothermic or exothermic:
a. if the boy is the system

509
 b. if the flagpole is the system

36. A man places some hamburgers on a charcoal grill. As the charcoal burns,
the heat cooks the burgers. Describe this change as endothermic or
exothermic:
a. if the charcoal is the system
b. if the burgers are the system

37. Find the total change in energy for these situations:

a. A gas absorbs 20 kJ of heat and has 30 kJ of work done to it.
b. A gas releases 30 kJ of heat and does 45 kJ of work on its surroundings.

38. Find the total change in energy for these situations:

a. A gas releases 53.2 kJ of heat and does 23.7 kJ of work on its
surroundings.
b. A gas absorbs 15.7 kJ of heat and has 42.3 kJ of work done to it.

39. Inside a small engine, the combustion of gasoline releases 47.0 kJ of

energy. If the sample does 15.0 kJ of work on the engine, how much
energy is lost as heat?

40. A sample of ethanol reacts, releasing 42.1 kJ of energy. If 25.0 kJ are

released as heat, how much work is done on the surroundings?

41. A military group tests a high-energy chemical explosive underground. As

the explosive is detonated, the ground around it shakes. The potential
energy has been released from the explosive, but where has it gone? Has
the total energy of the universe changed?

42. Miners extract coal from the mountains of West Virginia. The coal is
shipped to a power plant in New York, where it is burned to produce
electricity. What is the relationship between the chemical energy stored in
the coal, the heat released, and the electricity produced?

43. A bottle of frozen water is dropped into a cooler filled with water. As the
water inside the bottle melts, it absorbs 182,000 joules of heat energy from
the surrounding water. What happens to the temperature of the water as a
result? If the ice inside the bottle is the system, is this change endothermic
or exothermic? Use this table to complete your answer.

510
 Surroundings: Water in
System: Ice Bottle Cooler

Change +182,000 J

Result Ice melts, temperature

increases

Heat Absorbed or
Released?

44. A blacksmith immerses a red-hot iron rod in a large bucket of water. As

the iron cools, it releases 138 kJ of heat energy into the surrounding water.
What happens to the temperature of the water as a result? If the iron is the
system, is this change endothermic or exothermic? Use this table to
complete your answer.

System: Iron Rod Surroundings: Water in Bucket

Change −138 kJ

Result Iron cools

Heat Absorbed or Released?

8.2 Heat Energy and Temperature

45. What is the difference between heat and temperature?

46. A pot of water is heated on a stove. How does the motion of the molecules
in the liquid change as the temperature increases?

47. What is the difference between an object’s specific heat and its heat
capacity?

48. A bomb calorimeter contains a steel casing, insulation, and water

surrounding the inner bomb. When measuring temperature changes, does it
make more sense to use the specific heat for this calorimeter or the heat
capacity?

49. Zinc has a specific heat of 0.39 J/g·°C while iron has a specific heat of
0.45 J/g·°C. If a 100-g sample of each metal is cooled from 100 °C to
room temperature (25 °C), which one releases more heat energy?

511
50. The temperature of a spoonful of water increases by one degree for each 20
J of energy added. The temperature of a bathtub full of water increases one
degree for each 7.5 × 104 J of energy added. Which has the greater heat
capacity, the water in the spoon or the water in the bathtub? Which (if
either) has the greater specific heat?

51. An engineer tests the thermal properties of a metal alloy. Using a 50.0-g
sample, she finds that adding 485 J of heat energy to the alloy causes a
temperature change of 4.10 °C. What is the specific heat of this alloy?

52. A chemist tests the properties of a 230.0-g sample of an unknown metal.

He finds that 1,217 J of heat increases the temperature of the metal by 5.88
°C. What is the specific heat of this metal?

53. Aluminum has a specific heat of 0.91 J/g·°C. If a 4,000-g sheet of

aluminum absorbs 28,000 joules of energy, how much will the temperature
increase?

54. Tin has a specific heat of 0.21 J/g·°C. If a 3.2-kg sheet of tin absorbs 8.4 kJ
of energy, how much will the temperature increase?

55. Gold has a specific heat of 0.031 cal/g·°C. How many calories of heat are
required to raise the temperature of a 100.0-g gold bar by 15 °C?

56. Iron has a specific heat of 0.108 cal/g·°C. If a 2.1-kg iron bar cools from a
temperature of 180 °C to a temperature of 25 °C, how much heat does it
release?

57. An industrial reaction vessel is found to undergo a change in temperature

of 0.061 °C for each kilojoule of energy absorbed. What is the heat
capacity for this vessel in kJ/°C?

58. A boiler system requires 15,000 kJ to heat it from 25 °C to 200 °C. What is
the heat capacity for this system in kJ/°C?

59. A bomb calorimeter is a device used to measure heat changes. One model
of bomb calorimeter has a heat capacity of 14.5 kJ/°C. How much heat
would be required to raise the temperature of this calorimeter from 25.3 °C
to 33.7 °C?

512
60. When filled with water, a large reaction vessel in a chemical plant has a
heat capacity of 540,000 kJ/°C. How many kJ of heat are needed to heat
this entire vessel from 25 °C to 65 °C? How much energy will this require
in kilowatt hours?

61. An industrial reaction vessel has a heat capacity of 6.13 × 103 kJ/°C. If
7,702 MJ of heat is added to the system, by how much will the temperature
change?

62. An insulated reaction vessel has a heat capacity of 105.32 kJ/°C and an
initial temperature of 24.3 °C. If 1,000 kJ of heat energy is added to the
system, what temperature will the reaction vessel reach?

63. A house is built with a granite countertop. The heat capacity of the
countertop is 158.2 kJ/°C. A hot pan of water is placed on the countertop,
and 8,000 J of heat energy is transferred into the countertop. By how much
does the temperature of the countertop change?

64. A house is built with a concrete floor to help keep the house cool. How
many kJ of heat would be required to raise the temperature of a concrete
slab having a mass of 15,000 kg from 65 °F to 75 °F? The specific heat of
concrete is 0.880 kJ/kg·°C. How much energy will this require in kilowatt
hours?

65. How is coffee cup calorimetry different from bomb calorimetry?

66. In bomb calorimetry, the calorimetry is sealed shut so that no work can be
done by the reaction process. In light of this, how does the heat released in
a bomb calorimetry experiment compare to the total energy change?

67. A student places a block of hot metal into a coffee cup calorimeter
containing 150.0 g of water. The water temperature rises from 23.7 °C to
32.1 °C. How much heat (in calories) did the water absorb? How much
heat did the metal lose?

68. A student places a block of hot metal into a coffee cup calorimeter
containing 171.0 g of water. The water temperature rises from 23.3 °C to
42.5 °C. How much heat (in calories) did the water absorb? How much
heat did the metal lose?

69. In a coffee cup calorimetry experiment, a block of hot metal is placed in a

513
 calorimeter containing 100.0 g of water. The water temperature rises from
23.1 °C to 31.5 °C. Calculate qmetal and qwater for this experiment.

70. In a calorimetry experiment, a reaction takes place in 100.0 g of an

aqueous solution. The solution temperature drops from 22.8 °C to 15.5 °C.
Calculate qreaction and qsolution for this experiment. Assume the solution
has a specific heat identical to that of pure water.

71. A chemist conducts a calorimetry experiment on an unknown metal. The

mass of the sample is 24.4 g. The chemist heats the metal to 100 °C and
then places it in a coffee cup calorimeter containing 108.5 g of water with
an initial temperature of 21.3 °C. After some time, both the metal and
water reach an equal temperature of 25.0 °C.
a. How much heat was absorbed by the water?
b. How much heat was released by the metal?
c. What is the specific heat of the metal?
d. Based on Table 8.4, is the unknown metal most likely aluminum, lead,
nickel, or tin?

TABLE 8.4 Specific Heats for Several Metals

Material Specific Heat (cal/g·°C)

Aluminum 0.220

Iron 0.108

Titanium 0.125

Lead 0.031

Nickel 0.104

Tin 0.054

72. A chemist conducts a calorimetry experiment on an unknown metal. The

mass of the sample is 52.4 g. The chemist heats the metal to 100 °C and
then places it in a coffee cup calorimeter containing 95.4 g of water. The
temperature of the water rises from 21.9 °C to 23.2 °C.
a. How much heat was absorbed by the water?
b. How much heat was released by the metal?
c. What is the specific heat of the metal?
d. Based on Table 8.4, is the unknown metal most likely aluminum, lead,

514
 nickel, or tin?

73. A chemist dissolved an 11.6-g sample of KOH in 100.0 grams of water in

a coffee cup calorimeter. When she did so, the water temperature increased
by 25.5 °C. Based on this, how much heat energy was required to dissolve
the sample of KOH? Calculate the heat of solution for KOH in kJ/mol.
(Assume the specific heat of the solution is 4.184 J/g·°C.)

74. When a 20.8-g sample of CsBr was combined with 115.0 grams of water in
a coffee cup calorimeter, the water temperature decreased by 4.47 °C.
Based on this, how much heat energy was released when CsBr was
dissolved? Calculate the heat of solution for CsBr in kJ/mol. (Assume the
specific heat of the solution is 4.184 J/g·°C.)

75. A sample of jet fuel is tested in a bomb calorimeter. Before detonation, the
temperature of the calorimeter is 22.31 °C. If the fuel releases 558.1 kJ of
heat, and the heat capacity of the calorimeter is 13.12 kJ/°C, determine the
final temperature of the calorimeter.

76. A sample of charcoal is incinerated in a bomb calorimeter. Before

detonation, the temperature of the calorimeter is 22.56 °C. If the fuel
releases 33.58 kJ of heat, and the heat capacity of the calorimeter is 13.12
kJ/°C, determine the final temperature of the calorimeter.

77. A 1.0-g sample of a candy bar was placed in a bomb calorimeter with a
heat capacity of 3.54 kcal/°C. The sample was completely burned, causing
the temperature of the calorimeter to rise by 1.36 °C. How many
kilocalories of energy were stored in the candy bar?

78. A 2.0-g sample of a glazed donut was placed in a bomb calorimeter with a
heat capacity of 3.54 kcal/°C. The sample was completely burned, causing
the temperature of the calorimeter to rise by 2.02 °C. How many
kilocalories of energy were stored in the donut sample?

8.3 Heat Energy and Chemical Reactions

79. A synthetic fuel blend is tested by bomb calorimetry. It is found that an

80.10-g sample releases 28,096 kJ of heat. What is the fuel value of this
blend in kJ/g?

80. A 15.0-g sample of coal is tested by bomb calorimetry. Upon combustion,

515
 the sample releases 334.5 kJ of heat. What is the fuel value of this blend in
kJ/g?

81. Using data from Table 8.3 in the text, calculate the energy released by the
combustion of each of these compounds:
a. 100.0 g of methane
b. 15.0 mL of gasoline (density of gasoline = 0.70 g/mL)

82. Using data from Table 8.3 in the text, calculate the energy released by the
combustion of each of these compounds:
a. 10.0 kg of propane
b. 18.0-mL sample of ethanol (density of ethanol = 0.79 g/mL)

83. What mass of propane must be burned to produce 30,000 BTUs of heat
energy? (Refer to Tables 8.1 and 8.3 for the proper conversion factors and
fuel values. Report your answer to three significant digits.)

84. What mass of methane must be burned to produce 15,000 BTUs of heat
energy? (Refer to Tables 8.1 and 8.3 for the proper conversion factors and
fuel values. Report your answer to three significant digits.)

85. Recently scientists have explored the use of crop residues as a renewable
fuel. In one such test, a 21.05-g sample of dried cornstalks was placed
inside a bomb calorimeter with a known heat capacity of 28.72 kJ/°C.
When the stalks were burned, the temperature of the calorimeter rose from
24.10 °C to 36.93 °C. How much heat was released by the combustion of
the cornstalks? What is the fuel value of the cornstalks in kJ/g?

86. Recently scientists have explored the use of switchgrass as a renewable

fuel. In one such test, a 53.18-g sample of dried switchgrass (Figure 8.14)
was placed inside a bomb calorimeter with a known heat capacity of 30.31
kJ/°C. When the grass was burned, the temperature of the calorimeter rose
from 24.35 °C to 55.95 °C. How much heat was released by the
combustion of the switchgrass? What is the fuel value of the grass in kJ/g?

516
 Figure 8.14 Switchgrass is a hardy plant that grows easily in many places. Scientists
have explored the use of switchgrass as a renewable fuel.

87. Using the reaction enthalpy data shown, determine the amount of heat
released in each situation:
2 Ca (s)+O2 (g)→2 CaO (s) ΔHrxn=−1,269.8 kJ

a. If calcium reacts with 3 moles of O2

b. If 8 moles of calcium reacts with O2
c. If calcium and oxygen react to form 1 mole of CaO

88. Using the reaction enthalpy data shown, determine the amount of heat
released in each situation:
2 K (s)+Cl2 (g)→2 KCl (s) ΔHrxn=−873.0 kJ

a. If potassium reacts with 5 moles of chlorine

b. If 10 moles of potassium react with chlorine
c. If potassium and chlorine react to form 0.4 moles of KCl

89. Using the reaction enthalpy data shown, determine the heat change in each
situation:
Sn (s)+2 Cl2 (g)→SnCl4 (s) ΔHrxn=−511.3 kJ

a. If 23.13 g of tin reacts with chlorine gas

b. If tin reacts with 18.68 g of chlorine gas
c. If tin and chlorine react to form 11.83 g of tin(IV) chloride

517
90. Using the reaction enthalpy data shown, determine the heat change in each
situation:
MgCO3 (s)→MgO (s)+CO2 (g) ΔHrxn=+100.7 kJ

a. If 18.19 g of MgCO3 decomposes to form MgO and CO2

b. If MgCO3 decomposes to form 139.1 g solid MgO
c. If MgCO3 decomposes to form 12.1 g CO2 gas

91. Methane is the main component of natural gas. Using the reaction enthalpy
given here, calculate the heat energy produced by the combustion of one
kilogram of methane.
CH4 (g)+2 O2 (g)→CO2 (g)+2 H2O (l) ΔHrxn=−861.3 kJ

92. Calcium oxide is the major component of cement mix. Manufacturers

produce this compound by the decomposition of calcium hydroxide,
shown below. Using the reaction enthalpy, calculate the energy required to
produce one kilogram of calcium oxide by this process.
Ca(OH)2 (s)→H2O (g)+CaO (s) ΔHrxn=+108.5 kJ

93. The acid-base neutralization reaction shown here is an exothermic process.

How many grams of NaOH are neutralized if 218.5 kJ of heat are released
in the process?
HBr (aq)+NaOH (aq)→NaBr (aq)+H2O (l) ΔHrxn=−55.8 kJ

94. The acid-base neutralization reaction shown here is an exothermic process.

How many grams of HBr are neutralized if 119.0 kJ of heat are released in
the process?
HBr (aq)+KOH (aq)→KBr (aq)+H2O (l) ΔHrxn=−55.8 kJ

95. Sodium hydroxide dissolves in water in an exothermic process as shown

here. How much heat is released if 139.8 g of NaOH dissolve in water?
NaOH (s)→NaOH (aq) ΔHrxn=−44.0 kJ

518
96. When steam condenses to form liquid water, it releases heat to its
surroundings. How much heat is released when 100.0 g of steam returns to
the liquid state?
H2O (g)→H2O (l) ΔHrxn=−40.67 kJ

97. Thermite reactions, such as the reaction of iron(III) oxide with aluminum,
are used to produce intense heat (Figure 8.15). Based on the reaction
enthalpy below, how many grams of aluminum are needed to produce
1,000,000 kJ of heat by this reaction? Report your answer to three
significant digits.

Figure 8.15 Thermite reactions produce intense heat.

2 Al (s)+Fe2O3 (s)→Al2O3 (s)+2 Fe (s) ΔHrxn=−851.5 kJ

98. Butane, C4H10, is the fuel used in many handheld lighters (Figure 8.16).
Based on the reaction enthalpy below, how many grams of butane are
needed to produce 1,000 kJ of heat by this reaction? Report your answer to
three significant digits.

519
 Figure 8.16 Handheld lighters like this one use butane fuel.

2 C4H10 (g)+13 O2 (g)→8 CO2 (g)+10 H2O (l) ΔHrxn=−5,755 kJ

99. Magnesium reacts with oxygen in the very exothermic reaction shown
here. If 30.18 g of magnesium were reacted in a bomb calorimeter having
a heat capacity of 38.33 kJ/°C, by how much would the temperature rise?
2 Mg (s)+O2 (g)→2 MgO (s) ΔHrxn=−1,203.2 kJ

100. A chemical manufacturer needs to neutralize a mixture containing 25.2 kg

of sodium hydroxide by reacting it with excess hydrochloric acid. The
chemical engineers estimate that once filled, the insulated reaction vessel
will have a heat capacity of 318.5 kJ/°C. Based on this data, determine
the amount of heat released by this reaction. If the neutralization starts at
a temperature of 30.0 °C, how hot will the reaction vessel get?
HCl (aq)+NaOH (aq)→NaCl (aq)+H2O (l) ΔHrxn=−55.8 kJ

520
Chapter Nine
Covalent Bonding and Molecules

The Shortest Race

In April 2017, six teams competed in the shortest race in history. The race was
only 100 nanometers long—much too small to see with the naked eye. And the
cars that competed were each composed of a single molecule. Scientists call these
tiny vehicles nanocars (Figure 9.1).

Figure 9.1 (a) The first NanoCar Race featured single-molecule cars from six different
research teams. (b) Professor James Tour (right) works with students in his Rice
University laboratory. (c) This image is an artist’s rendition of an early nanocar on a
gold surface. (d) The NanoCar from Rice University ran on a silver surface, rather than
gold. It was piloted by students from the Grill research group at the University of Graz
in Austria. This image shows the nanocar (circled in red) at the top. The spots circled in

521
blue are single-atom obstacles that the car had to navigate.

Explore
Figure 9.1

The idea was pioneered by Dr. James Tour, a professor of chemistry and
materials science at Rice University. Inspired by the way nature builds structures,
Tour set out to create molecular-scale machines. In 2005, Tour and his students
reported building the first nanocar, and they showed that this single molecule could
roll across a smooth gold surface. In the years since, they’ve learned to power the
cars with “motors” that convert light energy to push the car along. They’ve built
single-molecule submarines that propel themselves through water. They’ve even
built “nanotrucks” and “nanotrains” that transport a tiny cargo from one place to
another.
Tour believes that, although it may take centuries for the technology to
develop, we’ll eventually use tiny machines to build large structures. “Right now,
we’re learning to control molecular motion,” he says.
His groundbreaking ideas are catching on. In 2017, several research teams met
in Toulouse, France, for the first ever NanoCar Race. The incredibly tiny course
was on a smooth gold surface inside an instrument called a scanning tunneling

522
microscope. On the surface, single atoms marked the obstacles and finish line. The
“drivers” used the instrument to guide the cars around the obstacles and across the
finish line.
So, how do these tiny machines work? The answer to this question begins with
the shape of the molecule. When atoms combine to form covalent bonds, they
produce structures with predictable shapes. Tour and his students take advantage of
this phenomenon. They use chemical reactions to connect groups of atoms as if
they were Lego® blocks, creating wheels, axles, and even motors for their
molecular vehicles.
In fact, molecular shape affects the properties of any covalent compound, from
simple structures like water to the complex machinery of living creatures. In this
chapter, we’ll see how atoms share electrons through covalent bonds, and look at
how electron arrangement determines the shape of a molecule. We’ll consider the
way charges spread across molecules and ions, and preview the impacts of shape
and charge distributions on the behavior of different compounds.
The concepts in this chapter are vitally important. Bonding, shape, and charge
form a link between the fundamental aspects of chemistry and the properties that
we see in the world around us. So fasten your nano-seatbelt; let’s take a drive.

Intended Learning Outcomes

After completing this chapter and working the practice problems, you should be
able to:

9.1 Covalent Molecules

Describe the electronic arrangements of covalent structures, including single,
double, and triple covalent bonds, lone pairs, and filled and expanded octets.
Draw Lewis structures for simple molecules.

9.2 Molecules and Charge

Calculate formal charge, and relate it to the overall charge of polyatomic
ions.
Draw Lewis structures for polyatomic ions.
Use resonance structures to describe bonding and charge distribution.

9.3 Shapes of Molecules

Apply the VSEPR model to predict the electronic and molecular geometry
for molecules having two, three, or four charge sets.

9.4 Polar Bonds and Molecules

Describe the trends in electronegativity across the periodic table.

523
Use differences in electronegativity to differentiate between covalent, polar
covalent, and ionic bonds.
Estimate molecular dipoles through the combination of polar covalent bonds
and molecular shape.

524
9.1 Covalent Molecules

Representing Covalent Structures

We first encountered chemical bonding in Chapter 5. Remember that there
are two broad classes of chemical bonds, ionic and covalent. Ionic bonds
result from the attraction between oppositely charged ions. In a covalent
bond, two atoms share electrons.
Covalent bonds occur between nonmetal atoms. By sharing electrons,
these atoms are often able to fill their valence levels and satisfy the octet
rule. Recall that the octet rule states that atoms are stabilized by
completely filled s and p sublevels—this corresponds to eight electrons in
the valence shell.
The octet rule: Atoms are stabilized by eight electrons in their valence level.

For example, consider chlorine gas, Cl2. Each neutral chlorine atom
has seven valence electrons. By forming a two-electron bond, both
chlorine atoms fill their valence levels. In Lewis structures, we use a dash
to represent two shared electrons:

Sometimes, atoms share two pairs of electrons. This is called a

covalent double bond. We can see an example of a covalent double bond
with molecular oxygen, O2. Each oxygen atom has six valence electrons
and therefore needs two additional electrons to fulfill the octet rule.

In a double bond, atoms share two pairs of electrons.

Similarly, atoms can share three pairs of electrons to form a covalent

triple bond. This type of bonding occurs in a molecule of nitrogen gas.
Each nitrogen atom has five valence electrons and therefore needs three
additional electrons to fill its valence. The nitrogen atoms form a triple
bond.

525
 In the Lewis structures just described, we also showed the unshared
valence electrons. Although unshared electrons do not form bonds, they
often affect the shape of a molecule.
Because electrons pair up with opposite spins, we typically represent
unshared electrons in pairs. For example, in the Lewis structure of
chlorine, each chlorine atom has four pairs of electrons. One pair of
electrons forms the covalent bond; the other three pairs of electrons are
unshared. Chemists refer to pairs of unshared valence electrons as lone
pairs (Figure 9.2). We can think of an octet as four pairs of electrons.

Figure 9.2 This image shows the Lewis structure for a chlorine molecule. The dash
between the two atoms represents a two-electron covalent bond. We refer to pairs of
unshared molecules as lone pairs.

An octet contains four pairs of electrons.

In most covalent molecules and polyatomic ions, all atoms follow the
octet rule. For example, consider the Lewis structure of carbon dioxide,
CO2 (Figure 9.3): In this molecule, the carbon atom forms four covalent
bonds. This means there are eight valence electrons around the carbon
atom. Each oxygen atom forms two covalent bonds and has two lone pairs.
Each oxygen atom also has eight valence electrons. Both carbon and the
two oxygen atoms fulfill the octet rule.

Figure 9.3 Each atom in a CO2 molecule fulfills the octet rule.

Figure 9.4 shows the Lewis structure for water. Notice that the oxygen
atom is surrounded by eight valence electrons—a complete octet. Each
hydrogen atom only has two electrons (the covalent bond). Because only
two electrons are needed to fill energy level 1, each hydrogen atom in this
structure also has a complete valence level.

526
Figure 9.4 In a water molecule, the oxygen atom’s valence level is filled with eight
electrons, and each hydrogen atom’s valence level is filled with two electrons.

Hydrogen needs just two electrons to fill its valence.

Example 9.1 Identifying the Valence Electrons in a Lewis

Structure
Urea (CH4N2O) is a covalent compound that is a major component of
urine. The Lewis structure for urea is shown below. Indicate the number of
covalent bonds and lone pairs around each carbon, hydrogen, nitrogen,
and oxygen atom. Do these atoms fulfill the octet rule?

The carbon atom in this molecule has four covalent bonds (two single
bonds and one double bond). This arrangement corresponds to eight
valence electrons and fulfills the octet rule. The two nitrogen atoms each
have three covalent bonds and one lone pair. This also adds up to eight
electrons. The oxygen atom also has eight valence electrons (two covalent
bonds and two lone pairs). So the carbon, nitrogen, and oxygen atoms all
fulfill the octet rule.
Each hydrogen atom has only two electrons in its valence shell. Because
energy level 1 holds only two electrons, the hydrogen atoms also have a
filled valence level. This table summarizes these common bonding patterns.
•
Atom Covalent bonds Lone pairs Valence electrons
C 4 0 8
N 3 1 8

527
 O 2 2 8
H 1 0 2

IT
TRY

1. The Lewis structure of hypochlorous acid, HClO, is shown here.

Identify the number of covalent bonds and lone pairs around each
atom. Do these atoms fulfill the octet rule?

Exceptions to the Octet Rule

Although most covalent molecules follow the octet rule, some exceptions
occur. For example, the element boron often forms covalent compounds in
which the boron atom has an incomplete valence level (Figure 9.5).

Figure 9.5 Many boron compounds have an incomplete valence level.

A more common exception occurs with larger elements. The nonmetals

in rows 3–7 of the periodic table sometimes form bonds that result in an
expanded octet containing more than eight electrons. This is possible
because the d sublevels in the higher energy levels can accommodate extra
electrons. Atoms with expanded octets often contain five or six covalent
bonds.

Atoms in rows 3–7 can form expanded octets with more than eight electrons.

For example, phosphorus commonly forms two compounds with

528
chlorine, PCl3 and PCl5 (Figure 9.6). The first compound, PCl3, adheres
to the octet rule. In the second compound, the phosphorus atom has an
expanded octet with 10 valence electrons.

Figure 9.6 PCl3 follows the octet rule. The phosphorus atom in PCl5 has an expanded
octet with 10 valence electrons.

Drawing Lewis Structures

To understand how covalent compounds behave, we often need to
understand their structures. The following four-step process will help you
draw Lewis structures from a molecular formula (Figure 9.7):
1. Add up all the valence electrons. Use the periodic table to determine
the number of valence electrons in each atom. Recall that the main-
group numbers (1A–8A) tell us the number of valence electrons in
each main-group element (Figure 9.8).

Figure 9.8 The main-group numbers 1A–8A tell us the number of valence
electrons in each main-group element.

2. Frame the structure, using single bonds. Typically the atom that is
nearer to the lower left of the periodic table is the central atom.
Because it can hold only two electrons, hydrogen is never a central
atom.
3. Fill the octets of the outer atoms first. Add lone pairs to the outer
atoms to complete their valence levels.
4. Fill the octet on the central atom. After filling in the octets on the

529
 outer atoms, place any remaining electrons on the central atom.
These electrons may be enough to fulfill the octet on the central
atom (and may even result in an expanded octet). If the central atom
does not have at least eight electrons, use double or triple bonds to
fulfill the octet.

Figure 9.7 Follow these steps to draw Lewis structures.

Examples 9.2 and 9.3 illustrate this important technique.

Example 9.2 Drawing a Lewis Structure

Draw a Lewis structure for nitrogen trichloride, NCl3.
We can solve this problem by carefully following the four-step process:
Step 1: Add up all the valence electrons. From the periodic table, we can
determine that the nitrogen atom has five valence electrons, and each
chlorine atom has seven valence electrons. Based on this information, there
are a total of 26 valence electrons in this structure.

Step 2: Frame the structure. We draw nitrogen as the central atom because
it is to the left of chlorine on the periodic table. We connect the central
nitrogen atom to the outer chlorine atoms with a single bond, which we
show as a dash.

Step 3: Fill the octets of the outer atoms first. The covalent bonds each
represent two electrons. To complete the octets on the outer atoms, we must
add six more electrons to each chlorine:

530
Step 4: Fill the octet on the central atom. We used 24 of the 26 valence
electrons to complete the octets around the chlorine atoms. We draw the
remaining two electrons on the central atom:

Notice that in this Lewis structure, we have accounted for all 26 valence
electrons, and each atom has a filled octet. •

Example 9.3 Drawing a Lewis Structure

Formaldehyde, CH2O, is a gas that is commonly used as a precursor to
manufacture plastics. Draw the Lewis structure for a formaldehyde
molecule.
As in the previous example, let’s follow the four-step process for drawing
this structure.
Step 1:
Carbon+Oxygen+two hydrogens=4+6+1(2)=12 valence electrons

Step 2: Draw carbon as the central atom, surrounded by the oxygen and
hydrogen atoms. We connect the carbon to each outer atom using single
bonds:

Step 3: We draw octets around the outer atoms. Because hydrogen is a row
1 element, its valence is filled by the single bond to carbon. We draw six
additional electrons around the oxygen atom to complete its octet:

531
 Step 4: Now we need to fill the octet on the central atom. However, we
have used all 12 valence electrons to complete the octets of the outer atoms.
Therefore, we need to share a pair of electrons between an outer atom and
the central carbon atom. To accomplish this, we move two electrons down
from the oxygen to form a double bond to the carbon, so that both atoms
have a filled octet. •

IT
TRY

2. Draw Lewis structures for these molecules:

a. NH3
b. SF6
c. CS2

3. Draw Lewis structures for these molecules:

a. SiCl4
b. HCN
c. HNO

532
9.2 Molecules and Charge

Polyatomic Ions and Formal Charge

In Chapter 5, we introduced polyatomic ions as groups of atoms with an
overall charge (see Section 5.2). The distribution of these charges across
the ions affects many of their chemical properties. Now that we have a
better understanding of covalent bonding, let’s look at the structures of
eight common polyatomic ions (Figure 9.9). Most of these ions follow the
octet rule, but two ions (sulfate and phosphate) have expanded octets.

Figure 9.9 The Lewis structures of several common polyatomic ions.

In Figure 9.9, notice that each polyatomic ion has at least one atom
with a positive or negative charge. These charges are called formal
charges. They allow us to locate charged sites within a molecule or
polyatomic ion. Formal charges are not a perfect indicator of the charge on
an atom, but they are a good tool for keeping track of the overall charge of
a molecule or ion. We can calculate the formal charge on any atom using
this equation:

Formal charge estimates the total electron charge around bonded

atoms. Formal charge assumes that atoms share electrons evenly. Because
two atoms share two electrons, each covalent bond contributes a charge
equivalent to one electron (Figure 9.10).

533
Figure 9.10 In a covalent bond, the charge from two electrons is spread over two atoms.
As a result, each nucleus feels the effective charge of one electron.

For example, the hydroxide ion has a charge of –1. Based on the
following Lewis structure, does the negative charge lie on the oxygen atom
or on the hydrogen atom?

To answer this, we calculate the formal charge on each atom. Let’s

begin with the hydrogen atom: A neutral hydrogen atom has one electron.
The hydrogen has one covalent bond and zero unshared electrons.
Therefore its formal charge is zero:

Now let’s examine the formal charge on the oxygen atom: A neutral
oxygen atom has six valence electrons. The oxygen in the OH– ion has six
unshared electrons plus an additional covalent bond. This means it has an
excess of electrons and an overall negative charge:
Formal charge (oxygen)=6−1−6=−1

When we draw the Lewis structure for hydroxide, we commonly show the
minus charge adjacent to the oxygen atom:

Let’s look at a second example. The thiocyanate ion, SCN–, has an

overall charge of minus one. If we draw the Lewis structure as shown, the

534
sulfur atom has a formal charge of minus one, and the carbon and nitrogen
atoms both have formal charges of zero:

The sum of the formal charges on the atoms equals the overall charge of the
molecule or ion.

Formal charges are a powerful tool and are especially important in

biological chemistry. Two elements that are critical for life, oxygen and
nitrogen, frequently carry formal charges. When covalently bonded, a
neutral oxygen atom has two bonds and four unshared electrons. If the
oxygen atom has three bonds and two unshared electrons, it will have a
formal charge of +1. On the other hand, if an oxygen atom has only one
bond and six unshared electrons, it will have a formal charge of –1 (Figure
9.11).

Figure 9.11 Oxygen and nitrogen both form species with different formal charges.

Nitrogen is similar: While a neutral nitrogen atom has three bonds and
two unshared electrons, it can gain one bond (giving it a formal charge of
+1) or lose one bond (giving it a formal charge of –1).

Example 9.4 Calculating Formal Charge

Automotive airbags contain sodium azide, NaN3. The Lewis structure for
the azide ion (without charges) is shown below. Calculate the formal
charge on each atom in this structure. What is the overall charge of the
azide ion?

535
 Because nitrogen is in group 5A, we know that a neutral nitrogen atom has
five valence electrons. The central nitrogen atom has four covalent bonds
and zero unshared electrons. Therefore, the formal charge of this atom is
+1:
Formal charge (central atom)=5−4−0=+1

The two outer nitrogen atoms each have two covalent bonds and four
unshared electrons. Therefore, the formal charges of these atoms are each –
1:
Formal charge (outer atoms)=5−2−4=−1

So the atoms in this ion have formal charges of –1, +1, and –1, as shown
below. Based on this, the overall charge of the ion is –1. The correct
formula for the azide ion is N3–. •

IT
TRY

4. Calculate the formal charge on each atom in these structures:

a.

b.

536
 c.

d.

5. Proteins are large molecules that affect almost every function of the
human body. Proteins are composed of smaller building blocks called
amino acids. The simplest amino acid is glycine. In water, two atoms
in a glycine molecule have a formal charge. In the structure of glycine
shown here, all bonds are drawn, but the unbonded valence electrons
are missing. Draw in the missing valence electrons, and locate both
sites that have a nonzero formal charge. Calculate the formal charge at
these sites.

Drawing Lewis Structures for Polyatomic Ions

Drawing Lewis structures for polyatomic ions is similar to drawing Lewis
structures for neutral molecules. However, we must consider the charge of
the ion when we determine the number of valence electrons present. If an
ion has a negative charge, it means that extra electrons are present.
For example, how many valence electrons are present in a hydroxide
ion? We know that hydroxide has the formula OH−. The oxygen has six
valence electrons, and the hydrogen has one valence electron. The charge
(–1) means that one additional electron is present:
Total electrons=6 (oxygen)+1 (hydrogen)+1 (charge)=8 valence electrons

Example 9.5 illustrates this idea further.

537
Example 9.5 Drawing a Lewis Structure for a Polyatomic Ion
Draw a Lewis structure for the nitrite ion, NO2–. Show all nonzero formal
charges.
We can solve this problem by carefully following the steps given earlier:
Step 1: Add up all the valence electrons. From the periodic table, we can
determine that the nitrogen atom has five valence electrons and that each
oxygen atom has six valence electrons. Because there is a –1 charge on the
ion, we know that one additional electron is present on this ion. Therefore,
this ion has a total of 18 valence electrons:

Step 2: Frame the structure. We draw nitrogen as the central atom because
it is to the left of oxygen on the periodic table:

Step 3: Fill the octets of the outer atoms first. Placing electrons around each
oxygen atom uses 16 of the 18 available valence electrons:

Step 4: Fill the octet on the central atom. We draw the remaining two
electrons on the central atom, as shown below. This leaves the nitrogen
atom two electrons short of a filled octet. To fulfill the octet, we must move
two electrons from one of the oxygen atoms to form a double bond between
the nitrogen and oxygen:

Show formal charges. As drawn, the oxygen atom on the left has a formal
charge of minus one. We indicate this with a minus charge, to give the final
answer. •

538
 IT
TRY

6. Draw a Lewis structure for the methylthiolate anion, CH3S–. Carbon is

the central atom. Show any nonzero formal charges.

Choosing the Best Lewis Structure

Formal charges are useful for identifying the best Lewis structures when
more than one structure is possible. For example, let’s draw the structure
of phosgene (COCl2), a highly toxic gas that was used as a chemical
weapon in World War I. We can draw a Lewis structure for this compound
by following the steps outlined earlier. This molecule has a total of 24
valence electrons. Placing the carbon atom in the center and filling in the
octets of the outer atoms, we get a structure like this:

Now we need to fill the octet on the central atom. We could do this
either with a double bond to the oxygen atom or a double bond to one of
the chlorine atoms, as shown below. Which structure is more correct?

In general, the best Lewis structures are those that minimize the charge
values or have zero values for all of the atoms. The best structure for
phosgene has a double bond to the oxygen, and single bonds to both
chlorines, because all atoms have a formal charge of zero:

539
Resonance
While Lewis structures and formal charges are helpful tools, they do not
always give us a complete picture of molecular structure. For example, the
Lewis structure of the nitrite ion, NO2–, shows a double bond to one
oxygen atom and a single bond to the other. Because double bonds are
shorter than single bonds, we might expect the nitrogen–oxygen bonds to
be different lengths (Figure 9.12).

Figure 9.12 In the nitrite ion, both oxygen atoms have bonds of equal length, and the
negative charge is shared evenly between the two oxygen atoms. We represent this
arrangement with resonance structures.

In fact, both of the bonds are the same length. What’s more, the
negative charge is shared equally between the two oxygens, so each atom
has about one-half the charge of an electron. How can we explain this?
The reality of the nitrite ion is difficult to show with a single Lewis
structure. To better describe the electronic structure, we draw a set of
Lewis structures, called resonance structures, that show the way
electrons are distributed around a molecule or ion. Figure 9.12 shows two
Lewis structures for the nitrite ion. Neither structure by itself is totally
correct, but the average of the two structures conveys the truth that the

540
negative charge is shared between the two oxygen atoms. Chemists use a
double-headed arrow to indicate resonance structures (Figure 9.13).

Figure 9.13 Chemists indicate resonance structures using a double-headed arrow.

Generally, ions that spread charge over multiple atoms are more stable
than those that do not. We can draw resonance structures for most
nonmetal oxyanions (including nitrate, nitrite, phosphate, sulfate,
carbonate, and acetate), and they all exhibit remarkable stability. In these
resonance structures, the outer oxygen atoms share the negative charge on
the ion (Figure 9.14).

Figure 9.14 The phosphate ion has four major resonance structures.

When drawing resonance structures, the first bonds between atoms

(that is, the single bonds) do not change. Resonance structures do not
completely break or form bonds between atoms. Resonance structures
involve changes in the second (or third) covalent bonds as well as changes
in the number of lone pairs. This concept is illustrated in Example 9.6.

Resonance structures involve second or third covalent bonds as well as lone

pairs.

There is an ancient parable about several blind men who touch a different part of
an elephant (the trunk, the side, the leg). Each describes the elephant differently

541
 —one as a snake, one as a wall, one as a tree. Similarly, each resonance
structure is a partial description; the actual bonding is a composite of these
individual pictures.

Example 9.6 Evaluating Charge Distribution from Resonance

Structures
The nitrate ion (NO3–) has three major resonance structures. Draw each
structure. Based on these structures, what is the average charge on each
oxygen atom?
Following the steps given in Section 9.1, we can first draw a single Lewis
structure for the nitrate ion, as shown in the left structure below. In this
structure, one of the three oxygen atoms contains a double bond. To draw
the other resonance structures, we shift the electrons so that the double
bond lies on each of the other oxygen atoms. We use a double-headed
arrow between the structures to indicate that these are resonance structures:

Notice in these structures that the first bonds between the atoms do not
change. The only changes involve the second bond and the unshared
electrons. The actual structure of the nitrate ion is an average of these three
structures.
Each oxygen atom has a negative charge in two of the three resonance
structures. Based on this, we can say that the charge on each oxygen atom

is approximately −23 .•

IT
TRY

7. The resonance structures for the carbonate ion are shown below,
including formal charges. Based on these structures, what is the
average charge on each oxygen atom?

542
8. The formate ion has the formula CHO2– (C is the central atom). Draw
two resonance structures for the formate ion. Include formal charges.

543
9.3 Shapes of Molecules
The shapes of molecules critically affect their properties. The nanocars
that we discussed at the beginning of this chapter are a dramatic example
of this, but shape affects many properties in ways that are more subtle. For
example, when we breathe (Figure 9.15), our bodies expel two
compounds, carbon dioxide (CO2) and water (H2O). Both are small,
simple molecules. But CO2 is a gas at room temperature, and water is a
liquid. Why the difference? As we’ll see in the pages to come, the shape of
each molecule plays a big part in determining these properties.

Figure 9.15 When we breathe, we emit carbon dioxide and water. The properties of
these molecules are partially determined by their shape.

Fortunately, we can use Lewis structures to predict the shapes of

molecules. To do this, we use an approach called the valence shell
electron pair repulsion (VSEPR) model. The VSEPR model is based on
the idea that electrons repel each other, and therefore they occupy regions
that are as far away from each other as possible. From this idea, a fairly
limited set of possible geometries emerges.
We describe geometries in two ways. First is the electronic geometry.
This is the arrangement of electrons around a central atom. The second
approach, called the molecular geometry, describes the arrangement of
atoms within a molecule (Figure 9.16). The molecular geometry
(sometimes simply called the shape) depends on the electronic geometry,
but it considers only the location of the atoms.

544
Figure 9.16 Electronic geometry considers all the electrons around the central atom.
Molecular geometry considers only the arrangement of atoms.

Two electron sets. Let’s begin with a simple molecule, hydrogen

cyanide. Figure 9.17 shows the Lewis structure for this molecule. There
are two “sets” of electrons around the central carbon atom—the set bonded
to the hydrogen atom (highlighted in green) and the set bonded to the
nitrogen atom (highlighted in orange). In the VSEPR model, we consider
the triple bond to be one “set” of electrons because these electrons all
occupy the area between the carbon and nitrogen atoms. The two electron
sets (orange and green) repel each other, so they will move as far away
from each other as possible—meaning they will end up on opposite sides
of the carbon atom. The electronic geometry around the central carbon is
linear; that is, the bond angle between the green set and the yellow set is
180°. As a result, the molecular geometry (the arrangement of atoms) is
also linear.

Figure 9.17 The electronic and molecular geometries of hydrogen cyanide are linear.
For simplicity, this figure and the ones that follow use black for the central atom and red
for all outer atoms. Additionally, these figures show single, double, and triple bonds as
a single set of electrons.

545
 Explore

Figure 9.17

In the VSEPR model, single, double, and triple bonds are all counted as one
electron “set.”

Let’s consider another molecule, carbon dioxide, CO2. Which of the

three geometries shown here would be the most stable?

Again, we need to consider the number of electron sets around the

central atom. Like our first example, this molecule has only two sets of
electrons around the central carbon atom (each double bond counts as one
set), and so the most stable arrangement is the one that keeps the electrons
as far away from each other as possible—that is, the geometry shown in
option (c). Any time the central atom has only two electron sets, the
electronic and molecular geometry around that atom are both linear, with a
bond angle of 180°.
Three electron sets. Next let’s examine a molecule that has three
electron sets around the central atom, such as formaldehyde, CH2O
(Figure 9.18). As before, these three electron sets repel each other,
pushing as far away from each other as possible. In this case, the electrons
adopt a geometry that we describe as trigonal planar. In a trigonal planar
geometry, the molecule is flat, and each set of electrons is separated by an
angle of 120°.

546
Figure 9.18 The electronic and molecular geometries of formaldehyde (CH2O) are
trigonal planar.

Explore
Figure 9.18

The nitrite ion, shown in Figure 9.19, also has three electron sets
around the central atom: the double bond to the oxygen atom (highlighted
in green), the single bond to the other oxygen atom (highlighted in blue),
and the lone pair (in orange). The lone pair takes up as much or more
space than a bonded set of electrons, and so we must consider it in
predicting the geometry of the molecule. Three electron sets adopt a
trigonal planar electronic geometry.

Figure 9.19 The nitrite ion has an electronic geometry (represented by the three shaded
clouds) that is trigonal planar. The molecular geometry, which considers only the
arrangement of atoms, is bent.

547
 Methane gas (CH4) is the main component of natural gas. This compound has a
tetrahedral geometry.

In this example, the electronic geometry and the molecular geometry

are not the same. Although the electronic geometry is trigonal planar, only
two of the three electron sets are connected to atoms, and so the molecular
geometry is better described as bent.
Four electron sets. Finally, let’s consider a molecule with four sets of
electrons around the central atom, such as CH4. What shape would this
molecule take? You might initially guess a cross shape, like a plus sign or
the letter x. In this geometry, the sets of electrons are separated by only
90°. By adopting a tetrahedral geometry, the electron sets create more
space between themselves. In a tetrahedral geometry, the sets of electrons
are separated by 109.5°. A tetrahedral geometry can be thought of as an X-
shape in which two of the branches have been turned sideways, as shown
in Figures 9.20 and 9.21.

Figure 9.20 Moving from an X-shape to a tetrahedron provides more space between the
electron sets.

Figure 9.21 The electronic and molecular geometries of methane (CH4) are tetrahedral.

548
 Explore
Figure 9.21

The tetrahedral geometry shown in Figure 9.21 occupies a three-dimensional

volume. To show this shape clearly, chemists sometimes draw bonds using
wedges and dashes. The wedge indicates that the atom is coming toward you;
the dash indicates it is going away from you.

The tetrahedral electronic geometry can also lead to more than one
molecular geometry. For example, look at the structure of ammonia shown
in Figure 9.22. Because there are four electron sets around the central
atom, ammonia has a tetrahedral electronic geometry. However, if we
consider only the atoms, the molecule forms a pyramid shape. We refer to
the molecular geometry of ammonia as trigonal pyramidal.

Figure 9.22 The central nitrogen atom of an ammonia molecule has three sets of
bonding electrons and one lone pair. This results in a molecular geometry that is
trigonal pyramidal.

The electronic geometry of a water molecule is also tetrahedral

(Figure 9.23). Notice that the water molecule has two bonding sets and
two lone pairs. The two lone pairs push the hydrogen atoms toward each
other. We refer to the molecular geometry of water as bent.

Figure 9.23 The oxygen atom of a water molecule has two sets of bonding electrons
and two lone pairs. The molecular geometry of water is bent.

Each electronic geometry produces one or more molecular geometries.

549
 Five and six electron sets. Atoms with expanded octets may have five
or six electron sets around them. However, we will not go into detail about
these geometries here. The three main geometries—linear, trigonal planar,
and tetrahedral—account for most of the shapes you will encounter in
chemistry. These geometries are summarized in Table 9.1.

TABLE 9.1 Electronic and Molecular Geometries

Non-
bonding
sets
Electonic Bonding (lone Molecular
Electronsets geometry Model sets pairs) geometry Examples

2 Linear 2 0 Linear

Trigonal Trigonal
3 3 0
planar planar

2 1 Bent

4 0 Tetrahedral

Trigonal
4 Tetrahedral 3 1
pyramidal

2 2 Bent

Remember that these geometries describe only the shape around one
central atom: If a molecule contains multiple atoms along the backbone,
the geometry of each atom must be considered. As an example, look at this
molecule:

550
 The atoms highlighted in green triangles have a trigonal planar
electronic geometry because they are each surrounded by three sets of
electrons. Those highlighted in yellow squares have a tetrahedral
electronic geometry because they are surrounded by four sets of electrons.
Notice that the peripheral atoms (those connected to only one other atom)
are not highlighted. Only the central atoms affect the geometry of the
molecule, and so we don’t consider the geometries of atoms bonded to
only one other atom.

Example 9.7 Determining the Molecular and Electronic

Geometry of a Molecule
Dihydrogen sulfide, H2S, is an awful-smelling gas produced by rotting
eggs. Draw the Lewis structure for this compound. What is the electronic
geometry of this compound? What is the molecular geometry?
To solve this problem, we must first draw a Lewis structure for H2S. A
correctly drawn structure shows that the sulfur atom has two bonds to
hydrogen atoms and two lone pairs:

Because this molecule has four electron sets around the central atom, its
electronic geometry is tetrahedral. However, its molecular geometry—the
arrangement of its three atoms—is bent. •

Example 9.8 Determining the Molecular and Electronic

Geometry of a Molecule
Lithium carbonate is a simple ionic compound that is widely used to treat

551
 bipolar disorder. What is the molecular geometry of the carbonate ion?
Although this question does not explicitly ask for a Lewis structure, it is
essential for solving the problem. The Lewis structure for the carbonate ion
is shown here:

Based on this structure, we can see that the central carbon atom has three
electron sets around it and therefore an electronic geometry that is trigonal
planar. Because all three electron sets are bonded to atoms, the molecular
geometry is also trigonal planar. •

Use the four-step technique from Section 9.1 to draw Lewis structures for
ions.

IT
TRY

9. Draw a Lewis structure for carbon disulfide. What is the electronic

geometry around the central atom?

10. Salicylic acid is found in willow bark and was used for centuries as
an anti-inflammatory and fever-reducing agent. Using the structure
of salicylic acid below, indicate the electronic and molecular
geometries on each of the numbered atoms. The first is done as an
example.

552
9.4 Polar Bonds and Molecules

Electronegativity and Polar Covalent Bonds

So far, we’ve described covalent bonds as electrons being “shared”
between two atoms. But electrons are often not shared evenly. For
example, consider the simple molecule hydrogen fluoride (Figure 9.24).
The fluorine pulls the electrons more strongly than the hydrogen, so the
electrons are, on average, closer to the fluorine than the hydrogen.
Consequently, the fluorine atom has a small negative charge while the
hydrogen atom has a small positive charge. We show this by writing δ+
(read as “delta-plus”) above the positive ion and δ– (“delta-minus”) above
the negative ion. (The lowercase Greek letter delta is commonly used in
math to show a very small increment.)

Figure 9.24 There are two ways to represent a polar covalent bond.

Polar covalent bonds are like two people sharing a bed when one of them hogs
the blanket. The one who gets a little more of the blanket is slightly warmer; the
one who gets less is slightly colder. They are sharing the blanket, but not evenly.

Alternatively, we can show this polarity by drawing an arrow pointing

in the direction that the electrons are pulled, with a small cross on the back
of the arrow to indicate the positive end. We refer to this type of bond as a
polar covalent bond.
How can we determine which bonds are polar covalent? To do this, we
use a property called electronegativity, which was first described by the
great American scientist Linus Pauling (Figure 9.25). Electronegativity is
a measure of how strongly atoms pull bonded electrons. The
electronegativities of the elements are shown in Table 9.2. This table
conveys one of the most important ideas in understanding chemical
bonding, and it is worth exploring in detail.

553
Figure 9.25 Linus Pauling (1901–1994) developed the electronegativity scale and
helped to develop the concepts relating to ionic and covalent bonds. He won the Nobel
Prize in Chemistry in 1954 and the Nobel Peace Prize in 1962.

TABLE 9.2 Pauling’s Table of Electronegativities

Notice that Table 9.2 does not include the noble gases. Electronegativity is a
measure of how strongly atoms pull electrons in a bond. Because noble gases
generally do not form compounds, they do not have electronegativity values.

Notice that electronegativity generally increases from left to right

across the periodic table. This pattern is consistent with what we already
know from ionic bonding: Metals tend to lose electrons easily, while
nonmetals tend to pull, or gain, electrons.
Electronegativity increases as we go up a group or column on the
periodic table. For example, iodine has a value of 2.5, while the elements
above it—bromine, chlorine, and fluorine—have increasingly higher
values. Fluorine is the most electronegative of the elements, meaning it has
the strongest pull for electrons. On the lower left-hand side, cesium and
francium are the least electronegative, meaning they have the weakest hold

554
on their electrons.
If a metal atom bonds with a nonmetal atom, the electronegative
nonmetal pulls the electrons away from the metal. This results in the
formation of positive and negative ions, and an ionic bond.
Polar covalent bonds lie somewhere between purely ionic and purely
covalent bonds. In fact there is a continuum between purely covalent (two
atoms having equal electronegativities) and purely ionic (Figure 9.26). We
measure just how covalent or how ionic a bond is by measuring the
difference between the electronegativities of the two elements (see Table
9.2).

Figure 9.26 Bonds may be classified by the differences in electronegativity of the two
bonded atoms.

In general, bonds are classified this way:

Covalent: Difference in electronegativity less than 0.5
Polar covalent: Difference between 0.5 and 2.0
Ionic: Difference greater than 2.0

Neither dog will let go of the rope, but the bigger dog has a much stronger hold
on it. Another example of a polar covalent bond.

Electronegativities help us understand how charge is distributed in a

bond. In a polar covalent bond, the more electronegative element has a
greater share of the electrons and therefore a slight negative charge. As
we’ll see in later sections, the polarity and charge differences described by
these simple rules have a huge impact on the chemical and physical
properties of substances.

555
 Example 9.9 Determining the Polarity of a Bond
Which bond is more polar, a C–O bond or an F–S bond? Show the
direction of polarity for both bonds.
To determine the polarity of a bond, we use the table of electronegativities
(Table 9.2). Carbon has an electronegativity of 2.5 while oxygen has a
value of 3.5. The difference between these values is 1.0, meaning this is a
polar covalent bond.
Similarly, the difference between F (4.0) and S (2.5) is 1.5. This bond is
also polar covalent. Because the difference is slightly larger in this case, we
can say that the F–S bond is more polar than the C–O bond.
The atom with the larger electronegativity has a slight negative charge, and
the atom with the smaller electronegativity has a slight positive charge. We
can show the polarities of these two bonds using either the arrow or the δ+/
δ– convention. •

IT
TRY

11. Find the difference in electronegativity between each of these atoms,

and describe the bond as covalent, polar covalent, or ionic:
a. C–O
b. N–H
c. Na–Cl
d. Fe–Cl
e. C–H

12. Use either the δ+ and δ– symbols or an arrow to show the direction
of bond polarity for each of these bonds:
a. O–H
b. Zn–S
c. C–Li

556
 13. Which bond is more polar, carbon–oxygen or carbon–nitrogen?

Molecules with Dipoles

Polar covalent bonds have a tremendous impact on a molecule’s
properties. Molecules with polar covalent bonds often exhibit an overall
polarity, called a molecular dipole (sometimes called a net dipole or
simply a dipole). A molecule with a dipole has one side with a slight
positive charge while the other side has a slight negative charge. As an
example, let’s consider one of the most important molecules in the
universe—water. Water’s properties make it uniquely capable of
supporting life (Figure 9.27). Many of these properties result from the
dipole (the polarity) of the water molecules (Figure 9.28).

Figure 9.27 Water’s extraordinary properties depend on its shape and the polarity of its
bonds.

Figure 9.28 Because of its shape and the polarity of its bonds, a water molecule
contains a dipole—one side of the molecule is slightly positive while the other side is
slightly negative.

Water’s dipole is a result of its polar covalent bonds and its shape. In

557
earlier sections, we saw that water has a bent molecular geometry because
of the two lone pairs on the oxygen atom. And because oxygen is more
electronegative than hydrogen, the oxygen side of the molecule has a slight
negative charge while the hydrogen side has a slight positive charge.

Molecular dipoles result from polar covalent bonds and the shape of the
molecules.

Identifying Molecules with a Net Dipole

The polarity of a molecule depends on its shape and the polarity of its
bonds. If we add together the effects of all polar covalent bonds in a
molecule, we can determine the net dipole for a molecule. Often this is
done graphically, by visual inspection.

Polar bonds add together to produce a net molecular dipole.

To see how this works, let’s compare two molecules, water and carbon
dioxide. Figure 9.29 shows the direction of the polar covalent bonds in
these molecules. In the water molecule, the polar covalent bonds pull
electron density toward the oxygen atom. To find the net dipole of this
molecule, we add together the pull of each polar covalent bond.

Figure 9.29 In a water molecule, the polar covalent bonds pull the electrons toward the
oxygen atom, producing a net dipole. Because carbon dioxide pulls the electrons in
opposite directions, the molecule does not have a net dipole—it does not have positive
and negative ends.

The CO2 molecule also contains polar covalent bonds. However, the
molecule has a linear geometry, so the two polar covalent bonds pull in
opposite directions. As a result, the molecule has no net dipole.
To determine whether a molecule has a net dipole, first identify any
polar covalent bonds using the table of electronegativities (see Table 9.2).
Then, based on the shape of the molecule, determine if the polar bonds
pull together or cancel each other out. Let’s consider three more examples,
illustrated in Figure 9.30. The first molecule, CH4, does not have a net
dipole. Because carbon and hydrogen have very similar electronegativities,

558
the bonds in CH4 are nonpolar. The second molecule, BCl3, contains polar
bonds; but the symmetrical pull of the three chlorine atoms cancel each
other out, so there is no overall dipole. In the third molecule, CH2F2, the
two carbon–fluorine bonds are polar, creating a net dipole in the direction
of the two fluorine atoms.

Figure 9.30 We can determine the net dipoles by adding together the pull of the polar
bonds. Individual bond dipoles are shown in red; net dipoles are shown in blue.

Explore

Figure 9.30

Example 9.10 Identifying a Molecular Dipole

Does sulfur dichloride have a net dipole? Draw a Lewis structure to
support your answer.
To answer this question, we must first draw a correct Lewis structure for

559
 this compound. Using the rules in Section 9.1, we obtain the following
Lewis structure:

The central sulfur atom has four electron sets, resulting in a tetrahedral
electronic geometry, and a bent molecular geometry. Because of the
difference in electronegativity, the sulfur–chlorine bonds are polar (shown
with the red arrows). And because the molecule has a bent molecular
geometry, these polar bonds add together to produce a small net dipole (the
thicker blue arrow). The chlorine side of the molecule is slightly negative
while the sulfur side of the molecule is slightly positive. •

Imagine two tractors pulling at a stump in different directions. The net pull on
the stump can be found by adding together the pull of the two tractors.
Similarly, we find the net dipole by adding together the pull of each polar
covalent bond.

IT
TRY

14. Draw Lewis structures for these molecules. Determine if the

molecules have a net dipole.
a. HCN
b. CS2
c. SiF4

560
How Dipoles Affect Properties—A Preview
The shape and polarity of molecules influence their behavior. For example,
let’s compare the properties of carbon dioxide and water. The two
molecules are about the same size, and both have polar covalent bonds.
But because of their different shapes, water has a net dipole while carbon
dioxide does not (see Figure 9.29).
Because of their dipoles, water molecules have an attraction for each
other. This causes the molecules to stick closely together (Figure 9.31).
Water is a liquid at room temperature because of the attraction between the
molecules. Water is also able to dissolve many ionic compounds because
charged particles (ions) are attracted to the polar water molecules.
On the other hand, carbon dioxide is a gas at room temperature.
Because its molecules do not have a net dipole, they do not stick together
as tightly as water molecules do.
In the chapters ahead, we will explore these ideas in much more detail
and see how the shape and polarity of molecules determines how they
interact with their surroundings.

Figure 9.31 Water’s net dipole causes the molecules to stick closely together, making it
a liquid at room temperature. Carbon dioxide has no net dipole, and it is a gas at room
temperature.

561
Summary
In this chapter, we’ve explored the bonding, shape, and polarity of
covalent molecules. In covalent bonds, two nonmetal atoms share
electrons. Atoms can share two electrons (a single bond), four electrons (a
double bond), or six electrons (a triple bond). We commonly represent
molecules with covalent bonds using Lewis structures.
In most covalent structures, atoms fulfill the octet rule by having eight
electrons in their valence level. Hydrogen, a row 1 element, can
accommodate only two electrons in its valence level. Atoms in rows 3
through 7 of the periodic table sometimes have expanded octets with more
than eight electrons.
Polyatomic ions are groups of atoms that are covalently bonded but
possess an overall charge. We often use formal charge to represent the
charges on atoms in a polyatomic ion. While formal charge is not a perfect
indicator of charge distribution, it does give us a tool for keeping track of
the total number of electrons as well as the total charge in a molecule or
ion. Some molecules and ions share electrons in a way that we cannot
adequately describe with a single Lewis structure. Resonance structures
are sets of Lewis structures that together describe the distribution of
electrons through a molecule or ion.
The shapes of molecules are critically important to their behavior. The
molecular geometry describes the arrangement of atoms in a molecule. A
related term is electronic geometry, which is the arrangement of electrons
around a bonded atom. Using the valence shell electron pair repulsion
(VSEPR) model, we can predict the electronic geometries of most
compounds. The three most common electronic geometries are linear,
trigonal planar, and tetrahedral, corresponding to two, three, or four sets of
electrons around a central atom. The molecular geometries are a function
of the basic electronic geometries.
Another factor that defines a substance’s behavior is the polarity of its
bonds. Pauling’s table of electronegativities defines how strongly an atom
pulls bonded electrons. By comparing the electronegativities of two atoms,
we can predict whether bonds between those atoms will be covalent, polar
covalent, or ionic.
The polarities of the individual bonds in a molecule add together to
produce a net dipole. If a molecule has a net dipole, one side of the
molecule has a slight positive charge, while the other side has a slight
negative charge. We can often estimate the direction of a net dipole by

562
considering the shape of the molecule and the direction of the individual
bond dipoles.
The type of bonds, shape, and polarity of a molecule profoundly affect
its physical properties. As we’ll see in upcoming chapters, properties such
as melting point and boiling point depend on these important factors.

563
Building a Nanocar

At the beginning of this chapter, we introduced Dr. James Tour, the Rice
University chemist behind the first “nanocar”—a single-molecule vehicle that can
propel itself across a smooth gold surface. Now that we’ve developed a better
understanding of molecular shape, let’s take a closer look at how the Rice
University team designed their molecular car (Figure 9.32).

Figure 9.32 (a) The complete nanocar features four wheels, two axles, a chassis, and a
paddlewheel-style motor. (b) Tour’s first nanocars used fullerene as the wheels. Newer
designs use adamantane. (c) The carbon–carbon triple bond produces a linear geometry
that is perfect for the axles. (d) The paddlewheel motor is powered by a carbon–carbon
double bond that changes shape when it absorbs light energy. (e) This image is a
simplified representation of the nanocar.

To build the wheels, the Tour group needed a spherical structure. Fortunately,
several common molecules have this type of shape. Their first nanocar used a
molecule called a fullerene as the wheels. This fascinating molecule is composed
of 60 carbon atoms that form a perfect sphere (Figure 9.32b). The fullerene wheels
rolled nicely on the gold surface, but they created problems when Tour and his
team added a light-driven motor. To avoid this issue, they had to change the wheel
design. Adamantane is another nearly spherical molecule with the molecular
formula C10H16. The team found that adamantane wheels also rolled smoothly on
the gold surface but didn’t cause the problems that the fullerene wheels did.
To build the axles, the Tour group used two carbons connected by a triple
bond. Each carbon has a linear geometry, producing a straight axle (Figure 9.32c).
A single bond connects the wheels to the axle. Because atoms can rotate freely
around a single bond, the wheels are able to spin.
Finally, the team designed a motor. Some double bonds change shape when
they absorb light (Figure 9.32d, e). The nanocar uses this change in shape to turn a

564
paddlewheel that pushes the molecule across the surface.

565
 Key Terms
9.1 Covalent Molecules
lone pair A pair of unshared valence electrons.
expanded octet A bonding arrangement in which an atom has 10 or 12
valence electrons; this is possible only with elements in rows 3–7 of the
periodic table.

9.2 Molecules and Charge

formal charge A method of identifying charged sites on a molecule or ion.
The formal charge of an atom is the number of electrons in the valence of that
atom in its neutral, unbonded state, minus the number of covalent bonds,
minus the number of unshared electrons.
resonance structures A set of Lewis structures that show how electrons are
distributed around a molecule or ion. Resonance structures are used when a
single Lewis structure cannot adequately depict the structure.

9.3 Shapes of Molecules

valence shell electron pair repulsion (VSEPR) model A way of predicting
the geometry of molecules based on the number of electron sets around a
central atom.
electronic geometry A description of the arrangement of electrons around a
central atom.
molecular geometry A description of the arrangement of atoms within a
molecule; the molecule shape.
linear A geometry in which two atoms or electron sets are separated by 180°
angles.
trigonal planar A geometry in which three atoms or electron sets are
separated by 120° angles.
tetrahedral A geometry in which four atoms or electron sets are separated by
109.5° angles.

9.4 Polar Bonds and Molecules

polar covalent bond A covalent bond in which the atoms do not share the
electrons evenly; in this type of bond, one atom has a slight positive charge
while the other has a slight negative charge.
electronegativity A measure of how strongly atoms pull bonded electrons.

566
molecular dipole An overall polarity in which different sides of a molecule
have slight positive and negative charges; sometimes called a net dipole or
simply a dipole.

567
 Additional Problems

9.1 Covalent Molecules

15. What is the octet rule? How do nonmetals fulfill the octet rule?

16. If an atom has a filled octet, it has eight electrons in its valence level. What
two energy sublevels are filled in a complete octet?

17. How many bonded electrons are represented in this Lewis structure? How
many nonbonded electrons are represented?

18. How many bonded electrons are represented in this Lewis structure? How
many nonbonded electrons are represented?

19. Complete the table to indicate the number of covalent bonds, lone pairs,
and valence electrons around each atom in the molecule shown. Do these
atoms fulfill the octet rule?

Atom Covalent bonds Lone pairs Valence electrons

H
O
O
H

20. Complete the table to indicate the number of covalent bonds, lone pairs,

568
 and valence electrons around each atom in the molecule shown. Do these
atoms fulfill the octet rule?

Atom Covalent bonds Lone pairs Valence electrons

C
O
Cl
H

21. Determine whether the central atom in each of these structures has an
incomplete octet, a complete octet, or an expanded octet:
a.

b.

c.

22. Determine whether the central atom in each of these structures has an
incomplete octet, a complete octet, or an expanded octet:
a.

b.

c.

23. Which of these elements—boron, carbon, nitrogen, or oxygen—commonly

569
 forms compounds with an incomplete octet?

24. Which of these elements—nitrogen, sulfur, or iodine—never forms an

expanded octet? Why not?

25. Find the total number of valence electrons in these molecules:

a. HCl
b. NH3
c. PCl3
d. CH3Cl

26. Find the total number of valence electrons in these molecules:

a. N2
b. C2H2
c. SF6
d. C2H4SO

27. Find the total number of valence electrons in these molecules:

a. PCl5
b. HBr
c. H2SO4
d. HNO3

28. Find the total number of valence electrons in these molecules:

a. BBr3
b. HCSN
c. C2H4
d. C6H12O6

29. Draw Lewis structures for each of these molecules:

a. H2
b. Cl2
c. HCl

30. Draw Lewis structures for each of these molecules:

a. HBr

570
 b. H2O
c. CCl4

31. Draw Lewis structures for each of these molecules:

a. N2
b. HNO (nitrogen is the central atom)
c. SCl2

32. Draw Lewis structures for each of these molecules:

a. CO2
b. PCl3
c. HCN (carbon is the central atom)

33. Draw Lewis structures for each of these molecules:

a. BrCl
b. SiF4
c. CS2

34. Draw Lewis structures for each of these molecules:

a. SeF4
b. PBr3
c. CH2Cl2 (carbon is the central atom)

9.2 Molecules and Charge

35. The carbonate ion (shown here) has an overall charge of –2. Calculate the
formal charge on each atom in this structure.

36. The sulfate ion (shown here) has an overall charge of –2. Calculate the
formal charge on each atom in this structure.

571
37. Show the formal charge on the nitrogen atom in each of the following:
a.

b.

c.

38. Show the formal charge on the sulfur atoms in each of the following:
a.

b.

c.

39. Identify the formal charge on each of these atoms:

a. an oxygen atom having one covalent bond and three lone pairs
b. an oxygen atom having two covalent bonds and two lone pairs
c. an oxygen atom having three covalent bonds and one lone pair

40. Identify the formal charge on each of these atoms:

a. a nitrogen atom having two covalent bonds and two lone pairs
b. a nitrogen atom having three covalent bonds and one lone pair
c. a nitrogen atom having four covalent bonds

572
41. In each of these Lewis structures, one atom has a nonzero formal charge.
Locate that atom, and indicate the charge.
a.

b.

c.

42. In each of these Lewis structures, two atoms have nonzero formal charges.
Locate those atoms, and indicate the charges.
a.

b.

c.

43. Find the total number of valence electrons in each of these ions:
a. NH4+
b. SO32−
c. BH4−
d. OH−

573
44. Find the total number of valence electrons in each of these ions:
a. H3O+
b. PO43−
c. IO3−
d. N3−

45. Draw Lewis structures for each of these ions. Show formal charges where
appropriate.
a. NH4+
b. NO2−
c. NO3−

46. Draw Lewis structures for each of these ions. Show formal charges where
appropriate.
a. BH4−
b. IO−
c. ICl4−

47. Draw Lewis structures for each of these molecules. Show formal charges
where appropriate.
a. SO
b. SO2
c. PF5

48. Draw Lewis structures for each of these ions. Show formal charges where
appropriate.
a. BrO−
b. CN−
c. PH4+

49. Draw Lewis structures for each of these oxyanions. Show formal charges
where appropriate.
a. SO42−

574
 b. SO32−
c. ClO3−

50. Draw Lewis structures for each of these oxyanions. Show formal charges
where appropriate.
a. PO43−
b. BrO2−
c. BrO4−

51. In this structure, all of the atoms have a formal charge of zero. Draw in any
missing electrons to complete the Lewis structure.

52. In this structure, the formal charges have been included, but the unshared
electrons have not. Complete the Lewis structure by drawing in any
additional valence electrons.

53. What are resonance structures? How do chemists represent resonance

structures?

54. When drawing resonance structures, which of the following do not

change? (Select all that apply.)
a. total number of electrons
b. location of the lone pairs
c. location of second and third bonds
d. location of first bonds

55. Thiocyanate, SCN−, can be drawn in two resonance structures, shown here.

575
 Calculate the formal charge on each atom of the two forms.

56. The thioformate ion, CHSO−, has two major resonance structures. The first
resonance structure is shown. Of structures a–d, which one is the correct
second resonance structure? For each other structure, indicate why it is not
a correct resonance structure for this ion.

a.

b.

c.

d.

57. The resonance structures for the chlorate ion are shown here with formal
charges. Based on these structures, what is the average charge on each
oxygen atom?

576
58. The resonance structures for the sulfite ion are shown here with formal
charges. Based on these structures, what is the average charge on each
oxygen atom?

59. These structures show only the bonds in the bicarbonate ion, HCO3−.
Complete the resonance structures by adding electrons and charges to
show how the negative charge is spread out in this ion.

60. These structures show only the bonds in the bisulfate ion, HSO4−.
Complete the resonance structures by adding electrons and charges to
show how the negative charge is spread out in this ion.

577
9.3 Shapes of Molecules

61. For each of the following, indicate the electronic geometry and the bond
angle that would occur around the central atom:
a. an atom with 4 charge sets
b. an atom with 2 charge sets
c. an atom with 3 charge sets

62. What is the difference between the electronic geometry and the molecular
geometry around an atom?

63. For each of these molecules or ions, indicate the number of charge sets
(using the VSEPR model) and give the electronic geometry around the
highlighted atom:
a.

b.

c.

64. For each of these molecules, indicate the number of charge sets (using the
VSEPR model) and give the electronic geometry around the highlighted
atom:

578
 a.

b.

c.

65. For each of these molecules, indicate the number of charge sets (using the
VSEPR model) and give the electronic geometry around the highlighted
atom:
a.

b.

c.

66. For each of these molecules or ions, indicate the number of charge sets
(using the VSEPR model) and give the electronic geometry around the
highlighted atom:
a.

b.

c.

67. For each of these structures, describe the electronic and molecular
geometry around the central atom:
a.

b.

579
 c.

68. For each of these structures, describe the electronic and molecular
geometry around the central atom:
a.

b.

c.

69. For each of these structures, describe the electronic and molecular
geometry around the central atom:
a.

b.

c.

70. For each of these structures, describe the electronic and molecular
geometry around the central atom:
a.

b.

c.

580
71. Draw Lewis structures for these molecules, and describe the electronic
geometry around the central atom:
a. boron trichloride, BCl3
b. phosphorus trichloride, PCl3
c. carbon dioxide, CO2

72. Draw Lewis structures for these molecules, and describe the electronic
geometry around the central atom:
a. carbon tetrachloride, CCl4
b. nitrogen trichloride, NCl3
c. carbon disulfide, CS2

73. Draw Lewis structures for these molecules, and describe the electronic
geometry around the central atom:
a. silicon tetrachloride, SiCl4
b. sulfur trioxide, SO3

74. Draw Lewis structures for these molecules, and describe the electronic
geometry around the central atom:
a. hydrogen cyanide, HCN (C is the central atom.)
b. hypochlorous acid, HOCl (O is the central atom.)

75. Draw Lewis structures for these two ions. Identify the electronic and
molecular geometry around the central atom in each ion.
a. sulfate ion
b. sulfite ion

76. There are four oxyanions of bromine: hypobromite, bromite, bromate, and
perbromate. Draw a Lewis structure for each ion, and identify the
electronic and molecular geometry around each bromine atom.

77. Identify the electronic geometry around each highlighted atom:

a.

581
 b.

78. Identify the electronic geometry around each highlighted atom:

a.

b.

79. Amino acids are naturally occurring compounds that are the building
blocks for many important biological molecules. One of the main amino
acids is asparagine, shown here. What is the electron geometry around
each of the highlighted atoms in this molecule?

80. Amino acids are naturally occurring compounds that are the building
blocks for many important biological molecules. One of the main amino
acids is cysteine, shown here. What is the electron geometry around each
of the highlighted atoms in this molecule?

582
9.4 Polar Bonds and Molecules

81. Using the periodic table as a guide, select the more electronegative atom
from each set:
a. Br or Cl
b. Si or C
c. N or O
d. K or Ca

82. Using the periodic table as a guide, select the more electronegative atom
from each set:
a. C or N
b. N or P
c. Ge or Si
d. Mg or Na

83. Using the periodic table as a guide, rank these atoms from most
electronegative to least electronegative:
a. Ca, O, Si, Ti
b. I, Cs, Br, O
c. K, In, N, F

84. Using the periodic table as a guide, rank these atoms from most
electronegative to least electronegative:
a. Sc, Fe, N, F
b. Po, Fr, Os, S
c. Ga, N, Al, O

85. Using Pauling’s table of electronegativities (Table 9.2), describe each of

these bonds as covalent, polar covalent, or ionic:
a. C–P
b. Ca–O
c. Al–Cl
d. N–H

86. Using Pauling’s table of electronegativities (Table 9.2), describe each of

these bonds as covalent, polar covalent, or ionic:
a. C–Cu

583
 b. Cs–Br
c. P–N
d. C–S

87. Based on Pauling’s table of electronegativities (Table 9.2), identify which

bond in each of these sets is more polar:
a. C–S or C–O
b. Ca–Cl or Ni–Cl
c. Ag–S or W–S

88. Based on Pauling’s table of electronegativities (Table 9.2), identify which

bond in each of these sets is less polar:
a. Se–Cl or Ti–Ge
b. H–N or H–O
c. Ca–N or Si–F

89. Use the δ+/δ– notation to show the more positive and negative ends of
each of these polar covalent bonds:
a. N–H
b. O–H
c. P–Cl

90. Use the δ+/δ– notation to show the more positive and negative ends of
each of these polar covalent bonds:
a. Br–Si
b. H–Cl
c. Cl–C

91. Use the arrow notation to show the more positive and negative ends of
each of these polar covalent bonds:
a. Sn–Cl
b. S–Ge
c. I–Cl

92. Use the arrow notation to show the more positive and negative ends of
each of these polar covalent bonds:
a. Al–Br
b. F–Cl

584
 c. Si–O

93. Each of the Lewis structures shown here contains one polar covalent bond.
Identify that bond, and use either the δ+/δ– notation or the arrow notation
to show the direction of polarity.
a.

b.

c.

94. Each of the Lewis structures shown here contains one polar covalent bond.
Identify that bond, and use either the δ+/δ– notation or the arrow notation
to show the direction of polarity.
a.

b.

c.

95. Using the symbol , show the direction of the polar bonds in each of
these molecules:
a.

b.

96. Using the symbol , show the direction of the polar bonds in each of
these molecules:

585
 a.

b.

97. Each of these molecules or ions has a net dipole. Show the direction of
polarity of the individual bonds and of the net dipole.
a.

b.

c.

98. Each of these molecules has a net dipole. Show the direction of polarity of
the individual bonds and of the net dipole.
a.

b.

c.

99. Draw Lewis structures for each of these molecules. Identify any polar
covalent bonds, and state whether the molecule has a net dipole. Show the
direction of the net dipole.
a. Cl2
b. HCN
c. PF3

100. Draw Lewis structures for each of these molecules. Identify any polar
covalent bonds, and state whether the molecule has a net dipole. Show
the direction of the net dipole.

586
 a. BCl3
b. FCl
c. SCl2

101. Draw Lewis structures for each of these molecules. Identify any polar
covalent bonds, and state whether the molecule has a net dipole. Show
the direction of the net dipole.
a. CS2
b. SF2
c. SOCl2

102. Draw Lewis structures for each of these molecules. Identify any polar
covalent bonds, and state whether the molecule has a net dipole. Show
the direction of the net dipole.
a. NCl3
b. PH2Cl
c. CHF3

Challenge Questions

103. The structure of aspirin is shown here. How many atoms in this molecule
have a tetrahedral electronic geometry? How many atoms have a trigonal
planar electronic geometry? How many polar covalent bonds are present?
Show the direction of polarity of these bonds.

104. The structure here is allicin, an active component in garlic. How many
atoms in this molecule have a tetrahedral electronic geometry? How
many atoms have a trigonal planar electronic geometry? How many polar
covalent bonds are present? Show the direction of polarity of these bonds.

587
588
Chapter Ten
Solids, Liquids, and Gases

The North Dakota Boom

In 2006, geologists discovered an enormous underground reserve of oil and natural
gas in western North Dakota. The impacts of the discovery were huge: While much
of the global economy staggered, the Dakotas boomed, creating thousands of jobs
and enormous wealth. In 2006, North Dakota was ranked 40th among U.S. states
for per capita income. By 2013, it was ranked second.
Coal, oil, and natural gas are collectively called fossil fuels. Composed mainly
of carbon and hydrogen, these fuels form by the decay of plant and animal matter
over long periods of time. Along with nuclear energy, fossil fuels are the
workhorse of the industrialized world: Each year, fossil fuels produce over 80% of
the energy used in the United States.
Among fossil fuels, natural gas has steadily grown in importance. In addition to
the environmental advantages of natural gas (it burns more cleanly than coal), the
surplus of oil and gas coming out of North Dakota and Montana has pushed down
the cost of fuel, which in turn reduces the cost of other commodities. Your gas bill
and your grocery bill are lower because of the massive 2006 discovery.
Collecting and transporting each type of fossil fuel has its own challenges.
Coal, a solid, has to be mined and then transported by ship or train. Companies
extract oil (a liquid) and natural gas from wells and often transport these
commodities through massive pipelines (Figure 10.1).

589
Figure 10.1 (a) Coal is the solid form of fossil fuels. It must be mined. (b) Oil is the
liquid form of fossil fuels. (c) An operating oil well. (d) A worker manages natural gas
lines. (e) Many newer buses run on natural gas. This fuel burns more cleanly than
gasoline.

Safely handling and transporting gases can be especially challenging. Workers

must follow special safety guidelines when working with natural gas. In fact, many
careers, from health care to mechanical work to restaurant management, involve
work with compressed gases. Understanding how gases behave will serve you well
in nearly any field.
In this chapter, we’ll examine the properties of matter in its three phases: solid,
liquid, and gas. We begin with solids and liquids, exploring how chemical bonds
affect a substance’s structure and properties. Then we’ll shift our attention to gases.
We will see how the pressure and volume of gases vary with temperature and with
the amount of gas present. We’ll explore chemical processes that involve gases,
such as combustion reactions or the production of beer.
At the end of this chapter, we will return to the challenge of safely storing and
transporting natural gas. We’ll look at new discoveries that are changing the way

590
scientists and engineers think about this problem and opening up exciting
opportunities for the future.

Intended Learning Outcomes

After completing this chapter and working the practice problems, you should be
able to:

10.1 Interactions between Particles

Describe the motion of particles in a solid, liquid, or gas

10.2 Solids and Liquids

Describe the bonding and arrangement of particles in ionic, metallic,
molecular, polymeric, and covalent-network substances.
Describe the different types of intermolecular forces, and relate these
differences to relative melting or boiling temperatures.

10.3 Describing Gases

Describe the key features of an ideal gas.
Describe how to use a liquid barometer to determine pressure.

10.4 The Gas Laws

Apply the combined gas laws to relate changes in the pressure, volume, and
temperature of a gas.
Relate the pressure, volume, number of moles, and temperature of a gas
using the ideal gas law.
Relate the temperature, volume, and pressure of a gas to atomic or molecular
motion.

10.5 Diffusion and Effusion

Describe the motion of larger and smaller gas particles at a given
temperature, and apply these concepts to the principles of diffusion and
effusion.

10.6 Gas Stoichiometry

Apply the principles of stoichiometry to solve problems involving reactions
of gases.

591
10.1 Interactions between Particles
In Chapter 1, we discussed the three common states of matter (also called
the phases of matter): solid, liquid, and gas. We saw that the macroscopic
(human-scale) properties are a function of their structure on the atomic or
molecular level. In a solid, particles pack closely together. Each atom is
held in a fixed position by the atoms around it. In a liquid, the particles are
close together, but they are not held in a fixed position; particles move
freely past each other. In a gas, the particles move independently of each
other—they are spaced far apart and have little or no interaction with the
other gas particles as they move. These concepts are summarized in Table
10.1.

TABLE 10.1 The States of Matter

Atomic/Molecular Arrangement Atomic/Molecular Arrangement

Solid Particles are close together and held in a fixed place. definite shape and volume

Liquid Particles are close together, but move freely past each definite volume adopts shape of
other. container

Gas Particles are far apart and have very little interaction. adopts shape and volume of container

The melting and boiling points of a substance depend on the forces holding the
particles together. The stronger the forces of attraction, the higher the phase
transition temperatures.

Substances with strong forces of attraction have high melting and boiling
points.

A transition from one state of matter to another is called a phase

change. For example, let’s look at the phase changes that take place with
water: In ice, the water molecules pack tightly together (Figure 10.2). If
we add heat (kinetic energy) to the ice, the molecules in the solid vibrate

592
more rapidly. If we continue to add heat, the molecules gain enough
energy to overcome the forces holding them in place. They break out of
the solid framework and begin to move freely past each other. The ice
melts—a phase change from solid to liquid. If we then heat the liquid
water, the molecules move faster and faster until they begin to break out of
the liquid phase and enter the gas phase. Particles in the gas phase move
about freely, interacting very little with each other.

Figure 10.2 These snow monkeys are enjoying a hot spring in northern Japan. In this
image, we see all three phases of water. The volcanic heat source melts the ice,
converting it to liquid water and then to steam. Each phase change requires heat energy
to overcome the forces of attraction between the water molecules.

593
 Explore
Figure 10.2

For a compound to change from solid to liquid or from liquid to gas,

there must be enough kinetic energy to overcome the forces of attraction.
The stronger the forces between the particles, the greater the energy
required to change phase. A substance with very strong forces between its
particles will have a high melting point and boiling point. We can use
phase transition temperatures to understand and compare the forces that
hold substances together.

IT
TRY

1. Methane, ammonia, and water are all covalent compounds that form
molecules. Based on the melting points shown, which compound
exhibits the strongest forces of attraction between molecules? Which
one exhibits the weakest forces of attraction?

594
Substance Formula Melting Point
Methane CH4 −182 °C
Ammonia NH3 −78 °C
Water H2O 0 °C

595
10.2 Solids and Liquids
In solids and liquids, forces of attraction hold particles close together.
These forces vary for different types of substances. In this section, we will
explore the different ways that particles can be arranged within solids and
liquids and learn about the properties that result from their unique
structures.

Ionic Substances
We first talked about ionic compounds in Chapter 5. In these compounds,
positive and negative ions pack tightly together in rigid frameworks, called
lattices (Figure 10.3). Because the interactions between the ions are so
strong, it takes a tremendous amount of energy to disrupt the lattices. As a
result, most ionic compounds have very high melting points (Table 10.2).

Figure 10.3 Ionic compounds form ordered lattices of alternating positive and negative
charges.

TABLE 10.2 Melting Points for Ionic Compounds

Compound Melting Point (°C)

NaCl 801

KCI 770

MgCI2 714

Cao 2,572

AI2O3 2,072

Metallic Substances

596
Elemental metals (as well as mixtures of metals, called alloys) usually
form ordered lattices of tightly packed atoms (Figure 10.4). Metallic
atoms have a loose hold on their outer electrons, and the electrons move
easily from one atom to the next. Because of this property, metals are good
conductors of electricity.

Figure 10.4 Metals pack into closefitting arrangements of atoms.

While atoms pack close together in metallic solids, the neutral metal
atoms do not bind as tightly to each other as the oppositely charged ions in
an ionic compound. Because of this, a metal lattice is not as rigid as an
ionic lattice, and the shapes of metallic solids are easily altered. Metals are
malleable, meaning they can be pounded into different shapes, and they
are ductile, meaning they can be stretched into wire.
Metals have moderately high melting points (Table 10.3). Metals like
iron, aluminum, and copper are commonly melted and molded into useful
shapes (Figure 10.5).

Figure 10.5 Workers pour molten (liquid) iron into a mold.

TABLE 10.3 Melting Points for Common Metals

597
Element Melting Point (°C)

Aluminum 660

Copper 1,085

Gold 1,064

lead 327

Iron 1,538

Recall that covalent bonds form between nonmetals.

Molecular Substances
Recall that molecules are discrete units of atoms held together by covalent
bonds. Covalent bonds within a molecule are very strong, but the forces
between the individual molecules—called intermolecular forces—are
much weaker (Figure 10.6). As a result, molecular compounds usually
have lower melting and boiling points than ionic compounds.

Figure 10.6 The forces between molecules (shown as dashed lines) are weaker than the
covalent bonds within a molecule.

The strength of intermolecular forces varies widely, depending on the

structure of the molecules. As a result, molecular materials exhibit a
stunning diversity of physical properties. Broadly, intermolecular forces
are divided into three key groups: dipole–dipole interactions, hydrogen
bonds, and dispersion forces. Each of these groups is described below.

Intra– versus Inter–

Be careful when using the prefixes intra– and inter–. Intra– means “within.”
Inter– means “between.” So “intramolecular forces” are the bonds within a
molecule—strong covalent bonds. “Intermolecular forces” are the weaker forces
between different molecules.

Dipole–Dipole Interactions

598
As we saw in Chapter 9, many molecules contain polar covalent bonds.
Depending on their shape, molecules with polar bonds often have a net
dipole, meaning different sides of the molecule have a slight positive and
negative charge (Figure 10.7). If a compound has a net dipole, the
molecules of that compound tend to “stick” together due to the attraction
of the positive and negative poles (Figure 10.8). This type of interaction is
called a dipole–dipole interaction.

Figure 10.7 Recall that we show polar covalent bonds and molecular dipoles with the δ
+ and δ- symbols, or with an arrow symbol that has a cross on the positive side and the
arrowhead on the negative side.

Figure 10.8 Molecules with net dipoles experience an attractive force between their
positive and negative ends.

In a polar covalent bond, the more electronegative atom has a slight

negative charge.

Dipole–dipole attractions affect a substance’s melting and boiling

points. To illustrate this idea, consider the molecules in Figure 10.9. The
molecules in Flask A have a molecular dipole, while those in Flask B do
not. Because of the forces of attraction between the dipolar molecules, it
takes more heat energy to pull them apart. Consequently, the dipolar
molecules have a higher melting point and boiling point than the nonpolar
molecules have.

599
Figure 10.9 The dipoles cause the molecules in flask A to stick together more tightly,
so more energy is needed for them to escape into the gas phase. As a result, molecules
with a dipole tend to have higher boiling points than those that do not.

Explore

Figure 10.9

Polar molecules tend to have higher melting and boiling points than
nonpolar molecules.

An example of this effect is shown in Table 10.4. Carbon dioxide has

no net dipole, sulfur dioxide has a small net dipole (owing to its bent
shape), and acetonitrile has a large net dipole. Notice that the larger the
size of the dipole, the higher the boiling point.

TABLE 10.4 Molecules with Larger Dipoles Exhibit Higher Boiling

Points

Carbon dioxide

600
 Sulfur dioxide

acetonitrile

Geometry Linear Bent Linear

Dipole Zero Small Large

Boiling Point −79°C* −10°C +82°C

* Carbon dioxide sublimes (transitions from solid directly to gas) at this temperature.

Melting and boiling points are a handy way to compare the strength of
different intermolecular forces. However, we must be careful in our
comparisons: Heavier molecules tend to have higher melting and boiling
temperatures than lighter molecules, so it is best to compare compounds
having a similar formula mass.

Dipole–dipole interactions are similar to the interactions between magnets—the

opposite ends are attracted to each other, and they tend to stick together.

Hydrogen Bonding
Look at the three small compounds in Table 10.5. The first compound,
methane (CH4), has no dipole. Because of this, CH4 molecules have very
weak interactions with each other, resulting in a very low boiling point
(−162 °C). The second compound, hydrogen cyanide, has a large dipole:
As expected, its boiling point (+26 °C) is significantly higher. But the third
compound, water, is surprising: Its dipole is smaller than that of hydrogen
cyanide, but its boiling point is much higher.

TABLE 10.5 A Comparison of the Properties of Methane, Hydrogen

601
 Cyanide, and Water

Hydrogen cyanide Water

Methane

Formula mass 16.0 u 27.0 u 18.0 u

Dipole strength* 0 2.98 1.85

Boiling point −162 °C +26 °C +100 °C

* These numbers convey the relative size of each dipole.

This surprising observation results from an especially strong type of

intermolecular force called hydrogen bonding. Hydrogen bonding is a
special dipole–dipole interaction that occurs only between molecules
containing H–F, H–O, or H–N bonds. Water exhibits very strong
hydrogen-bonding effects.

Hydrogen bonds are stronger than other dipole–dipole interactions.

To form a hydrogen bond, the hydrogen atom must be covalently bonded to one
electronegative atom (F, O, or N), but also attracted to an F, O, or N atom in a
neighboring molecule.

What causes the hydrogen-bonding effect? Recall that a hydrogen

atom contains only one electron. When hydrogen bonds to a more
electronegative element (F, O, or N), this one electron is pulled away,
leaving the positive nucleus exposed (Figure 10.10). This “naked”
positive charge attracts the negative poles of other molecules. Neighboring
molecules with small, slightly negative atoms (F, O, or N) can get
exceptionally close to the naked positive of the hydrogen atom—nearly as

602
close as a covalent bond. The result is an unusually strong intermolecular
force.

Figure 10.10 Hydrogen bonds arise when hydrogen is bonded to the electronegative
elements nitrogen, oxygen, or fluorine.

Water contains two O–H bonds, resulting in multiple hydrogen-

bonding interactions that cause water molecules to stick tightly to each
other (Figure 10.11). As a result, the melting and boiling points of water
are much higher than those of other small molecules. The interactions
between water molecules are strong enough that some insects are able to
walk on the surface of water, supported by the forces of attraction between
individual molecules. The tendency of water to form droplets in the air or
beads on a waxed surface also arises from this strong attraction of water
molecules for each other (Figure 10.12). Hydrogen bonding plays a
critical role in the chemistry of life, from the fundamental way water
molecules interact to the shapes of huge biological molecules like proteins
and DNA.

Figure 10.11 Water molecules are held tightly together by hydrogen bonds, as depicted
by a space-filling model (left) and Lewis structures (right).

603
Figure 10.12 Hydrogen bonding causes water molecules to stick tightly together, a
phenomenon called surface tension. These forces can (a) support the weight of a bug on
the water, and (b) cause water to form droplets in the air or (c) on a waxed surface.

London Dispersion Forces

So what about covalent compounds that don’t have dipoles? At first
glance, we might expect there to be no attraction between them. However,
this is not the case. Even molecules with no overall dipole exhibit a weak
attraction for each other (Figure 10.13). These forces of attraction are
called London dispersion forces (sometimes simply called dispersion
forces or London forces). London dispersion forces are related to dipole–
dipole forces but are much smaller in magnitude.

604
Figure 10.13 Although propane (C3H8) molecules do not have a net dipole, they
weakly attract each other through London dispersion forces.

London dispersion forces are the weakest intermolecular force.

Even if they are shared evenly between two atoms, electrons are
constantly moving. This motion produces slight, fleeting areas of positive
and negative charges called instantaneous dipoles. These rapidly changing
dipoles produce ripple effects in the surrounding molecules, as well: A
slight buildup of negative charge on one molecule will push the electron
density away on a neighboring molecule. The result is a dipole on the
secondary molecule and an instant of attraction between the two molecules
(Figure 10.14).

Figure 10.14 London dispersion forces are caused by the interaction of tiny
instantaneous dipoles.

Explore
Figure 10.14

605
 Instantaneous dipoles are like ripples on the surface of a pool: On average, the
surface of the pool is level, but the ripples make the water uneven. For an
instant, there is more water in one part of the pool than another. Similarly, the
“ripples” caused by electron motion produce a slight attraction between
neighboring molecules.

Summarizing Intermolecular Forces

In summary, there are three types of intermolecular forces: London
dispersion forces, dipole–dipole forces, and hydrogen bonds (Table 10.6).
Of these, dispersion forces are the weakest—resulting in molecules with
the lowest melting and boiling points. Hydrogen bonds are the strongest—
resulting in molecules with the highest melting and boiling points.

TABLE 10.6Types of Intermolecular Forces

Type Description Strength

Hydrogen bonding Molecules with H–F, H–o, or H–n bonds Strongest

Dipole–dipole forces Molecules with net dipoles

London dispersion forces All covalent molecules Weakest

The stars are always in the sky, but their faint light is often blocked by the light
of the moon or the bright light of the sun. In the same way, London dispersion
forces are always present—but their effects are slight in molecules that have
dipole–dipole or hydrogen-bonding interactions.

606
 Example 10.1 Predicting the Properties of Compounds
Classify these three compounds, LiCl, CH3F, and CH3OH, as ionic,
metallic, or covalent. Predict which compound would have the highest and
lowest boiling points.
The first compound, LiCl, contains both a metal and a nonmetal. Because
of this, we know that it is an ionic compound and therefore has high
melting and boiling points.
The other compounds, CH3F and CH3OH, are composed entirely of
nonmetal atoms and therefore are covalent (molecular) compounds. To
predict the relative boiling points of these compounds, we need to draw
Lewis structures and consider what polar bonds may be present:

Because F and O are much more electronegative than C or H, both of these

molecules contain polar bonds and an overall dipole. However, CH3OH
contains an O–H bond. This means that CH3OH molecules can form
hydrogen bonds with each other, resulting in a higher boiling point.
Based on this analysis, we would expect the ionic compound, LiCl, to have
the highest boiling point. Next highest is the compound that can form
hydrogen bonds (CH3OH), followed by the compound with only dipole–
dipole interactions (CH3F). •

IT
TRY

2. Classify the following as ionic, metallic, or covalent solids:

a. potassium nitrate
b. phosphorus tribromide
c. chromium

3. Which of these compounds would you expect to have the highest

melting point, and why?
a. hexane, C6H14

607
 b. sodium fluoride, NaF

4. Based on their polarity, which of these compounds would you expect

to have the highest boiling point?

a.

b.

Covalent Networks and Polymers

Some compounds contain covalent bonds but differ significantly from the
molecular solids described earlier. For example, elemental carbon exists in
two primary forms, diamond and graphite. Both of these are examples of
covalent networks—lattices of connected covalent bonds, forming one
giant molecule (Figure 10.15). In diamond, each carbon atom has four
single bonds and a tetrahedral geometry. The single bonds repeat
indefinitely in three dimensions, resulting in a rigid structure that is among
the hardest and most durable substances known. In graphite, the carbon
atoms each have a trigonal planar geometry. The atoms are arranged in
two-dimensional “sheets” that slide easily over each other. As a result,
graphite is softer and is used as the writing material in most pencils.

608
Figure 10.15 Diamond and graphite are both covalent networks of carbon. In diamond,
each carbon has a tetrahedral geometry; in graphite, each carbon has a trigonal planar
geometry.

Polymers are compounds that contain long chains of covalently

bonded atoms. There are many naturally occurring polymers, including
cellulose (the structural component of wood), starch (a major component
of many staple foods, such as rice and potatoes), and proteins. Plastics
such as poly(ethylene) and poly(vinyl chloride), or PVC, are synthetic
polymers (Figure 10.16).

Figure 10.16 Polyethylene is commonly used in disposable plastics like shrink wrap.
Poly(vinyl chloride), or PVC, is commonly used in plumbing applications.

Besides being used as a fuel source, fossil fuels provide the starting materials to
make most plastics.

Plastics or polymers are made up of long chains of covalent bonds.

Polymers are composed of small, repeating units that are bonded

together. For example, Figure 10.17 shows the structure of poly(vinyl
chloride). Each small unit is a single link in the longer chain. We’ll discuss
polymers in more detail in Chapter 15.

609
Figure 10.17 Polymers like PVC are composed of long chains of small, repeating units.
Each “link” of a PVC chain has the formula C2H3CI.

610
10.3 Describing Gases
In the previous section, we saw how the properties of solids and liquids
depend on the forces of attraction between particles. In the remaining
sections of this chapter, we’ll shift our attention to gases. In a gas, the
particles are spaced far apart. They move freely, interacting very little with
the particles around them. Gas particles travel in a straight line until they
bounce off another particle or off the walls of the container (Figure
10.18).

Figure 10.18 In a gas, particles move freely and have very little interaction with each
other.

Explore

Figure 10.18

When describing gases, we often assume that they are behaving as ideal
gases. An ideal gas has two key properties:
1. The volume of the particles is much, much less than the volume of
the container.

611
 2. The particles have no attraction for each other. When they pass each
other, they do not slow down. When they collide, they bounce off
each other like billiard balls.
When gases behave this way, we can predict their behavior mathematically
by using simple relationships between the temperature of the gas, the
volume the gas occupies, and the pressure it exerts.

Pressure
We commonly describe gases by the pressure they exert on their
surroundings. We can think of pressure as the “push” that particles exert
on everything around them. For example, consider the gas represented in
Figure 10.18. The particles move freely around the box. As they move,
they collide with the walls of the container. Collectively, these tiny
collisions create a larger push. This is the pressure on the container
(Figure 10.19).

We commonly describe gases by the pressure they exert.

Formally, pressure is defined as the force that is exerted divided by the

area over which it is applied:

Pressure=ForceArea
As we will see in the following sections, the pressure caused by the gas
depends on the volume, temperature, and amount of gas present.

Figure 10.19 The gas inside this rural propane tank exerts pressure on its walls.

Measuring Pressure
Gas pressure is a normal part of our lives. Earth’s atmosphere exerts a

612
pressure that is essential to life. Subtle variations in atmospheric pressure
cause wind and weather patterns (Figure 10.20). Pressurized air in tires
keeps vehicles rolling smoothly. The bounce of a basketball is caused by
the compressed air inside it. Air-conditioning systems rely on the
compression and expansion of gases under different pressures. Pneumatic
tools use compressed air to produce a powerful force (Figure 10.21).

Figure 10.20 A tornado occurs when areas of high and low air pressure come together.

Figure 10.21 From tires to air-conditioning systems and pneumatic tools, compressed
gas serves many useful purposes.

A barometer is a device used to measure atmospheric pressure. A

classic barometer is shown in Figure 10.22. In this device, a long tube is
filled with a liquid (usually mercury), then turned upside down in a

613
reservoir of the liquid. Gravity pulls the liquid down, creating a vacuum in
the top of the tube. At the same time, pressure from the outside air pushes
the liquid up into the tube. The higher the air pressure, the higher the liquid
is pushed into the tube. We measure the pressure of the atmosphere by
measuring the height of the mercury inside the tube. Because of this,
atmospheric pressure is often reported in millimeters of mercury (mm
Hg; also called torr). At sea level, the average atmospheric pressure is 760
mm Hg, or 760 torr. This is referred to as standard pressure.

1 torr = 1 mm Hg

Figure 10.22 A simple barometer is used to measure the pressure of the atmosphere.

Explore
Figure 10.22

614
 When you pull the air out of a straw, you lower the air pressure inside the straw.
Because the pressure outside is higher than inside, the liquid gets pushed up into
the straw. A barometer works the same way.

Why do barometers use mercury, rather than a less-toxic liquid like

water or oil? Because mercury is over 13 times denser than water.
Atmospheric pressure is enough to push a column of mercury 760
millimeters (0.76 meters) into the air. If we used water in a barometer, the
column would have to be over 33 feet (10 meters) high. Of course, there
are many other designs for pressure gauges. The shapes and styles of these
gauges differ based on their application (Figure 10.23).

Figure 10.23 Other uses for pressure gauges include (a) for measuring blood pressure,
(b) for an acetylene tank used in welding, (c) for a scuba diving regulator, and (d) for
measuring tire pressure.

For practical applications, we sometimes refer to the gauge pressure

of a compressed gas. The gauge pressure is the difference between the
compressed gas pressure and the atmospheric pressure. For example, if the
gauge on an air compressor reads zero, it doesn’t mean there is a vacuum
(meaning no air) inside the cylinder; it means that the air pressure inside
the cylinder is equal to the air outside. For the problems in this chapter,
you can assume that the pressures given are absolute pressures, not gauge
pressures.

615
 We can use many different units to describe gas pressure. A common
unit is the atmosphere (atm), based on the standard air pressure at sea
level. One atmosphere is defined as 760 mm Hg (29.92 inches Hg). In the
United States, pressure is often described in terms of pounds per square
inch (psi). A basketball has a gauge pressure of about 8 psi, while the tires
on your car likely have a gauge pressure of 30–40 psi. Other common units
of pressure include the kilopascal (kPa) and the bar. The relationships
between these units are summarized in Table 10.7.

TABLE 10.7 Common Pressure Conversions

1 atmosphere = 760 mm Hg (torr)

1 atmosphere = 101.3 kPa

1 atmosphere = 14.70 psi

1 atmosphere = 1.013 bar

IT
TRY

5. A tire has a maximum recommended pressure of 276 kPa. What is this

pressure in psi?

6. Use the relationships given in Table 10.7 to complete these

conversions:
a. Convert 2.4 atmospheres to psi.
b. Convert 0.892 kPa to atmospheres.
c. Convert 1500 kPa to psi.

616
10.4 The Gas Laws
The pressure, volume, and temperature of an ideal gas are related to each
other through simple mathematical relationships called the gas laws. These
relationships are very useful for describing common gases including air,
nitrogen, oxygen, helium, acetylene, and carbon dioxide.

A scientific law describes an observed behavior, but it doesn’t explain why the
behavior occurs. The gas laws are an example of this idea.

Boyle’s Law
Boyle’s law describes the relationship between the pressure and volume of
a gas. According to Boyle’s law, the pressure and the volume of a gas are
inversely related. That is, if the volume (V) goes up, the pressure (P) goes
down, and vice versa (Figure 10.24). As long as the temperature does not
change, the pressure times the volume of a gas is constant.

PV=constant

Figure 10.24 A firefighter carries an oxygen tank on his back. The oxygen gas is
compressed into a small volume, and so it has a high pressure.

Boyle’s law is useful when the pressure or the volume is changing, and we
wish to determine how the other unit will change. Because PV is constant,

617
we can say that

P1V1=P2V2
where P1 and V1 are the initial pressure and volume, and P2 and V2 are the
final pressure and volume.
When working with Boyle’s law, we can use any units of pressure and
any units of volume. Example 10.2 describes this type of problem.

Example 10.2 Using Boyle’s Law

A commercial compressor stores 2.8 liters of air at a pressure of 150 psi. If
this air is allowed to expand until the pressure is equal to 15 psi (just over
atmospheric pressure), what volume will the air occupy?
In this example, we are looking for the volume of air after the pressure is
decreased. P1 is 150 psi, P2 is 15 psi. V1 is 2.8 L, and we are trying to solve
for V2. To solve this problem, we rewrite the equation above to isolate V2
and then plug in the pressure and volume quantities to solve:
V2=P1V1P2=(150 psi)(2.8 L)15 psi=28 L

Notice that when we solved this problem, our units of pressure canceled
out. It is important to make sure that the units of P1 and P2 are the same.
The units for V1 and V2 will also be the same. •

IT
TRY

7. A balloon has a volume of 2.5 liters at a pressure of 1.0 bar. If the

pressure around the balloon is decreased to 0.80 bar, what is the new
volume of the balloon?

Charles’s Law
Charles’s law states that at constant pressure, the volume of a gas is

618
directly proportional to its temperature. If the temperature goes up, the
volume goes up. If the temperature goes down, the volume goes down.
Mathematically, we can represent this by the following equation:

V∝T
where the ∝ symbol means “is proportional to.” Alternatively, we can
write this as follows:

VT=constant

Using Charles’s Law to Find Absolute Zero

In the mid-19th century, William Thomson (who was later granted the title
Lord Kelvin) compared the plots of volume versus temperature for a
number of different gas samples, similar to the graph in Figure 10.25.
Although he could not measure gas pressures at very low temperatures,
Thomson observed that if the volume continued to decrease in a linear
fashion, all of the gases would reach a volume of zero at the same
temperature. This temperature, –273.15 °C, is absolute zero—the lowest
possible temperature. Recall that temperature is the measure of the kinetic
energy of the particles in a substance. At absolute zero, the particles in a
material have zero kinetic energy.
In the Kelvin temperature scale, absolute zero is given the value of 0

619
kelvin. A unit kelvin is the same size as a degree Celsius. We convert
between the Celsius and Kelvin scales by using the following relationship:

kelvin=°C+273.15

Figure 10.25 Absolute zero is calculated by extrapolating the volume–temperature

relationship of a gas back to a volume of zero.

Solving Volume–Temperature Problems Using Charles’s Law

We commonly use Charles’s law to predict how the volume of a gas will
change if the temperature changes, or vice versa. Because the ratio of
volume to temperature is constant, we can write Charles’s law using this
relationship:

V1T1=V2T2
where V1 and T1 are the initial volume and temperature, and V2 and T2 are
the final volume and temperature.
When working a Charles’s law problem, you can use any units of
volume; but you must express the temperature in kelvins. Use of Celsius in
a gas-law problem will give an incorrect answer. Example 10.3 illustrates
this type of problem.

For gas-law problems, temperatures must be in kelvins.

620
 Example 10.3 Using Charles’s Law
A balloon has a volume of 3.2 liters at room temperature (25 °C). The gas
inside the balloon is then heated to 100 °C. What is the new volume of the
balloon?
In this example, we are looking for the volume of the balloon after the
temperature has increased. V1 is 3.2 L. Before solving for V2, we must
convert the temperatures to the Kelvin scale:
T1=25 °C+273=298 KT2=100 °C+273=373 K

In these calculations, notice that we rounded 273.15 to 273 because we

knew the Celsius temperature only to the nearest degree. Finally, we can
rewrite the equation to isolate V2, then plug in the quantities to solve. •

V2=V1T2T1=(3.2 L)(373 K)298 K=4.0 L

IT
TRY

8. A gas occupies a volume of 15.2 L at 25 °C. At what temperature

would this gas expand to a volume of 30.4 L? Express your answer in
both kelvins and degrees Celsius.

The Combined Gas Law

In many situations, the pressure, volume, and temperature of a gas all
change together. For example, air-conditioning systems rely on the
compression and expansion of gases. As the pressure and volume of the
gas change, the temperature also changes. For situations like this, we use a
mathematical relationship called the combined gas law:

621
 P1V1T1=P2V2T2

The combined gas law can be used to solve any Boyle’s law or Charles’s law
problem.

As in Boyle’s and Charles’s laws, we can use any units of pressure and
volume, but the temperatures must be expressed in kelvins. Examples 10.4
and 10.5 illustrate the use of the combined gas law.

Example 10.4 Using the Combined Gas Law

A gas with a temperature of 280 K, a pressure of 200 kPa, and a volume of
25.8 L is compressed to 15.8 L, causing the pressure to increase to 350
kPa. What is the temperature of the gas under the new conditions?
In this example, we’re given P1, V1, and T1, as well as P2 and V2. We’re
asked to solve for T2. We can rearrange the equation above and insert the
values to reach the answer. •

T2=P2V2T1P1V1=(350 kPa)(15.8 L)(280 K)(200 kPa)(25.8 L)=300 K

Example 10.5 Using the Combined Gas Law

Under constant-pressure conditions, a sample of hydrogen gas initially at
88 °C and 1.62 L is cooled until its final volume is 942 mL. What is its final
temperature?
In this example, the pressure is constant, so P1 = P2. Because of this, it may
be canceled from the equation. Also, notice that V1 is given in liters, but V2
is in milliliters: For the volume units to cancel, they must be the same.
Therefore, let’s express V2 as 0.942 L. Converting the temperature to
kelvins and substituting the values into the equation, we obtain a final
answer of 210 K. •
T2=P2V2T1P1V1=(0.942 L)(361 K)(1.62 L)=210 K

622
 IT
TRY

9. On a cold Iowa morning when the temperature is –30 °C, a truck tire
is inflated to a pressure of 45.0 psi. The truck is then driven south until
the temperature reaches +33 °C. If the tire has not lost any air, what is
the pressure in the tire after it warms up?

Avogadro’s Law
Avogadro’s law states that, if pressure and temperature are constant, the
volume of a gas is proportional to the number of moles of gas present. The
more gas is present, the larger the volume it occupies. Mathematically, we
can say that

V∝n
where n is the number of moles of the gas.
The exact volume that a mole of gas occupies depends on its
temperature and pressure.
A temperature of 0 °C (273 K) and a pressure of 1.0 atmosphere is
called standard temperature and pressure, or STP. At STP, one mole of
gas occupies a volume of 22.4 liters (Figure 10.26).

623
Figure 10.26 At standard temperature and pressure (STP), 1 mole of gas occupies 22.4
L, the size of a large balloon.

IT
TRY

10. At STP, 1 mole of gas occupies a volume of 22.4 L. At the same

temperature and pressure, what volume does 5 moles of gas occupy?

The Ideal Gas Law

The gas laws described so far allow us to relate different properties of
gases. We can connect all of these together using the ideal gas law. This
extraordinarily useful law relates the amount of gas present to its pressure,
volume, and temperature. We usually write this law as

PV=nRT
where P is the pressure, V is the volume, T is the temperature, n is the
number of moles of gas, and R is a constant, called the gas constant, with a
value of 0.0821 L·atm/mol·K.
When using the ideal gas law, it is important to make sure that the
temperatures are expressed in kelvins and that the units of volume and
pressure align with the values included in the gas constant (for example,
liters and atmospheres). Examples 10.6 and 10.7 illustrate this law.

624
Remember, in an ideal gas the particles do not interact with each other at all.
Real gases will have some interactions, but most common gases behave in a
nearly ideal manner.

Example 10.6 Using the Ideal Gas Law

What volume does 1.00 mole of gas occupy at a temperature of 0.00 °C and
a pressure of 1.00 atmosphere?
To solve this problem, we rearrange the ideal gas law equation to isolate V
and then insert the values:
V=nRTP=(1.00 mol)(0.0821L⋅atmmol⋅K)(273.15 K)(1.00atm)=22.4 L

Notice that in the solution above, all of the units cancel except for liters,
and so the answer is given in liters. This problem corresponds to standard
temperature and pressure (STP). •

Example 10.7 Using the Ideal Gas Law

A portable oxygen tank has a volume of 2.40 L and a pressure of 243 psi at
a temperature of 22 °C. How many moles of oxygen are present in this
cylinder? What is the mass of the oxygen in grams?
To solve this problem, we first need to convert temperature to the Kelvin
scale (22 °C = 295 K). To align our units with the gas constant, we also
need to convert the pressure (243 psi = 16.5 atm). Using the ideal gas
equation, we can then find the number of moles:
n=PVRT=(16.5 atm)(2.40 L)(0.0821L⋅atmmol⋅K)(295 K)=1.64 moles

625
 Once we know the moles, we can then convert to grams of oxygen, as
described in Chapter 7. •

1.64 moles×32.00 g O21 mole=52.5 g O2

IT
TRY

11. At what temperature does 1.20 moles of hydrogen gas occupy a

volume of 28.1 L at a pressure of 121 kilopascals?

12. A room with a volume of 50 m3 has a pressure of 750 torr at a

temperature of 72 °F. How many moles of gas occupy the room?

Mixtures of Gases
We commonly encounter gases that contain several components. For
example, air is a mixture of gases, composed of about 78% nitrogen and
21% oxygen. The remaining 1% is a mixture of argon, carbon dioxide,
water vapor, and other trace components (Figure 10.27).

Figure 10.27 Air is a mixture of nitrogen, oxygen, argon, carbon dioxide, and smaller
amounts of other gases.

626
 Fortunately, the pressure and volume depend only on the amount of
gas present, not on the identity of the gas. One mole of oxygen gas has the
same pressure and volume as one mole of nitrogen gas or one mole of
carbon dioxide gas.
When working with mixtures of gases, we sometimes use the partial
pressure of the gases that are present. The partial pressure is the pressure
caused by one gas in a mixture. We can add up the partial pressures to find
the total pressure of the mixture. Example 10.8 illustrates this idea.

Adding up all the partial pressures gives the total pressure.

Example 10.8 Using Partial Pressures for a Mixture of Gases

If a 40.0-L cylinder is filled with 5.00 moles of nitrogen, 2.00 moles of
oxygen, and 3.00 moles of carbon dioxide at a temperature of 400 K, what
is the pressure inside the cylinder?
We could take two approaches in solving this problem. The first is to
calculate the pressure that is produced by each of the individual gases. The
partial pressures of nitrogen, oxygen, and carbon dioxide (labeled as PN2,
PO2, PCO2) are calculated as shown here:
PN2=nRTV=(5.00 mol)(0.0821L⋅atmmol⋅K)(400 K)(40.0 L)=4.11
atmPO2=nRTV=(2.00 mol)(0.0821L⋅atmmol⋅K)(400 K)(40.0 L)=1.64
atmPCO2=nRTV=(3.00 mol)(0.0821L⋅atmmol⋅K)(400 K)(40.0 L)=2.46 atm

Once we’ve done this, we can find the total pressure by adding together the
partial pressures:
Ptotal=PN2+PO2+PCO2=4.11 atm +1.64 atm +2.46 atm =8.21 atm

627
 The second approach to this problem simplifies the calculation. Because the
total pressure depends on the number of moles of gas (not the identity of
the gas), we can find the total number of moles present and then calculate
the pressure based on this combined value. •

5.00 moles N2+2.00 moles O2+3.00 moles CO2=10.00 moles total

Ptotal=nRTV=(10.00 mol)(0.0821L⋅atmmol⋅K)(400 K)(40.0 L)=8.21 atm

IT
TRY

13. An air sample contains 0.3% water vapor. If the total air pressure is
765 torr, what is the partial pressure due to the water?

14. A 24-liter cylinder contains 150.0 grams of NH3 gas and 600.0
grams of argon gas at a temperature of 300 K. What is the pressure
inside the cylinder?

15. While walking along the ocean, you take a deep breath. If your lung
capacity is 5.0 liters, the temperature is 87 °F, and the pressure is 765
torr, how many moles of air can you take in? Since air is composed
of about 21% oxygen (by mole percent), how many grams of oxygen
have you taken in?

A Molecular View of the Gas Laws

In the preceding sections, we used mathematical relationships to describe
how gases behave. By thinking about gases on the atomic or molecular
level, we can begin to understand why these mathematical relationships
are so, and we can also predict the behavior of gases in other situations.
Let’s ask three questions about the relationship between the particles in a
container and the pressure they exert:

628
 1. How does the pressure change if the temperature of the gas
increases? At higher temperatures, the particles move faster. Faster-
moving particles strike their surroundings with more force, exerting
more pressure. On the other hand, if the temperature decreases, the
pressure drops. This is why tires are sometimes partially flat on cold
mornings (Figure 10.28).

Pressure increases with temperature.

Figure 10.28 What happens to the pressure in bike tires on a cold morning? Why is this
so? What happens to the pressure if we pump more gas particles into the tire?

2. What happens to the pressure of a gas if the volume increases? If the

volume of a container increases, the particles have to travel farther
before reaching the walls of the container. As a result, the particles
collide with the container less frequently. An increase in volume
produces a decrease in pressure.
3. What happens to the pressure of a gas if the number of particles in a
container is increased? If you’ve ever pumped up a tire, you know
that increasing the amount of gas present increases the pressure. On a
molecular level, more gas particles mean the particles will strike the
sides of the container more often, creating more pressure. The
dependence of pressure on temperature, number of particles, and
volume of the container is shown in Figure 10.29.

629
Figure 10.29 How does the pressure change as the temperature, volume, and
number of particles change? (a) Increasing temperature increases pressure. (b)
Increasing the number of particles increases the pressure. (c) Increasing the
volume decreases the pressure.

Explore

630
 Figure 10.29

Increasing the amount of gas increases the pressure.

IT
TRY

16. How do these changes affect the pressure of a gas?

a. decreasing the temperature
b. decreasing the volume
c. decreasing the amount of gas

631
10.5 Diffusion and Effusion
Diffusion is the spread of particles through random motion. Diffusion can
refer to either the liquid or the gaseous state, but the principle is the same:
Particles randomly move, and as they do, they slowly spread from areas of
higher concentration to lower concentration. For example, gases emitted
from a smokestack diffuse out into the atmosphere (Figure 10.30).
Because lighter particles move more quickly than heavier particles, lighter
gases diffuse more quickly than heavier gases.

Figure 10.30 Gases spread from areas of high concentration to low concentration. This
process is called diffusion.

Effusion is the process of a gas escaping from a container. Like

diffusion, effusion depends on the velocity of the gas particles. For
example, Figure 10.31 shows two containers, one filled with helium, one
filled with air. Each container has a single small hole out of which the
gases can escape. The smaller helium particles move faster and therefore
collide with the container walls more frequently. As a result, they are more
likely to encounter the opening, and they leak out of the container more
quickly.

Figure 10.31 Because they move faster, lighter gases escape from a container more
quickly.

632
 Explore

Figure 10.31

Lighter gases diffuse and effuse faster.

We can see the principle of effusion in the way balloons lose air. If we
fill one rubber balloon with helium, another with nitrogen, and another
with argon, we’ll see that the balloon filled with helium (the lightest gas)
goes flat most quickly while the balloon filled with argon (the heaviest
gas) goes flat most slowly (Figure 10.32). For this reason, helium balloons
are typically made of heavier, less porous materials.

Figure 10.32 (a) When balloons are filled with different gases, (b) the one containing
the lightest gas goes flat most quickly.

633
 IT
TRY

17. Four containers of gas are opened to the air at the same time. One
contains carbon dioxide (CO2), one contains methane (CH4), one
contains propane (C3H8), and one contains carbon disulfide (CS2).
Which of these gases would mix with the air most quickly? Which
would mix most slowly?

634
10.6 Gas Stoichiometry
Sudden changes in gas pressure can have explosive results. These changes
often arise when gases form or are consumed in a chemical reaction. To
safely conduct reactions involving gases, we must know how much gas is
produced or consumed. For example, the process of fermentation, used in
making beer and bread (Figure 10.33), involves the reaction of glucose to
form ethanol and carbon dioxide gas:
C6H12O6 (s)→2 C2H6O (l)+2 CO2 (g)

Figure 10.33 These copper boiling and fermentation tanks are from the Samuel Adams
Brewhouse in Boston, Massachusetts. Because the fermentation reaction produces
carbon dioxide gas, brewers must monitor the pressure of gases inside the tanks.

In this fermentation reaction, how much carbon dioxide is produced for

each kilogram of glucose that reacts?
To answer this question, we must combine the rules of stoichiometry
(covered in Chapter 7) with the gas laws. At the center of these two
concepts is the mole. Using the ideal gas law, we can relate P, V, and T to
the number of moles of a gas:

In Chapter 7 we first talked about a “mole map” that showed the

relationships between grams, moles, particles, and the balanced equation
(see Chapter 7, Figure 7.14). Now we can expand that map to include
gases as well (Figure 10.34). Examples 10.9 and 10.10 demonstrate how

635
to solve these problems.

Figure 10.34 The mole is the central hub for conversion between units.

Example 10.9 Gas Stoichiometry

In the fermentation of glucose, how many moles of carbon dioxide are
produced for each kilogram of glucose that reacts? If the reaction takes
place in a sealed container and the gas occupies a volume of 8.10 liters at a
temperature of 21 °C, find the pressure of the carbon dioxide gas inside the
container.
C6H12O6 (s)→2 C2H6O (l)+2 CO2 (g)

To solve this problem, we first relate grams of glucose to moles of carbon

dioxide:
Grams C6H12O6→Moles C6H12O6→Moles CO2

We begin with 1 kg (1,000 g) of glucose:

Once we’ve found the moles of CO2, we use the ideal gas equation to find
the pressure. Be sure to convert the temperature to kelvins and then solve:

636
A pressure of 33.1 atm is a dangerous explosion risk. To prevent this risk,
brewers must be careful to release some of the CO2 formed and to monitor
the pressure inside fermentation vessels.
In this example, we first used the balanced equation to solve for moles; then
we used the ideal gas law to convert from moles to pressure. In other
problems, we may be given the pressure and volume of a gas and asked to
relate it to another reagent. In this case, we first convert to moles using the
ideal gas law and then solve the stoichiometry problem. Example 10.10
illustrates this process. •

Example 10.10 Gas Stoichiometry

Natural gas burns cleanly in air, according to the equation shown below
(Figure 10.35):
CH4 (g)+2 O2 (g)→CO2 (g)+2 H2O (g)

Figure 10.35 The characteristic blue flame of a natural gas stove.

If 13.1 liters of CH4 burn at a pressure of 1.00 atmosphere and a

temperature of 290 K, what mass of carbon dioxide gas is produced?
In this question, we’re given the pressure and volume of a gas. Our first
step is to convert this to moles, using the ideal gas equation:

Once we have the moles of CH4, we can use the balanced equation to relate
this to the moles of CO2. We can then convert from moles of CO2 to grams
of CO2, as shown here:

637
 Because these problems involve multiple steps, it will take some time
before you become proficient at them. I encourage you to try the following
problems. •

IT
TRY

18. In carbonated soft drinks, the carbon dioxide reacts with water to
produce carbonic acid. This gives the drinks a pungent flavor that is
balanced with a large amount of sugar. Over time, carbonic acid
converts back into carbon dioxide and water, as shown in the
equation below. What mass of H2CO3 would be required to produce
a volume of 2.0 liters of CO2 at a pressure of 1.0 atm and a
temperature of 298 K?
H2CO3 (aq)→H2O (l)+CO2 (g)

19. An airbag contains a solid fuel, which must react very rapidly to
produce a large amount of gas. Airbags typically use several
reactions in tandem, one of which is the decomposition of sodium
azide to give sodium and nitrogen:
2 NaN3 (s)→2 Na (s)+3 N2 (g)

In this reaction, what volume of gas would be produced by the

decomposition of 15 grams of NaN3 if the temperature is 350 K and
the pressure is 1.2 atm?

20. Acetylene torches burn at very high temperatures, using this reaction:
2 C2H2 (g)+5 O2 (g)→4 CO2 (g)+2 H2O (g)

At standard temperature and pressure, what volume of oxygen gas is

required to completely react with a 4.8-liter cylinder of acetylene

638
having an absolute pressure of 80 psi and a temperature of 20 °C?

639
Summary
In this chapter, we’ve explored the structure and properties of solids,
liquids, and gases. In a solid, the particles pack closely together in fixed
positions. In a liquid, the particles are close together but are not held in a
fixed position. In a gas, the particles are far apart and have little or no
interaction with the particles around them. The melting point and boiling
point of a substance depend on how strongly the particles are held
together.
The properties of a substance depend on its composition and structure.
Ionic compounds form rigid lattices with each particle strongly attracted to
those around it. Ionic compounds have very high melting and boiling
points. In metallic substances, the atoms pack closely together and share
electrons loosely between multiple atoms. This structure creates strong
interactions, but not the rigid framework of ionic compounds.
Molecular compounds contain discrete groups of atoms, called
molecules, that are connected by covalent bonds. The physical properties
of molecular compounds depend on the interactions between molecules.
Three major types of intermolecular forces are London dispersion forces,
dipole–dipole interactions, and hydrogen bonds. Dispersion forces are the
weakest interactions while hydrogen bonds are the strongest. Molecular
compounds typically have lower melting and boiling points than ionic
compounds.
Some substances, such as diamond and graphite, form extensive
networks of covalent bonds rather than small molecules. Polymers are
compounds that contain long chains of covalent bonds. Plastics are
synthetic polymers.
Gases have neither fixed shape nor fixed volume. In an ideal gas, the
particles do not interact with each other and occupy a negligible fraction of
the volume of the container.
Gases are typically measured by their pressure. A barometer is a device
that measures atmospheric pressure. Pressure is measured in several
different units, including atmospheres, torr (also called mm Hg), pounds
per square inch (psi), bars, and kilopascals (kPa).
Boyle’s law states that at constant temperatures, the product of the
pressure and volume of an ideal gas are constant. This means that as
pressure increases, the volume decreases, and vice versa.
Charles’s law states that the volume of a gas is directly proportional to
its temperature at constant pressure. By extrapolating the volume of a gas

640
down to zero, we can identify the lowest possible temperature, called
absolute zero. At this temperature, particles have no kinetic energy. The
Kelvin temperature scale measures the absolute temperature of a
substance: Absolute zero is given the value of 0 kelvin. When working
with the gas laws, we must convert temperatures to the Kelvin scale.
The combined gas law integrates Charles’s and Boyle’s laws into a
single law that relates the pressure, volume, and temperature of a gas. This
law is most commonly expressed as

P1V1T1=P2V2T2
where P, V, and T denote the pressure, volume, and temperature, and the
subscripts denote initial and final conditions.
Avogadro’s law states that at constant pressure, the volume of a gas is
proportional to the number of moles present. At standard temperature and
pressure (STP), one mole of gas occupies a volume of 22.4 liters.
The ideal gas law describes the relationship of the P, V, and T to the
number of moles of gas present. The relationship is given by the
expression

PV=nRT
where n is the number of moles of gas, and the gas constant, R, has a value
of 0.0821 L·atm/mol·K.
In a mixture of gases, each component contributes to the overall
pressure of the system. The total pressure of the system is simply the sum
of the partial pressures of the gases present.
We can understand the relationships described in the gas laws by
thinking about gases on a molecular level. Because the pressure inside a
container arises from the collisions of gases with the walls of the
container, an increase of pressure occurs as the amount of gas increases
(more particles striking the walls), as temperature increases (faster-moving
particles strike the walls more frequently and with more energy), or as the
volume decreases (the particles strike the walls more frequently).
Diffusion is the spread of particles through random motion, such as the
motion of gas particles. Because lighter gas particles move more quickly
than heavier particles, lighter gas particles diffuse more quickly. A related
term, effusion, refers to the escape of gas from a container. Because
smaller particles move faster, they also effuse more quickly than heavier
particles.

641
 As we saw in Chapter 7, we can use a balanced equation to predict the
grams and moles that will be consumed or produced in a reaction. By
combining stoichiometry calculations with the ideal gas law, we can
predict changes in pressure and volume that accompany reactions
involving gases.

642
Rethinking Gas Storage

Compressed gases are a part of modern life. From fountain drinks to air
conditioners and from mechanical work to health care, compressed gases are all
around us. But compressed gases can be hard to handle. Because of their high
pressures, gas cylinders require heavy steel walls. They can be difficult to
transport, and they are potentially dangerous. If a gas cylinder ruptures, the
pressure inside the container can fire the cylinder like a missile.
Omar Yaghi is a chemist at the University of California, Berkeley (Figure
10.36). For the past two decades, he has worked on a class of compounds called
metal-organic frameworks, or MOFs, that hold the potential to change the way
people store gases.

Figure 10.36 (a) The transport and storage of gas requires heavy steel containers. (b)
Omar Yaghi is a chemist at UC Berkeley who studies gas storage. (c) A sponge is an
example of a porous material. The empty spaces in these structures can trap other
materials. (d) Yaghi’s compounds, called MOFs, contain tiny pores. (e) These pores can
trap gas particles, reducing the number of free particles and decreasing the pressure in
the container. (f) Dr. Yaghi sits behind the wheel of a vehicle operating on a MOF
methane fuel tank.

Originally from Jordan, Yaghi arrived in the United States when he was 15.
Although he spoke very little English, he enrolled in classes at a community
college near Albany, New York. His abilities and hard work paid off, and he

643
eventually earned a Ph.D. in chemistry from the University of Illinois. As he
describes it, “I was drawn to the beauty of chemistry. I didn’t necessarily want to
change the world—I was driven by curiosity and passion for science.”
The MOF compounds that Yaghi creates are beautiful, but also very useful.
Like a sponge or a honeycomb, these substances contain many tiny pores—except
they are much smaller. Normally, gases move freely around a container, interacting
very little with their surroundings. A MOF is able to trap gases within the tiny
pores. As a result, there are fewer free particles, and the pressure of the gas
decreases. If a gas is placed in a cylinder that contains MOFs, the pressure is lower
than if the gas were placed in an empty cylinder!
MOF technology has created a surge of interest in gas storage. For example,
several companies are developing MOFs for the fuel tanks of vehicles powered by
natural gas. Vehicles using this technology can safely store more fuel, so they have
to be refueled less frequently.

644
Key Terms
10.2 Solids and Liquids
intermolecular forces The forces of attraction or repulsion that take place
between molecules.
dipole–dipole interaction An intermolecular force between two molecules
containing net dipoles.
hydrogen bond An unusually strong dipole–dipole interaction that occurs
between molecules containing H–F, H–O, or H–N bonds.
London dispersion forces Intermolecular forces that result from fluctuations
in charge density called instantaneous dipoles.
covalent networks Long two or three-dimensional sequences of covalent
bonds, resulting in very large single molecules.
polymer A compound composed of long chains of covalently bonded atoms.

10.3 Describing Gases

ideal gas A gas in which the volume of the particles is much less than the
volume of the container, and in which the particles have no attraction for each
other.
pressure The force that an object exerts divided by the area over which it is
applied; for gases, pressure describes the force that gases exert on their
surroundings.
barometer A device used to measure atmospheric pressure.
millimeters of mercury (mm Hg) A measure of gas pressure; this unit
originates from the height to which atmospheric pressure can push a column of
mercury in a barometer; 1 mm Hg = 1 torr.
gauge pressure The difference between a compressed gas pressure and
atmospheric pressure.
atmosphere (atm) A unit of gas pressure; 1 atm = 760 mm Hg.

10.4 The Gas Laws

gas laws Mathematical relationships between the pressure, volume, and
temperature of gases.
Boyle’s law The pressure and volume of an ideal gas are inversely related; the
product of PV is constant at constant temperature.
Charles’s law The volume of an ideal gas is directly proportional to its

645
temperature; the relationship between V and T is constant at constant pressure.
absolute zero The lowest possible temperature, corresponding to 0 K or –
273.15 °C; at this temperature, the particles in a substance have zero kinetic
energy.
combined gas law A combination of Boyle’s law and Charles’s law; it states
that for an ideal gas, the quantity PV/T is constant; usually expressed by the
equation P1V1/T1 = P2V2/T2, where the subscripts 1 and 2 denote two different
conditions.
Avogadro’s law If pressure and temperature are constant, the volume of a gas
is proportional to the number of moles of gas present.
ideal gas law The relationship between pressure, volume, temperature, and the
number of moles of an ideal gas; typically expressed in the form PV = nRT,
where R is the gas constant.
partial pressure The pressure caused by one gas in a mixture.

10.5 Diffusion and Effusion

diffusion The spread of particles through random motion; lighter gases diffuse
more quickly than heavier gases.
effusion The process of a gas escaping from a container; lighter gases effuse
more quickly than heavier gases.

646
Additional Problems
10.1 Interactions between Particles

21. Describe the arrangement and motion of particles in a solid, liquid, and
gas.

22. How does the motion of particles change as a substance transitions from
liquid to gas? In this transition, does the substance absorb or release heat
energy?

10.2 Solids and Liquids

23. How is the arrangement of particles different between an ionic and a

metallic solid?

24. Are there examples in which an ionic solid contains covalent bonds?

25. What are the intramolecular forces in a molecular solid? What types of
intermolecular forces can exist in a covalent solid?

26. What types of covalent bonds are necessary for an intermolecular

hydrogen bond to form?

27. Describe each of the following as ionic, metallic, or molecular solids:

a. calcium fluoride
b. glucose, C6H12O6
c. bronze, an alloy of copper and tin
d. table salt, NaCl

28. Determine whether the following properties broadly describe ionic,

metallic, or molecular solids. Some properties may describe more than one
group.
a. high melting point
b. malleable
c. low boiling point
d. loosely shared electrons within a network of atoms

647
29. These four compounds have very similar formula masses. Classify them as
ionic or covalent. Predict which of the compounds would have the highest
and lowest boiling points.
a. LiF
b. H2O
c. N2
d. HCl

30. These four compounds have very similar formula masses. Classify them as
ionic or covalent. Predict which of the compounds would have the highest
and lowest boiling points.
a. Li2S
b. HF
c. CH3F
d. O2

31. The Lewis structure of propane, C3H8, is shown here. What is the strongest
type of intermolecular force in liquid propane?

32. The Lewis structure of methylamine, CH3NH2, is shown here. What is the
strongest type of intermolecular force in liquid methylamine?

33. The Lewis structure of formaldehyde is shown here. What is the strongest
type of intermolecular force in liquid formaldehyde? Sketch two
molecules, using a dashed line (as in Figure 10.6) to show the positive
region of one molecule interacting with the negative region of another
molecule.

648
34. The Lewis structure for ammonia, NH3, is shown here. What is the
strongest type of intermolecular force in liquid ammonia? Sketch two
molecules, using a dashed line (as in Figure 10.6) to show the positive
region of one molecule interacting with the negative region of another
molecule.

35. What are polymers? How is a polymer different from a molecular solid?

36. How is a covalent network different from a molecular solid?

10.3 Describing Gases

37. Describe the motion of particles in a gas.

38. What are the two criteria for a gas to be considered an ideal gas?

39. Describe how the pressure on a container arises from the particles inside
and outside the container.

40. What happens if the pressure inside a sealed container is much greater than
the pressure outside the container? What happens if the pressure outside a
container is much greater than the pressure inside?

41. When drinking through a straw, you are able to control the height of the
liquid inside the straw by changing the pressure inside your mouth, as
shown in Figure 10.37. What happens if the pressure in your mouth is
lower than the air pressure outside? What happens if the pressure in your
mouth is higher than the air pressure outside?

649
 Figure 10.37 A straw is like a barometer.

42. Ignoring the fact that humans can’t survive in a vacuum, is it possible to
drink through a straw in the vacuum of space? Why or why not?

43. Measure the pressure shown on the barometer. Is this pressure higher or
lower than standard atmospheric pressure? (1 inch = 25.4 mm)

44. Measure the pressure shown on the barometer. Is this pressure higher or
lower than standard atmospheric pressure? (1 inch = 25.4 mm)

650
45. In August 1992, Hurricane Andrew slammed into the Miami area, causing
23 deaths and destroying over 25,000 homes. When it made landfall, the
pressure in the eye of the storm was 922 millibars. What is this pressure in
torr? Is this pressure higher or lower than standard atmospheric pressure?

46. In November 2013, Typhoon Haiyan struck the Philippine islands, causing
over 6,000 deaths. At one point, the pressure in the eye of the storm was
measured at 26.43 inches of mercury. What is this pressure in torr? What
is this pressure in millibars? Is this pressure higher or lower than standard
atmospheric pressure?

47. Express standard atmospheric pressure in millibars, torr, and inches of

mercury.

48. Express standard atmospheric pressure in mm Hg, kilopascals, and bars.

49. Convert the following pressures:

a. Express 698 mm Hg in millibars.
b. Express 3.2 atm in torr.
c. Express 1.42 bars in psi.

50. Convert the following pressures:

a. Express 1.2 atmospheres in bars and millibars.
b. Express 32.41 psi in torr.
c. Express 23.29 inches of mercury in atmospheres.

51. Gauge pressure refers to the difference in pressure between a compressed

gas and atmospheric pressure. If the atmospheric pressure is 15.0 psi, and
you inflate your bicycle tire to a gauge pressure of 110.0 psi, what is the

651
 absolute pressure in the tires?

52. Gauge pressure refers to the difference in pressure between a compressed

gas and atmospheric pressure. If the atmospheric pressure is 15.0 psi, and
you inflate your football to a gauge pressure of 10.0 psi, what is the
absolute pressure in the football?

10.4 The Gas Laws

53. How does the pressure change if the volume of a gas decreases? How does
it change if the temperature decreases?

54. If the volume of a gas increases, how does the pressure change? How does
the pressure change if the temperature of a gas increases?

55. A gas with an initial pressure of 1.0 atmosphere and a volume of 4.3 liters
is compressed to a pressure of 5.1 atmospheres. What is the new volume of
the gas?

56. A gas with an initial pressure of 780 torr and a volume of 150 liters is
compressed to a volume of 32 liters. What is the pressure of the
compressed gas?

57. A balloon with a pressure of 15.0 psi has a volume of 1.24 liters. If the
pressure drops, the balloon will expand. What will be the volume of the
balloon if the pressure decreases to 10.4 psi?

58. A pressurized gas is allowed to expand. If the gas originally occupied a

volume of 420 mL at a pressure of 300 psi, what volume will it occupy if
pressure is lowered to 150 psi?

59. A balloon with a volume of 1.41 liters at a temperature of 300.0 K is

heated to 350.0 K. What is the new volume of the balloon?

60. A balloon with a volume of 4,320 cm3 at a temperature of 25 °C is heated

to 95 °C. What is the new volume of the balloon?

61. An ideal gas is allowed to cool at a constant pressure. The gas occupies a
volume of 20.0 L at a temperature of 273.15 K (0 °C). At what
temperature will it occupy a volume of 10.0 L?

652
62. A gas occupies a volume of 800.0 mL at a temperature of 25 °C. At what
temperature would the gas occupy only half this volume?

63. What is absolute zero? Describe absolute zero with regard to the motion
and kinetic energy of molecules.

64. Is it possible for a substance to be a gas at absolute zero?

65. A gas has a pressure of 900.0 millibars at a temperature of 30.0 °C. If the
volume is unchanged but the temperature is increased to 80.0 °C, what is
the new pressure of the gas?

66. A gas occupying a volume of 43.0 liters has a pressure of 850 millibars at a
temperature of 35 °C. If the volume is unchanged, but the temperature is
increased to 120 °C, what is the new pressure of the gas?

67. A gas cylinder at a temperature of 0 °C has a pressure of 22.4 psi. At what

temperature would the pressure inside the cylinder increase to 35.0 psi?

68. A propane tank at a temperature of 0 °C has a pressure of 415 psi. At what

temperature would the pressure inside the tank increase to 600 psi?

69. A gas inside a piston initially has a pressure of 1.2 bars at a temperature of
25 °C and a volume of 20.0 mL. If the temperature is increased to 120 °C,
and the piston expands to a volume of 80 mL, what is the new pressure
inside the piston?

70. A gas inside a piston occupies a volume of 220 cm3 at a pressure of 1.03
bars and a temperature of 100 °C. The gas is then cooled to room
temperature (25.0 °C), and the volume of gas drops to 180 cm3. What is
the pressure of the gas under these new conditions?

71. A gas inside a balloon has a temperature of 293 K and a volume of 2.40
liters. The gas is cooled to 273 K, and pressure decreases from 790 mm Hg
to 750 mm Hg. What is the new volume of the balloon? Report your
answer to three significant digits.

72. In air-conditioning systems, compressed gases are allowed to expand, and

this expansion results in cooling. A gas with a volume of 5.00 mL at a
pressure of 8.0 bars at a temperature of 40 °C is allowed to expand to a
volume of 15.0 mL at a pressure of 2.0 bars. What is the temperature of the

653
 gas after it expands?

73. What is the volume of two moles of gas, calculated at standard temperature
and pressure?

74. What is the volume of 3.5 moles of gas, calculated at standard temperature
and pressure?

75. At STP, how many moles of gas occupy a room with a volume of 5.0 m3?
(1 m3 = 1,000 L)

76. At STP, how many moles of gas occupy a tank with a volume of 1.2 m3?
(1 m3 = 1,000 L)

77. What is the pressure of 2.31 moles of gas at a temperature of 400.0 K and a
volume of 3.5 liters?

78. What is the pressure of 12.5 moles of gas at a temperature of 360.0 K and a
volume of 5.02 liters?

79. At what temperature does 4.0 moles of gas under a pressure of 1.0
atmosphere occupy a volume of 120.3 liters?

80. At what temperature does 1.3 moles of gas under a pressure of 1.2
atmospheres occupy a volume of 108.4 L?

81. A gas cylinder contains 80.0 grams of helium gas, occupying a volume of
20.0 liters at a temperature of 280 K. What is the pressure inside the
cylinder?

82. What is the pressure of 14.0 moles of argon gas at a temperature of 0 °C

and a volume of 800 mL?

83. What volume is occupied by 2 moles of nitrogen gas at a temperature of 73

°F and a pressure of 238.0 psi?

84. What volume would be required to store 150 moles of argon gas at 70 °F
with a maximum pressure of 350.0 psi?

85. A gas cylinder contains 80.0 grams of carbon dioxide gas, occupying a

654
 volume of 20.0 liters at a temperature of 280 K. What is the pressure inside
the cylinder?

86. How many grams of helium gas can a 12.0-liter cylinder at a temperature
of 30 °C contain before the pressure exceeds 180.0 psi?

87. A 1.0-L cylinder of carbon dioxide has a pressure of 8.0 atmospheres and a
temperature of 373 K. How many CO2 molecules are in this cylinder?

88. What is the pressure caused by 1.5 × 1021 helium atoms occupying a
volume of 0.0023 L at a temperature of 398 K?

89. A gas cylinder contains an unknown gas. It is found that a 1.90-gram

sample of the gas occupies a volume of 2.30 liters at a pressure of 1.0
atmosphere and a temperature of 298 K.
a. How many moles are in the 100-gram sample?
b. What is the formula mass of this gas, in grams per mole?
c. Based on the formula mass, identify the gas as one of the following:
helium, neon, argon, nitrogen, oxygen, or carbon dioxide.

90. A gas cylinder at a temperature of 298 K contains an unknown gas. When

48.0 grams of the gas are released from the cylinder, the pressure of the
gas drops by 53.9 psi (3.67 atmospheres). How many moles of gas were
released from the tank? What is the formula mass of this gas, in grams per
mole? Identify the gas as one of the following: helium, neon, argon,
nitrogen, oxygen, or carbon dioxide.

91. What does the term partial pressure mean?

92. A gas mixture contains 50% oxygen, 30% nitrogen, and 20% carbon
dioxide by mole ratio. If the total pressure is 1.0 atmosphere, what is the
partial pressure of oxygen, nitrogen, and carbon dioxide in the sample?

93. A gas mixture contains 40% methane and 60% oxygen by mole ratio. If the
total pressure is 45 psi, what is the partial pressure of methane and oxygen
in the sample?

94. A cylinder of gas contains 100.0 moles of nitrogen gas and 25.0 moles of
oxygen gas. The overall pressure of the cylinder is 214.2 psi. Calculate the
partial pressure of oxygen in the cylinder.

655
95. A 24.0-liter gas cylinder contains 8.0 moles of nitrogen gas and 2.0 moles
of oxygen gas. If the temperature of inside the cylinder is 0 °C, what is the
total pressure inside the cylinder?

96. A balloon is filled with 10.9 grams of N2, 9.6 grams of O2, 2.2 grams of
CO2, and 0.04 grams of Ar. The pressure inside the balloon is 1.05 bars,
and the temperature is 70 °F. What is the volume of the balloon?

97. An air sample contains 0.5% water vapor. If the total air pressure is 750
torr, what is the partial pressure due to the water vapor?

98. Imagine you are sitting in a room with a volume of 25,000 liters and a
temperature of 72 °F. The total air pressure is 750 mm Hg, and the partial
pressure from oxygen is 21% of the total pressure. How many grams of
oxygen are in the room?

99. A mixture of nitrogen gas (N2) and oxygen gas (O2) has a pressure of 1.0
atmosphere at a temperature of 72 °F. At this temperature, which type of
molecules is moving with more velocity?

100. The pressure that a gas exerts on a container is determined by the

collisions the gas makes with the inside walls of the container. How
would these collisions change (in frequency or magnitude) under the
following conditions?
a. Keep the temperature constant and decrease the volume.
b. Decrease the temperature and decrease the volume.
c. Increase the temperature and decrease the volume.
d. Increase both the temperature and the volume.

10.5 Diffusion and Effusion

101. What is the difference between diffusion and effusion?

102. If you have a balloon filled with neon and a balloon filled with argon,
which one will deflate the fastest? How do you know?

10.6 Gas Stoichiometry

103. The reaction of potassium metal with water produces hydrogen gas, as in
the equation below. How many moles of hydrogen gas can be produced

656
 from the reaction of 5 moles of potassium? At STP, how many liters of
hydrogen gas can be produced?
2 K (s)+2 H2O (aq)→2 KOH (aq)+H2 (g)

104. Hydrogen and oxygen react explosively to form water, as in the reaction
below. If a balloon containing 1.5 liters of hydrogen gas at 25 °C and a
pressure of 1.0 atmosphere reacts with excess oxygen, how many grams
of water can be produced?

2 H2 (g)+O2 (g)→2 H2O (g)

105. Propane gas (C3H8) reacts with oxygen according to this balanced
equation:
C3H8 (g)+ 5 O2 (g)→3 CO2 (g)+4 H2O (g)

a. How many moles of water can be formed from 15.0 moles of propane?
b. How many moles of carbon dioxide can be formed from 15.0 moles of
propane?
c. At STP, what volume of CO2 would be produced from 15.0 moles of
propane?
d. Are more moles of gas produced or consumed in this reaction?

106. The Born–Haber process is used to manufacture ammonia (NH3) from

nitrogen gas and hydrogen gas:

3 H2 (g)+N2 (g)→2 NH3 (g)

In this reaction,
a. How many moles of nitrogen are needed to react with 15 moles of
hydrogen?
b. How many moles of ammonia can be produced from 15 moles of
hydrogen?
c. At a temperature of 800 K and a pressure of 4.00 atm, how many liters
of ammonia can be produced from 15.0 moles of hydrogen?
d. If this reaction proceeded to the right in a sealed container, would the
pressure inside the container increase or decrease? How do you know?

657
107. Natural gas, CH4, burns in oxygen as shown in the reaction below. A 1.0-
liter cylinder containing CH4 with a pressure of 2.1 atmospheres at a
temperature of 298 kelvin is completely used to power a portable stove.
CH4 (g)+2 O2 (g)→CO2 (g)+2 H2O (g)

a. How many moles of CH4 were in the cylinder?

b. How many moles of oxygen gas were necessary to react with this
amount of CH4?
c. How many moles of water were produced in this reaction?
d. How many grams of carbon dioxide were produced in this reaction?

108. When a beverage is carbonated, the carbon dioxide reacts with water to
produce carbonic acid, which gives soft drinks a bitterness that is
balanced with a large amount of sugar. Over time, carbonic acid converts
back into carbon dioxide and water:
H2CO3 (aq)→H2O (l)+CO2 (g)

In this reaction, what volume of CO2 can be produced from 2.0 grams of
H2CO3 at standard temperature and pressure? (Standard temperature and
pressure is 273 K and 1.0 atmosphere.)

109. Glucose fermentation takes place through the following reaction:

C6H12O6 (aq)→2 C2H6O (aq)+2 CO2 (g)

If 2.05 grams of C6H12O6 were placed in a sealed container having a

temperature of 200 °C and a volume of 5.0 liters, and if this reaction went
to completion, what would be the pressure from the carbon dioxide inside
the container?

110. Ammonia, NH3, is a colorless gas with a pungent odor. Its applications
range from cleaning supplies to fertilizer, and it is a building block for the
production of many industrial chemicals and consumer products. For
example, the first step in preparing nitric acid involves the reaction of
ammonia gas with oxygen gas to produce nitrogen monoxide and water,
as shown in this reaction:
4 NH3 (g)+5 O2 (g)→4 NO (g)+6 H2O (g)

658
 a. If this reaction takes place in a sealed container, and the temperature
inside the container is kept constant, will the pressure inside the
container increase or decrease? How do you know?
b. If 170 grams of NH3 react in this way, how many grams of H2O can
be produced?
c. If 170 grams of NH3 react in this way, how many liters of H2O can be
produced, given a temperature of 373 K and a pressure of 1.00
atmosphere?

111. Octane, a component of gasoline, burns according to this equation:

2 C8H18 (l)+25 O2 (g)→16 CO2 (g)+18 H2O (g)

Octane has a formula mass of 114.26 g/mol and a density of 0.703 kg/L.
At standard temperature and pressure, what volume of CO2 is produced
by the combustion of one liter of octane?

112. During photosynthesis, plants absorb sunlight and use this energy to
convert carbon dioxide and water into simple sugars and oxygen. The
process is exactly the reverse of combustion and is described by this
equation:
6 CO2 (g)+6 H2O (l)→C6H12O6+6 O2 (g)

At a temperature of 25 °C and a pressure of 1.00 atm, how many liters of

carbon dioxide gas are consumed by the production of 1.00 kg of
C6H12O6?

Challenge Questions

113. Propane gas reacts with oxygen according to this balanced equation:
C3H8 (g)+5 O2 (g)→3 CO2 (g)+4 H2O (g)

Assuming an air sample contains 21% oxygen (by volume), in which of

the following mixtures will the propane and oxygen almost completely
react, leaving almost no excess of either gas?
a. a mixture of 50% air and 50% propane

659
 b. a mixture of 74% air and 26% propane
c. a mixture of 82% air and 18% propane
d. a mixture of 96% air and 4% propane

114. Octane, a component of gasoline, burns according to this equation:

2 C8H18 (l)+25 O2 (g)→16 CO2 (g)+18 H2O (g)

In an engine cycle, 0.030 g of C8H18 is mixed with 0.30 L of air at a

pressure of 3.2 atmospheres and a temperature of 450 K inside a piston.
The spark plug detonates the gasoline, causing the explosive reaction to
take place. What mass of CO2 and what mass of H2O can be produced in
this engine cycle? Assume the air contains 21% oxygen. In what two
ways does this reaction increase the pressure and push the piston?

660
Chapter Eleven
Solutions

The Perfect Cup of Coffee

What makes a really good cup of coffee?
For Aaron Blanco, this is more than just a question—it’s an obsession. Aaron is
the founder of the Brown Coffee Company in San Antonio, Texas. For the past
decade, he has explored every facet of the coffee-making process, from growing
coffee plants to roasting and grinding the beans to brewing the perfect cup. He
makes amazing coffee (Figure11.1).

Figure 11.1 (a) Aaron Blanco works with his coffee roaster. (b) Ripe coffee cherries on
a tree in El Salvador. (c) Coffee beans drying in the sun. (d) Freshly ground coffee,
ready to brew. (e) A cappuccino at the Brown Coffee Company.

661
 Aaron started his business in 2005. He dreamed of opening a coffee shop but
lacked the funds to get it started. So instead, he bought a small oven and began
selling fresh-roasted beans to local restaurants. Roasting creates tiny pores in the
bean, so water can reach and dissolve the molecules inside. It also breaks complex
molecules into simpler, tastier structures. But it has to be done just right: Too little
heating leaves the flavors trapped; too much heating destroys them. By blending
the scientific method with the nuances of taste, Aaron explored how the heating
temperature and time affect the flavor of coffee.
After several years of roasting, he finally opened his first shop. Preparing
coffee for customers turned his attention to the grinding and brewing processes:
How finely should the beans be ground? How does the water temperature affect the
flavor? What volume of water is needed for each gram of coffee grounds? How
should the water and grounds be mixed?
As his business has grown, Aaron has returned to the beginning—to the bean
itself. Farmers grow two species of the tropical coffee plant: Coffea robusta grows
at any altitude and is well suited to massive farms and mechanized harvests.
Robusta coffee is bitter, but cheap. Its more delicate cousin, Coffea arabica,
produces a milder, tastier coffee—but it only grows well at high altitudes. Aaron
frequently travels to the highlands of Africa and Central America where arabica is
grown. He meets with farmers to discuss the soil, local weather, and subtle
differences in farming techniques.
Aaron’s passion and hard work have paid off. In 2015, Aaron opened his third
coffee shop, and his coffee was recognized as the “Best Coffee in San Antonio.”
Food Network star Alton Brown called it “The best cup of coffee I’ve ever had in
my life.” Aaron’s travels in search of the best coffee beans were even described in
the documentary Coffee Hunting: Kenya. As he puts it, “It took ten years to be an
overnight success.”
So what goes into a perfect cup of coffee? Aaron will tell you there is still a lot
he doesn’t know. Coffee is a complex mixture of hundreds of components
dissolved in water. But while we may never fully understand coffee, we can begin
by understanding how solutions behave. How do substances dissolve in water?
How do we measure the amount of a substance that is in a solution? Why do
different compounds exhibit different solubilities? How are the properties of
solutions different from the properties of pure liquids?
In this chapter, we’ll explore these questions. We’ll describe the composition
of solutions and see how different substances behave when dissolved in water.
We’ll explore chemical reactions that occur in solution and apply the principles of
stoichiometry to reactions in solution. Pour a cup of coffee, and let’s get started.

Intended Learning Outcomes

After completing this chapter and working the practice problems, you should be
able to:

11.1 Describing Concentration

662
 Calculate solution concentrations by percent, parts per million, parts per
billion, and molarity.
Convert between moles, volume, and molarity.
Quantitatively describe the preparation of solutions of a known molarity.

11.2 Electrolyte Solutions

Describe the behavior of electrolytes and nonelectrolytes in aqueous
solution.
Determine the molar concentrations of ions in solution.
Qualitatively describe the changes in freezing point, boiling point, and
osmotic pressure as a function of solute concentration.

11.3 Reactions in Solution—A Review and a Preview

Describe reactions that take place in aqueous solution using molecular and
ionic equations.

11.4 Solution Stoichiometry

Apply the principles of stoichiometry to solve problems involving solutions.

663
11.1 Describing Concentration
When sugar is added to coffee, the sugar dissolves (Figure11.2). That is,
the sugar particles disperse throughout the coffee, forming a homogeneous
mixture. This type of mixture is called a solution. The substance that
dissolves (for example, the sugar) is the solute. The liquid in which the
solute dissolves is the solvent.

Figure 11.2 Sugar dissolves in hot coffee. In a solution, the solute particles disperse
evenly throughout the solvent.

The amount of solute that is present in a solution is called its

concentration. If a solution contains a small amount of solute, we say it is
dilute. If it contains a large amount of solute, we say it is concentrated. A
solution that contains the maximum amount of dissolved solute is
saturated.
There are several different ways to describe concentration, and they
each have different applications. In this section, we will look at three
common ways of describing concentration: by percent, by fraction, or by
molarity.

Concentration by Percent
Percent by Mass and Volume
We frequently describe solutions by the percentage of solute present. One
way to do this is percent by mass:
Mass %=mass of solutemass of solution×100%

More commonly, we express concentration as percent by volume:

664
 Volume %=volume of solutevolume of solution×100%

We often use percent by volume when both the solute and solvent are
liquids. For example, instructions for cleaning a stove might call for a 5%
(v/v) solution of bleach in water. The abbreviation (v/v) indicates that both
the solute and the solution are measured by volume. In this solution, the
solute accounts for 5% of the total volume of the solution. We could
prepare this solution by adding 5 mL of bleach to enough water to make
100 mL of solution.

Cleaning solutions are often prepared based on volume percent.

Example 11.1 Using Percent Concentration

A cleaning solution contains 2% (v/v) ammonia in water. What volume of
ammonia is needed to prepare 10 liters of this solution?
To solve this problem, we substitute the volume of solution and the target
percentage into the equation shown above. •

2%=volume of solute10 L solution×100%0.02=volume of solute10 L

solutionVolume of solute=0.2 L or 200 mL

665
 IT
TRY

1. A chemist prepares a solution by adding 1.0 kg of salt to 5.0 kg of

water. What is the percent by mass of salt in this solution?

2. A common light beer contains 4.2% (v/v) alcohol. What is the volume
of alcohol (in ounces and in milliliters) in a 16-ounce bottle of beer?

Mass/Volume Percent
When the solute is a solid, we often express concentration by
mass/volume percent:
Mass/volume %=mass of solutevolume of solution×100%

We write this as “% (m/v),” and we frequently use units of g/mL. For

example, normal saline solution (used for intravenous fluids) is 0.9%
(m/v). This means that 0.9 grams of salt are present per 100 mL of solution
(Figure11.3).

666
Figure 11.3 Normal saline solution is used in hospital IV packs. It has a concentration
of 0.9% (m/v).

Seawater has a higher salt concentration than normal saline. Typically,

seawater has a salt concentration of about 3.5% (m/v). This is still far from
saturated: In the Middle East, the Dead Sea has a salt concentration of
approximately 34% (m/v); see Figure11.4.

Figure 11.4 These salt deposits lie along the Dead Sea, which has a salt concentration
of about 34% (m/v).

For aqueous solutions, mass/volume % is the number of grams of solute per 100
mL.

667
 Example 11.2 Using Percent Concentration
Normal saline solution is 0.90% NaCl (g/mL). How many grams of sodium
chloride are needed to prepare 0.50 L of normal saline solution?
Recall that a percentage is a fraction of 100. Therefore, if a solution is
0.90% (g/mL), this means it contains 0.90 g per 100 mL, or 9.0 grams per
liter:
0.90% (g/mL)=0.90 g100 mL=9.0 gL

To find the total mass of NaCl needed, we multiply the volume of solution
that we need by the ratio of grams to volume. This gives us a final answer
of 4.5 g of NaCl.
0.50 L×9.0 g NaClL=4.5 g NaCl

A different approach to this problem is to set up a ratio. •

9.0 g NaCl1 L=x0.50 Lx=4.5 g NaCl

IT
TRY

3. What mass of potassium chloride is present in 100 mL of a solution

that is 15.0% KCl (m/v)? How much more water would you need to
add to dilute this solution to 3.0% (m/v)?

Very Dilute Solutions: ppm and ppb

Clean water is vitally important. Toxic substances in the water supply can
cause widespread health issues or have drastic effects on ecosystems, even

668
when they are present in very dilute quantities. For example,
pharmaceutical compounds, cosmetics, pesticides, and components of
plastic containers often find their way into rivers, streams, and reservoirs,
endangering the wildlife and humans who drink from them (Figure 11.5).
This is why governmental agencies and watchdog groups closely monitor
the levels of pollutants in lakes and streams. Table 11.1 summarizes the
maximum allowable levels of several chemical contaminants in drinking
water.

Figure 11.5 Industrial, agricultural, and consumer waste contribute to the pollution of
waterways.

TABLE 11.1 Contaminants monitored in U.S. Drinking Water

Maximum
Concentration Level
in Drinking Water
Contaminant Source Health Effects (mg/L)

Polychlorinated Runoff from landfills Reproductive and 5 × 10–4

biphenyls nervous system
(PCBs) difficulties; increased
risk of cancer

Dioxin Emissions from waste incineration; Reproductive 3 × 10–8

(2,3,7,8- discharge from chemical factories difficulties; increased
TCDD) risk of cancer

Nitrate Runoff from fertilizer use; leaching Serious illness for 10

from septic tanks, sewage, erosion of infants below the age of
natural deposits six months

Mercury Erosion of natural deposits, Kidney damage 2 × 10–3

discharge from refineries and
factories; runoff from landfills and

669
 croplands

Atrazine Runoff from herbicides Cardiovascular and 3 × 10–3

reproductive problems

Source: Data from the U.S. Environmental Protection Agency (www.epa.gov).

Water monitoring is essential for public health and the protection of ecosystems.

We often express very dilute concentrations in parts per million

(ppm) or parts per billion (ppb). For example, if a pond contains 1 gram
of nitrate for every 1,000,000 grams of solution, we describe the
concentration as 1 part per million, or 1 ppm. Mathematically, we calculate
the parts per million using the following relationship:
Concentration (ppm)=mass of solutemass of solution×106

Even smaller concentrations are expressed using parts per billion, or

ppb. The relationship is similar:
Concentration (ppb)=mass of solutemass of solution×109

When calculating ppm or ppb, any unit of mass may be used, but the
units for the solute and the solution must be the same. By definition, 1 liter
of pure water at 4 °C has a mass of exactly 1 kg. Under these conditions, 1
ppm means 1 milligram of solute per liter of solution. Similarly, 1 ppb
means 1 microgram of solute per liter of solution. Example 11.3

670
demonstrates this principle.

For a dilute aqueous solution,

1 mg/L = 1 ppm
1 μg/L = 1 ppb

Because 1 μg = 10–6 g, we can write 45 μg as 45 × 10–6 g. While this is not

proper scientific notation, it is a simple way to convert from μg to g.

Example 11.3 Concentrations in ppm and ppb

Phthalates are a class of compounds that increase the flexibility and
durability of plastics. These compounds slowly leach out of plastic products
and may contaminate waterways. In a recent study, scientists found that
samples from a European wastewater treatment plant contained about 45
µg of phthalates per liter of solution. What is the concentration of this
compound in ppm? In ppb? (Assume the solution has the same density as
pure water at 4 °C.)
To solve this problem, let’s express both the mass of solute and the mass of
solution in grams: 45 μg = 45 × 10–6 g, and one liter of water has a mass of
1 kg, or 1,000 g:
Concentration (ppm)=45×10−6 g1,000 g×106=0.045 ppmConcentration
(ppb)=45×10−6 g1,000 g×109=45 ppb

So 45 μg of solute per liter of aqueous solution is the same as 45 ppb. •

IT
TRY

4. In 1999, scientists reported that groundwater from a waste site near

Atlantic City, New Jersey, had concentrations of diazepam (Valium®)
as high as 40 ppb. At this concentration, how many milligrams of

671
 diazepam were present in a 100-liter sample of this wastewater?

5. A 0.500-kg water sample contains 14.3 µg of contaminant. What is the

concentration of the contaminant in ppb? What is the concentration in
% g/mL?

Molarity
Concentrations based on the mass or volume of solute are useful for many
common measurements. However, when describing chemical reactions, we
are frequently concerned with the number of moles of a substance that are
consumed or produced. In these cases, it is helpful to express
concentration in a way that allows us to quickly convert to moles. For this
reason, chemists often describe concentration in terms of molarity (M),
which is the moles of solute per liter of solution:
Molarity (M)=moles of soluteliters of solution

We commonly write this in an abbreviated form:

M=molesV
where V is the volume in liters.
The molarity equation also allows us to find the number of moles
present, given the concentration and the volume. Because molarity is equal
to moles over volume, we can rearrange the equation to solve for moles:

moles=MV
Examples 11.4 and 11.5 illustrate these ideas.

672
For many solutions, such as acids and bases, it is important to label both the
ingredients and the concentration.

Example 11.4 Calculating Molarity

An aqueous solution with a volume of 9.41 L contains 412.3 grams of
dissolved potassium chloride. What is the molarity of this solution?
To solve this problem, we must first convert the amount of KCl from grams
to moles. The molar mass of KCl is 74.55 g/mol, so we can write:
412.3 g KCl×1 mole KCl74.55 g KCl=5.531 moles KCl

We can then find the molarity. •

Molarity=moles of soluteliters of solution=5.531 moles KCl9.41 L=0.588 M

KCl

Example 11.5 Finding Moles from Molarity and Volume

How many moles of HCl are in a 250-mL sample of 6.0 M HCl?

673
 To answer this question, we first convert the volume to liters (250 mL =
0.250 L). We then use the relationship given earlier to solve:
Moles=MV=(6.0molL)(0.250 L)=1.5 moles

Molarity is an important and useful tool for describing concentration.

Before moving on, I encourage you to attempt the following problems. •

IT
TRY

6. A 2.4-liter solution contains 23.9 grams of calcium chloride. What is

the molarity of calcium chloride in this solution?

7. How many moles of silver nitrate are present in 15.0 mL of a 1.2-M

aq. AgNO3 solution?

Preparing Solutions of Known Molarity

Here’s a question: If we dissolve 1.0 mole of sugar in 1.0 L of water, do
we have a 1.0-M solution? Not exactly. Because the sugar also takes up
space, the volume of the solution after mixing is a little more than 1.0 L.
The solution is almost 1.0 M, but not exactly. So, how do we prepare
solutions whose molarity is precisely known? Chemists routinely use this
three-step procedure (Figure 11.6):

674
Figure 11.6 To prepare a solution with a precisely known concentration: (a) Measure
out the correct amount of solute; (b) partially fill the volumetric flask, and mix until the
solid dissolves; and (c) slowly dilute the solution to the correct volume.

Explore

Figure 11.6

1. Measure out the desired amount of solute, and add it to a volumetric

flask (these flasks have long, slender necks for precisely measuring
volume).
2. Partially fill the flask with solvent, and mix until the solute dissolves
completely.
3. Dilute the solution, slowly adding solvent until the correct volume is

675
 reached.
By following this procedure, we know the amount of solute and the
volume of solution are both measured precisely.

IT
TRY

8. While working in a forensics lab, you need to prepare 2.00 L of a

1.40-M aq. sodium hydroxide solution. What mass of sodium
hydroxide is required to do this? Describe how you would prepare the
solution.

Preparing Dilute Solutions

Many laboratories store common chemicals, such acids and bases, as
concentrated solutions. While this saves space, chemists often need more
dilute solutions for routine use. For example, if we need a solution of 1.0
M aq. HCl, how do we prepare it from a concentrated 12.0-M solution?
Chemists prepare dilute solutions by mixing the concentrated solution
with additional solvent (Figure 11.7). Doing this increases the volume of
the solution, but the moles of solute do not change. That is, the number of
moles of solute in the initial (concentrated) solution is equal to the number
of moles in the final (dilute) solution:

molesinitial=molesfinal

Figure 11.7 Two glasses of orange drink. Both glasses contain the same amount of
drink mix, but the one on the right has been diluted by adding more water.

Because the moles of a solute are equal to the molarity times the
volume, we can rewrite this expression based on the molarity and volume
of the initial and final solutions:

676
 MiVi=MfVf
where Mi and Vi are the initial molarity and volume, and Mf and Vf are the
final molarity and volume.

Explore
Figure 11.7

Using this relationship, we can prepare dilute solutions of known

concentration, as illustrated in Examples 11.6 and 11.7.

MV = moles

Example 11.6 Preparing Dilute Solutions

You have 25.0 mL of a 10.0-M solution of potassium hydroxide. To this, you
add enough water to make 500.0 mL of a dilute solution. What is the
molarity of the final solution?
To solve this problem, we rearrange the equation above to find the final
molarity (Mf), then substitute the values in the question:
Mf=MiViVf=(10.0 M)(25.0 mL)500.0 mL=0.500 M KOH

Notice that, although we would need to convert the volume from milliliters
to liters in order to solve for the moles present, we did not have to convert
the volume units to solve this problem because the units cancel each other
out. •

Example 11.7 Preparing Dilute Solutions

Starting from a 12.0-M concentrated HCl solution, you need to prepare 180
mL of 1.0-M aq. HCl. What volume of concentrated HCl do you need?
In this example, we need to solve for the initial volume of the concentrated
solution, Vi. Rearranging the relationship to find Vi, we get

677
 Vi=MfVfMi=(1.0 M)(180 mL)12.0 M=15 mL conc.HCl

Notice again that units of molarity cancel out, and it was not necessary to
change the volume from milliliters to liters. We used milliliters for the
volume of Vf and were left with milliliters for the volume of Vi. •

Using Square Brackets to Represent Concentration

Sometimes chemists use square brackets to indicate the concentration of a
solute. For example, to represent the concentration of H2SO4 in a sample,
we might write [H2SO4], meaning “concentration of sulfuric acid.” If we
wish to say the concentration of sodium nitrate is 1.3 molar, we write this
as [NaNO3] = 1.3 M.

When diluting acid solutions, it is good practice to add small amounts of acid
slowly to water, rather than adding water to the acid. This method prevents the
solution from overheating and splattering hot, concentrated acid.

IT
TRY

9. You have 30.0 mL of a 7.80-M solution of MgBr2 that you wish to

678
dilute until [MgBr2] = 0.50 M. To what volume must you dilute the
sample?

679
11.2 Electrolyte Solutions
In previous chapters, we discussed solutions of ionic compounds in water.
Recall that when ionic compounds dissolve in water, they dissociate—that
is, they separate into positive and negative ions, as shown in Figure 11.8.
Because these compounds increase the ability of water to conduct
electricity, they are commonly called electrolyte solutions.

Figure 11.8 (a) When an ionic solid dissolves, the water molecules pull the ions away
from the solid and into solution. (b) In solution, the positive ions are surrounded by the
negative ends of the water molecules, while the negative ions are surrounded by the
positive ends of the water molecules.

Explore
Figure 11.8

To review electrolyte solutions, see Sections 5.7 and 6.5.

Covalent compounds generally do not ionize in water. An exception to

this occurs with acids, compounds that contain weak polar covalent bonds
to hydrogen. In water, these bonds break to produce an H+ cation and a
corresponding anion.
Sometimes we show the process of dissociation or ionization using
ionic equations. For example, solid KBr and gaseous hydrogen bromide
both produce ions when dissolved in water. We can write these as either
molecular equations or ionic equations:
Molecular equations: KBr (s)→KBr (aq)HBr (g)→HBr (aq)Ionic
equations: KBr (s)→K+ (aq)+Br− (aq)HBr (g)→H+ (aq)+Br− (aq)

680
Electrolyte Concentrations
Suppose one mole of magnesium chloride, MgCl2, is dissolved in enough
water to make one liter of solution. The concentration of MgCl2 is 1.0 M.
However, magnesium chloride is an ionic compound, so it dissociates
when dissolved in water:
MgCl2(s)→Mg2+(aq)+2 Cl−(aq)1 mole/L MgCl2→1 mole/L Mg2++2 moles/L Cl−

Notice that when the solid dissociates, two moles of chloride ions form for
every one mole of magnesium ions. This means that the concentration of
chloride ions in the solution is twice as much as that of magnesium. We
could describe this solution as having a 1.0-M concentration of MgCl2, or,
we could describe it as having a 1.0-M concentration of Mg2+ ions and a
2.0-M concentration of Cl– ions. Both descriptions are correct.
Let’s try a second example: What is the concentration of lithium ions
and sulfate ions in a 3.5-M solution of lithium sulfate? Again, we must
recognize that the ionic compound dissociates in solution:
Li2SO4 (s)→2 Li+ (aq)+SO42−(aq)

Notice that we get two moles of lithium ions for each mole of lithium
sulfate. As a result, the concentration of lithium ions is 2 × 3.5 M, or 7.0
M. Because there is only one mole of sulfate in each mole of lithium
sulfate, the concentration of sulfate ions is 3.5 M.

681
 A grilled cheese sandwich is made of two slices of bread and one slice of
cheese. If everyone at the table has one sandwich, you can also say that
everyone has one slice of cheese and two slices of bread.

Example 11.8 Determining Ion Concentrations

Write an ionic equation showing the ions formed when ammonium
phosphate dissolves in water. What is the concentration of ammonium and
phosphate ions in a 0.928-M ammonium phosphate solution?
Ammonium phosphate has the formula (NH4)3PO4. When dissolved in
water, this compound dissociates, producing three ammonium ions and one
phosphate ion from each unit of the compound:
(NH4)3PO4 (s)→3 NH4+ (aq)+PO43− (aq)

Because three ammonium ions are produced for each unit of compound, the
ammonium ion concentration is three times the concentration of the original
compound:
[ NH4+]=3×0.928 M=2.78 M

One unit of phosphate is formed for each unit of compound, so the

phosphate ion concentration is the same as that of the original compound. •

[ PO43−]=1×0.928 M=0.928 M

IT
TRY

682
 10. Find the concentration of each of the dissociated ions in these
solutions:
a. 5.0 M aq. AlBr3
b. 2.4 M aq. K2CO3
c. 7.3 M NH4Cl

11. If 0.50 L of 1.0-M aq. NaNO3 solution is combined with 2.40 L of

3.2-M aq. Fe(NO3)2 solution, how many moles of nitrate are in the
combined solution? What is the molarity of nitrate in the combined
solution?

Colligative Properties
The colligative properties are properties of a solution that do not depend
on the type of particles that are dissolved, but rather on the number of
dissolved particles in the solution. There are three common and important
colligative properties: freezing point depression, boiling point elevation,
and osmotic pressure.
Colligative properties depend on the number of dissolved particles in
solution.

Freezing Point Depression

In colder climates, salt trucks are a common sight in winter: These trucks
spread salt on roads to melt ice and snow (Figure 11.9). This works
because a saltwater solution has a lower freezing point than pure water. In
general, the presence of a solute (or a contaminant) lowers the freezing
point of the liquid. This property of aqueous solutions is called freezing
point depression.

683
Figure 11.9 A salt truck deposits salt on a road during a snowstorm. The salt lowers the
freezing temperature of water, preventing the roads from icing over.

How much the temperature is lowered depends on the total

concentration of dissolved particles. The more dissolved particles, the
lower the freezing temperature.
Figure 11.10 shows the freezing point of water containing sugar
(sucrose), sodium chloride, and calcium chloride. For each compound, the
freezing point of the solution drops as more solute is added. However, the
change in freezing point depends on the solute: It drops the most for
calcium chloride and the least for sucrose. Why is this?

Figure 11.10 A comparison of the freezing points of sucrose, sodium chloride, and
calcium chloride solutions at different concentrations.

Colligative properties depend on the number of particles, not the type.

Sucrose is a nonelectrolyte, meaning it doesn’t dissociate into ions (see
Figure 5.21). If one mole of sucrose is dissolved in water, then one mole of
solute particles is present. However, sodium chloride and calcium chloride
are both electrolytes. One mole of sodium chloride dissociates to give two
moles of dissolved particles:

684
 NaCl (s)→Na+(aq)+Cl− (aq)1 mole NaCl→1 mole Na++1 mole Cl−

The freezing point is the same thing as the melting point—at this temperature,
solid and liquid interconvert.

Similarly, one mole of calcium chloride dissociates to give three moles of

dissolved particles:
CaCl2 (s)→Ca2+ (aq)+2 Cl− (aq)1 mole CaCl2→1 mole Ca2++2 moles Cl−

Because calcium chloride dissociates into more particles, it drops the

freezing point of water more than sucrose or sodium chloride does at the
same concentration. Calcium chloride drops the freezing point more
effectively than sodium chloride, and it is usually the salt of choice for
melting ice on roadways.
Boiling Point Elevation
The presence of dissolved solids also raises the boiling temperature of a
liquid. This property is commonly called boiling point elevation. Like
freezing point depression, this property depends on the number (not the
type) of solute particles present.
One important example of both freezing point depression and boiling
point elevation is the use of mixtures as engine coolant (Figure 11.11).
Automobiles must have liquid coolant to carry excess heat away from the
engine, but the use of pure water in the coolant lines is risky: Because
water expands when it freezes, water lines may burst in cold weather. To
prevent this, ethylene glycol (commonly called antifreeze) is mixed with

685
water. A 40% (by mass) solution of ethylene glycol in water has a freezing
point of –45 °C. The ethylene glycol/water mix also has a higher boiling
temperature than pure water. This lowers the gas pressure inside the
coolant lines and allows the engine to run hotter without letting the coolant
escape as easily.

Figure 11.11 A mixture of ethylene glycol and water is used as engine coolant. This
mixture has a lower freezing point and a higher boiling point than pure water.

When making pasta, many people add a little salt to the water. Dissolved solids
do raise the boiling temperature of water, but a tiny amount of salt is not enough
to make an appreciable difference.

Example 11.9 Identifying Trends in Colligative Properties

Which solution has a lower freezing point, a 3.0-M solution of Na2CO3 or a
2.0-M solution of AlCl3?
To solve this problem, we must identify the total number of moles of ions
present in each solution. For 1 liter of the solution, 3.0 moles of sodium
carbonate dissociates to give a total of 9.0 moles of ions:
Na2CO3 (aq)→2 Na+ (aq)+CO32−(aq)3 moles/L→6 moles/L +3 moles/L

686
 (total ion concentration: 9 moles/L)

Similarly, 1.0 liter of 2.0-M AlCl3 solution dissociates to give a total of 8.0
moles of ions:
AlCl3(aq)→Al3+(aq)+3 Cl−(aq)2 moles/L→2 moles/L +6 moles/L (total ion
concentration: 8 moles/L)

Because the Na2CO3 solution has a greater concentration of dissolved ions,

it has a lower freezing point than the AlCl3 solution. •

IT
TRY

12. Which of these solutions has the lowest freezing point? Which one
has the highest freezing point?
a. 2.0 M aq. MgSO4
b. 2.0 M aq. K2CO3
c. 3.2 M aq. NH4Cl

Osmotic Pressure
Cell membranes are composed of a thin double layer of molecules called a
lipid bilayer. These membranes enclose the cell, holding its contents in
place. The lipid bilayer is a semipermeable membrane: This means that
water can pass back and forth through the membrane, but most molecules
and ions cannot (Figure 11.12).

687
Figure 11.12 Water molecules are able to move across a semipermeable membrane
(such as a cell membrane), but dissolved ions (like sodium and chloride) cannot.

When the solutions on either side of the membrane have the same
solute concentration, water flows equally through the membrane in both
directions. However, if one side of the membrane has a higher solute
concentration, the water will migrate toward the more concentrated side.
This tendency of water to move toward the more concentrated solution is
called osmotic pressure.
We can show this property in a laboratory by placing pure water and a
saltwater solution on either side of a synthetic membrane (Figure 11.13).
Over time, water will migrate toward the more concentrated side, causing
the volume of the saltwater solution to increase.

688
Figure 11.13 (a) Pure water and a saltwater solution are separated by a semipermeable
membrane. (b) The water molecules will flow through the membrane toward the solute.

Why does water do this? To help you understand this property, here’s
an analogy: Imagine two towns with equal populations (Figure 11.14).
Both towns are nice, and people like to live close to work. People with
jobs stay put, while those without jobs may move back and forth between
towns.

689
Figure 11.14 We can think about osmotic pressure in terms of two towns whose people
are moving back and forth between them. Top: The two towns have equal populations,
and people freely move between them. Middle: A new factory opens in Town A.
Suddenly, fewer people leave Town A, but people continue to move in from Town B.
Bottom: As a result, the population of Town A grows while that of Town B diminishes,
until the number of people moving to and from each town balances.

If the towns are the same size, and if each town has the same number
of jobs, then the same number of people would move in each direction
between Towns A and B. But what if a new factory opens in Town A?
Suddenly, more people have a job in Town A, and so fewer of them will
migrate to Town B, while people in Town B are still moving to Town A.
The result is that the population of Town A increases, and the population
of Town B decreases.
Just as jobs keep people in a town, solute ions form strong interactions
with water molecules, preventing them from moving across the membrane.
The more concentrated a solution is, the fewer “jobless” water molecules
migrate across the membrane. As a result, if two solutions are divided by a
membrane, there will always be a net migration of water toward the
solution with more ions—that is, toward the higher concentration (Figure
11.15).

690
Figure 11.15 Because water aggregates around ions (shown as green and violet
spheres), there are fewer “free” water molecules on the left-hand side. As a result, the
net migration of water will be from right to left, approaching a balance of free water
molecules on each side.

Explore
Figure 11.15

Human and animal cells are very sensitive to solute concentration and
osmotic pressure. For example, if cells are placed in a solution having low
solute concentration (called a hypotonic solution, or one with low osmotic
pressure), the water migrates into the cells, causing them to swell and
possibly burst.
On the other hand, if cells are placed in a solution having a high solute
concentration (called a hypertonic solution, or one with high osmotic
pressure), the water flows out of the cells, causing them to shrivel like a
raisin. Ideally, the solution outside the cell is isotonic, meaning the
concentration is the same as the concentration inside the cell (Figure
11.16).

691
Figure 11.16 In a hypertonic solution, water exits the cells, causing them to shrivel. In a
hypotonic solution, water enters the cells, causing them to swell and possibly burst. In
an isotonic solution, water flows into and out of the cells at the same rate.

One important application of this principle is in the administration of

intravenous (IV) fluids (Figure 11.17). Hospitals commonly use IVs to
keep patients hydrated or to administer medicine. However, use of pure
water in an intravenous drip can be fatal, because the resulting hypotonic
bloodstream can cause the red blood cells to swell and burst.

Figure 11.17 Normal saline solution is isotonic with the typical osmotic pressure of the
cells.

Example 11.10 Osmotic Pressure

A 1.0-M aq. NaCl solution and a 1.0-M aq. CaCl2 solution are separated
by a semipermeable membrane, as shown. Which solution has a higher
osmotic pressure? Will the water flow toward the NaCl solution or the

692
 CaCl2 solution?

As in Example 11.9, the key to solving this problem is to identify the total
concentration of dissolved ions.
1.0 M NaCl: [Na+]=1.0 M[Cl−]=1.0 M.Total ion concentration: [2.0 M]1.0 M
CaCl2: [Ca2+]=1.0 M[Cl−]=2.0 M.Total ion concentration: [3.0 M]

The CaCl2 solution has a higher total ion concentration and therefore a
higher osmotic pressure. The water will flow toward the CaCl2 solution. •

IT
TRY

13. Which solution has a higher osmotic pressure, 1.0 M magnesium

acetate or 2.0 M sodium acetate?

693
11.3 Reactions in Solution—A Review and a Preview
In Chapter 6, we looked at some common types of reactions that take place
in aqueous solution. Before moving on, let’s briefly review those reaction
types and preview another type of reaction (metal displacement reactions)
that also takes place in solution.

Precipitation Reactions
In precipitation reactions, two solutions are combined, and a solid product
is formed. For example, aqueous silver nitrate reacts with aqueous
potassium chloride to produce silver chloride as a white solid
(Figure11.18):

Figure 11.18 When silver and chloride ions are combined, a white precipitate forms.

AgNO3 (aq)+KCl (aq)→AgCl (s)+KNO3 (aq)

The driving force for this reaction is the formation of the precipitate. We
can also express this reaction using an ionic equation:
Ag+ (aq)+NO3− (aq)+K+ (aq)+Cl− (aq)→AgCl (s)+K+ (aq)+NO3− (aq)

In the equation above, the precipitating ions are shown in red. The

694
other ions, potassium and nitrate, are not changed. These are called
spectator ions. Net ionic equations do not include spectator ions—they
show only the ions directly involved in the reaction. Here is the net ionic
equation for this reaction:
Ag+ (aq)+Cl− (aq)→AgCl (s)

Precipitation reactions depend on the solubility of the products formed.

If two ions can combine to produce a product that is insoluble in water, a
precipitation reaction will take place. The solubilities of many ionic
compounds are given in Chapter 6 (see Table 6.3 and Figure 6.18). These
are useful references for determining whether a precipitation reaction will
take place.

Ion solubility (see Table 6.3) determines which precipitates form.

Acid-Base Neutralization Reactions

In an acid-base neutralization reaction, an acid combines with a hydroxide
base to produce water and an ionic compound (called a salt). For example,
hydrochloric acid reacts with magnesium hydroxide to produce water plus
magnesium chloride:
2 HCl (aq)+Mg(OH)2 (aq)→2 H2O (l)+MgCl2 (aq)

The driving force of a neutralization reaction is the formation of water.

Neutralization and precipitation reactions are both examples of double-
displacement reactions. We will discuss other reactions of acids in Chapter
12.

Metal Displacement Reactions

In a metal displacement reaction, a solution containing the ion of one
metal reacts with the elemental form of another metal. For example,
copper metal reacts with silver nitrate to produce copper nitrate and
elemental silver (Figure 11.19):

695
Figure 11.19 (a) When elemental copper is placed in a silver nitrate solution, (b) a
metal displacement reaction occurs, causing elemental silver to coat the copper surface.

Molecular equation:Cu (s)+2 AgNO3 (aq)→Cu(NO3)2 (aq)+2 Ag (s)Net ionic

equation:Cu (s)+2 Ag+ (aq)→Cu2+ (aq)+2 Ag (s)

These reactions are examples of single-displacement reactions. These

reactions will be described in more detail in Chapter 14.

IT
TRY

14. Predict the products of the following reactions. Refer to Table 6.3
and Figure 6.18 as needed.
Pb(NO3)2 (aq)+MgCl2 (aq)→H3PO4 (aq)+3 NaOH (aq)→

696
11.4 Solution Stoichiometry
In Chapter 7, we used balanced equations to answer the question, “How
much?” The first step in most stoichiometry problems is to convert the
amounts given to moles. When working with solutions, we commonly
know the concentration and the volume of a solution. From this, we can
determine the moles of solute present:

Once we have done this, we can predict the number of moles of any
other species consumed or produced in the chemical reaction. Let’s begin
by looking at Example 11.11.

Example 11.11 Solution Stoichiometry

Silver and chloride ions react as shown in the following net ionic equation.
Based on this reaction, how many grams of AgCl precipitate can be
produced if 10.0 mL of solution containing 0.0023 M chloride ions reacts
with excess Ag+?
Ag+ (aq)+Cl− (aq)→AgCl (s)

This question gives us the volume and the molarity of the chloride ion. To
solve the problem, we must first find the number of moles of chloride ion
present:
volume (chloride)=10.0 mL=0.0100 L

Once we have found the moles of Cl−, we can relate this to the moles of
AgCl, and then to the grams of AgCl, using the balanced equation. •

697
 IT
TRY

15. Calcium and carbonate ions combine in a precipitation reaction, as

shown in this net ionic equation. If 25.0 mL of a solution containing
0.0015 M Ca2+ reacts with excess carbonate, how many grams of
CaCO3 will form?
Ca2+ (aq)+CO32− (aq)→CaCO3 (s)

Gravimetric Analysis
To determine the concentration of ions in solution, chemists often use a
technique called gravimetric analysis. In this technique, the mass of a
precipitate is used to determine the concentration of a reactant.
For example, suppose a chemist needs to find the concentration of
iodide ions in a solution. How can she do this? Recall that iodide ions react
with lead(II) ions to produce lead(II) iodide as an insoluble precipitate:
Pb2+ (aq)+2 I− (aq)→PbI2 (s)

The chemist therefore takes a small sample of the unknown I– solution

and combines it with a second solution containing an excess of Pb2+. A
precipitate forms. She collects the precipitate by filtration and then allows
it to dry. From the mass of the precipitate, she can determine the number
of moles of iodide present, and therefore the concentration of the original
solution. This idea is illustrated in Figure 11.20 and Example 11.12.

698
Figure 11.20 Gravimetric analysis follows a four-step procedure.

Example 11.12 Gravimetric Analysis

The concentration of sulfate ion (SO42−) in a solution may be

measured using a precipitation reaction with barium chloride. The net
ionic equation for this reaction is
Ba2+ (aq)+SO42− (aq)→BaSO4 (s)

You need to measure the sulfate concentration in a large drum containing

aqueous sodium sulfate. From the drum, you measure a 0.200-L sample
and react it with excess barium chloride. The reaction produces BaSO4 as
a precipitate, which you carefully filter and dry. The BaSO4 precipitate has
a mass of 0.309 g. What is the concentration (in molarity) of sulfate ion in
the drum?
This question gives us the mass of BaSO4 and asks us to find the

concentration of (SO42−) in the drum. To do this, we must

convert from grams of BaSO4 to moles of BaSO4, then relate that to moles

of (SO42−) , and then solve for the concentration:

First, convert grams of BaSO4 to moles of BaSO4 and then to moles of

SO42–:

699
 This is the number of moles of sulfate present in the 0.200-L sample, so we
can calculate the molarity. •

Concentration (molarity)=1.32×10−3 moles0.200 L=6.60×10−3 M

IT
TRY

16. The presence of lead(II) in a water sample can be determined by

precipitating the lead(II) ions with a solution of potassium chromate,
K2CrO4. The reaction takes place according to this equation:
Pb2+ (aq)+K2CrO4 (aq)→PbCrO4 (s)+2 K+ (aq)

You have been assigned to find the concentration of lead(II) in the

waste stream of a manufacturing process. You therefore react a 1.00-
L sample from the waste stream with an excess of K2CrO4. The
resulting precipitate has a mass of 34.1 mg. What is the concentration
of Pb2+ in the wastewater?

Advanced Stoichiometry Problems

As we’ve explored stoichiometry over several chapters, we’ve developed
strategies to get from and to a number of quantities. The mole is central to
nearly all of these conversions. Most problems in chemistry involve a
conversion to moles at one or more steps.
Figure 11.21 shows a map for solving a wide range of problems. For
example, if solid sodium bicarbonate (NaHCO3, baking soda) is added to
aqueous HCl, the following reaction takes place:
NaHCO3 (s)+HCl (aq)→NaCl (aq)+H2O (l)+CO2 (g)

From this equation, we can ask a huge number of questions. For

example:

700
 1. Given the number of moles of NaHCO3, how many moles of H2O are
produced?
2. Given the molarity and volume of HCl, how many grams of CO2 are
produced?
3. What volume of 1.0 M HCl is required to react with 2.5 grams of
NaHCO3?
4. If this reaction produces 30.5 grams of NaCl at standard temperature
and pressure, what volume of CO2 gas forms?
5. If 200 mL of water is produced, how many moles of HCl are
consumed?

Each of these questions can be solved using the map in Figure 11.21.
There are many other possibilities—can you think of a few? How would
you solve them? Example 11.13 gives a similar problem.

Figure 11.21 The mole is the intermediate for most calculations in chemistry.

Example 11.13 Advanced Stoichiometry Problems

Calcium metal reacts with aqueous hydrobromic acid to produce calcium
bromide and hydrogen gas, as shown in the equation below. What volume
of 2.50-M aq. HBr is required to produce 200.0 L of hydrogen gas at a
temperature of 298 K and a pressure of 1.20 atm?
Ca (s)+2 HBr (aq)→CaBr2 (aq)+H2 (g)

We are given the volume of H2 gas and then asked to relate this to the

701
 volume of the aqueous HBr solution. Based on the mole map in Figure
11.21, we see that our pathway for solving this problem must be

The first step requires the ideal gas law, which we covered in Chapter 10:
n=PVRT=(1.20 atm)(200.0 L H2)(0.0821L⋅atmmol⋅K)(298 K)=9.81 mol H2

Following the pathway outlined above, we are able to relate this value to
the moles of HBr and reach a final answer of 7.85 L aq. HBr. •

IT
TRY

17. Ammonium phosphate reacts with barium chloride to form a white

precipitate, as shown here:
2 (NH4)3PO4 (aq)+3 BaCl2 (aq)→6 NH4Cl (aq)+Ba3(PO4)2 (s)

a. How many moles of ammonium phosphate are needed to completely

react with 96.2 mL of 2.10-M aq. BaCl2?
b. How many grams of barium phosphate could be produced in this
reaction?

18. Aqueous hydrochloric acid reacts with zinc metal according to this
reaction:
2 HCl (aq)+Zn (s)→ZnCl2 (aq)+H2 (g)

If 15.0 mL of 3.4 M aq. HCl reacts with excess zinc,

702
a. How many moles of ZnCl2 are produced?
b. How many grams of H2 gas are produced?
c. If the reaction takes place at standard temperature and pressure (273
K, 1.0 atm), what volume of H2 gas is produced?

703
Summary
A key characteristic of any solution is its concentration; that is, the amount
of solute present in the solution. Concentration is measured in a variety of
units, including percentage, parts per million (ppm), parts per billion (ppb),
and molarity. Molarity is defined as the moles of solute per liter of
solution; it is particularly useful because it allows us to easily relate
volume to moles.
When ionic compounds dissolve in water, the compounds dissociate
into positive and negative ions. Because aqueous ionic solutions conduct
electricity, water-soluble ionic compounds are sometimes called
electrolytes. When electrolytes dissociate, the concentration of the
individual ions may be higher than the overall concentration of the ionic
compound. Most covalent compounds are nonelectrolytes; they do not
dissociate in water.
Colligative properties are those properties of solutions that depend
only on the quantity, not the identity, of solute present. Colligative
properties include melting point depression, boiling point elevation, and
osmotic pressure.
A variety of chemical reactions involve aqueous solutions. They
include precipitation, acid-base neutralization, and metal displacement
reactions.
We can extend the principles of stoichiometry to solve problems
pertaining to solutions. Given the volume and molarity of a solution, we
can determine the number of moles present in a solution or in a reaction.
As noted in previous chapters, we can relate moles of any reactant or
product to those of any other component in the reaction. From moles, we
can convert to grams, atoms, pressure/volume measurements for gases, and
others.

704
Amazing Coffee—The Importance of Concentration

By Aaron Blanco, President of Brown Coffee Co., San Antonio, Texas

What you measure improves. This is true in any endeavor in life, and it is evident
in the world of specialty coffee.
For decades, most Americans bought their coffee already ground and in a can.
Grab a spoon, pile some—or a lot—of coffee into the filter; add some water from
the faucet. Close the hatch, flip the switch, and voilà!
“Coffee.”
“Joe.”
“Just a regular cuppa.”
Most coffee served in restaurants and diners was equally uninspired. However,
over the last decade or so, coffee has experienced a renaissance as people take a
more science-based approach to the art of coffee brewing. At its core is the
intersection of flavor and intensity, and how that is measured. What does a
particular coffee from a particular place taste like? How intense is the flavor?
These qualities can be thought of in terms of two parameters, strength and
extraction yield (Figure 11.22).

Figure 11.22 The strength (total dissolved solids, or TDS) and flavor (extraction yield)
both affect the quality of coffee. Data from CoffeeChemistry.com.

705
 Let’s tackle strength first. Coffee strength is measured in total dissolved solids
(TDS). This is typically expressed as a percent (m/v), or grams of solute per mL of
solution. The TDS is normally between 0.8% and 1.6%. The higher the TDS, the
“stronger” a cup of coffee is. But strength doesn’t taste like anything. It is merely a
measurement of intensity. You can detect something lovely in a cup that is very
faint, very powerful, or somewhere in between.
The second parameter, extraction yield, affects the flavor of the coffee. About
30% of the mass of a coffee bean is soluble in water. However, some compounds
are more soluble than others. If too little of the material is extracted, then the most
soluble compounds predominate, and the flavor is simple and bland. As the
extraction yield increases, the flavor becomes more complex. However, if too
much material is extracted, the coffee may take on a bitter flavor.
A number of factors affect both TDS and extraction yield. They include such
things as the size and uniformity of the grounds, the temperature of the water used
in brewing (hotter water is a more efficient solvent), external agitation (stirring),
freshness, and so on.
Historically, most American tastes have fallen between 1.15% and 1.35% TDS
and between 18% and 22% extraction yield (see Figure 11.22). But tastes change.
This chart, developed in the 1950s, was made in an era when lower-quality coffees
were the norm. People then used less than optimal grinding and brewing equipment
and often stale coffee from a giant can.
I personally prefer coffees at 1.4%–1.45% TDS and 22%–23% extraction yield
for regular filter coffee. Espresso is much stronger, with 10.5%–11% TDS and
24%–25% extraction yield. I expect that as technology helps us isolate and
improve each variable, we’ll start to see coffees regularly pushing 26–28%
extraction yields. But only time will tell if tastes will follow science, or if old
coffee-drinking habits will die hard.

706
 Key Terms
11.1 Describing Concentration
solution A homogeneous mixture; a solution is usually a liquid.
solute A substance that is dissolved in a solution.
solvent The major component of a solution.
concentration The amount of solute present in a solution.
percent by mass A measure of concentration; the mass of solute divided by
the total mass of the solution, expressed as a percentage.
percent by volume A measure of concentration; the volume of solute divided
by the total volume of the solution, expressed as a percentage.
mass/volume percent A measure of concentration; the mass of solute divided
by the total volume of the solution, expressed as a percentage; often measured
in grams of solute per milliliter of solution.
parts per million (ppm) A measure of concentration used for dilute solutions;
1 ppm = 1 g of solute per 1,000,000 grams of solution, or 1 mg of solute per
liter of solution.
parts per billion (ppb) A measure of concentration used for very dilute
solutions; 1 ppb = 1 g of solute per 1,000,000,000 grams of solution, or 1 μg of
solute per liter of solution.
molarity (M) A measure of concentration defined as the moles of solute per
liter of solution.

11.2 Electrolyte Solutions

electrolyte solutions An aqueous solution containing dissociated ions; this
type of solution conducts electricity more effectively than pure water.
colligative properties Properties that depend on the number of particles
dissolved in the solution, but not on the type of particles dissolved.
freezing point depression A colligative property of water; the presence of
solute in an aqueous solution lowers the freezing point below that of pure
water.
boiling point elevation A colligative property of water; the presence of solute
in an aqueous solution raises the boiling point above that of pure water.
osmotic pressure The tendency of water to move toward regions of greater
solute concentrations; an imbalance in concentration inside and outside of a
living cell can cause the cell to swell or shrink.

707
11.3 Reactions in Solution—A Review and a Preview
metal displacement reaction A reaction in which the ion of one metal reacts
with the elemental form of another metal.

11.4 Solution Stoichiometry

gravimetric analysis A technique that uses the mass of a precipitate to
determine the concentration of a reactant.

708
 Additional Problems

11.1 Describing Concentration

19. A solution contains 0.0025 grams of potassium nitrate dissolved in 2.0

liters of water. At 20 °C, one liter of water can dissolve over 300 grams of
this compound. Identify the solvent and the solute in this solution. Is this
solution best described as dilute, concentrated, or saturated?

20. What is the difference between a concentrated solution and a saturated

solution?

21. A calcium chloride solution contains 40.0 grams of calcium chloride

dissolved in 100 grams of water. What is the percent by mass of calcium
chloride in this solution?

22. You wish to prepare a solution that is 12% sodium chloride by mass. How
much sodium chloride is necessary to prepare 2.0 kg of this solution?

23. A cleaning solution calls for a 20% (v/v) mixture of cleaner in water. How
much cleaner and water are necessary to prepare 4.0 liters of the solution?

24. Antifreeze is a mixture of ethylene glycol and water. What volume of

ethylene glycol would be needed to prepare 8.0 liters of antifreeze that is
20% (v/v) ethylene glycol?

25. Wine typically contains about 12% (v/v) alcohol. What is the volume of
alcohol in a 200-mL glass of wine?

26. Which of these drinks has the greatest alcohol content?

a. two 12-ounce glasses of beer, which is 4.5% alcohol (v/v)
b. one 8-ounce glass of wine, which is 12% alcohol (v/v)
c. one 1.5-ounce shot of whiskey, which is 40% alcohol (v/v)

27. Calculate the percent by mass (g/mL) in these solutions:

a. an aqueous solution containing 52.0 g of potassium carbonate in a
solution of 1,000 mL
b. a saturated sugar solution with a volume of 1.0 liter, containing 2.0 kg
of dissolved sugar

709
 c. a solution containing 28.4 grams of salt dissolved in a 150-mL solution

28. How many grams of sodium nitrate are in 40 gallons of an 8.0% (g/mL)
sodium nitrate solution?

29. You need to prepare 4.0 liters of an aqueous solution that is 5% (m/v or
g/mL) sodium chloride. What mass of sodium chloride do you need to
prepare this solution?

30. You need to prepare 15.0 liters of an aqueous solution that is 0.4% (g/mL)
ammonium carbonate. What mass of ammonium carbonate do you need to
prepare this solution?

31. Express each of these concentrations in parts per million (ppm):

a. 5.0 milligrams per kilogram of water
b. 5.0 micrograms per kilogram of water
c. 5.0 nanograms per kilogram of water

32. Express each of these concentrations in parts per billion (ppb):

a. 8.0 milligrams per kilogram of water
b. 8.0 micrograms per kilogram of water
c. 8.0 nanograms per kilogram of water

33. The water from a river is found to contain 10 ppm of dissolved oxygen.
Assuming the density of the water is 1.0 g/mL, how many milligrams of
oxygen are in a liter of the river water?

34. A lake is found to contain 8.0 ppm of dissolved nitrates. Assuming the
density of the water is 1.0 g/mL, what mass of nitrates are dissolved in one
liter of the lake water?

35. Waterways in the Mississippi Valley region of the United States

commonly contain 0.03 mg of dissolved phosphorus per liter of water.
Assuming the density of the water is 1.0 g/mL, what is this concentration
in ppm?

36. Triazines are compounds commonly used as herbicides, which may leach
into waterways. A 1.0-L sample of lake water is found to contain 8.0
micrograms of triazines. What is the concentration of triazines in ppm?

710
37. Fluoride is commonly added to drinking water to promote bone health.
While a very small amount of fluoride can be beneficial, larger amounts
can be harmful or deadly. Don’t swallow your toothpaste. The U.S.
Environmental Protection Agency (EPA) currently sets the maximum
concentration level (MCL) of fluoride in drinking water at 4.0 mg/L. What
is this concentration in parts per million? What is this concentration in %
(g/mL)?

38. Arsenic is a highly toxic metal that enters the water supply through erosion
of natural deposits or as waste from agricultural and industrial
applications. The EPA sets the maximum concentration limit of arsenic in
drinking water as 0.010 mg/L. Given the density of water as 1.0 g/mL,
what is the maximum concentration of arsenic in ppm? In ppb?

39. Polychlorinated biphenyls (PCBs) were commonly used as plastic

additives, hydraulic fluids, and lubricating oils through the mid-twentieth
century, until they were banned because of their toxic effects. A recent
study around the Savannah River in Georgia showed PCB levels of 5 ng
per gram of sediment. What is this concentration in ppm? What is it in
ppb?

40. Mercury is a highly toxic metal that can cause nerve damage. The EPA
sets the maximum concentration level of mercury in drinking water as 2.0
× 10–6 grams per liter of water. What is this concentration level in ppm?

41. Tetrachloroethylene is sometimes discharged into the water supply from

factories and dry cleaners. The EPA considers the maximum concentration
level (MCL) of tetrachloroethylene in drinking water to be 5 ng/L. What is
this concentration in ppm? What is this concentration in ppb?

42. A river pollutant is found to have a concentration of 0.0025% m/v. If the

density of the river water is 1.03 g/mL, what is the concentration of this
pollutant in ppm?

43. Calculate the molarity of each of these solutions:

a. a 2.20-liter solution containing 48.3 grams of potassium iodide
b. a 4.9-liter solution containing 12.1 grams of dissolved carbon dioxide
c. a 500-mL solution containing 93.1 grams of magnesium acetate

44. Calculate the molarity of each of these solutions:

a. a 55.0-liter solution containing 2.3 kg of dissolved sugar (C12H22O11)

711
 b. a solution with a volume of 892 milliliters containing 1.27 grams of
dissolved sodium fluoride
c. a solution containing 2.53 g of ammonium acetate with a volume of
170.2 mL

45. Find the following values:

a. moles of NaOH in 2.0 L of a 5.0-M NaOH solution
b. moles of KCl in an 800-mL solution of 1.5-M KCl
c. moles of Na2CO3 in 300 µL of a 1.0-M Na2CO3 solution

46. Find each of the following:

a. moles of barium hydroxide in a 400-mL sample of 0.47-M Ba(OH)2
b. grams of K3PO4 in a 350-mL sample of a 1.4-M potassium phosphate
solution
c. kilograms of BeCl2 in a completely filled, 45-gallon drum containing
5.0 M aq. BeCl2

47. Using a 1.0-M aqueous solution of KOH, a chemist wishes to add 14.30
moles of potassium hydroxide to an acidic solution. What volume of KOH
solution does she need to add?

48. What volume (in milliliters) of 3.5-M aq. HCl contains 0.025 moles of
hydrochloric acid?

49. While working in a pharmaceutical laboratory, you need to prepare 3.0 L

of a 1.00-M NaCl solution. What mass of NaCl would be required to
prepare this solution? How would you go about preparing the solution?

50. You’ve just started a new job in the laboratory of the Food and Drug
Administration. Your supervisor asks you to prepare a 1.50-M solution of
aqueous potassium permanganate, KMnO4. What mass of KMnO4 will
you need to prepare 500 mL of this solution? Describe how you would
prepare the solution.

51. A 12.0-M solution of HCl is diluted from an initial volume of 25 mL to a

final volume of 800 mL. What is the new concentration of the HCl
solution?

52. A 150-mL sample of a solution of 4.80 M FeCl2 is diluted with enough

712
 water to raise the volume to 3.40 L. What is the new concentration of this
solution? How many moles of FeCl2 are present in this solution?

53. You need to prepare 100 mL of 0.100-M aq. CuSO4. Describe how you
would prepare this solution from a stock solution of 5.0-M aq. CuSO4.

54. You need to prepare 300 mL of 0.100-M aq. sodium acetate. Describe how
you would prepare this solution.
a. from solid sodium acetate
b. from a stock solution of 6.0-M aq. NaC2H3O2

11.2 Electrolyte Solutions

55. Indicate whether each of the following is or is not an electrolyte:

a. KCl
b. FeBr3
c. C2HO
d. CH5N

56. Indicate whether each of the following is or is not an electrolyte:

a. an ionic compound that dissolves readily in water
b. a nonionic compound that dissolves readily in water
c. a solution of magnesium chloride
d. a solution of acetone, a nonacidic covalent compound

57. Sketch the dissociation of a crystal of potassium bromide in water.

58. Sketch the dissociation of a crystal of sodium nitrate in water.

59. Rewrite these expressions as ionic equations:

a. Ba(OH)2 (s)→Ba(OH)2 (aq)

b. (NH4)2SO4 (s)→(NH4)2SO4 (aq)

60. Rewrite these expressions as ionic equations:

713
 a. Mg(NO3)2 (s)→Mg(NO3)2 (aq)

b. (NH4)2CO3 (s)→(NH4)2CO3 (aq)

61. Write ionic equations to show the dissociation of these ionic solids:
a. MgBr2
b. AlCl3
c. Ca(NO3)2

62. Write ionic equations to show the dissociation of these ionic solids:
a. CsBr
b. (NH4)3PO4
c. K3PO4

63. Find each of the following:

a. [OH–] in 1 M aq. NaOH
b. [Cl–] in 2 M aq. MgCl2
c. [Br–] in 3 M aq. FeBr3

64. Find the concentration of sodium ions in each of these solutions:

a. 0.60 M aq. NaCl
b. 0.60 M aq. Na2CO3
c. 0.60 M aq. Na3PO4

65. Find the concentration of all ions in each of these solutions:

a. 2.0 M aq. Li2S
b. 0.7 M aq. Li3PO4
c. 1.3 × 10–4 M aq. Mg(NO2)2

66. Find the concentration of all ions in each of these solutions:

a. 3.0 M aq. LiCl
b. 0.70 M aq. Fe(NO3)2
c. 1.7 × 10–4 M aq. NH4Cl

714
67. How many moles of chloride ion are in a 400.0-mL solution of 2.30-M aq.
MgCl2?

68. How many moles of sulfate ion are in a 350.0-mL solution of 1.20-M aq.
(NH4)2SO4?

69. A 300-mL solution of 1.3-M aq. NaCl is mixed with 500 mL of 2.5-M aq.
CaCl2. How many moles of chloride ion are in the combined solution?
What is the molarity of chloride in the combined solution?

70. A 2.10-L solution of 1.0-M aq. HBr is mixed with 1.25 L of 3.0-M aq.
BeBr2. How many moles of bromide ion are in the combined solution?
What is the molarity of bromide in the combined solution?

71. If 400 mL of 1.2-M aq. MgSO4 is mixed with 550 mL of 2.7-M aq.
K2SO4, what are the concentrations of magnesium, potassium, and sulfate
ions in the resulting solution?

72. If 0.040 L of 12.0-M KNO3 is mixed with 90 mL of 2.5 M Ca(NO3)2,

what is the concentration of potassium, calcium, and nitrate in the resulting
solution?

73. What is a colligative property?

74. If additional solute is added to a solution:

a. How does the freezing point change?
b. How does the melting point change?
c. How does the boiling point change?
d. How does the osmotic pressure change?

75. Which of these aqueous solutions has a lower freezing temperature: a 1.0-
M solution of MgCl2 or a 1.0-M solution of KBr? Explain your answer.

76. Which of these aqueous solutions has a higher boiling temperature: a 2.0-
M solution of CsBr or a 2.0-M solution of Ca(NO2)2? Explain your
answer.

77. Between a solution of 2.0 M aq. NaNO3 and 1.5 M aq. AlCl3:

715
 a. Which would have a lower freezing point?
b. Which would have a higher boiling point?
c. Which would have a higher osmotic pressure?

78. Between a solution of 1.1 M aq. K2CO3 and 1.5 M aq. CH4O (a
nonelectrolyte):
a. Which would have a lower freezing point?
b. Which would have a higher boiling point?
c. Which would have a higher osmotic pressure?

79. A 3.0-M aq. NaCl solution and a 1.0-M aq. Na2SO4 solution are separated
by a semipermeable membrane, as shown. Which solution has a higher
osmotic pressure? Will the water flow toward the NaCl solution or the
Na2SO4 solution?

80. A chemist prepares two aqueous solutions, each with a volume of 1.00 L.
The first solution contains 51.35 g of ammonium sulfate. The second
solution contains 73.25 g of sodium chloride. What is the total ion
concentration in each solution? Which solution has a higher osmotic
pressure?

81. Normal saline is a solution of aqueous sodium chloride with a

concentration of 0.90% (g/mL). Express this concentration in molarity;
then indicate whether the osmotic pressure of each of these solutions is
hypotonic, isotonic, or hypertonic compared to normal saline:
a. 1.0 M aq. KCl
b. 0.5 M aq. K2SO4
c. 4.3 M aq. NH4Cl
d. 0.15 M aq. sugar

82. In the figure shown, water molecules migrate from the area of lower solute

716
 concentration (right) to the area of higher solute concentration (left). Do
water molecules “know” which way to travel? If not, how can we explain
the migration of water molecules toward a higher solute concentration?

11.3 Reactions in Solution—A Review and a Preview

83. What is the driving force in a precipitation reaction?

84. What is the driving force in a neutralization reaction?

85. Write this precipitation reaction as a complete ionic equation and then as a
net ionic equation:
CaBr2 (aq)+Pb(NO3)2 (aq)→PbBr2 (s)+Ca(NO3)2 (aq)

86. Write this neutralization reaction as a complete ionic equation and then as
a net ionic equation:
Ca(OH)2 (aq)+2 HBr (aq)→CaBr2 (aq)+2 H2O (l)

87. Identify each of these reactions as precipitation, neutralization, or metal

displacement:
a. Mg (s)+NiCl2 (aq)→Ni (s)+MgCl2 (aq)

b. Mg(OH)2 (aq)+H2SO4 (aq)→MgSO4 (aq)+2 H2O (l)

c. MgCl2 (aq)+Pb(C2H3O2)2 (aq)→Mg(C2H3O2)2 (aq)+PbCl2 (s)

717
88. Identify each of these reactions as precipitation, neutralization, or metal
displacement:
a. Ba(OH)2 (aq)+FeCl2 (s)→BaCl2 (aq)+Fe(OH)2 (s)

b. Ba(OH)2 (aq)+2 HCl (aq)→BaCl2 (aq)+2 H2O (l)

c. 3 Ba (s)+2 Al(NO3)3 (aq)→3 Ba(NO3)2 (aq)+2 Al (s)

89. These reactions take place in aqueous solutions. Complete and balance
each reaction.

a. FeCl3 (aq)+ZnSO4 (aq)→

b. HBr (aq)+LiOH (aq)→

90. These reactions take place in aqueous solutions. Complete and balance
each reaction.

a. KOH (aq)+H2SO4 (aq)→

b. Ba(OH)2 (aq)+CuBr2 (aq)→

91. Write a balanced equation to show the precipitation reaction that occurs
when aqueous silver chlorate is combined with aqueous magnesium
chloride.

92. Write a balanced equation to show the neutralization reaction that occurs
between nitric acid and aqueous barium hydroxide.

11.4 Solution Stoichiometry

93. Magnesium nitrate dissociates in water, as shown in this equation. How

many moles of nitrate ion are in a 2.47-liter solution where [Mg(NO3)2] =
1.93 M?
Mg(NO3)2 (s)→Mg2+ (aq)+2 NO3− (aq)

718
94. Potassium phosphate dissociates in water, as shown in this equation. How
many moles of potassium ion are present in 25.0 mL of a 1.30-M aq.
potassium phosphate solution?
K3PO4 (s)→3 K+ (aq)+PO43− (aq)

95. Aluminum reacts with nitric acid, as in the equation below. How many
moles of aluminum are required to completely react with 100 mL of 12.0-
M HNO3?
2 Al (s)+6 HNO3 (aq)→2 Al(NO3)3 (aq)+3 H2 (g)

96. Carbonic acid decomposes according to the equation below. How many
moles of CO2 can be produced from 100 mL of a 1.3-M aq. H2CO3
solution?
H2CO3 (aq)→H2O (l)+CO2(g)

97. Tin metal reacts with hydrobromic acid according to the following
equation. How many moles of H2 gas can be produced if 500 mL of a 1.3-
M HBr solution is added to excess tin?
Sn (s)+2 HBr (aq)→SnBr2 (aq)+H2 (g)

98. Potassium hydroxide and iron(II) chloride combine in the precipitation

reaction shown below. How many moles of iron(II) hydroxide can be
produced in the reaction of 0.520 L of 3.4-M KOH with excess FeCl2?
2 KOH (aq)+FeCl2 (aq)→2 KCl (aq)+Fe(OH)2 (s)

99. Sodium reacts violently with water according to this equation:

719
 2 Na (s)+2 H2O (l)→2 NaOH (aq)+H2 (g)

If 0.50 grams of sodium metal is added to 2.0 liters of water, how many
moles of NaOH are produced? Assuming the volume of the resulting
solution is also 2.0 liters, what is the molarity of NaOH in the resulting
solution?

100. Zinc reacts with hydrochloric acid according to this equation:

Zn (s)+2 HCl (aq)→ZnCl2 (aq)+H2 (g)

If 0.50 grams of zinc metal is added to an excess of aqueous HCl, how

many moles of ZnCl2 are produced? If the volume of the resulting
solution is 0.750 liters, what is the molarity of ZnCl2 in the resulting
solution?

101. Zinc metal reacts with aqueous copper(II) chloride according to this
equation:
Zn (s)+CuCl2 (aq)→ZnCl2 (aq)+Cu (s)

In this reaction, what mass of copper metal can be produced from the
reaction of 500 mL of 1.20-M aq. CuCl2 with excess zinc?

102. Magnesium metal reacts with aqueous silver nitrate according to this
equation:
Mg (s)+2 AgNO3 (aq)→Mg(NO3)2 (aq)+2 Ag (s)

In this reaction, what mass of silver metal can be produced from the
reaction of 691 mL of 1.30-M aq. AgNO3 with excess magnesium?

103. The concentration of bromide ion may be determined by gravimetric

analysis, using this reaction:
Ag+ (aq)+Br− (aq)→AgBr (s)

720
 A 0.500-L aqueous solution containing bromide was reacted with excess
Ag+. If the AgBr precipitate formed in this reaction has a mass of 0.0035
g, how many moles of Br– were in the solution? What was the molarity
of Br– in the solution?

104. The concentration of mercury(II) ion in water was determined by

gravimetric analysis, using this reaction:
Hg2+ (aq)+H2S (aq)→HgS (s) +2 H+ (aq)

A chemist combined a 400.0-mL sample of aqueous solution with excess

H2S, resulting in the formation of 12.09 mg of mercury sulfide
precipitate. How many moles of mercury were in the original sample?
What was the molar concentration of Hg2+ in the original sample?

105. Concentration of calcium in a sample may be determined by precipitation

using the chromate ion, CrO42− :

Ca2+ (aq)+CrO42− (aq)→CaCrO4 (s)

A chemist combined 0.250 L of an unknown calcium solution with an

excess of ammonium chromate. This resulted in the precipitation of
calcium chromate. The mass of the precipitate was 314.1 mg. What was
the molar concentration of Ca2+ in the original sample?

106. The concentration of lead(II) ion in water may be determined by

gravimetric analysis, using this reaction:
Pb2+ (aq)+SO42− (aq)→PbSO4 (s)

A chemist treated an unknown lead(II) solution with a volume of 0.150 L

with an excess of sodium sulfate solution. The mass of the resulting
precipitate was found to be 0.00215 grams. What was the mass of lead(II)
in the original sample? What was the molar concentration of the original
sample?

721
Challenge Questions

107. Hydrochloric acid reacts with potassium hydroxide according to this

equation:
HCl (aq)+KOH (aq)→H2O (l)+K2SO4 (aq)

What volume of 3.34-M KOH is required to exactly react with 0.25 liters
of 1.35-M HCl?

108. Iron reacts with hydrochloric acid to produce aqueous iron(II) chloride, as
shown in this equation:
Fe (s)+2 HCl (aq)→H2(g)+FeCl2 (aq)

If 20.2 grams of Fe react with 100.0 mL of 2.1-M aq. HCl, how many
moles of H2 will be produced?

722
Chapter Twelve
Acids and Bases

Cocaine: Ruin and Recovery

Cocaine is a naturally occurring stimulant, produced in the leaves of the coca tree
that grows in Central and South America (Figure 12.1). Just as you or I might
drink coffee for a midday boost, the inhabitants of this region chew coca leaves to
provide energy for long workdays. The tradition dates back thousands of years.

Figure 12.1 (a) The leaves of the coca tree are the source of cocaine. (b) A Bolivian
woman chews on coca leaves. (c) An early advertisement for Coca-Cola® praises its
ability to boost energy. The manufacturer removed cocaine from the soft drink in 1904.
(d) The “crack” form of cocaine usually exists as small, hard pieces. (e) Dr. Susan
Hurley is a professor and substance abuse counselor in Dahlonega, Georgia. (f) The

723
University of North Georgia community clinic provides counseling to those struggling
with addictions and other issues.

But in the late 1800s, pharmaceutical companies began exploring the region in
search of new medicines. Drawing from native traditions, they extracted cocaine
from the coca leaves. They found that treating the compound with hydrochloric
acid produced a white crystalline powder that is easy to handle and ship, and stable
for long periods of time.
At first the dangerous properties of this substance were not well understood.
Cocaine was marketed as a stimulant in the United States and Europe. In Sir Arthur
Conan Doyle’s famous Sherlock Holmes stories, the brilliant detective used
cocaine occasionally. Cocaine was even a chief flavoring for the original Coca-
Cola® soft drink. (Although cocaine was removed from the formula in 1904, other
extracts from coca leaves are still used in cola flavoring.)
Cocaine abuse was rare until the 1960s and ’70s, when the rock ’n’ roll era saw
a surge in its use—largely in affluent circles. Then, in the mid-1980s, drug dealers
began converting the powder cocaine to a different form, called “crack.” This form,
which chemists refer to as the free-base form, was devastatingly addictive. In the
years that followed, untold numbers of lives were destroyed by crack cocaine use.
Dr. Susan Hurley has seen this damage firsthand. In her work as a substance
abuse counselor, she helps people overcome their addictions. “Crack is amazing to
me because it is so highly addictive,” she says. “The first time people use it, it
takes over their lives. Crack addicts will use their entire paycheck on the substance
instead of paying bills, buying groceries, or tending to their family.”
Both powder and crack cocaine produce a euphoric high, but the intensity and
duration of the high is very different. The powder produces a milder high that may
last for several hours. As Dr. Hurley observes, “Powder cocaine users can be quite
functional, partying, working, doing surgery, etc.” On the other hand, crack users
experience an intense high that lasts for only a few minutes. The euphoric high
(followed by a miserable low) leads crack users to binge on the drug. Addicts will
go for days without eating or sleeping, consuming as much as they can until their
money or supply runs out—or until they pass out from hunger and exhaustion.
Treatment strategies for recovering users are also different: Powder cocaine is
more common in affluent circles, and because its effects are less debilitating, many
addicts enter treatment with a support network of family and friends. But crack is
devastating. Many of the crack addicts Dr. Hurley works with have lost jobs,
families, and homes.
And yet they are not without hope: Through the vitally important work of
counselors like Dr. Hurley, addicts of all types can find new opportunities and
discover a life free of addiction.
So, what is the difference between powder and crack cocaine? And why do
they have such drastically different properties? To understand the differences
between the two drug forms, we must first understand two related types of
compounds, called acids and bases. In this chapter we’ll explore the structures and
properties of acids and bases. These compounds are all around us, and they are

724
vital to our lives. As we’ll see, the chemical differences between acids and bases
(like the two forms of cocaine) are slight, but they drastically affect how the
compounds behave. Subtle changes can have huge impacts.

Intended Learning Outcomes

After completing this chapter and working the practice problems, you should be
able to:

12.1 Introduction to Acids and Bases

Define acids and bases using both the Arrhenius and the Brønsted-Lowry
definitions.
Describe the process of acid ionization and base dissociation in aqueous
solution.

12.2 Acid-Base Equilibrium Reactions

Explain the difference between strong and weak acids.
Identify the acid, base, conjugate acid, and conjugate base in a chemical
equilibrium.

12.3 Reactions Involving Acids and Bases

Predict the reactions of metals with acid to form metal ions and hydrogen
gas.
Predict the products from acid-base neutralization reactions.

12.4 Acid and Base Concentration

Explain how the addition of acid or base to an aqueous solution affects the
concentration of H+ and OH– in an aqueous solution.
Relate the values of pH, pOH, [H+], and [OH–] in an aqueous solution.

12.5 Measuring Acid and Base Concentration

Identify common indicators for acid and base solutions.
Apply data from a titration experiment to find the concentration of an
unknown acid or base.

12.6 Buffers and Biological pH

Describe how buffers are able to stabilize the pH of solutions.

725
12.1 Introduction to Acids and Bases
Think for a moment about the term acidic. You might picture powerfully
corrosive materials that can disintegrate clothes, skin, or even metal. On
the other hand, people often describe foods like soda, oranges, and
barbeque sauce as acidic. What’s the difference between the nightmarishly
corrosive materials and the harmless flavorings of common foods? As we
will see in the coming sections, the behavior of an acid depends on two
factors: its strength and its concentration. We will carefully examine both
of these.
Bases are essentially the opposite of acids. Like acids, bases can be
very strong (like drain cleaner) or quite mild (like baking soda). Many
household products have acidic or basic properties (Figure 12.2). Bases
and acids neutralize each other, negating their powerful corrosive
properties and producing water and simple ionic compounds.

Figure 12.2 Many common household items have acidic or basic properties.

In this chapter we’ll explore the chemical properties of acids and bases,
and describe them in terms of their strength and concentration. We’ll look
at reactions of acids and bases, and discover how we can use neutralization
reactions in laboratory analysis. We’ll begin to see that acid-base behavior
is all around us: in every breath, in every drink, literally in every fiber of
our bodies.

The Arrhenius Definition

Let’s begin with a classic definition for acids and bases, called the
Arrhenius definition:
Acids are compounds that produce H+ ions in water.
Bases are compounds that produce OH– ions in water.

726
 Recall from earlier chapters that compounds that separate into ions in
an aqueous solution are called electrolytes. Acids and bases fall into this
category. Table 12.1 presents the names and formulas of some common
acids and bases.

TABLE 12.1 Common Acids and Bases

Acids

Formula Name

HF Hydrofluoric acid

HCl Hydrochloric acid

HBr Hydrobromic acid

HI Hydroiodic acid

H2CO3 Carbonic acid

HNO3 Nitric acid

HNO2 Nitrous acid

H2SO4 Sulfuric acid

H3PO4 Phosphoric acid

HC2H3O2 Acetic acid

Bases

Formula Name

NaOH Sodium hydroxide

KOH Potassium hydroxide

Mg(OH)2 Magnesium hydroxide

Ca(OH)2 Calcium hydroxide

In most acids, a hydrogen atom is bonded to a strongly electronegative

element, like oxygen or one of the halogens. In aqueous solution, these
compounds ionize to produce an H+ cation and a corresponding anion. For
example, hydrochloric acid ionizes to form H+ and Cl–:

HCl (aq)→H+ (aq)+Cl− (aq)

In an aqueous solution, the H+ ions are stabilized by water molecules.

We can represent this in two ways: Sometimes, we show the H+ as an
unattached cation with strong (but transient) forces of attraction to several

727
water molecules (Figure 12.3). Other times, we show the H+ attached to a
water molecule to form H3O+, called a hydronium ion:
HCl (aq)+H2O (l)→H3O+ (aq)+Cl− (aq)

Figure 12.3 When an acid dissolves in water, the H+ can be represented (a) by itself
(H+) or (b) bonded to a water molecule (H3O+). Both symbols are correct and used
routinely.

Explore
Figure 12.3

An aqueous H+ ion is also represented as H3O+.

Coffee and tea are both mildly acidic, due to a class of compounds called
tannins. These compounds also cause the dark coloring of these beverages.

The two forms of the acid equation mean the same thing: In aqueous
solution, the H+ separates from the chloride ion and is stabilized by the

728
water molecules around it. Because an H+ cation is composed of only one
proton, it is sometimes referred to simply as a proton, and the gain or loss
of an H+ is called a protonation or a deprotonation. For example, we could
say that the H2O molecule is “protonated” to form the hydronium ion
(Figure 12.4).

Figure 12.4 Acidic and basic solutions involve these ions.

When discussing acids and bases, a proton refers to an H+ ion.

A second example of an acid is nitric acid, HNO3. Again, acids release

H+ in aqueous solution. We can show this in two ways:
Water not included: HNO3 (aq)→H+ (aq)+NO3− (aq)Water included: HNO3
(aq)+H2O (l)→H3O+ (aq)+NO3− (aq)

Bases dissociate in water to form hydroxide (OH–) ions. Most common

bases are metal hydroxides, which dissociate to form the metal cation and
the hydroxide anion. For example, potassium hydroxide and magnesium
hydroxide dissociate in water, as shown here:
KOH (s)→K+ (aq)+OH− (aq)Mg(OH)2 (s)→Mg2+ (aq)+2 OH− (aq)

There are many other common bases of this type, such as sodium
hydroxide (NaOH), lithium hydroxide (LiOH), and calcium hydroxide
Ca(OH)2.

729
 IT
TRY

1. Complete the following equations to show the ions these acids would
form in aqueous solution:
HBr (aq)+H2O (l)→HClO4 (aq)+H2O (l)→

2. Write an ionic equation to show the ions that form when calcium
hydroxide is dissolved in water.

Polyprotic Acids
Polyprotic acids can release more than one H+ into aqueous solution. For
example, sulfuric acid, H2SO4, can release two protons into solution.
Release of one H+ forms the HSO4– ion (called the bisulfate ion). Release
of the second H+ forms the SO42– ion (the sulfate ion):
First ionization: H2SO4 (aq)→H+ (aq)+HSO4− (aq)Second ionization: HSO4−
(aq)→H+ (aq)+SO42– (aq)

Acids that can release only one proton into solution are monoprotic acids.
Hydrochloric acid (HCl), hydrobromic acid (HBr), and nitric acid (HNO3)
are all examples of monoprotic acids. Diprotic acids (such as sulfuric acid,
H2SO4, and carbonic acid, H2CO3) can release two protons. Triprotic
acids (such as phosphoric acid, H3PO4) can release three protons.

IT
TRY

3. Complete the second and third reactions to show the dissociation of

phosphoric acid to lose its first, second, and third protons.
First ionization: H3PO4 (aq)→H+ (aq)+H2 PO4− (aq)Second ionization:
H2PO4− (aq)→Third ionization:

730
The Brønsted-Lowry Definition
While the Arrhenius description of acids and bases is useful, it is limited to
reactions that occur in water. But compounds that behave as acids or bases
in water undergo very similar reactions in other solvents. For example,
consider the two reactions below. What do they have in common?
HCl+H2O→H3O++Cl−HCl+H3N→H4N++Cl−

In both examples, the HCl molecule transfers an H+ to the other

species to produce a cation (hydronium or ammonium) and a chloride
anion. We describe HCl as an acid in the first reaction, so why not in the
second? The Arrhenius definition is too narrow. A broader description of
acids and bases is the Brønsted-Lowry definition. According to this
definition,
Acids are compounds that donate H+ ions.
Bases are compounds that accept H+ ions.
In the reactions above, HCl is the acid—it donates the H+ to the base.
Water and ammonia (NH3) are the bases—they receive the H+ from the
acid. These reactions are shown structurally in Figure 12.5.

731
Figure 12.5 HCl reacts similarly with H2O and with NH3. The blue arrows show the
electron motion: The oxygen and nitrogen atoms donate electrons to form a new
covalent bond with H+. When this happens, the electrons in the H–Cl bond move to the
chlorine atom, producing a chloride ion.

Let’s look at another example of a Brønsted-Lowry acid-base reaction.

In the reaction below, ammonia and water combine in an acid-base
reaction. Which compound behaves as the acid? Which one behaves as the
base?
NH3 (aq)+H2O (l)→NH4+ (aq)+OH− (aq)

Acids are proton donors; bases are proton acceptors.

In this example, the NH3 gains an H+ to form NH4+. That is, NH3 is
the base. On the other hand, H2O loses an H+ to form OH–. Water behaves
as the acid in this reaction. Also notice that in this reaction, ammonia
combines with water to produce a hydroxide ion. Substances that are basic
in the Brønsted-Lowry sense (H+ acceptors) also behave as bases in the
Arrhenius sense (produce hydroxide in aqueous solutions). Figure 12.6
provides a summary of the acid-base definitions.

732
Figure 12.6 Acid-base definitions

Ammonia, NH3, is a weakly basic substance that is commonly used as a

fertilizer.

Example 12.1 Identifying Acids and Bases

Identify the acid and base in the following reaction:
H2S (aq)+H2O (l)→HS− (aq)+H3O+ (aq)

This reaction does not contain any of the familiar acids and bases in Table
12.1, but we can still identify the acid and base from the changes that take
place. Notice that as we go from reactants to products, the H2O gains an H+
to become H3O+. This means that H2O is the proton acceptor—the base.
Similarly, as we go from reactants to products, H2S loses a hydrogen ion
and a charge of +1 to become HS–. This means that H2S is the proton donor
—the acid. •

IT
TRY

733
4. Identify the acid and the base in these reactions:
HCl+C2H6O→C2H6OH++Cl−HNO2+CH5N→NO2−+CH6N+

5. Water (H2O) can behave as both an acid and as a base.

a. Draw the Lewis structure for the ion formed when water acts as an
acid.
b. Draw the Lewis structure for the ion formed when water acts as a base.

734
12.2 Acid-Base Equilibrium Reactions
Strong acids completely ionize in water. The examples of acids given
earlier (HCl and HNO3) are both strong acids. In fact there are only six
common strong acids, as shown in Table 12.2.

TABLE 12.2 Common Strong Acids

Formula Name

HCl Hydrochloric acid

HBr Hydrobromic acid

HI Hydroiodic acid

HNO3 Nitric acid

H2SO4 Sulfuric acid

HClO4 Perchloric acid

Most acids only partially ionize in water—these are weak acids. For
example, hydrofluoric acid (HF) is a weak acid. In solution, only about 2%
of HF breaks into ions; the remaining 98% remains intact.
HF (aq)+H2O (l) ⇌H3O+ (aq)+F− (aq)98% 2%

The ionization of weak acids is an example of an equilibrium

reaction. This type of reaction occurs in both the forward and backward
directions. Hydrofluoric acid can ionize to form H3O+ and F–, but the ions
can also combine to re-form HF. To indicate this movement, we use an
equilibrium arrow (⇋), which shows the reaction going in both directions.
The concept of equilibrium will be covered further in Chapter 13.
Let’s consider the forward and backward reactions of this equilibrium
in more detail (Figure 12.7). Going from left to right, the HF donates the
H+ to a water molecule. This means that HF is the acid and H2O is the
base. What about the reverse reaction? Going from right to left, the
hydronium (H3O+) gives the H+ back to the fluoride ion. In the reverse
reaction, H3O+ is the acid and F– is the base. When describing an acid-
base equilibrium, we refer to the acid and base on the right-hand side of

735
the chemical equation as the conjugate acid and the conjugate base.

Figure 12.7 In the forward reaction (blue arrows), HF is the acid and H2O is the base.
In the reverse reaction (green arrows), H3O+ is the acid and F− is the base.

Let’s look at another example: Trichloroacetic acid (HC2Cl3O2) is a

weak acid. In water, the ionization of this acid is written as follows:

Trichloroacetic acid is a weak acid that is used in the treatment of warts and
removal of tattoos.

In the forward direction, HC2Cl3O2 donates the H+. It is the acid. In

the reverse direction, C2Cl3O2– accepts the H+; it is the conjugate base.

736
Similarly, H2O is the base in the forward direction, but H3O+ is the acid in
the reverse direction.
Notice that the only difference between the acid and conjugate base is
the H+. The acid form has the H+; the base form does not. Because of this,
we refer to the acid and conjugate base as an acid-base pair. The base and
conjugate acid are also a pair—the only difference between them is the H+.

In this acid-base equilibrium, the dog is the acid. Can you identify the
conjugate-base dog? Of course. A conjugate base is identical to the acid, but
without the H+.

Many drug molecules (both legal and illegal) can undergo acid-base
reactions. For example, drug producers commonly ship cocaine in the acid
form, called the acid salt or powder form (Figure 12.8). In the 1980s, drug
dealers began treating acid cocaine with base. This acid-base reaction
converts cocaine to the conjugate base (also called the free-base form),
commonly known as crack.

Figure 12.8 Powder and crack cocaine are an acid-conjugate base pair. Notice that the
difference between them is the presence of an H+ (and a counter ion, the Cl−). The line
structures are a simplified way to show carbon atoms. To learn more about line
structures, check out Section 15.3.

737
 The free-base form of cocaine has much lower melting and boiling points than
the ionic acid form. This is why it can be smoked rather than snorted. When
cocaine enters the body through the lungs, it reaches the brain much more
quickly, leading to a faster and more intense high.

It was not a new trick. For years, cigarette companies have treated
tobacco with ammonia. Since ammonia is basic, it converts nicotine from
its acid form to its conjugate base form (Figure 12.9). This practice
increases nicotine intake when smoking, fueling addiction and increasing
the consumption of cigarettes.

Figure 12.9 Like cocaine, nicotine exists as an acid salt or a free base.

Example 12.2 The Components of an Acid-Base Equilibrium

Identify the acid-base pairs in the following equilibrium reaction:
HBr+CH4O⇌CH5O++Br−

The key to identifying acid-base pairs is to realize that they are the same

738
 species, with or without an H+. In the forward reaction, notice that HBr
donates an H+, leaving the Br– alone. Therefore, HBr is the acid, and Br– is
the conjugate base. By extension, CH4O is the base. Its conjugate acid is
the same species, but with an H+ attached. Therefore, CH4O and CH5O+
are a base/conjugate acid pair. •

IT
TRY

6. Label the acid, base, conjugate acid, and conjugate base in this
reaction:

Are Conjugate Bases Basic?

Let’s take one more look at the reaction of HF with water:
HF (aq)+H2O (l)⇌H3O+ (aq)+F− (aq)

In this equilibrium, we say that HF is the acid, and F– is the conjugate

base. Does this mean that F– is basic? Actually, yes. In a solution of
fluoride ions, the ions act as a weak base, removing H+ from water to
produce hydroxide:
F− (aq)+ H2O (l)⇌OH− (aq)+HF (aq)

Because we wrote the F– on the left-hand side of this equation, F– is

the base, and HF is the conjugate acid.
Let’s look at another example: hypochlorous acid, HClO, is a weak
acid. This means that its conjugate, ClO–, is basic. We can write two
reactions, showing how both the acid and the conjugate base react with
water:

The conjugates of weak acids are basic.

739
 The conjugates of weak bases are acidic.

Acidic reaction of HClO: HClO (aq)+H2O (l)⇌H3O+ (aq)+ClO− (aq)Basic

reaction of ClO– : ClO− (aq)+H2O (l)⇌OH− (aq)+HClO (aq)

As a general rule, the conjugates of weak acids are basic, and the
conjugates of weak bases are acidic.

IT
TRY

7. Carbonic acid, H2CO3, is weakly acidic. Write two equilibria

equations, one showing the acidic reaction of H2CO3 in water and the
other showing the basic reaction of HCO3– in water.

8. Hypobromite, BrO–, is a weak base. Write two equilibria equations,

one showing the reaction of BrO– in water and the other showing the
reaction of its conjugate acid in water.

740
12.3 Reactions Involving Acids and Bases

Neutralization Reactions
Acid-base neutralization reactions were introduced in Chapter 6. In these
reactions, an acid and a hydroxide base combine to produce water and an
ionic compound, commonly referred to as a salt:

Acid+Base→Water+Salt
For example, hydrofluoric acid reacts with sodium hydroxide:
HF (aq)+NaOH (aq)→H2O (l)+NaF (aq)

Neutralization reactions were introduced in Section 6.3. See that section for
other examples.

The H+ from the acid combines with the OH– from the base, forming
water. The remaining ions end up together as NaF. Similarly, perchloric
acid reacts with potassium hydroxide:
HClO4 (aq)+KOH (aq)→H2O (l)+KClO4 (aq)

Again, the H+ from the acid reacts with the OH– from the hydroxide base
to form water. The remaining ions combine to produce a salt.
What drives a neutralization reaction to take place? To understand this
question, it’s helpful to rewrite the reaction as a complete ionic equation:
H+ (aq)+ClO4− (aq)+K+ (aq)+OH– (aq)→H2O (l)+K+ (aq)+ClO4− (aq)

Notice that the K+ and the ClO4– ions are unchanged in this reaction.
They are spectator ions. The net ionic reaction for neutralization is the
formation of water from H+ and OH–. The formation of water drives
neutralization reactions.

H+ (aq)+OH– (aq)→H2O (l)

The formation of water is the driving force for neutralization reactions.

741
 Example 12.3 Predicting the Products from Acid-Base
Neutralization Reactions
Write a balanced equation to show the reaction of aq. nitric acid with aq.
calcium hydroxide.
In this reaction, the H+ from the acid and the OH– from the base combine to
produce water. The remaining ions, Ca2+ and NO3–, combine to produce an
ionic compound, Ca(NO3)2.
HNO3 (aq)+Ca(OH)2 (aq)→H2O (l)+Ca(NO3)2 (aq) not balanced

A simple way to balance neutralization reactions is to balance the number

of H+ and OH– ions that combine. Because Ca(OH)2 contains two
hydroxide ions, two hydrogen ions (and therefore two HNO3 molecules)
are required to react with it. This produces two water molecules and leads
us to the balanced form. •
2 HNO3 (aq)+Ca(OH)2 (aq)→2 H2O (l)+Ca(NO3)2 (aq) balanced

IT
TRY

9. Predict the products for these neutralization reactions. Make sure the
equations are balanced.
HBr (aq)+NaOH (aq)→HNO2 (aq)+Al(OH)3 (aq)→H2SO4 (aq)+Mg(OH)2

(aq)→

10. Write a complete ionic equation and a net ionic equation for this
neutralization reaction:
Ba(OH)2 (aq)+2 HNO3 (aq)→Ba(NO3)2 (aq)+2 H2O (l)

742
Reactions of Acids with Metal
Most metals react with acid to produce metal cations and hydrogen gas
(Figure 12.10). For example, iron metal reacts with hydrobromic acid to
produce iron(II) bromide and hydrogen gas. We can write this as either a
molecular equation or as a net ionic equation:
Molecular equation: Fe (s)+2 HBr (aq)→FeBr2 (aq)+H2 (g)Net ionic
equation: Fe (s)+2 H+ (aq)→Fe2+ (aq)+H2 (g)

Similarly, aluminum reacts with nitric acid to produce aluminum

nitrate and hydrogen gas:
Molecular equation: 2 Al (s)+6 HNO3 (aq)→2 Al(NO3)3 (aq)+3 H2 (g)Net
ionic equation: 2 Al (s)+6 H+ (aq)→2 Al3+ (aq) +3 H2 (g)

The reactions of metals with acid are single-displacement reactions. The

reactants begin with one element (metal) and produce another element
(hydrogen).

Precious metals (silver, gold, platinum) do not react with acid. In fact,
the reason these metals are “precious” is that they are rare, beautiful, and
durable. They are not easily destroyed or chemically changed the way
more common metals are.

743
Figure 12.10 Most metals react with acid to produce metal cations and hydrogen gas.
Precious metals like gold, silver, and platinum do not.

The gold in this Incan figurine has endured for centuries. Interestingly, the lump
on his left cheek is thought to represent a wad of coca leaves.

IT
TRY

11. Potassium, calcium, and aluminum all react with acid. Complete and
balance these reactions to show the products that form:
K (s)+HBr (aq)→Ca (s)+H2SO4 (aq)→Al (s)+HNO3 (aq)→

744
Formation of Acids from Nonmetal Oxides
As their name implies, nonmetal oxides are compounds or ions that
contain a nonmetal covalently bonded to one or more oxygen atoms. Many
nonmetal oxides react with water to form acids. For example, carbon
dioxide reversibly reacts with water to form carbonic acid:
CO2 (g)+H2O (l)⇌H2CO3 (aq)

Nonmetal oxides react with water to form acids.

In carbonated beverages, like beer and soda, the dissolved CO2 forms
carbonic acid through this reaction (Figure 12.11). As we’ll see in Section
12.6, this reaction also occurs as carbon dioxide is dissolved in the
bloodstream—a process that helps regulate the acid-base balance in the
blood.

Figure 12.11 Dissolved CO2 in soda forms carbonic acid, giving soda its pungent
quality.

Sulfur oxides and nitrogen oxides also react with water to form acidic
compounds. For example, sulfur trioxide reacts to form sulfuric acid, while
dinitrogen pentoxide reacts to form nitric acid:
SO3 (g)+H2O (l)→H2SO4 (aq)N2O5 (g)+H2O (l)→2 HNO3 (aq)

745
 These reactions are important because sulfur and nitrogen are common
in fossil fuels, such as coal and oil. Burning these “dirty fuels” can
produce sulfur oxides (sometimes called SOx emissions) and nitrogen
oxides (called NOx emissions). These oxides combine with moisture in the
air, leading to the effect known as acid rain. In heavily industrialized
areas, acid rain accelerates erosion and is harmful to plant and animal life
(Figure 12.12). To minimize the effects of acid rain, most first-world
nations have developed strategies for removing nitrogen oxides and sulfur
oxides from fossil fuel emissions. For example, coal-burning power plants
use “scrubbers” to trap SOx and NOx gases, preventing their emission into
the atmosphere. While these technologies are not perfect, they have
dramatically improved air quality over the past half-century.

Figure 12.12 The erosion of this statue was caused by acid rain.

746
In wet cement, calcium oxide reacts with water to produce calcium hydroxide,
Ca(OH)2. Wet cement is basic.

747
12.4 Acid and Base Concentration
Strongly acidic solutions can dissolve lead. Weakly acidic solutions, like
buffalo wing sauce and Coca-Cola, taste great during football season.
What is the difference between the two? How can we measure the strength
of acids and bases? To answer these questions, we must begin with pure
water and the two species, H+ and OH–, that define acids and bases.

Concentrations of H+ and OH– in Aqueous Solutions

Pure water always contains a tiny amount of both H+ and OH–. In its
liquid form, water molecules constantly collide with each other. Once in a
while these collisions result in a transfer of an H+ from one water molecule
to another:
H2O (l)+H2O (l)⇌H3O+ (aq)+OH− (aq)

This process is called the self-ionization reaction of water. For

simplicity, we sometimes show it as a single water molecule breaking into
two ions:

H2O (l)⇌H+ (aq)+OH− (aq)

In pure water, one in every 560,000,000 water molecules is ionized.
This means that the concentrations of H+ and OH− in pure water are both
equal to 1.0 × 10–7 M. Mathematically, we say that in pure water,

[H+][OH−]=1.0×10−14

Recall that [H+] means “concentration of H+.”

What happens if we add acid to pure water? Suddenly, the

concentration of H+ jumps up. This excess H+ reacts with the small
amount of OH– that is present, and [OH−] decreases. On the other hand, if
we add excess base, the opposite occurs: The [OH–] jumps up, and [H+]
decreases. The concentrations of OH– and H+ are like a seesaw: If one
goes up, the other goes down (Figure 12.13).

748
Figure 12.13 The [H+] and [OH−] are like two sides of a seesaw: If one goes up, the
other goes down by the same factor.

Interestingly, even if [H+] and [OH–] change, the product of the two
concentrations remains constant. For liquid water near 25 °C, [H+] times
[OH–] always equals 1.0 × 10–14.

Math Review: Exponential and Scientific Notation

The values of [H+] and [OH–] are usually expressed in scientific or in

exponential notation. We’re going to use these notations quite a bit in the
coming sections, so a quick review may be helpful.
Scientific notation and exponential notation are very similar in that
they both involve ten raised to some power. The difference is that
scientific notation uses a coefficient times a multiplier (10 raised to an
integer power), whereas exponential notation raises 10 to some non-integer
power and does not use a coefficient (Figure 12.14). The value of the
exponent in exponential notation is always within one unit of properly
written scientific notation.

Figure 12.14 Scientific notation uses a coefficient, a multiplier, and an integer

exponent. In exponential notation, there is no coefficient, and the exponent may be a
non-integer. The three forms shown here represent the same value.

Remember that when multiplying exponents, we always add the

exponents together. Mathematically, this is represented by the following
generic equation:

749
 Xm Xn=X(m+n)

For example, we could find the product of 52 and 57:

(52)(57)=59

Because [H+][OH–] = 10–14, this means that if we express the

concentration as exponents of 10, the two exponents will always add up to
equal –14.

Connecting [OH–] and [H+]

Table 12.3 shows the H+ and OH– concentrations in solutions ranging

from strongly acidic to strongly basic. Notice that the product of [H+] and
[OH–] always equals 1.0 × 10–14. At the neutral point, the concentration of
both species is 1.0 × 10–7 M. The [H+] is greater than this in acidic
solutions, but less in basic solutions.

In water, [H+][OH–] = 10–14.

TABLE 12.3 [H+] and [OH–] in Solutions Ranging from Strongly

Acidic to Strongly Basic
[H+] (mol/L) [OH–] (mol/L)

Strongly acidic 1 × 10° 1 × 10–14

1 × 10–1 1 × 10–13

1 × 10–3 1 × 10–11

Weakly acidic 1 × 10–5 1 × 10–9

Neutral 1 × 10–7 1 × 10–7

Weakly basic 1 × 10–9 1 × 10–5

1 × 10–11 1 × 10–3

1 × 10–13 1 × 10–1

Strongly basic 1 × 10–14 1 × 10°

Example 12.4 Relating [H+] and [OH–

750
 The concentration of H+ in an aqueous solution is 1.0 × 10–6 M. What is
the concentration of hydroxide? Is this solution acidic or basic?
We begin with the relationship between [H+] and [OH–] and then substitute
in the value of [H+]:
[H+][OH−]=1.0×10−14(1.0×10−6)[OH−]=1.0×10−14

Rearranging this equation to solve, we find that [OH–] = 1.0 × 10–8 M:

[OH−]=1.0×10−141.0×10−61.0×10−8 M

In this example, [H+] = 1.0 × 10–6 M, which is a larger amount than in

neutral water (remember that in neutral water, [H+] = [OH–] = 1.0 × 10–7
M). Therefore, this solution is acidic. •

IT
TRY

12. For each of these solutions, find the H+ concentration. Label the
solution as acidic, basic, or neutral.
a. In a sodium hydroxide solution, [OH–] = 1.0 × 10–4 M.
b. In a calcium hydroxide solution, [OH–] = 4.3 × 10–7 M.
c. In a titanium chloride solution, [OH–] = 6.1 × 10–9 M.

The pH Scale
As we saw in Table 12.3, the acidity or basicity of a solution depends on
the concentration of H+ or OH–. These concentrations vary widely but
usually range from about 1.0 (that is, 1.0 × 100) all the way down to
0.00000000000001 (1 × 10–14) molar. To describe these vastly different
concentrations, chemists use the pH scale. The pH of a solution is defined

751
as the negative log of the H+ concentration:

pH=−log[H+]
The log of a number is defined as the exponent, when a number is
expressed in exponential notation with a base of 10 (Table 12.4). For
example, the log of 1018 is 18. The log of 10–2.3 is –2.3. What is the log of
100? Because 100 = 102, the log of 100 = 2.

TABLE 12.4 Logarithmic Notations

Value Exponential Notation Log

1,000. 103 3

100. 102 2

10. 101 1

1. 100 0

0.1 10–1 –1

0.01 10–2 –2

0.001 10–3 –3

The pH scale reverses the sign of the log of H+ concentrations. This is

because [H+] is generally small, and so the exponent is negative. Taking
the negative log allows us to work with positive numbers:
If [H+] = 10–7 M, the pH is 7.
If [H+] = 10–13 M, the pH is 13.
For a neutral aqueous solution, [H+] = [OH–] = 10–7 M. Based on this,
the neutral point on the pH scale is 7. Acidic solutions have pH values
lower than 7, and basic solutions have pH values higher than 7 (Table
12.5).

TABLE 12.5 The pH and pOH Values Always Add Up to 14

[H+](mol/L) pH [OH–](mol/L) pOH

Strongly acidic 1 × 100 0 1 × 10–14 14

1 × 10–1 1 1 × 10–13 13

1 × 10–3 3 1 × 10–11 11

752
Weakly acidic 1 × 10–5 5 1 × 10–9 9

Neutral 1 × 10–7 7 1 × 10–7 7

Weakly basic 1 × 10–9 9 1 × 10–5 5

1 × 10–11 11 1 × 10–3 3

1 × 10–13 13 1 × 10–1 1

Strongly basic 1 × 10–14 14 1 × 100 0

A pH of 7 is neutral. Lower pH values are acidic; higher pH values are

basic.

It is also possible to measure the pOH, defined as the negative log of

the hydroxide ion concentration. Because the product of [H+] and [OH–] =
10–14, we can take the negative log of both sides of the equation, which
gives us another helpful relationship:

pH+pOH=14
However, pOH is seldom used. Because we can describe both acidic
and basic solutions using pH, we do not need to use pOH in most
applications. Before moving on, I encourage you to look at the following
examples of conversions between pH, pOH, and the ion concentrations in
aqueous solutions.

Example 12.5 Finding pH from [H+]

A chemist measures the acidity of a solution and finds that [H+] = 0.01 M.
What is the pH of this solution? Is the solution acidic or basic?
To solve this problem, we first convert the concentration to scientific
notation and then to exponential notation:

Now we can convert from exponential notation to pH:

pH=−log[H+]=−log(10−2)=2

753
Because the pH is less than 7, the solution is acidic. •

Example 12.6 Finding pH from [H+]

A solution of hydrochloric acid in water is found to have [H+] = 0.0030 M.
What is the pH of this solution?
Estimate: Before using a calculator, let’s use the definition of the log to find
the approximate value: We know that 0.0030 M is slightly greater than
0.001, or 10–3. It’s also less than 0.01, or 10–2. In light of this, the exponent
will have a value between –2 and –3. Therefore the pH will be somewhere
between 2 and 3 (Figure 12.15).

Figure 12.15 The pH tells how many spaces lie between the decimal point and the
first nonzero digit. With practice, you can approximate pH values without a
calculator.

Practice
Figure 12.15 Can you estimate pH from the H+ concentration? Try this
interactive to test your skills.

Being able to estimate the log value is very helpful: Quickly approximating
the concentration allows you to rule out incorrect answers, and it is a fast
way to check that you’ve done the problem correctly. It’s also fun to do at
parties, if things get slow.
Calculator: To calculate the pH, we first find the log[0.0030] and then take
the negative of that value:
pH=−log[H+]=−log(0.0030)=2.52

754
 One of the most common problems students encounter with pH problems is
entering the numbers into the calculator correctly. Different calculators
require slightly different ways to enter data for these problems; I encourage
you to try each of the sample problems in this section on your specific
calculator to make sure you get answers that agree with those listed here. •

IT
TRY

13. If the [H+] of a solution is 10–4 M, what is the pH? Is the solution
acidic or basic?

14. The pH of a solution is 9.3. Is the solution acidic or basic? What is

the pOH of this solution?

Example 12.7 Finding [H+] and [OH–] from pH

Carbonated soft drinks are fairly acidic and typically have a pH of about 3.
What are [H+] and [OH–] at this pH?
Because the pH = 3, this means that [H+] = 10–3 M, or 0.001 M.
Next we find [OH–]: We know that [H+][OH–] = 10–14, so we can say that
[OH−]=10−14[H+]=10−1410−3=10−11 M

We could also solve this problem using the pOH. Since the pH = 3, the
pOH must equal 11. Therefore, [OH–] = 10–11 M. •

Example 12.8 Finding [H+] from pH

What is the [H+] in an aqueous solution with a pH of 9.43?

755
 Again, using the definition of pH, [H+] = 10–9.43 M. We typically enter this
in the calculator as 10^ –9.43, or as the inverse log of –9.43. This
calculation should give an answer of

10−9.43=3.7 × 10−10 M
In answering this question, notice that we’ve gone from exponential
notation (10–9.43) to scientific notation (3.7 × 10–10). •

IT
TRY

15. If the hydronium concentration of a solution is 0.00001 M, what is

the pH? What is the pOH?

16. A solution of baking soda in water has a pH of 8.3. What is the [H+]
in this solution?

17. A solution has an H+ concentration of 0.041 M. Is it strongly acidic

or strongly basic? Without using a calculator, estimate the pH of this
solution. Then check your answer on a calculator.

756
12.5 Measuring Acid and Base Concentration

Determining pH in the Laboratory

We often need to determine the pH of aqueous solutions. One way of
doing this is with pH indicators. These are compounds that change color
depending on the pH; the color changes indicate the approximate pH of the
solution. Several common indicators are shown in Figure 12.16.

Figure 12.16 pH indicators change colors depending on the pH of the solution. (a)
Litmus turns blue in base. (b) Phenolphthalein is colorless in acid, but pink in base. (c)
pH paper contains a blend of indicators, giving a spectrum of colors that chemists use to
approximately measure pH. (d) Pool test kits also use pH indicators.

Explore
Figure 12.16

One of the oldest and most common indicators is litmus. Litmus turns
blue in the presence of base, and red in the presence of acid. Chemists
sometimes test for acids or bases using litmus paper, small strips of paper
that are pretreated with litmus. The paper comes in red or blue, and its
color changes or stays the same depending on the pH.
A slightly more advanced form of litmus paper is pH paper. This
paper contains a blend of indicators and produces a series of colors ranging
from deep red to deep blue. Rather than simply indicate whether the
solution is acidic or basic, pH paper allows us to approximate the pH to the
nearest integer.
A third common indicator is a compound called phenolphthalein. In
acidic solution, phenolphthalein is colorless. In basic solution, it is bright

757
pink.
For more precise pH measurements, chemists use an analytical
instrument called a pH meter (Figure 12.17).

Figure 12.17 A pH meter measures the pH of a solution more precisely than simple
chemical indicators.

Acid-Base Titrations
Titration is an analytical technique used to precisely determine the
concentration of an acid or base by measuring the volume required for a
neutralization to occur. For example, if a chemist wishes to determine the
concentration of a solution of hydrochloric acid, she will follow this
procedure:

1. Prepare a basic solution (such as sodium hydroxide) with a known

concentration.
2. Precisely measure a small volume of the acid. Place the acid in an
Erlenmeyer flask, and add 1–2 drops of phenolphthalein indicator.
(Remember, this indicator is colorless in acidic solution, but bright
pink in base.)
3. Fill a buret with the known basic solution. A buret is a tall dispensing
tube with volume markings (Figure 12.18). Measure the initial
volume in the buret.

758
 Figure 12.18 Titrations use a long tube with volume markings called a buret.

4. Begin adding base to the acid, swirling the Erlenmeyer flask

occasionally to mix the solutions. As more base is added, a pink color
begins to form and then disappears. Slowly add base, drop by drop,
until a faint pink color remains (Figure 12.19). Measure the volume
of the added base.

Figure 12.19 An acid-base titration. Base is slowly added to the solution until it
turns pink.

759
 Explore

Figure 12.19

By adding the solution slowly until the faintest pink color remains, the
chemist adds just enough base to neutralize the acid. For the example
above, the neutralization reaction is as follows:
HCl (aq)+NaOH (aq)→NaCl (aq)+H2O (l)

In this reaction, one mole of HCl reacts with one mole of NaOH.
Therefore the chemist has combined the same number of moles of acid and
base in the solution:

molesHCl=molesNaOH
Because the molarity (moles/L) times the volume (L) equals the moles,
we can rewrite this equation as

MaVa=MbVb
where Ma and Va are the molarity and volume of the acid, and Mb and Vb
are the molarity and volume of the base.
In a titration, we measure the volume of the known, the molarity of the
known, and the volume of the unknown. Based on these measures, we can
solve for the molarity of the unknown. Examples 12.9 and 12.10 illustrate
this process further.

Titrations are used to determine the molarity of an unknown acid or base.

Example 12.9 Titration

760
 A 25.0-mL sample of HCl is neutralized by reaction with 17.4 mL of 2.00-M
NaOH. What is the concentration of HCl in the unknown sample?
At the neutralization point, the number of moles of acid equals the number
of moles of base. Therefore we can say that MaVa = MbVb. We know the
values of Va (25.0 mL), Vb (17.4 mL), and Mb (2.00 M). By rearranging the
equation and inserting these values, we can solve this problem:
Ma=MbVbVa=(2.00 M)(17.4 mL)25.0 mL=1.39 M HCl

Notice that because the volumes of the acid and base cancel, it was not
necessary to convert from milliliters to liters. •

Example 12.10 Titration

In a titration, a 50.0-mL sample of LiOH is neutralized with an HCl
solution having a concentration of 4.18 M. If 18.1 mL of HCl are required
to complete the neutralization, what was the concentration of the lithium
hydroxide solution?
In this example, the base is the unknown. Since the acid and base react in a
one-to-one ratio in the balanced equation, the number of moles of acid is
equal to the number of moles of base at the neutralization point. Therefore
we can say that MaVa = MbVb. We know the value of Vb (50.0 mL), the
value of Va (18.1 mL), and the value of Ma (4.18 M). We can therefore
rearrange the equation and solve for Mb. •
Mb=MaVaVb=(4.18M)(18.1 mL)50.0 mL=1.51 M LioH

Acid-Base Titrations with Different Coefficients

Sometimes we need to do titrations for reactions in which the coefficients
of acid and base are not equal. For example, consider the neutralization of
barium hydroxide with hydrochloric acid:

761
 2 HCl (aq)+Ba(OH)2 (aq)→BaCl2 (aq)+2 H2O (l)

Based on the balanced equation, the neutralization reaction requires

twice as many moles of HCl as of Ba(OH)2, so we cannot use the simple
assumption that moles of acid and moles of base are equal, as we did
earlier. What can we do? To set the moles of each reactant equal, we can
divide by the coefficient of each reactant in the balanced equation. That is,
we can say

Ma Vaca=Mb Vbcb
where ca and cb are the coefficients of the acid and base in the balanced
equation. Notice that if ca and cb = 1, this simplifies to the relationship
given above for a 1:1 ratio of acid to base. Example 12.11 illustrates this
principle.

Example 12.11 Titration with Unequal Coefficients

A 50.0-mL sample of H2SO4 is neutralized with a KOH solution having a
concentration of 0.400 M. If 22.6 mL of KOH are required to complete the
neutralization, what is the concentration of the sulfuric acid solution?
The balanced equation for this neutralization is
H2SO4 (aq)+2 KOH (aq)→K2SO4 (aq)+2 H2O (l)

In the balanced equation the coefficient for the acid (ca), H2SO4, is equal to
one; the coefficient for the base (cb), KOH, is equal to two. Va = 50.0 mL,
Mb = 0.400 M, Vb = 22.6 mL, and we’re trying to find Ma. Rearranging to
solve for Ma, and substituting in the values, we get
Ma=MbVbCacbVa=(0.400 M)(22.6 mL)(1)(2)(50.0 mL)=0.0904 M H2SO4

We also could have solved this as a traditional stoichiometry problem.

Starting with the molarity and volume of KOH, we could have found the

762
 concentration of sulfuric acid by following the strategy shown below. •

IT
TRY

18. A solution containing nitric acid, HNO3, is analyzed by titration. A

volume of 19.5 mL of 1.50-M NaOH solution is required to
neutralize 50.0 mL of the acid. What is the original concentration of
the nitric acid solution?

19. What volume of 2.4-M aq. NaOH is required to completely

neutralize 200 mL of 6.0-M aq. HBr?

20. A drop of phenolphthalein is added to a solution of 1.0-M H2SO4

with a volume of 10.0 mL. The solution is mixed with 8.2 mL of
0.70-M NaOH. After the reaction is complete, is the solution acidic,
basic, or neutral? What color is the indicator?

763
12.6 Buffers and Biological pH
The cells that make up your skin, muscle, blood, and organs must remain
in a very narrow pH range in order to function properly. The fluid inside
cells typically maintains a pH of about 6.8, while the bloodstream
maintains a pH very near 7.4. Given the number of acidic and basic
compounds present in the bloodstream and inside the cells, how do our
bodies remain within these ranges?
Our bodies use buffer systems to keep the pH at healthy levels. A
buffer is a solution containing a mixture of acidic and basic components
that resists changes in pH. In the absence of a buffer, adding a small
amount of acid or base can drastically affect the pH of a solution.
However, in a buffered solution, adding the same amount of acid or base
causes a much smaller pH change (Figure 12.20).

Figure 12.20 The pH of water changes drastically when an acid or base is added. Buffer
solutions resist changes in pH.

764
 How do buffer solutions work? Buffer solutions contain one of the
following two combinations:
A mixture of a weak acid and its conjugate base, or
A mixture of a weak base and its conjugate acid
Both species must be present in large quantities. For example, animal
cells contain a buffer consisting of H2PO4– (a weak acid) and HPO4–2 (its
conjugate base). If acid is added to the cells, the H+ reacts with the
conjugate base. If base is added to the cells, the OH– is consumed by the
acid. The ratio of H2PO4– to HPO4–2 will change, but the amount of H+
and OH– present changes only slightly. This is shown in Figure 12.21.

Figure 12.21 The phosphate buffer is composed of H2PO4 – (a weak acid, in red) and
HPO4 2– (its conjugate base, in blue). If base is added, the acid neutralizes it. If acid is
added, the base consumes it.

765
 Explore

Figure 12.21

Chemically, we represent this buffer using the equilibrium shown here:

A buffer solution can be thought of as two guards watching opposite

directions. One component guards against an influx of acid, the other
guards against an influx of base.
The bloodstream and the fluids between cells use a different buffer
system, based on carbonic acid and its conjugate base, the bicarbonate ion.
When carbon dioxide dissolves in the bloodstream, it reacts with water to
form carbonic acid. This acid partially ionizes to form the bicarbonate ion
(HCO3–):
H2CO3 (aq)+H2O (l)⇌H3O+ (aq)+HCO3− (aq)carbonic acidbicarbonate ion

This buffer typically maintains blood pH in the range of 7.35 to 7.45.

766
 It’s hard to sneak up on two guard dogs facing in opposite directions. Either way
you approach, one of the dogs will see you. Similarly, with a buffer solution, the
acid guards against excess base, and the conjugate base guards against excess
acid.

Certain conditions can cause blood pH to fluctuate. For example,

hyperventilation (Figure 12.22) is a condition in which a person breathes
abnormally fast, taking in more oxygen than the body needs. This
condition is usually brought on by extreme stress or panic, and it causes
dissolved CO2 levels in the bloodstream to drop. This in turn reduces the
amount of H2CO3 present in the bloodstream, causing the pH of the blood
to quickly rise. While this situation may cause other symptoms, it can
usually be brought into balance after normal breathing is restored.

Figure 12.22 Hyperventilation is often brought on by stress and causes a rapid rise in

767
blood pH.

It is possible to overload a buffer. For example, if we add a large

amount of acid to a phosphate buffer, eventually all of the base will be
consumed. At that point, the pH of the solution will change sharply.
Buffers can be adjusted to specific pH values. The pH of a buffer
solution depends on the specific combination of acid and conjugate base,
and on the concentration of each component present in solution.

IT
TRY

21. A common buffer is prepared from citric acid and its conjugate base,
the citrate ion. Which of these combinations would be the most
effective buffer solution?
a. 0.01 M citric acid and 0.01 M citrate ion
b. 1.0 M citric acid and 1.0 M citrate ion
c. 1.0 M citric acid and 0.01 M citrate ion
d. 0.01 M citric acid and 1.0 M citrate ion

22. A buffer solution contains a mixture of acetic acid (HC2H3O2) and

acetate ion (C2H3O2–). What reaction will take place if H+ is added
to this solution? What reaction will take place if OH– is added to the
solution?

768
Summary
Acids and bases are all around us, and they are very important to our air,
water, and living systems. In the Arrhenius definition, acids produce H+
ions in water, which may also be represented by the hydronium ion, H3O+.
Bases produce hydroxide (OH–) ions in water.
The Brønsted-Lowry definitions for acids and bases are broader and
enable us to describe reactions outside of an aqueous solution. A Brønsted-
Lowry acid is any compound that gives up an H+ (called a proton) in a
chemical reaction. Conversely, a Brønsted-Lowry base is any compound
that accepts an H+.
There are six common strong acids; nearly all other acids are weak
acids. Strong acids completely ionize in water, but weak acids only partly
ionize. Weak acids and bases exchange protons (that is, H+ ions) in
equilibrium reactions. Equilibrium reactions occur in both the forward and
reverse directions and are shown by two arrows pointing in opposite
directions (⇋).
On the left-hand side of an acid-base equilibrium reaction, the proton
donor is the acid, and the proton acceptor is the base. On the right-hand
side of the equation, the proton donor and acceptor are called the
conjugate acid and the conjugate base. A compound that gives a proton
(the acid) will gain the proton back in the reverse reaction. As a result, the
acid and the conjugate base are the same species, differing only by one H+.
Acids and bases react with each other in neutralization reactions. Acids
react with hydroxide bases to form water and ionic compounds, called
salts. Acids also react with many metals, producing metal cations and
hydrogen gas. Nonmetal oxides often react with water to produce acidic
compounds.
Pure water undergoes a self-ionization reaction to produce a small
amount of H+ and OH– ions. These ions are always present, even in pure
water. At 25 °C, the product of the molar H+ and OH– concentrations
equals 10–14. That is, [H+][OH–] = 1.0 × 10–14.
Acidic solutions have a higher concentration of H+ than of OH–. In
basic solutions, the opposite is true. The concentration of H+ in an acidic
or basic solution is measured using the pH scale. The pH is the negative
log of the H+ concentration. In neutral solutions, [H+] = [OH–] = 1.0 × 10–

769
7 M, and so the pH of a neutral solution is 7. Acidic solutions have a lower
pH, and basic solutions have a higher pH.
Indicators are compounds that change color depending on the pH of
their surroundings. Two common indicators are litmus (which turns blue in
basic solution and red in acidic solution) or phenolphthalein (which is
colorless in acid and bright pink in base). Combinations of indicators are
used to make pH paper. Instruments called pH meters measure pH more
precisely than simple indicators.
Titration is an analytical technique for measuring the concentration of
an acidic or basic solution. In this technique, chemists use an indicator to
determine the exact amount of base needed to react with an unknown acid
(or vice versa). Based on the volumes of the known and unknown
solutions, and on the molarity of the known solution, the molarity of the
unknown may be determined. The equation used for this determination is

MaVaca = MbVbcb
where M is molarity, V is volume, and c is the coefficient values of acid
and base in the balanced equation.
Living systems use buffers to maintain the pH of cells and intercellular
fluid (such as blood) within a narrow pH range. Buffers use a weak acid
and its conjugate base (or a weak base and its conjugate acid) in high
concentrations. In a buffer, the acidic component will react with any
excess base, and the basic component will react with any excess acid. The
body uses a mixture of H2PO4– and HPO42– in the cells, and it uses a
mixture of H2CO3 and HCO3– in the bloodstream.

770
The War on Drugs, Then and Now

In 1986, the United States was in a panic over crack cocaine. Across the country,
countless lives were devastated by the powerfully addictive drug. The “crack
epidemic” produced a spike in theft, violent crime, and gang activity. Things
seemed to be spiraling out of control.
In this context, Congress passed the Anti-Drug Abuse Act of 1986. The bill
included mandatory sentencing for possession and trafficking of drugs, including
cocaine and marijuana, and imposed especially harsh sentences for crack (Figure
12.23).

Figure 12.23 The Anti-Drug Abuse Act of 1986 led to a spike in incarceration rates.

The law mandated a 5-year minimum prison sentence for the possession of 500
grams of powder cocaine—and the same minimum for just 5 grams of crack. In the
years that followed, the 100:1 ratio was widely criticized as unfair. In 1995, the
U.S. Sentencing Commission reported that this gap was “a primary cause of the
growing disparity between sentences for black and white federal defendants.” In
addition to racial inequities, the sentencing guidelines disproportionately targeted
low-level offenders. Due to the stability of the powder form, major drug traffickers
nearly always import cocaine as the acid salt. Local dealers converted the cocaine
to the crack form, but these small-time dealers faced harsher penalties.
In the decades following the Anti-Drug Abuse Act, the problems inherent in
the law became clear, and policy makers struggled to formulate more just
sentencing guidelines. Finally, in 2010, the Fair Sentencing Act was passed by a
unanimous vote in the Senate and signed into law by the president. This law

771
narrowed the gap in cocaine sentencing and eliminated minimum sentencing for
low-level offenders.

772
 Key Terms
12.1 Introduction to Acids and Bases
Arrhenius definition Describes an acid as a compound that produces H+ or
H3O+ ions in water, and a base as a compound that produces OH– ions in
water.
acid A compound that produces H+ or H3O+ ions in water (Arrhenius
definition); a compound that donates H+ ions (Brønsted-Lowry definition).
base A compound that produces OH– ions in water (Arrhenius definition); a
compound that accepts H+ ions (Brønsted-Lowry definition).
hydronium ion H3O+, the ion formed in acidic aqueous solutions.
polyprotic acids Acids that can release more than one H+ into aqueous
solution.
Brønsted-Lowry definition Describes an acid as a compound that donates H+
ions, and a base as a compound that accepts H+ ions.

12.2 Acid-Base Equilibrium Reactions

strong acid An acid that completely ionizes in water.
weak acid An acid that only partially ionizes in water.
equilibrium reaction A reaction that occurs in both the forward and backward
directions.
conjugate acid The acid on the right-hand side of a chemical equation in an
acid-base equilibrium; the acid formed when a base reacts with H+.
conjugate base The base on the right-hand side of a chemical equation in an
acid-base equilibrium; the base formed when an acid releases an H+.

12.3 Reactions Involving Acids and Bases

neutralization reaction A reaction in which an acid and a base combine to
produce water and an ionic compound (called a salt).
nonmetal oxide A compound or ion containing a nonmetal covalently bonded
to one or more oxygen atoms.
acid rain The effect observed when nonmetal oxides combine with moisture
in the atmosphere to produce acidic rainfall.

773
12.4 Acid and Base Concentration
self-ionization A process that occurs in water when a water molecule
fragments to produce H+ and OH– ions.
pH scale A scale that indicates the relative concentration of acid or base in an
aqueous solution; pH is defined as the negative log of the H+ concentration.

12.5 Measuring Acid and Base Concentration

pH indicator A compound that changes color depending on the pH; the color
changes indicate the approximate pH of the solution.
litmus A pH indicator that turns blue in the presence of base and red in the
presence of acid.
pH paper Paper containing a blend of indicators that can be used to estimate
pH based on color.
phenolphthalein A pH indicator that is bright pink in base, but colorless in
acid.
titration An analytical technique that can precisely measure the concentration
of an acid or base by measuring the volume required for a neutralization to
occur.

12.6 Buffers and Biological pH

buffer A solution containing a mixture of acidic and basic components; it
resists changes in pH.

774
 Additional Problems

12.1 Introduction to Acids and Bases

23. Complete the following acid dissociation reactions:

a. HBr (aq)→____________+Br− (aq)

b. HBr (aq)+H2O (l)→____________+Br− (aq)

c. H2SO4 (aq)→____________+HSO4− (aq)

d. H2CO3 (aq)+H2O (l)→H3O+ (aq)+____________

24. Complete the following acid dissociation reactions:

a. HNO3 (aq)→+NO3− (aq)

b. HNO3 (aq)+H2O (l)→____________+NO3− (aq)

c. HClO4 (aq)→____________+ClO4− (aq)

d. H3PO4 (aq)→H+ (aq)+____________

25. Complete the following base dissociation reactions:

a. NaOH (s)→Na+ (aq)+____________

b. Ca(OH)2 (s)→____________+2 OH− (aq)

26. Complete the following base dissociation reactions:

a. CsOH (s)→____________+____________

775
 b. ____________→Ba2+ (aq)+2 OH− (aq)

27. Write ionic equations to show the dissociation of these acids in water:
a. HCl
b. HNO3
c. HClO4

28. Write ionic equations to show the dissociation of these acids in water:
a. HBr
b. HI
c. H2SO4

29. Write ionic equations to show the dissociation of these bases in water:
a. NaOH
b. Mg(OH)2
c. CsOH

30. Write ionic equations to show the dissociation of these bases in water:
a. KOH
b. Ca(OH)2
c. Ba(OH)2

31. Identify these acids as monoprotic, diprotic, or triprotic:

a. H2CO3
b. HCl
c. HClO4
d. H2SO4

32. Identify these acids as monoprotic, diprotic, or triprotic:

a. H3PO4
b. HBr
c. HNO3
d. H2C2O4

776
33. What is the difference between the Arrhenius definition of an acid and base
and the Brønsted-Lowry definition?

34. Is it possible to be an acid or base in the Arrhenius sense without being an

acid or base in the Brønsted-Lowry sense? Is it possible to be a Brønsted-
Lowry acid or base without being an Arrhenius acid or base?

35. Draw a sketch showing the dissociation of HBr in water. Explain why it is
reasonable to show the dissociation as forming H+ (aq) or H3O+ (aq).

36. According to the Arrhenius definition, a base is anything that produces

hydroxide (OH−) in aqueous solution. Although ammonia (NH3) does not
dissociate to form an OH−, it is still an Arrhenius base. Why is this so?
Write a chemical equation to support your answer.

37. Identify the acid and the base in these reactions:

a. KOH (aq)+HF (aq)→KF (aq)+H2O (l)

b. HBr (aq)+H2O (l)→H3O+ (aq)+Br− (aq)

c. H3N (aq)+H2SO4 (aq)→H4N+ (aq)+HSO4− (aq)

38. Identify the acid and the base in these reactions:

a. H2O (l)+HCl (aq)→H3O+ (aq)+Cl− (aq)

b. NH4+ (aq)+H2O (l)→NH3 (aq)+H3O+(aq)

c. HNO3 (aq)+CH3NH2 (aq)→NO3− (aq)+CH3NH3+ (aq)

12.2 Acid-Base Equilibrium Reactions

39. What is the difference between a strong acid and a weak acid?

777
40. Consider the following two acid-base equations. Why is one arrow used
for the first expression, but two arrows for the second? What do the
different arrows mean?
HNO3 (aq)+H2O (l)→H3O+ (aq)+NO3− (aq)HNO2 (aq)+H2O (l) H⇌3O+
(aq)+NO3− (aq)

41. Write the molecular formulas for each of these acids. Which one is not a
strong acid?
a. hydrochloric acid
b. hydrobromic acid
c. sulfuric acid
d. phosphoric acid

42. Write the molecular formulas for each of these acids. Which of them is not
a strong acid?
a. hydroiodic acid
b. nitric acid
c. carbonic
d. perchloric acid

43. Identify these compounds as strong or weak acids:

a. H3PO4
b. H2CO3
c. HSO4–
d. HNO3

44. Identify these compounds as strong or weak acids:

a. HCl
b. HF
c. HClO4
d. HCN

45. What is an equilibrium?

778
46. Which of the following are examples of an equilibrium?
a. water flowing between the deep end and shallow end of a pool
b. people moving back and forth between two towns
c. water flowing over a waterfall
d. oxygen gas flowing back and forth between different rooms in a house
e. a candle burning to produce carbon dioxide and water

47. Ammonia (NH3) can behave as both an acid and a base.

a. Draw the Lewis structure for the conjugate acid of ammonia.
b. Draw the Lewis structure for the conjugate base of ammonia.

48. Methyl alcohol has the molecular formula CH4O. This compound can
behave as both an acid and a base.
a. Draw the Lewis structure for the conjugate acid of methyl alcohol.
b. Draw the Lewis structure for the conjugate base of methyl alcohol.

49. In the following acid-base equilibria of weak acids in water, label the acid
(A), the base (B), the conjugate acid (CA), and the conjugate base (CB):
a. HClO2 (aq)+H2O (l)⇌H3O+ (aq)+ClO2−

b. H2CO3 (aq)+H2O (l)⇌H3O+ (aq)+HCO3− (aq)

c. H2O (l)+CH3NH3+ (aq)⇌CH3NH2 (aq)++H3O+ (aq)

50. In the following acid-base equilibria, label the acid, the base, the conjugate
acid, and the conjugate base:
a. CH3CO2 H (aq)+H2O (l)⇌H3O+ (aq)+CH3CO2−(aq)

b. H2O (l)+H2O (l)⇌H3O+ (aq)+OH− (aq)

c. HNO2 (aq)+HCl (aq)⇌H2NO2+ (aq)+Cl− (aq)

51. Label the acid, base, conjugate acid, and conjugate base in these acid-base

779
 equilibria:
a. HCl (g)+CH3OH (g)⇌CH3OH2+ (g)+Cl- (g)

b. CH3OH+PBr3⇌CH3O−+HPBr3+

c. H2S (g)+HF (aq)⇌H3S+ (g)+F− (g)

52. Label the acid, base, conjugate acid, and conjugate base in these acid-base
equilibria:
a. H3CN+HF⇌H4CN++F−

b. C2H6O (l)+HCl (aq)⇌C2H7O+ (aq)+Cl− (aq)

c. CH3O− (aq)+H2O (l)⇌CH3OH (aq)+OH− (aq)

53. HF is a weak acid. Fill in the missing species in these equations to show
the acidic behavior of HF and the basic behavior of F–.
HF (aq)+H2O (l)⇌____________F− (aq)+H2O (l)⇌____________

54. HCN is a weak acid. Fill in the missing species in these equations to show
the acidic behavior of HCN and the basic behavior of CN–.
____________ (aq)+H2O (l)⇌H3O+ (aq)+CN− (aq)____________
(aq)+____________ (l)⇌OH− (aq)+HCN (aq)

55. Complete these acid-base reactions to show the missing species:

H2CO3 (aq)+H2O (l)⇌H3O+ (aq)+____________(aq) HNO2 (aq)+H2O

780
 (l)⇌____________+____________NO2− (aq)+H2O
(l)⇌____________+____________

56. Complete these acid-base reactions to show the missing species:

HClO3 (aq)+H2O (l)⇌H3O+ (aq)+____________ (aq)HF (aq)+H2O
(aq)⇌____________+____________N3− (aq)+H2O
(l)⇌____________+____________

57. Complete these acid-base reactions to show the missing species:

H3PO4 (aq)+H2O (l)⇌H3O+ (aq)+____________ (aq)H2PO42− (aq)+H2O
(l)⇌H3O+ (aq)+____________ (aq)____________+H2O (l)⇌H3O+ (aq)+PO43−
(aq)

58. Complete these acid-base reactions to show the missing species:

H2SO3 (aq)+H2O (l)⇌H3O+ (aq)+____________(aq)____________+H2O
(l)⇌H3O+ (aq)+SO32− (aq)

59. Hypochloric acid (HClO) is a weak acid. The conjugate base of this acid is
the hypochlorate ion (ClO–).
a. Write a balanced equation showing the reaction of HClO with water.
b. Write a balanced equation showing the reaction of ClO– with water.

60. Methylamine (CH3NH2) is a weakly basic compound. The conjugate acid

781
 of this compound is CH3NH3+.
a. Write a balanced equation showing the reaction of CH3NH2 with water.
b. Write a balanced equation showing the reaction of CH3NH3+ with
water.

61. Sometimes acid-base reactions show only a single product. For example,
the compound methylamine (CH5N) reacts with HCl to produce a single
solid product. What is the conjugate acid and the conjugate base in this
reaction? (Hint: Write the products as ions rather than as a neutral
compound).
HCl (g)+CH5N(g)→CH6NCl(s)

62. Sometimes acid-base reactions show only a single product. For example, in
the gas phase, ammonia reacts with hydrofluoric acid to form solid
ammonium fluoride. What is the conjugate acid and the conjugate base in
this reaction? (Hint: Write this as an ionic equation).

NH3 (g)+HF (g)→NH4F (s)

63. Water can act as both an acid and as a base. Draw the conjugate acid and
the conjugate base of a water molecule.

64. Ammonia, NH3, can act as both an acid and as a base. Draw the conjugate
acid and the conjugate base of an ammonia molecule.

65. Vinegar is a mixture of acetic acid and water. Label the acid (A), base (B),
conjugate acid (CA), and conjugate base (CB) in the acid-base equilibrium
of acetic acid shown below.

66. Sodium naproxide is the active component in the anti-pain and anti-
inflammatory compound Aleve®. In aqueous solution, this compound
undergoes the acid-base equilibrium reaction shown. Label the acid (A),

782
 base (B), conjugate acid (CA), and conjugate base (CB) for this reaction.

12.3 Reactions Involving Acids and Bases

67. Nonmetal oxides often react with water to form acids. Which of these
compounds are nonmetal oxides?
a. CO2
b. BaO
c. NH3
d. SO3

68. Complete these reactions to show the formation of carbonic acid and
sulfuric acid from water and a nonmetal oxide:
____________+H2O (l)⇌H2CO3 (aq)____________+H2O (l)⇌H2SO4 (aq)

69. Each of the following metals reacts with hydrochloric acid to form the
metal chloride and hydrogen gas. Complete and balance the reactions to
show the formation of these products.

a. Sn (s)+HCl (aq)→

b. Li (s)+HCl (aq)→

c. Fe (s)+HCl (aq)→

d. Al (s)+HCl (aq)→

70. Each of the following metals reacts with nitric acid to form the metal
chloride and hydrogen gas. Complete and balance the reactions to show
the formation of these products.

a. Zn (s)+HNO3 (aq)→

783
 b. Pb (s)+HNO3 (aq)→

c. Na (s)+HNO3 (aq)→

d. Al (s)+HNO3 (aq)→

71. Complete and balance these reactions. One of the metals does not react.

a. Be (s)+HF (aq)→

b. Sr (s)+H2SO4 (aq)→

c. Au (s)+HNO3 (aq)→

72. Complete and balance these reactions. One of the metals does not react.

a. Ni (s)+HNO3(aq)→

b. Pt (s)+HClO4 (aq)→

c. Ba (s)+HBr (aq)→

73. Show the products from these acid-base neutralization reactions:

a. HCl+NaOH→

b. 2 HBr+Mg(OH)2→

c. 3 HNO3+Al(OH)3→

74. Show the products from these acid-base neutralization reactions:

a. HBr+KOH→

b. H2SO4+2 NaOH→

c. 4 HNO3+Ti(OH)4→

75. Complete and balance these acid-base neutralization reactions:

a. HCl (aq)+Ca(OH)2 (s)→

b. Mg(OH)2 (aq)+H2SO4 (aq)→

784
 c. Al(OH)3 (s)+HF (aq)→

76. Complete and balance these acid-base neutralization reactions:

a. CsOH (aq)+HBr (aq)→

b. HBr (aq)+Fe(OH)2 (s)→

c. H3PO4 (aq)+KOH (aq)→

77. Write a complete ionic equation and a net ionic equation for this
neutralization reaction:
Mg(OH)2 (aq)+2 HNO3 (aq)→Mg(NO3)2 (aq)+2 H2O (l)

78. Write a complete ionic equation and a net ionic equation for this
neutralization reaction:
H2SO4 (aq)+2 KOH (aq)→K2SO4 (aq)+2 H2O (l)

79. Citric acid has the formula H3C6H5O7. It is a triprotic acid, meaning it can
lose three H+ ions, and the citrate anion has a charge of minus three
(C6H5O73–). Write a balanced equation showing the reaction of citric acid
with sodium hydroxide to produce sodium citrate and water.

80. Phosphoric acid has the formula H3PO4. It is a triprotic acid, meaning it
can lose three H+ ions, and the phosphate anion has a charge of minus
three (PO43–). Write a balanced equation showing the reaction of
phosphoric acid with potassium hydroxide to produce potassium
phosphate and water.

81. Nonmetal oxides often react with water to form acids. Write a balanced
equation showing the reaction of gaseous sulfur trioxide with water to
form aqueous sulfuric acid.

785
82. Write a balanced equation showing the reaction of carbon dioxide with
water to produce aqueous carbonic acid.

12.4 Acid and Base Concentration

83. The self-ionization reaction of water is shown below. In this reaction, one
water molecule acts as an acid; the other acts as a base. Label the
conjugate acid and conjugate base in this reaction.
H2Oacid(l)+H2Obase (l)⇌H3O+ (aq)+OH− (aq)

84. How do the H+ and OH– concentration change

a. when an acid is added to water?
b. when a base is added to water?

85. Find the concentration of H+ ions in each of these situations:

a. a strongly acidic solution where [OH–] = 1.0 × 10–12 M
b. a weakly acidic solution where [OH–] = 5.23 × 10–9 M
c. a basic solution where [OH–] = 4.28 × 10–5 M

86. Find [OH–] in each of these situations:

a. a strongly acidic solution where [H+] = 1.0 × 10–2 M
b. a neutral solution where [H+] = 1.0 × 10–7 M
c. a basic solution where [H+] = 4.28 × 10–11 M

87. Calculate the pH for each of the following. Identify whether the solution is
acidic, basic, or neutral.
a. [H+] = 1.0 × 10–3 M
b. [H+] = 1.0 × 10–7 M
c. [H+] = 5.23 × 10–11 M

88. Calculate the pH for each of the following. Identify whether the solution is
acidic, basic, or neutral.
a. [H+] = 1.0 × 10–8 M

786
 b. [H+] = 2.5 × 10–12 M
c. [H+] = 6.91 × 10–5 M

89. Calculate the pH for each of the following. Identify whether the solution is
acidic, basic, or neutral.
a. [OH–] = 1.0 × 10–7 M
b. [OH–] = 1.0 × 10–9 M
c. [OH–] = 1.26 × 10–2 M

90. Calculate the pH for each of the following. Identify whether the solution is
acidic, basic, or neutral.
a. [OH–] = 1.0 × 10–3 M
b. [OH–] = 1.0 × 10–7 M
c. [OH–] = 8.21 × 10–11 M

91. Complete this table to show the pH and pOH of each solution:

[H+] [OH–] pH pOH

Solution A 1.0 × 10–5 5

Solution B 1.0 × 10–3 3

Solution C 1.0 × 10–9

Solution D 8

92. Complete this table to show the pH and pOH of each solution:

[H+] [OH–] pH pOH

Solution A 1.0 × 10–8 8

Solution B 1.0 × 10–3 11

Solution C 1.0 × 10–10

Solution D 6

93. Identify each solution as acidic, basic, or neutral:

787
 a. an aqueous solution with a pH of 7.0
b. an aqueous bleach solution with a pH of 8.2
c. a solution of monosodium phosphate with a pH of 6.1
d. a solution with [H+] = 1.5 × 10–5 M
e. a solution with [H+] = 5.2 × 10–10 M

94. Identify each solution as acidic, basic, or neutral:

a. an aqueous solution where the hydroxide ion concentration is greater
than the hydronium ion concentration
b. a solution of vinegar in water, with a pH of 5.2
c. a solution with [OH–] = 5.2 × 10–2 M
d. a solution where the –log[H+] = 7.0
e. a solution with a pOH of 4.0

95. Find the concentration of hydronium ions in each of these solutions:

a. a solution with a pH of 4.2
b. a solution with a pH of 11.3
c. a solution with a pOH of 6.7
d. a solution with a pOH of 3.66

96. Find the concentration of hydroxide ions in each of these solutions:

a. a solution with a pOH of 4.7
b. a solution with a pOH of 10.3
c. a solution with a pH of 6.2
d. a solution with a pH of 13.4

97. In a dilute sodium bicarbonate solution, [H+] = 1.3 × 10–8 M. Without

using a calculator, what integer value will this pH be near? How can you
tell? Estimate the pH and then use a calculator to verify your answer.

98. A solution has a hydronium ion concentration of 4.1 × 10–11 M. Without

using a calculator, what integer value will this pH be near? How can you
tell? Estimate the pH and then use a calculator to verify your answer.

99. Which has a lower pH, a 0.010-M solution of HCl, or a 0.010-M solution
of HF? How do you know?

788
100. Which has a lower pH, a 1.0 × 10–4 M solution of HCl, or a 1.0 × 10–4 M
solution of H2SO4? How do you know?

12.5 Measuring Acid and Base Concentration

101. Between litmus paper, pH paper, and a pH meter, which gives the most
precise reading of pH? Which gives the least precise reading?

102. Describe the color of litmus and the color of phenolphthalein in both
acidic and basic solutions.

103. Describe the color of the indicator in each of these solutions:

a. litmus at a pH of 3.0
b. litmus at a pH of 9.3
c. phenolphthalein in a solution where [H+] = 1.5 × 10–4 M
d. phenolphthalein in a solution where [H+] = 1.5 × 10–10 M

104. Describe the color of the indicator in each of these solutions:

a. phenolphthalein at a pH of 12.1
b. phenolphthalein at a pH of 9.3
c. litmus in a solution with a pOH of 3.4
d. litmus in a solution where [OH–] = 1.0 × 10–2 M

105. While working in an analytical laboratory, you are asked to find the
molarity of a sodium hydroxide solution. Describe how you would do
this using a titration.

106. Your research laboratory uses a sample of hydrochloric acid and obtains
unexpected results. Your research advisor asks you to measure the
concentration of this solution using a titration. Describe how you would
go about this process.

107. In a titration experiment, a solution of nitric acid is analyzed. It is found

that 28.3 mL of 1.50-M aqueous sodium hydroxide is required to
neutralize 25.0 mL of the nitric acid solution. What is the molarity of the
nitric acid solution?

108. In a titration experiment, 85.3 mL of 2.00-M aq. KOH is required to

neutralize 20.0 mL of a hydrochloric acid solution. What is the molarity

789
 of the HCl solution?

109. In a titration experiment, two drops of phenolphthalein are added to 25.0

mL of aq. NaOH solution, causing it to turn bright pink. A solution of
3.05-M hydrochloric acid is added to the solution. It is found that 5.2 mL
of the acid solution is required to turn the solution colorless. What is the
concentration of the original base solution?

110. A solution of cesium hydroxide is analyzed by titration. It is found that

0.0824 L of 1.50-M aqueous nitric acid is required to neutralize 0.0250 L
of the base solution. What is the concentration of the original base
solution?

111. In a titration experiment, 12.42 mL of a 0.105-M aq. sodium hydroxide

solution is required to neutralize a 5.00-mL sample of sulfuric acid. Write
a balanced equation for this neutralization reaction. What is the
concentration of the sulfuric acid solution?

112. In a titration experiment, 15.3 mL of a 0.010-M hydrochloric acid

solution is required to neutralize a 0.500-L solution of barium hydroxide.
Write a balanced equation for this neutralization reaction. What is the
concentration of the barium hydroxide solution?

113. What volume of 5.0-M aq. NaOH would be required to completely

neutralize 0.100 liters of 5.32-M aq. HI solution?

114. What volume of 1.4-M aq. KOH would be required to completely

neutralize 0.500 liters of 4.23-M aq. H2SO4 solution?

115. Describe the color of the indicator in each of these solutions:

a. litmus in an acidic solution
b. phenolphthalein in a solution where [OH–] > [H3O+]
c. phenolphthalein in a solution where [H+] = 0.0043 M
d. litmus in a solution with a pH of 3

116. Describe the color of the indicator in each of these solutions:

a. litmus in a basic solution
b. phenolphthalein in a solution where [OH–] < [H3O+]
c. phenolphthalein in a solution where [H+] = 3.2 × 10–11 M

790
 d. litmus in a solution with a pH of 9.4

117. A drop of phenolphthalein is added to a solution of 0.042-M HBr with a

volume of 10.0 mL. The solution is mixed with 11.7 mL of 0.70-M
NaOH. After the reaction is complete, is the solution acidic, basic, or
neutral? What color is the indicator?

118. If 4.0 mL of 0.014-M aq. HBr is reacted with 1.0 mL of 0.010-M

Ba(OH)2, will the solution be acidic, basic, or neutral? How many moles
of the excess product will remain?

12.6 Buffers and Biological pH

119. What are buffers?

120. What are the two components of a buffer solution? How do these
components neutralize excess acid and base?

121. What buffer system is found inside animal cells? What are the acid and
(conjugate) base species in this buffer?

122. What buffer system is used to regulate the pH of the bloodstream? What
are the acid and (conjugate) base species in this buffer?

123. A mixture of acetic acid and sodium acetate makes an effective buffer.
Here is the equilibrium equation for these two species:
HC2H3O2 (aq)+H2O (l)⇌C2H3O2− (aq)+H3O+ (aq)

a. What reaction would take place if acid were added to this buffer?
b. What reaction would take place if base were added to this buffer?

124. The carbonic acid/bicarbonate buffer that is found in the bloodstream is

described by this equation:
H2CO3 (aq)+H2O (l)⇌HCO3− (aq)+H3O+ (aq)

a. What reaction would take place if acid were added to this buffer?
b. What reaction would take place if base were added to this buffer?

791
Challenge Questions

125. If 100.0 mL of 1.21-M HCl is reacted with 50.0 mL of 1.06-M NaOH,

how many moles of HCl solution will remain unreacted? What will be the
H+ concentration in the resulting solution? What will be the pH of the
resulting solution?

126. Albuterol is commonly used in breathing treatments. Albuterol can be

converted from the free-base form to the acid salt form by reacting it with
HCl, as shown in the reaction below. What volume of 0.100-M HCl
would be required to completely react with 500.0 grams of albuterol?

792
Chapter Thirteen
Reaction Rates and Equilibrium

The Haber-Bosch Process

Before 1800, the world population was less than 1 billion. But the dawn of the
nineteenth century was a turning point: Advances in science and medicine reduced
infant mortality rates and increased life spans. More people were being born, and
fewer were dying young. The population began to rise. Today the world has seven
times as many people, and the number continues to climb (Figure 13.1).

Figure 13.1 (a) Today there are far more people in the world than ever before. (b) This
graph shows the dramatic jump in the world’s population since 1800. (c) Ammonia
fertilizer enhances crop growth. (d) Farmers today produce much more food per acre
than ever before. (e) Fritz Haber developed the process used to produce ammonia.

793
 For many, the population jump was a nightmare scenario: How could the world
grow enough food to keep up with the population? Would there be mass
starvation? Would the wars of the future be clashes over food?
Fortunately, this scenario has not come to pass, largely due to a technological
breakthrough that changed agriculture and multiplied the world’s food-production
capability. Today it is estimated that a third of the world’s population has food
because of this technology. The breakthrough, called the Haber-Bosch process,
enabled mass production of nitrogen fertilizer.
Nitrogen is the fourth most common element in your body (after carbon,
hydrogen, and oxygen). Nitrogen atoms are in every bit of muscle tissue and every
fragment of DNA. We get nitrogen from our diet, which means that ultimately it
comes from plant life. Despite the abundance of nitrogen in the atmosphere, N2 gas
is unreactive, and plants can’t absorb it directly. Rather, they draw nitrogen from
the soil, where tiny but vital microbes perform a process called nitrogen fixation—
converting atmospheric nitrogen into usable ammonia (or its conjugate acid, the
ammonium ion). The problem is, microbes produce a limited amount of “fixed”
nitrogen, and the amount of nitrogen in the soil limits how well crops can grow.
In the early 1900s, a German scientist named Fritz Haber tackled the problem
of producing ammonia from nitrogen gas. In theory, the reaction should be easy—
combine nitrogen with hydrogen to form ammonia:

N2 (g)+3 H2 (g)→2 NH3 (g)

In practice, it was much harder. Even though ammonia is a stable product, the
reaction doesn’t occur at room temperature. And when heated, the reaction
produces only a tiny amount of ammonia, leaving mostly unreacted nitrogen and
hydrogen. Why wouldn’t this reaction work?
Haber analyzed the changes in energy that controlled this reaction. Further, he
recognized that this was an equilibrium—a reaction that occurred in both the
forward and reverse directions. By analyzing these two closely related concepts,
energy and equilibrium, he developed a new process for producing large amounts
of ammonia.
Haber’s process changed the face of agriculture. Farmers today can grow four
times more food per acre than was possible in 1900, making them better able to
deal with the needs of a growing population. It has been estimated that 80% of the
nitrogen in our bodies was once ammonia, produced through the Haber-Bosch
process.
In this chapter, we’re going to look at the concepts that were essential to
unlocking the nitrogen problem. We’ll see how energy changes drive chemical
reactions to take place and determine how quickly these changes occur. Building
on this concept, we’ll turn our attention to equilibrium reactions. We’ll describe
equilibria conceptually as well as mathematically. Finally, we’ll see how Haber
was able to overcome the nitrogen problem and impact the world so profoundly.

794
 Intended Learning Outcomes
After completing this chapter and working the practice problems, you should be
able to:

13.1 Reaction Rates

Describe the effects of concentration and temperature on reaction rates.
Describe the energy changes that accompany chemical reactions using a
reaction diagram.
Describe the relationship between activation energy and rate of reaction.

13.2 Equilibrium Reactions

Describe equilibrium in terms of the rates of opposite reactions.
Given a reaction energy diagram, predict whether an equilibrium will favor
the reactants or the products.

13.3 Equilibrium Expressions

Write equilibrium expressions for equilibria involving solutions, solids, and
gases.
Relate the magnitude of the equilibrium constant, K, to equilibrium
concentrations.

13.4 Le Chatelier’s Principle

Apply Le Chatelier’s principle to describe how the addition or removal of
reactants or products affects an equilibrium.
Describe the effect of temperature and pressure changes on an equilibrium.

795
13.1 Reaction Rates
Some reactions, like the explosion of rocket fuel, happen very quickly.
Other reactions, like the gradual rusting of iron, occur much more slowly
(Figure 13.2). The speed of a chemical reaction is called its reaction rate.
To understand why reactions take place at different rates, we must look
more carefully at how they occur.

Figure 13.2 Different chemical reactions occur at different rates: Some are very fast
while others take place much more slowly..

Chemical reactions occur when atoms or molecules collide. For

example, consider the reaction of hydrofluoric acid with water to produce
hydronium and fluoride ions:
HF (aq)+H2O (l)→F− (aq)+H3O+ (aq)

For this reaction to happen, the HF and H2O molecules must collide
with each other. However, not every collision produces a reaction. The
molecules must collide with enough energy and at an orientation that
allows the rearrangement to take place (Figure 13.3).

796
Figure 13.3 Chemical reactions occur when molecules collide. For a reaction to take
place, the molecules must be oriented correctly relative to each other.

Explore
Figure 13.3

Three major factors that affect how collisions occur and the rates of
chemical reactions are concentration, temperature, and energy changes.
We’ll discuss each of these in the sections that follow.

797
 When a bat hits a baseball, the result depends on both the angle and the energy
at which the bat and ball collide. In much the same way, the angle and energy of
molecular collisions determine the outcome. Not every hit is a home run, and
not every collision results in a reaction.

Reactions occur when atoms or molecules collide.

How Concentration and Temperature Affect Reaction Rates

Because chemical reactions require the collision of molecules, the rate of
reactions depends in part on how often collisions occur. When molecules
collide more frequently, reactions occur at a faster rate.
One way to increase the number of collisions is to increase the
concentration of reactants. We can see this experimentally by reacting a
metal (such as zinc) with hydrochloric acid (Figure 13.4). The dilute acid
reacts slowly with the zinc, gradually forming a few bubbles of hydrogen
gas. By contrast, the concentrated acid reacts violently, quickly producing
a large volume of hydrogen gas.

Increasing concentration increases the reaction rate.

798
Figure 13.4 Increasing the concentration increases the reaction rate. Zinc reacts slowly
with 1 M HCl, but quickly with 6 M HCl.

Explore
Figure 13.4

A second way to increase the rate of a reaction is to increase the

temperature. Recall that at higher temperatures, molecules move more
quickly. Whether the reaction takes place in a solid, liquid, or gas, this
means that the molecules collide more frequently (Figure 13.5). It also
means that when particles do collide, they do so more forcefully, making
the molecules more likely to rearrange.

Figure 13.5 Increasing the temperature increases the velocity of the molecules in a gas.
The faster the molecules move, the more often they collide.

Explore
Figure 13.5

How Changes in Energy Affect Reaction Rates

Energy changes also affect reaction rates. Recall that chemical reactions
can be endothermic (absorbing heat energy from their surroundings) or
exothermic (releasing heat energy to their surroundings). At first glance,
the relationship between energy change and reaction rate may seem
unclear. For example, many exothermic reactions happen very quickly, but
others are much slower (Figure 13.6). How can we explain this?

Increasing the temperature increases the reaction rate.

799
Figure 13.6 Reaction rates vary for different compounds. (a) The reaction between
sodium and chlorine is exothermic. The two elements react immediately and violently.
(b) The reaction of wood with oxygen is also exothermic, but the reaction takes place
very slowly unless heat is added first.

Keeping food in the refrigerator slows down the chemical reactions that cause
food to spoil.

Describing Energy Changes in Chemistry: Reaction Energy Diagrams

Although reaction rates are intimately connected to energy changes, the
relationship is more nuanced than simple “endothermic” or “exothermic”
descriptions. To understand this relationship, it’s helpful to map the
energetic factors that make reactions happen (or prevent them from
happening). This type of map, called a reaction energy diagram, shows
the energetic changes that accompany a chemical reaction.
For example, Figure 13.7 shows a reaction energy diagram for the
reaction of hydrochloric acid with hydroxide:

800
 HCl (aq)+OH− (aq)→Cl− (aq)+H2O (l) ΔH°=−55.8 KJ/mol

Figure 13.7 A reaction energy diagram shows energy changes as a reaction progresses.

Explore
Figure 13.7

We know that the reaction is exothermic, as indicated by the negative

enthalpy value. But does this reaction also require energy for the changes
to occur? Let’s think carefully about the reaction that takes place. For the
hydroxide to take the hydrogen away from the chloride, there must be
some instant when the hydrogen ion is pulled between the OH and the Cl,
but it is not fully bonded to either. This instant, called the transition state,
is higher in energy than either the reactants or the products.

The transition state is the highest-energy arrangement of atoms that

occurs during a chemical reaction. This arrangement lasts only an instant,
but it determines the energy “hill” that a reaction must get over. This
energy barrier is called the activation energy. Along with concentration,
the activation energy determines how quickly a reaction occurs. Even if a
reaction will form a more stable product, if the activation energy is too

801
high, the reaction will not take place.

Remember your first date? Or your first breakup? Even if it gets you to a more
stable place, making or breaking bonds always involves an uncomfortable, high-
energy transition.

This idea also explains why many reactions require energy to get
started. For example, the combustion of charcoal with oxygen is
exothermic, but it doesn’t happen until you add a match—and probably
some lighter fluid (Figure 13.8). The match provides the activation
energy, in the form of heat, to initiate the reaction. After that, the heat
released by the combustion enables the reaction to continue.

Figure 13.8 The match provides enough heat to get over the activation energy and
initiate the exothermic reaction.

As a rule of thumb, reactions with an activation energy less than about

40 kJ/mol (or 10 kcal/mol) occur at room temperature. Reactions with a
higher activation energy require heating before they take place.

Reaction rates depend on the concentration and the activation energy.

802
 Have you ever wanted to start a business? If so, you’ve probably thought about
the initial investment. For example, to open a restaurant, you first need to build
or rent a building, secure permits, hire staff, and develop a menu. The cost to do
this is the activation energy for getting the business started. Regardless of how
lucrative the business could be, if the resources are not available to get over the
activation barrier, the business cannot get started. In the same way, even
exothermic reactions don’t take place if the activation energy is too high.

Catalysts
A catalyst is a species that is not part of the balanced equation, but causes
the reaction to go more quickly. A catalyst stabilizes the transition state in
the reaction. This reduces the activation energy required to go from
reactant to product. As a result, reactions are able to go more quickly
(Figure 13.9).

Figure 13.9 A catalyst reduces the activation energy, causing the reaction to occur more
quickly.

803
 Catalysts are critically important in a wide variety of settings. In living
creatures, compounds called enzymes act as catalysts to make complex
reactions proceed quickly and with minimal energy cost. Manufacturers
use catalysts to reduce the time, energy, and cost necessary to produce
goods.

Example 13.1 Interpreting Reaction Energy Diagrams

Consider a simple reaction where compound A is converted to compound
B. Based on the energy diagram shown, is this reaction endothermic or
exothermic? Find both the activation energy and the net energy change for
this reaction.

The reaction diagram shows the energy of the substances involved. Because
the energy of the product is higher than the energy of the starting material,
the reaction is endothermic.
The activation energy is the energy barrier that the reaction must get over in
order to take place. The starting material has an energy of 40 kJ/mol, and
the transition state has an energy of 140 kJ/mol. Therefore,
Activation energy=140 kJ/mol−40 kJ/mol=100 kJ/mol

Because the activation energy is greater than 40 kJ/mol, this reaction will
not take place without heating.
The net energy change for this reaction is the difference in energy between
the starting material and product. •

Net energy change=Eproduct−Ereactant=120 kJ/mol−40 kJ/mol=80 kJ/mol

804
 IT
TRY

1. Compare the two reactions diagrammed below. What is the net energy
change for each reaction? Which one is more exothermic? Which
reaction will occur more quickly?

805
13.2 Equilibrium Reactions
Many reactions are reversible—that is, they can occur in both the forward
and the reverse directions. For example, consider how hydrofluoric acid
reacts in water:

HF (aq)⇌H+ (aq)+F− (aq)

The reaction can go in the forward direction (the acid ionizes to form
H+ and F–), but it can also go in the backward direction (the ions combine
to reform HF). This type of back-and-forth reaction is called an
equilibrium reaction. In a chemical equation, we use a double arrow (⇌)
to show that the reaction proceeds in both directions.

At equilibrium, the forward and reverse reactions occur at the same rate, so the
concentrations of reactants and products do not change.

Consider what happens if pure hydrogen fluoride (HF) dissolves in

water: For the first instant, there is a large concentration of HF, but
essentially no H+ or F– ions. But as the HF molecules begin to dissociate,
the concentration of H+ and F– ions increases (Figure 13.10).

Figure 13.10 This graph shows how the concentrations of reactants and products
change in an equilibrium reaction. Once a reaction reaches equilibrium, the
concentrations of reactants and products no longer change.

As their concentrations increase, these ions begin to encounter each

806
other in solution, and they react to form HF. Although HF molecules
continue to ionize, the reverse reaction is also taking place. Soon the
amount of each species stabilizes. Both the forward and reverse reactions
are taking place, but because they are happening at the same rate, the
amounts of each component are not changing. This situation is called
equilibrium.
At equilibrium, the rates of the forward and reverse reactions are the same.

Imagine two towns, A and B, that are isolated from all other towns. If the towns
are in equilibrium, the number of people who are moving in each direction is the
same, so the overall populations of the towns don’t change.

Some equilibrium reactions favor the reactants, while others favor the
products. For example, recall that hydrofluoric acid, HF, is a weak acid.
This means that at equilibrium, there is much more of the reactant (HF)
than there is of the products (H+ and F–):
HF (aq)⇌H+ (aq)+F− (aq)98%2%

We can analyze equilibrium reactions like this one by using reaction

energy diagrams (Figure 13.11). In this reaction, the activation energy hill
is small enough that the reactants can get over the hill to form products,
but the products can also get over the hill in the reverse direction.

807
Figure 13.11 In an equilibrium reaction, the reaction proceeds in both the forward and
reverse directions.

In this energy diagram, the reactants are lower in energy than the
products. In general, the side of an equilibrium that is lower in energy will
have a greater concentration, and the side that is higher in energy will
have a smaller concentration (Figure 13.12). This is why in this reaction,
98% of the HF exists in the acid form, and only 2% exists as the conjugate
base (F–).

Figure 13.12 At equilibrium there is a larger concentration of the lower energy species,
but the rates of the forward and reverse reactions are the same.

Remember, the rate of a reaction depends on both the concentration

and the activation energy. The low-energy side of an equilibrium has a
greater concentration but a higher activation energy. The high-energy side

808
has a smaller concentration but a lower activation energy. Together, the
balance of concentration and activation energy causes the forward and
reverse reactions to occur at the same rate.
In general, equilibria favor the lower-energy species.

In the English football system, the top 20 teams play in the Premier League.
Below this league is the 72-team Football League. After each season, the top
three teams in the lower league are promoted to the higher league, and the
bottom three teams in the higher league are relegated to the lower league. This is
an example of an equilibrium: Because the rates in both directions are the same,
the number of teams in each league doesn’t change. And much like the Premier
League, which has fewer teams, the higher energy levels in a chemical
equilibrium have a lower concentration of species.

Example 13.2 Energy Diagrams and the Position of an

Equilibrium
The equilibrium reaction and energy diagram given below represent the
ionization of acetic acid, HC2H3O2. At equilibrium, is there a higher
concentration of reactants (left) or products (right) in solution?

809
Notice in the figure that the potential energy is lower for the reactants than
for the products. Because of this, the equilibrium favors the reactants. •

Example 13.3 Finding Activation Energies in the Forward and

Reverse Directions
The energy diagram for an equilibrium reaction is shown here. What is the
activation energy for the forward reaction? What is the activation energy
for the reverse reaction? Will this equilibrium favor the reactants or the
products?

In this reaction, the reactants have an energy of 50 kJ/mol The transition

state has an energy of 60 kJ/mol This means that in the forward reaction,
the activation energy is 10 kJ/mol:
Activation energy (forward)=Energy (transition state)−Energy (reactants)=60
kJ/mol−50 kJ/mol=10 kJ/mol

810
 The products have an energy of 20 kJ/mol Again, the transition state has an
energy of 60 kJ/mol In the reverse reaction, the activation energy is 40
kJ/mol
Activation energy (reverse)=Energy (transition state)−Energy (products)=60
kJ/mol−20 kJ/mol=40 kJ/mol

Because the products are lower in energy than the reactants, there will be a
higher concentration of products than reactants at equilibrium. •

IT
TRY

2. The following equilibrium is slightly endothermic:

C6H5OH (aq)⇌H+ (aq)+C6H5O− (aq)

At equilibrium, is there a higher concentration of reactants or of

products?

3. A chemical reaction has the energy profile shown below. For this
reaction, what is the energy of activation in the forward reaction?
What is the energy of activation in the reverse reaction? At room
temperature, will the reactants and products reach an equilibrium?

811
812
13.3 Equilibrium Expressions
Some equilibria strongly favor the products, while others strongly favor
the reactants. For example, hydrochloric acid (HCl) is a strong acid that
almost completely ionizes in aqueous solution. On the other hand, HF is a
weak acid; only about 2% of HF ionizes in aqueous solution.

When describing equilibria, chemists often need to describe how far to

the left or the right an equilibrium lies. To do this we write an equation
called an equilibrium expression, which describes the balance between
the reactants and products in an equilibrium. To see how equilibrium
expressions work, let’s begin with a generic chemical equation, where A
and B are reactants, C and D are products, and the lowercase letters are the
coefficients for each species.

aA+bB⇌cC+dD
For this equation, we write the equilibrium expression as follows:

K=[C]c [D]d[A]a [B]b

Put simply, equilibrium expressions show the concentrations of the products

over the reactants:

K=productsreactants

Recall that the square brackets around each species mean the
concentration of that species. For solutions, we express this as molarity
(moles/L). In this equation, K is the equilibrium constant for the reaction.
The value of K is very important because it tells us whether the
equilibrium favors the reactants (left-hand side) or the products (right-

813
hand side).
[A] means “concentration of A.”

For example, the ionization of HCl strongly favors the products, so the
equilibrium constant has a large value. (In fact, this equilibrium favors the
products so completely that it is normally written with just a single arrow.)
On the other hand, the K value for the HF equilibrium is small, since the
equilibrium favors the starting materials (Figure 13.13). The equilibrium
constant does not have units.

Figure 13.13 Equilibria that favor the products have very large K values; those that
favor the reactants have small K values.

If K is large, the equilibrium favors the products. If K is small, the

equilibrium favors the reactants.

Examples 13.4 and 13.5 illustrate the use of equilibrium expressions.

The equilibrium constant shows the balance between the left and right side of an
equation.

814
Example 13.4 Writing an Equilibrium Expression
When heated, molecular bromine breaks to form individual bromine atoms.
In the reverse reaction, these atoms recombine to form molecular bromine.
Write an equilibrium expression for this reaction:

Br2⇌2 Br
When writing an equilibrium expression, we place the products in the
numerator and the reactants in the denominator. The concentration of each
species is raised to the power of its coefficient. We therefore write the
equilibrium expression as follows:

K=[Br]2[Br2]
In this example, we don’t know the value of the equilibrium constant, K. To
find it, we must know the concentrations of each component in the
equilibrium. Example 13.5 illustrates this process. •

Example 13.5 Finding the Value of the Equilibrium Constant

Nitrous acid ionizes in water, as shown in the equilibrium reaction below.
At equilibrium, aqueous nitrous acid has these concentrations: [H+] =
0.0140 M; [NO2–] = 0.0140 M; and [HNO2] = 0.493 M. What is the value
of K for this equilibrium? Does this equilibrium lie to the right or the left?
HNO2 (aq)⇌H+ (aq)+NO2− (aq)

First, notice that the equilibrium concentration of HNO2 is much higher

than the concentrations of the products. Therefore the equilibrium lies to
the left, and we can expect the value of K to be small.
To find the value of K, we set up the equilibrium expression and solve:
K=[H+][NO2−][HNO2]=(0.0140 M)(0.0140 M)0.493 M=3.98×10−4

815
 Indeed, we find a very small value of K, indicating that the equilibrium lies
far to the left. Notice that the value of K does not have units. •

IT
TRY

4. Write an equilibrium expression for these reactions:

AlCl3 (aq)+Cl− (aq)⇌AlCl4− (aq)Zn2+ (aq)+4 NH3 (aq)⇌[Zn(NH3)4]2+ (aq)

5. The ionization of acetic acid is shown in the equilibrium below. In

water, a solution of acetic acid is found to have these concentrations:
[H+] = 0.0040 M; [C2H3O2–] = 0.0040 M; and [HC2H3O2] = 0.889
M. What is the value of K for this equilibrium? Is this a strong or a
weak acid?
HC2H3O2 (aq)⇌H+ (aq)+C2H3O2− (aq)

Equilibrium Expressions Involving Solvents

In some situations, special rules apply for writing equilibrium expressions.
The three most important of these situations occur when an equilibrium
involves a solvent, a solid, or gases.
When a chemical equilibrium involves a pure liquid or a solvent, this
species is not included in the equilibrium expression. Because there is
typically so much more of the solvent than there is of the other species, the
concentration of the solvent doesn’t change significantly. For this reason,
we do not include the solvent in the equilibrium expression. The
concentration of the solvent is considered to be part of the equilibrium
constant, K. For example, we can write the ionization of HF in water in
two different ways:

816
 HF (aq)⇌H+ (aq)+F− (aq) K=[H+][F−][HF]

or
HF (aq)+H2O (l)⇌H3O+ (aq)+F− (aq) K=[H3O+][F−][HF]

Solvents are not included in equilibrium expressions.

Because water is the solvent, we don’t include the concentration of

water in the expression; so, no matter which way we decide to write the
expression, the value of K is the same. This type of expression is
particularly important for describing the behavior of acids and bases.

Example 13.6 Writing Equilibrium Expressions When a Solvent

Is Involved
When ammonia dissolves in water, it produces small amounts of hydroxide
and ammonium ions through the equilibrium shown below. Write an
equilibrium expression for this reaction:
NH3 (aq)+H2O (l)⇌OH− (aq)+NH4+ (aq)

Because water is the solvent in this reaction, it is not included in the

equilibrium expression. •

K=[OH−][NH4+][NH3]

Example 13.7 Finding Concentrations of an Aqueous Acid

Hydrocyanic acid dissolves in water to produce H3O+ and CN– as in the

817
 equilibrium below. The equilibrium constant for this reaction is 4.9 × 10–10
If [H3O+ and [CN–] both equal 7.3 × 10–6 M, what is the concentration of
HCN in the solution?
HCN (aq)+H2O (l)⇌H3O+ (aq)+CN− (aq)

As before, we do not include water in the equilibrium expression:

K=[H3O+][CN−][HCN]

We know the values of K, [H3O+ and [CN–]. We therefore rearrange the

equation to solve for [HCN] and then substitute the values given above.
From this, we find the concentration of HCN. •

[HCN]=[H3O+][CN−]K=(7.3×10−6 M)(7.3×10−6 M)4.9×10−10=0.11 M

IT
TRY

6. The following equilibria take place in water. Write equilibrium

expressions for each one.
a. CrO4− (aq)+H2O (l)⇌HCrO4 (aq)+OH− (aq)

b. 2 H2O (l)⇌H3O+ (aq)+OH− (aq)

c. NH4+ (aq)+H2O (l)⇌NH3 (aq)+H3O+ (aq)

7. Buffer solutions contain a mixture of a weak acid and their conjugate

base. A chemist prepares an acetate buffer solution where [HC2H3O2]
= 0.100 M and [C2H3O2–] = 0.200 M at equilibrium. Based on the
equilibrium data below, what is the concentration of H3O+in this

818
 solution?
HC2H3O2 (aq)+H2O (l)⇌H3O+ (aq)+C2H3O2− (aq) K=1.8×10−5

Equilibrium Expressions Involving Solids

Equilibria involving solids are very common. For example, think about
making a pitcher of sweet tea. A modest amount of sugar dissolves in the
tea. But if we add too much sugar, we reach a saturation point where no
more sugar can dissolve. When this happens, the undissolved sugar settles
in the bottom of the pitcher (Figure 13.14).

Figure 13.14 Once a pitcher of sweet tea is saturated, adding more sugar does not
change the sweetness. The amount of solid present does not affect the equilibrium..

In this delicious situation, the sugar reaches an equilibrium with the

aqueous solution. Some sugar is in the solid form, and some is dissolved.
The individual sugar molecules are able to move in and out of solution:
C12H22O11 (s)⇌C12H22O11 (aq)

Think about this for a moment: Once the solution is saturated, can we
make the tea sweeter by adding more sugar? No. If we add more sugar, it
simply falls to the bottom of the pitcher. The presence of additional solid
sugar does not change the concentration of dissolved sugar.
This leads us to an important general rule: When equilibria involve
solids, the amount of solid present doesn’t affect the concentration of the
other components. Because of this, materials in the solid phase are not
included in equilibrium expressions.

819
 Solids are not included in equilibrium expressions.

Solubility Products
We often encounter equilibria involving solids in the laboratory. For
example, suppose a chemist mixes solid calcium hydroxide with water.
Some of the solid dissolves, but not all of it. Once the solution reaches
equilibrium, the excess solid sits on the bottom of the container (Figure
13.15). We can write this equilibrium as follows:
Ca(OH)2 (s)⇌Ca2+ (aq)+2 OH− (aq)

Figure 13.15 This saturated solution has undissolved solid at the bottom..

How would we write the equilibrium expression for this equation?

Since Ca(OH)2 is a solid, we do not include it in the expression; instead,
we put a value of one in the denominator:
K=[products][reactants]=[Ca2+][OH−]21=[Ca2+][OH−]2

In this expression, the equilibrium constant, K, tells us the amount of

calcium and hydroxide ions that can dissolve in solution. The equilibrium
constant for the solution of an ionic compound is called a solubility
product. These constants are often written as Ksp.

820
 As with other equilibrium expressions, a small Ksp value means the
equilibrium lies far to the left. The smaller the Ksp value, the less soluble
the compound is. The solubility products for several common compounds
are shown in Table 13.1.

TABLE 13.1 Solubility Products for Some Ionic Compounds at 25 °C

Silver Halides Ksp Lead(II) Halides Ksp

AgCl 1.77 × 10–10 PbCl2 1.70 × 10–5

AgBr 5.35 × 10–13 PbBr2 6.60 × 10–6

AgI 8.52 × 10–17 PbI2 9.8 × 10–9

Hydroxides Ksp Carbonates Ksp

Ca(OH)2 5.02 × 10–6 BaCO3 2.58 × 10–9

Mg(OH)2 5.61 × 10–12 FeCO3 3.13 × 10–11

Fe(OH)2 4.87 × 10–17 MgCO3 6.82 × 10–6

Fe(OH)3 2.79 × 10–39 ZnCO3 1.46 × 10–10

Solubility products are normally used only for slightly soluble

compounds. More soluble compounds are typically described by their
solubility in moles/liter or in grams/liter.

In Chapter 6 we used solubility rules to broadly classify compounds as “soluble”

and “insoluble.” While these rules are helpful, many compounds are very
slightly soluble. The solubility product allows us to predict the composition of
solutions much more precisely than if we use the broad solubility rules.

Example 13.8 Determining the Solubility Product

In a saturated solution of manganese(II) carbonate at 25 °C both [Mn2+]
and [CO32 – are equal to 4.73 × 10–6 M. Write an equilibrium expression
and calculate the value of Ksp for this compound.
We begin by writing the solubility equilibrium for this compound:
MnCO3 (s)⇌Mn2+ (aq)+CO32− (aq)

821
Based on this, we can write the following equilibrium expression:

Ksp=[Mn2+][CO32−]
Substituting in the concentrations for each ion, we obtain a Ksp value of
1.91 × 10–11. •

Ksp=[Mn2+][CO32−]=(4.37×10−6 M)(4.37×10−6 M)=1.91×10−11

Example 13.9 Determining the Concentration from the

Solubility Product
The solubility product of barium sulfate is 1.08 × 10–10. If barium sulfate is
dissolved in pure water, what is the concentration of barium ions?
We begin by writing the solubility equilibrium for this compound:
BaSO4 (s)⇌Ba2+ (aq)+SO42− (aq)

When it dissociates, each unit of barium sulfate produces one barium ion
and one sulfate ion. Because of this, the concentrations of the two ions must
be the same. If we call the unknown concentration x, we can say that

X=[Ba2+]=[SO42−]
Now let’s substitute x for the concentrations of the two ions, and substitute
the value of Ksp given in the problem. Taking the square root of both sides,
we find that x = 1.04 × 10–5 Therefore, [Ba2+] = 1.04 × 10–5 M. •

Ksp=[Ba2+][SO42−]1.08×10−10=X21.08×10−10=X1.04×10−5 M=X=[Ba2+]

822
 IT
TRY

8. Write an equilibrium expression for each of these solubility equilibria:

a. NaCl (s)⇌Na+ (aq)+Cl− (aq)

b. MgBr2 (s)⇌Mg2+ (aq)+2 Br− (aq)

c. Fe(OH)3 (s)⇌Fe3+ (aq)+3 OH− (aq)

9. The solubility products for three compounds are shown below. Based
on the Ksp values given, which compound is most soluble? Which
compound is least soluble?
a. BaSO4 (s)⇌Ba2+ (aq)+SO42− (aq) Ksp=1.1×10−10

b. LiF (s)⇌Li+ (aq)+F− (aq) Ksp=3.8×10−3

c. Ni(OH)2 (s)⇌Ni2+ (aq)+2 OH− (aq) Ksp = 2.0 × 10−15

Equilibrium Expressions Involving Gases

Many equilibria involve gases. Equilibrium expressions for gases are
identical to those for aqueous solutions, except that we use the partial
pressures of the gases rather than molar concentrations. For example, at
high temperatures, Br2 and Cl2 gases react to form BrCl. We write the
equilibrium and the equilibrium expression for this reaction as shown
below, where PBr2, PCl2 and PBrCl represent the partial pressure from each
gas:
Br2 (g)+Cl2 (g)⇌2 BrCl (g) K=PBrCl2PBr2PCl2

823
 Example 13.10 Finding Gas Pressures Using an Equilibrium
Expression
A common method of manufacturing hydrogen gas involves heating a
mixture of methane and steam to produce hydrogen and carbon monoxide,
as shown in the equilibrium below. Given the following equilibrium data,
what is the partial pressure of CO in an equilibrium mixture if PCH4=1.0

atm, PH2O=1.0 atm,

and PH2=0.10 atm?

CH4 (g)+H2O (g)⇌3 H2 (g)+CO (g) K=1.5×10−6

To solve this problem, we write the equilibrium expression using partial

pressures:

K=PH23PCOPCH4PH2O
To solve for the pressure of CO, we rearrange the equation and substitute
the values given above. Under these conditions, we find that 1.5 × 10–3 atm
of CO are present. •
PCO=KPCH4PH2OPH23=(1.5×10−6)(1.0 atm)(1.0 atm)(0.10
atm)3=1.5×10−3 atm

IT
TRY

10. Ammonia is an important fertilizer, cleaning agent, and raw material

for many nitrogen-containing products. Ammonia is typically
prepared through the gas-phase reaction of nitrogen and hydrogen.
N2 (g)+3 H2 (g)⇌2 NH3 (g)

At 600 °C and high pressure, the equilibrium constant, K, for this

824
process is 1.5 × 10–3 If a gas mixture at equilibrium contains H2 with
a partial pressure of 10.0 atmospheres and N2 with a partial pressure
of 83.0 atmospheres, find the partial pressure of ammonia in the
mixture.

825
13.4 Le Chatelier’s Principle
At the beginning of this chapter, we introduced ammonia (NH3) a
compound of vital importance to the world’s food supply. Above about
400 °C, ammonia forms by the reversible combination of nitrogen and
hydrogen gas:

N2 (g)+3 H2 (g)⇌2 NH3 (g)

However, there is a problem. At higher temperatures, the equilibrium
for this reaction strongly favors the starting materials. This raises a crucial
question: Is it possible to control an equilibrium reaction? Can we drive an
equilibrium to one side or another?
Fortunately, it is possible to do this. To understand how, let’s begin
with another simple equilibrium: Imagine two pools, connected by a pipe
running along the bottom (Figure 13.16). Because water can pass back
and forth between the pools, it is an equilibrium system. What happens if
we add more water to the pool on the left? Water flows to the right,
causing the level in the right pool to rise, until the pools level off. What
happens if we remove water from the left pool? The water will flow
toward the left pool, and again the pools will level off. You intuitively
know that at equilibrium, the height of the water in both pools will always
be the same.

Figure 13.16 What happens if we add water to one of the pools? What if we remove
water?.

826
 Explore

Figure 13.16

The same idea applies to chemical systems: When some change takes
place in concentration, temperature, pressure, or other key factor, the
equilibrium shifts to minimize that change, and a new equilibrium is
established. This idea is called Le Chatelier’s principle. In the sections
that follow, we’ll explore this important concept in detail.

Equilibrium and Concentration

Changes in concentration can push the equilibrium to the right or to the
left. For example, consider the equilibrium of HF with water:
HF (aq)+H2O (l)⇌H3O+ (aq)+F− (aq)

What happens if we add more HF to this mixture? The additional HF

reacts with the water, forming more F−, until the levels of HF and F– again
reach equilibrium. Much like the pool analogy, increasing the
concentration on one side also increases the concentration on the other
side. Similarly, decreasing the concentration also affects both sides.
Another useful example is the evaporation of water. At room
temperature, water interconverts between the liquid phase and the gas
phase:

827
 H2O (l)⇌H2O (g)
Imagine a sealed container, partially filled with water (Figure 13.17).
Inside the container are water molecules in both the liquid and gas phases.
The water molecules are constantly moving. Some molecules transition
from liquid to gas while others move from gas to liquid. At equilibrium,
the number of molecules moving in either direction is the same.

Figure 13.17 Water molecules can move back and forth between the liquid and gas
phases. (a) At equilibrium, the same number of molecules move from liquid to gas as
from gas to liquid. (b) If the lid is removed, the water vapors are pulled away from the
liquid. The system can’t reach equilibrium, and eventually all of the water evaporates.

But what if we take the cap off the container? This allows the
molecules in the vapor phase to drift away from the liquid, and the system
never reaches equilibrium. As you might expect, the water will continue to
evaporate until there is no liquid left.

Adding to an equilibrium pushes it toward the other side.

As a general rule, if we add something to an equilibrium, it pushes the

equilibrium toward the other side. On the other hand, if we remove
something, it pulls the equilibrium toward the side it was removed from.
The example of Le Chatelier’s swimming pool, shown again in Figure
13.18, is a good way to understand and remember this concept.

828
Figure 13.18 Le Chatelier’s principle says that adding to an equilibrium pushes it to the
other side, but removing from an equilibrium pulls it back.

Example 13.11 Predicting Equilibrium Changes Using Le

Chatelier’s Principle
A reaction vessel has the following compounds present in equilibrium:
CH3CO2H+CH3OH⇌CH3CO2CH3+H2O

a. Would adding water shift this equilibrium to the left or to the right?
b. What would be the effect of adding CH3OH to the mixture?
Any time we add something to an equilibrium, the reaction shifts toward
the other side until balance is restored. Adding water would push the
equilibrium away from water in the equation—that is, it would push the
equilibrium to the left. On the other hand, adding CH3OH would push the
equilibrium toward the right. •

IT
TRY

11. Carbonated beverages (such as soda pop) are packaged under

pressurized carbon dioxide. The CO2 reacts with water in an
equilibrium to produce carbonic acid (H2CO3) a weak acid that gives
these drinks their sharp flavor. After the soda pop is opened, the CO2
escapes. How does removing the CO2 affect the concentration of
H2CO3 in the soda?

CO2 (g)+H2O (l)⇌H2CO3 (aq)

829
Equilibrium and Temperature
Le Chatelier’s principle also applies to temperature changes. For example,
we could write the evaporation of water like this:

Heat+H2O (l)⇌H2O (g)

The evaporation of water is an endothermic change—heat is being added
in. Heating the water drives the equilibrium to the right, toward the gas.
On the other hand, removing heat (cooling the system down) makes the
equilibrium shift to the left, toward the liquid.
Another example of this principle is the gas-phase equilibrium reaction
between N2O4 and NO2. N2O4 is colorless. However, in the gas phase,
this compound forms an equilibrium with NO2, which is brown. The
reaction is endothermic. We can write the equilibrium like this:
Heat+N2O4 (g)⇌2 NO2 (g)colorlessbrown

Figure 13.19 shows the effects of temperature on this equilibrium. Heating

the mixture shifts the equilibrium toward the brown NO2. Cooling the
mixture shifts the equilibrium toward the colorless N2O4.

830
Figure 13.19 N2O4 and NO2 exist in equilibrium. The three containers shown hold the
same amount of these materials. Heating the mixture shifts the equilibrium toward the
brown NO2. Cooling the mixture shifts the equilibrium toward the colorless N2O4.

Explore
Figure 13.19

Heating shifts an equilibrium toward the higher-energy side.

Equilibrium and Pressure

The pressure of a chemical system can also affect the equilibrium.
Remember that gas molecules exert a pressure on their surroundings.
Increasing the amount of gas increases the pressure. For a change that
produces gas molecules, we can even include pressure in the balanced
equation, like this:

NH3 (l)⇌NH3 (g)+pressure

Again, Le Chatelier’s principle applies. Adding pressure shifts the
equilibrium toward the liquid. Removing pressure shifts the equilibrium
toward the gas. This idea has many practical applications. For example,
ammonia manufacturers use high pressure to force the ammonia into liquid
form, making it easier to store and transport (Figure 13.20).

831
Figure 13.20 Under high pressure, ammonia is a liquid. Decreasing the pressure shifts
the equilibrium toward the gas phase.

Fritz Haber used this important concept to drive the synthesis of

ammonia. Recall that ammonia forms in an equilibrium with nitrogen and
hydrogen:

The left-hand side of the equation contains four moles of gas, while the
right-hand side contains only two moles of gas. Haber realized that high
pressures shifted the equilibrium toward the side with fewer moles of gas
—that is, toward the ammonia. As a general rule, increasing the pressure
on a chemical system shifts the equilibrium toward the side with fewer
moles of gas.

High pressure shifts equilibria toward the side with fewer moles of gas.

The Ammonia Problem and the Haber-Bosch Process

N2 (g)+3 H2 (g)⇌2 NH3 (g)+heat

The synthesis of ammonia from nitrogen posed a significant challenge.

Although the reaction is exothermic, the activation energy is too high for the
reaction to take place at room temperature. Heating the mixture allows the
reaction to take place, but it shifts the equilibrium toward the starting material so
that very little product forms at high temperature.
Fritz Haber used three techniques to solve this problem. First, he used an
iron catalyst to lower the activation energy. Second, he ran the reaction at high
pressure. Because there are fewer moles of gas in the products than in the
reactants, high pressure pushes the equilibrium toward the product. Third, he
removed ammonia from the mixture as it formed to pull the reaction toward the

832
 formation of products. Today the manufacture of ammonia relies on all three of
these techniques.

Example 13.12 Predicting Equilibrium Changes Using Le

Chatelier’s Principle
Nitrosyl chloride, NOCl, is a toxic gas that forms in equilibrium with
nitrogen monoxide and chlorine gas, as shown in this equation:
2 NO (g)+Cl2 (g)⇌2 NOCl (g) ΔH=−75.54 kJ/mol

Based on this information, determine whether the following changes would

shift the equilibrium to the left or to the right:
a. adding Cl2 to the mixture
b. increasing the temperature
c. increasing the pressure
Any time we add something to an equilibrium, it pushes the equilibrium
toward the other side. Adding Cl2 will shift the equilibrium to the right, and
so the amount of NOCl will also increase.
Notice that the ΔH value for this reaction is negative. This means that the
reaction is exothermic. In other words, we could say that heat is a product
for this reaction:
2 NO (g)+Cl2 (g)⇌2 NOCl (g)+heat

Therefore, increasing the temperature for this reaction should drive the
equilibrium to the left.
Finally, notice that the reactants are three moles of gas while the products
are only two. Increasing the pressure will shift the equilibrium toward the
products because there are fewer moles of gas on that side. •

IT
TRY

12. The Haber-Bosch process for making ammonia from nitrogen and
hydrogen was introduced at the beginning of this chapter. Indicate
whether each of the following changes would increase or decrease
the amount of ammonia produced:

833
 N2 (g)+3 H2 (g)⇌2 NH3 (g)+heat

a. adding more N2 to the mixture

b. removing NH3 from the mixture
c. adding NH3 to the mixture
d. adding a catalyst
e. increasing the reaction temperature from 300 °C to 400 °C

834
Summary
In this chapter we’ve worked through many important ideas connecting
energy, reaction rates, and equilibrium reactions. The rate of a reaction
depends on the energy and frequency of molecular collisions. Reaction
rates increase as the concentration and temperature increase.
Reaction rates also depend on the specific energetic changes that
accompany a reaction. We commonly map these energy changes using
reaction energy diagrams. These diagrams typically show the energy
barrier, called the activation energy, that substances must overcome for a
reaction to occur. The rate of a reaction depends on the activation energy.
Catalysts are materials that lower the activation energy for a reaction,
enabling the reaction to happen more quickly and at lower temperature.
Equilibrium reactions can proceed in both the forward and reverse
directions. When a system is in equilibrium, the forward reaction and the
reverse reaction take place at the same rate.
We use equilibrium expressions to describe equilibrium reactions. The
equilibrium constant, K, describes how far to the left or right the
equilibrium lies. A large K indicates the products are favored; a small K
means the reactants are favored. When writing equilibrium expressions,
solvents, pure liquids, and solids are not included. Equilibrium expressions
may be written using concentrations, or for gas-phase reactions, using
partial pressures.
Le Chatelier’s principle states that when some change in concentration,
temperature, pressure, or other key factor takes place, the equilibrium
shifts to minimize that change, and a new equilibrium is established.
Chemists commonly use Le Chatelier’s principle to drive an equilibrium
reaction toward the formation of desired products.

835
Miracles and Monstrosities: The Brutal Ironies of
Fritz Haber

By nature, humans have the capacity for both goodness and evil. Most of us live
between the two extremes: We are neither saints nor sociopaths—our lives are
colored with love, hard work, and creative impulses, but marred with the wreckage
of past mistakes and smeared with selfish and destructive desires. As St. Paul
wrote in his letter to the early Christian church, “The good that I love I don’t do,
but the evil that I hate—that is exactly what I do!”
Seldom is the contrast between dark and bright as stark as it was in the life of
Fritz Haber (Figure 13.21). His process for ammonia production profoundly
increased the world’s food supply and earned him the 1918 Nobel Prize in
chemistry. But there’s more to the story.

Figure 13.21 Fritz Haber unlocked the secret of ammonia production. He is also
remembered as the father of chemical weapons.

Haber was at the peak of his career when World War I broke out in Europe. A
patriotic German, Haber eagerly applied his scientific ability to the war effort. It
was the era of trench warfare, and Germany was locked in prolonged and costly
trench battles. Haber proposed the use of gases to clear out enemy trenches.
Working closely with the German army, he supervised the implementation of
chlorine gas on the battlefield in April 1915 (Figure 13.22). His wife, who was
opposed to his work with chemical weapons, committed suicide a month later.
Despite this, Haber continued his efforts in chemical warfare, developing and

836
testing several other gases for use on the battlefield.

Figure 13.22 These World War I German soldiers wore gas masks to protect
themselves from exposure to chemical weapons.

After World War I ended, Haber resumed his work as director of the Kaiser
Wilhelm Institute for Physical Chemistry and Electrochemistry in Berlin. He
worked closely with great scientists in an era of extraordinary development. He
helped develop a model for understanding the stability of crystal lattices. He was a
good friend and neighbor of Albert Einstein.
Like Einstein, Haber was Jewish, and the rise of the Nazi party in the 1930s
brought an end to his tenure. In 1933 he fled Germany; three months later, he died.
In a final, horrible irony, one of the gas mixtures he had developed as an
insecticide was modified and used in the murders of millions of Jews in the
concentration camps, including several of Haber’s close relatives.
How should history remember Fritz Haber? He is the father of chemical
warfare, one of the horrors of the modern era. Yet his nitrogen fixation process has
helped feed billions of people and is among the most important scientific findings
of all time. His legacy is a testament to the strengths and weaknesses of human
nature, and a reminder that the power of science must be coupled with wisdom and
decency.

837
 Key Terms
13.1 Reaction Rates
reaction rate The speed at which a chemical reaction takes place.
reaction energy diagram A depiction of the energetic changes that
accompany a chemical reaction.
transition state The highest-energy arrangement of atoms that occurs during a
chemical reaction.
activation energy The energy barrier for a reaction; this energy determines
how quickly a reaction occurs.
catalyst A species that is not part of a balanced equation but causes a reaction
to go more quickly.

13.2 Equilibrium Reactions

equilibrium reaction A reaction that occurs in both the forward and reverse
directions.
equilibrium The state in which forward and reverse reactions take place at the
same rate, so the concentrations of reactants and products do not change.

13.3 Equilibrium Expressions

equilibrium expression An equation that describes the balance between
reactants and products in an equilibrium.
equilibrium constant (K) The ratio of products to reactants when a reaction is
at equilibrium.
solubility product (Ksp) The equilibrium constant for the solution of a slightly
soluble ionic compound.

13.4 Le Chatelier’s Principle

Le Chatelier’s principle When some change takes place in concentration,
temperature, pressure, or other key factor, the equilibrium shifts to minimize
that change, and a new equilibrium is established.

838
 Additional Problems

13.1 Reaction Rates

13. How do molecular collisions relate to the rates of chemical reactions?

14. Do all molecular collisions result in a chemical reaction? What other

factors play a role in determining whether a reaction takes place?

15. What are two steps a chemist can take to make a reaction go more slowly?

16. At temperatures above about 1300 °C, nitrogen and oxygen gas react
quickly to form nitrogen oxides, such as nitrogen monoxide. A simplified
reaction scheme can be written this way:

N2 (g)+O2 (g)→2 NO (g)

Why does this reaction occur at high temperature, but not at room
temperature? How does the elevated temperature affect the collisions that
occur between gas molecules?

17. A chemist wishes to convert iron into iron(II) bromide using this reaction:
Fe (s)+2 HBr (aq)→FeBr2 (aq)+H2 (g)

She mixes small pieces of iron with 0.2 M aqueous HBr at room
temperature and discovers that the reaction proceeds slowly. List two ways
she could speed up the reaction.

18. A biologist monitoring fish populations collects tissue samples for

analysis. However, the tissue samples decompose as they react with
oxygen in the air, producing foul-smelling by-products. How can he slow
down or stop this reaction from occurring?

19. What is a reaction energy diagram? What information does this type of
diagram provide?

20. What is the difference between the activation energy and the net energy
change for a reaction?

839
21. What is a transition state? How can you identify a transition state on an
energy diagram?

22. What energetic factor determines the rate of a reaction?

23. A single-step reaction has an activation energy of +80.0 kJ/mol and a net
energy change of +30.0 kJ/mol Create an energy diagram for this reaction.

24. A single-step reaction has a transition state with an energy change of +5.0
kJ/mol and a net energy change of –18.0 kJ/mol Create an energy diagram
for this reaction.

25. A single-step reaction has an activation energy of +20.0 kJ/mol and a net
energy change of +5.0 kJ/mol Is this reaction endothermic or exothermic?
Will the reaction occur at room temperature?

26. A single-step reaction has an activation energy of +7.5 kJ/mol and a net
energy change of −30.2 kJ/mol Is this reaction endothermic or exothermic?
Will the reaction occur at room temperature?

27. Describe each of the following reactions as endothermic or exothermic.

Also determine the activation energy and the net energy change for each
reaction.

840
28. Describe each of the following reactions as endothermic or exothermic.
Also determine the activation energy and the net energy change for each
reaction.

841
29. For each reaction diagram, determine the activation energy and the net
energy change:

842
30. For each reaction diagram, determine the activation energy and the net
energy change:

843
31. The energy profiles of three reactions are listed below. If the three
reactions are run side by side at the same concentrations and same
temperature, which reaction will proceed most quickly? How can you tell?

Activation Energy Net Energy Change

Reaction
(kilocalories/mole) (kilocalories/mole)

A 2.8 –8.4
B 4.8 –20.7
C 10.4 +4.2

32. These two reactions are both endothermic. Which reaction will proceed
more quickly? How can you tell?

844
33. While camping, you need to get a fire started. You know that the
combustion of wood is exothermic, but it won’t start without overcoming
the activation energy. How can you overcome the activation energy to get
this reaction started?

34. Broadly speaking, how does a catalyst work? How does a catalyst affect
the net energy change of a reaction? How does it affect the activation
energy?

13.2 Equilibrium Reactions

35. Consider a simple reaction, given by the equation A ⇌ B If this system is

in equilibrium, is A converted into B? Is B converted into A?

36. These two reactions contain two different types of arrows. What is the
difference in meaning between the two?
HCl (aq)→H+ (aq)+Cl− (aq)HF (aq)⇌H+ (aq)+F− (aq)

37. On a winter day, a lake contains patches of ice and patches of liquid water.
The temperature is 0 °C, and the ice and water are in equilibrium. In this
situation, are any liquid water molecules freezing? Are any molecules in
the ice melting away from the solid? How do the rates of these changes
compare?

38. Many ionic compounds, like lead(II) chloride, are slightly soluble in water.
This means that most of the ions are undissolved solid, but a few ions are
dissociated in water. Describe the motion of ions between the solid and

845
 liquid phase in this equilibrium.

39. Consider two towns, A and B. Each day, 100 people move from Town A to
Town B, and 50 people move from Town B to Town A. Are the
populations of the two towns in equilibrium? Why or why not?

40. From 1961 to 1989, the Berlin Wall divided the city of Berlin into
communist East Berlin and democratic West Berlin. Although the
populations of East and West Berlin lived side by side, essentially no one
moved from East to West, or vice versa. Would it be appropriate to say
that the populations of East and West Berlin were in equilibrium? Why or
why not?

41. Does this reaction represent an equilibrium? How can you tell?
CH4 (g)+2 O2 (g)→CO2 (g)+2 H2O (g)

42. Does this reaction represent an equilibrium? How can you tell?

N2 (g)+3 H2 (g)⇌2 NH3 (g)

43. The following reaction is endothermic. At equilibrium, is there a higher

concentration of reactants or products?

N2O4 (g)⇌2 NO2 (g)

44. The following reaction is slightly exothermic. At equilibrium, is there a

higher concentration of reactants or products?

2 Hl (g)⇌H2 (g)+l2 (g)

45. Trifluoroacetic acid, HC2F3O2, ionizes in water. This reaction is described

by the equilibrium reaction and energy diagram shown. The energy levels
of the starting materials and products are very nearly the same. Describe
the balance of starting materials and products at equilibrium.
HC2F3O2 (aq)+H2O (l)⇌H3O+ (aq)+C2F3O2− (aq)

846
46. Boric acid, B(OH)3, undergoes an unusual equilibrium in water. The
equilibrium and reaction diagram are shown. At equilibrium, will there be
a higher concentration of starting materials (left) or products (right) in
solution?
B(OH)3 (aq)+H2O (l)⇌B(OH)4− (aq)+H+ (aq)

47. Consider this reaction diagram. What is the activation energy of this
reaction in the forward direction? What is the activation energy in the
reverse direction?

847
48. Consider this reaction diagram. What is the activation energy of this
reaction in the forward direction? What is the activation energy in the
reverse direction?

49. An equilibrium between starting materials and products can proceed

through an uncatalyzed pathway (1) or a catalyzed pathway (2), as shown
in the energy diagram. What effect does the catalyst have on the
equilibrium?

50. Indicate whether each of these factors will affect whether an equilibrium
lies to the right or to the left:
a. net energy change for a reaction
b. presence of a catalyst
c. activation energy of a reaction

13.3 Equilibrium Expressions

848
51. Write equilibrium expressions for each of these equations:
a. A+2 B⇌C
b. 3 A+B⇌C+2 D

52. Write equilibrium expressions for each of these equations:

a. A+B⇌2 C
b. 2 A⇌B

53. Write equilibrium expressions for each of these equations:

a. HNO2 (aq)⇌H+ (aq)+NO2− (aq)

b. Cr3+ (aq)+4 OH− (aq)⇌[Cr(OH)4]−(aq)

54. Write equilibrium expressions for each of these equations:

a. HBrO2 (aq)⇌H+ (aq) +BrO2− (aq)

b. Ag+ (aq)+2 CN− (aq)⇌[Ag(CN)2]− (aq)

55. For each of these reactions, use the equilibrium constant to determine
whether the reaction favors the starting materials (left) or the products
(right):
a. NH4HS (g)⇌H2S (g)+NH3 (g)K = 0.00012

b. Pb2+ (aq)+2 F− (aq)⇌PbF2 (s)K = 2.8×108

c. CH3SO3H (aq)⇌CH3SO3− (aq)+H+ (aq)K = 340

849
56. For each of these reactions, use the equilibrium constant to determine
whether the reaction favors the starting materials (left) or the products
(right):
a. HClO4 (aq)⇌H+ (aq)+ClO4− (aq)

K = 10,000,000,000
b. BaF2 (s)⇌Ba2+ (aq)+2 F− (aq)

K = 1.7×10−6
c. NH3 (aq)+H2O (l)⇌NH4+ (aq)+OH− (aq)K = 1.8×10−5

57. At equilibrium, an aqueous solution of formic acid (HCHO2) has these

concentrations: [H+] = 0.0031 M; [CHO2−] = 0.0031 M; and [HCHO2] =
0.0535 M. What is the value of K for this equilibrium? Does the
equilibrium lie to the right or to the left?
HCHO2 (aq)⇌H+ (aq)+CHO2−(aq)

58. At equilibrium, an aqueous solution of iodic acid (HIO3) has these

concentrations: [H+] = 0.0203 M; [IO3–] = 0.0203 M; and [HIO3] =
0.0024 M. What is the value of K for this equilibrium? Does the
equilibrium lie to the right or to the left?
HIO3 (aq)⇌H+ (aq)+lO3− (aq)

59. Write equilibrium expressions for each of these equations:

850
 a. H2O2 (aq)+H2O (l)⇌H3O+ (aq)+HO2− (aq)

b. B(OH)3 (aq)+H2O (l)⇌B(OH)4− (aq)+H+ (aq)

60. Write equilibrium expressions for each of these equations:

a. 2 C6H12O6 (aq)⇌H2O (l)+C12H22O11(aq)

b. NH3 (aq)+H2O (l)⇌NH4+ (aq)+OH− (aq)

61. Hydrazoic acid, HN3, ionizes in water. At equilibrium, this solution is

found to have these concentrations: [H3O+] = 0.0043 M; [N3−] = 0.0043
M; and [HN3] = 0.994 M. Write an equilibrium expression, and calculate
the value of K for this reaction:
HN3 (aq)+H2O (l)⇌H3O+ (aq)+N3− (aq)

62. Cyanate ion, NCO–, is weakly basic. A solution containing cyanate ion is
found to have these concentrations: [NCO–] = 0.14 M; [HCNO] = 2.0 ×
10–6 M; and [OH–] = 2.0 × 10–6 M. Write an equilibrium expression, and
calculate the value of K for this reaction:
NCO− (aq)+H2O (l)⇌HNCO (aq)+OH− (aq)

63. Hydrofluoric acid dissociates in water as shown below. The equilibrium

constant for this reaction is 6.8 × 10–4. Given the values shown in the table
below, find the concentration of HF in this solution:

HF (aq)⇌H+ (aq)+ F− (aq)

Species Concentration

H+ 1.0 × 10–5 M

851
 F– 1.0 × 10–5 M
HF ?

64. The equilibrium constant for the acid-base reaction of ammonia in water
(shown below) is 1.8 × 10–5. Given the values shown in the table below,
find the concentration of OH– in this solution:
NH3 (aq)+H2O (l)⇌NH4+ (aq)+OH− (aq) K=1.8× 10−5

Species Concentration

NH4+ 9.5 × 10–4 M

OH– ?
NH3 0.050 M

65. Nitrous acid reacts with water as shown in the equilibrium below. The
equilibrium constant for this reaction is 4.5 × 10–4. A chemist prepares an
aqueous solution using both nitrous acid and potassium nitrite, so that
[HNO2] = 0.050 M and [NO2–] = 0.035 M. What is the concentration of
H3O+ in this solution?
HNO2 (aq)+H2O (l)⇌H3O+ (aq)+NO2− (aq)

66. Hypochlorous acid reacts with water as shown in the equilibrium below.
The equilibrium constant for this reaction is 3.0 × 10–8. A chemist
prepares an aqueous solution using both hypochlorous acid and sodium
hypochlorate. If [HClO] = 0.100 M and [ClO–] = 0.0025 M, what is the
concentration of H3O+ in this solution?
HClO (aq)+H2O (l)⇌H3O+ (aq)+ClO− (aq)

67. Write equilibrium expressions for each of the following:

a. PbI2 (s)⇌Pb2+ (aq)+2 I− (aq)

852
 b. Fe3+ (aq)+3 OH− (aq)⇌Fe(OH)3 (s)

68. Write equilibrium expressions for each of the following:

a. Ca(OH)2 (s)⇌Ca2+ (aq)+ 2 OH− (aq)

b. Ag+ (aq)+Cl− (aq)⇌AgCl(s)

69. What is a solubility product? What does a small solubility product

indicate?

70. Solubility products are usually reported at a given temperature, such as 25

°C How do you think the solubility product changes as the temperature
increases?

71. In a saturated solution of cadmium carbonate at 25 °C both [Cd2+] and

[CO32–] = 1.0 × 10–6 M. Write an equilibrium expression and calculate the
value of Ksp for this compound.

72. In a saturated solution of barium fluoride, [Ba2+] = 3.58 × 10–3 M and [F–]
= 7.16 × 10–3 M. Write an equilibrium expression and calculate the value
of Ksp for this compound.

73. The solubility product of silver chloride at 25 °C is 1.77 × 10–10. If silver

chloride is mixed with pure water, what is the concentration of silver ions
in solution?

74. The solubility product of iron(II) carbonate is 3.13×10–11 at 25 °C. If a

chemist mixes iron(II) carbonate with pure water, what will be the
equilibrium concentration of iron(II) ions?

75. Using the data in Table 13.1, find the maximum concentration of Pb2+ and
I– in solution when PbI2 is mixed with pure water at 25 °C.

853
76. Using the data in Table 13.1, find the maximum concentration of Pb2+ and
Cl– in solution when PbCl2 is mixed with pure water at 25 °C.

77. Using the partial pressures of each gas, write equilibrium expressions for
these gas-phase equilibria:

a. N2O4 (g)⇌2 NO2 (g)

b. 4 HCl (g)+O2(g)⇌2 Cl2 (g)+2 H2O (g)

78. Using the partial pressures of each gas, write equilibrium expressions for
these gas-phase equilibria:
a. PCl3 (g)+Cl2 (g)⇌PCl5 (g)

b. H2 (g)+I2 (g)⇌2 HI (g)

79. At 25 °C, NO2 and N2O4 exist in the equilibrium shown below. Find the

equilibrium pressure of N2O4 if PNO2 = 1.5 atm

2 NO2 (g)⇌N2O4 (g) K=7.0

80. At 25 °C, gaseous ClF exists in equilibrium with elemental chlorine and
fluorine, as shown below. If PClF = 1.3 atm, find the pressure of the two

elemental gases. Assume Pcl2=PF2.

2 CIF (g)⇌Cl2 (g)+F2 (g) K=2.9×10−11

13.4 Le Chatelier’s Principle

81. What is Le Chatelier’s principle?

82. Consider a U-shaped tube filled with mercury, as shown here. If additional
mercury were added to the left-hand side of the tube, would the mercury

854
 inside the tube move to the left, to the right, or show no change?

83. For the equilibrium A+B⇌C, indicate whether each of

the changes given will shift the equilibrium to the right or to the left:
a. adding compound A to the mixture
b. removing compound B from the mixture
c. adding compound C to the mixture
d. removing compound C from the mixture

84. For the equilibrium D+E⇌F+G, indicate

whether each of the changes given will shift the equilibrium to the right or
to the left:
a. removing compound D from the mixture
b. adding compound E to the mixture
c. adding compound F to the mixture
d. removing compound G from the mixture

85. At high temperatures, bromine and chlorine gas can rearrange to form
BrCl, as shown here:

Br2 (g)+Cl2 (g)⇌2 BrCl (g)

How would the following changes affect this equilibrium?
a. adding Br2 to the mixture
b. removing Cl2 from the mixture
c. removing BrCl from the mixture
d. adding catalyst

86. In aqueous solution, simple sugars (called monosaccharides) can couple

855
 together to form disaccharides:
2 C6H12O6 (aq)simple sugar⇌H2O (l)+C12H22O11 (aq)disaccharide

How does the addition of water affect this equilibrium? How does the
removal of water affect this equilibrium?

87. Methanol, CH3OH, reacts reversibly with hydroiodic acid, as shown here:

CH3OH+HI⇌CH3I+H2O
Based on Le Chatelier’s principle, will more CH3I be produced if water is
the solvent or if methanol is the solvent?

88. Methyl acetate is a compound commonly used in fingernail polish. When

combined with water, it slowly decomposes according to the equilibrium
below. If you wanted to decompose methyl acetate, which solvent would
be a better choice, water or methanol? Why is this so?
C3H6O2methyl acetate+H2O⇌C2H3O2Hacetic acid+CH3OHmethanol

89. Hydroxylamine (NH2OH) is weakly basic, as shown in the equilibrium

below. Would the addition of more OH– shift this equilibrium toward the
left or the right?
NH2OH (aq)+H2O (l)⇌NH3OH+ (aq)+OH− (aq) K=6.6×10− 9

90. Lead(II) chloride is slightly soluble in water, as shown in the equilibrium

below. If more chloride ion is added to the solution, will [Pb2+] increase or
decrease?
PbCl2 (s)⇌Pb2+ (aq)+2 Cl− (aq) K=5.9×10− 5

91. At equilibrium, water molecules move back and forth between the liquid
and gas phase, as shown below. If more heat is added, does the equilibrium

856
 shift to the left or the right? What if heat is removed?

Heat+H2O (l)⇌H2O (g)

92. Barium hydroxide is only slightly soluble in water at room temperature.

Based on the equilibrium below, how does the solubility change if the
solution is heated?
Heat+Ba(OH)2 (s)⇌Ba2+ (aq)+2 OH− (aq)

93. Consider the equilibrium and reaction enthalpy shown below. Would
heating this mixture shift the equilibrium to the left or to the right?
NH4+ (aq)+OH− (aq)⇌NH3 (aq)+H2O (l) ΔH°=−2.8 kJ/mol

94. Consider the equilibrium and reaction enthalpy shown below. Would
heating this mixture shift the equilibrium to the left or to the right?
AgCl (s)⇌Ag+ (aq)+2 Cl− (aq) ΔH°=+65.1 kJ/mol

95. When heated, PCl3 and Cl2 combine to form PCl5 in equilibrium, as
shown below. Would increasing the pressure on this system shift the
equilibrium to the left or to the right?

PCl3 (g)+Cl2 (g)⇌PCl5 (g)

96. When heated, hydrogen and iodine gas combine to form hydrogen iodide
in equilibrium. Would increasing the pressure on this system shift the
equilibrium to the left, to the right, or cause no change?

H2 (g)+l2 (g)⇌2 Hl (g)

97. A sealed container of methanol (CH3OH) contains both the liquid and gas
in equilibrium:
CH3OH (l)⇌CH3OH (g) ΔH°=+37.6.1 kJ/mol

857
 Indicate whether these changes would shift the equilibrium to the left or to
the right:
a. opening the container so that CH3OH (g) can escape
b. increasing the pressure in the container
c. increasing the temperature

98. Silver chloride is a slightly soluble compound. It dissociates to form ions

as shown in this equilibrium:
AgCl (s)⇌Ag+ (aq)+2 Cl− (aq) ΔH°=+65.1 kJ/mol

Indicate whether these changes would shift the equilibrium to the left or to
the right:
a. adding Cl– ions
b. removing Ag+ ions
c. increasing the temperature

Challenge Questions

99. A chemist prepares a 1.00-M solution of hypochlorous acid. This weak

acid ionizes in solution, as shown here:
HClO (aq)⇌H+ (aq)+ClO− (aq) K=3.0×10− 8

To determine the pH of weakly acidic solutions, chemists often construct a

table like the one below to summarize the initial concentration of each
species, the change in concentration of each species, and then the
equilibrium concentration. In this table, the change in concentration is
given by the unknown x. Because K is so small, we can assume that the
concentration of HClO does not change significantly. Construct an
equilibrium expression based on this table, and find the concentrations of
H+ and ClO– in this solution. Finally, calculate the pH of this solution.

[HClO] [H+] [ClO–]

Initial 1.00 M − −
Change –x +x +x
Equilibrium ˜1.00 M x x

858
100. Buffer solutions resist changes in pH when acid or base is added to them.
A common buffer solution is composed of acetic acid and sodium acetate.
Here is the equilibrium data for the conversion of this acid-conjugate
base pair:
HC2H3O2 (aq)+H2O (l)⇌H3O+ (aq)+C2H3O2− (aq) K=1.8×10− 5

A chemist prepares a buffer in which [HC2H3O2] and [C2H3O2] both

equal 0.200 M. She then adds a base, which converts some of the acetic
acid into acetate ion. After this reaction, the new concentrations are
[HC2H3O2] = 0.105 M and [C2H3O2–] = 0.295 M. Using the table below,
determine the concentration of [H3O+] and the pH both before and after
the addition of base.

[HC2H3O2] [C2H3O2–] [H3O+] pH

Before adding base 0.200 M 0.200 M

After adding base 0.105 M 0.295 M

859
Chapter Fourteen
Oxidation-Reduction Reactions

Volta’s Marvel
The year 1800 changed the world. An Italian scientist, Alessandro Volta, was
working to understand the nature of electricity. As part of this work, he began to
study electrical interactions between different metals. Volta discovered that when
he placed plates of different metals—such as zinc and copper—on either side of a
piece of cardboard soaked in dilute sulfuric acid or saltwater, he could produce an
electric current. Volta called his invention a “voltaic pile.” Today we know it as an
electrochemical cell, or more simply as a battery (Figure 14.1).

Figure 14.1 (a) Batteries are a common and essential part of modern life. (b) The first
battery, the voltaic pile, consisted of plates of copper and zinc separated by cardboard

860
soaked in dilute acid. (c) In this print, based on a painting by Giuseppe Bertini, Volta
shows his battery to the French Emperor Napoleon Bonaparte in 1801. (d) A battery like
Volta’s can be made at home using aluminum foil, zinc washers, copper pennies, and
thick paper or cardboard that has been soaked in saltwater or vinegar. The electrical
potential energy can be measured with an electrical meter or simply with a small light
bulb. (e) Both the chemistry of life and modern power applications are driven by
oxidation-reduction reactions.

Volta’s invention was simple to reproduce—you can do it at your kitchen table

—and many scientists quickly followed his lead. A few weeks after Volta
announced his discovery, two English scientists, William Nicholson and Anthony
Carlisle, built their own voltaic pile and used the electric current to separate water
into hydrogen and oxygen. Other scientists improved on Volta’s design, producing
long-lasting batteries that resemble many batteries we still use today.
Within just a few years, another English scientist, Humphry Davy, built a
massive battery at the Royal Institution in London. Through experiments with this
battery, he discovered the common elements sodium, potassium, barium, calcium,
magnesium, and boron.
The Volta battery gave scientists a controlled source of electricity and a way to
study the behavior of electricity and charged particles more carefully than they had
ever done before. These advances led to many critical inventions, including
Michael Faraday’s electrical generator and Thomas Edison’s light bulb.
Ultimately, Volta played a critical role in ushering in the modern era.
Today, batteries are everywhere. Advanced batteries power our phones,
computers, power tools, and even cars. But how do they actually work?
The electrical current produced by a battery arises from a type of chemical
reaction called an oxidation-reduction reaction, in which electrons are transferred
from one substance to another. Oxidation-reduction reactions power the chemistry
of life: Plants use this type of reaction to store the energy of the sun. With every
breath, our bodies use oxidation-reduction reactions to convert fuel into energy.
These reactions also power the world around us: From fossil fuels to solar cells and
batteries, we rely on oxidation-reduction reactions to provide energy for nearly
everything we do.

Explore
Figure 3.20

In this chapter, we’ll explore these reactions in detail. We’ll examine different
types of oxidation-reduction reactions and consider different ways of describing
these reactions. We’ll also survey some of the applications that arise from these
reactions. We’ll see how batteries are built and explore some other useful and
important applications of oxidation-reduction chemistry. Let’s get started.

Intended Learning Outcomes

861
After completing this chapter and working the practice problems, you should be
able to:

14.1 Oxidation and Reduction

Determine oxidation numbers for atoms in ions and molecules.
Identify species that are oxidized or reduced in a chemical reaction.

14.2 Types of Redox Reactions

Describe the oxidation-reduction reactions that occur between metals and
nonmetals.
Predict the products of combustion reactions.
Use the activity series to predict the products of metal displacement
reactions.
Predict the products of reactions of metals with water or with acid.

14.3 Half-Reactions and Batteries

Describe the transfer of electrons in a chemical change using half-reactions.
Describe the key components of electrochemical cells.

14.4 Balancing Redox Equations

Balance redox equations by balancing charge and by balancing half-
reactions.

14.5 Other Applications of Redox Reactions

Describe how electroplating is used to coat surfaces.
Describe the basic properties of a hydrogen fuel cell.

862
14.1 Oxidation and Reduction
The invention of the battery changed the world. Volta’s discovery kicked
off the field of electrochemistry—the study of chemical processes that
involve the movement of electrons. To understand electrochemistry, we
must begin with a class of reactions called oxidation-reduction reactions
(often simply called redox reactions).
We looked briefly at oxidation-reduction reactions in Chapter 6. In this
type of reaction, one species loses electrons, while another species gains
electrons. The loss of electrons is called oxidation. When a species loses
electrons, we say it has been oxidized. The gain of electrons is called
reduction. When a species gains electrons, we say it has been reduced.

Electrons are negative, so if an atom gains electrons, its charge is reduced.

Some of the simplest redox reactions involve metals and nonmetals.

For example, magnesium reacts with oxygen to produce magnesium oxide:

2 Mg (s)+O2 (g)→2 MgO (s)

In this reaction, each magnesium atom loses two electrons to form Mg2+.
Each oxygen atom gains two electrons to form O2–. Magnesium is
oxidized, while oxygen is reduced.

The oxygen causes the magnesium to be oxidized in this reaction. A

species that pulls electrons away from other atoms is sometimes called an
oxidizing agent. Similarly, a species that loses electrons is sometimes
called a reducing agent.

An oxidizing agent takes electrons; a reducing agent loses electrons.

Oxidation numbers are a bookkeeping tool to help us track the electron

changes.

Oxidation Numbers
When we are dealing with simple monatomic ions, the gain or loss of

863
electrons is easy to determine. However, many redox reactions involve
polyatomic ions or neutral molecules, and the change that takes place on
each atom can be harder to measure. For example, consider the reaction of
sodium metal with water, written as an ionic equation:
2 Na (s)+2 H2O (l)→ 2 Na+ (aq)+2 OH− (aq)+H2 (g)

In this reaction, the sodium is oxidized (it loses electrons). But which atom
gains the electrons? At first glance, this question may be difficult to
answer.
Scientists use oxidation numbers to keep track of electron changes in
redox reactions. Oxidation numbers do not necessarily reflect the actual
charge on atoms. Rather, they are a bookkeeping tool to ensure that the
overall electronic changes are accounted for. We use the following rules to
assign oxidation numbers to atoms:

1. The oxidation number of any species in its elemental form is zero.

2. For monatomic ions, the oxidation number is simply the charge of the
ion.
3. For polyatomic ions, the oxidation numbers of the atoms must add up
to the overall charge of the ion. For neutral compounds, the oxidation
numbers of the atoms must add up to zero (Figure 14.2).

Figure 14.2 The oxidation numbers for a compound or polyatomic ion are added
up to find the overall charge.

4. For atoms in molecules or polyatomic ions (Table 14.1):

a. The oxidation number of fluorine is –1.
b. The oxidation number of oxygen is –2.
c. The lower halogens (Cl, Br, I) have an oxidation number of –1, unless
they are bonded to oxygen or fluorine.
d. If hydrogen is bonded to a nonmetal, its oxidation number is +1.
e. If hydrogen is bonded to a metal, its oxidation number is –1.

TABLE 14.1 Oxidation Numbers for Atoms in Molecules and

864
 TABLE 14.1 Oxidation Numbers for Atoms in Molecules and
Polyatomic Ions
Atom Oxidation Number

F –1

O –2

Cl, Br, l –1

H +1 or –1

To see how oxidation numbers work, let’s analyze the reaction of

sodium metal with water. We begin by finding the oxidation number for
each atom present. Na and H2 are both elemental forms, so their oxidation
numbers are zero. For the monatomic sodium ion, Na+, its oxidation
number equals its charge (+1). For H2O and OH−, we assign oxygen a
value of −2 and hydrogen a value of +1:

The oxidation number for elemental forms is zero.

For a neutral compound or polyatomic ion, the oxidation numbers must add
up to the total charge.

The oxidation number of sodium changes from 0 to +1. This means that
sodium has lost electrons—it has been oxidized. Notice that the oxidation
number of hydrogen changes from +1 (in water) to 0 (in H2). The
hydrogen atoms have gained electrons—they have been reduced.
It takes some practice to get the hang of assigning oxidation numbers. I
encourage you to work through Examples 14.1 and 14.2 as well as the
sample problems that follow them.

Example 14.1 Identifying Oxidation Numbers in Compounds

Identify the oxidation number for sulfur in the following three species: S2–,

865
SOCl2, HSO4− .
For monatomic ions, the oxidation number is simply the charge: The
oxidation number for sulfur in S2– is –2.
SOCl2 is a neutral compound, so the oxidation numbers must total zero.
Based on the rules in Table 14.1, oxygen has an oxidation number of –2.
The chlorine atoms are bonded to sulfur in this compound, and each
chlorine atom has an oxidation number of –1. This means the oxidation
number of sulfur must be +4. It may help you to set this up as an algebra
equation, where x is the oxidation number for sulfur:

The hydrogen sulfate ion, HSO4–, has an overall charge of –1, so the
oxidation numbers must all add up to –1. Based on the rules given above,
the oxidation number of oxygen is –2, and hydrogen is +1. This means that
the oxidation number of sulfur must be +6. Again, we can set this up as an
algebra equation. •

Example 14.2 Identifying Oxidation Numbers in Reactions

Find the oxidation number of each atom in the following reaction. Identify
the species that is oxidized or reduced.
CH4 (g)+2 O2 (g)→CO2 (g)+2 H2O (g)

Based on the rules for oxidation numbers, the oxygen in O2 (the elemental
form of oxygen) has an oxidation number of zero. In the compounds CO2
and H2O, oxygen has a value of –2. Each hydrogen in CH4 and in H2O has
a value of +1. The only remaining atom is carbon: In CH4, the carbon must
have a value of –4. In CO2, carbon must have a value of +4:

866
 In this reaction, O is reduced while C is oxidized. •

When a substance reacts with oxygen, its oxidation number increases. The
substance has been oxidized.

IT
TRY

1. Identify the oxidation number of sulfur in each of these species:

a. S8
b. H2S
c. H2S2

d. SO32−

2. Identify the atoms that are oxidized and reduced in each of these
reactions:
a. 4 Fe (s)+3 O2 (g)→2 Fe2O3 (s)

b. Mg (s)+Br2 (g)→MgBr2 (s)

c. 2 HCl (aq)+Zn (s)→ZnCl2 (aq)+H2 (g)

d. 8 H+ (aq)+MnO4− (aq) − 5 Fe3+ (aq)+Mn2+ (aq)+4 H2O (l)

867
14.2 Types of Redox Reactions
Redox reactions take many different forms. In this section, we’ll briefly
review some of the reactions you’ve seen before, and then we’ll look at
several new types of redox reactions.

Reactions of Metals with Nonmetals

The reactions between metals and nonmetals were covered in Chapter 6.
Recall that when metals and nonmetals react, they form ionic compounds
(Figure 14.3). In these reactions, the metals are oxidized to form cations,
and the nonmetals are reduced to form anions. The following three
reactions illustrate this type of change:

Figure 14.3 Rust is the compound iron(III) oxide, formed by the redox reaction of iron
metal with oxygen gas.

2 Fe (s)+3 O2 (g)→Fe2O3 (s) Zn (s)+Cl2 (g)→ZnCl2 (s)2 Ag (s)+F2 (g)→2 AgF (s)

Combustion Reactions
Combustion reactions are reactions that involve molecular oxygen. Most
elements react with oxygen to form oxides (Figure 14.4). In these
reactions the oxygen is reduced, and the other element is oxidized.

868
Figure 14.4 Minerals are ionic compounds found in rock formations. Many minerals
are metal oxides. This image shows the mineral cassiterite, which is composed of
tin(IV) oxide.

2 Zn (s)+O2 (g)→2 ZnO (s) Si (s)+O2 (g)→SiO2 (s)2 S (s)+3 O2 (g)→2 SO3 (s)

Compounds also react with oxygen. Some of the most important

combustion reactions involve hydrocarbons—compounds composed of
carbon and hydrogen (Figure 14.5). Hydrocarbons react in combustion
reactions to produce two compounds, carbon dioxide and water. For
example, the following two reactions describe the combustion of acetylene
(C2H2) and propane (C3H8):
2 C2 H2 (g)+5 O2 (g)→4 CO2 (g)+2 H2O (g) C3H8 (g)+5 O2 (g)→3 CO2 (g)+4
H2O (g)

For more on combustion reactions, see Section 6.4.

869
Figure 14.5 The white-hot flame of an acetylene torch results from the reaction of
acetylene with oxygen.

Animals (including humans) breathe in oxygen gas, which combines with

carbon and hydrogen to produce carbon dioxide and water as well as energy.
Overall this process, called cellular respiration, is the same as a combustion
reaction.

Metal Displacement Reactions

In a metal displacement reaction, an elemental metal reacts with an ionic
metal. For example, consider the reaction of aqueous copper(II) sulfate
with zinc metal, as shown in Figure 14.6. Aqueous solutions of copper(II)
ions have a blue color. However, when we drop a bar of zinc into the
solution, we observe several changes: The blue color of the solution
becomes fainter. The zinc rod begins to dissolve, and deposits of copper
metal accumulate in the bottom of the beaker and on the zinc rod.

870
Figure 14.6 The reaction of zinc metal with aqueous copper(II) ions is a metal
displacement reaction.

Explore

Figure 14.6

Metal displacement reactions are single-displacement reactions.

In this reaction, the zinc metal is oxidized—it loses electrons to

become Zn2+ ions. At the same time, the copper(II) ion is reduced—it
gains electrons to become elemental copper. The anion in solution, sulfate,
is a spectator ion. We can describe this single-displacement reaction as a
net ionic equation or as a complete molecular equation:

871
 Net ionic equation: Zn (s)+Cu2+ (aq)→Zn2+ (aq)+Cu (s)Molecular
equation: Zn (s)+CuSO4 (aq)→ZnSO4 (aq)+Cu (s)

There are many reactions of this type (Figure 14.7). For example,
elemental calcium reacts with iron(II) ions to produce calcium ions and
elemental iron:

Figure 14.7 Metal displacement reactions occur for many combinations of metals and
ions.

Explore
Figure 14.7

Net ionic equation: Ca (s)+Fe2+ (aq)→Ca2+ (aq)+Fe (s)Molecular

equation: Ca (s)+FeCl2 (aq)→CaCl2 (aq)+Fe (s)

872
Similarly, elemental magnesium reacts with aluminum ions to produce
magnesium ions and elemental aluminum:
Net ionic equation: 2 Al3+ (aq)+3 Mg (s)→3 Mg2+ (aq)+2 Al (s)Molecular
equation: 2 Al(NO3)3 (aq)+3 Mg (s)→3 Mg(NO3)2 (aq)+2 Al (s)

The Activity Series

So, how do we know when an oxidation-reduction will occur? For
example, if we place an iron nail into a solution of nickel(II) chloride, will
a reaction take place? Will the opposite reaction occur if we place a nickel
screw in a solution of iron(II) chloride? To answer these questions, we
could run two side-by-side experiments (Figure 14.8). If we did this, we
would find that iron reacts with nickel ions, but the opposite reaction does
not occur:

Figure 14.8 (a) An iron nail reacts in a nickel(II) chloride solution, (b) but a nickel
screw does not react in an iron(II) chloride solution.

Explore
Figure 14.8

Fe (s)+Ni2+ (aq)→Fe2+ (aq)+Ni (s)Fe2+ (aq)+Ni (s)→no reaction

Because of the profound importance of metals in so many applications,

information about metal reactions like these is readily available. A helpful
tool for predicting metal displacement reactions is the activity series

873
(Table 14.2). The activity series ranks the reducing powers of metals. The
metals at the top of the table release their electrons very easily. These are
the most reactive metals and the strongest reducing agents. Metals at the
bottom of the series are the least reactive—it takes much more energy to
remove electrons from these atoms.

TABLE 14.2 The Activity Series

For example, let’s compare the information on the activity series with
our observations about the reactions of iron and nickel in Figure 14.8. Iron
is higher on the activity series than nickel. This means that iron reduces
Ni2+, but nickel does not reduce Fe2+. Figure 14.9 shows this concept,
and Example 14.3 describes it further.

874
Figure 14.9 A metal can react with a cation below it on the activity series.

The metals at the bottom of the activity series are the precious metals. Elements
like gold and platinum are valued not just for their exquisite beauty, but for their
durability. Unlike metals such as iron, which will oxidize (rust) over time,
metals like gold and platinum retain their elemental form and beauty.

Example 14.3 Predicting Metal Displacement Reactions

Based on the activity series in Table 14.2, which of the following metals
will react with an aqueous solution of nickel(II) chloride: gold, tin, or zinc?
Write a balanced molecular equation for the reaction that occurs.
By examining Table 14.2, we can see that Zn lies above Ni2+ on the
activity series, while Sn and Au both lie below it. Based on this ranking, Zn
is the only one of these elements that will react with Ni2+. We can write a
balanced molecular equation for this reaction. •

Zn (s)+NiCl2 (aq)→ZnCl2 (aq)+Ni (s) balanced

875
 The activity series is like college football rankings: Higher-ranked teams are
expected to beat teams ranked lower on the list. Similarly, a metal higher on the
activity series will reduce a cation that is lower on the series.

IT
TRY

3. Based on the activity series, which of these elements can reduce

iron(II) to elemental iron?
a. calcium
b. magnesium
c. chromium
d. silver

4. Complete and balance each of the following reactions. If no reaction

occurs, indicate this.

a. Zn (s)+MgSO4 (aq)→

b. CO (s)+NiCl2 (aq)→

Reactions of Metals with Acid and Water

Most metals react with acid to produce metal cations and hydrogen gas.
For example, zinc metal reacts with hydrochloric acid to produce zinc
chloride and hydrogen gas (Figure 14.10). We could write this as a
molecular equation or as a net ionic equation:

876
Figure 14.10 Zinc metal reacts with hydrochloric acid to produce zinc chloride and
hydrogen gas. The bubbles are caused by the hydrogen gas produced in this reaction.

Explore
Figure 14.10

Molecular equation: Zn (s)+2 HCl (aq)→ZnCl2 (aq)+H2 (g)Net ionic

equation: Zn (s)+2 H+ (aq)→Zn2+ (aq)+H2 (g)

Similarly, magnesium reacts with nitric acid to produce magnesium nitrate

and hydrogen gas:
Molecular equation: Mg (s)+2 HNO3 (aq)→Mg(NO3)2(aq)+H2 (g)Net ionic
equation: Mg (s)+2 H+ (aq)→Mg2+ (aq)+H2 (g)

Not all metals react with acid. The precious metals (silver, gold, platinum)
do not react with acid. Based on these observations, chemists have fit H+
into the activity series in Table 14.2. The metals above H+ on the activity
series react with acids, and those below it do not.

Precious metals do not react with acid.

The most active metals react with water. These reactions are similar to

877
the reactions with acid in that the metal is oxidized to the cation, while H+
is reduced to form hydrogen gas. For example, sodium and potassium both
react violently with water to produce a metal hydroxide and hydrogen gas
(Figure 14.11).
2 Na (s)+2 H2O (l)→2 NaOH (aq)+H2 (g) 2 K (s)+2 H2O (l)→2 KOH (aq)+H2 (g)

Figure 14.11 (a) Alkali metals like sodium and potassium react violently with water. (b)
To prevent reaction with water, these elements are commonly stored under a layer of
oil.

Explore
Figure 14.11

As a general rule, alkali metals react violently with water. Alkaline

earth metals also react with water, but the reactions take place more
slowly. Some alkaline earth metals, like calcium, react slowly with water
at room temperature. Others, like magnesium, require heating in order for
the reaction with water to take place. These trends are summarized in
Table 14.3.

TABLE 14.3 Reactivity of Metals with Acid and Water

React Violently with Water React with Water or Steam React with Acid

878
 Alkali metals ✔ ✔ ✔
(Li, na, K)

Alkaline earth metals ✔ ✔

(Mg, ca, Ba)
Most transition metals* ✔
(Fe, co, ni)

Precious metals
(cu, au, ag, Pt)

* The most reactive transition metals, such as Zn, also react with steam.

Because of their extreme reactivity, sodium, potassium, and the other

alkali metals must be carefully stored to keep them from coming in contact
with moisture. Chemists frequently store these metals under oil to prevent
exposing them to water or air.

Alkali metals react violently with water; alkaline earth metals react slowly
with water.

Example 14.4 Predicting an Acid-Metal Reaction from the

Activity Series
Will aluminum metal react with hydrochloric acid? If so, write a balanced
molecular equation to describe this reaction.
We see from Table 14.2 that Al is higher than H+ on the activity series, so
we know that this reaction can occur. Aluminum forms a +3 cation, and so
aluminum chloride has the formula AlCl3. After balancing the equation, we
arrive at the following solution. •

2 Al (s)+6 HCl (aq)→2 AlCl3 (aq)+3 H2 (g) balanced

IT
TRY

5. Using the activity series (Table 14.2), indicate whether these metals
will react with acid:
a. gold
b. nickel

879
 c. platinum
d. chromium

6. Use the information in Tables 14.2 and 14.3 to complete and balance
each of the following reactions. Also indicate if no reaction occurs.

a. Zn (s)+HBr (aq)→

b. Co (s)+HNO3 (aq)→

c. Ag (s)+HCl (aq)→

d. Li (s)+H2O (l)→

880
14.3 Half-Reactions and Batteries

Half-Reactions
Oxidation-reduction reactions take place in two distinct parts. On the one
hand, an atom or ion gains electrons (reduction). On the other, an atom or
ion loses electrons (oxidation). We sometimes write these two separate
processes as half-reactions. Half-reactions show the reduction without
indicating where the electrons came from, or they show the oxidation
without indicating where the electrons went.
For example, let’s look again at the reaction of zinc metal and
copper(II) ions. We can write this process as two half-reactions. In one
half-reaction, zinc is oxidized. In the other, copper(II) is reduced. In the
reactions below, the symbol e– represents an electron:
Oxidation half-reaction: Zn (s)→Zn2+ (aq)+2 e−Reduction half-
reaction: 2 e−+Cu2+ (aq)→Cu (s)Net ionic equation:2 e + Cu2+ (aq)+Zn
(s)→Cu (s)+Zn2+ (aq)+2 e¯

Half-reactions use e– to represent electrons in either the reactants or the

products.

Notice that the two electrons released by zinc are consumed by

copper(II). We can add the two half-reactions together to produce the net
ionic equation. Because the two electrons appear on the reactant and
product sides, we can cancel them out.
The activity series (see Table 14.2) is actually a series of half-
reactions. We can use the activity series to write the oxidation and
reduction half-reactions for a chemical change. For example, consider the
reaction of nickel metal with tin(II) ions. As Figure 14.12 shows, nickel is
above tin on the activity series. We write the half-reaction for the
oxidation of nickel the same way it appears on the activity series:

881
Figure 14.12 Nickel falls above tin on the activity series. When the two react, nickel is
oxidized while Sn2+ is reduced.

Oxidation half-reaction: Ni(s)→Ni2+ (aq)+2 e−

The other half-reaction that takes place is the reduction of Sn2+. Because
the activity series shows oxidations, we must reverse the reactants and
products to show the reaction as a reduction:
Reduction half-reaction: 2 e−+Sn2+ (aq)→Sn (s)

As a general rule, when writing half-reactions from the activity series, the
upper half-reaction proceeds in the forward direction; the lower half-
reaction will go in the reverse direction. In some instances, we must
balance the half-reactions to produce a correctly balanced equation. We’ll
look at this process in more detail in Section 14.4.

Example 14.5 Writing Half-Reactions

Write a net ionic equation for the redox reaction shown below. Then show
the oxidation and reduction half-reactions that take place.
SnCl2 (aq)+Fe (s)→FeCl2 (aq)+Sn (s)

We begin by writing the complete ionic equation:

Sn2+ (aq)+2 Cl− (aq)+Fe (s)→Fe2+ (aq)+2 Cl− (aq)+Sn (s)

Chloride is a spectator ion, so we remove it from the equation to give the

net ionic equation:
Sn2+ (aq)+Fe (s)→Fe2+ (aq)+Sn (s)

882
Next, we break the reaction into oxidation and reduction half-reactions.
Iron is oxidized from Fe to Fe2+, meaning it loses two electrons. Tin is
reduced from Sn2+ to Sn, so it gains two electrons. We can write the two
half-reactions as follows:
Oxidation: Fe (s)→Fe2+ (aq)+2 e−Reduction: 2 e− +Sn2+ (aq)→Sn (s)

Adding the two half-reactions together results in the net ionic equation.
Note also where these two reactions are on the activity series. Iron falls
higher on the activity series. It is oxidized in this reaction. Tin falls lower
on the activity series, so it is reduced. •

When two football players collide, the player exerting the stronger force
continues to move forward, while the player exerting the weaker force is
knocked backward. Similarly, the half-reaction for the stronger reducing metal
on the activity series proceeds in the forward direction (oxidation). The half-
reaction for the weaker metal gets reversed (reduction).

Example 14.6 Writing Half-Reactions

Show the oxidation and reduction half-reactions that take place when zinc
reacts with nitric acid to produce zinc nitrate and hydrogen gas.
To begin, we write a balanced equation for this reaction. Zinc forms a +2
cation, so the formula for zinc nitrate is Zn(NO3)2. This leads us to the
balanced equation:

883
 Zn (s)+2 HNO3 (aq)→Zn(NO3)2 (aq)+H2 (g)

Writing the dissociated ions and removing the spectator ions (NO3–) gives
us the net ionic equation:
Zn (s)+2 H+ (aq)→Zn2+ (aq)+H2 (g)

Zinc is oxidized from Zn to Zn2+, meaning it loses two electrons. The H+

ions are reduced: Adding two electrons to two H+ ions results in H2 gas, as
shown below. •

Oxidation: Zn (s)→Zn2+ (aq)+2 e−Reduction: 2 e−+2 H+ (aq)→H2 (g)

IT
TRY

7. Write half-reactions for each of these chemical changes:

a. Ca (s)+Cl2 (aq)→Ca2+ (aq)+2 Cl− (aq)

b. Al (s)+Cr(NO3)3 (aq)→Al(NO3)3 (aq)+Cr (s)

8. Write two half-reactions to describe the reaction of calcium with

hydrochloric acid to produce calcium chloride and hydrogen gas.

Batteries
Batteries (also called electrochemical cells) are powered by chemical
reactions in which oxidation occurs in one location, and reduction occurs
at another. For the reaction to take place, the electrons must travel from the
oxidation site to the reduction site, resulting in an electric current (Figure

884
14.13). We use half-reactions to describe the changes taking place at each
site.

Figure 14.13 Batteries are powered by oxidation-reduction reactions in which half-

reactions occur at separate sites. The flow of electrons between the two sites produces
an electric current.

Figure 14.14 shows a classic electrochemical cell. The cell consists of

two chambers (in this case, two beakers). One beaker contains a strip of
copper immersed in a solution of copper(II) sulfate. The other beaker
contains a strip of zinc immersed in a solution of zinc sulfate. If we
connect these two strips with a conducting wire, the result is an electric
current. We call the copper and zinc strips electrodes. An electrode is a
site where an oxidation or reduction half-reaction takes place. The
electrode where oxidation occurs is the anode. The electrode where
reduction occurs is the cathode.

885
Figure 14.14 In an electrochemical cell, an oxidation takes place in one chamber while
the reduction takes place in another. The transfer of electrons between the two
halfreactions produces an electric current. The salt bridge balances the charges between
the two solutions.

886
 Explore

Figure 14.14

Reduction occurs at the cathode; oxidation occurs at the anode.

The current arises from a redox reaction at the two electrodes. On the
zinc electrode (the anode), an oxidation takes place: The zinc atoms lose
two electrons to form Zn2+ ions. These electrons travel through the
conducting wire toward the copper electrode (the cathode). At the copper
electrode, Cu2+ ions gain two electrons to form elemental copper.
Although the oxidation and reduction half-reactions take place in
separate containers, they must occur together. For the reaction to happen,
the electrons must move from one atom or ion to another.

887
 Although batteries vary widely in shape, size, and potential energy, they all rely
on oxidation-reduction reactions to produce electric currents.

Notice that the electrochemical cell in Figure 14.14 also contains a U-

shaped tube, called a salt bridge. This bridge is essential to the function of
the cell. A salt bridge contains an ionic compound like sodium nitrate. As
the electrons move from the anode to the cathode, the anode cell builds up
a positive charge, while the cathode builds up a negative charge. The salt
bridge provides spectator ions that can balance these charges. For example,
if the bridge contains sodium nitrate, the sodium cations will migrate
toward the negative charges, while the nitrate anions migrate toward the
positive charges. If a salt bridge is not present, the buildup of charge on the
electrodes limits the flow of electrons.

A salt bridge balances the charges between the cathode and anode.

Example 14.7 Predicting the Direction of Electron Flow in an

Electrochemical Cell
A chemist assembles an electrochemical cell as shown. One side of the cell
contains an iron electrode immersed in aqueous iron(II) nitrate solution.
The other side of the cell contains a magnesium electrode immersed in
aqueous magnesium nitrate solution. Using the activity series, identify the
electrode where oxidation takes place (the anode) and the electrode where
reduction takes place (the cathode). Write half-reactions to describe the
change that takes place at each electrode.

888
Because magnesium lies above iron on the activity series, we know that Mg
reacts with Fe2+:

This means that oxidation takes place at the magnesium electrode, while
reduction takes place at the iron electrode. We can obtain the two half-
reactions from the activity series. The reaction that is higher on the list
proceeds as an oxidation reaction (that is, as written). For the reaction that
is lower on the list, we reverse the reactants and products to show the
reduction half-reaction.
Oxidation: Mg (s)→Mg2+ (aq)+2 e−Reduction: 2 e−+Fe2+ (aq)→Fe

(s)
We also can describe this process by the net ionic reaction:
Fe2+ (aq)+Mg (s)→Fe (s)+Mg2+ (aq)

889
 Because oxidation takes place at the anode, the magnesium strip is the
anode, and the iron strip is the cathode. •

IT
TRY

9. A battery contains a zinc anode immersed in an aqueous zinc chloride

solution and a platinum cathode immersed in an aqueous platinum(II)
chloride solution. What half-reactions take place at the anode and
cathode of this cell?

10. An electrochemical cell contains two chambers. The first consists of a

nickel electrode immersed in aqueous nickel(II) sulfate. The second
consists of a zinc electrode immersed in aqueous zinc sulfate. If these
two chambers are connected by a salt bridge and a conducting wire, in
which direction will the electrons flow? Write half-reactions and a net
ionic equation to describe this chemical reaction.

890
14.4 Balancing Redox Equations
Sometimes we need to balance a redox equation. As with other balanced
equations, this means that the total number of each type of atom must be
equal on both sides of the equation. When describing redox reactions, we
must also balance the charge. Stated differently, the same number of
electrons lost in the oxidation will be gained in the reduction.

For redox equations, we must balance both the atoms and the charge.

For example, consider the reaction of calcium metal with aqueous

chromium(III): In this reaction, elemental calcium is oxidized to Ca2+
while Cr3+ is reduced to elemental chromium. We could write an
unbalanced equation to describe this change:
Ca (s)+Cr3+ (aq)→Ca2+ (aq)+Cr (s)not balanced (changenot equal)

How should we go about balancing it? Two approaches are discussed

below.
Approach 1. In this equation, the number and type of the atoms are the
same on both sides, but the charge is not. By adding coefficients to the
ions, we can balance the charge (the charge is now +6 on both sides):
Ca (s)+2 Cr3+ (aq)→3 Ca2+ (aq)+Cr (s)not balanced (changeequal, but not atoms)

Then, by adding coefficients to the elemental forms, we ensure that both

the charge and the atoms/ions are balanced.
3 Ca (s)+2 Cr3+ (aq)→3 Ca2+ (aq)+2 Cr (s) balanced

Approach 2. Another way to tackle this problem is to make sure that the
same number of electrons are released in the oxidation and acquired in the
reduction. We begin by writing two unbalanced half-reactions:
Oxidation: Ca→Ca2++2 e−Reduction: 3 e−+Cr3+→Cr

The oxidation releases two electrons, but the reduction requires three.

891
For an oxidation-reduction reaction to balance, the electrons lost in the
oxidation must equal the electrons gained in the reduction. To make the
electrons equal, we multiply each species in the oxidation by a factor of
three and each species in the reduction by a factor of two. Now six
electrons are lost in the oxidation, and six electrons are gained in the
reduction:

Finally, we add the oxidation and reduction half-reactions together to

obtain the net ionic reaction. Because the number of electrons present on
the left and right sides of the equation is the same, we can cancel them out,
leaving the correctly balanced net ionic equation:
Oxidation half-reaction: 3 Ca→3 Ca2++6 e−Reduction half-
reaction: 6 e−+2 Cr3+→2 CrNet ionic quation:6 e− +3 Ca (s)+2
Cr3+ (aq)→3 Ca2+ (aq)+2 Cr (s)+6 e−¯

Example 14.8 provides another problem of this type.

Example 14.8 Balancing a Redox Equation

Silver metal undergoes a single-displacement reaction with platinum(II)
nitrate to form silver nitrate and elemental platinum. Write a balanced net
ionic equation for this single-displacement reaction.
Let’s solve this problem using both of the approaches discussed above.
First, let’s do it by balancing the atoms and the charge in the net ionic
equation. Nitrate is a spectator ion, so it doesn’t need to be included. This
leaves the following species involved with this reaction:
Ag+Pt2+→Ag++Pt not balanced

In this case, we can place a coefficient (2) in front of the Ag+ to balance the
charge. We then put a 2 in front of elemental Ag to balance the number and
type of atoms. Finally, we insert phase symbols to give the complete
answer:

892
 2 Ag (s)+Pt2+ (aq)→2 Ag+ (aq)+Pt (s) balanced

The alternative approach is to balance the electron flow in the oxidation and
reduction. We begin by writing two half-reactions:
Oxidation: Ag→Ag++e−Reduction: 2 e−+Pt2+→Pt

To balance the electrons in each half-reaction, we multiply each species in

the oxidation by two. We then combine the two half-reactions. There are
two electrons on both the reactant and product side, so we can cancel them
out, leaving a balanced equation. •

IT
TRY

11. Elemental barium reacts with a solution of cobalt(II) nitrite to

produce barium chloride and elemental cobalt. Write half-reactions
to show the oxidation and reduction components of this change.

12. Elemental aluminum reacts with a solution of iron(II) chloride to

produce aluminum chloride and elemental iron. Describe this change
using half-reactions, using a net ionic equation, and using a
molecular equation.

893
14.5 Other Applications of Redox Reactions

Electroplating
Electroplating is a technique that produces a thin layer of a metal such as
gold, silver, copper, or chromium on the outside surface of another
material. For example, “chrome” motorcycle parts are typically composed
of steel, which is protected and beautified by thin outer layers of copper,
nickel, and chromium (Figure 14.15).

Figure 14.15 The brilliant chrome surfaces on this motorcycle are produced by
electroplating each of the steel parts. The parts are coated with thin layers of copper,
nickel, and chromium.

Electroplating is actually very similar to what happens in an

electrochemical cell. In an electrochemical cell, a redox reaction creates an
electric current. In electroplating, the reverse occurs: An electric current
drives a redox reaction to take place. Figure 14.16 illustrates the process
of applying a nickel coating to an iron pipe. The pipe is immersed in an
aqueous solution containing dissolved nickel(II) ions. A nickel electrode is
also immersed in the solution. The two electrodes (the pipe and the nickel)
are then connected to a battery. The electrical potential energy of the
battery drives a redox reaction to take place. The reduction occurs on the
surface of the pipe. As nickel(II) ions are reduced to elemental nickel, they
adhere to the metal surface. On the other electrode, nickel atoms are
oxidized to form nickel(II) ions, which dissolve in the solution and
replenish the supply.

894
Figure 14.16 Electroplating uses electrical potential energy (voltage) to drive a redox
reaction to take place. In this nickel electroplating apparatus, nickel is oxidized from the
electrode at left and reduced on the electrode at right, forming a thin layer of nickel.

Historically, pennies were made almost completely of copper. However,

because of the rising cost of copper, in 1983 the U.S. Mint began electroplating
zinc pennies with a thin copper surface.

Following Volta’s invention of the battery, the English scientist Humphry Davy

895
 discovered many of the main-group metals using a technique similar to that used
in electroplating. By immersing both electrodes into a solution containing ionic
compounds, he was able to reduce the metal ions to their elemental form.

Fuel Cells
In the past few decades, fuel cells have emerged as an exciting new
approach to storing and releasing energy. Fuel cells convert the potential
energy of a combustion reaction directly into electrical energy by
separating the oxidation and reduction half-reactions (Figure 14.17). The
most common fuel cells involve the very exothermic reaction of hydrogen
with oxygen:

2 H2 (g)+O2 (g)→2 H2O (l)

Figure 14.17 A hydrogen fuel cell divides the combustion of hydrogen into two half-
reactions and uses the transfer of electrons to produce an electric current.

Recall from acid-base chemistry that we commonly refer to H+ ions as

protons.

A hydrogen fuel cell consists of two electrodes, separated by a substance

called a proton exchange membrane. H+ ions can flow through the
membrane, but electrons cannot. At the anode, H2 undergoes an oxidation
to H+ ions:
Oxidation: H2 (g)→2 H++2 e−

896
At the cathode, oxygen is reduced, reacting with electrons and H+ ions to
form water:
Reduction: O2 (g)+4 H++4 e−→2 H2O (l)

Because the proton exchange membrane allows only H+ ions to flow

through it, the electrons must travel through an external route in order for
the reaction to take place. The result is an electric current.
Fuel cells were developed by NASA for use in space travel. More
recently, design improvements have made fuel cells a viable option for
many power applications. In 2016, the automaker Hyundai released the
first mass-produced vehicle powered by a hydrogen fuel cell (Figure
14.18).

Figure 14.18 In 2016, Hyundai launched the Tucson Fuel Cell, the first mass-produced
vehicle powered using a hydrogen fuel cell.

897
Summary
Oxidation-reduction (or redox) reactions involve the gain and loss of
electrons. An atom, molecule, or ion that loses electrons is oxidized; a
species that gains electrons is reduced. Oxidation-reduction reactions are
central to the field of electrochemistry, the study of chemical processes
that involve the movement of electrons.
There are several different classes of redox reactions. Metals and
nonmetals combine in redox reactions to produce ionic compounds. Metals
and nonmetals alike undergo combustion reactions, in which an element or
compound reacts with oxygen to produce new compounds called oxides.
These reactions often release a great deal of energy as heat or light.
Hydrocarbons (compounds composed of hydrogen and carbon) are an
important class of compounds because of their use as fossil fuels.
Combustion reactions of hydrocarbons produce two products, carbon
dioxide and water.
Metal displacement reactions are another common type of redox
reaction. In these reactions, an elemental metal reacts with a metal ion,
typically in aqueous solution. The elemental metal is oxidized to the ion
while the ion is reduced to the elemental form. These reactions are
examples of single-displacement reactions. Some metals are more active or
more strongly reducing than others. The activity series is a ranking of
metals by their reducing power; it allows us to predict which combinations
of metals and ions will react.
Most metals react with acids to produce metal ions and hydrogen gas.
The most active metals undergo a similar reaction with water.
Sometimes, it is helpful to write redox reactions as oxidation and
reduction half-reactions. To balance a redox reaction, the number of
electrons lost in the oxidation half-reaction must equal the number gained
in the reduction half-reaction. Stated differently, a balanced redox reaction
must not only balance the number and type of atom present, it must also
balance the overall charge.
Batteries (also called electrochemical cells) are devices that produce
electric current through a redox reaction in which the two half-reactions
occur at separate sites. Simple electrochemical cells contain two electrodes
connected by a salt bridge.
Batteries are just one of many important applications in the field of
electrochemistry. A related technology is electroplating, in which electric
potential energy is used to coat a material with a thin layer of a metal such

898
as chromium, nickel, or gold.
Another important application of electrochemistry is the fuel cell.
These devices convert the energy of a combustion reaction directly into
electrical energy by separating the oxidation and reduction half-reactions.
Fuel cells hold the potential to significantly impact the transportation
industry in the future.

899
Charging Ahead: Batteries Today and Tomorrow

Chainsaws are amazing tools (Figure 14.19). With a good saw, you can turn a
fallen tree into a stack of firewood in an hour. But they can also be difficult to
work with. They are heavy and dangerous. The small gasoline engines in most
chainsaws require a mixture of gas and oil, and they are notoriously hard to start.
But thanks to recent advances in battery technology, all of that is poised to change.

Figure 14.19 Chainsaws must be portable but also very powerful.

To power a chainsaw, a battery must be very strong but also lightweight. It

must last for a good length of time, and it must be rechargeable. For years, batteries
simply couldn’t meet these requirements. But a new generation of batteries called
lithium-ion batteries has changed the market for chainsaws, as well as other
products.
Lithium-ion batteries have been in use for over a decade in laptop computers
and other portable electronic devices. Like Volta’s original voltaic pile, these
batteries rely on oxidation-reduction reactions. But innovative new materials and
designs have maximized their performance. And unlike early batteries, lithium-ion
batteries are rechargeable.
Figure 14.20 illustrates the design of these remarkable devices. In this type of
battery, electrons are stored on graphite sheets. Lithium ions are able to move
freely between the sheets, balancing the overall charge. When the two electrodes
are connected, electrons flow through the circuit from the graphite sheets to a metal
oxide surface, such as Mn2O42–. The result is a net reduction:

Mn2O4−+e−→Mn2O42−

900
Figure 14.20 In a lithium-ion battery, electrons are stored on graphite sheets. The
lithium ions lie between the sheets and balance the overall charge. When the battery is
used, electrons flow through a circuit while lithium ions flow through a porous
membrane. Recharging the battery pushes the electrons and lithium ions back onto the
graphite sheets.

While the electrons flow through the circuit, the lithium ions migrate through a
porous membrane and into the Mn2O42– framework. This balances the overall
charge and allows the current to continue until the supply of extra electrons on the
graphite sheets is depleted.
Recharging a battery pushes the electrons back from the metal oxide surface to
the graphite surface. This process oxidizes the metal while reducing the graphite.
Once the electrons have been pushed back to the graphite surface, the battery is
recharged. Pop the battery back in, and you’re ready to go.

901
 Key Terms
14.1 Oxidation and Reduction
electrochemistry The study of chemical processes that involve the movement
of electrons.
oxidation-reduction reaction A chemical change in which one species loses
electrons (oxidation) while another gains electrons (reduction); also called a
redox reaction.
oxidation The loss of electrons.
reduction The gain of electrons.
oxidation number A bookkeeping tool for tracking electron changes in
oxidation-reduction reactions.

14.2 Types of Redox Reactions

combustion A reaction in which oxygen gas combines with elements or
compounds to produce oxide compounds.
metal displacement reaction A reaction between two metals in which one
metal is oxidized to its ionic form while the other metal ion is reduced to its
elemental form.
activity series A table that lists metals by their reducing powers; used to
predict whether metal displacement reactions will occur.

14.3 Half-Reactions and Batteries

half-reaction A chemical equation that shows only the reduction or oxidation
component of a reaction.
battery A device that produces electric current through a redox reaction in
which the two half-reactions occur at separate sites; also called an
electrochemical cell.
electrode In electrochemical cells and related devices, a site where an
oxidation or reduction half-reaction takes place.

14.5 Other Applications of Redox Reactions

electroplating A technique that uses electrical potential energy (such as from
a battery) to produce a thin layer of a metal such as gold, silver, copper, or
chromium on the outside surface of another material.
fuel cell A device that converts the energy of a combustion reaction directly
into electrical energy by separating the oxidation and reduction half-reactions.

902
 Additional Problems

14.1 Oxidation and Reduction

13. In each of the following changes, determine whether the element or ion is
oxidized or reduced:
a. A silver atom loses an electron to form an Ag+ cation.
b. A fluorine atom gains an electron to form a fluoride ion.
c. Oxygen reacts to form oxide ions.
d. Tin reacts to form tin(IV) ions.

14. In each of the following changes, determine whether the element or ion is
oxidized or reduced:
a. Elemental hydrogen forms hydrogen (H+) ions.
b. Elemental hydrogen forms hydride (H–) ions.
c. Sulfur reacts to form sulfide ions.
d. Magnesium reacts to form magnesium ions.

15. Why do scientists use oxidation numbers?

16. In Chapter 9, we examined the concept of formal charge. How do

oxidation numbers differ from formal charges?

17. Find the oxidation number for the metal ion in each of these compounds:
a. CoBr2
b. CuI
c. PbO2
d. K2S

18. Find the oxidation number for the metal ion in each of these compounds:
a. NaBr
b. PtCl4
c. NiCl2
d. AlF3

903
19. Determine the oxidation number for each element in these compounds:
a. CuCl
b. CuCl2
c. SiO2
d. H2O

20. Determine the oxidation number for each element in these compounds:
a. TiCl4
b. H2CO
c. H2CO3
d. SiBr4

21. Determine the oxidation number for each element in these compounds:
a. BeBr2
b. BBr3
c. NH3
d. LiH

22. Determine the oxidation number for each element in these compounds:
a. PCl3
b. PCl5
c. POCl3
d. PH3

23. Determine the oxidation number for each element in these ions:
a. Pb2+
b. Br−
c. BrO−
d. H2N−

24. Determine the oxidation number for each element in these ions:
a. Mg2+
b. ClO2−
c. ClO3−

904
 d. ClO4−

25. Determine the oxidation number for each element in these ions:
a. Cr3+
b. CO32–
c. IO3–
d. BF4–

26. Determine the oxidation number for each element in these ions:
a. NH4+
b. NO2–
c. HCO3–
d. CCl3–

27. In these reactions, identify the element that is oxidized and the element that
is reduced:
a. 2 Mg (s)+O2 (g)→2 MgO (s)

b. Fe (s)+S (s)→FeS (s)

c. Br2 (l)+Ca (s)→CaBr2 (s)

28. In these reactions, identify the element that is oxidized and the element that
is reduced:
a. 2 Cu (s)+O2 (g)→2 CuO (s)

b. S (s)+Hg (l)→HgS (s)

c. Sn (s)+2 Cl2 (g)→SnCl4 (s)

29. In these reactions, identify the species that is oxidized and the species that
is reduced:

905
 a. 2 H2 (g)+O2 (g)→2 H2O (g)

b. H2CO (aq)+O2 (g)→H2CO3 (aq)

c. H2S2 (g)+H2 (g)→2 H2S (g)

30. In these reactions, identify the species that is oxidized and the species that
is reduced:
a. 2 NaN3 (s)→2 Na (s)+3 N2 (g)

b. Mg (s)+CuBr2 (aq)→MgBr2 (aq)+Cu (s)

c. Ni (s)+2 HCl (aq)→NiCl2 (aq)+H2 (g)

31. In these reactions, identify the oxidizing agent:

a. 2 Mg (s)+O2 (g)→2 MgO (s)

b. CH4 (g)+2 O2 (g)→CO2 (g)+2 H2O (g)

c. Sn (s)+2 F2 (g)→SnF4 (s)

32. In these reactions, identify the reducing agent:

a. H2 (g)+Cl2 (g)→2 HCl (g)

b. 2 Na (s)+2 H2O (l)→2 NaOH (aq)+H2 (g)

c. Ca (s)+FeCl2 (aq)→CaCl2 (aq)+Fe (s)

33. In these reactions, identify the reducing agent:

906
 a. Ca (s)+2 H2O (l)→Ca(OH)2 (aq)+H2 (g)

b. F2 (g)+Be (s)→BeF2 (s)

c. S (s)+O2 (g)+H2O (g)→H2SO3 (g)

34. In these reactions, identify the oxidizing agent:

a. Si (s)+O2 (g)→SiO2 (s)

b. O2 (g)+4 Ag (s)→2 Ag2O (s)

c. 2 H2S (aq)+H2O2 (aq)→H2S2 (aq)+2 H2O (l)

14.2 Types of Redox Reactions

35. What types of compounds are formed by the reaction of metals with
nonmetals?

36. When a metal reacts with a nonmetal, which species is oxidized? Which
species is reduced?

37. Write molecular equations to describe the following oxidation-reduction

reactions. Include phase symbols, and identify the species that is oxidized
and reduced in each reaction.
a. Solid iron metal reacts with chlorine gas to form solid iron(II) chloride.
b. Liquid mercury reacts with oxygen gas to form solid mercury(II) oxide.

38. Write molecular equations to describe the following oxidation-reduction

reactions. Include phase symbols, and identify the species that is oxidized
and reduced in each reaction.
a. Solid calcium metal reacts with liquid bromine to form solid calcium
bromide.
b. Solid cobalt reacts with aqueous iodine to form aqueous cobalt(II)
iodide.

907
39. Predict the products from these reactions, and balance the equations:

a. Be (s)+Cl2 (g)→

b. K (s)+Cl2 (g)→

c. Co (s)+Cl2 (g)→

d. Cu (s)+O2 (g)→

40. Predict the products from these reactions, and balance the equations:

a. Ca (s)+O2 (g)→

b. Sr (s)+Br2 (l)→

c. Pb (s)+O2 (g)→

d. Al (s)+O2 (g)→

41. What element is always involved in combustion?

42. What are hydrocarbons? What two products always result from the
combustion of hydrocarbons?

43. Predict the products from these combustion reactions, and balance the
equations:

a. Mg (s)+O2 (g)→

b. C2H4 (g)+O2 (g)→

c. C4H10 (g)+O2 (g)→

44. Predict the products from these combustion reactions, and balance the
equations:

a. Zn (s)+O2 (g)→

b. C5H12 (l)+O2 (g)→

c. C8H18 (l)+O2 (g)→

908
45. Write molecular equations to describe the following chemical reactions.
Include phase symbols.
a. Solid cobalt reacts with aqueous tin(II) chloride to produce aqueous
cobalt(II) chloride and solid tin.
b. Aqueous gold nitrate reacts with silver metal to form solid gold metal
and aqueous silver nitrate.

46. Write molecular equations to describe the following chemical reactions.

Include phase symbols.
a. When barium metal is dropped into a solution of aqueous calcium
acetate, solid calcium and aqueous barium acetate are formed.
b. Aqueous nickel(II) chloride reacts with zinc metal to form aqueous zinc
chloride and solid nickel.

47. Nickel(II) chloride forms a green solution when dissolved in water (Figure
14.21). If a strip of aluminum is placed in the solution, the green color
begins to fade, and a coating can be observed on the surface of the
aluminum. Why does the color fade? What is the coating on the
aluminum?

Figure 14.21 Nickel(II) chloride forms a green solution in water.

48. When a copper wire is placed in a solution of silver chloride, the wire
slowly builds up a layer of silver (Figure 14.22). What is being oxidized
in this reaction? What is being reduced? If all of the silver reacts, what
ions are left in solution? Why is the solution blue?

909
 Figure 14.22 The changes shown occur when a copper wire is placed in a solution of
silver nitrate.

49. Rewrite these metal displacement reactions as complete ionic and net ionic
equations:
a. Mg (s)+Cu(NO3)2 (aq)→Mg(NO3)2 (aq)+Cu (s)

b. Al (s)+CrCl3 (aq)→AlCl3 (aq)+Cr (s)

c. 2 Cr (s) 3 PtBr2 (aq)→3 Pt (s)+2 CrBr3 (aq)

50. Rewrite these metal displacement reactions as complete ionic and net ionic
equations:
a. Ni (s)+SnSO4 (aq)→NiSO4 (aq)+Sn (s)

b. 2 Al (s)+3 CuSO4 (aq)→3 Cu (s)+Al2(SO4)3 (aq)

c. Al (s)+3 AgNO3 (aq)→Al(NO3)3 (aq)+3 Ag (s)

51. The following complete ionic equations represent metal displacement

reactions. Rewrite them as molecular equations.
a. Mg (s)+Pb2+ (aq)+2 NO3− (aq)→Mg2+ (aq)+2 NO3− (aq)+Pb (s)

b. 2 Al (s)+3 Pt2+(aq)+6 Br− (aq)→3 Pt (s)+2 Al3+ (aq)+6 Br− (aq)

910
52. The following complete ionic equations represent metal displacement
reactions. Rewrite them as molecular equations.
a. Fe (s)+2 Au+ (aq)+2 F− (aq)→Fe2+ (aq)+2 F− (aq)+2 Au (s)

b. 2 Cr (s)+3 Ni2+ (aq)+6 ClO4− (aq)→2 Cr3+ (aq)+6 ClO4− (aq)+3 Ni (s)

53. Based on the activity series (Table 14.2), which of these metals are able to
reduce copper(II) to elemental copper?
a. aluminum
b. calcium
c. platinum
d. gold

54. Based on the activity series (Table 14.2), which of these metals are able to
reduce chromium(III) to elemental chromium?
a. lithium
b. silver
c. nickel
d. tin

55. Based on the activity series (Table 14.2), which of these ions can be
reduced by elemental aluminum?
a. K+
b. Mg2+
c. Fe2+
d. Pb2+

56. Based on the activity series (Table 14.2), which of these ions can be
reduced by elemental iron?
a. Mg2+
b. Ni2+
c. Co2+
d. Cu2+

57. Based on the activity series (Table 14.2), determine whether these

911
 reactions would take place spontaneously:
a. Zn2+ (aq)+Mg (s)→Zn (s)+Mg2+ (aq)

b. Fe (s)+Sn2+ (aq)→Sn (s) +Fe2+ (aq)

c. 3 Ag (s)+Al3+ (aq)→3 Ag+ (aq)+Al (s)

d. Cr (s)+3 LiCl (aq)→3 Li (s)+CrCl3 (aq)

58. Based on the activity series (Table 14.2), determine whether these
reactions would take place spontaneously:
a. Ca2+ (aq)+Ni (s)→Ca (s)+Ni2+ (aq)

b. Ba (s)+Cu2+ (aq)→Ba2+ (aq)+Cu (s)

c. 2 Au (s)+Zn2+ (aq)→2 Au+ (aq)+Zn (s)

d. Pb (s)+2 KNO3 (aq)→2 K (s)+Pb(NO3)2 (aq)

59. Use the activity series (Table 14.2) to complete each of these reactions.
Also indicate if no reaction occurs.

a. Sn2+ (aq)+Ni (s)→

b. Ni (s)+Fe2+ (aq)→
c. CO (s)+Pb(C2H3O2)2 (aq)→

60. Use the activity series (Table 14.2) to complete each of these reactions.
Also indicate if no reaction occurs.

912
 a. Zn2+ (aq)+Mg (s)→

b. Zn (s)+Mg2+ (aq)→

c. Ba (s)+Cu(NO3)2 (aq)→

61. Which of these metals are oxidized by reaction with water?

a. Li
b. Na
c. Ca
d. Cu

62. Which of these metals are oxidized by reaction with water?

a. V
b. Pt
c. In
d. K

63. Complete these equations to show how each element reacts with water:

a. Na (s)+H2O (l)→

b. K (s)+H2O (l)→

c. Ca (s)+H2O (l)→

64. Each of these reactions occurs slowly with liquid water, but much more
quickly if the water is heated to the vapor phase (steam). Complete the
reactions to show the products formed.

a. Mg (s)+H2O (g)→

b. Zn (s)+H2O (g)→

c. Al (s)+H2O (g)→

65. Which of these metals are oxidized by reaction with acid?

a. aluminum
b. lithium

913
 c. nickel
d. platinum

66. Which of these metals are oxidized by reaction with acid?

a. zinc
b. silver
c. gold
d. iron

67. Complete and balance these equations to show how each element reacts
with hydrochloric acid:

a. Mg (s)+HCl (aq)→

b. Zn (s)+HCl (aq)→

c. Fe (s)+HCl (aq)→

68. Complete and balance these equations to show how each element reacts
with nitric acid:

a. Ni (s)+HNO3 (aq)→

b. Co (s)+HNO3 (aq)→

c. Al (s)+HNO3 (aq)→

69. Write net ionic equations to show the chemical reactions that occur in the
following situations. In some cases, no reaction will occur.
a. Calcium metal is combined with aqueous iron(II) ions.
b. Calcium metal is combined with aqueous acid.
c. A solution containing iron(II) ions is combined with copper metal.
d. A solution containing iron(II) ions is combined with zinc metal.

70. Write net ionic equations to show the chemical reactions that occur in the
following situations. In some cases, no reaction will occur.
a. Platinum metal is combined with aqueous acid.
b. Aluminum metal is combined with aqueous acid.
c. A solution containing nickel(II) ions is combined with aluminum metal.
d. A solution containing tin(II) ions is combined with chromium metal.

914
14.3 Half-Reactions and Batteries

71. Write two half-reactions to describe each of these metal displacement

reactions:
a. Mg (s)+CoBr2 (aq)→MgBr2 (aq)+Co (s)

b. Cu (s)+2 AuCl (aq)→CuCl2 (aq)+2 Au (s)

72. Write two half-reactions to describe each of these metal displacement

reactions:
a. Fe (s)+SnI2 (aq)→FeI2 (aq)+Sn (s)

b. 3 Ba (s)+2 AlCl3 (aq)→3 BaCl2 (aq)+2 Al (s)

73. Using the activity series (Table 14.2) as a guide, write two half-reactions to
show the reaction of iron metal with cobalt(II) ions.

74. Using the activity series (Table 14.2) as a guide, write two half-reactions to
show the reaction of Zn2+ with Mg.

75. Write two half-reactions to describe each of these metal displacement

reactions:
a. Elemental nickel reacts with silver ions to produce nickel(II) ions and
elemental silver.
b. Zinc reacts with cobalt(II) chloride solution to produce zinc chloride
and elemental cobalt.

76. Write two half-reactions to describe each of these metal displacement

reactions:
a. Elemental aluminum reacts with tin(II) ions to produce elemental tin
and aluminum ions.
b. Lead reacts with copper(II) sulfate solution to produce lead(II) sulfate
and elemental copper.

915
77. What are the roles of the anode, the cathode, and the salt bridge in an
electrochemical cell?

78. The salt bridge in an electrochemical cell contains sodium and nitrate ions.
As the redox reaction takes place, which ions flow toward the cathode?
Which ions flow toward the anode?

79. A chemist assembles an electrochemical cell as shown in Figure 14.23.

One side of the cell contains a zinc electrode immersed in aqueous zinc
nitrate solution. The other side of the cell contains an iron electrode
immersed in aqueous iron(II) nitrate solution. In this reaction, zinc is the
anode and iron is the cathode. Write equations showing the half-reactions
that take place at each electrode.

Figure 14.23 This electrochemical cell contains zinc and iron electrodes.

80. A chemist assembles an electrochemical cell as shown in Figure 14.24.

One side of the cell contains a silver electrode immersed in aqueous silver
nitrate solution. The other side of the cell contains a zinc electrode
immersed in aqueous zinc nitrate solution. In this reaction, zinc is the
anode and silver is the cathode. Write equations showing the half-reactions
that take place at each electrode.

916
 Figure 14.24 This electrochemical cell contains zinc and silver electrodes.

81. The electrochemical cell in Figure 14.25 is called a standard

electrochemical cell. In this type of cell, the concentration of dissolved
ions is 1.0 M in each solution. If we replaced the Cu2+ solution with pure
water, would the electrochemical cell still produce an electric current?
What if we replaced the Zn2+ solution with pure water?

Figure 14.25 A standard electrochemical cell contains 1.0-M solutions of dissolved ions.

917
82. After an electrochemical cell has been in operation for some time, the
appearance of the electrodes changes. For example, after a standard
copper-zinc cell (such as the one shown in Figure 14.26) has run for a
while, the zinc anode becomes much smaller, while the copper cathode
becomes much larger (Figure 14.26). At the same time, the blue color of
the copper(II) solution fades. How can you explain these three
observations?

Figure 14.26 This zinc-copper electrochemical cell is almost depleted. The anode has
nearly disappeared while the cathode is much larger. Why does this happen?

14.4 Balancing Redox Equations

83. Balance this ionic equation to show the reaction of aluminum metal with
aqueous silver ions:
Al (s)+Ag+ (aq)→Al3+ (aq)+Ag (s)

84. Balance this ionic equation to show the reaction of aqueous iron(II) ions
with chromium metal:
Fe2+ (aq)+Cr (s)→Cr3+ (aq)+Fe (s)

918
85. Balance this ionic equation to show the reaction of lithium metal with
dilute acid:
Li (s)+H+ (aq)→Li+ (aq)+H 2 (g)

86. Balance this ionic equation to show the reaction of aluminum with acid:
Al (s)+H+ (aq)→Al3+ (aq)+H2 (g)

87. Calcium metal reacts with aqueous chromium(III) chloride, as shown in

the half-reactions here. Write a balanced net ionic reaction based on these
two half-reactions.
Oxidation: Ca (s)→Ca2+ (aq)+2 e−Reduction:3 e−+Cr3+ (aq)→Cr (s)

88. Aluminum metal reacts with aqueous lead(II) chloride, as shown in the
half-reactions here. Write a balanced net ionic reaction based on these two
half-reactions.
Oxidation: Al (s)→Al3+ (aq)+3 e−Reduction:2 e−+Pb2+ (aq)→Pb (s)

89. Barium metal reacts with aqueous iron(III) chloride, as shown in the half-
reactions here. Write a balanced net ionic reaction based on these two half-
reactions.
Oxidation: Ba (s)→Ba2+ (aq)+2 e−Reduction:3 e−+Fe3+ (aq)→Fe (s)

90. Lead metal reacts with aqueous gold nitrate, as shown in the half-reactions

919
 here. Write a balanced net ionic reaction based on these two half-reactions.
Oxidation: Pb (s)→Pb2+ (aq)+2 e−Reduction:e−+Au+ (aq)→Au (s)

91. Zinc reacts vigorously with aqueous hydrobromic acid to produce zinc
bromide and hydrogen gas. What are the oxidation and reduction half-
reactions for this change? Write a balanced, net ionic equation to describe
this reaction.

92. Aluminum reacts with concentrated nitrous acid to produce aluminum

nitrite and hydrogen gas. What are the oxidation and reduction half-
reactions for this change? Write a balanced, net ionic equation to describe
this reaction.

93. Elemental chromium reacts with aqueous lead(II) nitrate in a single-

displacement reaction. What are the oxidation and reduction half-reactions
for this change? Describe this change using a balanced, complete ionic
reaction.

94. Aluminum metal reacts with aqueous nickel(II) bromide in a single-

displacement reaction. What are the oxidation and reduction half-reactions
for this change? Describe this change using a balanced, complete ionic
reaction.

14.5 Other Applications of Redox Reactions

95. What is electroplating? How is electroplating different from the reactions

that take place in an electrochemical cell (that is, a battery?)

96. Electroplating uses a battery to drive non-spontaneous reactions to take

place. Using the activity series (Table 14.2), determine whether the
following reactions would occur spontaneously or require an electrical
voltage for them to occur.
a. Zn2+ (aq)+Mg (s)→Zn (s)+Mg2+ (aq)

b. Fe (s)+Ni2+ (aq)→Ni (s)+Fe2+ (aq)

920
97. A manufacturer uses an electric current to gold-plate jewelry, using a gold
solution and zinc as a counter electrode (Figure 14.27). In this reaction, is
the gold oxidized or reduced? What is the anode in this reaction? What is
the cathode?

Figure 14.27 Jewelry can be gold-plated using an apparatus like the one shown here.

98. Consider the setup in Figure 14.27. What would happen if the direction of
current were reversed, so that electrons flowed toward the zinc and away
from the nickel loop?

99. What is a fuel cell? What are the key components of a fuel cell?

100. What is the overall reaction that takes place in a hydrogen fuel cell? What
half-reactions take place at the anode and at the cathode of a fuel cell?

Challenge Questions

101. Classify these oxidation-reduction reactions as synthesis, decomposition,

single-displacement, or combustion reactions:

a. Hg (l)+S (s)→HgS (s)

b. Fe (s)+Cu(C2H3O2)2 (aq)→Cu (s)+Fe(C2H3O2)2 (aq)

c. C8H16 (l)+12 O2 (g)→8 CO2 (g)+8 H2O (g)

921
102. Classify these oxidation-reduction reactions as synthesis, decomposition,
single-displacement, or combustion reactions:
a. Fe (s)+2 AgC2H3O2 (aq)→Fe(C2H3O2)2 (aq)+2 Ag (s)

b. Fe (s)+2 HCl (aq)→FeCl2 (aq)+H2 (g)

c. 2 K (s)+Br2 (g)→2 KBr (s)

103. Animals convert glucose into energy through a process called cellular
respiration. Based on the equation for cellular respiration shown here,
identify the oxidation state of each atom in both the reactants and the
products. In this reaction, which element is oxidized? Which element is
reduced?
C6H12O6 (s)+6 O2 (g)→6 CO2 (g)+6 H2O (l)+energy

104. In Chapter 15 we will examine the process of photosynthesis, which green

plants use to store energy from the sun in chemical bonds. Based on the
equation for photosynthesis shown here, identify the oxidation state of
each atom in both the reactants and the products. In this reaction, which
element is oxidized? Which element is reduced?
Energy+6 CO2 (g)+6 H2O (l)→C6H12O6 (s)+6 O2 (g)

105. Write a balanced equation showing the reaction of metallic iron with
aqueous nitric acid. If 25.35 grams of iron react with 500.0 mL of 1.20 M
nitric acid, how many grams of iron(II) nitrate can form?

106. What volume of 1.0-M gold nitrate is needed to completely react with
1.63 grams of zinc in a single-displacement reaction? What mass of gold
could be produced by this reaction?

922
Chapter Fifteen
Organic Chemistry and Biomolecules

Forming New Bonds: The Grubbs Catalyst

As a kid, Robert Grubbs loved putting things together. The son of a schoolteacher
and a diesel mechanic, he was fascinated with how things were built. While most
kids spent money on candy, he bought nails to connect pieces of scrap wood. As he
grew, so did his interests in construction and mechanics. He worked with his father
and on his uncles’ farms, fixing cars, installing plumbing, and building houses.
In 1961, Robert left his Kentucky home and enrolled at the University of
Florida as an agricultural chemistry major. He quickly found work in an animal
nutrition lab. It was nasty work—analyzing the remnants from steer feces.
Fortunately, a friend worked in a chemistry lab on campus, and Robert began
to help out at night. It was less smelly than his day job, and he found the work
fascinating. As he later described it, “Building new molecules was even more fun
than building houses.” The hands-on skills he had developed as a farm kid served
him well, and he excelled in the lab. Six years later, he completed his Ph.D. in
chemistry from Columbia University.
Young Dr. Grubbs was fascinated with the way metal atoms interact with
carbon. He explored new metal-carbon structures and unearthed new patterns of
chemical reactions. As he honed his interests, he began to focus on the transition
metal ruthenium. He discovered that compounds containing this metal could
catalyze the formation of new carbon–carbon bonds. Instead of using metal nails to
join boards, Grubbs now used metal atoms to connect carbon atoms (Figure 15.1).

923
Figure 15.1 (a) Professor Robert Grubbs works in his research laboratory at Caltech. (b)
The molecular structure of one of the Grubbs catalysts is shown here. (c) The durable
plastic on this combine was produced using the Grubbs catalyst. (d) This factory in
Indonesia produces materials for cleaning, lubricants, polymer, and personal care
applications from renewable resources like palm oil, using the Grubbs catalyst. (e)
Professor Grubbs received the Nobel Prize from the King of Sweden in 2005.

For years—first at Michigan State and then at Caltech—Grubbs tirelessly

studied these reactions that formed carbon–carbon bonds, refining his techniques
and exploring new uses. Today, the ruthenium catalysts he and his students
developed (commonly called “Grubbs catalysts”) produce a wide variety of
materials, including medicines, personal care products, automotive lubricants, and
even the tough plastic on John Deere tractors. In 2005, Grubbs and others were
awarded the Nobel Prize in chemistry for their development of new approaches to
building carbon–carbon bonds.
Why are carbon bonds so important? As we’ll see in the pages that follow,
carbon provides the framework for all living organisms, both plant and animal. By
bonding carbon with hydrogen, nitrogen, oxygen, and a handful of other elements,
nature forms a sweeping array of compounds that range from small, simple

924
molecules like methane gas to massive, complex structures like proteins.
And we can emulate nature. Starting from the simple carbon compounds in
fossil fuels, chemists use reactions like those Grubbs developed to synthesize new
compounds with innovative structures and exciting new properties.
In this chapter, we’ll survey the stunning diversity of carbon-based compounds.
Building on earlier chapters, we’ll reexamine the way carbon and other nonmetal
atoms bond. We’ll explore bonding patterns in familiar molecules and observe
simple reactions that take place with many different compounds. Finally, we’ll
look into the chemistry that takes place in living organisms and discover the
elegant ways that nature builds structures, stores energy, and even stores and
copies information. The chemistry of carbon—the chemistry of life—is
breathtaking. Let’s get started.

Intended Learning Outcomes

After completing this chapter and working the practice problems, you should be
able to:

15.1 Organic Chemistry and the Carbon Cycle

Broadly describe the exchange of carbon among the atmosphere, land, sea,
and plant and animal life.

15.2 Covalent Bonding with Carbon and Other Nonmetals

Draw neutral covalent structures for nonmetal compounds.
Identify compounds that are isomers, and draw isomers for a given molecular
formula.

15.3 Drawing Covalent Structures

Convert between Lewis, condensed, and skeletal representations for
molecular structures.

15.4 Major Functional Groups

Identify the structures and names of major functional groups.
Describe the products of condensation reactions in the formation of ethers,
esters, and amides.

15.5 Polymers and Plastics

Qualitatively describe the structural features of synthetic polymers.

15.6 Biomolecules—An Introduction

Identify the major structural features and functions of major classes of

925
biomolecules, including carbohydrates, proteins, and DNA.

926
15.1 Organic Chemistry and the Carbon Cycle
Organic chemistry is the chemistry of carbon-containing molecules. The
term organic refers to living organisms: All living creatures, whether plant
or animal, are composed of carbon-containing molecules. And much of the
carbon on Earth—whether it is found in the land, sea, or atmosphere—has
been part of a plant or animal at some point in Earth’s history.
The carbon cycle describes how Earth’s carbon moves among rocks,
sediments, water, atmosphere, plants, and animals (Figure 15.2). At its
core, the carbon cycle involves two key reactions. The first is
photosynthesis, the sequence used by green plants to harvest the energy of
the sun. In photosynthesis, plants combine carbon dioxide from the
atmosphere with water from the soil to produce oxygen gas and new
compounds like glucose (C6H12O6) that are composed of carbon,
hydrogen, and oxygen. The second reaction is the reverse of the first one:
Animals and plants combine oxygen with carbon compounds to produce
carbon dioxide and water and release energy. This process is called
cellular respiration.

Figure 15.2 Photosynthesis converts CO2 in the atmosphere into larger carbon-based
molecules in living things. In combustion or cellular respiration, the process is reversed.

Photosynthesis:Energy+6 CO2+6 H2O→C6H12O6+6 O2Cellular

respiration:C6H12O6+6 O2→6 CO2+6 H2O+energy

927
 Notice that the cellular respiration reaction is identical to a combustion
reaction: When we say that your body “burns” calories, this is a fairly
correct description. Both combustion and cellular respiration consume
oxygen and produce carbon dioxide and water.
When plants and animals die, the complex carbon structures begin to
decompose. The fossil fuels—coal, oil, and natural gas—form by the
decay of plant and animal matter over long periods of time (Figure 15.3).
These fuels contain a mixture of compounds that are composed mainly of
carbon and hydrogen.

Figure 15.3 This exposed coal seam formed from the gradual decay of plant and animal
matter.

Carbon is also central to aquatic life. The ocean exchanges carbon

dioxide with the atmosphere. In water, carbon dioxide reacts to form
carbonic acid and then bicarbonate ions:
CO2 (aq)+H2O (l)⇌H2CO3 (aq)H2CO3 (aq)⇌H+ (aq)+HCO3− (aq)

Many creatures produce hard calcium carbonate shells from the

bicarbonate ions dissolved in the water (Figure 15.4). Over time, these
shells accumulate on the seafloor, forming sediments that are rich in
carbon.

928
Figure 15.4 Many sea organisms create hard outer shells using calcium carbonate.

The carbon cycle is an intricate dance—a complex set of equilibria that

govern Earth’s composition, climate, and economic and natural resources.
In this chapter we will focus on the chemistry of carbon: How does carbon
bond? What kinds of carbon structures are most important to our lives?
And how do the structures of carbon compounds affect their properties?

929
15.2 Covalent Bonding with Carbon and Other Nonmetals
In Chapter 5, we explored chemical bonding. Recall that the nonmetal
atoms (Figure 15.5) bond by “sharing” electrons through covalent bonds.
Covalent bonds enable nonmetals to fill their valence level with eight
electrons, thereby fulfilling the octet rule.

Figure 15.5 Nonmetals form covalent bonds.

The octet rule: An atom is stabilized by having eight electrons in its valence
level.

The number of covalent bonds that an atom forms depends on the

number of electrons needed to complete its valence level. For example,
hydrogen needs only one electron to complete its valence, so it forms one
covalent bond. The halogens (F, Cl, Br, I) also need just one electron to fill
their valence levels, so they also form just one bond. Oxygen and sulfur
need two electrons, and so they usually form two bonds. Nitrogen and
phosphorus need three electrons and therefore form three bonds. Finally,
carbon needs four electrons to fulfill its valence, so it forms four covalent
bonds. Table 15.1 summarizes the bonds required in neutral atoms.

TABLE 15.1 Covalent Bonds in Neutral Atoms

Atom Valence Electrons Electrons Needed Covalent Bonds Formed

H 1 1 1

C 4 4 4

930
 N 5 3 3

O 6 2 2

F 7 1 1

For more details about covalent bonding, see Chapter 9.

Living tissue is built primarily of four atoms: carbon, hydrogen,

nitrogen, and oxygen. Using the simple bonding rules given above, these
four atoms can produce an infinite combination of compounds, shapes, and
properties. For example, consider the three small molecules shown in
Figure 15.6. Each molecule has a different structure—and therefore
different properties. But each molecule follows the same basic rules: four
bonds to carbon, three bonds to nitrogen, two to oxygen, and one to
hydrogen.

Figure 15.6 These three small molecules are composed of C, H, N, and O.

The structures in Figure 15.6 contain only single covalent bonds.

However, recall that double covalent bonds (two pairs of electrons) and
triple covalent bonds (three pairs of electrons) are also possible. For
example, Figure 15.7 shows four simple molecules that are commonly
used as raw materials for chemical manufacturing. Notice that each of
them has a double or triple covalent bond in its structure, but the
fundamental bonding rules are fulfilled.

931
Figure 15.7 These common industrial chemicals all have double or triple covalent
bonds.

Most molecules contain many covalent bonds. As the number of atoms

increases, many possible ways of combining the atoms arise. For example,
a molecule with the formula C3H7Cl may have two different bonding
arrangements:

In the structure on the left, the chlorine is attached to the central carbon
atom. In the structure on the right, the chlorine is attached to one of the
end carbons. Compounds with the same molecular formula but different
bonding sequences are called isomers. Because their structures are
different, isomers have different physical properties. For example, the two
molecules above have different melting and boiling points.

Isomers have different physical properties.

When distinguishing isomers, we must pay close attention to the way

the atoms are connected. Molecules may be drawn in many different
shapes (called conformations) and from many different perspectives. For
example, each of the structures shown in Figure 15.8 represents the same
molecule; there are three carbon atoms in a row, and a chlorine attached to
the carbon on the end. The molecules may be arranged in different
conformations—like a snake that coils or uncoils—but the sequence of
atoms is the same. Therefore, they are not isomers; they are different
conformations of the same molecule.

932
Figure 15.8 Much like a snake, many molecules coil into different shapes. These four
structures all represent the same molecule.

Like people, most molecules can fold into different shapes. When drawing
isomers, it’s important to focus on the connectivity. Molecules may be drawn in
different ways but still have the same connectivity.

Because it can form four bonds, carbon forms the backbone of most
organic structures. For example, consider the structure of octane, a key
component of gasoline (Figure 15.9). The structure of this molecule is
based on the arrangement of the carbon atoms.

Figure 15.9 Carbon forms the backbone in the structure of octane.

933
Example 15.1 Drawing Isomers
There are two isomers with the molecular formula C2H4F2. Draw Lewis
structures for each of them.
Hydrogen and fluorine both form only one bond, but carbon forms four
bonds. Because of this, the carbon atoms are the backbone for these
molecules. We can draw a carbon-only framework like this, using lines to
represent bonds to the carbon atoms:

Each of the six peripheral spaces is occupied by a hydrogen atom or a

fluorine atom. These can be arranged in two different ways: In one
arrangement, each carbon atom is attached to one fluorine atom. In the
other arrangement, both fluorine atoms are attached to the same carbon
atom. •

Example 15.2 Drawing Isomers

Draw the three isomers having the molecular formula C5H12.
The first isomer is simple: We draw five carbon atoms in a row, as shown.
To help visualize the structure, it’s helpful to number the carbon atoms:

To draw another isomer, we must create a different bonding arrangement.

Let’s remove carbon 5, and connect it to a different atom. If we connect it
to carbon 1, we get the same structure (five in a row). But if we connect it
to carbon 2 or 3, we get a new structure:

934
 Next, let’s remove carbon 4 and attach it to another atom. If we attach it to
carbon 1 (or to carbon 5) in the structure above, we get the same
arrangement—four in a row, with a branch at the second carbon atom.
However, if we attach it to carbon 2, we get a different structure:

These are the only three possible combinations for these atoms. If you have
drawn other structures, compare them to the ones shown here. If the
structure contains the correct number of bonds to each atom, it will match
one of the structures shown above, even if it is drawn in a different shape. •

IT
TRY

1. There are four possible isomers with the formula C4H9F. Draw
structures for each of them.

935
15.3 Drawing Covalent Structures
Lewis structures are incredibly useful for describing small molecules, but
they are cumbersome for larger molecules. For example, the anticancer
drug Taxol (introduced in Chapter 1) contains 113 atoms. But that is still
small: Later in this chapter, we’ll encounter molecules that contain
thousands or even hundreds of thousands of atoms. For large molecules,
we need simpler, more elegant ways to represent structures.
The first simplification is to draw Lewis structures without showing
nonbonded electrons. For example, the “alcohol” in alcoholic beverages is
ethyl alcohol (also called ethanol). Figure 15.10 shows this molecule as a
full Lewis structure and as a simplified structure in which the nonbonded
valence electrons are not drawn (but understood to be present). As you
continue to practice drawing molecules, you will get more comfortable
with the inferred meaning in simplified structures.

Figure 15.10 The alcohol in beer has the formula C2H6O. In a full Lewis structure
(top), all unshared electrons are shown. In a simplified structure (bottom), the
nonbonded electrons are not drawn, but inferred from the structure.

Condensed Structures
Scientists use condensed structures to show how atoms are connected. In
condensed structures, most covalent bonds are not shown, but atoms are
listed in their order of connectivity. For example, we could use condensed
structures to represent the compound C4H10:

936
 A second example is shown below. In this case we use parentheses to
indicate that a larger group (the —NH2 group) is connected to the second
carbon from the left.

Sometimes chemists use partially condensed structures to simplify

Lewis structures. In the molecule below, the C—H bonds are shown in
condensed form, but all other bonds are drawn explicitly.

Example 15.3 Representing Structures in Different Styles

From these condensed structures, draw simplified Lewis structures that
show each bond but do not show unshared electrons.
CH3OCH(CH3)2HOCH2CH(CH3)NHCH2CH2OH

We can draw fully expanded structures for these two molecules as follows:

Although we did not explicitly draw the unshared valence electrons on the
oxygen or nitrogen atoms, remember that these atoms do have unshared
electrons present. •

937
 IT
TRY

2. Write condensed representations for each of these Lewis structures:

Skeletal Structures
While less cumbersome than Lewis structures, condensed structures are
still most useful for fairly small molecules. Many molecules have larger,
more complex structures. And some arrangements, such as cyclic
structures (called rings), are hard to represent using condensed structures.
In these cases, chemists use a simplified drawing technique called skeletal
structures, as shown in Figure 15.11. For molecules containing only
carbon and hydrogen, there are three rules for drawing skeletal structures:

1. Carbon skeletons are represented using a line structure. Each vertex

(corner) of the line and each end point of the line represents a single
carbon atom.
2. The hydrogen atoms bonded to carbon are not drawn, but inferred
from the number of lines present.
3. Double and triple bonds are represented by double and triple lines.

938
Figure 15.11 For complex structures, chemists often use skeletal structures. The
structures above are the human female and male sex hormones. The two structures are
very similar, each having four rings. The upper structure is one of several closely related
structures that collectively are referred to as estrogens.

For example, Figure 15.12 shows the structures of two substances

found in crude oil. Each of them contains only carbon and hydrogen. The
Lewis structures are cumbersome and hard to read, but the skeletal
structures clearly and elegantly depict each arrangement.

Figure 15.12 These two components of crude oil contain only carbon and hydrogen.
Using skeletal structures makes it much easier to see how the atoms are connected.

If skeletal structures involve atoms other than hydrogen and carbon, two
other rules apply:

939
 4. Atoms other than carbon and hydrogen are shown with their atomic
symbol.
5. Hydrogen atoms bonded to atoms other than carbon are written
explicitly as condensed structures.

For example, Figure 15.13 shows the structure of the common pain
medicine acetaminophen, represented both as a full Lewis structure and as
a skeletal structure. Notice that the hydrogen atoms on the oxygen and the
nitrogen are shown in condensed form, but those attached to carbon are
not.

Figure 15.13 Acetaminophen is the active component of the commercial pain medicine
Tylenol®.

In the classic comic strip Calvin and Hobbes, artist Bill Watterson used simple
lines to depict characters, actions, and emotions. Similarly, line structures
convey a wealth of structural meaning in a very simple way.

940
Example 15.4 Converting between Skeletal and Lewis Structures
The structure shown below is the commercial pain medicine ibuprofen.
Redraw this molecule using a complete Lewis structure.

In this structure, each end point or vertex that does not have an explicit
atom label represents a carbon atom. One helpful way to visualize the
carbon atoms is to draw a dot on each end point. Then, because each carbon
atom must have four bonds, any dot that does not have four bonds shown
must have hydrogens attached. We draw in the “understood” hydrogens on
each dot:

Finally, we can redraw our structure using atom labels for each carbon and
including unshared electrons on each oxygen atom. •

Example 15.5 Converting between Skeletal, Condensed, and

Lewis Structures
Food manufacturers sometimes add propylene glycol to packaged foods to

941
 keep them moist. The condensed structure for propylene glycol is
CH2(OH)CH(OH)CH3. Draw a Lewis structure and a skeletal structure for
this compound.
The first part of the structure has the formula CH2(OH). This indicates that
the first carbon is bonded to two hydrogens. The (OH) in parentheses
means that the carbon is attached to an oxygen, which in turn is attached to
another hydrogen. The second carbon is attached to one H, and also to an
OH. Finally, the third carbon is connected to three hydrogens. We draw the
complete Lewis structure and the corresponding skeletal structure as shown
here. •

IT
TRY

3. Compounds called alpha-glucosidase inhibitors are used to treat

diabetes. In 2015, a team of French scientists reported the preparation
of a new alpha-glucosidase inhibitor using a Grubbs catalyst. Based on
its skeletal structure (shown below), what is the molecular formula of
this compound? Redraw this structure as a full Lewis structure.

942
15.4 Major Functional Groups
Imagine you’re driving across the United States. As you head west through
rural Colorado, you see a sign for an exit one mile ahead. You’re feeling
hungry, so you move over into the right lane and prepare to exit. How did
you know to move into the right lane?
The answer is simple: Nearly all rural exits in the United States are on
the right-hand side of the road. You’ve driven on other roads before, so
you know what to expect on this one. Similarly, if you encounter a stop
sign, a traffic light, or a railroad crossing, you know what to expect:
You’ve seen these structural features before.
In much the same way, molecules have structural features that behave
in consistent ways. These structural features are called functional groups
—small groups of atoms within a molecule that behave in a characteristic
way. These behavior patterns allow us to predict the physical and chemical
properties of molecules. In this section, we’ll survey the common
functional groups.

Hydrocarbon Functional Groups

Alkanes and Cycloalkanes
Hydrocarbons are compounds that contain only carbon and hydrogen.
Alkanes are hydrocarbons that contain only single bonds. The simplest
alkane, CH4, is the major component of natural gas. Larger alkanes are
also used as fuels. For example, butane (C4H10) is the fuel in lighters,
while alkanes having six, seven, or eight carbon atoms are all components
of gasoline. Larger alkanes burn more slowly: Kerosene and the wax in
candles are composed of larger alkanes (Figure 15.14).

Alkanes contain only C and H, and all single bonds.

Figure 15.14 Alkanes are commonly used as combustion fuels. Lighter fluid, gasoline,

943
kerosene, and candle wax are all composed of alkanes.

Chemists name alkanes based on the number of carbons in the chain.

Each alkane name uses a prefix that indicates the number of carbons. The
suffix –ane indicates that the structure contains only single bonds. The
names and structures of the straight-chain alkanes are given in Table 15.2.
Notice from this table that each straight-chain alkane has two hydrogens
per carbon atom, plus one extra hydrogen on each end of the chain.
Because of this, alkanes have the general formula CnH2n+2.

TABLE 15.2 The Alkanes from C1 to C10

Condensed
Formula Lewis Structure Structure Name

CH4 CH4 Methane

C2H6 CH3CH3 Ethane

C3H8 CH3CH2CH3 Propane

C4H10 CH3(CH2)2CH3 Butane

C5H12 CH3(CH2)3CH3 Pentane

C6H14 CH3(CH2)4CH3 Hexane

C7H16 CH3(CH2)5CH3 Heptane

944
C8H18 CH3(CH2)6CH3 Octane

C9H20 CH3(CH2)7CH3 Nonane

C10H22 CH3(CH2)8CH3 Decane

At a dinner table, seats are placed across from each other, and two additional
seats are at the ends of the table. Similarly, straight-chain alkanes contain two
hydrogen atoms per carbon plus two more on the ends. This structure results in
the formula CnH2n+2.

Alkanes commonly have branched structures. For instance, we saw

earlier in this chapter (Example 15.2) that the formula C5H12 has three
isomeric forms (Figure 15.15). The presence of branches in the chains
does not affect the number of hydrogen atoms present.

945
Figure 15.15 Three isomers have the formula C5H12.

Cycloalkanes are alkanes that form a cyclic structure (commonly

called a ring). Cycloalkanes are named based on the number of carbon
atoms in the ring, but the prefix cyclo– is added to the alkane name to
indicate that it is a cyclic structure (Figure 15.16).

Figure 15.16 Skeletal structures of four cycloalkanes.

The molecular formulas of cycloalkanes differ from those of linear

alkanes. For example, compare the structures of propane and cyclopropane
in Figure 15.17: The cyclic structure has two fewer hydrogen atoms than
the straight-chain structure. Cycloalkanes have the general formula CnH2n.

946
Figure 15.17 Linear alkanes have the general formula CnH2n+2. Cycloalkanes have the
general formula CnH2n.

Aside from their flammability, alkanes are fairly unreactive. Organic

chemists often consider alkane and cycloalkane structures as the
foundation for organic molecules, but not one of the major functional
groups.

Example 15.6 Formulas for Alkanes and Cycloalkanes

Dodecane is an alkane that contains a 12-carbon chain. How many
hydrogen atoms are in dodecane? How many hydrogen atoms are in
cyclododecane?
For linear alkanes (those without a ring), the molecular formulas follow the
general formula CnH2n+2. For dodecane, n = 12, so
2n+2=2(12)+2=26 hydrogen atoms

Therefore, the molecular formula of dodecane is C12H26.

For cyclododecane, the formula is CnH2n, or C12H24. •

IT
TRY

4. What are the molecular formulas for heptane and for cycloheptane?
Draw a skeletal structure for each of these molecules.

Alkenes and Alkynes

947
A carbon–carbon double bond is called an alkene functional group.
Chemists also use this term for simple molecules that contain a C=C bond.
The simplest alkene is ethene (also called ethylene), and it is shown in
Figure 15.18. This small molecule, which is a gas at room temperature, is
an important building block for making plastics (we’ll discuss plastics in
Section 15.5). Larger molecules also contain alkene functional groups. For
example, limonene (Figure 15.19) is the major component of citrus oil,
and it contains two alkene functional groups.

Figure 15.18 Ethylene is a precursor for polyethylene plastic.

Figure 15.19 Limonene is the major component of citrus oil and is widely used in
household cleaners.

If a molecule contains only one major functional group, the functional group
name is sometimes used to describe the entire molecule. For example, alkene

948
 refers both to the specific functional group (the C=C bond) and to a simple
molecule that contains this functional group.

Alkynes are hydrocarbons that contain a carbon–carbon triple bond.

The simplest alkyne, called acetylene (also called ethyne) is a hot-burning
combustion fuel used for welding (Figure 15.20). Acetylene is also a
precursor for making plastics that conduct electricity. Although some
naturally occurring compounds contain alkynes, they are much less
common than alkenes.

Three types of hydrocarbons:

Alkanes—all single bonds
Alkenes—double bond
Alkynes—triple bond

Figure 15.20 Acetylene is used as a fuel for welding torches and as a precursor in
building plastics that conduct electricity.

Aromatic Compounds
The molecule in Figure 15.21 is called benzene. This simple cyclic
structure contains six carbons, and it is commonly drawn with alternating
single and double bonds around the ring. However, this is no ordinary
alkene: Benzene is an unusually stable molecule because of its unique
structure. In a double bond, the first set of electrons occupies the space
between the two atoms; the second set of electrons, called pi electrons,
occupies the space above and below. Because of the ring structure, the pi
electrons in benzene actually occupy the entire region above and below the
six carbon atoms (Figure 15.22). To show this unique behavior, chemists

949
sometimes represent benzene as two separate Lewis structures (called
resonance structures) or as a single six-carbon skeletal structure with a
circle in the middle (Figure 15.22).

Figure 15.21 Benzene contains six carbon atoms in a ring with alternating single and
double bonds.

Figure 15.22 (a) In benzene, electron waves occupy the region above and below the
carbon ring. Sometimes this is shown (b) as two separate structures, called resonance
structures, or (c) as a single six-atom ring with a circle inside it.

Benzene is an example of an aromatic ring, a ring structure that

contains alternating single and double bonds and is generally less reactive
than simple alkenes. Although benzene contains only carbon and
hydrogen, aromatic rings can contain other atoms. For example, the
molecule pyridine (Figure 15.23) is structurally very similar to benzene
and also exhibits aromatic behavior.

Figure 15.23 Pyridine is similar to benzene and is also aromatic.

950
 Aromatic rings are present in many molecules, both naturally occurring
and human-made. For example, the explosive TNT and the pain
medication aspirin both contain a benzene ring (Figure 15.24). If you look
back at Figure 15.1, you’ll notice that the Grubbs catalyst also contains a
benzene ring. Although benzene itself is very toxic, more complex
structures that contain benzene may be perfectly safe. The benzene ring
gives shape and structural stability to many types of molecules.

Figure 15.24 Both TNT and aspirin contain a benzene ring in their structures.

IT
TRY

5. Identify the alkene, alkyne, and aromatic functional groups in these

molecules:

Oxygen-Containing Functional Groups

Alcohols and Ethers
Many organic molecules contain oxygen. Perhaps the most abundant
functional group of all is the alcohol group. In an alcohol, an oxygen atom
is singly bonded to a carbon atom and to a hydrogen atom, giving the
bonding sequence C—O—H.
Alcohols are structural analogs of water (Figure 15.25). These
molecules form strong hydrogen bonds with water and with other alcohol
molecules. As a result, small alcohols mix easily with water. There are

951
many common small alcohols, including ethyl alcohol (the alcohol in beer
and wine), isopropanol (commonly called rubbing alcohol), and ethylene
glycol (used to prepare antifreeze mixtures). There are millions of
examples of larger molecules that contain alcohol groups.

Figure 15.25 Alcohols contain C—O—H bonds.

The ether functional group is composed of an oxygen atom that is

singly bonded to two carbon atoms; it contains the bonding sequence C—
O—C. Ethers are common in natural molecules, and they are important
solvents for many manufacturing processes. Chemists commonly prepare
ethers by combining two alcohols to form an ether and water. This is an
example of a condensation reaction—a reaction in which two smaller
molecules combine to produce water plus a larger molecule.
For example, diethyl ether is a common but very flammable solvent. It
is formed by the combination of two molecules of ethyl alcohol:
CH3CH2OH+HOCH2CH3→CH3CH2OCH2CH3+H2OTwo alcohols→Etherwater

Condensation reactions like this one are common both in nature and in the
laboratory and factory. In Section 15.6 we’ll see how nature uses
condensation reactions to form molecules like proteins and sugars.

In a condensation reaction, two molecules combine to form water plus a

larger molecule.

952
 Example 15.7 Drawing Condensation Products
The industrial solvent MTBE is produced by a condensation reaction of two
alcohols to form an ether. Given the precursors shown below, draw the
structure of this product.

In a condensation reaction, two molecules combine with the elimination

of water. In this reaction we can remove the OH from one alcohol and
the H (on the oxygen atom) from the other. The carbon atom that was
attached to the OH forms a new bond to the oxygen atom in the other
molecule. This reaction gives the products shown here. •

Carbonyl Groups
After the alcohol group, perhaps the next most common functional group
is the carbonyl group. A carbonyl is a carbon–oxygen double bond. The
behavior of a carbonyl group depends largely on the other two atoms that
are bonded to the carbon atom. For this reason, chemists use more specific
functional group names that also include the atoms surrounding the
carbonyl.
In an aldehyde, the carbonyl carbon bonds to a carbon atom on one
side and a hydrogen atom on the other:

Small aldehydes are often quite fragrant, and they are welcome additions
in the kitchen. For example, the main component of both cinnamon and
vanilla is aldehydes, in addition to other functional groups (Figure 15.26).

953
Figure 15.26 The active ingredients of both cinnamon and vanilla contain aldehyde
functional groups. What other functional groups are present in these molecules?

In a ketone, the carbonyl carbon is bonded to carbon atoms on both sides:

Ketones are more stable than aldehydes. The simplest ketone is

acetone, CH3COCH3. This compound is a common solvent. Ketones are
found across a wide variety of substances. For example, carvone is the
“minty” component of spearmint oil (Figure 15.27).

Figure 15.27 Carvone is the flavorful component in spearmint oil.

In a carboxylic acid, a carbonyl is coupled with an alcohol (OH):

954
In condensed structures, we commonly represent this structure as —
COOH or as —CO2H. Carboxylic acids are present throughout plant and
animal systems. For example, “fatty acids” are carboxylic acids that
contain long carbon tails (Figure 15.28). These acids are abundant in
animal fat. Vegetable oils are also composed of fatty acids, and fatty acids
are a critical component of cell structure.

Figure 15.28 Cooking oils contain a mixture of fatty acids.

As their name implies, carboxylic acids are acidic. For example, acetic
acid ionizes in solution to produce H+ and its conjugate base, the acetate
ion:
CH3CO2H (aq)⇌CH3CO2− (aq)+H+ (aq)

At the pH levels in biological systems, a large percentage of carboxylic

acids are deprotonated (that is, the equilibrium lies toward the conjugate
base, —COO−, rather than the acid, —COOH). The conjugate bases of
carboxylic acids are called carboxylate ions. We’ll encounter several
examples of carboxylate ions in Section 15.6.

955
 Vinegar is a solution of acetic acid in water.

The ester functional group is similar to the carboxylic acid group,

except that the O—H is replaced with an O—C bond:

Esters commonly form through a condensation reaction involving a

carboxylic acid with an alcohol. For example, ethyl acetate is a widely
used solvent that is often present in fingernail polish remover. This ester
may be prepared by the condensation reaction of acetic acid with ethyl
alcohol:

A Summary of Oxygen-Containing Groups

So far in this section, we’ve looked at six main oxygen-containing groups:
alcohols, ethers, aldehydes, ketones, carboxylic acids, and esters. These six
structures are summarized in Figure 15.29. The oxygen-containing
functional groups are all around us. Many complex molecules contain
multiple oxygen-containing functional groups. I encourage you to practice
identifying these functional groups in the examples that follow and in the
end-of-chapter practice questions.

956
Figure 15.29 A summary of the oxygen-containing functional groups.

Example 15.8 Identifying Functional Groups

When wood burns, it produces a compound called syringol. Food
manufacturers use this pleasant-smelling substance to produce “smoky”
flavors. What functional groups are present in this syringol molecule?

This molecule has three major functional groups: An aromatic ring

(benzene), an alcohol group (the OH at the top), and two ether groups (the
oxygen atoms on the left and right side, which are bonded to both the CH3
and to a carbon atom on the benzene ring). •

IT
TRY

6. Curry is a delicious spice blend that is used in dishes across Southeast

Asia. One of the key compounds in curry is called curcumin. A
molecule of curcumin is shown below. What functional groups can
you identify in this molecule?

957
 7. Butyl acetate is a sweet-smelling compound that occurs naturally in
many fruits. This compound can be manufactured by the condensation
of a carboxylic acid (acetic acid) with an alcohol (butyl alcohol).
Given the reactant structures below, draw the structures of the two
products, butyl acetate and water.

Nitrogen-Containing Functional Groups

Amines contain a nitrogen atom with three single bonds. In an amine, the
nitrogen atom is bonded to one, two, or three carbon atoms. Amines are
very common in biological molecules. For example, amino acids, as their
name implies, contain both an amine group and a carboxylic acid group.
(We’ll look at amino acids in detail in Section 15.6.) Many other
biologically important compounds also contain amines: For example, the
molecules dopamine and serotonin carry nerve signals in the brain. These
important molecules are critical for motor function, mood modulation,
perceptions of hunger and pain, and so on. Both of these molecules contain
an amine functional group (Figure 15.30).

958
Figure 15.30 Dopamine and serotonin regulate brain function. Both contain amine
functional groups.

Amines are slightly basic compounds. Recall that ammonia reacts with
water to produce the ammonium ion and hydroxide ion:
NH3 (aq)+H2O (l)⇌NH4+ (aq)+OH− (aq)

Amines like methylamine (CH3NH2) react very similarly in water:

CH3NH2 (aq)+H2O (l)⇌CH3NH3+ (aq)+OH− (aq)

Amides are nitrogen-containing analogs of acids and esters. These

compounds contain a carbonyl adjacent to a nitrogen atom:

Amides are less reactive than carboxylic acids or esters. Amides form the
linkages between building blocks in protein molecules (we’ll discuss
proteins in Section 15.6). Amides are also found in many smaller
molecules, such as the insect repellant DEET and the molecule capsaicin,
the substance that gives chili peppers their “hot” flavor (Figure 15.31).
Amide bonds also make up the key link in materials like nylon.

959
Figure 15.31 The insect repellant DEET and the hot flavoring capsaicin both contain
amide functional groups.

Finally, the nitrile functional group is composed of a carbon–nitrogen

triple bond. The simplest nitrile, HCN, is a highly toxic gas. However, as
with other functional groups, the biological effects depend on the structure
of the complete molecule. Citalopram (Figure 15.32) is a substance used
to treat depression, anxiety, and obsessive-compulsive disorder. It contains
both an amine and a nitrile functional group.

Figure 15.32 The antidepressant drug citalopram has several functional groups,
including an ether, an amine, and a nitrile.

Example 15.9 Identifying Functional Groups

Dentists commonly use lidocaine to numb a patient’s mouth before
treatment. The structure of lidocaine is shown below. Identify and name the

960
 two nitrogen-containing functional groups present in lidocaine.

The nitrogen atom on the right-hand side of this molecule is surrounded by

three carbon atoms. This is an amine group. The nitrogen atom on the left-
hand side of the molecule is adjacent to a carbonyl—this is an amide group.
•

IT
TRY

8. Tranilast is an anti-allergenic drug used in parts of Asia. This

compound contains several of the functional groups we’ve discussed.
Can you identify five different functional groups in this molecule?

961
15.5 Polymers and Plastics
Polymers are molecules containing simple repeating units linked together
in long chains of covalent bonds. Biopolymers are naturally occurring
polymers. Many vitally important natural compounds have polymer
structures. Biopolymers give wood its rigid structure. Proteins and even
DNA are biopolymers. We will look at the structures of these compounds
in more detail in the next section.
Plastics are synthetic polymers. From garbage bags to car parts to
water bottles, plastics are a seemingly indispensable part of life. Plastics
are produced from simple small molecules called monomers that assemble
to form long chains (Table 15.3).

TABLE 15.3 Common Monomers and Polymers

Polymer Monomer(s) Uses

Grocery bags, packaging, etc.

Toys, furniture, fabrics, bottles,

etc.

PVC piping (plumbing)

962
 Synthetic fabrics, resins, etc.

Electronic components, DVDs,

automotive headlights, bulletproof
glass, etc.

Chemists normally represent polymers by drawing the structure of the

repeating unit, surrounded by parentheses and the subscript n to indicate
that this unit occurs over and over. For example, polypropylene is a
common plastic used in toys, furniture, kitchen storage containers, and a
host of other applications (Figure 15.33).

Figure 15.33 Plastics like polypropylene are composed of long molecular chains
containing simple repeating units.

It is composed of repeating three-carbon units. We represent the structure

of polypropylene this way:

963
 Polyester is a polymer made by combining a molecule with two
carboxylic acids (terephthalic acid) with another molecule containing two
alcohols (ethylene glycol) in a condensation reaction (Figure 15.34). This
produces a repeating structure connected by ester groups. Water is a side
product in this reaction.

Figure 15.34 Polyester is a polymer that forms through a condensation reaction.

964
Explore
Figure 15.34

965
15.6 Biomolecules—An Introduction
So far, we’ve seen how atoms combine to form molecules. We’ve also
looked at how different arrangements of atoms within a molecule, called
functional groups, contribute to the overall structure and behavior of the
substance. In this section, we’ll briefly introduce the structures and
properties of some of the most important biological molecules. These
molecules, which form in living organisms, are essential to their structure
and function.

Carbohydrates are composed of carbon, hydrogen, and oxygen.

Hydrocarbons are composed of carbon and hydrogen only.

Carbohydrates
Carbohydrates are compounds composed of carbon, hydrogen, and
oxygen. These molecules play many important roles in plant and animal
life. Most importantly, they store energy: Recall from Section 15.1 that
plants store energy through the process of photosynthesis. This energy is
stored in the form of carbohydrate molecules.
The term carbohydrate means “carbon and water.” Carbohydrates have
the general formula Cm(H2O)n, where the subscripts m and n denote the
number of carbon atoms and H2O units, respectively (Figure 15.35).

Figure 15.35 Table sugar is a compound called sucrose. This carbohydrate has the
molecular formula C12H22O11, also written as C12(H2O)11.

Simple carbohydrate molecules usually have between one and six

carbon atoms. Glucose, with the formula C6H12O6, is perhaps the most
important simple carbohydrate molecule. Glucose exists in three primary
forms: a linear form and two closely related cyclic forms. Figure 15.36
depicts these forms, using numbers for each carbon atom to clarify how
the structures interconvert. In the linear form, glucose contains five alcohol

966
groups and one aldehyde. To convert to the cyclic form, the alcohol on
carbon 5 reacts with the aldehyde (carbon 1) to form a new six-membered
ring. In the cyclic form, carbon 1 is singly bonded to two oxygen atoms—a
newly formed ether linkage and an alcohol group.

Figure 15.36 Glucose converts between an open form (shown in green) and two closed,
or cyclic, forms. The arrangement of atoms on carbon 1 is the only difference between
the two closed forms. In these skeletal structures, the darker lines indicate the front of
the ring—the part of the molecule that is closest to you. The bond that forms during the
cyclization process is shown in red.

Explore

Figure 15.36

The two cyclic forms differ only in the position of the alcohol on
carbon 1: In the alpha form (called α-glucopyranose), the OH on carbon 1
points down from the ring. In the beta form (β-glucopyranose), the OH on
carbon 1 points out from the ring.

967
 The alpha and beta forms of glucopyranose are closely related compounds
called geometric isomers. These compounds have the same bonding sequence,
but they are locked into different spatial arrangements that give them slightly
different properties.

Carbohydrate molecules link together through condensation reactions.

A molecule composed of two linked carbohydrate molecules is a
disaccharide. For example, Figure 15.37 shows the formation of maltose
(commonly called malt sugar) from two glucopyranose molecules. In this
condensation, two alcohols combine to produce an ether linkage, plus
water. The ether link between two simple sugars is called a glycosidic
bond.

Figure 15.37 Maltose is a disaccharide formed from two glucopyranose molecules.

Explore
Figure 15.37

Glycosidic bonds often form between carbon 1 of the first molecule

and carbon 4 of the second molecule; this is called a 1,4-linkage. In the
same way that the — OH group can be in the alpha or beta positions, the
glycosidic bond also forms in the alpha or beta positions. These are
referred to as alpha-linkages and beta-linkages. In the example of maltose
above, the alpha alcohol on carbon 1 connects with the alcohol on carbon 4
of the second molecule. This is called an α(1→4) linkage (read as “alpha
1-4”).

Polysaccharides are naturally occurring polymers of carbohydrates.

Polysaccharides are naturally occurring polymers composed of many

simple carbohydrates that are connected by glycosidic bonds. Two
important polysaccharides, cellulose and starch, are built entirely from
glucose molecules (Figure 15.38). In cellulose, the glucose molecules

968
connect using only β(1→4) bonds. Plants create cellulose and use it in
rigid cell walls. Cellulose also gives wood its rigid quality. In starch, the
glucose molecules connect by α(1→4) bonds. This material is less rigid
than cellulose. Starches are found in many foods, such as potatoes, wheat,
corn, and rice.

Figure 15.38 Our bodies can break down the α(1→4) bonds in starch, but not the
β(1→4) bonds in cellulose. Because of this, we eat the potato but not the cutting board.

Our bodies contain enzymes that slowly break starch down into
glucose molecules, then convert the glucose into energy through cellular
respiration. However, humans cannot break down the β(1→4) bonds in
cellulose. Because of this, wood, grass, and most leaves do not serve as
nutrients.

The carbohydrates in food that our bodies can’t break down are called fiber or
roughage. While we don’t gain nutritional value from these compounds, they do
keep the digestive tract working smoothly.

Amino Acids and Proteins

Proteins are large molecules that are vital to life. Some proteins function

969
as enzymes: They catalyze very specific reactions. Other proteins transport
molecules. Proteins are essential to the replication of our genetic code.
And every move we make—every muscle contraction—arises from a
change in the shape of a protein.
Proteins are biopolymers composed of building blocks called amino
acids. Unlike most polymers, which contain simple repeating units, each
protein contains a specific sequence of amino acid building blocks. This
sequence gives the protein its unique properties.
Amino Acids
There are 20 fundamental amino acids (Table 15.4). Each amino acid
contains a backbone consisting of an amine, a central carbon, and a
carboxylic acid. The central carbon of each amino acid connects to a
unique side chain (shown in blue).

TABLE 15.4 The 20 Amino Acids

970
971
 The English alphabet contains 26 letters that we use to create words, sentences, and
even books. The amino acids are the “letters” from which complex proteins are
made.

Amino acids are sometimes represented in their neutral form, as shown

on the left-hand side of the equilibrium below. However, because the
amine is basic and the carboxylic acid is acidic, amino acids undergo an
internal acid-base reaction to produce two charged sites. This equation
shows the equilibrium for the simplest amino acid, glycine:

In water (and in cells), amino acids exist almost completely in the ionic
form. Because of the two charged sites on the ionic form, amino acids
dissolve easily in water.

Peptides and Peptide Bonds

Amino acids link together through condensation reactions. In these
reactions, the carbonyl of one amino acid connects with the amine of
another to form an amide. The bond formed between amino acids is called
a peptide bond, and groups of amino acids connected in this way are
called peptides. For example, glycine and alanine can bond through this
condensation reaction:

The molecule formed from two amino acids is a dipeptide. Similarly, a

chain of three amino acids is a tripeptide. Proteins are polypeptides—they

972
are built from many amino acids. In the dipeptide structure above, notice
that there is still an amine group and a carboxylic acid group on either end
of the molecule. The amine end is called the N-terminus. The carboxylic
acid end is called the C-terminus (Figure 15.39).

Figure 15.39 This two-amino-acid structure contains an N-terminus, a C-terminus, and

a peptide bond connecting the two amino acids.

Amino acids connect through peptide bonds.

Longer peptide sequences have a huge number of atoms present. For

these molecules, using Lewis structures and even skeletal structures is
cumbersome. For simplicity, chemists and biologists often describe
peptides using three-letter designations to represent the amino acid
sequence. In this convention, the structure is always written with the N-
terminus on the left-hand side. For example, consider a tripeptide
containing lysine (on the N-terminus), connected to glutamine, connected
to tryptophan (the C-terminus). We can represent this complex structure
using the designation Lys-Gln-Trp. Example 15.10 gives an additional
example of this type of representation.

Example 15.10 Drawing Peptide Chains

Draw the tripeptide Cys-Ala-Thr.
Using Table 15.4, we draw the structures of the three amino acid building
blocks, orienting the N-terminus of each molecule to the left:

Next we connect the three structures. N is connected to the C=O of the

973
 structure to the left:

Again, notice that the first amino acid named in the sequence contains the
N-terminus, and the last one named contains the C-terminus. •

IT
TRY

9. Draw the structure of the tripeptide Thr-Ala-Cys. How does this

compare with the tripeptide Cys-Ala-Thr that we drew in Example
15.10?

10. Use the three-letter abbreviations to represent the tripeptide sequence

shown here:

Proteins are made of hundreds or even thousands of amino acids linked

together through peptide bonds. These chains contain many O—H and N
—H bonds that participate in strong hydrogen bonds between different
parts of the chains, causing the proteins to fold in certain predictable
patterns. The exact sequence of amino acids determines the way a protein
chain will fold and therefore the shape of the protein. In turn, the shape of
the protein determines how it can function.
For example, hemoglobin (Figure 15.40) is the protein that transports
oxygen from the lungs and through the bloodstream to every cell in the
body. A hemoglobin protein is composed of four different polypeptide
chains. Each chain contains either 141 or 146 amino acid pieces. These

974
chains bind with structures called heme units that contain Fe2+ ions. The
Fe2+ ions bind to oxygen, transporting it throughout the body.

Figure 15.40 This image shows the structure of hemoglobin, the protein that transports
oxygen throughout the body. The complex structure is composed of four peptide chains
and four heme units. The colored ribbons in this structure represent the shapes of the
polypeptide chains, as shown in the magnification at right. The space occupied by the
L90 protein is represented by a semitransparent surface.

Explore
Figure 15.40

DNA
We all begin life as a single cell. Embedded in that cell is a blueprint—a
code for every feature and function of our bodies. This code, called our
genome or simply our genes, is passed down from our parents. As we
grow, this blueprint is duplicated. Every cell of our bodies—every strand
of hair or scrape of skin—contains this blueprint in full.
Plants and animals store their genetic information in a set of polymeric
molecules called deoxyribonucleic acids (DNA). These molecules are
massive: The human genetic code contains 6 billion different bits of coded
information that are stored in just 46 molecules, called chromosomes.
The DNA coding in our chromosomes is built on four small molecules,
called bases. The four bases are adenine, thymine, guanine, and cytosine
(Figure 15.41). Each base is symbolized by the first letter of its name: A,
T, G, and C. These four bases encode all of the information needed to
produce a living organism.

975
Figure 15.41 These four structures store the genetic code in DNA.

976
 When you send a text message, you use letters to create words. But your phone
uses a series of ones and zeros—a binary code—to store those letters. Much like
letters are the building blocks for words, amino acids are the building blocks for
proteins. And much like your phone stores letters in binary code, your DNA
code stores the amino acid sequence for every protein of your body.

In a DNA molecule, these four bases bond to a simple sugar called

deoxyribose and a phosphate linker. The base, the sugar, and the phosphate
form a nucleotide—a single monomer in the DNA polymer (Figure
15.42). The nucleotides link together through condensation reactions to
form a sugar–phosphate backbone (Figure 15.43). A single strand of DNA
is composed of millions of these nucleotides, each representing a piece of
the genetic code.

Figure 15.42 A single nucleotide contains a base that is connected to a backbone

composed of deoxyribose and a phosphate linker.

Figure 15.43 The phosphate linker and the deoxyribose provide the structural backbone
for the DNA strand. The sequence of bases provides the “code” from which our bodies
are built.

This elegant method for storing information is amazing, but it gets

977
even better. Our bodies make copies of the DNA blueprint. Partial copies
travel within the cells, and when cells divide, they produce a complete
copy of the master blueprint. Here’s how it works: Each base couples with
one of the other bases by forming tight hydrogen bonds. Thymine bonds
with adenine, and guanine bonds with cytosine (Figure 15.44). A
complete DNA molecule contains two strands; each base in one strand is
specifically paired to the matching base in the other strand. Because of
these hydrogen bonds, DNA forms a unique shape called a double helix:
The two backbones twist around each other with base pairs in between
(Figure 15.45).
DNA bases pair together:
T pairs with A.
G pairs with C.

Figure 15.44 In a DNA molecule, the bases selectively pair through hydrogen bonds:
Thymine bonds with adenine, and cytosine bonds with guanine.

978
Figure 15.45 A DNA molecule contains two matching strands, wound together in a
double helix. The image at left shows the detailed molecular structure, and the structure
at right is a simplified representation.

Explore
Figure 15.45

Because the two strands are complementary pairs, and each strand
encodes the complete set of genetic instructions, the double-helix structure
allows the cells to duplicate the genetic code. Figure 15.46 illustrates how
this works: When a cell is ready to divide, an enzyme (that is, a protein
that serves a specific function) begins to “unzip” the double-helix
structure. Other enzymes then build two new backbones with bases that
match the two unzipped strands. When the unzipping is complete, two
identical double-helix structures have been produced.

979
Figure 15.46 The double helix structure of DNA enables replication of the genetic
code.

Explore
Figure 15.46

In order for the cell to use the information stored on its blueprint, the
DNA code is copied onto a related molecule called ribonucleic acid
(RNA). RNA relays segments of information from the DNA to other parts
of the cell, where it is used to assemble amino acids into proteins.

In 1984, the U.S. government launched the human genome project, a massive
research effort involving laboratories across the world, with the goal of mapping
every base pair in a human DNA sample. In 2003 the project was declared
complete, and in 2007 scientists published the entire 6-billion-nucleotide
sequence of a single person. Today, advanced genome sequencing techniques
have become much faster and cheaper, and scientists routinely analyze gene
sequences for living creatures past and present.

980
Summary
Carbon compounds form the basis for life. Organic chemistry is the
chemistry of carbon-based compounds. Organic chemistry is founded in
the chemistry of living organisms, but it also includes carbon compounds
from other sources.
Carbon is present in our atmosphere (as carbon dioxide), in the oceans
(as bicarbonate ions), in plants and animals, and even in fossil fuels buried
deep underground. The carbon cycle is a model that describes how carbon
migrates between these different regions through processes such as
photosynthesis, cell respiration, and exchange of carbon dioxide between
the atmosphere and the sea.
Organic chemistry begins with covalent bonds. The number of
covalent bonds a nonmetal atom forms depends on the number of electrons
needed to complete its valence electron level. Hydrogen typically forms
one bond, oxygen forms two, nitrogen three, and carbon four. All plant and
animal life is composed primarily of these four atoms. Because of its
ability to form four covalent bonds, carbon forms the backbone for most
biological molecules.
We often represent simple molecules using Lewis structures. A
complete Lewis structure depicts all the valence electrons, both bonded
and unbonded. As molecules increase in complexity, Lewis structures
quickly get cumbersome. We therefore use simplified structures to convey
bonding information in ways that are less cluttered. In simplified Lewis
structures, chemists show the bonded electrons as dashes but do not
include nonbonded electrons. In condensed structures, atoms are listed in
the sequence of their bonding, and the bonds are inferred from the
arrangement.
In skeletal structures, lines represent carbon chains. Unless labeled
otherwise, each vertex or end point in a skeletal structure represents a
carbon atom. Hydrogen atoms bound to carbon are not drawn, but are
inferred from the structures. Hydrogen atoms bonded to other atoms (such
as oxygen and nitrogen) are shown explicitly. Scientists often use these
different drawing styles interchangeably, drawing structures with both
skeletal and condensed elements.
Alkanes and cycloalkanes are composed only of carbon and hydrogen,
and they contain all single bonds. In addition to C–H single bonds, most
organic molecules contain groupings of atoms or bonds that behave in
distinct patterns. These patterns of atoms and bonds are called functional

981
groups. Molecules that are composed only of carbon and hydrogen are
called hydrocarbons.
There are several hydrocarbon functional groups: Carbon–carbon
double bonds are called alkenes. Carbon–carbon triple bonds are called
alkynes. Aromatic compounds contain rings with alternating single and
double bonds. Aromatic compounds are less reactive than simple alkenes.
Benzene is the most common aromatic compound.
There are six major functional groups containing C, H, and O:
alcohols, ethers, aldehydes, ketones, carboxylic acids, and esters. Each of
these arrangements produces distinctive properties, and these functional
groups are common in natural and synthetic molecules. The nitrogen-
containing functional groups include amines, amides, and nitriles.
Alcohols, carboxylic acids, and amines commonly undergo
condensation reactions. In these reactions, two smaller molecules combine
to produce water plus a larger molecule.
Polymers are long-chain molecules composed of simple repeating
units. Polymers that form in living systems are called biopolymers.
Synthetic polymers are commonly called plastics.
In this chapter we briefly examined some of the major classes of
biomolecules. Carbohydrates are compounds composed of carbon,
hydrogen, and oxygen. Plants create carbohydrates through the process of
photosynthesis. Simple carbohydrates often link together through
glycosidic bonds. Cellulose and starch are two examples of
polysaccharides—biopolymers made from simple carbohydrate monomers.
Proteins are large compounds that perform a wide variety of biological
functions. Proteins are composed of 20 distinct building blocks, called
amino acids. The sequence of amino acids helps determine the shape of
proteins, which in turn determines how the protein functions.
The genetic information of every living creature is stored in its DNA.
DNA molecules are massive, and they are composed of units of genetic
code called nucleotides. Each nucleotide contains a base (that carries the
specific genetic information), a sugar, and a phosphate linker. DNA uses
only four bases: adenine, thymine, guanine, and cytosine. These bases
form specific hydrogen-bonded pairs: Adenine binds with thymine, and
guanine binds with cytosine. The DNA molecule contains a double-helix
structure, in which the sugar and phosphate backbone circles around each
base pair. The unique double-helix structure allows DNA to be duplicated,
and it allows related molecules called RNA to transport fragments of the
genetic code throughout the cells. Cells use the information stored in DNA
to construct proteins.

982
 This chapter has been a brief survey. Carbon and the other nonmetals
combine in unique ways that enable an infinite combination of atoms and
structures. These structures range from very small to astoundingly large
molecules. The study of organic and biological chemistry is far too rich
and vast for one person to understand. And yet this sea of knowledge
revolves around a few simple rules: two bonds to oxygen, three bonds to
nitrogen, and four bonds to carbon.

983
How Catalysts Work

At the beginning of this chapter, we introduced Robert Grubbs, a Nobel Prize–

winning chemist who developed ruthenium catalysts that produce carbon–carbon
bonds. His work led to cleaner and safer ways to manufacture medicines, plastics,
lubricants, and many other valuable products. But how does the catalyst actually
work?
Catalysts are like ministers who preside over weddings. They are not getting
married, but they facilitate the bond between the bride and groom. Symbolically,
the minister may even join the couple’s hands together (Figure 15.47). Similarly,
catalysts help new chemical bonds to form, but the catalysts are not changed in the
overall process.

Figure 15.47 In a wedding ceremony, the minister joins the bride and groom together.

For example, chemists use ruthenium catalysts to prepare cyclic structures

containing carbon–carbon double bonds, as shown in Figure 15.48a. Notice that
the catalyst is not present in the starting material or in the products, but it is written
over the arrow.

984
Figure 15.48 (a) This reaction uses a ruthenium catalyst. (b) This reaction takes place in
the step-by-step process shown.

985
 Explore

Figure 15.48

Figure 15.48b shows the step-by-step process through which the ruthenium
catalyst works. In the first step, the catalyst (structure 1) reacts with a carbon–
carbon double bond to form a four-atom ring (structure 2). This ring then breaks to
form a new carbon–ruthenium double bond (structure 3). This cycle repeats, as
ruthenium forms a new four-atom ring (structure 4) and then releases its bond to
the larger structure, producing a new double bond. After the new bond forms, the
catalyst returns to its original state. The bond has been formed; the ring is
produced. You may kiss the bride.

986
 Key Terms
15.1 Organic Chemistry and the Carbon Cycle
organic chemistry The chemistry of carbon-containing molecules.
carbon cycle A description of how Earth’s carbon moves between rock and
sediment, water and atmosphere, and plants and animals.
photosynthesis The sequence of chemical reactions by which green plants
harvest the energy of the sun.
cellular respiration The sequence of chemical reactions by which animals
release energy stored in the chemical bonds of substances they consume.

15.2 Covalent Bonding with Carbon and Other Nonmetals

isomers Compounds having the same molecular formula but different bonding
sequences.

15.3 Drawing Covalent Structures

condensed structure A way of representing chemical bonds that does not
show most covalent bonds, but lists atoms in order of their connectivity.
skeletal structure A simplified representation for chemical structures in
which the end of each line segment denotes a carbon, and C—H bonds are
inferred rather than drawn explicitly.

15.4 Major Functional Groups

functional group A small group of atoms within a molecule that behaves in a
characteristic manner.
hydrocarbon A compound that contains only carbon and hydrogen.
alkane A hydrocarbon composed entirely of single bonds.
cycloalkane An alkane that forms a cyclic structure (commonly called a ring).
alkene A functional group consisting of a carbon–carbon double bond; this
term also refers to a simple molecule containing this functional group.
alkyne A functional group consisting of a carbon–carbon triple bond; this term
also refers to a simple molecule containing this functional group.
benzene A very stable compound having the formula C6H6, in which the six
carbon atoms form a ring with alternating single and double bonds; benzene is
one of the simplest examples of an aromatic ring.
aromatic ring A ring structure that contains alternating single and double

987
bonds and is generally less reactive than simple alkenes.
alcohol A functional group consisting of an oxygen atom that is singly bonded
to a carbon atom and a hydrogen atom, giving the bonding sequence C—O—
H.
ether A functional group composed of an oxygen atom singly bonded to two
carbon atoms, giving the bonding sequence C—O—C.
condensation reaction A reaction in which two smaller molecules combine to
produce water plus a larger molecule.
carbonyl A functional group consisting of a carbon–oxygen double bond.
aldehyde A functional group consisting of a carbonyl connected to a hydrogen
atom.
ketone A functional group containing a carbonyl connected to two carbon
atoms.
carboxylic acid A functional group containing a carbonyl bonded to an
alcohol, commonly represented by the condensed formula —COOH.
ester A functional group containing a carbonyl bonded to an oxygen atom that
is bonded to another carbon atom.
amine A functional group consisting of a nitrogen atom with three single
bonds, usually to hydrogen or carbon atoms.
amide A functional group consisting of a carbonyl group bonded to a nitrogen
atom.
nitrile A functional group consisting of a carbon–nitrogen triple bond.

15.5 Polymers and Plastics

polymer A molecule containing simple repeating units that are linked together
in long covalent chains.
plastic A synthetic polymer.
monomer A small molecule that can connect with other molecules to form a
polymer.

15.6 Biomolecules—An Introduction

carbohydrate A naturally occurring molecule composed of carbon, hydrogen,
and oxygen and having the general formula Cm(H2O)n.
disaccharide A carbohydrate composed of two simpler carbohydrates that are
linked together through a condensation reaction.
polysaccharide A carbohydrate composed of many simpler carbohydrates that
are linked together.
protein A biopolymer composed of building blocks called amino acids; this

988
type of molecule has many functions in living creatures.
amino acid Small molecules having both amine and carboxylic acid functional
groups; plant and animal cells use 20 fundamental amino acids to create
proteins.
peptide bond The carbon–nitrogen bond that connects two amino acids
together; peptide bonds are formed by the condensation of a carboxylic acid
and an amine to form an amide.
deoxyribonucleic acids (DNA) Massive molecules containing the genetic
code of living creatures.

989
 Additional Problems

15.1 Organic Chemistry and the Carbon Cycle

11. What are the two chemical products from cellular respiration?

12. What are the two chemical products from photosynthesis?

13. Is photosynthesis an endothermic or exothermic process? What about

cellular respiration?

14. Write two chemical reactions to show how carbon dioxide reacts in water
to produce the bicarbonate ion.

15. Describe a pathway in the carbon cycle by which carbon dioxide in the
atmosphere becomes part of a coal deposit deep underground.

16. Describe a pathway in the carbon cycle by which the carbon atoms in a
squirrel eventually become part of the carbon atoms in a tree.

17. Describe a pathway in the carbon cycle by which carbon dioxide from the
atmosphere becomes trapped in the sediment on the ocean floor.

18. Describe two different pathways in the carbon cycle through which carbon
atoms in a tree become part of the carbon dioxide in the atmosphere.

15.2 Covalent Bonding with Carbon and Other Nonmetals

19. Indicate the number of covalent bonds typically formed by each of these
nonmetal atoms:
a. C
b. N
c. O
d. F

20. Indicate the number of covalent bonds typically formed by each of these
nonmetal atoms:
a. Si

990
 b. P
c. S
d. Cl

21. Draw proper Lewis structures for these compounds:

a. CH4
b. CCl4
c. C2H5Cl

22. Draw proper Lewis structures for these compounds:

a. CF4
b. CH2Cl2
c. C3H8

23. Draw proper Lewis structures for the following compounds. Use double
and triple bonds as needed.
a. C2H4
b. C2H4O
c. C2H5N
d. CHN

24. Draw proper Lewis structures for the following compounds. Use double
and triple bonds as needed.
a. C2H2
b. C3H4
c. CH2O
d. HNO

25. Three of the following four structures have identical bonding sequences.
Which of these compounds is not like the others—that is, which one is an
isomer of the other three?
a.

991
 b.

c.

d.

26. Three of the following four structures have identical bonding sequences.
Which of these compounds is not like the others—that is, which one is an
isomer of the other three?
a.

b.

c.

d.

27. Two isomeric structures have the molecular formula C2H6O. Draw these
two isomers.

28. Draw the four possible isomers having the structure C4H9Br.

29. Draw the three possible isomers having the molecular formula C3H8O.

992
30. Draw the four possible isomers having the structure C3H9N.

15.3 Drawing Covalent Structures

31. Show these condensed structures as Lewis structures:

a. CH3CH2OCH3
b. CH3CH2CH(CH3)2
c. CH3(CH2)4CHCl2

32. Show these condensed structures as Lewis structures:

a. H2NCH2CH2OH
b. CH3CH(OH)CH3
c. CH3(CH2)2CHBrCH3

33. Redraw these structures as condensed structures:

a.

b.

34. Redraw these structures as condensed structures:

a.

b.

35. Write the molecular formula for each of these skeletal structures:
a.

b.

993
 c.

36. Write the molecular formula for each of these skeletal structures:
a.

b.

c.

37. Write the molecular formula for each of these skeletal structures:
a.

b.

c.

38. Write the molecular formula for each of these skeletal structures:
a.

b.

c.

39. Draw these Lewis structures as skeletal structures:

a.

994
 b.

c.

40. Draw these Lewis structures as skeletal structures:

a.

b.

c.

41. Draw these skeletal structures as Lewis structures:

a.

b.

c.

42. Show these skeletal structures as Lewis structures:

a.

b.

995
 c.

43. Show these condensed structures as Lewis structures and as skeletal

structures:
a. CH3CH2NHCH3
b. CH2CHCH2OH
c. CH3CH2CN

44. Show these condensed structures as Lewis structures and as skeletal

structures:
a. CF3CHCH2
b. (CH3CH2)2NH
c. CHC(CH2)3Cl

45. The skeletal structure below belongs to a class of compounds called

penicillins, which were the first widely used antibiotics. What is the
molecular formula of this compound?

46. Nicotine is the highly addictive substance found in tobacco. The skeletal
structure of nicotine is shown below. What is the molecular formula for
this compound?

15.4 Major Functional Groups

47. How many hydrogen atoms are in a linear alkane with 8 carbon atoms?

48. How many hydrogen atoms are in a linear alkane with 12 carbon atoms?

996
49. How many hydrogen atoms are in a cycloalkane having 4 carbon atoms?

50. How many hydrogen atoms are in a cycloalkane having 8 carbon atoms?

51. What are the molecular formulas for pentane and for cyclopentane? Draw a
skeletal structure for each of these molecules.

52. What are the molecular formulas for octane and for cyclooctane? Draw a
skeletal structure for each of these molecules.

53. Name each of these alkanes:

a. CH3CH2CH2CH3

b.

c.

54. Name each of these alkanes or cycloalkanes:

a. CH3(CH2)5CH3

b.

c.

55. Draw three isomers of a linear alkane with the formula C6H14.

56. Draw three isomers of a linear alkane with the formula C7H16.

57. Draw three isomers of a cycloalkane with the formula C7H14. (Hint: Try
different ring sizes.)

997
58. Draw three isomeric cycloalkanes having the formula C8H16. (Hint: Try
different ring sizes.)

59. A compound has a molecular formula C3H6. Draw isomers of this

compound in which
a. a the structure is a cycloalkane.
b. the structure is an alkene.

60. Draw isomers of the compound C5H8 in which

a. the compound is an alkyne.
b. the compound has a cyclic structure and an alkene.

61. Identify these hydrocarbons as alkanes, alkenes, alkynes, or aromatic

compounds:
a.

b.

c.

d.

62. Identify these hydrocarbons as alkanes, alkenes, alkynes, or aromatic

compounds:
a.

b.

c.

998
 d.

63. Identify the oxygen-containing functional groups in these small molecules:

a.

b.

c.

d.

64. Identify the oxygen-containing functional groups in these small molecules:

a. CH3CH2CO2H

b.

c.

d.

65. Draw a structure having the formula C4H8O that contains:

a. an alcohol group
b. an aldehyde group
c. a ketone group
d. an ether group

66. Draw a structure having the formula C3H6O2 that contains:

999
 a. a carboxylic acid group
b. an ester group
c. both a ketone and an alcohol

67. The structure below is that of ascorbic acid, better known as vitamin C.
Identify the major functional groups present in this molecule.

68. “Agent Orange” was an herbicide used in the Vietnam War to remove
foliage that shielded enemy positions. The compound shown below made
up 50% of the Agent Orange mixture. Identify the major functional groups
present in this molecule.

69. Identify the nitrogen-containing functional groups in these small

molecules:
a.

b.

c.

70. Identify the nitrogen-containing functional groups in these small

molecules:

1000
 a.

b.

c.

71. Albuterol is commonly used in breathing treatments to treat conditions

such as bronchitis and asthma. Identify the major functional groups present
in this molecule.

72. Strychnine is an infamous poison. This naturally occurring compound was

extracted from certain shrubs and used for many centuries to kill animal
pests. History records many cases of murder or attempted murder by
strychnine poisoning; some have even postulated that this poison caused
the death of Alexander the Great. The elaborate structure of this compound
is shown below. What functional groups are present in this poison?

73. Draw the structure of the ether produced in this condensation reaction:

1001
74. Draw the structure of the ester produced in this condensation reaction:

75. Draw the structure of the amide produced in this condensation reaction:

76. Draw the structure of the amide produced in this condensation reaction:

77. As their name implies, carboxylic acids are acidic. Show the products of
the acid-base reactions that take place when these acids are mixed with
water:
a. CH3CO2H+H2O⇌

b. CH3CH2CO2H+H2O⇌

78. Amines are basic compounds. Show the products of the acid-base reaction
that takes place when these amines are mixed with water:

1002
 a. CH3NH2+H2O⇌

b. (CH3)3N+H2O⇌

79. Show the products from the acid-base reactions below. Remember that
carboxylic acids are acidic and amines are basic.

a. HCO2H+NH3⇌

b. CH3CO2H+CH3NH2⇌

80. Show the products from the acid-base reactions below. In these reactions,
the alcohols behave as acids, while the amines behave as bases.

a. CH3NH2+CH3OH⇌
b. CH3CH2OH+NH3⇌

15.5 Polymers and Plastics

81. What are polymers? Give two examples of naturally occurring polymers
and two examples of synthetic (human-made) polymers.

82. In the table below, the first column shows a portion of a polymer structure.
In the second column, draw a representation that shows the repeating unit
of the polymer in parentheses. The first one is done as an example.

Partial Polymer Structure Representation

1003
15.6 Biomolecules—An Introduction

83. What three elements are present in a carbohydrate?

84. What is the major function of carbohydrates in living structures?

85. What is the difference between alpha-glucopyranose and beta-

glucopyranose?

86. What is the major difference between cellulose and starch?

87. Sucrose, or table sugar, is a disaccharide formed by the combination of two

simple sugars, glucose and fructose, through a condensation reaction. The
two alcohols that condense are highlighted in the structure below. Draw
the structure of the sucrose molecule that results from this condensation.

88. Lactose, or milk sugar, is a disaccharide formed by the combination of two

simple sugars, galactose and glucose, through a condensation reaction. The
two alcohols that condense are highlighted in the structure below. Draw
the structure of the lactose molecule that results from this condensation.

89. What types of roles do proteins play in living organisms?

90. What two functional groups are present in every amino acid?

91. How many of the amino acids in Table 15.4 have a benzene ring in their
side chain?

1004
92. How many of the amino acids in Table 15.4 have a second amine group in
their side chain?

93. A tripeptide contains serine (the N-terminus), connected to proline,

connected to arginine (the C-terminus). Give the representation of this
structure using the three-letter amino acid abbreviations. Refer to Table
15.4 as needed.

94. A tripeptide contains lysine (the N-terminus), connected to glutamic acid,

connected to isoleucine (the C-terminus). Give the representation of this
structure using the three-letter amino acid abbreviations. Refer to Table
15.4 as needed.

95. Use the three-letter amino acid abbreviations to represent the tripeptide
sequences shown here:
a.

b.

96. Use the three-letter amino acid abbreviations to represent the tripeptide
sequences shown here:
a.

1005
 b.

97. Draw the structures for these dipeptides:

a. Cys-Thr
b. Thr-Cys

98. Draw the structures for these dipeptides:

a. Ser-Glu
b. Glu-Ser

99. Draw the structures for these tripeptides:

a. Thr-Phe-Asp
b. Lys-Gln-Gly

100. Draw the structures for these tripeptides:

a. Ser-Tyr-Thr
b. Leu-Arg-Val

101. What types of interactions cause peptides to fold into different shapes?

102. Broadly, how does the amino acid sequence determine the properties of a
protein?

103. Why is DNA important within an organism?

104. What are the three components of a nucleotide?

105. How do the components of a nucleotide link to other nucleotides to form a

single strand of DNA?

106. How are the two strands of DNA in a double helix bound together?

107. The image shown represents the bases on a fragment of DNA. Fill in the

1006
 second strand with bases that match those on the first strand.

108. The image shown represents the bases on a fragment of DNA. Fill in the
first strand with bases that match those on the second strand.

109. In this chapter we discussed several types of biopolymers, including

polysaccharides, proteins, and DNA. Which of these classes of polymers
contain only one repeating monomer? Which classes contain multiple
monomers with information coded in the monomer sequence?

110. Cellulose, starch, proteins, and DNA are all examples of biopolymers. In
the table below, list the monomers from which each one is built, identify
the type of reaction used to join the monomers together, and state the
function of each polymer.

1007
Biopolymer Monomer Reaction That Joins Monomers Function
Starch Condensation

Cellulose
Protein
DNA Nucleotides

1008
Chapter Sixteen
Nuclear Chemistry

Fukushima
On March 11, 2011, an earthquake struck the Tōhoku region of northern Japan.
Centered about 70 kilometers off Japan’s east coast, the quake produced a tsunami
—an immense wave over 40 meters high. Less than an hour after the earthquake,
the tsunami slammed into the coastal areas, destroying nearly everything in its path
and killing over 15,000 people (Figure 16.1).

Figure 16.1 (a) On March 11, 2011, an earthquake struck northern Japan. The
earthquake triggered a tsunami that slammed the coastline. (b) The epicenter of the
earthquake was about 70 km from the coast of Japan. (c) The Fukushima plant before
the disaster: The four square buildings each housed a reactor core. (d) After the
accident, emergency workers wore special protective gear. (e) A volunteer stands beside

1009
a radiation meter near the Fukushima site, four years after the disaster. The meter alerts
residents to health risks resulting from the accident.

Along the coast, nuclear power plants produced energy for the region. The
plants were engineered to survive an earthquake, but they were not adequately
prepared for the tsunami. As the waters rose around one of the power facilities—
the Fukushima Daiichi plant—a new disaster began to unfold.
The heart of a nuclear power plant is the reactor core. The atomic changes that
take place in the core generate huge amounts of heat energy, but they also produce
very toxic by-products. Nuclear plants continuously pump cooling water through
the core. Both the core and cooling water are sealed inside a central reactor
building.
When the tsunami hit Fukushima Daiichi, rapidly rising water knocked out
power to the cooling pumps as well as multiple backup systems. Inside the reactor
building, temperatures and pressures began to rise. As engineers battled to restore
cooling systems, authorities evacuated the surrounding cities. The temperatures
continued to rise. Steam and hydrogen gas began to fill the reactor building. The
ultrahot nuclear fuel melted through the containment vessel, carving a path deep
into the building’s foundation. Then the hydrogen in the reactor exploded, ripping
the top off the concrete building and spewing toxins into the air.
To cool the toxic fuel, emergency workers flooded the structures with water.
For weeks following the accident, the hot fuel boiled water out of the buildings,
releasing steam and toxic matter into the atmosphere. Cracks in the foundation
released contaminated water into the sea. It was nine months before the fuel finally
cooled, and the release of toxins subsided.
The Fukushima accident stirred fears around the world. In Tokyo, reports of
water contamination sparked a run on bottled water. Within days of the accident,
the United States Food and Drug Administration blocked the import of some foods
from Japan. Germany shut down half of its nuclear plants and drastically changed
its long-term energy policies.
Seven years after the accident, contamination around the Fukushima site is still
a problem, and many people have not returned to their homes. Yet despite global
panic over the Fukushima meltdown, nobody died from exposure to the toxic
materials after the accident. In fact, there’s no clear evidence that anyone even got
sick.
The events at Fukushima raise many important questions: What are the real
risks of nuclear energy? What are the benefits? What level of exposure to these
materials is acceptable? What levels are harmful?
In this chapter, we’ll survey the key concepts of nuclear chemistry. We’ll see
how fundamental changes to atomic structure occur, and why structural changes
are accompanied by such enormous changes in energy. From the dramatic and
massive reactions of the stars to small and gradual changes on Earth, we’ll explore
many of the naturally occurring nuclear reactions that affect our lives. We’ll also
explore applications of nuclear chemistry, from commercial power generation to
medicine to archaeology, and some of the public health and safety issues that arise

1010
when humans come in contact with the powerful forces stored inside the nucleus of
the atom.

Intended Learning Outcomes

After completing this chapter and working the practice problems, you should be
able to:

16.1 Nuclear Changes

Write balanced nuclear equations to describe nuclear changes.

16.2 Radioactivity
Define the different types of radioactive decay, and predict the products
formed by these decays.
Describe the rate of radioactive decays using half-lives.
Describe how radioactive decay and human exposure are measured.
Qualitatively describe the use of radioactive nuclides in medicine, geology,
and archaeology.

16.3 Energy Changes in Nuclear Reactions

Describe the interconversion of matter and energy in nuclear changes.
Describe how mass defect and binding energy relate to the stability of a
nucleus.

16.4 Nuclear Power: Fission and Fusion

Describe the processes of nuclear fission and fusion.
Broadly describe the risks, benefits, and technical challenges associated with
the use of fission and fusion for power generation.

1011
16.1 Nuclear Changes
We first looked at atomic structure in Chapter 3. Recall that atoms are
composed of a very dense nucleus that contains protons and neutrons,
surrounded by a cloud of negatively charged electrons (Figure 16.2).

Figure 16.2 A nucleus is composed of protons and neutrons. The nucleus is surrounded
by the electron cloud.

Most everyday chemistry involves the electrons: Covalent and ionic

bonds, acid-base behavior, and oxidation-reduction reactions all involve
changes in electron structure. Atoms may gain, lose, or share electrons, but
the atomic nuclei remain unchanged.
Nuclear chemistry describes a different type of change—those
involving the nucleus of an atom. Nuclear changes involve massive
amounts of energy. The heat produced by the sun and the stars, the very
hot temperatures of the Earth’s core, and the devastating power of the
atomic bomb arise from changes in the atomic nucleus. Although most of
the chemistry that occurs around us is non-nuclear in nature, nuclear
changes are constantly occurring, and they are essential to life on Earth.

The Nucleus—A Review

Atoms are defined by the number of protons and neutrons in their nucleus.
Collectively, protons and neutrons are called nucleons. Recall that the
number of protons is the atomic number: This number gives the atom its
unique properties. The mass number is the number of nucleons (the sum of
protons and neutrons) in a particular nucleus. When writing atomic
symbols, we show the mass number on the upper left-hand side of the
atomic symbol. The atomic number is shown on the lower left-hand side

1012
(Figure 16.3).

Figure 16.3 The mass number is written at the upper left of the symbol. The atomic
number is written at the lower left.

A nucleon is a proton or a neutron.

An atom or nucleus containing a particular number of protons and

neutrons is referred to as a nuclide. Isotopes are nuclides that have the
same atomic numbers. That is, isotopes are the same type of atom, but with
different mass numbers. For example, lithium has two nuclides: 6Li and
7Li. These two nuclides are isotopes. The symbol 6Li is read as “lithium-
6,” and it is commonly written this way as well. We will use both styles to
represent nuclides in this chapter.

Nuclides and Isotopes

Three unique nuclides are shown here. The two with the same atomic number
are isotopes.

Example 16.1 Identifying Isotopes

Which of the following nuclides is an isotope of B511 ?

a. a nuclide with mass number 11 and atomic number 6
b. a nuclide with mass number 10 and atomic number 5
c. a nuclide with mass number 12 and atomic number 6

1013
 Isotopes have the same atomic number but different mass numbers.
Therefore answer (b) is correct. •

Nuclear Reactions
In a nuclear reaction, the structure of an atomic nucleus changes. There
are many types of nuclear reactions: Sometimes nuclei spontaneously
decompose, releasing energy and smaller particles. Other reactions
produce larger nuclei: For example, in the upper atmosphere, high-energy
particles from the sun collide with oxygen and nitrogen, transforming
these atoms into different elements. In some laboratories, scientists
emulate this process by firing high-energy particles into target elements to
produce new nuclides for use in medicine. The heat from the sun, the heat
within Earth’s core, and the energy produced from nuclear power plants all
arise from nuclear changes. We’ll look at these different changes in the
sections that follow.

Nuclear equations usually show both the mass numbers and the atomic
numbers.

We often describe nuclear reactions using nuclear equations. In a

balanced nuclear equation, the total mass numbers and the total atomic
numbers are the same on both sides of the equation. For example, bismuth-
212 slowly decomposes to form two new nuclides: thallium-208 and
helium-4. We can write a balanced nuclear equation for this process as
follows:

B83212i→ 81208Ti+H24e
Notice in this nuclear equation that the total atomic number (that is, the
total number of protons) is the same on the left- and right-hand side of the
equation. Likewise, the total mass number is also the same on both sides of
the equation (Figure 16.4).

1014
Figure 16.4 In a balanced nuclear equation, the sums of the mass numbers and the
atomic numbers are the same on both sides of the equation.

In many nuclear equations, a nucleus gains or loses a fundamental

atomic particle. In these cases we use the symbol p11 to represent a

proton, n01 to represent a neutron, and e−10 to represent an

electron (Table 16.1).

TABLE 16.1 Nuclear Symbols for Subatomic Particles

For example, some medical tests use the nuclide indium-111 (111In).
Scientists produce this isotope in a laboratory using a device called a
cyclotron (Figure 16.5). Using magnetic fields, this device accelerates a
proton to very high speeds and then smashes it into a cadmium-111
nucleus. The result of this collision is a nucleus of indium-111 and a free
neutron. We can depict the nuclear change using this equation:

p11+C48211d→l49211n+n01

1015
Figure 16.5 This cyclotron uses magnetic fields to accelerate charged particles. The
collision of these high-energy particles results in a nuclear reaction, producing new
isotopes for medicinal applications.

When writing nuclear equations, we can sometimes predict one product

from the other products. Example 16.2 illustrates how this is done.

Example 16.2 Completing a Nuclear Equation

When molybdenum-100 is bombarded with a proton, it produces a new
nuclide and emits two neutrons, as shown in the equation below. Determine
the atomic number, mass number, and atomic symbol for the unidentified
product.
p11+M42100o→n01+n01+ _

The total mass numbers on the left-hand side of the equation sum to 101.
Because the right-hand side of the equation has two neutrons (each with a
mass number of 1), the unknown mass number must be 99. Similarly, the
atomic numbers on the left-hand side sum to 43. Since the neutrons have an
atomic number of zero, the unknown atomic number must be 43. Atomic
number 43 corresponds to the element technetium (Tc). Based on this, we
can write a complete balanced equation. •

p11+M42100o→n01+n01+T4399c

1016
 IT
TRY

1. In Earth’s upper atmosphere, a high-energy neutron collides with a

nitrogen-14 nucleus, forming a carbon-14 nucleus and releasing one
proton. Describe this process using a balanced nuclear equation.

2. When a proton collides with an 127I nucleus in a cyclotron, it forms a

new nucleus, 123Xe, along with several neutrons. How many neutrons
are released in this process? Write a balanced nuclear equation to
describe this process.

1017
16.2 Radioactivity
Some of the most common nuclear reactions are radioactive decays. In
these reactions, nuclei spontaneously transition to a more stable state. In
some decays, nuclei eject particles from the nucleus. In other decays, the
nucleus shifts slightly, releasing energy. The spontaneous release of
particles and/or energy from the nucleus is called radioactivity. The
energy released by these reactions is called radiation.

When radiation strikes zinc sulfide, it causes the compound to glow. After the
discovery of radium in the early 1900s, trace amounts of this material were
combined with zinc sulfide to create watch faces that glowed in the dark. Today,
watch lights use much safer technologies.

For example, the element radium slowly decomposes over thousands

of years to form a new element, radon. Like bursting popcorn kernels, the
radium atoms individually “pop,” transforming into radon atoms (Figure
16.6). Each time an atom decays, the substance releases a small amount of
energy. Because so many atoms are present, a sample of radium releases
this energy over thousands of years.

1018
Figure 16.6 This mineral contains the radioactive element radium. Radium releases
particles and energy as it slowly decomposes to form a new element.

The discovery of radioactivity was closely tied to the discovery of

atomic structure introduced in Chapter 3. In 1897 Antoine Becquerel
discovered that when he placed uranium near certain substances, such as
zinc sulfide, the substances glowed. In the years that followed, Marie
Curie and her husband Pierre (Figure 16.7) worked extensively to isolate
and characterize radioactive substances. Through these experiments, they
discovered the elements radium and polonium. Marie Curie was able to
classify three types of radioactive decays, which she termed alpha, beta,
and gamma radiation.

Figure 16.7 Marie and Pierre Curie were jointly awarded the 1903 Nobel Prize in
Physics. After Pierre’s tragic death in 1906, Marie continued their work; in 1911 she
received the Nobel Prize in chemistry.

1019
Types of Radioactive Decay
Alpha Decay
In an alpha (α) decay, a nucleus ejects an alpha particle composed of two
protons and two neutrons. An alpha particle is a helium nucleus. When a
nucleus emits an alpha particle, its mass number decreases by four, and its
atomic number decreases by two.

For example, when a U92238 atom undergoes an alpha decay

(Figure 16.8), its mass number decreases by four, forming a new nuclide
with a mass of 234. Its atomic number decreases by two, changing the
atom from uranium (atomic number 92) to thorium (atomic number 90).

Figure 16.8 In an alpha decay, a larger nucleus ejects two protons and two neutrons (a
helium nucleus) to form a smaller, more stable nucleus.

We can represent this change using a nuclear equation:

U92238→T90234h+H24e
Notice in this nuclear equation that the total mass numbers and the total
atomic numbers are the same on both sides of the equation.

An alpha decay ejects H24e from a nucleus.

Example 16.3 Nuclear Equations and Alpha Decays

Write a nuclear equation showing the alpha decay of a 210Po nucleus.
Polonium is atomic number 84. After undergoing an alpha decay, its mass
decreases by four, and its atomic number decreases by two. The resulting
nucleus has a mass number of 206 and an atomic number of 82. From the
periodic table, we see that atomic number 82 is lead. Therefore we can

1020
 write the complete equation as follows. •

P84210o→P82206b+H24e

Beta Decay
In a beta (β) decay, a neutron decays into two particles: a proton and an
electron (Figure 16.9). The proton remains in the nucleus, while the
electron is ejected. In a nuclear equation, we represent this as follows:

n01→p11+e −10

Figure 16.9 In a beta decay, a neutron transforms into a proton and an electron.

Notice in this equation that our total mass numbers did not change (the
tiny electron has a mass number of zero), and our total atomic number
(that is, our total charge) also remained constant. The neutral particle split
into two particles, one with a +1 charge and the other with a –1 charge.
When a nucleus undergoes a beta decay, the electron exits the nucleus,
but the proton remains. The mass number of the nucleus does not change,
but the atomic number increases by one. For example, radium-228
undergoes a beta decay to form actinium-228:

R88228a→A89228c+e−10

A beta decay increases the atomic number by one.

Example 16.4 illustrates this idea further.

Example 16.4 Nuclear Equations and Beta Decays

Write a nuclear equation showing the beta decay of thallium-209.
Thallium is atomic number 81. A beta decay increases the atomic number
by one, but it doesn’t change the mass number. The result is a nucleus with

1021
 atomic number 82 (lead) and mass number 209. We can write the complete
equation as follows. •

T81209i→P82209b+e−10

Gamma Decay
In a gamma (γ) decay, a nucleus releases energy in the form of
electromagnetic waves called gamma rays. Gamma decays do not involve
the release of particles, so they do not change the atomic number or the
mass number. A gamma decay may be thought of as the nuclear particles
adjusting their arrangement to form a slightly more stable arrangement.

A gamma decay does not change the atomic number or mass number.

Radioactive Decay Series

Radioactive nuclei often decay multiple times before reaching a stable
isotope. These decays occur through common sequences called a decay
series or a decay chain. For example, radon is a naturally occurring gas
that sometimes accumulates in low levels in basements or other low-lying
areas. Radon-218 is a radioactive nuclide. Over time, the Rn nucleus
undergoes a series of decays, forming several intermediates before
eventually reaching a stable nucleus, lead-206. We can represent this series
of decays as follows:

Three common decay series, known as the thorium, uranium, and

actinium series (Figure 16.10), account for most of the abundant
radioactive nuclides on Earth. The remaining naturally occurring
radioactive nuclides result from cosmic radiation—high-energy particles
from the sun and other stars that react with elements on Earth to produce
new radioactive nuclides.

1022
Figure 16.10 The uranium decay series begins with uranium-238 and ends with lead-
206. This graphic shows one of several possible pathways.

In the 1980s, homes began to be tested for radon. These tests show elevated
levels of radon in many basements and poorly ventilated areas.

Example 16.5 Predicting the Products from a Decay Series

Plutonium-241 is a radioactive substance produced in nuclear power
plants. Over time it decays to form radium-225 through the sequence β, α,
α, β, α, α. What intermediates are formed in this sequence of decays?
Each alpha decay decreases the mass number by four and the atomic
number by two. In a beta decay the mass number is unchanged, but the
atomic number increases by one. Based on this, we expect the mass
numbers and atomic numbers to change as follows:

1023
 Using the periodic table, we can identify the atomic symbols that
accompany each atomic number. This gives us a complete picture of the
decay series, as shown. •

IT
TRY

3. Write balanced equations showing these processes:

a. alpha decay of 40Ca
b. beta decay of 67Cu

4. Radium-224 decays through the pathway α, α, α, β, α, β. What

intermediates are formed in this decay series? What is the final stable
nucleus formed?

Half-Life
Radioactive decays occur at different rates. Some radioactive nuclides
decay quickly; others are more stable and hang around for years, centuries,
or millennia before they decay (Table 16.2). The half-life of a nuclide is
the amount of time required for one-half of a sample of a radioactive
substance to decay into something else (Figure 16.11). For example,
cesium-137 has a half-life of about 30 years. If you were given a 1.000-
gram sample of cesium-137 today, in 30 years 0.500 grams would still be
Cs-137—the rest would have decayed to form other substances. In 60
years (two half-lives), you would have 0.250 grams of Cs-137. In 90 years,
the amount would be 0.125 g.

TABLE 16.2 Half-Lives of Common Radioactive Nuclides

Nuclide Half-Life Occurrence/Use
3H 12.3 years Energy (from the sun)

10Be 1.38 million years Archaeology/geology

14C 5,730 years Archaeology

1024
 60Co 5.3 years Radiation therapy

99mTc* 6.0 hours Medical imaging

125I 59 days Radiation therapy

131Cs 9.7 days Radiation therapy

222Rn 3.8 days Produced naturally from 238U

238U 4.5 billion years Nuclear power

239Pu 24,000 years Nuclear power

*The “m” in the superscript means that this nuclide undergoes a gamma decay to form a
more stable form of the same nuclide, 99Tc.

Figure 16.11 As a nuclide decays, it forms a new substance. (a) Ruthenium (Ru)
becomes rhodium (Rh) through a beta decay. (b) The half-life is the amount of time it
takes for one-half of a substance to decay to another substance. The half-life of
ruthenium-106 is just over one year.

1025
 Explore

Figure 16.11

Consider another example: Iodine-131 is a radioactive nuclide that was

released into the atmosphere during the Fukushima accident. However,
iodine-131 has a short half-life of about 8 days. This means that 8 days
after the initial release, one-half of the iodine-131 had decayed into
something else. Mathematically, we can describe half-lives using the
equation

Nt=N0(12)tt1/2
where Nt is the amount of a nuclide that remains, N0 is the amount of
nuclide originally present, t is the amount of time that has passed, and t1/2
is the half-life of the nuclide. In this equation we can use any unit of
measure for the amount, but the time passed and the half-life must use the
same units. Example 16.6 illustrates this concept.

Example 16.6 Using Half-Lives

Samarium-153 EDTMP is the common name for a compound used to treat

1026
 bone cancer. The compound contains samarium-153, which has a half-life
of two days. If your laboratory received a sample containing 16.0 µg of
153Sm, how many µg would be present after 10 days?

Each half-life decreases the amount of samarium present by one-half. After

10 days, five half-lives would have passed, so the amount of 153Sm would
divide in half five times:

Based on this, 0.5 µg of 153Sm would be present after 10 days. We also

could have used the half-life equation above to solve this problem. In this
case, t = 10 days, t1/2 = 2 days, and N0 = 16.0 μg:
Nt=N0(12)tt1/2=(16.0 μg)(12)10 days2 days=(16.0 μg)(12)5=0.5 μg

Because the half-life of this drug is so short, health professionals typically

administer it within about 48 hours of its production. •

IT
TRY

5. Cobalt-58 is a radioactive isotope with a half-life of 71 days. If your

laboratory received a sample containing 20.0 µg of 58Co, how much
58Co would be left after 7 months (213 days)?

6. Carbon-14 has a half-life of 5,730 years. If you have a 20-ng sample

of 14C today, how much will remain in 8,000 years? Estimate your
answer using the definition of half-lives; then use the equation above
to obtain a more precise answer.

Health Effects of Radiation Exposure

Radioactive decay releases a tremendous amount of energy. Like tiny
bullets, these high-energy waves and particles damage human cells.
Exposure to large amounts of radiation can result in sickness or even

1027
death. Long-term exposure to smaller doses of radiation can lead to the
formation of tumors. Because of this, we must be careful to limit our
exposure to radioactive substances.
At the same time, radiation is all around us. The sun and stars
constantly emit radiation. Most of this is absorbed by the atmosphere, but
some still reaches us. Earth also contains naturally radioactive substances.
In fact, everything around us—rocks, trees, air, and even our own bodies
—contains radioactive elements that emit these harmful rays. The risks of
radiation exposure depend on the type of radiation as well as the amount of
radiation we’re exposed to.
The three major classes of radiation—alpha, beta, and gamma—have
very different properties. Alpha particles are much heavier than beta
particles, and they can be very damaging to living tissue. However, these
particles are easily blocked by clothing or even by a sheet of paper
(Figures 16.12 and 16.13).

Figure 16.12 Alpha particles are the most damaging radiation, but they are also the
easiest to block.

1028
Figure 16.13 Workers in highly radioactive sites, like the Fukushima responders shown
here, wear protective gear. This gear shields the workers from nearly all alpha particles
and most beta particles.

Beta radiation consists of fast-moving electrons that can go through

clothing and into living tissue. A thin sheet of metal or other hard
substance can block most beta radiation. Gamma radiation is the hardest to
block. Heavy lead shields are often used to block gamma radiation.
However, gamma rays are also less damaging than alpha or beta particles.
Perhaps the most important factor in radiation safety is the need to
understand how much radiation exposure we experience. Our bodies can
easily absorb a small amount of radiation. In fact, radioactive substances
are very useful—for power, for medicine, and even for archaeology. The
important things are to monitor the amount of radiation exposure and to
understand the health effects of different exposure levels.
Measuring Radiation
There are many devices that measure radiation. One of the most common
is the Geiger counter. A Geiger counter holds a small amount of a gas
(such as helium) inside a metal tube containing positive and negative
electrodes (Figure 16.14). When alpha, beta, or gamma radiation enters
the tube, it strikes a gas atom, knocking off an outer electron of the atom.
This ionized atom then migrates toward the negative electrode, creating a
slight electric current that is amplified and measured as a “count.” The
counter measures the number of particles that strike the electrode in a
given time period. Geiger counters often produce a clicking sound each
time they register a count—this sound gives the user some idea of how
much radiation is present.

1029
Figure 16.14 In a Geiger counter, incoming radiation ionizes atoms in a gas tube. The
ions travel toward an electrode, where they produce a slight electrical current.

Explore
Figure 16.14

A similar device is a scintillation counter. This type of device uses a

material that produces a tiny flash of light when radiation strikes it. A
detector counts the number of flashes and creates a count of total radiation
exposure.
A more advanced instrument for measuring radiation is a
semiconductor counter. This type of detector not only measures the
number of decays but also determines which nuclides produced the decay.
Different nuclides produce unique energy signatures when they decay. A
semiconductor counter is able to measure the different energies of
radiation absorbed and therefore identify which nuclides are present
(Figure 16.15).

Figure 16.15 A semiconductor counter measures the energies of the radiation absorbed,
and it can distinguish between different nuclide decays.
Data from Claus Grupen, Introduction to Radiation Protection: Practical Knowledge
for Handling Radioactive Sources (Berlin: Springer-Verlag, 2010).

A dosimeter is a radiation detector that measures human exposure to

radiation. In addition to Geiger counters and scintillation counters, many
smaller dosimeters are available. People working in areas with high

1030
radiation risks often wear a dosimeter on their work uniform to monitor
(and limit) their exposure to harmful radiation (Figure 16.16).

Figure 16.16 A worker wears a personal dosimeter that tracks his total radiation
exposure.

Common Exposure Levels

Scientists use several units to measure radioactivity and radiation
exposure. The simplest unit, the becquerel (Bq), is the number of decays
that occur each second. For example, a sample of radioactive material
might undergo 300 decays each second; we express this rate as 300 Bq.
While becquerels are useful in describing how quickly a substance
decays, this unit does not differentiate between alpha, beta, and gamma
decays. Because each type of decay produces a different energy, the
becquerel gives limited information on how these decays affect humans.
When describing human exposure to radiation, scientists commonly
use a different unit, the sievert (Sv). One sievert is one joule of energy per
kilogram of mass:

1 sievert (Sv)=1 J/kg

Because a sievert measures the actual amount of energy absorbed, it is a
better indicator of radiation exposure than the becquerel. A similar unit to
the sievert is the rem:

1 Sv=100 rem

1031
 Health physicists study the health effects of radiation. Industries from medicine
to mining involve exposure to radioactive substances, and professionals trained
to address these safety concerns are often in high demand.

The effect of radiation exposure depends on the type of biological

tissue exposed. For example, bone is less susceptible to radiation damage,
while softer tissue like gonad or breast tissue is more susceptible. When
assessing radiation exposure, health professionals take these differences
into account to determine a total effective dose.

The units sievert and rem measure human exposure to radiation.

Many countries have guidelines for how much radiation people should
be exposed to each year, especially in the workplace. For example, U.S.
law requires employers to limit the total effective dose to 50 millisieverts
(mSv) per year (Table 16.3).

TABLE 16.3 Approximate Levels of Radiation Exposure*

Exposure

Event mSv mrem

Background radiation (cosmic rays, radiation in food, radon in the air, 3 300
etc.; per year)
Dental X-ray 0.005 0.5

Plane flight (2 hours) 0.01 1

Mammogram 0.4 40

Radiation exposure near the Fukushima reactor site (2 km away, 1 10.0 1,000
month)**

Annual Limits

U.S. maximum annual dose 50 5,000

1032
 Clear link to increase cancer risk (per year) 100 10,000

Short-Term Limits

Short-term dose limit for emergency workers 250 25,000

Potential for radiation sickness (short-term levels) 400 40,000

Potential for death (short-term levels) 2,000 200,000

* Amounts will vary based on diet, location, and other factors.

** Based on measurements obtained in December 2015.

In 2014, Japanese researchers published a major study of radiation levels around

the Fukushima site. Although radiation in many places has decreased to safe
levels (< 20 mSv/yr), other areas near the reactor will remain uninhabitable for
years.

Uses of Radioactive Nuclides

Although it is important to limit exposure to radioactivity, radioactive
nuclides also have several practical applications in fields ranging from
medicine to geology and archaeology. In this section, we’ll survey some
applications of these nuclides.
Uses in Medicine

1033
Nuclear medicine deals with the use of radioactive nuclides for medicinal
purposes. The two primary applications for nuclear medicine are imaging
and radiation therapy.
In radiation imaging, a patient is administered a dose of a compound
containing a radioactive nuclide. As this material travels through the body,
it emits gamma radiation. This radiation is detected using a scintillation
camera, an instrument that produces an image based on the intensity of
gamma radiation. For example, doctors sometimes use radiation imaging
to locate internal infections. This involves “tagging” a patient’s white
blood cells with a radioactive isotope such as indium-111. The white blood
cells congregate around infected sites. As they do, they provide a gamma
ray “beacon” that pinpoints the location of the infection (Figure 16.17).

Figure 16.17 The dark spots in the rib cage are produced by gamma radiation emitted
from white blood cells that have been tagged with radioactive indium. An image like
this allows doctors to locate an internal infection.
Source: this research was originally published in JNmt. love c and Palestro cJ.
radionuclide imaging of infection. J Nucl Med Technol. 2004; 32(2):47—57. Figure 3.

Radiation therapy is a common treatment for cancer. This technique

uses the destructive properties of radiation to destroy cancer cells. In one
variation of this technique, doctors implant “seeds” containing radioactive
nuclei (such as Cs-131) near the tumor site (Figure 16.18). Radiation
therapy is a messy technique: Although it may destroy cancer cells, it
damages nearby healthy cells as well, and many patients battle radiation
sickness throughout the treatment process.

1034
Figure 16.18 These “seeds” containing Cs-131 are implanted in a patient near the site
of a tumor.

Uses in Geology and Archaeology

Because radioactive nuclei decompose at predictable rates, they are useful
as a “clock” for dating archaeological artifacts and understanding Earth’s
history. Many different nuclei are used for this type of application, but
we’ll look at just a few examples here.
Carbon-14 Dating of Plant and Animal Remains
When high-energy radiation from the sun enters the upper atmosphere, it
produces several nuclear reactions. In one such reaction, a neutron collides
with nitrogen-14 to produce carbon-14 and an additional proton. In Earth’s
atmosphere, a tiny fraction of the total carbon (about 1.5 atoms per trillion)
is carbon-14.

n01+N714→C614+p11

Unlike the main isotopes of carbon (12C and 13C), carbon-14 is

radioactive, with a half-life of about 5,730 years. As plants grow, they take
in carbon dioxide from the atmosphere, including carbon-14. After a plant
dies, the carbon-14 begins to decay. Based on the amount of carbon-14 left
in a sample, scientists can determine the age of substances that were
derived from plants. This includes animals (which directly or indirectly
consume plant matter), paper, leather, or even food residues found in
ancient pots (Figure 16.19).

1035
Figure 16.19 In 1948, a young shepherd discovered a collection of scrolls in the caves
near Qumran, along the Dead Sea in Israel. The scrolls, including this text of the Book
of Isaiah, were analyzed by carbon dating and found to be written approximately 2,200
years ago.

By dating fragments from a known time period (for example, a

document from the Roman era), scientists have confirmed the validity of
this process. However, carbon dating is considered reliable only for events
occurring in the past 50,000 years.
Dating of Exposed Rock Surfaces Using Be-10 and Al-26
Quartz is a type of rock with the empirical formula SiO2. When high-
energy radiation strikes quartz, it produces small amounts of the
radioactive nuclides beryllium-10 and aluminum-26. If quartz is exposed
on the Earth’s surface, it is bombarded by solar radiation, and these
nuclides gradually accumulate. If the quartz is buried, solar radiation is
blocked, and the isotopes begin to decompose. Geologists use the amounts
of Be-10 and Al-26 present in a quartz sample to determine how long a
rock surface has been exposed or how long it has been buried. This
technique enables geologists to track the movement of glaciers over long
periods of time (Figure 16.20).

Figure 16.20 Tiny amounts of Be-10 and Al-26 trapped in the quartz structures help
geologists determine how long this rock surface has been exposed.

1036
 Example 16.7 Using Half-Lives in Radiocarbon Dating
An archaeologist excavating an ancient city finds a sample of grain in a
pot. Carbon-14 analysis shows that the grain has a 14C concentration that
is one-fourth of the amount in modern grain. How old is the grain in the
pot?
Each half-life reduces the concentration of a radioisotope by one-half.
Since only one-fourth of the amount of 14C is present, two half-lives have
passed:

12×12=14

Because the half-life of 14C is 5,730 years, the sample will be

approximately 11,460 years old. •

IT
TRY

7. Iodine-131 decays by beta emission and has a half-life of about 8

days. Iodine-129 also decays by beta emission, but it has a half-life of
about 16 million years. Which of these isotopes presents the greater
radiation hazard?

8. The concentration of 14C is usually measured in the number of decays

observed. Living plants have a 14C concentration of about 240 Bq/kg
C. If a tree dies today, what concentration of 14C would still be
present after 3 half-lives?

1037
16.3 Energy Changes in Nuclear Reactions
In the nucleus, positively charged protons and uncharged neutrons pack
tightly together. Although the nucleus holds nearly all the mass of the
atom, it is only a tiny fraction of the atom’s volume (Figure 16.21).

Figure 16.21 Compared to the atom’s volume, its nucleus is about the size of an insect
inside a football stadium.

Think about this for a moment: You know that positively charged
particles repel each other, but in the nucleus, these particles pack tightly
together. For example, a single atom of lead has 82 protons crammed into
its tiny nuclear space. How is this possible?
The force of repulsion between charged particles is overcome by
another force, called the nuclear force, that holds the nucleus together.
Like other fundamental forces, such as gravity or the attraction and
repulsion between charged particles, the nature of the nuclear force is not
well understood. But just as we can measure energy changes caused by
gravity or the attraction between charges, we can measure energy changes
that occur in the nucleus. However, understanding these energy changes
requires us to rethink some basic assumptions about the nature of matter.

Mass Defect, Binding Energy, and Einstein’s Famous Equation

Earlier in this book, we introduced the law of conservation of mass. This
law states that in chemical changes, the total mass does not change—the
mass before a chemical reaction is the same as the mass afterward. In a
chemical reaction, the number and type of atom do not change—only the
arrangement of electrons changes.

Conservation of mass applies to chemical changes but not to nuclear

changes.

1038
 The law of conservation of mass does not apply to nuclear changes.
For example, a helium nucleus contains two protons and two neutrons. The
masses of protons and neutrons have been precisely measured and are
listed in Table 16.4. Based on the mass of the protons and neutrons, we
would expect the mass of the helium nucleus to be 4.031880 atomic mass
units (u):
Expected mass of He nucleus=2(proton mass)+2(neutron mass)=2(1.007276
u )+2(1.008664 u)=4.031880 u

TABLE 16.4 Subatomic Particles

Particle Mass (u) Charge Location

Proton 1.007276 +1 Nucleus

Neutron 1.008664 none Nucleus

Electron 0.000549 –1 Electron cloud

However, the mass of a helium nucleus has also been precisely measured:
It is 4.001503 u. The actual mass is slightly less than the sum of the
masses of the protons and neutrons! The difference between the masses of
the individual particles and the mass of the complete nucleus is called the
mass defect:
Expected mass:4.031880 u– Actual mass:4.001503 u¯Mass defect: 0.030377 u

So what does the mass defect actually mean? In 1905 Albert Einstein
(Figure 16.22) answered this question when he introduced the concept of
mass-energy equivalence. This groundbreaking idea states that mass can
be converted into energy, and energy can be converted into mass. The two
are related by his famous equation:

E=mc2

1039
Figure 16.22 Einstein’s discoveries shook the way people think about matter and
energy. His equation is vital to understanding the energy changes in nuclear reactions.

This equation states that energy (E) is equal to mass (m) times the
speed of light (c) squared. When two protons and two neutrons combine to
form a helium nucleus, a small fraction of the mass (the mass defect) is
actually converted into energy. This energy, called the binding energy, is
released when the protons and neutrons form a new nucleus. Stated
differently, the binding energy is the energy required to break a nucleus
into its individual particles.
For nuclear equations, the law of conservation of mass is replaced by a
more complete law, called the law of conservation of matter and energy,
which states that in any change, the total matter and energy in the universe
remain constant.

In exothermic nuclear reactions, some mass is converted into energy.

In this brief survey, we will not work examples using Einstein’s

equation. However, it is important to recognize that nuclear reactions
always involve a conversion between matter and energy. As we’ll see in
the sections that follow, these energy changes are vital to understanding
and harnessing nuclear reactions.

Nuclide Stability
Every nuclide has a unique mass defect, and therefore a unique binding
energy. Some nuclides have higher potential energy, while others are more
stable. Nuclear scientists compare the stability of different nuclei using the
binding energy per nucleon. The greater the binding energy per nucleon,
the more stable the nucleus is. The most stable nucleus is 56Fe. Nuclei that
are near this size tend to be stable, but very large or very small nuclei are

1040
less stable (Figure 16.23). As we’ll see in coming sections, these
differences in stability create the potential for high-energy nuclear
changes.

Figure 16.23 The most stable compounds have the highest binding energy per nucleon.
In this graph, the negative binding energy is plotted so that the most stable isotope is at
the bottom of the energy curve.

The stability of nuclides also depends on the ratio of protons to

neutrons. Smaller atoms often have a 1:1 ratio of protons and neutrons in
their nuclei. For example, most carbon atoms contain 6 protons and 6
neutrons. Larger atoms have more neutrons than protons, although the
ratio is only slightly higher than 1:1 (Table 16.5). Notice in Figure 16.24
that only a narrow band of neutron–proton ratios form stable nuclei.

TABLE 16.5 Proton–Neutron Ratios

Atom Protons Neutrons

Hydrogen 1 0

Helium 2 2

Carbon 6 6

Potassium 19 20

Iron 26 30

Gold 79 118

Lead 82 126

Uranium 92 146

1041
Figure 16.24 Most stable nuclei have slightly more neutrons than protons.

IT
TRY

9. The alpha decay of polonium-210 (shown below) releases a

tremendous amount of energy. Based on the law of conservation of
matter and energy, which has the greater mass—the reactants or the
products?
P84210o→P82206b+H24e+energy

1042
16.4 Nuclear Power: Fission and Fusion

Fission
Nuclear power plants such as the Fukushima reactor depend on a process
called fission to produce power. In a fission reaction, a large nucleus
shatters into several smaller nuclei, releasing huge amounts of energy.
Only a few nuclides are susceptible to fission reactions; the most common
and important of these is uranium-235. When a high-energy neutron
strikes a U-235 nucleus, the uranium shatters into smaller fragments
(Figure 16.25).

Figure 16.25 In a fission reaction, a heavy nucleus like 235U shatters into smaller
fragments, releasing a tremendous amount of energy.

1043
 Explore

Figure 16.25

Fission reactions create many different products. The following

nuclear equations show two of the possible fragmentations:
n01+U92235→X54138e+S3895r+3 n01 →B56144a+K3690r+2 n01

The fission of a uranium nucleus produces several more high-energy

neutrons. If there are enough other U-235 atoms nearby, these neutrons
produce a chain reaction, releasing tremendous amounts of energy.
Because each nuclear fission produces multiple neutrons, the number
of atoms involved in a fission chain reaction can multiply rapidly. The
devastating power of atomic bombs, such as those unleashed on Japan at
the end of World War II, rely on this unrestrained reaction (Figure 16.26).
For peaceful applications, nuclear scientists and engineers carefully guard
against runaway reactions. Two keys to avoiding runaway reactions are
fuel composition and reactor design. We’ll discuss each of these
approaches next.

1044
Figure 16.26 At the close of World War II, a weaponized nuclear chain reaction
decimated the Japanese city of Hiroshima.

Uranium Enrichment
Naturally occurring uranium consists of two primary isotopes. Only the
lighter isotope, U-235, is fissile (that is, it undergoes fission reactions).
The heavier isotope, U-238, does not undergo fission reactions. The non-
fissile U-238 is the major isotope: 99.3% of Earth’s uranium is U-238.
Only 0.7% of naturally occurring uranium is the fissile U-235 (Figure
16.27).

Figure 16.27 Uranium-235 undergoes nuclear fission, but it is only 0.7% of naturally
occurring uranium.

To sustain a nuclear chain reaction, scientists take uranium through an

enrichment process that increases the percentage of U-235 in a sample.
Here’s how it works: Uranium is mined as uranium oxide (U3O8). Purified
U3O8—sometimes called yellowcake uranium because of its color and
texture (Figure 16.28)—is converted to pure uranium metal and then to
uranium hexafluoride, UF6:

1045
 U3O8→3 U+4 O2U+3 F2→UF6

Figure 16.28 Purified U3O8 is called yellowcake because of its bright color.

A liquid at room temperature, UF6 has a low boiling point (57 °C) and
is easily converted into gas. The UF6 gas is then passed through a gas
centrifuge (Figure 16.29). The centrifuge spins the gas samples, pushing
the heavier 238UF6 to the edges, while the lighter 235UF6 collects in the
middle. Because the mass difference is so small, this approach produces
only a slight enrichment of UF6. However, by repeating the process
thousands of times, scientists eventually can obtain highly purified
235UF . This material can then be converted back into elemental uranium.
6

1046
Figure 16.29 A gas centrifuge separates UF6 based on molecular mass. When the
centrifuge spins, the heavier 238UF6 presses to the outside, while the lighter 235UF6
concentrates in the middle.

Nuclear reactors do not use highly purified uranium-235. The fuel rods
used in nuclear reactors contain about 5% 235U and 95% 238U. The excess
uranium-238 serves two purposes: First, it absorbs some of the neutrons,
preventing a runaway reaction. Second, after absorbing a neutron, this
nuclide undergoes two beta decays to form plutonium:
n01+U92238→U92239U92239→β N93239p+e−10N93239p→β P94239u+e−10

Plutonium also undergoes nuclear fission. So even though U-238 does not

1047
undergo fission directly, it does slowly produce plutonium fuel.

Peaceful nuclear power requires only about 5% enriched 235U. Nuclear

weapons require highly purified 235U.

In contrast to peaceful applications, weaponization of nuclear power

requires highly purified U-235 or Pu-239. In recent years two nations,
North Korea and Iran, have aggressively developed nuclear weapons
programs. Despite international opposition, these nations installed large
groups of centrifuges to carry out the enrichment process. North Korea
became a nuclear power around 2009. In 2015, Iran reached an agreement
with the international community, ostensibly delaying its acquisition of
nuclear weapons. It is unclear what impacts these programs will have in
years to come.
Fission Reactor Design
A nuclear fission reactor produces heat that can be converted into
electrical power. Because of the hazards of nuclear fission, reactors are
designed to closely regulate the chain reaction and isolate the radioactive
by-products. Nuclear reactors have three distinct stages, or loops: the
reactor loop, the power loop, and the cooling loop (Figure 16.30).

Figure 16.30 A schematic of a nuclear reactor. Water from the reactor core heats water
in the power loop, converting it to steam, which turns a turbine to produce electricity.
The cooling loop condenses the water in the power loop so it can be recycled. Excess
heat energy is released through large cooling towers.

The reactor loop contains the reactor core, where fission takes place.
The reactor core is immersed in water; pumps circulate this water to carry

1048
heat energy away from the core. A series of graphite control rods may be
raised or lowered into the core. The control rods are able to absorb
neutrons produced in the fission reaction. Inserting the control rods slows
down the reaction; withdrawing the rods speeds it up.
Water in the reactor loop is contaminated with radioactive nuclides
produced in the fission process. Because of this the reactor loop is
completely sealed off from the outside environment and housed in a
reinforced containment structure. Inside the containment structure, the
reactor loop comes in contact with water in the power loop. The reactor
loop heats the water in the power loop, converting it to steam. The steam
turns turbines, producing electric current.
After passing through the turbines, the steam is cooled and condensed
by water in the third loop—the cooling loop. This water is drawn from a
nearby source, such as an ocean or large river. The excess heat absorbed
by the cooling loop is released through massive cooling towers (Figure
16.31). Although the towers appear to emit smoke, they release only water
vapor into the air.

Figure 16.31 The cooling towers of a nuclear plant release steam into the atmosphere.

Notice in Figure 16.30 that no water from inside the containment

structure is released into the environment. The use of three water loops
prevents radioactive material from escaping into the environment.
Waste from Nuclear Fission
Generating power through fission reactions has both benefits and
drawbacks. On one hand, nuclear fission plants normally do not release
toxic products into the environment. On the other hand, the by-products
from nuclear fission reactions are highly radioactive and remain so for

1049
thousands of years. There is no way to get rid of these materials, so they
must be stored in huge underground caverns (Figure 16.32). And, as the
Fukushima disaster illustrates, there is always the risk of an accident with
terrible environmental consequences.

Figure 16.32 Spent nuclear fuel is stored deep underground.

Fusion
Energy produced by the sun and other stars arises from a nuclear reaction
called fusion, in which two smaller nuclei (usually hydrogen or helium)
combine to form larger nuclei. Like fission reactions, fusion reactions
produce huge amounts of heat energy (Figure 16.33). Two of the major
events that take place in our sun are shown in these nuclear reactions:
H11+H12→H23eH23e+H23e→H24e+2 H11

Figure 16.33 In a fusion reaction, two smaller nuclei combine to produce a larger one.

Explore
Figure 16.33

Replicating Fusion on Earth

1050
Since the dawn of the nuclear age, scientists have dreamed of harnessing
fusion power for use on Earth. From an environmental perspective, fusion
power is the ultimate in clean energy: The fuel source (hydrogen) is nearly
limitless, and the products (helium) are benign.
Fusion reactions require three conditions: First, there must be high
pressure, so the small nuclides are compressed into a tiny space. Second,
there must be very high temperature—in excess of 10 million degrees
Celsius. Third, there must be a long contact time—that is, the pressure and
temperature must be sustained long enough for fusion to take place
(Figure 16.34). Of the possible fusion reactions, the one that requires the

lowest density and temperature is the reaction of a H12 nuclide (called

deuterium) with a H13 nuclide (called tritium):

H12+H13→H24e+n01

Figure 16.34 For fusion to occur, three key parameters must be met.

In the stars, strong gravitational forces hold atoms tightly together, and
fusion occurs easily. On Earth, obtaining the extremely high temperature
and pressure requirements has proven very difficult—at least for peaceful
purposes.
By the 1950s and 1960s, the United States and the Soviet Union had
mastered the use of fusion for weapons purposes. Thermonuclear
weapons, sometimes called hydrogen bombs, use a devastating two-stage
detonation: The first stage is a fission reaction that compresses hydrogen
fuel into a tiny area. The resulting heat, pressure, and contact time initiate
a second detonation—a fusion reaction. Figure 16.35 shows the
mushroom cloud from a hydrogen bomb over 500 times more powerful
than the fission bomb that was dropped on Hiroshima.

1051
Figure 16.35 This mushroom cloud was caused by a hydrogen bomb test in 1952 that
was over 500 times more powerful than the bomb dropped on Hiroshima. This type of
weapon initiates a primary fission reaction that produces enough energy to generate a
secondary fusion reaction.

In 1961, the Soviet Union detonated the most powerful hydrogen bomb ever
tested, releasing an energy equivalent to 50 million tons of TNT.

Are We There Yet?

Despite advances in weapons technology, generations of scientists have
struggled to harness fusion power for peaceful purposes. The core
challenge is plasma containment. Plasma is a gas-like state that is
produced at very high temperatures, and it is often considered the fourth
state of matter. In the plasma state, electrons have such high energies that
they do not remain bound to any one atom. The result is a gaseous “soup”
of positive and negative charges.
Plasma is far too hot to contain in a solid structure. Instead it must be
trapped inside a very strong electromagnetic field, sometimes called a
magnetic bottle. To date the most effective magnetic bottle design is the
tokamak design, developed by the Soviet Union in the 1950s. In this
design, powerful magnetic fields create a donut-shaped pathway for
plasma to circulate, as shown in Figure 16.36.

1052
Figure 16.36 The tokamak design traps plasma inside a powerful magnetic coil. This
image is a simulation of plasma movement within these coils.

Over the past 70 years, scientists have made steady improvements in

tokamak design. The plasma temperature, density, and contact time have
steadily increased. Nuclear scientists have observed fusion reactions
within tokamak reactors, but they have not yet reached the break-even
point, where the energy output from a tokamak reactor exceeds the energy
required to power the plasma containment vessel.
But that may be about to change. In 2013, scientists began construction
of the massive International Thermonuclear Experimental Reactor (ITER)
in southern France. Nuclear scientists expect ITER, which is scheduled to
begin operating in 2027, to exceed the break-even point by a factor of 10
—producing 500 megawatts of power for every 50 megawatts of power
consumed (Figure 16.37). In the coming years, it will be interesting to see
if this international effort is, in fact, able to meet this goal.

IT
TRY

10. At first it might seem surprising that both breaking nuclei (fission)
and forming nuclei (fusion) could release so much energy. Based on
Figure 16.23, can you explain how both fission and fusion processes
are energetically favorable? When does each of them occur?

1053
Figure 16.37 (a) The ITER reactor depicted here is currently under construction in
southern France. (b) This photograph shows the construction of the massive reactor
building.

1054
Summary
Nuclear changes are all around us. Although most of the events we
observe each day do not involve nuclear changes, these reactions are a
vital part of our existence. Nuclear changes occur in a variety of settings,
from fusion in the sun to reactions in Earth’s upper atmosphere to
synthetic changes produced in laboratories and power plants.
We use nuclear equations to represent changes that take place in
atomic nuclei. Because the identity of the atom is determined by the
number of protons in a nucleus, many nuclear changes convert atoms from
one element to another.
Radioactive nuclei decompose over time, releasing particles and
energy to produce a more stable species. There are three major types of
radioactive decay: alpha, beta, and gamma. An alpha decay is the emission
of two protons and two neutrons—a helium nucleus. A beta decay is the
emission of an electron from the nucleus—a process that converts a
neutron into a proton. A gamma decay is a release of energy as the nucleus
shifts to a more stable configuration. The high-energy particles and energy
released by radioactive decay are called radiation.
Some radioactive nuclei decompose more quickly than others. The
half-life is a measure of the rate of nuclear decay; it is the time required for
one-half of a radioactive sample to decay to other substances. While
humans are constantly exposed to low levels of radiation, exposure to
higher levels of radiation can be harmful or fatal. Many nations have
guidelines for acceptable exposure levels.
Although radioactive substances carry some risks, they also are useful
for many applications. Radioactive nuclei are used for medical imaging
and treatment of cancer. Radioactive substances are also used in geology
and archaeology to identify the ages of different artifacts. Carbon-14 is
used to date materials derived from plant life. Be-10 and Al-26 are used to
determine how long rock samples (particularly quartz) have been exposed
to the atmosphere.
Nuclear changes involve a huge amount of energy. When protons and
neutrons combine to form a new nucleus, a portion of the mass is
converted into energy. The relationship between mass and energy is
governed by Einstein’s famous equation E = mc2. The difference between
the mass of a nucleus and the masses of the individual particles is the mass
defect, and the energy released in a nuclear change is the binding energy.
Nuclear fission is the process by which larger nuclei, such as uranium

1055
and plutonium, are split into smaller nuclei, releasing huge amounts of
heat energy. Nuclear power plants use the fission process. Nuclear fusion
is the process by which smaller nuclei, such as hydrogen and helium, bind
together to form larger ones. Nuclear fusion powers the sun and the other
stars.
Harnessing the nuclear fusion reaction for energy production has been
a major goal of nuclear scientists for many years. A massive fusion reactor
currently under construction in Europe may finally pass the break-even
point, where the power produced by the fusion reaction equals the power
required by the magnetic containment system, and begin to move toward
producing commercial power.

1056
Powering the Future

At the beginning of this chapter, we discussed the 2011 earthquake and tsunami
that struck Japan, claiming over 15,000 lives and causing the biggest nuclear
disaster in nearly 30 years. The nuclear meltdown contaminated the land and
waters around the power plant. Much of the area around the Fukushima plant is
now uninhabitable.
The Fukushima disaster provoked fears across the world. In the weeks that
followed, policy experts debated the role of nuclear power in the world’s future.
Even today, that future is unclear.
What is clear is that the world needs power—lots of it. Over the past 80 years,
world consumption of energy has steadily increased. As major countries like China
and India continue to develop, the global appetite for power will only grow. But
where will this power come from?
The traditional answer has been fossil fuels. Coal, oil, and natural gas are the
workhorses of global energy (Figure 16.38). Recent discoveries of oil and natural
gas ensure that we’ll have enough fossil fuel energy for many years to come. But
burning fossil fuels produces carbon dioxide, and many people worry about the
effects of increased CO2 levels in the atmosphere.

Figure 16.38 Global energy consumption continues to increase. Fossil fuels are the
primary sources of energy, although hydroelectric and other renewables are having an
increased impact in the past few years. Notice that nuclear energy usage has decreased
since the Fukushima disaster.
Data from BP Statistical Review of World Energy, June 2015; retrieved August 15,
2017, from https://www.bp.com.

1057
 Nuclear fission solves this problem, but creates another one. Fission produces
only a small amount of waste, but the waste produced is highly radioactive. This
waste is not released into the atmosphere, but it must be stored somewhere—
typically in secure underground locations. There is no way to get rid of nuclear
waste. And in the event of an accident like Fukushima, the environmental effects
can be devastating.
Many people tout renewable energy sources like solar, wind, hydroelectric, and
geothermal energies. But these sources are expensive, geographically limited, and
currently supply only a tiny fraction of the world’s energy needs.
Perhaps nuclear fusion will provide a long-term solution to humanity’s energy
needs, but this resource is years away. Until then, the world will continue to
wrestle with energy issues. It’s an intricate dance of science, politics, and
economics. There is no perfect answer.

1058
Key Terms
16.1 Nuclear Changes
nuclear chemistry The study of the properties and changes of atomic nuclei.
nucleon A particle that resides in the nucleus; either a proton or a neutron.
nuclide An atom or nucleus containing a particular number of protons and
neutrons.
isotopes Nuclides having the same atomic numbers but different mass
numbers.
nuclear reaction A reaction in which the structure of an atomic nucleus
changes.
nuclear equations Equations that describe nuclear changes; in a balanced
nuclear equation, the total mass numbers and total atomic numbers are the
same on both sides of the equation.

16.2 Radioactivity
radioactive decay A nuclear change in which the nucleus spontaneously
transitions to a more stable state; decays involve the release of energy and
sometimes mass.
radioactivity The spontaneous release of particles and/or energy from the
nucleus.
alpha (α) decay A radioactive decay in which the nucleus ejects an alpha
particle consisting of two protons and two neutrons.
beta (β) decay A radioactive decay in which the nucleus ejects an electron (a
beta particle); this change transforms a neutron into a proton.
gamma (γ) decay A radioactive decay in which the nucleus releases energy
but no particles.
decay series A naturally occurring sequence of radioactive decays; also called
a decay chain.
half-life The amount of time required for one-half of a radioactive substance to
decay.
Geiger counter A device that measures radiation, including alpha- and beta-
emissions and gamma rays.
scintillation counter A device used for measuring radiation; a scintillation
counter measures the total number of radiative particles and waves striking the
detector.

1059
semiconductor counter A device that measures radiation; some
semiconductor counters can distinguish between the decays of different
elements based on the energy of the radiation.
dosimeter A radiation detector that measures human exposure to radiation.
becquerel (Bq) The number of radioactive decays that occur per second for a
particular substance.
sievert (Sv) A measure of radiation useful for measuring human exposure; 1
Sv = 1 joule of energy per kilogram of mass.
nuclear medicine The use of radioactive nuclei for medicinal purposes.

16.3 Energy Changes in Nuclear Reactions

nuclear force The force that holds the nucleus together.
mass defect The difference between the masses of the individual particles and
the mass of the complete nucleus.
binding energy The energy released when protons and neutrons combine to
form a nucleus, or the energy required to break a nucleus into its component
particles.
law of conservation of matter and energy In any change, the total matter and
energy in the universe remain constant.

16.4 Nuclear Power: Fission and Fusion

fission A nuclear change in which a large nucleus shatters into several smaller
nuclei, releasing large amounts of energy.
fusion A nuclear change in which two smaller nuclei (usually hydrogen or
helium) combine to form larger nuclei.
plasma A gas-like state produced at very high temperatures and often
considered the fourth state of matter; in this state, electrons do not remain
bound to a single atom.

1060
 Additional Problems

16.1 Nuclear Changes

11. Which subatomic particle determines the elemental identity of an atom?

12. What is the difference between the atomic number and mass number?

13. What is the difference between nucleons and nuclides?

14. What is the difference between nuclides and isotopes?

15. Identify the element that corresponds to each of these nuclides:

a. a nuclide with atomic number 15
b. a nuclide with atomic number 37
c. a nuclide with 6 protons and 7 neutrons
d. a nuclide with 9 protons and 10 neutrons

16. Identify the element that corresponds to each of these nuclides:

a. a nuclide with atomic number 18
b. a nuclide with atomic number 4
c. a nuclide with 7 protons and 8 neutrons
d. a nuclide with 21 protons and 24 neutrons

17. Determine the number of protons and neutrons in each of these nuclides:

a. C2040a

b. 2041Ca

c. N23a

d. C52r

18. Determine the number of protons and neutrons in each of these nuclides:

a. F2656e

1061
 b. F2654e

c. T98c

d. T159b

19. Write atomic symbols, including atomic number and mass number, for
these nuclides:
a. a nuclide with 15 protons and 17 neutrons
b. a nuclide with 4 protons and 4 neutrons
c. a nuclide with 9 protons and a mass number of 19
d. a nuclide with an atomic number 13 and mass number 28

20. Write atomic symbols, including atomic number and mass number, for
these nuclides:
a. a nuclide with 31 protons and 38 neutrons
b. a nuclide with 26 protons and 30 neutrons
c. a nuclide with 46 protons and a mass number of 106
d. a nuclide with atomic number 16 and mass number 32

21. Write atomic symbols, including atomic number and mass number, for
these nuclides:
a. helium-4
b. carbon-14
c. uranium-235

22. Write atomic symbols, including atomic number and mass number, for
these nuclides:
a. magnesium-24
b. silicon-29
c. ytterbium-172

23. Indicate whether or not each of these nuclides is an isotope of carbon-12:

a. a nuclide with atomic number 5 and mass number 12
b. a nuclide in which 6 of the 13 nucleons are positively charged
c. a nuclide with 8 protons and 9 neutrons
d. a nuclide with 6 protons and 8 neutrons

1062
24. Indicate whether or not each of these nuclides is an isotope of B511:

a. a nuclide with atomic number 5 and mass number 12

b. a nuclide with atomic number 6 and mass number 13
c. a nuclide containing 15 nucleons, including 7 protons
d. a nuclide with 6 protons and 8 neutrons

25. Write a nuclear equation to describe this process: When samarium-152 is

struck by a neutron, it forms samarium-153.

26. Write a nuclear equation to describe this process: A thorium-232 nucleus

decomposes to form radium-228 and helium-4.

27. Fill in the missing nuclide to complete each nuclear equation:

a. n01+M4298o→_____

b. R86222n→_____+P84218o
c. U92235+n01→_____+R3797b+2 n01

28. Find the nuclide necessary to complete each nuclear equation:

a. n01+T52130e→_____

b. P94240u→H24e+_____
c. U92235+n01→M4299o+_____+2 n01

29. Find the nuclide necessary to complete each nuclear equation:

a. R88226a→H24e+_____
b. p11+C48111d→I49111n+_____

c. U92235+n01→S62156m+_____+3n01

1063
30. Find the nuclide necessary to complete each nuclear equation:

a. P84214o→_____+P82210b
b. p11+M42100o→_____+2 n01

c. P94239u+n01→_____+X54140e+2 n01

31. Palladium-103 is commonly used in treating prostate cancer. This isotope

is produced by striking 102Pd with a fast-moving neutron. Write a
balanced nuclear equation to describe this process.

32. When a proton collides with C48111d in a particle accelerator, it

forms a new nucleus, I49111n, and emits a neutron. Write a

balanced nuclear equation to describe this change.

33. When an alpha particle collides with S51121b in a particle

accelerator, it forms a new nucleus, 123I, plus several neutrons. How many
neutrons are released in this process? Write a balanced nuclear equation to
describe this process.

34. When an alpha particle collides with A47109g in a particle

accelerator, it forms a new nucleus, 111In, plus several neutrons. How
many neutrons are released in this process? Write a balanced nuclear
equation to describe this process.

35. A fast-moving neutron strikes a 235U nucleus. The nucleus shatters,

producing 155Sm, 78Zn, and three neutrons. Describe this change using a
balanced nuclear equation.

36. Among the reactions that take place in the sun is the combination of tritium

1064
 (3H) and deuterium (2H) to produce 4He plus a neutron. Describe this
change using a balanced nuclear equation.

16.2 Radioactivity

37. What are the differences between alpha, beta, and gamma radiation?

38. Identify each of these events as alpha, beta, or gamma decays:

a. A cobalt-60 nucleus emits energy, but no mass.
b. Plutonium-239 ejects a helium nucleus to form uranium-235.
c. A nitrogen-16 nucleus ejects an electron to form oxygen-16.

39. Identify these reactions as alpha, beta, or gamma decays:

a. A79201u→H80201g+e−10

b. M4299mo→M4299o+energy

c. U92233→H24e+T90229h

40. Identify these reactions as alpha, beta, or gamma decays:

a. A89222c→F87218r+H24e

b. A89229c→T90229h+e−10

c. R86222n→R86222n+energy

41. Show the products formed by these alpha decays:

a. R88226a→α

b. T90228h→α

c. P84210o→α

42. Show the products formed by these alpha decays:

1065
 a. U92238→α

b. T90218h→α

c. F87223r→α

43. Show the products formed by these beta decays:

a. L38i→β

b. N1125a→β

c. F87223r→β

44. Show the products formed by these beta decays:

a. A1328l→β

b. T81207l→β

c. T90231h→β

45. Write balanced equations showing these processes:

a. alpha decay of T90231h

b. beta decay of A89228c

c. gamma decay of T4399mc

46. Write balanced equations showing these processes:

a. alpha decay of T90232h

b. beta decay of C48120d

1066
 c. gamma decay of X54124me

47. Astatine-219 sometimes undergoes this series of decays: α, β, β. What

elements are formed in this decay series? What nucleus is formed after
these three decays occur?

48. Radon-222 is a radioactive gas that decays through this pathway: α, α, β, α,

β. What nuclides form in this decay series? What nucleus is formed after
these five decays occur?

49. Thorium-227 is a radioactive isotope that sometimes decays through this

pathway: α, α, α, β, α, β. What nuclides form in this decay series? What
stable nucleus results from this decay?

50. Lead-210 is a radioactive isotope that sometimes decays through this

pathway: β, α, β, β, β, α. What nuclides form in this decay series? What
stable nucleus results from this decay?

51. Why is alpha radiation more damaging to tissue than gamma radiation?
Why can we shield ourselves more easily from alpha radiation than from
gamma radiation?

52. List two factors that determine how damaging radiation is to human tissue.

53. List three devices commonly used to measure radiation.

54. How are a Geiger counter and a scintillation counter similar? How are they
different?

55. What advantage does a semiconductor counter have over a traditional

Geiger counter?

56. Although becquerels (Bq) and sieverts (Sv) are both used to measure
radiation, they are important in very different circumstances. What is the
difference between these units? When is each of them important?

57. Convert these radiation exposure measurements to rem:

a. exposure for a full-body CT scan (0.01275 Sv)
b. exposure from living in a brick house for one year (0.0007 Sv)

1067
 c. radiation from outer space for a person living near sea level on the
Atlantic Coast of the United States (0.41 mSv)

58. Convert these radiation exposure measurements to millisieverts (mSv):

a. exposure from a chest X-ray (10 mrem)
b. living within 50 miles of a coal-fired power plant (3 × 10–5 rem)
c. exposure from smoking a pack of cigarettes each day (36 mrem)

59. Some people worry about the cancer risk associated with a dental X-ray.
The X-ray provides an exposure of 0.005 mSv while the threshold
exposure for an increased risk of cancer is 100 mSv. Ignoring other
sources of radiation, how many dental X-rays would be required before the
radiation exposure reached the threshold for a cancer risk?

60. Health officials encourage women over age 50 to get a mammogram (an
X-ray test for breast cancer) every two years. While X-rays may detect
tumors, they also can cause them. The radiation exposure from a
mammogram is 40 mrem, and the lowest annual exposure that has been
clearly linked to an increased risk of cancer is 100 mSv. Ignoring other
sources of radiation, how many mammograms would be required in a year
before the radiation exposure reached the threshold for a cancer risk?

61. According to U.S. guidelines, companies cannot expose employees to

more than 50 mSv per year while they are working. For people flying in an
airplane, radiation exposure is 0.005 mSv/hour. At this rate, what is the
annual radiation exposure for a flight attendant who spends 1,200 hours in
the air?

62. The region around Naples, Italy, contains a labyrinthine network of

underground cemeteries, called catacombs. These cemeteries are adorned
with fantastic artwork (Figure 16.39) and are common tourist destinations.
However, radon levels in the catacombs must be monitored. Radon is a
naturally occurring radioactive gas that is produced in rock and can
accumulate in poorly ventilated underground areas. The Italian
government limits radiation exposure to 3 mSv/year. A recent study
estimated the radon exposure in the San Gennaro Superiore catacombs to
be 0.37 mSv/hour. At this level, what is the risk to tourists who take the 1-
hour tour of the catacombs? How many hours can a guide spend in the
catacombs each year and still be below the acceptable risk level?

1068
 Figure 16.39 This fresco adorns the San Gennaro Superiore catacombs near Naples,
Italy.

63. In nuclear medicine, radioactive nuclei are used primarily for imaging and
for radiation therapy. Give an example of a nucleus used for each
procedure, and briefly describe the role it plays.

64. Most radioisotopes used in medicinal imaging are gamma emitters rather
than alpha or beta emitters. Why do you think this is so?

65. How is carbon-14 formed in Earth’s atmosphere? How are scientists able
to use this process to determine the age of plant matter?

66. Although carbon-14 is taken up by plants, scientists also use it to

determine the age of animal remains, such as bone or leather. Why are
they able to do this?

67. Scientists measure the number of carbon-14 decays to determine the

percentage of carbon-14 in a sample. The carbon-14 in living plant matter
decays at a rate of about 240 Bq (that is, 240 decays/second). What will
the rate of decay be after one half-life? After two half-lives?

68. A sample of leather is analyzed by carbon-14 dating. The concentration of

14C in this sample is found to be about 1/16th of what a modern leather

sample would contain. About how old is the sample? The half-life of 14C
is 5,370 years.

69. How are beryllium-10 and aluminum-26 formed in quartz rock (SiO2)?
What information can geologists mine from the concentration of these
isotopes in quartz?

70. Recently, a team of geologists measured the levels of beryllium-10 in

1069
 quartz rock from Mt. Darling, Antarctica (Figure 16.40). They found that
the concentrations of Be-10 decreased as they sampled farther down the
mountain. What does this finding suggest about the ice on Mt. Darling?

Figure 16.40 An aerial view of Mt. Darling, Antarctica.

16.3 Energy Changes in Nuclear Reactions

71. What is the mass defect of a nucleus?

72. Chemical reactions (those involving the gain, loss, or sharing of electrons)
follow the law of conservation of mass. Nuclear reactions (those involving
changes in the nuclear structure) do not. Why is this so?

73. What is the relationship between the mass defect and the energy released in
a nuclear reaction?

74. What is the binding energy of a nucleus?

75. The most stable nucleus (in terms of binding energy per nucleon) is 56Fe.
Based on this, what type of nuclear reactions would you anticipate from
nuclei that are much lighter, and those that are much heavier, than this
nucleus?

76. In general, how does the neutron–proton ratio change from small nuclides
to larger nuclides?

77. The nuclear reaction below absorbs energy. Based on the law of
conservation of matter and energy, which has more mass, the reactants or
the products?
Energy+H12+N714→N713+H13

1070
78. The alpha decay below releases a tremendous amount of energy. Based on
the law of conservation of matter and energy, which has more mass, the
reactants or the products?
P94240u→H24e+U92236+energy

16.4 Nuclear Power: Fission and Fusion

79. In a nuclear fission reaction, a heavy nucleus splits into several lighter
nuclei. Complete these nuclear equations to show the missing products
from the fission of a uranium-235 nucleus:
a. n01+U92235→S3895r+_____+2 n01

b. n01+U92235→_____+X54138e+3 n01

c. n01+U92235→R3793b+_____+2 n01

80. In a nuclear fission reaction, a heavy nucleus splits into several lighter
nuclei. Complete these nuclear equations to show the missing products
from the fission of a plutonium-239 nucleus:
a. n01+P94234u→Z40100r+_____+3 n01

b. n01+P94239u→_____+T52134e+3 n01

c. n01+P94239u→I53135+_____+3 n01

81. What is a nuclear chain reaction? What subatomic particles are responsible
for propagating a chain reaction?

82. Describe the isotope distribution of naturally occurring uranium. In nuclear

1071
 reactions, how do these isotopes differ?

83. Approximately 95% of the uranium in a nuclear fuel pellet is the non-
fissile isotope 238U. Why don’t nuclear reactors use highly purified 235U
for fuel? What purposes does the 238U serve?

84. Describe the differences in the composition of uranium fuel used for
nuclear weapons and uranium fuel used for peaceful nuclear power.

85. Write two balanced chemical reactions showing the conversion of uranium
oxide (U3O8) to uranium hexafluoride.

86. What is a gas centrifuge? How do these machines separate the isotopes of
uranium?

87. In a nuclear reactor, uranium-238 nuclei absorb neutrons to become U-239.

This unstable isotope undergoes two beta decays to produce plutonium-
239. Write two balanced nuclear equations showing the products formed in
this decay sequence.

88. In fission reactors, what feature controls the chain reaction to prevent it
from becoming too hot and melting through the containment vessel?

89. Describe the steps required to convert nuclear energy to electrical energy
in a power generator.

90. Describe the role of the reactor loop, the power loop, and the cooling loop
in the function of a nuclear reactor.

91. Nuclear reactions produce highly radioactive by-products. How does the
reactor design prevent the cooling water from being contaminated with
radioactive by-products?

92. What is the difference between nuclear fission and nuclear fusion?

93. What three conditions must be met for a fusion reaction to occur?

94. Fill in the missing reactants or products to complete these fusion reactions:

a. _____+H12→H23e

1072
 b. H23e+H23e→_____+2 H11

c. H12+H13→H24e+_____

95. One of the most common fusion reactions involves tritium (H13).
Nuclear scientists produce tritium by bombarding lithium-6 nuclei with
neutrons. When a neutron strikes lithium-6, the nucleus splits to produce
helium-4 and tritium. Write a balanced nuclear equation for this reaction.

96. What major technical challenge has hindered the development of nuclear
fusion?

97. What is a plasma? How does a plasma differ from a gas?

98. What is a tokamak? What potential does the tokamak hold for controlled
nuclear fusion?

99. What is the break-even point in nuclear fusion?

100. The nuclear fission reaction below releases a tremendous amount of

energy. Based on this, which has the greater mass—the starting materials
or the products?
n01+U92235→B56144a+K3690r+2 n01+energy

1073
Answers to Odd-Numbered Problems
Chapter 1: Foundations
11. Matter is anything that has mass and takes up volume.
a. Air is made of matter.
b. A football is made of matter.
c. Sunlight is not made of matter.
d. Water is made of matter.
13. Composition refers to the simple components that make up the
material. Structure refers to both the composition and arrangement of
those simpler substances.
15. Mixtures in which the components are evenly blended throughout are
homogeneous mixtures. In contrast, heterogeneous mixtures contain
regions with significantly different composition.
a. Earth’s atmosphere is an example of a homogeneous mixture.
b. Fluorine gas is an example of an element.
c. Carbon monoxide is an example of a compound.
17. a. element
b. heterogeneous mixture
c. homogeneous mixture
19. Particles in a solid vibrate within their spaces in the ordered
framework. As a solid is heated to a liquid, the atoms vibrate faster.
The more heat we add, the faster the atoms move, until eventually they
begin to break out of the rigid framework and travel freely past each
other. The substance is now in the liquid state. The particles in a liquid
move randomly but remain close to each other. If we continue heating
the liquid, the atoms move faster and faster, until they begin to break
out of the liquid phase and enter the gas phase. Particles in the gas
phase move about freely, interacting very little with each other.
21. a. solid
b. gas
c. solid

1074
23. a. physical
b. physical
c. chemical
d. chemical
e. physical
25. a. physical
b. chemical
c. physical
27. physical change
29. physical change
31. When a substance is heated, the particles within the substance vibrate
or move faster and faster.
33. a. The rock on the top of the hill has more potential energy due to its
position. The rock at the bottom of the hill is more stable because it
has no more potential energy to release.
b. Wood has more potential energy because it has more stored
chemical energy in it than the ashes have. As such, the ashes are
more stable because they have no more potential energy to release.
c. A tightly wound spring has more potential energy because it has
stored energy. The relaxed spring is more stable because it has no
potential energy to release.
35. a. a mixture
b. a compound
c. chemical change
d. The product, in container (b), has less potential energy and is more
stable.
37. Because it absorbs energy from the sun, the growth is endothermic.
39. A hypothesis is a tentative explanation that has not been tested,
whereas a theory is an idea supported by experimental evidence.
41. To test the hypothesis, you could assess the students several times
during the 9–11 a.m. time frame and at other time frames during the
day. Then, you would review the results and compare how the students
did in each time frame. Did the students do better during the 9–11 a.m.
slot or not? If they did, then your hypothesis was proven correct. To

1075
 increase the reliability of your results, you should test the students
several times during each time frame and make sure the assessment is
of similar difficulty each time.

Chapter 2: Measurement
23. a. 1.5 × 106 km
b. 3.98 × 10–4 g
c. 1.2 × 109 J
d. 2.019 × 10–2 s
25. a. 0.0005192 kg
b. 1,230 J
c. 420,000,000 °C
d. 602,000,000,000,000,000,000,000 atoms
27. a. 5.21 × 1010 min
b. 8.3 × 108 g
c. 4.35 × 10–7 m
29. a. 5.3 × 10–8
b. 1.2 × 102
c. 1.2 × 109
d. 1.94 × 10–5
31. a. ampere
b. Kelvin
c. millisecond
d. milligram
e. kiloampere
33. a. 1,000,000 g
b. 1,000 calories
c. 1,000,000 microseconds
d. 1,000,000,000 A
35. A number rounded farthest to the right from the decimal is the more
precise measurement. Therefore, a number rounded to the nearest
nanometer would be more precise than a number rounded to the

1076
 nearest gigameter.
37. Since we don’t know the true value of the compound’s mass, we
cannot say with confidence which balance is more accurate. However,
the second balance (the one that measured the compound as 11.1925
grams) is more precise, since it measured the mass to the ten-
thousandths place as compared to the first balance (hundredths place).
39. Since the ruler is marked in increments of 1 cm, our estimation needs
to go only as low as the tenths place. Since the test tube is slightly
longer than 5 cm, we can safely estimate a measurement of 5.1 cm.
41. a. 4 significant digits
b. 4 significant digits
c. 5 significant digits
d. 2 significant digits
43. 20.31 meters has a total of 4 significant digits.
45. a. 5.200 × 106 ng
b. 3.920 × 10–4 nL
c. $1.80000 × 106
47. Measured numbers have a degree of uncertainty, but exact numbers do
not. For example, if you had 4 quarters in your pocket, you counted an
exact number. There was nothing to measure.
49. a. exact
b. measured
c. measured
d. exact
51. a. 6.12 × 103 ft2
b. 31.43 cm
c. 7.0 g/cm3
53. Each person should receive $250.00 (a sum of money reported in 5
significant digits).
55. 50.4 cm
57. 103.5 g

1077
59. a. 2.13 × 107 ng
b. 17.3974 km
c. 310 μL
61. a. 2.321 × 10–5 L
b. 5 × 101 kg
c. 5.40 × 102 nm

63. 5.4 × 10–1 MΩ

65. a. 38.6 cm
b. 1.976 × 105 J
c. 1.09 × 103 kg
d. 5.71 km
67. 13 mm

69. 2.41 × 102 meters/second

71. 14.6 km/L
73. 0.17 g/lb

75. 0.7 g/m3

77. 1 × 106 m2

79. 27 ft3

81. 1 × 106 mL

83. 8.90 × 103 kg/m3

85. 0.643 g/mL
87. 1.0 g/mL
89. a. The solid will sink in water since its density is 1.05 g/cm3
b. The solid will float in water since its density is 0.96 g/cm3
91. a. 0.259 cm3
b. 2.3 × 102 cm3
93. a. Iron: 1.6 × 102 g

1078
 b. Aluminum: 54 g
c. Lead: 2.3 × 102 g
95. Absolute zero is the lowest possible temperature at which particles
have zero kinetic energy. Absolute zero is 0 K. Absolute zero is –
273.15 °C. Absolute zero is –460 °F.
97. 2.5 °F is the larger temperature change, since 1.8 × 1.3 = 2.34 °F.
99. 22 °C; 295 K
101. Carpet is cheaper at $0.644/ft2.
For an area of 600 ft2, carpet would cost $386 and
hardwood would cost $840.
Hardwood will cost more.
103. The mass of the liquid that fills the volume to full capacity: 2.4 × 104
kg
The empty mass of the truck: 4.0 × 104 kg
The sum of these is 44 tons.
The truck will not exceed the weight limit.

Chapter 3: Atoms
15. The atom is the fundamental unit of matter.
17. a. Democritus
b. Antoine Lavoisier
c. John Dalton
19. Because the bottle is sealed (nothing can escape), the mass of the
substances before the reaction is the same as the mass of the substances
at the end of the reaction. This follows the law of conservation of
mass, which states that matter is neither created nor destroyed in a
chemical reaction.
21. According to the law of conservation of mass, matter cannot be created
nor destroyed, so the mass of the products will be equal to the mass of
the reactants. A total of 300 g of sodium chloride and water will be
produced from the 200 g of sodium hydroxide and 100 g hydrogen
chloride.
23. 40.3 g magnesium oxide.

1079
25. a. possible
b. not possible
c. not possible
d. possible
27. Reaction (a) does not follow the law of conservation of mass. The
reactants contain four orange spheres and eight green spheres, whereas
the products contain three orange spheres and nine green spheres. In
the other two reactions, the number of orange and green spheres is the
same for the reactants and products.
29. a. cesium
b. titanium
c. potassium
d. boron
e. xenon
31. a. tungsten
b. boron
c. uranium
d. sulfur
e. carbon
33. a. F
b. Ge
c. Sn
d. Mg
35. a. I
b. Cu
c. Ag
d. H
37. Elements within the same column (group) on the periodic table have
similar properties.
39. a. metalloid
b. main-group metal
c. transition metal
d. main-group metal

1080
 e. inner-transition metal
f. nonmetal
g. main-group metal
h. transition metal
41. The blocks of elements on the left and right sides of the periodic table
are the main-group elements.
43. a. metal
b. nonmetal
c. metal
d. metal
45. These are Group 1A elements known as the alkali metals.
47. a. halogen
b. noble gas
c. alkali metal
d. alkaline earth metal
49. Beryllium, because it is located within the same group (column) as
magnesium.
51. Argon and sulfur, because these two elements are not found in the
same group as chlorine.
53. a. V
b. Cl
c. Ne
55. positive, negative, and neutral
57. 1. b
2. c
3. d
4. a

59. In 1909, Ernest Rutherford and his students were studying the behavior
of positively charged particles called alpha particles. They were
interested in the pattern these heavy particles made as they passed
through a thin gold film. Rutherford expected the alpha particles to
pass through the film and hit behind it. However, a small number of

1081
 the particles reflected off the film and back toward the alpha particle
source.
61. The electron cloud occupies a larger volume than the nucleus, but the
nucleus has more mass.
63. Dalton’s idea that atoms could not be broken down into smaller pieces
was proven incorrect with the discovery of subatomic particles.
65. The proton and the neutron each have a mass of about one atomic mass
unit. Both of these particles are located in the nucleus. The proton has
a +1 charge, and the electron has a –1 charge.
67. a. Ti = 22
b. S = 16
c. Te = 52
d. Li = 3
69. a. Al
b. Rn
c. Es
71. The atomic number is equal to the number of protons. The mass
number is equal to the number of protons plus the number of neutrons.
a. atomic number = 14; mass number = 30
b. atomic number = 27; mass number = 59
c. atomic number = 20; mass number = 46
73.
a. K1939

b. A1840R

c. F100257m
75. 28
77.

Atom Symbol Protons Neutrons Atomic Number Mass Number

Hydrogen H 1 0 1 1
Sulfur S 16 16 16 32
Tellurium Te 52 76 52 128

1082
 Helium He 2 2 2 4
Zirconium Zr 40 51 40 91

79. Total value = $1.48

Weighted average value of coins = $0.07

81. G3169a and G3171a

83. Both isotopes of silver have 47 protons. However, 107Ag has 60

neutrons and 109Ag has 62 neutrons.
85. Average atomic mass Br = 80.0 u
87. Average atomic mass Pb = 207.2174 u
89. The mass number of an isotope is the sum of the protons and neutrons.
The mass that is usually displayed on periodic tables is the average
atomic mass. This is a weighted average of the masses of the different
isotopes of an element.
91. The plum pudding model described small, negatively charged electrons
spread throughout a positive atomic substance. The Bohr model
correctly described the dense, positively charged nucleus with orbiting
electrons. The Bohr model described the electrons in fixed orbits
around the nucleus, like planets orbiting the sun.
93. a. 1
b. 2
c. 6
d. 34
95. a. Li number of protons = 3; number of electrons = 3
b. Ge number of protons = 32; number of electrons = 32
c. Bi number of protons = 83; number of electrons = 83
d. Ba number of protons = 56; number of electrons = 56
97. Beryllium ion = +2 charge
99. a. –3 charge
b. +2 charge
c. 0 (neutral) charge
101. A neutral silver atom has 47 protons and will therefore have 47

1083
 electrons. A +1 charged silver ion will have one less electron than its
neutral counterpart; 46 electrons.
103. Titanium +2 has 22 protons and therefore 20 electrons.
Titanium +4 has 22 protons and therefore 18 electrons.
105. Gold +1 has 79 protons and therefore 78 electrons.
Mercury +2 has 80 protons and therefore 78 electrons.
107. F –1 has 9 protons and therefore 10 electrons.
Cl –1 has 17 protons and therefore 18 electrons.
Br –1 has 35 protons and therefore 36 electrons.

Chapter 4: Light and Electronic Structure

13. a. Ultraviolet is higher in energy than infrared.
b. Green light is higher in energy than red light.
c. Microwave is higher in energy than radio.
15. a. The low-frequency wave has less energy.
b. Infrared light is lower in energy than yellow light.
17. a. Red light has a lower frequency than green light.
b. The 400 nm light does not have a lower frequency than green light.
c. Light with a wavelength of 1.0 × 10–6 m has a lower frequency
than green light.
19. a. radio wave
b. microwave
c. infrared light
21. a. 400 m
b. 1.5 × 107 m
c. 1,200 m
23. Green.

25. 2.76 × 10–19 J

27. 9.47 × 10–15 J

29. v = 7.04 × 1013 s–1

λ = 4.26 × 10–6 m

1084
 This radiation falls in the infrared range.
31. Rather than producing a continuous spectrum of colors, gas lamps
produce colors of only certain energies, called spectral lines. For
example, the light from a hydrogen lamp contains only four lines of
color: red, light blue, deep blue, and violet. These distinctive patterns
are called line spectra. Scientists often use line spectra as
“fingerprints” to identify elements. For example, when scientists
analyze light from the Sun, they find lines that correspond to the
spectral lines of hydrogen and helium. From this, they know that the
Sun and other stars are composed largely of these two elements.
33. When an electron absorbs light (i.e., energy), it jumps to a higher
energy level producing an excited state electron.
35. Electrons absorb energy that causes them to jump to high energy
levels. As these electrons return to the ground state, they release
energy in the form of light.
37. a. absorbed
b. emitted
c. absorbed
d. emitted
39.

Transition Wavelength Region

5→4 4,057 nm infrared
4→2 487 nm visible
2→1 122 nm ultraviolet

41. In quantum mechanics, we never talk about the exact position of an

electron. According to Heisenberg, this is impossible for us to know.
Instead, we talk about the most probable locations of the electrons or
the energies that the electrons possess. These probable locations are
orbitals.
43. 2
45. a. Level 1: The s orbital is present.
b. Level 2: The s and p sublevel orbitals are present.
c. Level 3: The s, p, and d sublevel orbitals are present.

1085
 d. Level 4: The s, p, d, and f sublevel orbitals are present.
47. A total of 6 electrons can reside in a p sublevel.
49. An orbital is the region where the electrons are most likely to be found.
Each sublevel contains a different number of these orbitals.
51. Two electrons can occupy the lowest energy level since it’s an s
orbital. The second lowest energy level holds the s and p orbitals,
which can hold up to 8 electrons.
53. a. This is possible since level 1 is the s orbital.
b. This is possible since level 4 includes the s orbital.
c. This is not possible since the p orbital is not part of level 1.
55.

57. a. 1s1
b. 1s22s2
c. 1s22s22p2
d. 1s22s22p6
59. The valence level is the highest-occupied energy level. The s and p
sublevels are involved with the atom’s valence.
61. The octet rule states that an atom is stabilized by having its highest-
occupied (valence) energy level filled. The noble gas family illustrates
the octet rule.
63. a. [He]2s1
b. [Ne]3s2
c. [Ne]3s23p5

1086
65. F: [He]2s22p5
Cl: [Ne]3s23p5
Br: [Ar]4s23d104p5
I: [Kr]5s24d105p5

67. The electronic configuration for Mg is 1s22s22p63s2. It has 2 valence

electrons (3s2) and 10 inner-shell electrons (1s22s22p6).
69. a. The inner-shell electrons are 1s2 and the valence electrons are
2s22p3.
b. The inner-shell electrons are 1s22s22p63s23p6 and the valence
electrons are 4s1.
c. The inner-shell electrons are [Ar] and the valence electrons are
4s23d104p1.
d. The inner-shell electrons are [Xe] and the valence electrons are
6s24f 143d4.
71. For anions, you add the same electrons to the outer sublevels as the
charge. For cations, you remove the same number from the valence
electrons.
a. O2– 1s22s22p6
b. K+ 1s22s22p63s23p6
c. Br– 1s22s22p63s23p64s23d104p6
d. N3– 1s22s22p63s23p6
73. a. O would need to gain 2 electrons and be O2–.
b. F would need to gain 1 electron and be F–.
c. Na would need to lose 1 electron and be Na+.
d. Mg would need to lose 2 electrons and be Mg2+.

75. Li+, Be2+, and H–

77. Fluorine has an electronic configuration of 1s22s22p5 and only needs

to gain one electron to have the noble gas–like configuration of neon
(1s22s22p6). Sodium has an electron configuration of 1s22s22p63s1
and only needs to lose an electron to have neon’s electronic
configuration. F gaining one electron and Na losing one electron gives
a filled 2s2 and 2p6 sublevel set (i.e., obeys the octet rule).

1087
79. K+ and S2– will both have the stability of a filled valence level having
an electronic configuration of Ar (1s22s22p63s23p6).
81. Their outer electrons are found in an s orbital.
83. Si, P, and Ar all have valence electrons in energy level 3 (all row 3
elements).

85. The inner electron configuration of elements in row 2 is 1s2.

87. Two, from their s valence orbital.
89. a. 2
b. 3
c. 7
d. 8
91. a. Mg: 3s2
b. N: 2s22p3
c. P: 3s23p3
d. As: 4s24p3
e. Cl: 3s23p5
f. Ca: 4s2
93. a. K
b. N
c. Pd
95. a. Following the 4s is the 3d subshell.
b. Following the 4d is the 5p subshell.
c. Following the 3p is the 4s subshell.
97. a. 4p2
b. 5p4
c. 4d5
d. 2p1
99. a. group 2A, the alkaline earth metals
b. group 8A, the noble gases
c. group 2A, the alkaline earth metals

1088
101. The valence electron configuration of group 6A (the chalcogens) is
s2p4.
103. Group 12 is the last column of the d block. The electronic
configuration is d10.

Chapter 5: Chemical Bonds and Compounds

15. Li has 1 valence electron.
C has 4 valence electrons.
Si has 4 valence electrons.
Kr has 8 valence electrons.
Se has 6 valence electrons.
17.

19. Mg: 1s22s22p63s2; the 3s2 are the valence electrons.

N: 1s22s22p3; the 2s22p3 are the valence electrons.
P: 1s22s22p63s23p3; the 3s23p3 are the valence electrons.
I: 1s22s22p63s23p64s23d104p65s24d105p5; the 5s25p5 are the valence
electrons.
21. a. It does not. A single electron occupies the 3s orbital.
b. It does. It’s isoelectronic with Ne.
c. It does not. It needs one electron to complete an octet.
d. It does. It’s isoelectronic with Ne.
23. the alkali metals

25. It is isoelectronic with Ar: 1s22s22p63s23p6

27. a. 1s22s1
b. 1s2
c. 1s22s22p63s1
d. 1s22s22p6
29. a. +2
b. +2
c. +1
d. +1

1089
31. +1 and +2
33. a. sodium ion
b. magnesium ion
c. chromium(II) ion
d. chromium(III) ion
35. a. iron(II) ion
b. iron(III) ion
c. rubidium ion
d. barium ion
37. a. Sr2+
b. Zn2+
c. Cu2+
d. Mn3+
39. Group 7A, the halogens, forms only –1 ions. Group 6A, the
chalcogens, forms only –2 ions.

41. One more electron would make it [He]2s22p6.

43. a. 1s22s22p63s23p5
b. 1s22s22p63s23p6
c. 1s22s22p63s23p64s23d104p5
d. 1s22s22p63s23p64s23d104p6
45. a. lose 1 electron to fulfill its octet
b. gain 2 electrons to fulfill its octet
c. lose 2 electrons to fulfill its octet
d. gain 1 electron to fulfill its octet
47. a. lose 2 electrons to fulfill its octet
b. gain 1 electron to fulfill its octet
c. gain 2 electrons to fulfill its octet
d. lose 2 electrons to fulfill its octet
49. a. F–
b. I–
c. O2–

1090
 d. Se2–
51. a. fluoride ion
b. sulfide ion
c. oxide ion
d. iodide ion
53. a. +2
b. –2
c. –1
55. a. –1
b. +1
c. –1
57. a. potassium, K+
b. rubidium, Rb+
c. chloride, Cl–
d. bromide, Br–
59. a. F–
b. Sr2+
c. Be2+
d. P3–
61. a. monatomic
b. polyatomic
c. polyatomic
d. monatomic
63. The commonly used suffixes for oxyanions are –ite and –ate.
65.
a. NH4+

b. CO32−
c. OH–

d. C3H3O2−
67.
a. ClO3−

1091
 b. SO32−
c. ClO–

d. MnO4−
69. a. Sn4+
b. Cu2+
c. F–

d. SO42−
71. a. Zn2+

b. CrO42−

c. SO32−
d. P3–
73. a. LiCl
b. CaBr2
c. CaO
d. Fe3P2

75. a. AlCl3
b. FeS
c. CaSO4
d. Al2O3

77. a. Cr(C2H3O2)3
b. Zn(ClO3)2
c. AgNO3
d. PbCO3

79. a. Cr(ClO)3
b. KMnO4
c. NaCN
d. Pb(ClO4)2

1092
81. a. +2 since the anion is –1 each
b. +4 since the anion is –1 each
c. +2 since the anion is –1 each
d. +2 since the anion is –2
83. a. sodium bromide
b. potassium oxide
c. iron(III) bromide
d. copper(II) sulfide
85. a. iron(II) carbonate
b. aluminum nitrite
c. barium nitrate
d. ammonium sulfate
87. Since they do not gain or lose electrons to fulfill their valence, the
nonmetals will share electrons with other nonmetals to fulfill their
octet.
89. Two electrons are shared per covalent bond. In drawing structures we
typically represent these two electrons as a line.
91. There are five water molecules in this figure. Each water molecule has
two covalent bonds (O—H twice per molecule).
93. The molecular formula for acetic is C2H4O2. The empirical formula is
CH2O.

95. The molecular formula for freon 112 is C2Cl4F2. The empirical
formula is CCl2F.

97. Prefixes are necessary when naming binary covalent compounds

because they often combine in more than one ratio.
99. a. sulfur dichloride
b. nitrogen trifluoride
c. dinitrogen tetroxide
d. tetraphosphorus decoxide
101. Binary ionic compounds are composed of a metal and nonmetal,
whereas binary covalent compounds are comprised of two nonmetals.
103. a. ionic bonds

1093
 b. covalent bonds
c. ionic bonds
105. a. ionic lattice
b. discrete molecule
c. discrete molecule
d. ionic lattice
107. a. ionic compound (metal + nonmetal); sodium bromide
b. covalent compound (only nonmetals); phosphorus tribromide
c. ionic compound (metal + nonmetal); magnesium bromide
d. covalent compound (only nonmetals); sulfur dibromide
109. a. covalent compound (only nonmetals); silicon tetrachloride
b. ionic compound (metal + nonmetal); aluminum chloride
c. covalent compound (only nonmetals); boron tribromide
d. ionic compound (metal + polyatomic ion); sodium sulfite
111. Electrolyte is the term used for a compound that, when dissolved in
water, will conduct electricity. Ionic compounds are likely to be
electrolytes.
113. When sodium sulfate dissociates in water, it produces Na+ ions and
SO42− ions.
115. a. KCl is likely to dissociate in water because it’s an ionic compound.
Ionic compounds tend to dissociate when dissolved in water.
b. CaBr2 is likely to dissociate in water because it’s an ionic
compound. Ionic compounds tend to dissociate when dissolved in
water.
c. CO2 will not dissociate into ions when dissolved in water since it’s
not an ionic compound (i.e., it’s a covalent compound).
d. C2H6O will not dissociate into ions when dissolved in water since
it’s not an ionic compound (i.e., it’s a covalent compound).
117. Acids are covalent compounds that produce H+ ions in aqueous
solution. Most acids contain a covalent bond between hydrogen and a
species that can form a stable anion. When dissolved in water, this
bond breaks to produce a hydrogen cation and a corresponding anion.
Their ability to dissociate makes them differ from most covalent
compounds.

1094
119. a. hydrochloric acid
b. hydrobromic acid
c. hydroiodic acid
121. a. nitric acid
b. nitrous acid
c. perchloric acid
d. chlorous acid
123. formic acid
125. a. ionic; sodium nitrite
b. covalent; dinitrogen tetroxide
c. acid; nitrous acid
d. ionic; potassium nitrite

Chapter 6: Chemical Reactions

19. H2CO3→H2O+CO2
21. 2 CH4S+H2O2→C2H6S2+2 H2O

23. a. H2+Br2→2 HBr

b. 2 Na+F2→2 NaF
25. 30 units of A will produce 15 units of C.
27. a. 30 molecules of H2 will yield 60 molecules of HCl.
b. 6 molecules of Cl2 are needed to make 12 molecules of HCl.

29. a. 15 molecules of Cl2 are needed to react with 10 Al atoms.

b. 10 Al atoms will produce 10 units of AlCl3.

31. a. 2 PCl3+3 F2→2 PF3+3 Cl2

b. 2 SO2+O2→2 SO3
c. 2 B+3 F2→2 BF3

1095
33. a. PCl3+3 H2O→H3PO3+3 HCl

b. 3 O2+2 H2S→2 SO2+2 H2O

c. 4 HCl+MnO2→MnCl2+2 H2O+Cl2

35. a. Hg(NO3)2+2 KCl→2 KNO3+HgCl2

b. MgBr2+2 NaOH→2 NaBr+Mg(OH)2

c. 2 AgC2H3O2+BaCl2→2 AgCl+Ba(C2H3O2)2

37.
a. N2+3 H2→2 NH3
b. 2 C2H6+7 O2→4 CO2+6 H2O

c. 6 HCl+2 Al→2 AlCl3+3 H2

39. a. 4 Na+O2→2 Na2O

b. Pb(NO3)2+2 KCl→PbCl2+2 KNO3

c. 2 C2H2+5 O2→4 CO2+2 H2O

41. a. Zn (s)+CuCl2 (aq)→Cu (s)+ZnCl2 (aq)

b. C3H8 (g)+5 O2 (g)→3 CO2 (g)+4 H2O (g)

c. Fe (s)+Cl2 (g)→FeCl2 (s)

1096
43. a. aqueous hydrogen chloride (usually called hydrochloric acid) and
aqueous sodium carbonate
b. solid iron and aqueous nitric acid
c. aqueous zinc chloride and solid lead(II) perchlorate
45. a. decomposition
b. synthesis
c. double displacement
47. a. synthesis
b. single displacement
c. double displacement
49. Ionic compounds are formed when metals and nonmetals react.
51. a. Mg is oxidized; O is reduced.
b. Fe is oxidized; S is reduced.
c. Ca is oxidized; Br is reduced.
53. a. Be+Cl2→BeCl2

b. 2 K+Cl2→2 KCl

c. Co (s)+Cl2 (g)→CoCl2

d. 2 Cu (s)+O2 (g)→2 CuO

55. a. 2 Mg+O2→2 MgO
b. C2H4+3 O2→2 H2O+2 CO2

c. 2 C4H10 (g)+13 O2 (g)→10 H2O+8 CO2

57. Oxygen (O2) is always involved in combustion reactions.

59. Hydrocarbons are compounds made of only hydrogen and carbon

atoms. Combustion of hydrocarbons produces water vapor and carbon
dioxide gas.
61. a. 2 C2H2+5 O2→2 H2O+4 CO2

1097
 b. C4H8+6 O2→4 H2O+4 CO2

c. 2 C9H18+27 O2→18 H2O+18 CO2

63. a. KCl (s)→K+ (aq)+Cl− (aq)

b. Li2SO4 (s)→2 Li+ (aq)+SO42− (aq)

c. (NH4)2 CO3 (s)→2 NH4+ (aq)+CO32− (aq)

65. a. NaHCO3 (s)→Na+ (aq)+HCO3− (aq)

b. AlCl3 (s)→Al3+ (aq)+3 Cl− (aq)

c. Cr(ClO2)3 (s)→Cr3+ (aq)+3 ClO2− (aq)

67. a. Compounds containing the ammonium ion are always soluble.

b. Compounds containing ions with a charge of 2 or 3 tend to be
insoluble.
c. Compounds containing the nitrate ion are always soluble.
69. a. This compound is soluble because it contains the ammonium and
chlorate ions.
b. Compounds containing ions with a charge of 2 or 3 tend to be
insoluble.
c. This compound is soluble because it contains the perchlorate ion.
71. a. soluble
b. soluble
c. insoluble
d. soluble
e. soluble

1098
 f. soluble
73. Generally a salt with a +1 cation and a –2 anion is more likely to be
soluble than a salt with a +2 cation and a –2 anion due to the larger
charge on the +2 cation.
75. a. Ag+ (aq)+NO3− (aq)+K+ (aq)+Cl− (aq)→K+ (aq)+Cl− (aq)+AgCl
(s)

b. Ba2+ (aq)+2 ClO4− (aq)+2 K+ (aq)+SO42− (aq)→BaSO4 (s)+2

K+ (aq)+2 ClO4− (aq)

NO3−
77. The spectator ions are and Cs+.
NO3−
79. The spectator ions are Na+ (aq) and (aq).
The net ionic equation is Br− (aq)+Ag+ (aq)→AgBr (s)

81. a. Fe2+ (aq)+2 Cl− (aq)+2 K+ (aq)+2 OH− (aq)→Fe(OH)2 (s)+2 K+

(aq)+2 Cl− (aq)

b. Cl− (aq) and K+ (aq)

c. Fe2+ (aq)+2 OH− (aq)→Fe(OH)2 (s)

d. The driving force is the formation of Fe(OH)2 (s).

83. a. 2 KCl (aq)+Pb(NO3)2 (aq)→PbCl2 (s)+2 KNO3 (aq)

b. 3 KOH (aq)+FeCl3 (aq)→Fe(OH)3 (s)+3 KCl (aq)

c. 3 BaCl2 (aq)+2 K3PO4 (aq)→Ba3(PO4)2 (s)+6 KCl (aq)

85. a. 2 Fe(C2H3O2)3 (aq)+3 BaS (aq)→Fe2S3(s)+3 Ba(C2H3O2)2 (aq)

1099
 b. 2 Li3PO4 (aq)+3 CuSO4 (aq)→3 Li2SO4 (aq)+Cu3(PO4)2 (s)

c. no reaction
d. FeSO4 (aq)+Ba(OH)2 (aq)→BaSO4 (s)+Fe(OH)2 (s)

(Note that the reaction in part d forms two insoluble products.)

87. a. LiI (aq)+AgNO3 (aq)→LiNO3 (aq)+AgI (s)Li+ (aq)+I− (aq)+Ag+
(aq)+NO3− (aq)→Li+ (aq)+AgI (s)+NO3− (aq)I− (aq)+Ag+
(aq)→AgI (s)

b. Pb(C2H3O2)2 (aq)+ZnSO4 (aq)→Zn(C2H3O2)2 (aq)+PbSO4

(s)Pb2+ (aq)+2 C2H3O2− (aq)+Zn2+ (aq)+SO42− (aq)→PbSO4
(s)+Zn2+ (aq)+2 C2H3O2− (aq)Pb2+ (aq)+SO42− (aq)→PbSO4 (s)

c. Pb(NO3)2 (aq)+CaI2(aq)→PbI2 (s)+Ca(NO3)2 (aq)Pb2+ (aq)+2

NO3− (aq)+Ca2+ (aq)+2 I− (aq)→PbI2 (s)+2 NO3− (aq)+Ca2+
(aq)Pb2+ (aq)+2 I−(aq)→PbI2 (s)

89. Acids are covalent compounds that ionize to produce H+ ions and a
stable anion when dissolved in water. Bases are compounds that
produce hydroxide (OH–) ions in aqueous solution.
91. a. HCl (aq)→H+ (aq)+Cl− (aq)

b. HNO3 (aq)→H+ (aq)+NO3− (aq)

93. a. H2SO4+2 NaOH→NaSO4+2 H2O

b. 3 HNO3+Al(OH)3→Al(NO3)3+3 H2O

1100
 c. 3 H2SO4+2 Cr(OH)3→Cr2(SO4)3+6 H2O

95. a. HCl (aq)+KOH (aq)→KCl (aq)+H2O (l)

b. H2SO4 (aq)+2 LiOH (aq)→Li2SO4 (aq)+2 H2O (l)

c. Ba(OH)2 (aq)+2 HNO2 (aq)→Ba(NO2)2 (aq)+2 H2O (l)

97. NaOH (aq)+HI (aq)→NaI (aq)+H2O (l) Na+ (aq)+OH− (aq)+H+

(aq)+I− (aq)→Na+ (aq)+I− (aq)+H2O (l)

99. Pb2+ (aq)+2 I− (aq)→PbI2 (s)

101. a. synthesis, oxidation-reduction

b. single displacement, oxidation-reduction
c. double displacement, precipitation
d. single displacement, oxidation-reduction
e. double displacement, acid-base neutralization
f. combustion, oxidation-reduction
103. a. AgC2H3O2 (aq)+NaCl (aq)→AgCl (s)+NaC2H3O2 (aq)

b. 2 Na (s)+Cl2 (g)→2 NaCl (s)

c. 2 HCl (aq)+Ba(OH)2 (aq)→2 H2O (l)+BaCl2 (aq)

d. C10H20 (l)+15 O2 (g)→10 CO2 (g)+10 H2O (g)

1101
Chapter 7: Mass Stoichiometry
17. a. Formula mass for BaSO4: 233.39 u
b. Formula mass for K2S: 110.26 u
c. c. Formula mass for C6H12O6: 180.18 u
d. Formula mass for FeCl3: 162.20 u

19. a. Percent by mass for C6H12O6: 53.28%

b. Percent by mass for C2H6O: 34.72%
c. Percent by mass for C16H19N3O5S: 21.89%

21. a. % carbon = 59.99%

b. % hydrogen = 4.49%
c. % oxygen = 35.52%
23. % silver = 63.50%; mass is 63.50 g Ag.
25. Scientists determine formula mass using a technique called mass
spectrometry.
27. Nitroglycerin and trinitrotoluene have a formula mass of 227 u, and
both could be the unknown that was analyzed.
29. The sample is morphine because its elemental analysis (%C and %H)
matches the percent composition of morphine (71.55% C and 6.72%
H).
31. a. 1 mole C3H8 = 44.14 g
b. 1 mole CaCl2 = 110.98 g
c. 1 mole C2H6O = 46.08 g
d. 1 mole C12H22O11 = 342.34 g

33. a. Report in grams/mole because this is a measured quantity.

b. Because it is a single molecule, it’s more appropriate to use atomic
mass units (u).
35. a. 1.13 × 103 g KCl
b. 38.4 g MgSO4
c. 833 g Ne
37. a. 295 g Au

1102
 b. 1.65 × 103 g BF3
c. 0.524 g He
39. a. 7.507 moles Mg
b. 0.413 moles Cl2
c. 1.468 moles Pb
41. a. 0.226 moles Fe
b. 0.864 moles LiF
c. 1.2 × 10–5 moles Br2

43. a. 1.02 moles Na; 6.14 × 1023 atoms Na

b. 0.588 moles P; 3.54 × 1023 atoms P
c. 16.01 moles C; 9.64 × 1024 atoms C
45. a. 0.627 moles H2O; 3.77 × 1023 molecules H2O
b. 0.563 moles PCl3; 3.39 × 1023 molecules PCl3
c. 2.94 × 103 moles CO2; 1.77 × 1027 molecules CO2

47. a. 1.2 × 1024 molecules HCl

b. 7.2 × 1022 atoms Au
c. 1.291 × 1024 atoms Zn

49. 0.241 moles Cu; 1.45 × 1023 atoms Cu

51. 1.235 × 1026 molecules H2O; 3,698 g H2O

53. 0.05302 moles Zn; 3.467 g Zn

55. 0.2279 moles H2; 0.4604 g H2

57. 3.988 × 1024 molecules C12H22O11

59. 20 moles O2 consumed

20 moles PbO2 produced

61. 6.0 moles HCl consumed

3.0 moles ZnCl2 produced
3.0 moles H2 produced

1103
63. a. 1 mole H2O consumed
b. 0.5 mole H2 produced
c. 28.0 moles K required
d. 3,014.2 moles KOH produced
65. a. 0.217 moles Zn consumed
b. 0.434 moles HCl consumed
c. 0.217 moles ZnCl2 consumed
d. 29.6 g ZnCl2 produced

67. a. 72.6 g CuBr consumed

b. 32.2 g Cu produced
c. 46.6 g MgBr2 produced

69. 91.1 g FeBr2 produced

71. 212.8 g NH3 required

73. 52.6 g CaO produced

75. 84.04 g CaO produced; 65.96 g CO2 produced

77. 1.4 × 102 g Fe(OH)2 produced

79. a. 35.9 g Cl2 consumed

b. 3.05 × 1023 molecules Cl2 consumed
c. 65.9 g SnCl4 produced

81. Limiting reagent is SO2. Excess reagent is O2.

83. a. Compound A is the limiting reagent.

b. 2.7 moles of C are expected to form.
85. Because a lake will have a large abundance of water, CaO is likely to
be the limiting reagent.
87. The limiting reagent is Fe, and a total of 0.10 mole FeS can form in
this reaction.
89. The limiting reagent is K, and a total of 0.29 g KOH can form in this
reaction.

1104
91. a. The limiting reagent is Na, producing 10.0 moles of NaI.
b. 5.0 moles I2 used
c. 1.0 moles I2 excess

93. a. The limiting reagent is SOCl2.

b. The amount of excess reagent used in the reaction is 0.420 moles.
c. The amount of excess reagent left over is 0.69 moles H2O.

95.
Si + O2 →
SiO2
Starting 1.0 4.0 mol 0 mol
Moles mol
Change –1.0 –1.0 +1.0
mol mol mol
Ending 0 mol 3.0 mol 1.0 mol
Moles
Ending 0.0 g 96 g 60 g
Grams

97. a. The limiting reagent is KOH.

b. The mass of water produced in the reaction is 48.2 g.
c. The amount of excess reagent left over is 9.0 g H3PO4.

99. Sometimes part of the material adheres to the container walls. Other
times, product is lost during the purification process. Sometimes a
competing reaction forms unwanted side products in a reaction. All of
these can lower the percent yield.
101. 92.30%
103. a. The limiting reagent is A.
b. The theoretical yield of D is 1.5 moles D.
c. 80%
105. a. Theoretical yield = 28.5 g C6H5Br
b. Percent yield = 57.2%
107. a. The limiting reagent is Na3PO4.
b. The theoretical yield of Mg3(PO4)2 is 2.75 g.

1105
 c. Percent yield = 77.8%
109. The law of conservation of mass only appears not to be followed. All
gases have mass. When the wood is burned, water vapor and carbon
dioxide are produced in addition to the ashes. The remaining 463 g
(500 g – 37 g ashes = 463 g) is lost to the environment in the form of
water vapor and carbon dioxide, common products from combustion
reactions. The law of conservation of mass is still obeyed, and if we
were to capture these gases and measure their masses, we would be
able to verify this law.
111. a. 3.219 × 105 g C2H6O; 408.0 L C2H6O
b. 6.156 × 105 g C4H8O2; 615.6 kg C4H8O2; 682.5 L C4H8O2

Chapter 8: Energy
13. Energy is the ability to do work. Changes in energy take place in two
forms: heat and work. Work is the transfer of energy from one form to
another, whereas heat causes the particles within the substance to
vibrate or move faster.
15. a. Energy is released when the anvil falls off the cliff.
b. The chemical energy stored in the spaghetti is released when
digested.
c. As the string releases its tension, the energy is released. If the
string has an arrow on it, it will transfer this energy to the arrow.
d. The chemical energy stored in dynamite is released when the
dynamite detonates.
17. a. 4.61 × 107 J
b. 1.085 × 106 J
c. 1.02 × 106 BTU

19. 2.46 × 106 BTU; 2.60 × 106 kJ

21. 4.39 × 105 J

23. The two ways in which energy transfers take place are through heat
and work.
25. This is an exothermic reaction because heat is released with the
products. Some of the potential energy of the reactants was converted

1106
 to heat energy in the reaction. Therefore the products have less
potential energy than the reactants.
27. a. endothermic
b. exothermic
c. exothermic
d. endothermic
29.
a. H2O (l)→H2O (s)+heat
b. 2 Mg (s)+O2 (g)→2 MgO (s)+heat

c. Ca(OH)2 (s)+heat→CaO (s)+H2O (l)

31. a. exothermic
b. exothermic
c. endothermic
33. If the calcium sulfate is the system, then water is the surroundings.
Exothermic.
35. a. exothermic
b. endothermic
37. a. ΔE=q+w, so 20 kJ+30 kJ=50 kJ

b. ΔE=q+w, so 30 kJ+45 kJ=75 kJ

39. ΔE=q+w, so q=ΔE − w=47.0 kJ − 15.0 kJ=32.0 kJ

lost as heat
41. The potential energy was released as heat, which warmed the
surroundings. It was also lost as work, which caused the ground to
shake. The total energy of the universe has been left unchanged.
43.

System: Ice Bottle Surroundings: Water in

1107
 Cooler
Change +182,000 J –182,000 J
Result Ice melts + temperature Water cools, temperature
increases decreases
Heat Absorbed or Absorbed Released
Released?

45. Heat refers to the total kinetic energy transferred from one substance
or object to another. Temperature is a measure of the average kinetic
energy of the particles in a substance.
47. The specific heat is the amount of heat required to raise the
temperature of one gram of material by one degree Celsius. Heat
capacity is the amount of heat required to raise the temperature of an
object regardless of its mass.
49. Because iron has the greater specific heat, it releases more heat energy.
51. 2.37 J/(g·°C)
53. 7.7 °C
55. 46.5 cal
57. 16 kJ/°C
59. 122 kJ
61. 1260 °C
63. 0.05 °C

65. A coffee cup is made of Styrofoam®. Because Styrofoam is a very

good insulator, nearly all of the heat involved in the change is absorbed
or released by the water. By measuring the temperature change for the
water, we can determine the amount of heat that is gained or lost by the
system. In bomb calorimetry, the test substance is placed in a heavy
steel container. Outside this container is a second container, which is
filled with water and equipped with a thermometer. A pair of ignition
wires detonates the sample. The gases produced cannot expand, so all
of the energy is released as heat into the bomb calorimeter. A more
precise measurement is obtained with a bomb calorimeter than with a
coffee cup.

1108
67. 1,300 calories were absorbed by the water. This same amount of
energy was released by the metal.

69. qwater=3,500 Jqmetal=−3,500 J

71.

a. qwater=1,700 Jqmetal=−1,700 J

b. sm=0.22 calg⋅°C
c. The unknown metal has a specific heat of 0.220 cal/(g·°C).
d. The unknown metal is most likely aluminum.
73. qsolid=−11.9 kJHeat of solution=− 57.5 kJ/mol

75. 64.85 °C.

77. 4.81 kcal in a 1.0-g sample
79. The value of the fuel blend (in kJ/g) = 350.8 kJ/g
81. a. 5.550 × 103 kJ released
b. 488 kJ released
83. 629 g propane
85. 17.51 kJ/g
87. a. –3,809.4 kJ
b. –5,079.2 kJ
c. –634.90 kJ
89. a. –99.62 kJ
b. –67.36 kJ
c. –23.22 kJ

91. –5.370 × 104 kJ

1109
93. 157 g NaOH
95. –154 kJ

97. 6.34 × 104 g Al

99. 19.49 °C

Chapter 9: Covalent Bonding and Molecules

15. The octet rule states that atoms are stabilized by completely filled s and
p sublevels. This corresponds to 8 electrons in the valence shell.
Nonmetals can fulfill the octet rule by sharing valence electrons.
17. There are a total of 6 bonded electrons in this molecule (2 electrons per
bond) and a total of 6 nonbonded electrons.
19. These atoms fulfill the octet rule.
Atom Covalent Bonds Lone Pairs Valence Electrons
H 1 0 2
O 2 2 8
O 2 2 8
H 1 0 2

21. a. complete octet (central atom has 8 electrons around it)

b. incomplete octet (central atom has less than 8 electrons around it)
c. expanded octet (central atom has more than 8 electrons around it)
23. boron
25. a. 8
b. 8
c. 26
d. 14
27. a. 40
b. 8
c. 32
d. 24
29. a.

1110
 b.

c.

31. a.
b.

c.

33. a.
b.

c.

35. a. Carbon: 0
b. Oxygen (double bonded): 0
c. Oxygen (single bonded): –1 each
37. a. –1
b. 0
c. +1
39. a.

b.

c.

41. a.

b.

1111
 c.

43. a. 5 + 4(1) – 1 = 8
b. 6 + 3(6) + 2 = 26
c. 3 + 4(1) + 1 = 8
d. 6+1+1=8
45. a.

b.

c.

47. a.
b.

c.

49. a.

b.

c.

1112
51.

53. Resonance structures are a set of Lewis structures that show how
electrons are distributed around a molecule or ion. Resonance
structures are used when a single Lewis structure cannot sufficiently
show the distribution of the electrons in the molecule or ion. Chemists
use a double-headed arrow to indicate resonance structures.
55.

57. Because there are four oxygen atoms per Lewis structure and the
charge is distributed evenly across all four oxygen atoms, each O atom
contributes ¼ negative charge.
59.

61. a. tetrahedral, 109.5°

b. linear, 180°
c. trigonal planar, 120°
63. a. charge sets = 2; electronic geometry = linear
b. charge sets = 3; electronic geometry = trigonal planar
c. charge sets = 4; electronic geometry = tetrahedral
65. a. charge sets = 3; electronic geometry = trigonal planar
b. charge sets = 2; electronic geometry = linear
c. charge sets = 4; electronic geometry = tetrahedral
67. a. electronic geometry = trigonal planar; molecular geometry = bent
b. electronic geometry = tetrahedral; molecular geometry = bent

1113
 c. electronic geometry = tetrahedral; molecular geometry =
tetrahedral
69. a. electronic geometry = tetrahedral; molecular geometry = trigonal
pyramidal
b. electronic geometry = linear; molecular geometry = linear
c. electronic geometry = tetrahedral; molecular geometry = trigonal
pyramidal
71. a.

b.

c.

73. a.

b.

75. a.

b.

77. a.

1114
 b.

79. Both C=O carbons have a trigonal planar electronic geometry. The
highlighted N atom has a tetrahedral electronic geometry.
81. a. Cl
b. C
c. O
d. Ca
83. Ordered from most to least electronegative:
a. O, Si, Ti, Ca
b. O, Br, I, Cs
c. F, N, In, K
85. a. covalent
b. ionic
c. polar covalent
d. polar covalent
87. a. C—O
b. Ca—Cl
c. W—S
89. a.
b.

c.

91. a.
b.

c.

93.

1115
 a.

b.

c.

95. a.
b.

97. a.

b.

c.

99. a.
b.

c.

a. This bond is not polar and has no net dipole.

b. The C—N bond is polar covalent.
c. All P—F bonds are polar covalent.
101. a.

1116
 b.

c.

a. The C—S bonds are not polar. This molecule has no net dipole.
b. The S—F bonds are polar covalent.
c. The S—Cl bonds and the S=O bond are all polar covalent.
103.

Chapter 10: Solids, Liquids, and Gases

21. In a solid, particles pack closely together. Each atom is held in a fixed
position by the atoms around it. In a liquid, the particles are close
together, but they are not held in a fixed position; particles move freely
past each other. In a gas, the particles move independently of each
other—they are spaced far apart and have little or no interaction with
the other gas particles as they move.
23. In an ionic compound, alternating cations and anions are packed tightly
together in rigid frameworks called lattices. The interactions between
the oppositely charged ions are strong, so it takes a tremendous amount
of energy to disrupt the lattice. Metallic solids usually form ordered
lattices of tightly packed neutral atoms of the same elemental metal or
alloy. The neutral atoms do not bind as tightly to each other as the ions
in the ionic compound.
25. Intramolecular forces are the bonds within a molecule. Covalent bonds
are the intramolecular forces present in a molecular solid.
Intermolecular forces are the weaker forces between different
molecules. Dipole–dipole interactions, hydrogen bonds, and dispersion

1117
 forces are all intermolecular forces in a molecular solid.
27. a. ionic
b. molecular
c. metallic
d. ionic
29. a. ionic
b. covalent
c. covalent
d. covalent
N2 will have the lowest boiling point; it is a covalent
compound with no dipole. LiF will have the highest boiling
point; it is an ionic compound.
31. London dispersion is the strongest type of intermolecular force in this
molecule.
33.

The strongest type of intermolecular interaction in formaldehyde is

dipole–dipole.
35. Polymers are compounds that contain long chains of covalently bonded
atoms. Polymers are composed of small, repeating units that are
bonded together, whereas a molecular solid is composed of molecules
held together by intermolecular forces.
37. In a gas, the particles are spaced far apart. They move freely,
interacting very little with the particles around them. Gas particles
travel in a straight line until they bounce off another particle or off the
walls of the container.
39. Pressure inside a container results from the force the gas particles exert
on the inside of the container when they collide with the walls of the
container. Pressure from outside a container also results from gas
molecules colliding with the outside walls of the container.
41. If the pressure in your mouth is less than the outside pressure, the
outside pressure will force the liquid up the straw and into your mouth.
If the pressure in the mouth is greater than the outside pressure, the

1118
 mouth pressure will force the liquid down through the straw.
43. 765.3 mm Hg; lower than standard atmospheric pressure.
45. 692 torr; lower than standard atmospheric pressure.

47. 1 atmosphere = 1.013 × 103 mbars = 760 torr = 29.9 inches Hg

49. a. 930 mbars
b. 2.4 × 103 torr
c. 20.6 psi
51. 95.0 psi
53. As the volume decreases, pressure increases. If the temperature
decreases, pressure decreases.
55. 0.84 L
57. 1.79 L
59. 1.65 L
61. 137 K
63. Absolute zero is the lowest possible temperature, corresponding to 0 K
or –273.15 °C; at this temperature, all motion stops and the particles in
a substance have zero kinetic energy.
65. 1,050 mb
67. 427 K
69. 0.40 bars
71. 2.36 L
73. 44.8 L

75. 2.2 × 102 moles

77. 22 atm

79. 3.7 × 102 K

81. 23.0 atm
83. 3.00 L

1119
 85. 2.09 atm

87. 1.57 × 1023 molecules CO2

89. a. 0.094 moles

b. 20.2 g/mol
c. This formula mass corresponds to neon gas.
91. the pressure caused by one gas in a mixture
93. The partial pressure of methane = 18 psi
The partial pressure of oxygen = 27 psi
95. 9.34 atm
97. 4 torr
99. Nitrogen gas (N2) would move faster.
101. Diffusion is the spread of particles through random motion; lighter
gases diffuse more quickly than heavier gases. In contrast, effusion is
the process of a gas escaping from a container; lighter gases effuse
more quickly than heavier gases.
103. 2.5 moles H2; 56 L H2
105. a. 60.0 moles H2O
b. 45.0 moles CO2
c. 1.01 × 103 L CO2
d. More moles of gas are produced (7 moles total versus 6 moles).
107. a. 0.086 mole CH4
b. 0.17 mole O2
c. 0.17 mole H2O
d. 3.8 g CO2
109. 0.18 atm
111. 1.10 × 103 L CO2
113. A mixture containing 96% air and 4% propane (answer a) would
completely react with almost no excess of either gas.

1120
Chapter 11: Solutions
19. Potassium nitrate is the solute because it is dissolved into the solvent
(water). The small amount of potassium nitrate dissolved in 2.0 liters
of water, in comparison to what can dissolve in one liter (300 g), tells
us this solution is dilute.
21. 28.6% CaCl2

23. 0.80 L cleaner; 3.2 L water

25. 24 mL alcohol
27. a. 5.20% g/mL
b. 200% g/mL
c. 18.9% g/mL
29. 200 g NaCl
31. a. 5.0 ppm
b. 0.0050 ppm
c. 5.0 × 10−6 ppm
33. 10 mg oxygen
35. 0.03 ppm
37. 4.0 ppm; 0.00040% (g/mL)
39. 0.005 ppm; 5 ppb

41. 5 × 10−6 ppm; 5 × 10−3 ppb

43. a. 0.132 M KI
b. 0.056 M CO2
c. 1.31 M Mg(C2H3O2)2

45. a. 10 moles NaOH

b. 1.2 moles KCl
c. 0.0003 moles Na2CO3

47. 14 L

49. 1.8 × 102 g NaCl

Measure out the desired amount of NaCl and add it to a 3.0-L

1121
 volumetric flask. Partially fill the flask with water and mix until the
NaCl dissolves completely. Dilute the solution, slowly adding water
until the correct volume is reached.
51. 0.38 M HCl
53. Add 2.0 mL of the stock solution to a 100-mL volumetric flask and
dilute with water to the correct volume.
55. a. electrolyte
b. electrolyte
c. not an electrolyte
d. not an electrolyte
57.

59. a. Ba(OH)2 (s)→Ba2+ (aq)+2 OH− (aq)

b. (NH4)2SO4 (s)→2NH4+ (aq)+ SO42− (aq)

61. a. MgBr2 (s)→Mg2+(aq)+2 Br−(aq)

b. AlCl3 (s)→Al3+ (aq)+3 Cl− (aq)

c. Ca(NO3)2 (s)→Ca2+ (aq)+2 NO3− (aq)

1122
63. a. 1 M
b. 4 M
c. 9 M
65. a. 2.0 M
b. 0.7 M
c. 2.6 × 10−4 M

67. 1.84 moles Cl−

69. 2.89 moles Cl−; 3.6 M Cl−

71. 0.50 M Mg2+; 3.1 M K+; 2.1 M SO42−

73. A colligative property is one that depends on the number of particles

dissolved in the solution, but not on the type of particles dissolved.
75. MgCl2 has a total ion concentration of 3.0 moles/L; KBr has a total ion
concentration of 2.0 moles/L. Therefore 1.0 M MgCl2 has the lower
freezing point.
77. NaNO3 has a total ion concentration of 4.0 moles/L; AlCl3 has a total
ion concentration of 6.0 moles/L. Because the AlCl3 solution has the
higher total ion concentration, it has the lower freezing point, the
higher boiling point, and the higher osmotic pressure.
79. A 3.0-M NaCl (aq) solution will provide the greater number of ions
(mole/L), therefore it will have the higher osmotic pressure. Water will
flow from the 1.0-M aq. Na2SO4 solution into the 3.0-M NaCl (aq)
solution in order to dilute it.
81. 0.154 M
a. hypertonic
b. hypertonic
c. hypertonic
d. isotonic
83. The driving force for a precipitation reaction is the formation of the
precipitate.

1123
85. Complete ionic equation: Ca2+ (aq)+2 Br− (aq)+Pb2+ (aq)+2 NO3−
(aq)→PbBr2 (s)+Ca2+ (aq)+2 NO3− (aq)

Net ionic equation:2 Br− (aq)+Pb2+ (aq)→PbBr2 (s)

87. a. metal-displacement reaction

b. neutralization reaction
c. precipitation reaction
89. a. 2 FeCl3 (aq)+3 ZnSO4 (aq)→3 ZnCl2 (aq)+Fe2(SO4)3 (s)

b. HBr (aq)+LiOH (aq)→H2O (l)+LiBr (aq)

91. 2 AgClO3 (aq)+MgCl2 (aq)→2 AgCl (s)+Mg(ClO3)2 (aq)

93. 9.53 moles NO3−

95. 0.400 moles Al

97. 0.33 moles H2

99. 0.011 M NaOH

101. 38.1 g Cu
103. 1.9 × 10−5 moles Br−; 3.8 × 10−5 M Br−
105. 0.00805 M Ca2+
107. 0.10 L KOH

Chapter 12: Acids and Bases

23. a. HBr (aq)→H+ (aq)+Br− (aq)

1124
 b. HBr (aq)+H2O (l)→H3O+ (aq)+Br− (aq)

c. H2SO4 (aq)→H+ (aq)+HSO4− (aq)

d. H2CO3 (aq)+H2O (l)→H3O+ (aq)+HCO3− (aq)

25. a. NaOH (s)→Na+ (aq)+OH− (aq)

b. Ca(OH)2 (s)→Ca2+ (aq)+2 OH− (aq)

27. a. HCl (aq)+H2O (l)→H3O+ (aq)+Cl− (aq)

b. HNO3 (aq)+H2O→H3O+ (aq)+NO3− (aq)

c. HClO4 (aq)+H2O→H3O+ (aq)+ClO4− (aq)

29. a. NaOH (s)→Na+ (aq)+OH− (aq)

b. Mg(OH)2 (s)→Mg2+ (aq)+2 OH− (aq)

c. CsOH (s)→Cs+ (aq)+OH− (aq)

31. a. Diprotic because it has two available H+.

b. Monoprotic because it has one available H+.
c. Monoprotic because it has one available H+.
d. Diprotic because it has two available H+.

33. In the Arrhenius definition, acids produce H+ ions in water, which may

1125
 also be represented by the hydronium ion, H3O+. Bases produce
hydroxide (OH–) ions in water. A Brønsted-Lowry acid is any
compound that gives up an H+ ion (called a proton) in a chemical
reaction. Conversely, a Brønsted-Lowry base is any compound that
accepts an H+ ion.
35.

The acid in solution ionizes to produce an H+ cation and a

corresponding anion. In the aqueous solution, the H+ ion is stabilized
by multiple water molecules. This can be shown as the H+ ion
attached to a water molecule to form H3O+. Both mean the same
thing.
37. a. HF is the acid and KOH is the base.
b. HBr is the acid and H2O is the base.
c. H2SO4 is the acid and H3N is the base.

39. Strong acids completely ionize in water. Most acids only partially
ionize in water—these are weak acids.
41. a. HCl
b. HBr
c. H2SO4
d. H3PO4 (weak acid)

43. a. weak acid

b. weak acid
c. weak acid
d. strong acid
45. The ionization of weak acids is an example of an equilibrium reaction.
This type of reaction occurs in both the forward and backward

1126
 directions. The acid can dissociate into its ions, but the dissociated ions
can also re-form the acid.
47. a.

b.

49. a. acid = HClO2; base = H2O; conjugate acid = H3O+; conjugate base
= ClO2–
b. acid = H2CO3; base = H2O; conjugate acid = H3O+; conjugate
base = HCO3–
c. acid = CH3NH3+; base = H2O; conjugate acid = H3O+; conjugate
base = CH3NH2

51. a. acid = HCl; base = CH3OH; conjugate acid = CH3OH2+; conjugate

base = Cl–
b. acid = CH3OH; base = PBr3; conjugate acid = HPBr3+; conjugate
base = CH3O–
c. acid = HF; base = H2S; conjugate acid = H3S+; conjugate base =
F–
53. a. HF (aq)+H2O (l)⇌H3O+ (aq)+F− (aq)

b. F− (aq)+H2O (l)⇌HF (aq)+OH− (aq)

55. a. H2CO3 (aq)+H2O (l)⇌H3O+ (aq)+HCO3− (aq)

b. HNO2(aq)+H2O (l)⇌H3O+ (aq)+NO2− (aq)

c. NO2− (aq)+H2O (l)⇌HNO2 (aq)+OH− (aq)

1127
57. a. H3PO4 (aq)+H2O (l)⇌H3O+ (aq)+H2PO4− (aq)

b. H2PO4− (aq)+H2O (l)⇌H3O+ (aq)+HPO42− (aq)

c. HPO42– (aq)+H2O (l)⇌H3O+ (aq)+PO43− (aq)

59. a. HClO (aq)+H2O (l)⇌H3O+ (aq)+ClO− (aq)

b. ClO− (aq)+H2O (l)⇌HClO (aq)+OH− (aq)

61. Conjugate acid: CH6N+; conjugate base: Cl–

63.

65.

67. a. nonmetal oxide

b. metal oxide
c. not an oxide (nonmetal or otherwise)
d. nonmetal oxide
69. a. Sn (s)+2 HCl (aq)→H2 (g)+SnCl2 (aq)

b. 2 Li (s)+2 HCl (aq)→H2 (g)+2 LiCl (aq)

c. Fe (s)+2 HCl (aq)→H2 (g)+FeCl2 (aq)

1128
 d. 2 Al (s)+6 HCl (aq)→3 H2 (g)+2 AlCl3 (aq)

71. a. Be (s)+2 HF (aq)→H2 (g)+BeF2(aq)

b. Sr (s)+H2SO4 (aq)→H2(g)+SrSO4 (aq)

c. Au (s)+HNO3 (aq)→Precious metals (silver, gold, platinum) do

not react with acid.

73. a. HCl+NaOH→H2O (l)+NaCl (aq)

b. 2 HBr+Mg(OH)2→2 H2O (l)+MgBr2 (aq)

c. 3 HNO3+Al(OH)3→3 H2O (l)+Al(NO3)3 (aq)

75. a. 2 HCl (aq)+Ca(OH)2 (s)→2 H2O (l)+CaCl2 (aq)

b. Mg(OH)2 (aq)+H2SO4 (aq)→2 H2O (l)+MgSO4 (aq)

c. Al(OH)3 (aq)+3 HF (aq)→3 H2O (l)+AlF3 (aq)

77. Complete ionic equation:2 H+ (aq)+2 NO3−(aq)+Mg2+ (aq)+2 OH−

(aq)→2 H2O (l)+Mg2+ (aq)+2 NO3− (aq)Net ionic equation:H+
(aq)+OH− (aq)→H2O (l)

79. H3C6H5O7 (aq)+3 NaOH (aq)→Na3C6H5O7 (aq)+3 H2O (l)

1129
81. SO3 (g)+H2O (l)⇌H2SO4 (aq)

83. H2O (l)+acidH2O (l)base⇌H3O+ (aq)+conjugate acidOH−(l)conjugate

base

85. a. 1.0 × 10–2 M

b. 1.91 × 10–6 M
c. 2.34 × 10–10 M
87. a. 3.00; acidic
b. 7.00; neutral
c. 10.28; basic
89. a. 7.00 – neutral
b. 5.00 – acidic
c. 12.100 – basic
91.

[H+] [OH–] pH pOH

Solution A 1.0 × 10–5 1.0 × 10–9 5 9

Solution B 1.0 × 10–11 1.0 × 10–3 11 3

Solution C 1.0 × 10–9 1.0 × 10–5 9 5

Solution D 1.0 × 10–8 1.0 × 10–6 8 6

93. a. neutral
b. basic
c. acidic
d. acidic
e. basic
95. a. 6 × 10–5 M
b. 5 × 10–12 M

1130
 c. 5 × 10–8 M
d. 5 × 10–11 M

97. It should be near a pH of 8 because the –log of 10–8 = 8. pH = 7.88

99. The 0.010-M solution of HCl will have the lower pH because it is a
strong acid and will fully ionize in solution.
101. A pH meter gives the most precise pH reading, whereas litmus paper
gives the least precise pH information.
103. a. red
b. blue
c. colorless
d. pink
105. To determine the concentration of a solution of sodium hydroxide, you
would follow this procedure:

1. Prepare an acidic solution (such as hydrochloric acid) with a known

concentration.
2. Precisely measure a small volume of the base solution. Place the base
solution in an Erlenmeyer flask and add 1−2 drops of phenolphthalein
indicator. (Remember, this indicator is pink in basic solution, but
colorless in acid.)
3. Fill a buret with the known acidic solution. Measure the initial
volume in the buret.
4. Begin adding acid to the base, swirling the Erlenmeyer flask
occasionally to mix the solutions. As more acid is added, the pink
color will begin to disappear, then disappear fully. Slowly add acid,
drop by drop, until a faint pink color completely disappears.
By adding the solution slowly until the faintest pink color disappears,
you add just enough acid to neutralize the base.
5. In this reaction, one mole of base reacts with one mole of acid.
Therefore, at the end point the number of moles of acid and base is
equal. The equation MaVa = MbVb can be used to determine the
molarity of the base because the volume of the base and acid
solutions are known, and so is the concentration of the acid solution.

107. 1.70 M HNO3

109. 0.63 M NaOH

1131
111. 2 NaOH (aq)+H2SO4 (aq)→2 H2O (l)+NaSO4 (aq) 0.130 M H2SO4

113. 1.1 × 102 mL NaOH

115. a. red
b. pink
c. colorless (pH = 2.4)
d. red
117. The solution is basic after the reaction is complete. This will cause the
phenolphthalein to be pink.
119. A buffer is a solution containing a mixture of acidic and basic
components that resists changes in pH.
121. Animal cells contain a buffer consisting of H2PO4– (a weak acid) and
HPO42– (its conjugate base).
123. a. If acid were added to the buffer solution, it would react with the
C2H3O2–:
H+(aq)+C2H3O2− (aq)→HC2H3O2 (aq)

b. If base were added to this buffer, it would react with HC2H3O2:

OH− (aq)+HC2H3O2 (aq)→H2O (l)+C2H3O2− (aq)

125. The 0.0530 moles of NaOH will consume 0.0530 mole of HCl, leaving
0.068 mole HCl unreacted. The final [H+] will be 0.45 M. Finally, the
pH 0.35.

Chapter 13: Reaction Rates and Equilibrium

13. Because the collision of molecules is required for chemical reactions to
occur, the rate of reactions depends in part on how frequently
collisions occur. When molecules collide more frequently, reactions
occur at a faster rate.
15. A decrease in the temperature or a decrease in the concentration of the
reactants makes a reaction go more slowly.
17. She could heat the reaction mixture, or she could react the iron with a

1132
 more concentrated HBr solution.
19. A reaction energy diagram shows energy changes as a reaction
progresses. It provides information as it pertains to the energetic
changes that accompany a chemical reaction.
21. The transition state is the highest-energy arrangement of atoms that
occurs during a chemical reaction. It is the point of highest energy on a
reaction energy diagram.
23.

24.

Endothermic. This reaction should occur at room temperature.

27. a. endothermic
activation energy: +30 kJ/mol; net energy change: +20 kJ/mol
b. exothermic
activation energy: +10 kJ/mol; net energy change: –30 kJ/mol
29. a. activation energy:

1133
 +40 kJ/mol; net energy change: +20 kJ/mol
b. activation energy:
+20 kJ/mol; net energy change: –50 kJ/mol
31. Reaction A will proceed most quickly because it has the lowest
activation energy of the three reactions.
33. Add enough energy in the form of heat to overcome the activation
energy. After that, the heat released by the reaction enables it to
continue.
35. Because the reaction is at equilibrium, both the forward and reverse
reactions are occurring simultaneously, meaning that A is being
converted to B and B is being converted to A. Because these reactions
are happening at the same rate, the concentrations of A and B do not
change.
37. Because the ice and water are in equilibrium, there are liquid water
molecules freezing as well as molecules in the ice melting away from
the solid. Because this process is at equilibrium, the rates of the liquid
molecules freezing and molecules in the ice melting are equal.
39. The populations of the two towns are not at equilibrium, because the
number of people moving between the two towns is not constant.
There are more people moving from Town A to Town B than from
Town B to Town A. Eventually, all of the people will have moved to
Town B. For the populations to be at equilibrium, the number of
people moving between the two towns each day would need to be the
same.
41. This reaction does not represent an equilibrium because it uses a single
forward arrow, and not a double arrow.
43. Because the reaction is endothermic, the energy of the reactants is less
than the energy of the products. In an equilibrium reaction, the side
that is lower in energy will have a greater concentration. Therefore, the
reactants are present in a higher concentration.
45. In general, the side of an equilibrium that is lower in energy will have
a greater concentration. For this reaction involving trifluoroacetic acid
in water, equilibrium will favor neither the reactants nor the products
because both have an almost identical potential energy. Therefore, the
concentrations of the reactants and products will be nearly equal at
equilibrium.

1134
47. Activation energy (forward direction): 60 kJ/mol
Activation energy (reverse direction): 30 kJ/mol
49. A catalyst speeds up the rate of the forward and reverse reactions by
reducing the activation energy, but it does not affect the equilibrium
position of the reaction.
51.
a. K=[C][A][B]2

b. K=[C][D]2[A]3[B]
53.
a. K=[H+][NO2−][HNO2]

b. K=[[Cr(OH)4]−][Cr3+][OH−]4
55. a. The reaction favors the reactants because the K value is less than 1.
b. The reaction favors the products because the K value is greater
than 1.
c. The reaction favors the products because the K value is greater
than 1.

57. K = 1.8 × 10–4 (equilibrium lies to the left)

59.
a. K=[H3O+][HO2−][H2O2]

b. K=[H+][B(OH)4−][B(OH)3]
61. K=[H3O+][N3−][HN3]=[0.0043][0.0043][0.994]=1.9×10−5

63. [HF]=1.5×10−7 M

1135
65. [H3O+]=6.4×10−4 M
67.
a. K=[Pb2+][I−]2

b. K=1[Fe3+][OH−]3
69. The solubility product is the equilibrium constant for a solution of an
ionic compound. A small solubility product indicates the ionic
compound is not very soluble in the solution.
71. KSP=[Cd2+][CO32−]=1.0×10−12

73. [Ag+]=1.33×10−5 M
75. [Pb2+]=1.4×10−3 M and [I−]=2.7×10−3 M.

77.

a. K=PNO22PN2O4

b. K=PCl22PH2O2PHCl4PO2
79. 16 atm
81. Le Chatelier’s principle states that when a change in concentration,
temperature, pressure, or other key factor takes place, the equilibrium
shifts to minimize that change, and a new equilibrium is established.
83. a. Adding more A will shift the reaction equilibrium to the right.
b. Removing B will shift the reaction equilibrium to the left.
c. Adding C will shift the reaction equilibrium to the left.
d. Removing C will shift the reaction equilibrium to the right.
85. a. Adding Br2 will shift the reaction equilibrium to the right.
b. Removing Cl2 will shift the reaction equilibrium to the left.

1136
 c. Removing BrCl will shift the reaction equilibrium to the right.
d. Adding a catalyst will speed up the reaction rate, but it will not
affect the equilibrium.
87. More CH3I will be produced if methanol is the solvent because it will
be present in a large amount and drive the reaction to the right. If water
is the solvent, it will drive the reaction to the left.

89. Adding OH– shifts the equilibrium to the left to accommodate the
increased amount of OH–.
91. This is an endothermic reaction because heat is on the reactant side.
Increasing the heat in an endothermic reaction will shift the
equilibrium to the right. Removing heat will shift the equilibrium to the
left.
93. This is an exothermic reaction because of the negative enthalpy value.
Therefore, heating the mixture will shift the equilibrium to the left.
95. The reactants contain two moles of gas, and the products contain one
mole of gas. Increasing the pressure will cause the equilibrium to shift
toward the side with fewer moles of gas. Therefore the reaction will
shift to the right.
97. a. Equilibrium shifts to the right.
b. Equilibrium shifts to the left.
c. Equilibrium shifts to the right.
99. K=[H+][ClO−][HClO][H+]=[ClO−]=1.7×10−4 MpH=3.76

Chapter 14: Oxidation-Reduction Reactions

13. a. oxidized
b. reduced
c. reduced
d. oxidized

1137
15. Scientists use oxidation numbers to keep track of electron changes in
redox reactions.
17. a. +2
b. +1
c. +4
d. +1
19. a. Cu = +1, Cl = –1
a. Cu = +2, Cl = –1
b. Si = +4, O = –2
c. H = +1, O = –2
21. a. Be = +2, Br = –1
b. B = +3, Br = –1
c. N = –3, H = +1
d. Li = +1, H = –1
23. a. +2
b. –1
c. O = –2, Br = +1
d. H = +1, N = –3
25. a. +3
b. O = –2, C = +4
c. O = –2, I = +5
d. F = –1, B = +3
27. a. oxidized = Mg; reduced = O
b. oxidized = Fe; reduced = S
c. oxidized = Ca; reduced = Br
29. a. oxidized = H2; reduced = O
b. oxidized = C; reduced = O
c. oxidized = H2; reduced = S

31. a. oxidizing agent = O2

b. oxidizing agent = O2
c. oxidizing agent = F2

1138
33. a. reducing agent = Ca
b. reducing agent = Be
c. reducing agent = S
35. Ionic compounds are formed from the reaction of metals and
nonmetals.
37. a. Fe (s)+Cl2 (g)→FeCl2 (s)oxidized = Fe; reduced = Cl2

b. O2 (g)+2 Hg (l)→2 HgO (s)oxidized = Hg; reduced = O2

39. a. Be (s)+Cl2 (g)→BeCl2 (s)

b. 2 K (s)+Cl2 (g)→2 KCl (s)

c. Co (s)+Cl2 (g)→CoCl2 (s)

d. 2 Cu (s)+O2 (g)→2 CuO (s)

41. oxygen
43. a. 2 Mg (s)+O2 (g)→2 MgO (s)

b. C2H4 (g)+3 O2 (g)→2 CO2 (g)+2 H2O (g)

c. 2 C4H10 (g)+13 O2 (g)→8 CO2 (g)+10 H2O (g)

45. a. Co (s)+SnCl2 (aq)→CoCl2 (aq)+Sn (s)

1139
 b. AuNO3 (aq)+Ag (s)→Au (s)+AgNO3 (aq)

47. The color fades due to an oxidation-reduction reaction between

nickel(II) chloride and solid aluminum:
3 NiCl2 (aq)+2 Al (s)→2 AlCl3 (aq)+3 Ni (s).

As the reaction proceeds, the green nickel(II) chloride is replaced by

AlCl3 and the aluminum gets a coating of solid nickel on it.

49. a. Complete ionic equation:

Mg (s)+Cu2+ (aq)+2 NO3− (aq)→Mg2+ (aq) 2 NO3− (aq)+Cu
(s)Net ionic equationMg (s)+Cu2+ (aq)→Mg2+ (aq)+Cu (s)

b. Complete ionic equation:

Al (s)+Cr3+ (aq)+3Cl− (aq)→Al3+ (aq)+3Cl− (aq)+Cr (s) Net
ionic equationAl (s)+Cr3+→Al3+ (aq)+Cr (s)

c. Complete ionic equation:

2 Cr (s)+3 Pt2+ (aq)+6 Br− (aq)→2 Cr3+ (aq)+6 Br− (aq)+3 Pt
(s)Net ionic equation2 Cr (s)+3 Pt2+ (aq)→2 Cr3+ (aq)+3 Pt (s)

51. a. Mg (s)+Pb(NO3)2 (aq)→Mg(NO3)2+Pb (s)

b. 2 Al (s)+3 PtBr2 (aq)→3 Pt (s)+2 AlBr3 (aq)

53. Aluminum and calcium can reduce Cu2+ to elemental copper.

55. Aluminum can reduce Fe2+ and Pb2+.

1140
57. a. This reaction will occur spontaneously.
b. This reaction will occur spontaneously.
c. This reaction will not occur.
d. This reaction will not occur.
59. a. Sn2+ (aq)+Ni (s)→Ni2+ (aq)+Sn (s)

b. No reaction occurs.
c. Co (s)+Pb(C2H3O2)2 (aq)→Pb (s)+Co(C2H3O2)2 (aq)

61. a. oxidized by reaction with water

b. oxidized by reaction with water
c. oxidized by reaction with water
d. not oxidized by reaction with water
63. a. 2 Na (s)+2 H2O (l)→2 NaOH (aq)+H2 (g)

b. 2 K (s)+2 H2O (l)→2 KOH (aq)+H2 (g)

c. 2 Ca (s)+2 H2O (l)→Ca(OH)2 (aq)+H2 (g)

65. Lithium, aluminum, and nickel are oxidized by reaction with acid.
67. a. Mg (s)+2 HCl (aq)→MgCl2 (aq)+H2 (g)

b. Zn (s)+2 HCl (aq)→ZnCl2 (aq)+H2 (g)

c. Fe (s)+2 HCl (aq)→FeCl2 (aq)+H2 (g)

69. a. Ca (s)+Fe2+ (aq)→Fe (s)+Ca2+ (aq)

b. Ca (s)+2H+ (aq)→Ca2+ (aq)+H2 (g)

1141
 c. No reaction will occur.
d. Zn (s)+Fe2+ (aq)→Fe (s)+Zn2+ (aq)

71. a. Oxidation:Mg (s)→Mg2+ (aq)+2e− Reduction:Co2+ (aq)+2e

−→Co (s)

b. OxidationCu (s)→Cu2+ (aq)+2e− Reduction:2 Au+ (aq)+2e−→2

Au (s)

73. Oxidation:Fe (s)→Fe2+ (aq)+2e− Reduction:Co2+ (aq)+2e−→Co (s)

75. a. Oxidation:Ni (s)→Ni2+ (aq)+2e− Reduction:2 Ag+ (aq)+2e−→2

Ag (s)

b. Oxidation:Zn (s)→Zn2+ (aq)+2e− Reduction:Co2+ (aq)+2e−→Co

(s)
77. The electrode where oxidation occurs is the anode. The electrode
where reduction occurs is the cathode. The current in an
electrochemical cell arises from a redox reaction at the two electrodes.
The salt bridge is essential to the function of the cell. A salt bridge
contains an ionic compound like sodium nitrate. As the electrons move

1142
 from the anode to the cathode, the anode cell builds up a positive
charge while the cathode builds up a negative charge. The salt bridge
provides spectator ions that can balance these charges.
79. AnodeZn (s)→Zn2+ (aq)+2e− CathodeFe2+ (aq)+2e−→Fe (s)

81. For this reaction to occur, two half-reactions must take place:
Anode:Zn (s)→Zn2+ (aq)+2e− Cathode:Cu2+ (aq)+2e−→Cu (s)

If the Zn2+ solution were replaced with pure water, the reaction would
still take place.
However, if the Cu2+ solution were replaced with pure water, the
reaction would not take place, since Cu2+ is a reactant.
83. Al (s)+3 Ag+ (aq)→Al3+ (aq)+3 Ag (s)

85. 2 Li (s)+2 H+ (aq)→2 Li+ (aq)+H2 (g)

87. 3 Ca (s)+2 Cr3+ (aq)→2 Cr (s)+3 Ca2+ (aq)

89. 3 Ba (s)+2 Fe3+ (aq)→2 Fe (s)+3 Ba2+ (aq)

91. Oxidation:Zn (s)→Zn2+ (aq)+2e−Reduction:2 H+ (aq)+2 e −→ H 2

(g)Zn (s)+2 H+ (aq)→H2 (g)+Zn2+ (aq)

1143
93. Oxidation:Cr (s)→Cr3+ (aq)+3e−Reduction:Pb2+ (aq)+2e−→Pb (s)2
Cr (s)+3 Pb2+ (aq)+6 NO3− (aq)→3 Pb (s)+2 Cr3+ (aq)+6 NO3− (aq)

95. Electroplating is a technique that uses electrical potential energy (such

as from a battery) to produce a thin layer of a metal such as gold,
silver, copper, or chromium on the outside surface of another material.
In an electrochemical cell, a redox reaction creates an electric current.
In electroplating, the reverse occurs: an electric current drives a redox
reaction to take place.

97. The gold in solution (Au+) is reduced to Au metal on the ring.

Therefore, the ring is the cathode. The zinc counter electrode is the
anode.
99. A fuel cell is a device that converts the energy of a combustion
reaction directly into electrical energy by separating the oxidation and
reduction half-reactions. The most common fuel cells involve the
reaction of hydrogen with oxygen. A hydrogen fuel cell consists of two
electrodes separated by a proton exchange membrane that allows
protons, but not electrons, to flow through it.
101. a. synthesis
b. single displacement
c. combustion
103. Reactants:
H (in C6H12O6) oxidation number = +1
O (in C6H12O6) oxidation number = –2
C (in C6H12O6) oxidation number = 0
O (in O2) oxidation number = 0
Products:
O (in CO2) oxidation number = −2
C (in CO2) oxidation number = +4
H (in H2O) oxidation number = +1
O (in H2O) oxidation number = –2
In this reaction, carbon is oxidized from an oxidation number of 0 to
an oxidation number of +4. Also, oxygen (in O2) is reduced from an

1144
 oxidation state of 0 to an oxidation state of –2.
105. 54.0 g Fe(NO3)2

Chapter 15: Organic Chemistry and Biomolecules

11. Cellular respiration produces carbon dioxide and water.
13. Photosynthesis is an endothermic process, and cellular respiration is an
exothermic process.
15. Carbon dioxide is absorbed by plants to make glucose and oxygen. The
plants eventually die and begin to decay. After a long period of decay,
the carbon-containing compounds of the dead plants are converted into
coal.
17. The ocean exchanges carbon dioxide with the atmosphere. In water,
carbon dioxide reacts to form carbonic acid and then bicarbonate ions.
Many creatures produce hard calcium carbonate shells from the
bicarbonate ions dissolved in the water. Over time, these shells
accumulate on the seafloor, forming sediments that are rich in carbon.
19. a. 4
b. 3
c. 2
d. 1
21. a.

b.

c.

23. a.

b.

1145
 c.

d.
25. The compound in (b) is an isomer. The F atom is attached to a different
carbon.
27.

29.

31. a.

b.

c.

33. a. CH3CH2CH3
b. CH3CCl2CH2CH3

35. a. C5H12
b. C5H12
c. C5H11Cl

37. a. C4H10O

1146
 b. C3H9NO
c. C4H4O

39. a.
b.
c.

41. a.

b.

c.

43. a.

b.

c.

45. C11H15O4SN

47. 18
49. 8

1147
51.

53. a. butane
b. hexane
c. heptane
55.

57.

59. a.
b.
61. a. alkane
b. aromatic
c. alkene
d. alkyne
63. a. alcohol
b. ether
c. ketone
d. ester
65. a.

b.

c.

d.

67. alcohol, alkene, and ester

1148
69. a. amine
b. amide
c. nitrile
71. alcohol, aromatic, amine
73. CH3CH2CH2OCH3

75.

77. a. CH3CO2H+H2O⇌CH3CO2−+H3O+

b. CH3CH2CO2H+H2O⇌CH3CH2CO2−+H3O+

79. a. HCO2H+NH3⇌NH4++HCO2−

b. CH3CO2H+CH3NH2⇌CH3CO2−+CH3NH3+

81. Polymers are molecules containing simple repeating units linked

together in long chains of covalent bonds. Naturally occurring
polymers include proteins and DNA. Two examples of synthetic
polymers are polypropylene and polyester.
83. carbon, hydrogen, and oxygen
85. The two cyclic forms differ only in the position of the alcohol on
carbon 1: In the alpha form (called α-glucopyranose), the OH on
carbon 1 points down from the ring. In the beta form (β-
glucopyranose), the OH on carbon 1 points out from the ring.
87.

1149
89. Some proteins function as enzymes that catalyze very specific
reactions. Other proteins transport molecules. Proteins are essential to
the replication of our genetic code. And every move we make—every
muscle contraction—arises from a change in the shape of a protein.
91. Phenylalanine, tryptophan, and tyrosine have the benzene ring in their
side chain.
93. Ser-Pro-Arg
95. a. Met-Gly-Ile
b. Cys-Phe-Ser
97. a.

b.

99. a.

1150
 b.

101. Hydrogen bonding interactions affect the shape of proteins.

103. DNA stores all of our genetic information needed for replication and
reproduction.
105. The nucleotides link together through condensation reactions to form
a sugar–phosphate backbone.
107.

109. Polysaccharides contain one repeating monomer. Proteins and DNA

contain multiple monomers with coded information in the monomer
sequence.

Chapter 16: Nuclear Chemistry

11. The number of protons determines the elemental identity of an atom.
13. Collectively, protons and neutrons are called nucleons. A nuclide is an
atom or nucleus containing a particular number of protons and

1151
 neutrons.
15. a. P
b. Rb
c. C
d. F
17. a. protons = 20 and neutrons = 20
b. protons = 20 and neutrons = 21
c. protons = 11 and neutrons = 12
d. protons = 24 and neutrons = 28
19.
a. P1532

b. B48e

c. F919

d. A1328l
21.
a. H24e

b. C614

c. U92235
23. a. This is not an isotope of carbon-12.
b. This is not an isotope of carbon-12. (has the same number of
protons = 6).
c. This is not an isotope of carbon-12.
d. This is an isotope of carbon-12 (has the same number of protons =
6).

25. n01+S62152m→S62153m
27.
a. n01+M4298o→M4299o

b. R86222n→H24e+P84218o

1152
 c. U92235+n01→C55137s +R3797b + 2n01

29.
a. R88226a→H24e + R86222n
b. P11+ C48111d→I49111n+n01

c. U92235+n01→S62156m+Z3077n+3n01

31. P46102d+n01→P46103d

33. H24e+S51121b→I53123+2n01
35. U92235+n01→S62155m+Z3078n+3n01

37. In an alpha (α) decay, a nucleus ejects an alpha particle composed of

two protons and two neutrons. An alpha particle is a helium nucleus. In
a beta (β) decay, a neutron decays into two particles: a proton and an
electron. The proton remains in the nucleus, while the electron is
ejected. In a gamma (γ) decay, a nucleus releases energy in the form of
electromagnetic waves called gamma rays. Gamma decays do not
involve the release of particles, so they do not change the atomic
number or the mass number.
39. a. beta (β) decay
b. gamma (γ) decay
c. alpha (α) decay
41.
a. 88226Ra →α 24He+R86222n

b. 90228Th →α 24He+R88224a

c. P84210o →α H24e+P82206b

1153
43.

a. L38i →β B48e+e−10

b. N1125a →β M1225g+e−10

c. F87223r →β R88223a+e−10
45.
a. T90231h →α H24e+R88227a

b. A89228c →β T90228h+e−10
c. T4399mc →γ T4399c +energy

47. Astatine-219 first undergoes alpha decay to produce

B83215i:A85219t→αH24e+B83215iNext, 83215Bi undergoes beta
decay to produce 84215Po:B83215i→βP84215o+e−10Finally, the
second beta decay produces 84215At:P84215o →β A85215t+e−10

49. The first three alpha decays of thorium-227 produce

R88223a,R86219n, and finally
P84215o:T90227h→αH24e + R88223aR88223a→αH24e + R86219nR86219n→αH
−10The final alpha decay yields B83211i:A85215t→αH24e+B83211iThe
final beta decay produces P84211oB83211i→βP84211o+e−10

1154
51. Because alpha particles are much heavier than gamma particles, they
can be very damaging to our body tissue. However, these particles are
easily blocked by clothing or even by a sheet of paper. Gamma
radiation is the hardest to block. Heavy lead shields are often used to
block gamma radiation. Even so, gamma rays are less damaging than
alpha particles.
53. Geiger counter, scintillation counter, semiconductor counter, dosimeter
55. A semiconductor counter is able to measure the different energies of
radiation absorbed, and therefore it can identify which nuclides are
present. A Geiger counter cannot do this.
57. a. 1.275 rem
b. 0.07 rem
c. 0.041 rem
59. 20,000 x-rays
61. 6 mSv/yr
63. In radiation imaging, a patient is administered a dose of a compound
containing a radioactive nuclide. As this material travels through the

1155
 body, it emits gamma radiation. This radiation is detected using a
scintillation camera, an instrument that produces an image based on the
intensity of gamma radiation. Indium-111 has been used in radiation
imaging to detect internal infections. The patient’s white blood cells
are tagged with indium-111. The white blood cells congregate around
the infected sites. The gamma rays that are emitted help pinpoint the
location of the infection. Radiation therapy is a common treatment for
cancer. This technique uses the destructive properties of radiation to
destroy cancer cells. In one variation of this technique, doctors implant
“seeds” containing radioactive nuclei (such as Cs-131) near the tumor
site.
65. When high-energy radiation from the sun enters the upper atmosphere,
it produces several nuclear reactions. In one such reaction, a neutron
collides with nitrogen-14 to produce carbon-14 and an additional
proton:

n01+N714→C614+P11
Carbon-14 is radioactive and has a half-life of about 5,730 years. As
plants grow, they take in carbon dioxide from the atmosphere,
including carbon-14. After a plant dies, the carbon-14 begins to decay.
Based on the amount of carbon-14 left in a sample, scientists are able
to determine the age of substances that were derived from plants.
67. After one half-life, the rate of decay will be one-half the original rate
of decay (120 Bq). After two half-lives, the rate of decay will be one-
quarter of the original rate of decay (60 Bq).
69. When high-energy radiation strikes quartz, it produces small amounts
of the radioactive nuclides beryllium-10 and aluminum-26. If quartz is
exposed on Earth’s surface, it is bombarded by solar radiation, and
these nuclides gradually accumulate. If the quartz is buried, solar
radiation is blocked and the isotopes begin to decompose. Geologists
use the amounts of Be-10 and Al-26 present in a quartz sample to
determine how long a rock surface has been exposed, or how long it
has been buried.
71. The mass defect is the difference between the masses of the individual
particles and the mass of the complete nucleus. The difference lies in a
small fraction of the mass that is converted into energy.
73. When the individual particles combine to form a nucleus, a small

1156
 fraction of the mass (the mass defect) is converted into energy. This
energy is released when the protons and neutrons form a new nucleus.
75. Lighter nuclei will combine to form heavier, more stable nuclei, and
heavier nuclei will break apart to form lighter, more stable nuclei.
77. In this reaction, energy (reactants) is converted into matter (products).
Therefore, the mass of the products is greater.
79. a. n01+U92235→S3895r+X54139e+2n01

b. n01+U92235→S3895r+X54138e+3n01

c. n01+U92235→R3793b+C55141s+2n01

81. When a high-energy neutron strikes a U-235 nucleus, the uranium

shatters into smaller fragments. The fission of a uranium nucleus
produces several more high-energy neutrons. If there are enough other
235U atoms nearby, these neutrons produce a chain reaction, releasing
tremendous amounts of energy. Because each nuclear fission produces
multiple neutrons, the number of atoms involved in a fission chain
reaction can multiply rapidly.
83. The use of highly purified uranium-235 in a nuclear reactor would
cause a runaway reaction. The excess uranium-238 serves two
purposes: First, it absorbs some of the neutrons released in the fission
of uranium-235, preventing a runaway reaction. Second, after
absorbing a neutron, U-238 undergoes two beta decays to form
plutonium, which also undergoes nuclear fission.

85. U3O8→3 U+4 O2U+3 F2→UF6

87. U92239→βN93239p+e−10N93239p→βP94239o+e−10

1157
89. Fission takes place in the reactor core, which generates a tremendous
amount of heat. The reactor core is immersed in water; pumps circulate
this water through the reactor loop to carry heat energy away from the
core. The reactor loop comes in contact with water in the power loop.
The reactor loop heats the water in the power loop, converting it to
steam. The steam turns turbines, producing electric current.
91. The reactor loop, which contains the water contaminated with the
radioactive by-products, is completely sealed off from the outside
environment and housed in a reinforced containment structure.
93. First, there must be high pressure, so the small nuclides are
compressed into a tiny space. Second, there must be very high
temperature—more than 10 million degrees Celsius. Third, there must
be a long contact time—that is, the pressure and temperature must be
sustained long enough for fusion to take place.

95. n01+L36i→H24e+H13
97. Plasma is a gas-like state that is produced at very high temperatures,
and it is often considered the fourth state of matter. In the plasma state,
electrons have such high energies that they do not remain bound to any
one atom. The result is a gaseous “soup” of positive and negative
charges.
99. The break-even point is the level at which the energy output from a
tokamak reactor exceeds the energy required to power the plasma
containment vessel.

1158
Glossary
A
absolute zero The lowest possible temperature, corresponding to 0 K or –
273.15 °C; at this temperature, the particles in a substance have zero kinetic energy.
accuracy A measure of how reliable measurements are—that is, how closely
they reflect the true value.
acid A compound that produces H+ or H3O+ ions in water (Arrhenius definition);
a compound that donates H+ ions (Brønsted-Lowry definition).
acid rain The effect observed when nonmetal oxides combine with moisture in
the atmosphere to produce acidic rainfall.
activation energy The energy barrier for a reaction; this energy determines
how quickly a reaction occurs.
activity series A table that lists metals by their reducing powers; used to predict
whether metal displacement reactions will occur.
actual yield The amount that a chemist actually recovers from an experiment.
alcohol A functional group consisting of an oxygen that is singly bonded to a
carbon and a hydrogen, giving the bonding sequence C–O–H.
aldehyde A functional group consisting of a carbonyl connected to a hydrogen atom.
alkali metals Metal elements in column 1 (or 1A) of the periodic table; these
metals are very reactive.
alkaline earth metals Metal elements in column 2 (or 2A) of the periodic
table; these metals are very reactive.
alkane A hydrocarbon composed entirely of single bonds.
alkene A functional group consisting of a carbon–carbon double bond; this term
also refers to a simple molecule containing this functional group.
alkyne A functional group consisting of a carbon–carbon triple bond; this term
also refers to a simple molecule containing this functional group.
alpha (α) decay A radioactive decay in which the nucleus ejects an alpha
particle consisting of two protons and two neutrons.
amide A functional group consisting of a carbonyl group bonded to a nitrogen atom.
amine A functional group consisting of a nitrogen atom with three single bonds,
usually to hydrogen or carbon atoms.
amino acid Small molecules having both amine and carboxylic acid functional
groups; plant and animal cells use 20 fundamental amino acids to create proteins.
anion A negatively charged ion.
aromatic ring A ring structure that contains alternating single and double bonds
and is generally less reactive than simple alkenes.

1159
Arrhenius definition Describes an acid as a compound that produces H+ or
H3O+ ions in water, and a base as a compound that produces OH− ions in water.
atmosphere (atm) A unit of gas pressure; 1 atm = 760 mm Hg.
atomic mass unit (u or amu) A unit of mass is equal to 1.66 × 10−27 kg.
atomic number The number of protons in an atom; also the number of
electrons in a neutral atom.
atomic theory A theory describing matter in terms of fundamental units called atoms.
atoms The fundamental units of matter.
Avogadro’s law If pressure and temperature are constant, the volume of a gas
is proportional to the number of moles of gas present.
Avogadro’s number The number of particles in a mole; 6.02 × 1023.

B
balanced equation A chemical equation in which the number and type of atom
is the same for the reactants and the products.
barometer A device used to measure atmospheric pressure.
base A compound that produces OH− ions in water (Arrhenius definition);
compounds that accept H+ ions (Brønsted-Lowry definition).
battery A device that produces electric current through a redox reaction in
which the two half-reactions occur at separate sites; also called an electrochemical cell
becquerel (Bq) The number of radioactive decays that occur per second for a
particular substance.
benzene A very stable compound having the formula C6H6, in which the six
carbon atoms form a ring with alternating single and double bonds; benzene is one
of the simplest examples of an aromatic ring.
beta (β) decay A radioactive decay in which the nucleus ejects an electron (a
beta particle); this change transforms a neutron into a proton.
binding energy The energy released when protons and neutrons combine to
form a nucleus, or the energy required to break a nucleus into its component particles.
Bohr model An early model of atomic structure that treated the atom like a tiny
solar system, with the nucleus at the center, and the electrons orbiting the nucleus.
boiling point elevation A colligative property of water; the presence of solute
in an aqueous solution raises the boiling point above that of pure water.
bomb calorimetry A technique for measuring heat changes using a sealed
container; commonly used to measure high-energy reactions.
Boyle’s law The pressure and volume of an ideal gas are inversely related; the
product of PV is constant.
Brønsted-Lowry definition Describes an acid as a compound that donates H+
ions, and a base as a compound that accepts H+ ions.
buffer A solution containing a mixture of acidic and basic components; it resists

1160
changes in pH.

C
calorimetry An experimental technique used to measure heat changes.
carbohydrate A naturally occurring molecule composed of carbon, hydrogen,
and oxygen and having the general formula Cm(H2O)n.
carbon cycle A description of how Earth’s carbon moves between rock and
sediment, water and atmosphere, and plants and animals.
carbonyl A functional group consisting of a carbon– oxygen double bond.
carboxylic acid A functional group containing a carbonyl bonded to an alcohol,
commonly represented by the condensed formula –COOH.
catalyst A species that is not part of a balanced equation but causes a reaction to
go more quickly.
cation A positively charged ion.
cellular respiration The sequence of chemical reactions by which animals
release energy stored in the chemical bonds of substances they consume.
Celsius scale (°C) A temperature scale commonly used throughout the world.
On the Celsius scale, water freezes at 0 °C and boils at 100 °C. Sometimes called
the Centigrade scale.
charge A characteristic property of subatomic particles that affects how particles
interact with each other.
Charles’s law The volume of a gas is directly proportional to its temperature;
the relationship between V and T is constant.
chemical changes Changes that produce new substances; also called chemical reactions
chemical equation A symbolic representation of a chemical change. Such
equations consist of reactants and products separated by an arrow.
chemical formula A representation of the type and amount of each element
present in a compound.
chemical properties Properties of a substance that cannot be measured without
changing the identity of a substance.
chemical reactions Changes that produce new substances; also called chemical changes
chemistry The study of matter and its changes.
coefficient In a chemical formula, the numbers written before each reagent or
product to indicate the ratios in which components of the reaction are consumed or produced.
coffee cup calorimetry A technique for measuring heat changes that uses an
insulated container (such as a Styrofoam™ coffee cup) to measure heat changes.
colligative properties Properties that depend on the number of particles
dissolved in the solution, but not on the type of particles dissolved.
combined gas law A combination of Boyle’s law and Charles’s law; it states
that for an ideal gas, the quantity PV/T is constant; usually expressed by the
equation P1V1/T1 = P2V2/T2, where the subscripts 1 and 2 denote two different conditions.
combustion A reaction in which oxygen gas combines with elements or

1161
compounds to produce oxide compounds.
complete ionic equation An equation that shows all ions present in a solution.
composition The components that make up a material.
compounds Pure substances composed of more than one element in a fixed ratio.
concentration The amount of solute present in solution.
condensation A transition from the gas phase to the liquid phase.
condensation reaction A reaction in which two smaller molecules combine to
produce water plus a larger molecule.
condensed structure A way of representing chemical bonds that does not
show most covalent bonds, but lists atoms in order of their connectivity.
conjugate acid The acid on the right-hand side of a chemical equation in an
acid-base equilibrium; the acid formed when a base reacts with H+.
conjugate base The base on the right-hand side of a chemical equation in an
acid-base equilibrium; the base formed when an acid releases an H+.
conversion factors Fractions that are used to convert from one unit to another.
A conversion factor contains equivalent amounts of different units in the numerator
and the denominator.
covalent bond A bond in which two electrons are shared between atoms;
covalent bonds typically form between nonmetals.
covalent compounds Compounds formed by covalent bonds; these compounds
form discrete groups of atoms called molecules.
covalent networks Long two or three-dimensional sequences of covalent
bonds, resulting in very large single molecules.
cycloalkane An alkane that forms a cyclic structure (commonly called a ring).

D
decay series A naturally occurring sequence of radioactive decays; also called a
decay chain.
decomposition A reaction in which a single reactant forms two or more products.
density A physical property of a substance, defined as the mass per unit volume.
deoxyribonucleic acids (DNA) Massive molecules containing the genetic
code of living creatures.
diffusion The spread of particles through random motion; lighter gases diffuse
more quickly than heavier gases.
dipole–dipole interaction An intermolecular force between two molecules
containing net dipoles.
disaccharide A carbohydrate composed of two simpler carbohydrates that are
linked together through a condensation reaction.
dissociation The process by which ions are pulled apart from a solid lattice
when an ionic compound dissolves in water.
dosimeter A radiation detector that measures human exposure to radiation.

1162
double displacement A reaction in which two compounds swap cation-anion
pairs to form two new compounds.
d sublevel A sublevel that contains five orbitals and can hold up to 10 electrons;
the d sublevel is present in energy levels 3 and higher.

E
effusion The process of a gas escaping from a container; lighter gases effuse
more quickly than heavier gases.
electrochemistry The study of chemical processes that involve the movement
of electrons.
electrode In electrochemical cells and related devices, a site where an oxidation
or reduction half-reaction takes place.
electrolyte solution An aqueous solution containing dissociated ions; this type
of solution conducts electricity more effectively than pure water.
electromagnetic radiation A form of energy produced when charged particles
move or vibrate relative to each other; electromagnetic radiation exists as waves.
electromagnetic spectrum All forms of electromagnetic energy, ranging from
low-energy waves (TV and radio) to visible light to high-energy waves such as
gamma rays.
electron A negatively charged subatomic particle; the electrons occupy the space
around the nucleus.
electron cloud The space around the nucleus; the electron cloud accounts for
nearly the entire volume of the atom.
electron configuration The number of electrons in each energy level and sublevel.
electronegativity A measure of how strongly atoms pull bonded electrons.
electronic geometry A description of the arrangement of electrons around a
central atom.
electroplating A technique that uses electrical potential energy (such as from a
battery) to produce a thin layer of a metal such as gold, silver, copper, or chromium
on the outside surface of another material.
element A substance made of only one type of atom.
elemental analysis A technique used to determine the percent composition of a substance.
empirical formula A chemical formula that gives the smallest whole-number
ratio of atoms in a compound.
endothermic change A change that absorbs energy.
energy The ability to do work.
equilibrium The state in which forward and reverse reactions take place at the
same rate, so the concentrations of reactants and products do not change.
equilibrium constant (K) The ratio of products to reactants when a reaction is
at equilibrium.
equilibrium expression An equation that describes the balance between
reactants and products in an equilibrium.

1163
equilibrium reaction A reaction that occurs in both the forward and backward directions.
ester A functional group containing a carbonyl bonded to an oxygen that is
bonded to another carbon.
ether A functional group composed of an oxygen singly bonded to two carbon
atoms, giving the bonding sequence C–O–C.
exact numbers Numbers for which there is no uncertainty. Counted integers
and defined relationships (such as metric prefixes) are exact numbers.
excess reagent In a chemical reaction, a reagent that is present in larger
stoichiometric quantities than the other reagents; an excess reagent is not
completely consumed.
exothermic change A change that releases heat energy.
expanded octet A bonding arrangement in which an atom has 10 or 12 valence
electrons; this is possible only with elements in rows 3–7 of the periodic table.

F
Fahrenheit scale (°F) A temperature scale commonly used in the United
States. On the Fahrenheit scale, water freezes at 32 °F and boils at 212 °F.
fission A nuclear change in which a large nucleus shatters into several smaller
nuclei, releasing large amounts of energy.
formal charge A method of identifying charged sites on a molecule or ion. The
formal charge of an atom is the number of electrons in the valence of that atom in
its neutral, unbonded state, minus the number of covalent bonds, minus the number
of unshared electrons.
formula mass The mass of a molecule or formula unit.
formula unit In ionic compounds, the smallest number of ions necessary to
form a compound; the combination of atoms described by an empirical formula.
freezing A transition from the liquid phase to the solid phase.
freezing point depression A colligative property of water; the presence of
solute in an aqueous solution lowers the freezing point below that of pure water.
frequency (v) The number of waves that pass through a point in one second;
typically measured in hertz.
f sublevel A sublevel that contains seven orbitals and can hold up to 14
electrons; the f sublevel is present in energy levels 4 and higher.
fuel cell A device that converts the energy of a combustion reaction directly into
electrical energy by separating the oxidation and reduction half-reactions.
fuel value The amount of heat energy that can be released by a combustion
reaction of a certain substance.
functional group A small group of atoms within a molecule that behaves in a
characteristic manner.
fusion A nuclear change in which two smaller nuclei (usually hydrogen or
helium) combine to form larger nuclei.

1164
G
gamma (γ) decay A radioactive decay in which the nucleus releases energy but
no particles.
gas A state of matter that does not have a definite shape or a definite volume. The
particles in a gas move freely with very little interactions.
gas laws Mathematical relationships between the pressure, volume, and
temperature of gases.
gauge pressure The difference between a compressed gas pressure and
atmospheric pressure.
Geiger counter A device that measures radiation, including alpha- and beta-
emissions and gamma rays.
gravimetric analysis A technique that uses the mass of a precipitate to
determine the concentration of a reactant.
group A vertical column on the periodic table; also called a family. Elements
within a group exhibit similar behaviors.

H
half-life The amount of time required for one-half of a radioactive substance to decay.
half-reaction A chemical equation that shows only the reduction or oxidation
component of a reaction.
halogens Nonmetal elements in column 17 (or 7A) of the periodic table; these
elements form many different types of compounds.
heat capacity The amount of heat required to raise the temperature of a given object.
heat energy A type of kinetic energy involving the movement of particles
within a substance; heat energy specifically refers to the total kinetic energy
transferred from substance to another.
hertz (Hz) A unit of frequency equal to one wave cycle per second.
heterogeneous mixtures Mixtures in which the components are not evenly
blended throughout.
homogeneous mixtures Mixtures in which the components are evenly blended throughout.
hydrocarbon A compound that contains only carbon and hydrogen.
hydrogen bond An unusually strong dipole–dipole interaction that occurs
between molecules containing H–F, H–O, or H–N bonds.
hydronium ion H3O+, the ion formed in acidic aqueous solutions.
hypothesis A tentative explanation that has not been tested.

I
ideal gas A gas in which the volume of the particles is much less than the
volume of the container, and in which the particles have no attraction for each other.
ideal gas law The relationship between pressure, volume, temperature, and the
number of moles of an ideal gas; typically expressed in the form PV = nRT, where

1165
R is the gas constant.
intermolecular forces The forces of attraction or repulsion that take place
between molecules.
ion An atom or group of atoms with an overall charge.
ionic bond A force of attraction between oppositely charged ions.
ionic compound A compound of oppositely charged ions.
ionic equation A chemical equation that shows dissociated ions as separate species.
ionic lattice A tightly packed array of alternating positive and negative charges;
the characteristic arrangement of ions in an ionic solid.
isoelectronic Describes atoms that have the same electron configuration.
isomers Compounds having the same molecular formula but different bonding sequences.
isotopes Nuclides having the same atomic numbers but different mass numbers.

J
joule (J) The IUPAC unit of energy; 1 J = 1 kg·m2/s2.

K
Kelvin scale (K) The official SI scale for measuring temperature. The Kelvin
scale sets absolute zero at 0 K, the freezing point of water at 273.15 K, and the
boiling point of water at 373.15 K.
ketone A functional group containing a carbonyl connected to two carbon atoms.
kinetic energy The energy of motion; the faster an object is moving, the greater
kinetic energy it has.

L
law of conservation of energy In a chemical or physical change, the total
energy of the universe remains constant.
law of conservation of mass In chemical reactions, matter is neither created
nor destroyed.
law of conservation of matter and energy In any change, the total matter
and energy in the universe remain constant.
Le Chatelier’s principle When some change in concentration, temperature,
pressure, or other key factor takes place, the equilibrium shifts to minimize that
change and establishes a new equilibrium.
Lewis dot symbol A method of representing the valence structure of an atom or
ion that involves using dots around the atomic symbol to indicate valence electrons.
Lewis structure A method of showing the arrangement of valence electrons in
a molecule or polyatomic ion, in which the Lewis symbols for atoms are shown
connected by dashes representing covalent bonds.
limiting reagent In a chemical reaction, a reagent that is completely consumed

1166
and limits the amount of product that can be formed.
linear A geometry in which two atoms or electron sets are separated by 180° angles.
line spectrum A pattern of light energies called spectral lines; they are formed
when gas-phase elements release energy. Each element has a characteristic line spectrum.
liquid A state of matter having definite volume but no definite shape. The
particles in a liquid are close together but move freely past each other.
liter (L) A common unit of volume, defined as 1 cubic decimeter (1 dm3).
litmus A pH indicator that turns blue in the presence of base and red in the
presence of acid.
London dispersion forces Intermolecular forces that result from fluctuations
in charge density called instantaneous dipoles.
lone pair A pair of unshared valence electrons.

M
main-group elements The elements in columns 1–2 and 13–18 (or 1A–8A) of
the periodic table.
mass defect The difference between the masses of the individual particles and
the mass of the complete nucleus.
mass number The sum of protons and neutrons in an atom.
mass spectrometry A technique used to measure the mass of a molecule.
mass/volume percent A measure of concentration; the mass of solute divided
by the total volume of the solution, expressed as a percentage; often measured in
grams of solute per milliliter of solution.
matter Anything that has mass and takes up volume.
melting A transition from the solid phase to the liquid phase.
metal displacement reaction A reaction between two metals in which one
metal is oxidized to its ionic form while the other metal ion is reduced to its
elemental form.
metalloids Elements whose properties lie between those of metals and nonmetals.
metals The elements on the left-hand side of the periodic table; these elements
can be molded into different shapes, and they conduct heat and electricity.
millimeters of mercury (mm Hg) A measure of gas pressure; this unit
originates from the height to which atmospheric pressure can push a column of
mercury in a barometer; 1 mm Hg = 1 torr.
mixtures Substances containing more than one substance.
molarity (M) A measure of concentration defined as the moles of solute per
liter of solution.
molar mass The formula mass of an element or compound, expressed in grams
per mole.
mole A quantity consisting of 6.02 × 1023 units.
molecular dipole An overall polarity in which different sides of a molecule
have slight positive and negative charges; sometimes called a net dipole or simply

1167
a dipole.
molecular equation A chemical equation in which all species are written as
neutral compounds.
molecular formula A formula that gives the actual number of atoms in the molecule.
molecular geometry A description of the arrangement of atoms within a
molecule; the molecule shape.
molecules Groups of atoms that are held tightly together.
monomer A small molecule that can connect with other molecules to form a polymer.

N
net ionic equation An equation in which the only ions shown are those directly
involved in the chemical change, and spectator ions are omitted.
neutralization reaction A reaction in which an acid and a base combine; acids
and hydroxide bases combine to produce water and an ionic compound, commonly
called a salt.
neutron A subatomic particle having no charge that resides in the nucleus of the atom.
nitrile A functional group consisting of a carbon–nitrogen triple bond.
noble gases The nonmetal elements in column 18 (or 8A) of the periodic table;
these elements usually do not form compounds.
nonmetal oxide A compound or ions containing a nonmetal covalently bonded
to one or more oxygen atoms.
nonmetals Elements on the right-hand side of the periodic table; these elements
have widely varying properties and form many different compounds.
nuclear chemistry The study of the properties and changes of atomic nuclei.
nuclear equations Equations that describe nuclear changes; in a balanced
nuclear equation, the total mass numbers and total atomic numbers are the same on
both sides of the equation.
nuclear force The force that holds the nucleus together.
nuclear medicine The use of radioactive nuclei for medicinal purposes.
nuclear reaction A reaction in which the structure of an atomic nucleus changes.
nucleon A particle that resides in the nucleus; either a proton or a neutron.
nucleus The tiny, dense center of an atom; the nucleus contains protons and neutrons.
nuclide An atom or nucleus containing a particular number of protons and neutrons.

O
octet rule The principle that an atom is stabilized by having its highest-occupied
(valence) energy level filled.
orbital A region where electrons are most likely to be found; each orbital can
hold up to two electrons.
organic chemistry The chemistry of carbon-containing molecules.
osmotic pressure The tendency of water to move toward regions of greater

1168
concentrations; an imbalance in concentration inside and outside of a living cell
can cause the cell to swell or shrink.
oxidation The loss of electrons.
oxidation number A bookkeeping tool for tracking electron changes in
oxidation-reduction reactions.
oxidation-reduction reaction A chemical change in which one species loses
electrons (oxidation) while another gains electrons (reduction); also called a redox reaction
oxyacid A compound that dissociates in aqueous solution to form H+ and an oxyanion.
oxyanion A polyatomic ion containing oxygen.

P
partial pressure The pressure caused by one gas in a mixture.
parts per billion (ppb) A measure of concentration used for very dilute
solutions; 1 ppm = 1 g of solute per 1,000,000,000 grams of solution, or 1 μg of
solute per liter of solution.
parts per million (ppm) A measure of concentration used for dilute solutions;
1 ppm = 1 g of solute per 1,000,000 grams of solution, or 1 mg of solute per liter of solution.
peptide bond The carbon–nitrogen bond that connects two amino acids
together; peptide bonds are formed by the condensation of a carboxylic acid and an
amine to form an amide.
percent by mass A measure of concentration; the mass of solute divided by the
total mass of the solution, expressed as a percentage.
percent by volume A measure of concentration; the volume of solute divided
by the total volume of the solution, expressed as a percentage.
percent composition The percentage (by mass) of each element in a compound.
percent yield A measure of the efficiency of a reaction; the actual yield divided
by the theoretical yield, expressed as a percentage.
period A horizontal row on the periodic table; a period encompasses a range of
behavior from metallic to nonmetallic.
periodic table of the elements A chart that organizes all the known elements
based on their masses and properties.
phenolphthalein A pH indicator that is bright pink in base, but colorless in acid.
pH indicator A compound that changes color depending on the pH; the color
changes indicate the approximate pH of the solution.
photon A small increment or packet of electromagnetic energy (often visible light).
photosynthesis The sequence of chemical reactions by which green plants
harvest the energy of the sun.
pH paper Paper containing a blend of indicators that can be used to estimate pH
based on color.
pH scale A scale that indicates the relative concentration of acid or base in an
aqueous solution; pH is defined as the negative log of the hydronium concentration.
physical changes Changes that do not alter the identity of the substance.

1169
physical properties The properties of a substance that can be measured
without changing the identity of the substance.
plasma A gas-like state produced at very high temperatures and often considered
the fourth state of matter; in this state, electrons do not remain bound to a single atom.
plastic A synthetic polymer.
polar covalent bond A covalent bond whose atoms do not share electrons
evenly; in this type of bond, one atom has a slight positive charge while the other
has a slight negative charge.
polyatomic ion A group of bonded atoms with an overall charge.
polymer A molecule containing simple repeating units that are linked together in
long covalent chains.
polyprotic acids Acids that can release more than one H+ into aqueous solution.
polysaccharide A carbohydrate composed of many simpler carbohydrates that
are linked together.
potential energy Energy that is stored.
precipitate A solid product formed from the combination of two solutions.
precipitation reaction A type of chemical change in which two aqueous
solutions combine to produce an insoluble product.
precision A measure of how finely a measurement is made, or how close a
group of measurements are to each other. Precision is often denoted by significant digits.
pressure The force that an object exerts divided by the area over which it is
applied; for gases, pressure describes the force that gases exert on their surroundings.
principal quantum number An integer number that identifies the energy
level an electron occupies.
product The compounds produced in a chemical change, shown on the right-
hand side of a chemical equation.
protein A biopolymer composed of building blocks called amino acids; this type
of molecule has many functions in living creatures.
proton A positively charged subatomic particle that resides in the nucleus of the atom.
p sublevel A sublevel that contains three orbitals and can hold up to six
electrons; the p sublevel is present in energy levels 2 and higher.
pure substances Substances composed of only one element or only one compound.

Q
quantum model The modern description of electronic behavior that treats
electrons as particles and as waves.

R
radioactive decay A nuclear change in which the nucleus spontaneously
transitions to a more stable state; decays involve the release of energy and
sometimes mass.

1170
radioactivity The spontaneous release of particles and/or energy from the nucleus.
reactant The starting material in a chemical change, shown on the left-hand side
of a chemical equation.
reaction energy diagram A depiction of the energetic changes that
accompany a chemical reaction.
reaction enthalpy (ΔHrxn) The amount of heat energy that is absorbed or
released in a chemical reaction at constant pressure.
reaction rate The speed at which a chemical reaction takes place.
reduction The gain of electrons.
resonance structures A set of Lewis structures that show how electrons are
distributed around a molecule or ion. Resonance structures are used when a single
Lewis structure cannot adequately depict the structure.

S
scientific law A statement that describes observations that are true in widely
varying circumstances. Scientific laws often describe mathematical relationships.
However, they do not explain why something occurs; they only observe that it occurs.
scientific method A cyclical process of making observations, formulating new
ideas, and then testing those ideas through experiments.
scientific notation A way to show very large and very small numbers in a
concise format. Scientific notation expresses numbers as the product of two values,
called the coefficient and the multiplier.
scintillation counter A device used for measuring radiation; a scintillation
counter measures the total number of radiative particles and waves striking the detector.
self-ionization A process that occurs in water when a water molecule fragments
to produce H+ and OH− ions.
semiconductor counter A device that measures radiation; some
semiconductor counters can distinguish between the decays of different elements
based on the energy of the radiation.
sievert (Sv) A measure of radiation useful for measuring human exposure; 1 Sv
= 1 joule of energy per kilogram of mass.
significant digits The digits contained in a measured value. The number of
significant digits indicate how precisely a measurement is made. Also called
significant figures.
single displacement A reaction in which one element replaces another element
in a compound.
skeletal structure A simplified representation for chemical structures in which
the end of each line segment denotes a carbon, and C–H bonds are inferred rather
than drawn explicitly.
solid A state of matter having a definite shape and a definite volume. The
particles in a solid are held in fixed positions.

1171
solubility product (Ksp) The equilibrium constant for the solution of a slightly
soluble ionic compound.
soluble Having the ability to be dissolved in a liquid.
solute A substance that is dissolved in a solution.
solution A homogeneous mixture; for example, a solid mixed in a liquid.
solvent The major component of a solution.
specific heat The amount of heat required to raise one gram of a substance by
one degree Celsius; sometimes called specific heat capacity.
spectator ion An ion that is present in a solution but not directly involved in a
chemical change.
s sublevel A sublevel that contains one orbital and can hold up to two electrons;
the s sublevel is present in every energy level.
states of matter The classification of matter as a solid, liquid, or gas (also
called the phases of matter).
stoichiometry problem A problem that relates the amount of one reagent or
product to another in a chemical reaction, using a balanced equation.
strong acid An acid that completely ionizes in water.
structure The arrangement of simple units within a substance. In chemistry,
structure refers to both the composition and arrangement of simple units within a substance.
subatomic particles The particles from which atoms are composed. The three
major subatomic particles are protons, neutrons, and electrons.
sublevel A set of electron orbitals that occurs in an electron energy level; the
four main sublevels are s, p, d, and f.
subscript In a chemical formula, subscript numbers show the number of each
atom or ion present.
surroundings In thermodynamics, everything that exists around the system
being studied.
synthesis A reaction in which two reactants join together to form a single
product; also called a combination reaction.
system In thermodynamics, the part of the universe being studied.

T
temperature A measure of the average kinetic energy of the molecules in a substance.
tetrahedral A geometry in which four atoms or electron sets are separated by
109.5° angles.
theoretical yield The amount of product that can form in a chemical reaction,
based on the balanced equation and the amount of starting materials present.
theory An idea that has been tested and refined; also a way of thinking about a
particular topic.
thermodynamics The scientific field that deals with energy and temperature changes.
titration An analytical technique that can precisely measure the concentration of

1172
an acid or base by measuring the volume required for a neutralization to occur.
transition metals The metals in columns 3–12 of the periodic table; these
metals are harder and less reactive than those in columns 1–2.
transition state The highest-energy arrangement of atoms that occurs during a
chemical reaction.
trigonal planar A geometry in which three atoms or electron sets are separated
by 120° angles.

U
uncertainty principle The idea that it is impossible to know the exact velocity
and location of a particle; this principle becomes important when studying electrons.
units of measurement Quantities with accepted values.

V
valence level The highest-occupied electron energy level in an atom.
valence shell electron pair repulsion (VSEPR) model A way of
predicting the geometry of molecules based on the number of electron sets around
a central atom.
vaporization A transition from the liquid phase to the gas phase.
visible spectrum The narrow range of electromagnetic energy that we perceive
as light.

W
wavelength (λ) The distance from a point on one wave to the same point on the
next wave.
weak acid An acid that only partially ionizes in water.
work The transfer of energy from one form to another.

1173
Index
Note: Page numbers followed by f and t indicate figures and tables, respectively.

A
Absolute zero, 283, 296
Accuracy, 29, 48
meaning, 30f
Acetaminophen, 434f
Acetic acid (HC2H3O2), dissociation, 374, 378
Acetylene (C2H2), 439
hydrogen gas, reaction, 227
torches
burn temperatures, 293
usage, 402f
usage, 439f
Acidic compounds, formation, 344
Acidic, term (usage), 335
Acid rain, 344, 358
Acids, 134–135, 137, 334–338, 358
acid-base chemistry, protons, 414
acid-base definitions, 338f
acid-base equilibrium, 339
components, 340
reactions, 338–341
acid-base neutralization reactions, 167–169, 196, 319
double-displacement reactions, 168
product prediction, 168, 342
acid-base pair, 339
acid-base reactions, 340
prediction, activity series (usage), 406
acid-base titrations, 350–352
coefficients, differences, 351–352
acid-conjugate base pair, 340f
acid salt/powder, 340
Arrhenius definition, 335
binary acids, 134
Brønsted-Lowry definition, 337
concentration, measurement, 349–352
conjugate acid, 339

1174
 corrosivity, 134
diprotic acids, 336
formation, nonmetal oxides (impact), 344
identification, 338
ionization, 312
ions, aqueous solution, 336
metals
reactions, 405–407
reactivity, 406t
monoprotic acids, 336
naming, 134–135
oxyacids, 134–135
polyprotic acids, 336–337
proton donors, 337
range, 346t
reactions, 341–344
strong acids, types, 338t
triprotic acids, 336
types, 167t, 335t
water dissolution, 335f
weak acids, 338–339
Actinide series, 63
Actinium-228, formation, 475
Actinium, decay series, 475
Activation energy, 369, 371, 389
determination, 374–375
heat (creation), match (usage), 370f
reduction, catalyst (impact), 370f
Activity series, 403–405, 407, 417
strength, ranking, 404t
usage, 406
Actual yield, 197, 200
Adenine, 454–455
Adrenaline, 136f
Air
mixture, 287, 287f
sample, water vapor (presence), 288
Airbag, solid fuel (reaction), 293
Alanine, glycine (bonding), 452
Alcohol (C2H6O), 432f, 457, 460
C-O-H bonds, 441f
group, 441
oxygen-containing functional group, 441
Aldehyde, 442, 460

1175
 functional groups, 442f
Alkali metals, 64, 78
+1 ions, formation, 118
behavior, 117
reaction, 64f, 406f
Alkaline earth metals, 64, 78
+2 ions, formation, 118
types, 64f
Alkanes, 436–438, 457, 460
branched structures, 438
formula, 438
linear alkanes, formula, 438f
single bonds, 439
straight-chain alkanes, 438
types, 437t
Alkenes, 438–439, 457, 460
double bonds, 439
identification, 440
Alkynes, 438–439, 460
identification, 440
triple bonds, 439
Alloys, 272
Alpha(1→4) bonds, 450–451, 450f
Alpha (α) decay, 474, 474f, 493
Alpha (α) form, 449
Alpha-glucopyranose (α-glucopyranose), 449
Alpha-glucosidase inhibitors, usage, 436
Alpha-linkages, 450
Alpha (α) particles
damage, 478f
source, 68
Alpha (α) radiation, 473
Aluminum
+3 ions, formation, 157
aluminum-26, usage, 483, 491
foil, usage, 67f
heating, 9
metal
hydrochloric acid, reaction, 406
nitric acid, reaction, 193
Aluminum oxide (Al2O3), moles (calculation), 186
Amalgam, 55
Amides, 445, 460
functional groups, 446f

1176
Amine functional groups, 445f
Amines, 457, 460
Amino acids, 450–454, 458, 460
neutral form representation, 452
structures, drawing, 453
two-amino-acid structure, components, 452f
types, 451t
Ammonia (NH3)
base, 337
covalent compounds, 272
cylinder, pressure, 288
fertilizer, usage, 338, 365, 382
importance, 382–383
making, Haber-Bosch process, 387
molecular geometry, 251
molecule
Lewis structure, 129
nitrogen atom, bonding electrons/lone pair, 251f
pressure, container, 385f
problem, 386
synthesis, 385
Ammonium chloride (NH4Cl)
dissolving, 215, 221
water, combination, 221–222
Ammonium ions, production, 378
Ammonium phosphate [(NH4)3PO4]
barium chloride, reaction, 323
Ammonium phosphate [(NH4)3PO4], ionic equation, 313
Ammonium sulfide, empirical formula, 126
Anions, 119–120, 137
in double displacement reactions, 154
naming, 120
oxyanions, 121
types, 120t
Anode, oxidation (occurrence), 409
Anti-Drug Abuse Act, 357
Antiseptics, 65
Aqueous acid, concentrations, determination, 379
Aqueous calcium hydroxide, aqueous nitric acid (reaction), 342
Aqueous copper(II) chloride, zinc metal (reaction), 192
Aqueous copper(II) ions, zinc metal (reaction), 402f
Aqueous copper(II) sulfate, zinc metal (reaction), 402
Aqueous HCl, sodium bicarbonate (addition), 322

1177
Aqueous H+ ion (H3O+), 336
Aqueous hydrobromic acid, calcium metal (reaction), 323
Aqueous hydrochloric acid, zinc metal (reaction), 323
Aqueous nitric acid, aqueous calcium hydroxide (reaction), 342
Aqueous solutions, 132–134, 153
concentrations, 344–346
hydrochloric acid, reaction, 154–155
mass/volume percent, 306
molarity, calculation, 309
ppm/ppg calculation, 308
reactions, 160–169
Archaeology, radioactive nuclide usage, 482–483
Area, calculation, 37f
Argon
gas, cylinder pressure, 288
noble gas, 100
Aromatic compounds, 439–440
Aromatic functional groups, identification, 440
Aromatic ring, 440, 460
Arrhenius definition, 335–336, 337, 355, 358
Arsenic, usage, 108f
Aspirin, benzene ring, 440, 440f
-ate suffix, 134
Atmosphere (atm), 281, 296
Atomic arrangements, visualization (ability), 60
Atomic mass, 71–73
Atomic mass unit (u) (amu), 66, 72, 78
mass, description, 185
Atomic number, 69–71, 76 78, 471
atomic symbols, writing, 70
display, 472
periodic table location, 69f
protons, number, 69
sum, 472f
usage, 70
Atomic ratio, 187
Atomic structure
clues, 57
discovery, 66–69
light/color, connection, 87f
models, development, 74t
Atomic symbols
representation, 62

1178
 writing, 70
Atomic theory, 15, 78
framework, 58–59
Atoms, 18
analysis, 57–60
arrangement, 7
building blocks, 57–60, 64
charge, balance, 411
close-fitting arrangements, 272f
collision, 367
covalent bond, formation, 126f
depiction, 59f
description, 69–73
electronegativity, difference, 255f, 256
electrons
configurations, 100–101
holding, 100
loss/gain, 74f
sharing, 127
identification, proton number (usage), 69
manipulation, scanning tunneling microscopy (usage), 60f
oxidation numbers, 399t
region, orbital description, 95
representation, 5f
root name, 130
stabilization, 239
type/arrangement, 57f
visualization, 60
Autopsy, site, 23f
Average atomic mass, 71–73
usage, 181
Average mass, calculation, 72
Avogadro’s law, 285, 294, 296
Avogadro’s number, 184–185, 200
usage, 186–187

B
Baking soda. See Sodium bicarbonate
Balance
calibration, 30
mass measurement, 43f
Balanced chemical equations, ratios (prediction), 149
Balanced equation, 148, 172, 229
mole concept, 187–196

1179
 usage, 189
Balanced molecular equation, 166
Balanced nuclear equation, 472f
Ball, diameter (measurement), 29
Balloon
gases, presence, 290f
volume, determination, 284
Barclay, Arthur, 1
Barium chloride
solution, ions (presence), 165
sulfate ion, concentration, 321
Barium hydroxide (neutralization), hydrochloric acid (usage), 351–352
Barium sulfate (BaSO4)
reaction, 421
solubility product, 381
Barometer, 280, 296
usage, 280f
Bases, 167, 172, 334–338, 358
acid-base chemistry, protons, 414
acid-base definitions, 338f
acid-base equilibrium, 340
reactions, 338–341
acid-base neutralization reactions, 167–169, 197, 319
double-displacement reactions, 168
product prediction, 168, 342
acid-base pair, 339
acid-base reactions, 340
acid-base titrations, 350–352
coefficients, differences, 351–352
acid-conjugate base pair, 340
Arrhenius definition, 335
Brønsted-Lowry definition, 337
concentration, measurement, 349–352
conjugate base, 339
dissociation, 336
hydroxide bases, 167t
identification, 338
proton acceptors, 337
range, 346t
reactions, 341–344
types, 335t
Batteries (electrochemical cells), 67f, 397f, 407–411, 415, 417
advancement, 416
electrical energy, 67f

1180
 invention, 76
lithium-ion batteries, 416, 416f
power, oxidation-reduction reactions (impact), 409f
usage, 67
Beaker, usage (volume measurement), 31f
Becquerel (Bq), 480, 493
Becquerel, Antoine, 473
Becquerel, Edmond, 85
Bent, geometry description, 251
Benzene (C6H6), 439–440, 460
carbon atoms, ring shape, 439f
electron waves, 440f
Beryllium
alkaline earth metal, 64f
beryllium-10, usage, 483, 491
electron configuration, 98
Beta(1→4) bonds, 450–451
Beta(β) decay, 475
neutron, transformation, 475f
Beta-glucopyranose (β-glucopyranose), 449
Beta-linkages, 450
Beta (β) radiation, 473, 478
Bike, energy states (decline), 12f
Binary acids, 134
Binding energy, 484–485, 493
per nucleon, 485, 485f
Biological pH, 353–355
Biomolecules, 448–456
Biopolymers, 447, 457–458
Bismuth-212, decomposition, 472
Blanco, Aaron, 303, 325
Bleach, 65, 65f
Blood pH, maintenance, 354
Blue waves, wavelength, 89f
Bohr model, 73, 76, 78, 90–93
electron orbit, 92f
Bohr, Niels, 73
Boiling point, 271
elevation, 314–315, 326
Bomb calorimeter, function, 222f
Bomb calorimetry, 222–223, 230
usage, 223
Bonded atoms, electronegativity (differences), 255f

1181
Bond-forming reaction, representation, 199f
Bonds
classification, 255
polarity, determination, 255
Boron
compounds, valence level (incompleteness), 240f
electron sublevels, 98
Boyle’s law, 282, 294, 296
usage, 282
Break-even point, 490
British thermal unit (BTU), 212
Bromide ions
formation, 158
formula, 124
Bromine
copper, reaction, 191
electron configuration, 103
molecular bromine, breakage, 377
properties, similarities, 103f
tin metal, reaction, 158
Brønsted-Lowry definition, 337–338, 355, 358
Bronze, copper/tine mixture, 6f
Brown, Alton, 303
Brown Coffee Company, cappuccino, 303f
Buffers, 353–355, 358
adjustment, 355
preparation, citric acid/conjugate base (usage), 355
representation, 354
solutions, mixture, 379
Buret, 350
Butane gas, burning (heat release), 217
Butyl acetate, reactant structures, 445

C
C-O bond, 255
C-terminus, 452, 452f
Cade, John, 114f, 115–116, 136
Calcium
alkaline earth metal, 64f
alpha decay, 476
hydrochloric acid, half-reactions, 408
ionic compound production, 124
iron(II) ions, reaction, 403

1182
 metal, aqueous hydrobromic acid (reaction), 323
oxide, water (reaction), 344
Calcium carbonate (CaCO3), 145
chemical change, 147
production, 171
Calcium chloride (CaCl2)
freezing point, 314f
molarity, calculation, 310
sodium phosphate, precipitation reaction, 166
solution, sodium chloride solution (separation), 318
Calcium hydroxide [Ca(OH)2]
base, 336
production, 344
water, mixture, 380
Calcium ions
carbonate ions, combination, 320
production, 403
Calcium oxide (CaO)
conversion, 154f
formation, 145
production, 171
Calculations
multiple, 36
significant digits, usage, 34–35
Calculators, indications, 34
Calendar, organization, 61f
Calibrating, 30
Caliper, precision, 29f
Calorie, 212
Calorimetry, 219–223, 228, 230
bomb, 222–223
coffee cup, 219–222
Candle wax, 436f
Capsaicin, 445
Car battery, 67f
Carbohydrates, 448–450, 460
formula, 449
Carbon
average mass, 72
carbon-14
concentration, measurement, 483
dating, 482–483
half-life, 478
usage, 491

1183
 carbonate formation, 121
carbon–carbon bonds, 425
carbon–carbon triple bonds, 439, 457
carbon–oxygen double bond, 441–442
covalent bonding, 428–431
cycle, 427–428, 460
electrons, sublevels, 98
hydrogen atoms, bonding, 433
importance, 427–428
nonmetal type, 64
skeletons, 433
Carbonated beverages, pressurized carbon dioxide (usage), 385
Carbonated soft drinks, carbon dioxide/water reaction, 293
Carbonate ion
calcium ion, combination, 320
resonance structures, 249
Carbon–carbon triple bond, 260
Carbon dioxide (CO2)
carbonic acid, formation, 344
conversion, photosynthesis (impact), 427
exhalation, 249
formation, 59f, 145, 148
gas mixture, 290
geometries, 250
molecule, octet rule fulfillment, 239f
net dipole, absence, 258f, 274
production, 10f, 180
properties, 59
water, reaction, 293
Carbon disulfide
Lewis structure, drawing, 253
mixture, 290
Carbonic acid (H2CO3)
acidic reaction, 341
diprotic acid, 336
formation, 344
reaction, 153f
Carbonyl, 460
groups, 441–443
Carboxylic acid, 443, 448, 452, 457, 460
Carlisle, Anthony, 397
Carvone (spearmint oil component), 442f
Cassiterite, 401t
Catalysts, 370–371, 389, 459–460

1184
 Grubbs catalyst, 425–436
ruthenium catalysts, usage, 459–460, 459f
Cathode, reduction (occurrence), 409
Cations, 117–123, 137
metal, reaction, 404f
naming, 118–119
Cellular respiration, 402, 427, 460
Cellulose, 450
beta(1→4) bonds, breakdown (inability), 450f
Celsius scale, 45, 45f, 48
reporting, 46
Central atom
electronic geometry, 250f
octet fulfillment, 241, 242, 247
shape, 252
Central science, 3
Cesium-131, implanting, 482f
Chainsaws, portability, 416
Chalcogens, 119
Changes, 9–11
endothermic changes, 13, 18
exothermic changes, 13, 18
Charcoal, oxygen (reaction), 59, 59f, 370
Charged particles
discovery, 66–67
flow, 67f
studying, 67f
Charges, 66, 78
atoms, balance, 411
balancing, 124
distribution, evaluation, 248
formal charges, 243–244
calculation, 244–245
sum, 244
molecules, relationship, 243–249
zero value, 124
Charles’s law, 282–284, 294, 296
usage, 283–284
Chemical behavior, patterns, 169t
Chemical bonding, visualization, 117
Chemical bonds, outer electron involvement, 101
Chemical change, 10, 18, 146
energy, changes, 12
example, 11, 14

1185
 heat/work, impact, 213
results, 10f
Chemical equations, 147–153, 172
balanced chemical equations, ratios (prediction), 149
energy changes, expression, 215
writing, 148
Chemical formula, 137
usage, 123
Chemical properties, 10, 18, 243
example, 11
Chemical reactions, 10, 18
classification, 154–155
energy profile, 375
heat energy
relationship, 223–227
release, 216
occurrence, 367, 367f
Chemical weapons, protection, 388f
Chemistry, 3, 18
energy changes, description, 369–370
examples, 3f
organic chemistry, 427–428
Chlorine
electron configuration, 103
electrons, levels/sublevels, 100
oxyanion formation, 121
properties, similarities, 103f
Chlorine gas (Cl2), 386
Lewis structure, 239f
magnesium, reaction, 157f
oxygen, reaction, 194
phosphorus, reaction, 195
Chromium, heat flow, 219
Cinnamon, aldehyde functional groups, 442f
Citalopram, 445, 446f
Citric acid, usage, 355
Cleaning solution, ammonia volume (determination), 305
Coal
combustion, 158f
energy, release, 224–225
fossil fuel, 269f
seam, 427f
sulfur, presence, 160f
Cobalt-58, half-life, 478

1186
Cobalt(II) nitrite, elemental barium (reaction), 412
Cocaine (C17H22NO4), 333–334, 357
crack cocaine, 333, 333f, 357
formula mass, 183
free-base form, 333, 340
presence, indication, 183f
Coefficients, 25, 147–148, 172
adjustment, 150
differences, 351–352
fractional, usage, 152
subscript, difference, 148f
usage, 189, 345f
Coffee
concentration, importance, 325
creation, 303–304
Coffee arabica,, 303
Coffee cup calorimetry, 219–222, 230
usage, 220f, 221–222
Coffee Hunting: Kenya (documentary), 303
Colligative properties, 313–318, 324, 326
trends, identification, 315
Color, 90–93
absorption, 92
line spectra, 91f
relationship, 90–91
Combined gas law, 284–285, 294, 296
usage, 285
Combustion, 172, 401, 417
fuels, value, 224t
usage, 184
Combustion reactions, 158–160, 169, 401–402
products, prediction, 159
Community mines, 55f
Complete ionic equation, 164, 172
Complex structures, skeletal structures (usage), 433f
Composition, 2, 4–5
similarities, 5f
Compounds, 5, 7f, 18, 57
atomic view, 8
covalent compounds, usage, 127
formula mass, calculation, 181
ionic bonds, 123–126
ionic, 123
ionization, 134

1187
 nonionic, dissolving, 133f
oxidation numbers, identification, 400
percent composition, finding, 181–182
properties, prediction, 277
Compressed gas, examples, 280f
Concentrated, term (usage), 305
Concentration, 326
by percent, 305–308
description, 305–312
equilibrium, relationship, 383–385
impact, 368
representation, square brackets (usage), 311
Condensation, 8, 18
products, drawing, 441
reaction, 441, 448f, 460
Condensed structures, 432–433, 460
conversion, 435
Lewis structures, 433
Conformations, 429–430
Conjugate acid, 339, 355, 358
Conjugate base, 339, 355, 358
basic characteristic, determination, 341
Constant-pressure conditions, 285
Contact time, fusion reaction condition, 489
Containment structure, 487
Contaminants
concentration, calculation, 308
monitoring (U.S. drinking water), 307t
Control rods, 487
Conversion factors, 48
multiple conversions, problems, 39
selection, 38
usage, 37
Conversions, density (usage), 44
Cooking oils, fatty acids, 443f
Cooling loop, 487
Copper
beta decay, 476
boiling/fermentation tanks, carbon dioxide production, 291f
bromine, reaction, 191
element, 6f
elemental copper, silver nitrate solution (metal displacement), 319f
pennies, usage, 67f
transition metals, 63f

1188
Copper chloride (CuCl2), naming, 125
Copper(II) chloride, potassium metal (reaction), 150
Copper(II) reduction, 407
Corn ethanol, debate, 209–210
Cosmic radiation, 475
Counted numbers, exact numbers, 33
Counted values, exact numbers, 33–34
Covalent bonding, 126–130
carbon/nonmetals involvement, 428–431
Covalent bonds, 137, 239, 255, 273
electrons, charges, 243f
formation, 126f
neutral atoms, involvement, 428t
nonmetals, involvement, 428f
structure complexity, 128
Covalent compounds, 137
atoms, naming order, 130
ionic compounds, distinguishing, 131–134
naming, 130–132
prefixes, usage, 130t
phosphorus/oxygen, presence, 129t
usage, 127, 128
Covalent double bond, 239
Covalent molecules, 239–243
Covalent networks, 277–278, 296
Covalent structures
condensed structures, 432–433
drawing, 431–436
representation, 239–240
skeletal structures, 433–436
Covalent triple bond, 239
Crack cocaine, 333, 333f, 357
acid-conjugate base pair, 340f
Crude oil, components, 434f
Cube, volume (equivalence), 40f
Cubic centimeter, measurement, 41
Cubic meter, size, 41
Cubic units, conversion, 41
Cupric ions, 118–119
Cuprous ions, 118–119
Curcumin, molecule, 444
Curie, Marie/Pierre, 473, 474f
Currencies, exchange rates (display), 37f

1189
Cyclic form, 449
Cycloalkanes, 436–438, 457, 460
formula, 438, 438f
skeletal structures, 438f
Cycloheptane, molecular formulas, 438
Cyclotron, 472, 472f
proton, collision, 473
Cys-Ala-Thr (tripeptide), drawing, 453
Cytosine, 454

D
d orbital, 97
d sublevel, 109
orbitals, 96f
partial filling, 101
Dalton, John, 58–59, 66, 73, 75
Davy, Humphry, 397, 413
Dead Sea, salt deposits/concentration, 306f
Decay series, 475–476, 493
products, prediction, 476
Decimal place, estimated value, 29
Decimal point
absence, 33
presence, 33
Decomposition, 171, 172
reactions, 154
DEET, amide functional groups, 445, 446f
Delta-minus (δ−), 254
Delta-plus (δ+), 254
Delta (Δ), usage, 213
Democritus, 57, 57f, 75
Denominator, unit conversions, 39
Density, 43–45, 48
definition, 43
ranking, 43t
usage, 43, 44
Deoxyribonucleic acid (DNA), 454–456, 458, 460
bases, pairings, 455
double helix structure, 455f, 456f
genetic code, structures, 454f
molecule
bases (pairing), hydrogen bonds (usage), 455f
matching strands, 455f

1190
 strand, structural backbone, 455f
Deoxyribose, 455
structural backbone, 455f
Detector, usage, 183
Deuterium, 489
Diameter, measurement, 29, 34
Diamond, carbon forms/networks, 278, 278f
Diatomic molecules (two-atom molecules), 5f, 127
formation, 127f
Diazepam, concentrations, 308
Diffraction grating, 91
Diffusion, 290, 290f, 294, 296
Dihydrogen sulfide (H2S), Lewis structure, 253
Dilute, term (usage), 305
Dilute solutions, preparation, 311
Dimensional analysis, 37–38
Dinitrogen tetroxide (N2O4), naming, 130
Dipeptide, 452
Dipoles, 256–258
dipole–dipole interactions, 273–274, 296
impact, 258
instantaneous dipoles, 276
Diprotic acids, 336
Dirty fuels, burning, 344
Disaccharide, 449, 460
maltose, 449f
Dispersion forces, 273, 276
Displacement reactions, 319–320
Dissociation, 133–134, 137, 169, 312
bases, 336
Dissolved compounds, presence, 132f
Distillation, 209f
Dollar bill surface, mass spectrum, 183f
Dopamine, 136, 136fe
amine functional groups, 445f
Dosimeter, 479, 480f, 493
Double bonds, 247, 457
covalent double bond, 239
electrons, sharing, 239
representation, 433
Double displacement, 153, 172
Double-displacement reaction, 154, 155, 168
Double-headed arrow, usage, 248

1191
Drace, Kevin, 55
Ductile, 272

E
E15 mixtures, 229
Edison, Thomas, 397
Effusion, 290, 294, 296
Einstein, Albert, 388
equation, 484–485
Ek’Balam, plaster sculptures, 145f
Electrical current, 66
Electrical potential energy, usage, 413f
Electric current, creation/production, 108f, 414f
Electric field, usage, 183
Electricity
production, 158f
water, conductor, 133f
Electric meter, energy measurement, 212f
Electrochemical cells (batteries), 66, 409, 415
chambers, 411
electron flow, direction (prediction), 410
oxidation/reduction, occurrence, 409f
Electrochemistry, 399, 417
Electrodes, 409, 417
immersion, 413
Electrolytes, 133, 324, 335
concentrations, 312–313
solutions, 133, 137, 312–318, 326
Electromagnetic radiation, 87, 109
Electromagnetic spectrum, 87–89, 109
Electron configurations, 99f, 100–103, 109
description, 97–103
identification, 106
periodic table, relationship, 103–107
representation, 98f
writing, 99, 102
Electronegative atom, 273
Electronegativity, 261
polar covalent bonds, relationship, 254–256
table (Pauling), 254t
Electronic geometry, 250–252, 252t, 259, 261
Electronic structure, basis, 104f
Electrons, 68t, 73–75

1192
 capacity, energy levels/sublevels (relationship), 96t
cloud, 68–69, 78
discovery, 66–67
energy absorption, 92
energy levels, 94, 94t, 97
principal quantum number representation, 94f
flow, 67
direction, prediction, 410
gain/loss, 74
inner electrons, 100
location, uncertainty, 93
loss, 118, 411
number, indication, 117f
orbit
Bohr model predictions, 92f
Bohr model proposal, 91f
orbitals, 93–97
outer electrons, 100–101
oxidation release, 411–412
particles, Bohr model description, 95
pi, 440
probable location, 94
sets, 250, 252
tetrahedron, X-shape, 251f
sharing, 127, 239
spin, 95
transfer, 409f, 414f
valence, 99–100
valence level, 100, 239
wave descriptions, 94f
wave nature, 93–94
waves, 440f
behavior, 94
Electroplating, 413, 417
electrical potential energy, usage, 413f
Elemental aluminum
iron(II) chloride, reaction, 412
production, 403
Elemental analysis, 184, 200
combustion, usage, 184f
Elemental barium, cobalt(II) nitrite (reaction), 412
Elemental forms
balancing order, 151
oxidation number, 400
Elemental iron, production, 403

1193
Elemental oxygen, existence, 5f
Elements, 5, 7f, 18
atomic view, 8
average atomic mass, 71
average mass, calculation, 72
colors, emission, 90f
families (groups), 104
identification, line spectra (fingerprints), 91
inner transition elements, 63
list, 62t
main-group elements, 63, 78
noble gases, reaction (absence), 65f
percent composition, 181
periodic table, 61–66, 78
presentation, 61f
transition elements, 63
valence configurations, 104
Empirical formula, 123, 131, 137
usage, 126
Empty orbitals, energy, 99f
Enchilada, ingredients (measurement), 27f
Endothermic change, 13, 18, 214–215, 230
Energy, 12–14, 18, 211–217, 230
ability, 211
absorption, 92, 224, 226
activation, 369
changes, 212, 225
example, 14
expression, 215
forms, 12
impact, 368–371
conservation, law, 216–217, 230, 493
diagram, 98f, 369–370, 369f
equilibrium position, relationship, 374
differences, 97f
heat, 12
kinetic, 12, 18, 211
levels, 91, 94–97
sublevels/electron capacity, relationship, 96t
measurements, 212f
potential, 12, 18, 211
range, 87f
release, 224
shells, 95
states, decline, 12f, 13

1194
 storage/release, 12f
sublevels, 94–97, 94t
energy differences, 97f
filling sequence (determination), periodic table (usage), 105f
subshells, 95
total change, measurement, 213
transfer, forms, 213
units, 212t
conversion, 212
wavelength, relationship, 89
English units, metric units, relationships, 28t
English yew, 2
Enthalpy
reaction (ΔHrxn), 225–227
usage, 227
Enzymes, action, 370
Ephedrine, 136f
Equations, balancing, 149–151
calculation, 150–151
elemental forms, balancing order, 151f
fractional coefficients, usage, 152
polyatomic ions, usage, 152
step-by-step process, 150f
strategies, 151–152
Equilibrium (equilibria), 366, 372, 389
addition, 383, 384f
arrow, 339
changes, prediction, 384
Le Chatelier’s principle, usage, 386
concentration, relationship, 383–385
constant (K),, 376, 387, 389
value, finding, 377
energy species, concentration, 373f
example, 374
position, energy diagrams (relationship), 374
pressure, relationship, 385–386
products, K values, 376f
temperature, relationship, 385
Equilibrium (equilibria) expressions, 376–382, 389
gas involvement, 382
solids, involvement, 379–381
solvents, involvement, 378
usage, 382
writing, 377
solvent, involvement, 378

1195
Equilibrium reaction, 339, 358, 372–375, 389
analysis, 373
process, forward/reverse directions, 373f
reactants/products, concentrations (change), 372f
Erlenmeyer flask, 350
Ester, 443, 445, 457, 460
Estimated digit, value, 34
Ethanol (C2H6O), 432
combustion, reaction enthalpy, 225–226
debate, 209–210
fuel value, 229
Lewis structure, 432f
Ether, 457, 460
oxygen-containing functional group, 441
Ethyl alcohol (alcohol), 209, 432
C-O-H bonds, 441f
Ethylene, 438, 439f
Ethylene glycol (antifreeze), 448
C-O-H bonds, 441f
water, mixture, 315
Ethylene oxide, behavior, 13, 13f
Ethyne, 439
Exact numbers, 34f, 48
multiplication, 35
usage, 33–34
Excess reagent, 194, 200
Excited state, 91
Exothermic change, 13, 18, 214–215, 230
reaction enthalpy, 225
Exothermic nuclear reactions, 484
Expanded octet, 240, 261
Exponent
indications, 25
negative exponent, 26
Exponential notation, 345
Exposed rock surfaces (dating), Be-10/Al-26 (usage), 483
Extensive property, 224
Extraction yield, 325, 325f

F
f orbital, 97
F-S bond, 255
f sublevel, 109

1196
 orbitals, 96f
partial filling, 101
Factor-label method, 37
Fahrenheit scale, 45, 45f, 48
reporting, 46
Fair Sentencing Act, 357
Families (groups), similarities, 104
Faraday, Michael, 397
Fennell, Jared, 179
Fermentation, 209f
glucose fermentation, gas stoichiometry, 292
process, 291
Ferric ions, 118–119
Ferrous ions, 118–119
Fiber (roughage), 450
Filling sequence rules, 105
Filtration, 7
Fire
energy, release, 224f
light production, reason, 92
retardants, 65, 65f
Fissile, term (usage), 486
Fission, 486–488, 493
nuclear fission, waste, 488
reactor design, 487–488
Flame, element color, 90f
Flame test, 90
Floatation, example, 44, 44f
Fluorescence, 93
Fluoride ions, negative charge, 75
Fluorine
bonding, occurrence, 127
diatomic molecule, 127
electron configuration, 103
electron gain, 75, 119f
properties, similarities, 103f
Fluorine gas
sulfur, reaction, 197
uranium, reaction, 195
Food, energy content label, 212f
Forces of attraction, 271, 271f
Ford, Henry (Model T), 209f
Formal charges, 243–244, 261
calculation, 244–245

1197
 sum, 244
Formaldehyde (CH2O)
electronic/molecular geometries, 250f
Lewis structure, drawing, 242
Formula mass, 181–184, 200
calculation, 181f
chemist measurement, 182–184
Formula unit, 123, 131, 137
Forward directions, activation energies (determination), 374–375
Forward reaction, 339f
occurrence, 372
reverse reaction, equivalence, 373f
Fossil fuels, 397
collection/transportation, 269
fuel source, 278
Fractional coefficients, usage, 152
Free-base form, 333, 340
Freezing, 8, 18
Freezing point
calculation, 315
depression, 314, 326
Frequency, 87–88, 109
speed/wavelength, relationship, 88f
symbolization, 88
wavelength/light energy, relationship, 89
Fuel cells, 414, 417
Fuel mixture, heat/work production, 213f
Fuel value, 224–225, 230
usage, 224–225
Fukushima Daiichi
accident, risk, 488
radiation levels, study, 481
responders, 479f
tsunami, impact, 469–470, 492
Fullerene, 260
Functional groups, 436–446, 457, 460
amine functional groups, 445f
aromatic functional groups, identification, 440
hydrocarbon functional groups, 436–440
identification, 444–445, 446
nitrile functional group, composition, 445
nitrogen-containing functional groups, 445–446
oxygen-containing functional groups, 441–445
Fundamental units, 28

1198
Fusion, 486, 488–491, 493
occurrence, parameters, 489f
reaction, 489f
conditions, 489

G
Gallium, usage, 108, 108f
Gamma (γ)
decay, 475, 493
radiation, 473, 478
rays, 475, 478
Gas centrifuge, UF6 separation, 487f
Gases, 8, 18, 279-293
amount, increase, 289
combined gas law, 284–285
compressed gas, examples, 280f
description, 279–281
gas-law problems, 284
ideal gas, 279
law, 286–287
involvement, equilibrium expressions, 382
lamps, color production, 90f
laws, 282–289, 296
molecular law, 289
mixtures, 287–288
partial pressures, usage, 287–288
mole, volume occupation, 285
noble gases, 65f, 78
particles, movement, 279f
pressure, 279
change, particle increase (impact), 289
determination, equilibrium expression (usage), 382
increase, volume increase (impact), 289
measurement, 279–281
spread (diffusion), 290f
stoichiometry, 291–293
glucose fermentation, 292
natural gas, 292
storage, 295
stove, flame (results), 10f
transition, 9f
transportation/storage, requirements, 295f
volume, 282
Gasoline (C8H18), 229, 436f

1199
 explosion, 215
percent composition, 181–182
potential energy
conversion, 213
release, 212
total energy, release, 213
Gas stoves, methane gas combustion (usage), 159f
Gauge pressure, 281, 296
Geiger counter, 479, 479f, 493
Genetic code, duplication, 456
Geology, radioactive nuclide usage, 482–483
Geometric isomers, 449
Glass prism, usage, 90, 90f
Global energy
consumption, increase, 492f
future, 492
Glucopyranose molecules, 449f
Glucose
conversion, 449f
fermentation, gas stoichiometry, 292
forms, 449
Glutamine, connection, 452
Glycine, alanine (bonding), 452
Glycosidic bond, 449
Gold
film, particles (impact), 68
foil experiment, 73
mercury, similarities, 56
sample, 343
small-scale gold mining, mercury contamination, 55–56
transition metals, 63f
valuation, 404
Graduated cylinder, usage, 43
volume measurement, 31f
Grams
conversion, 185–187
gram-to-gram conversion, 191
gram-to-gram questions, 190–191
moles (relationship), balanced equation (usage), 190
Graphite, carbon forms/networks, 278, 278f
Graphite control rods, 487
Gravimetric analysis, 321–322, 326
four-step procedure, 321f
Green light, wavelength, 88

1200
Ground state, 91
Groundwater, diazepam concentrations, 308
Group, 1A elements, +1 ions (formation), 118f
Group, 2A elements, +2 ions (formation), 118f
Groups (families), 62
Grubbs catalyst, 425, 436
Grubbs, Robert, 425–426, 425f, 459
Guanine, 454–455
Guitar, string (vibration), 94f

H
H+ ions
concentration, 344–346, 356
determination, 346
pH, usage, 348
production from acids, 167
relationship to OH− concentration, 345f
Haber-Bosch process, 365–366, 386
ammonia production, 387
Haber, Fritz, 365–366, 386, 388
Half-lives, 476–478, 493
usage, 478, 483
Half-reactions, 407–411, 417
electrons, transfer, 409f
writing, 408
Halogens, 65, 78, 335
–1 ions, formation, 120f
behavior, 117
reactive nonmetals, 65f
Heat capacity, 217–219, 227, 230
equation, 218
mass, involvement (absence), 218
usage, 219
Heat energy (q), 12, 211–217, 230
changes (prediction), reaction enthalpy (usage), 225–226
chemical reactions, relationship, 223–227
flow, 214f
gain/loss (measurement), coffee cup calorimetry (usage), 221
measurement, 38
measurement, calorimetry (usage), 219–223
production, 213f
release, 216
reliance, 211f

1201
 specific heat, 217–219
temperature, relationship, 217–223
transfer, 211f, 213
travel, 215f
Heisenberg, Werner, 93–94
Helium
electron level/sublevel, 98
nucleus, expected mass, 484
valence energy level, filling, 104
Heme units, 454
Hemoglobin, structure, 454f
Heptane, molecular formulas, 438
Hertz (Hz), 88, 109
Heterogeneous mixture, 6f, 7, 7f, 18
Hieroglyphs, 145
evidence (Palenque), 171
High-energy neutron, nitrogen-14 (collision), 473
High-energy particles, collision, 472f
Highest-shell electron configuration, identification, 106
Hindenburg airship, hydrogen/oxygen (reaction), 188
Hiroshima, decimation, 486f
Homogeneous mixture, 6f, 7, 7f, 18
Household items, acidic/basic properties, 335f
Human Genome Project, 456
Hund’s rule, 98, 99f
Hurley, Susan, 333–334
Hydrazine (N2H4), usage, 128, 128f
hydro- (prefix), 134
Hydrobromic acid (HBr), 134
monoprotic acid, 336
Hydrocarbons, 159t, 415, 448, 457, 460
functional groups, 436–440
involvement, 402
types, 439
Hydrochloric acid (HCl), 134
aluminum metal, reaction, 406
aqueous solution, reaction, 154–155
calcium, half-reactions, 408
dissociation, 376
H+ value, 347
iron (reaction), moles (calculation), 188–189
moles, calculation, 309
monoprotic acid, 336

1202
 neutralization, titration (usage), 351
sodium hydroxide (NaOH), combination, 196
volume, calculation, 311
water/ammonia, reaction, 337f
zinc metal, reaction, 405f
Hydrocyanic acid, dissolving, 379
Hydrofluoric acid (HF), 134
Hydrogen
atoms, carbon (bonding), 433
bombs, 489
bonding, 274–275
impact, 275f
cations (H+), production, 170
diatomic molecule, 127
electron, level/sublevel, 97
fuel cells, 414f
isotopes, 69f
line spectrum
explanation, 93
transitions, 92t
nitrogen, reaction (balanced equation), 149
nonmetal type, 64
oxygen, reaction (chemical equation), 147, 148f, 187
reaction, 147f
valence, fulfillment, 239
Hydrogen atoms
bonding, 335
oxygen atom, binding, 5f
transitions, visible light production, 92
valence level, 127–128
Hydrogen bonds, 273, 296
formation, 275
impact, 275f
strength, 275
usage, 455f
Hydrogen cyanide
electronic/molecular geometries, 250f
properties, 274t
Hydrogen fluoride (HF)
dissolving, 372
equilibrium, 383
Hydrogen gas
acetylene, reaction, 227
constant-pressure conditions, 285

1203
 impact, 226
manufacture method, 382
production, 343f, 405f
temperature/volume, 287
Hydrogen sulfate ion (HSO4-), charge, 400
Hydroiodic acid (HI), 134
Hydronium concentration, pH determination, 349
Hydronium ion, 335, 358
Hydroxide bases, 167t
Hydroxide (OH−)ion
charge, 243–244
production, 167, 338, 378
Hypertonic solution, 318
water, exit, 318f
Hypobromite (BrO−), equilibria equations, 341
Hypochlorous acid (HClO), Lewis structure, 240
Hypothesis, 15, 18
Hypotonic solution, 318

I
Ibuprofen, Lewis structure, 434–435
-ic acid (suffix), 134
Ice
melting, 226
volcanic heat, impact, 271f
pack, heat energy flow, 214f
Ideal gas, 279, 296
law, 286–287, 294, 296
Imaging therapy, 482
Infrared (IR) energy, emission, 93f
Infrared (IR) radiation, 88
Ingredients, measurement, 27f
Inner electrons, 100
identification, 101
Inner transition elements, 63
Instantaneous dipoles, 276
analogy, 276
Intermolecular forces, 273, 276–277, 296
types, 276t
International Thermonuclear Experimental Reactor (ITER), 490, 490f
Inter- prefix, 273
Intra- prefix, 273

1204
Iodine-131 decay, beta emission/half-life, 483
Ionic bonds, 123–126, 137, 239, 255
Ionic compounds, 123, 137
charge, zero value, 124
covalent compounds, distinguishing, 131–134
dissociation, 133f
formulas
prediction, 123–125
writing, 124
melting points, 272t
naming, 125–126, 131–132
ordered lattices, formation, 272f
production, 124
solubility products, 380t
types, 125t
Ionic equations, 168, 172, 312
Ionic lattice, 123, 137
three-dimensional framework, 123f
Ionic solid, dissolving, 133f, 312f
Ionic substances, 272
Ionizing chamber, usage, 183
Ions, 74
charge, 74, 122f
finding, 75
prediction, 120
spread, 248
subatomic particles, relationship, 75
concentrations, determination, 313
dissociation, concentration (determination), 313
electron configuration, 102–103
writing, 102
formation, 74–75, 74f, 117–120, 118f, 120f
formulas, 122f
Lewis structure, four-step technique, 253
monatomic ions, 120
names, 122f
information, gathering, 122
naming, 119t
negative change (anions), 119–120
packing, 123f
polyatomic ions, 120–121
positive charge (cations), 117–123
prediction, 120
production, metal-nonmetal reactions, 157
summary, 121–123

1205
 water, aggregation, 317f
Iron
metal, volume, 44
molten (liquid) iron, pouring, 273f
nail, nickel(II) chloride solution reactions, 404f
oxygen gas, reaction, 193
reaction, 188f
moles, calculation, 188
transition metal, 63f
Iron(II), 119
reduction, 405
Iron(II) chloride
elemental aluminum, reaction, 412
formation, 189
Iron(III), 119
ions, formula, 124
Iron(III) oxide, formation, 149, 401f
Iron oxide (Maya usage), 149
Isoelectronic, term (meaning), 102, 109, 118, 119f
Isomers, 429, 438f, 460
drawing, 430–431
geometric isomers, 449
Lewis structures, 430
Isopropanol (rubbing alcohol), C-O-H bonds, 441f
Isotonic solution, 318
Isotopes, 69, 78, 471, 493
identification, 471
weighted average, 71

J
Jáidar, Yareli, 172f
Joules, 89, 230
defining, 212

K
Kelvin scale, 45–46, 45f, 48
reporting, 46
Kelvin temperature scale, 294
Kerosene, 436f
Ketones, 442, 457, 460
Kiefer, Adam, 55, 56
Kilopascal (kPa), 281
Kilowatt hour (kWh), 212

1206
Kinetic energy, 12, 18, 211, 227, 230
addition, 271
King Tikal, crypt, 145f
Kraner, James, 23–24, 37
Krypton, noble gas, 100

L
Lambda (λ), 87
Lanthanide series, 63
Large-scale batch reactions, manufacturer usage, 184f
Lavoisier, Antoine, 57
Law of conservation of energy. See Energy
Law of conservation of mass. See Mass
Law of conservation of matter. See Matter
Lead(III), presence (determination), 322
Le Chatelier’s principle, 382–387, 389
equilibrium, addition, 384f
usage, 386
Length measurements, 37
Lewis dot symbols, 117, 137
Lewis structures, 137, 239, 275f, 440
complexity, 457
condensed representations, 433
conversion, 434–435
drawing, 241–246
steps, 241f
four-step technique, 253
interpretation, 129
selection, 247
valence electrons, identification, 240
Lewis symbols, 117
Lidocaine, nitrogen-containing functional groups, 446
Light
energy, wavelength/frequency (relationship), 89
packet, 87, 87f
production, 92
separation, 91
speed (c), 88
travel, photons, 87f
Light beer, alcohol volume/concentration, 305
Lighter fluid, 436f
Lime cycle (Maya), 171–172
Lime kilns, 154f

1207
Limestone, conversion, 154f, 171f
Limiting reagents, 200
calculations, 193–195
consumption, 194f
finding, 194–195
Limonene, 439f
Linear alkanes, formula, 438f
Linear form, 449
Linear geometry, 257, 260, 261
Line spectrum (line spectra), 90–93, 109
color, relationship, 90–91
patterns, 91
Lipid bilayer, 316
Liquids, 8, 18, 272–278
transition, 9f
Liter, 48
cubic decimeter, 40f
volume unit derivation, 40
Lithium
electron configuration, 103
electron level/sublevel, 98
ions
concentration, 313
positive charge, 74
nuclides, 471
Lithium carbonate
effects, 115
ionic compound, usage, 129
molecular geometry, 253
Lithium chloride (LiCl), classification, 277
Lithium hydroxide (LiOH)
base, 336
neutralization (HCl solution usage), titration (usage), 351
nitric acid, reaction, 168
Lithium-ion batteries, 416, 416f
Lithium sulfate, ions (concentration), 313
Litmus, 349, 349f, 358
paper, 349
Logarithmic notations, 346t
London dispersion forces, 276, 296
causes, 276f
Lone pairs, 239, 248, 261
Lower-energy species, favoring, 373

1208
M
Magnesium
alkaline earth metal, 64f
chlorine gas, reaction, 157f
electronic configuration, 118
ions, production, 402
metallic magnesium, oxygen gas (reaction), 156f
Magnesium acetate, osmotic pressure (contrast), 319
Magnesium chloride (MgCl2), naming, 125
Magnesium oxide (MgO), moles (formation), 189
Magnetic bottle, 490
Magnetic coil, 490f
Magnetic fields, cyclotron usage, 472f
Magnets, interactions, 274
Main-group elements, 63, 78
Main-group metals, discovery, 413
Main-group numbers, 106f
electron number indication, 117f
1A-8A, valence electron numbers, 241f
Malleable, 272
Maltose, disaccharide, 449f
Manganese(II) carbonate, equilibrium expression, 380–381
Manufacturing reaction, 179f
Mass
atomic mass unit (u) (amu), 66
conservation, law, 57, 58f, 78
application, 484
usage, 58
defect, 484–485, 493
description, 185
kilograms, 38
mass-energy equivalence, 484
measurement, 30
balance, usage, 43f
reaction, 58f
spectrometer, forensic chemist usage, 183f
spectrometry, 183, 200
volume
conversion, 43–44
relationship, 43–45
Mass number, 69–71, 78, 471
atomic symbols, writing, 70
display, 472
sum, 472f

1209
 usage, 70
writing, 471f
Mass/volume percent, 306–307, 326
Materials
densities, ranking, 43t
specific heat, 218t
Matter, 7f, 18
classification, 7
conservation, law, 484–485, 493
describing, 4–11
phases, 271
states, 8–9, 18, 271t
Maya
lime cycle, 171–172
lost cities, 145
Measurement, 25–36
accuracy, 25
fundamental (SI) units, 28t
precision (indication), significant digits (usage), 32
quality, description, 29–30
significant digits, number (determination), 32–33
units, 27–29, 48
volume measurement, beaker/graduated cylinder (usage), 31f
Medicine, radioactive nuclide usage, 482
Melting, 8, 18
points, 271, 272t, 314
Mendeleev, Dmitri, 61, 76
Mercury, 77f
contamination, 55–56
gold, mixture, 55f
volume, 44
Metal hydroxides, 167
Metallic magnesium, oxygen gas (reaction), 156f
Metallic substances, 272–273
Metalloids, 63–64, 78
semiconductors, 64
silicon, importance, 64f
Metal-organic frameworks (MOFs), 295, 295f
Metals, 63–64, 78
acid-metal reaction (prediction), activity series (usage), 406
acids, reactions, 343, 405–407, 415
metal cations/hydrogen gas, production, 343f
acids, reactivity, 406t
alkali metals, 64, 78

1210
 alkaline earth metals, 64, 78
atoms, close-fitting arrangements, 272f
cations, production, 343f
displacement reactions, 319–320, 326, 402–405, 417
occurrence, 403f
prediction, 405
ductility, 272
heat loss, 220
malleability, 272
melting points, 273, 273t
metal-nonmetal reactions, 157
nonmetals
ionic bond, 131
reactions, 156–158
oxides, 401f
p-block metals, ions (formation), 118f
reaction products, prediction, 157
specific heat, 221
measurement, coffee cup calorimetry (usage), 220f
substances, 134
transition metals, 78
ions, formation, 118f
types, 63f
water
reactions, 405–407
reactivity, 406t
Methamphetamine, 136f
Methane (CH4)
burning, chemical equation, 148
covalent compounds, 272
gas
combustion, 159f
production, hydrogen gas (impact), 226
limiting reagent, consumption, 194
mixture, 290
natural gas component, 158, 251
oxygen, combination, 10f, 148
properties, 274t
tetrahedral geometry, 251
Methylenedioxypyrovalerone (MDPV)
identification, 23–24
levels, 47
presence, 25
Methylthiolate anion (CH3S−), Lewis structure (drawing), 246
Metric prefixes, 28t

1211
Metric system, 28
Metric units, English units (relationships), 28t
Milliliter, measurement, 41
Millimeters of mercury (mm Hg) (torr), 280, 296
Minerals, ionic compounds, 401f
Mixtures, 6
analysis, mass spectrometer (usage), 183f
atomic view, 8
blending, 7
separation, 7
Molarity (M), 309–312, 326
calculation, 309
Molar mass, 185, 200
Molecular bromine, breakage, 377
Molecular compounds, 293
Molecular dipoles, 256, 261
δ+/δ− symbols, 273f
identification, 257
Molecular equations, 160–161, 172, 312
acids/metal, reactions, 343
Molecular formula, 128, 137
Molecular geometry, 250–252, 252t, 261
Molecular substances, 273–277
Molecules, 5, 131
charge, relationship, 243–249
collision, 367f
composition, 429f
covalent molecules, 239–243
dipoles
impact, 274f
presence, 256–258
forces, 273f
formation, covalent compounds (usage), 127
identification, net dipole (usage), 257
Lewis structures, drawing, 257
molecular/electronic geometry, determination, 253
net dipoles, 273f
oxidation numbers, 399t
polar molecules, 254–258
properties, dipoles (impact), 258
ratio, 188f
shapes, 237, 249–253, 430
variation, 429f
Moles, 184–185, 200

1212
 calculation, 310
molarity/volume, usage, 309
concept
atom/gram connection, 185f
usage, 187–196
conversion, 185–187
coefficients, usage, 189
hub, 291f
display, 185f
equivalence, 185
grams (relationship), balanced equation (usage), 190
intermediate, function, 322f
map, 192f, 291
ratio, 187, 188f
relationship, balanced equation (usage), 189
volume occupation, 285, 285f
Molybdenum-100, protons (impact), 473
Monatomic ions, 120
Monomers, 447, 447t, 460
Monoprotic acids, 336
Moonshine, usage, 209f
Mousetrap, energy, 13, 13f
MTBE (production), condensation reaction (usage), 441
Multiple conversions, problems, 39–40
Multiplication, scientific notation (usage), 26–27
Multiplier, 25
usage, 26t
Mushroom cloud (hydrogen bomb), 489f
Musk, Elon, 85

N
N-H bonds, 453–454
n-type semiconductor, 108
Nanocar
building, 260
features, 260f
NanoCar Race, 237
Nanolime, 172
Natural gas
combustion, 158f, 193f
energy release, 225
gas stoichiometry, 292
importance, increase, 269
methane, component, 158

1213
Neon
electron configuration, 102
isoelectronic relationship, 119f
noble gas, 100
sodium, similarity, 118f
Net dipole, 256, 273f
absence, 258f
determination, 257
usage, 257
Net ionic equation, 164, 172
acids/metal, reactions, 343
solution stoichiometry, 320
Neutral atoms, covalent bonds, 428t
Neutral compound, oxidation numbers, 400
Neutralization reactions, 168, 172, 341–342, 359
ionic equation/net ionic equation, 342
Neutrons, 68, 68t, 78, 471f
proton-neutron ratios, 485t
transformation, 475f
Nicholson, William, 397
Nickel
activity series, 407
electroplating apparatus, 413f
specific heat, 220
Nickel(II) chloride, solution, 403
iron nail, reaction, 404
Nicotine
acid salt/free base, 340f
conjugate base form, 340
Nitrate, ionic compound production, 124
Nitrate ion (NO3-), resonance structures, 248
Nitric acid (HNO3)
aluminum metal, reaction, 193
analysis, titration (usage), 352
dissociation, 377
ionization, 167
lithium hydroxide, reaction, 168
monoprotic acid, 336
zinc, reaction, 408
Nitrile functional group, composition, 445
Nitriles, 457, 460
Nitrite ion (NO2-)
electronic/molecular geometry, 251f
electron sets, 251

1214
 Lewis structure, 247, 247f
Nitrogen
−3 ions, formation, 120f
electrons, donation, 337f
fixation, 365
formal charges, 244f
hydrogen, reaction (balanced equation), 149
nitrogen-14, high-energy neutron (collision), 473
nitrogen-containing functional groups, 445–446, 457
nonmetal type, 64
Nitrogen dioxide (NO2), naming, 130
Nitrogen monoxide (NO), 386
Nitrogen oxides (NOx), water (reaction), 344
Nitrogen trichloride (NCl3), Lewis structure (drawing), 241–242
Nitrosyl chloride (NOCl), equilibrium, 386
Noble gases, 65, 78
behavior, 117
color, 90f
electron configurations, 100t
notation, 100
reaction, absence, 65f
Nonionic compounds, dissolving, 133f
Nonmetals, 63–64, 78
atoms, covalent bonds (formation), 126
covalent bonding, 428–431
covalent bonds, 273, 428f
elements, oxygen (reaction), 159
metal-nonmetal reactions, 157
metals, reactions, 156–158, 401
nonmetal-nonmetal bonds, 126–127
nonmetals, covalent bond, 131
oxides, 358
impact, 344
oxyanions, resonance structures (drawing), 248
periodic table presence, 64
reaction products, prediction, 157
reactive nonmetals, 65f
Nonzero digits, significance, 32, 33
North Dakota boom, 269
N-terminus, 452, 452f
Nuclear chain reaction, sustaining, 487
Nuclear changes, 471–473, 491
Nuclear chemistry, 471, 493
Nuclear equations, 493

1215
 alpha decays, 474
beta decays, 475
completion, 473
mass/atomic numbers, display, 472
Nuclear fission, 491
waste, 488
Nuclear force, 484, 493
Nuclear fuel, spent rods (storage), 488f
Nuclear fusion reaction, control, 491
Nuclear medicine, 482, 493
Nuclear power
fission/fusion, 486–491
Nuclear power plants, uranium (usage), 70
Nuclear reactions, 471–473, 493
energy changes, 483–486
exothermic nuclear reactions, 484
Nuclear reactor, schematic, 488f
Nuclear structure (determination), atomic/mass number (usage), 70
Nucleons, 471, 493
binding energy per nucleon, 485, 485f
Nucleotide, 455
base, connection, 455f
Nucleus (nuclei), 78
composition, 471f
discovery, 67–69
electrons, orbit, 73f
fission/fusion, 491
review, 471
stability, 485f
Nuclides, 471, 493
decay, 477f
half-life, 476–477
radioactive nuclides, half-lives, 477t
stability, 485–486
nu (ό), frequency symbol, 88
Numbers/units, multiplication, 37f

O
O-H bonds, 453–454
Ocean water, dissolved compounds (presence), 132f
Octane (C8H18)
fuel value, 229
gasoline component, 159

1216
 molecule, composition, 128f
structure, 430, 430f
Octet
electron pairs, 239
expanded octet, 240, 240f
fulfillment, 246
Octet rule, 99–100, 109, 117
atoms, stabilization, 239
exceptions, 240
fulfillment, 239f
covalent compounds, usage, 128
main-group elements, impact, 117
OH− concentrations, 344–346
determination, pH (usage), 348
H+ concentration, relationship, 346
OH− ions (production), bases (impact), 167
Oil
crude oil, components, 434f
fossil fuel, 269f
shortage (1973), 209
OPEC, oil shipment blockage, 209–210
Orange juice, dilution, 311
Orbitals, 95, 109
composition, 95f
empty orbitals, energy, 99f
representation, 97
Orbits, 91
Ordered lattices, formation, 272f
Organic chemistry, 427–428, 460
Osmotic pressure, 316–318, 326
analogy, 317f
solutions, separation, 318
-ous acid (suffix), 134
Outer atoms, octet fulfillment, 241, 242, 246
Outer electrons, 100–101
identification, 101
involvement, 101
Oxidation, 172, 399–401, 417
electrons, loss, 411
half-reaction, 407
numbers, 399–401, 399t, 417
identification, 400–401
occurrence (anode location), 409

1217
Oxidation-reduction (redox) reaction, 156, 172, 397, 407, 417
applications, 413–414
balancing, 411–412
battery power, 409f
net ionic equation, 408
types, 401–407
Oxide ion, electron configuration, 102
Oxides, formation, 401–402
Oxidized, term (usage), 156
Oxidizing agent, impact, 399
Oxyacids, 134–135, 137
Oxyanions, 121
acids, naming, 135
formation, 121
Oxygen (O2)
−2 ions, formation, 120f
charcoal, reaction, 59, 59f
electrons
donation, 337f
gains, 102, 119f
excess reagent, 194f
formal charges, 244f
gas, iron (reaction), 193
hydrogen, reaction (chemical equation), 147, 148f
methane, combination, 10f
molecule, combination, 147f
nonmetal, type, 64
oxygen-containing functional groups, 441–445
oxygen-containing groups, 443–445
presence, 129t
reaction, 156
sulfur, reaction, 160f
tank
ideal gas law, usage, 286
usage, 282f
transport, 454f

P
p-block metals, ions (formation), 118f
p-n junction, 108
p orbital, 97
p sublevel, 95–96, 109
filling, 100f
orbital composition, 95f

1218
p-type semiconductor, 108
Pacific yew, 1
Taxol, discovery, 14
Palladium compound, usage, 199
Paradigm, 15
Partial pressures, 296
usage, 287–288
Particles
arrangement, 9
charged, 66–67
conversion, 185–187
high-energy, collision, 472f
interactions, 271–272
absence, 286
movement, 279f
passage, 68f
subatomic, 66, 68t, 78
velocity/location, 93
Parts per billion (ppb), 307–308, 324, 326
concentrations, 308
Parts per million (ppm), 307–308, 324, 326
concentrations, 308
Pauling, Linus, 254, 254f
Pennies, composition, 413
Peptides
bonds, 452–454, 452f, 460
chains, drawing, 453
types, 452
Percent by mass, 305, 326
Percent by volume, 305, 326
Percent composition, 181–184, 200
chemist measurement, 182–184
usage, 182
Percent concentration, usage, 305, 306
Percent yield, 196–198, 200
calculation, 197
finding, 197
Period, 61, 78
Periodic table, 78
average atomic masses, 181f
electron configuration
identification, 106
relationship, 103–107
elements

1219
 average atomic mass, 71
presentation, 61f
filling sequence rules, 105
groups (families), 62
highest-shell electron configuration, identification, 106
left-hand/right-hand blocks, main-group elements, 63
left side, 64
navigation, 65
numbering systems, 63f
organization, 103
electronic structure basis, 104f
regions, 63
right side, 64
rows, numbering, 63f
usage, 105f
Periodic, term (usage), 61
Phase, 153, 271
changes, 9
notations, 153
symbols, 153t
Phenolphthalein, 349, 358
addition, 352
pH, 346–356
indicators, 349, 349f, 358
meter, 350, 350f, 356
paper, 349, 349f
scale, 346–349, 347t, 358
decimal point spaces, 348f
determination, H+ (usage), 347
Phosgene (COCl2), structure (drawing), 247
Phosphate
buffer, composition, 354f
linker, 455f
Phosphate ion, resonance structures, 248f
Phosphoric acid (H3PO4)
dissociation, 337
triprotic acid, 336
Phosphorus
-3 ions, formation, 120f
atom, octet expansion, 240f
chlorine gas, reaction, 195
electrons, levels/sublevels, 100
phosphate formation, 121
presence, 129t

1220
Phosphorus pentachloride (PCl5), naming, 130
Phosphorus trichloride (PCl3), naming, 130
Photoelectric effect, 91
Photons, 87f, 109
energy, 89
Photosynthesis, 427, 460
impact, 427
Phthalates, concentration, 308
Physical changes, 10, 18, 226
description, enthalpy (usage), 227
energy, changes, 12
example, 11
Physical properties, 9, 18, 43
example, 11
Pi electrons, 440
Planck, Max, 89
Planck’s constant, 89
Plant/animal life, atomic building blocks, 64f
Plant/animal remains, carbon-14 dating, 482–483
Plasma, 490, 493
containment, 490
Plastics, 278, 447–448, 457, 460
composition, 448f
covalent bond chains, 278
Platinum
atomic number, 69f
cathode, aqueous platinum(II) chloride solution immersion, 411
valuation, 404
Platinum(II) nitrate, single-displacement reaction, 412
Plum pudding model, 68, 68f, 73
Plutonium
nuclear fission, 487
plutonium-241, decay, 476
pOH values, 347t
Polar bonds, 254–258
addition, 257
Polar covalent bonds, 261, 273
δ+/δ− symbols, 273f
electronegativity, relationship, 254–256
location, 255
Polar molecules, 254–258
melting/boiling points, 274
Polonium

1221
 alpha decay, nuclear equation, 474
polonium-210, alpha decay, 486
Polyatomic ions, 120–121, 137, 243
Lewis structures, 243f
drawing, 245–246
naming, 121
oxidation numbers, 399t, 400
types, 121t
usage, 152
Polyester (polymer), 448, 448f
Polyethylene
plastic, 439f
usage, 278
Polymers, 277–278, 447–448, 447t, 458, 460
covalent bond chains, 278
repeating units, 278f
Polypeptide, 452
Polypropylene, composition, 448f
Polyprotic acids, 336–337, 358
Polysaccharides, 450, 457–458, 460
Polyvinyl chloride (PVC), usage, 278
Positive charges, negative charges (equivalence), 123–124
Potassium
electron configuration, 103
metal, copper(II) chloride (reaction), 150
reaction, 406f
Potassium carbonate (K2CO3), formula mass, 181
Potassium chloride, mass (presence), 307
Potassium chlorine gas, reaction, 194
Potassium hydroxide, dilution (molarity determination), 311
Potassium sulfate (K2SO4), mass calculation, 186
Potential energy, 12, 18, 211, 230
decline, 13f
Pounds per square inch (psi), 281
Power loop, 487
water, heating, 488f
Powers of ten. See Ten
Precipitate, 163, 172
formation, 319
Precipitation reactions, 172, 319
balancing, 166
calcium/carbonate ions, combination, 320
prediction, 165, 166
solubility rules, usage, 165

1222
Precision, 29, 48
description, 30–36
meaning, 30
Prefixes, usage, 130
Pressure, 279, 296
change, 289f
gas temperature increase, impact, 289
constancy, 282
conversions, 281t
fusion reaction condition, 489
gauge pressure, 281
gauges, types, 281f
increase, 289
measurement, 279–281
osmotic pressure, 316–318, 326
partial pressures, usage, 287–288
standard pressure, 280
standard temperature and pressure (STP), 285
temperature, correlation, 289
Principal quantum number, 94, 109
electron representation, 94f
Prism, white light (passage), 90f
Process chemist, role, 179
Process chemistry, 199
Process development, 179–180, 199
Products, 147, 172
concentrations, 376
change, 372f
equilibrium, favoring, 376
theoretical yield, 196–198
total mass, 148f
Propane (C3H8)
formula, 159
mixture, 290
tank, gas containment, 279f
Properties, 9–11
chemical, 10
physical, 9
Propylene glycol [CH2(OH)CH(OH)CH3], addition, 435
Prosthetic limbs, strength, 43f
Proteins, 450–454, 458, 460
Protium, 69f
Protons, 68, 68t, 78, 414, 471f
collision, 473

1223
 donors/acceptors, 337
exchange membrane, 414
H+ ion, 336
number, impact, 69
proton-neutron ratios, 485t
Pure substances, 6, 7f, 18
Pure water
conductor, 133f
saltwater solution, separation, 316f
Pyridine, 440, 440f

Q
q (see heat)
Quantum mechanics, 93
Quantum model, 73, 78, 93–97, 109
Qumran, Dead Sea scrolls (carbon dating), 482f

R
Radiation, 473
alpha, 473
beta, 473. 478
cosmic, 475
energies, semiconductor counter measurement, 480f
exposure
levels, 480–481, 481t
exposure, health effects, 478–481
gamma (γ), 473, 478
measurement, 479–480
therapy, 482
Radioactive decays, 473, 493
series, 475–476
types, 474–475
Radioactive nuclei (seeds), 482
Radioactive nuclides
contamination, 487
half-lives, 477t
uses, 482–483
Radioactivity, 473–483, 493
Radiocarbon dating, half-lives (usage), 483
Radium
particles/energy, release, 474f
-224, decay, 476
-228, beta decay, 475

1224
Radon, noble gas, 100
Rare earth metals, 63
Reactants, 172
concentration, change, 372f
favoring, 373
K values, 376f
total mass, 148f
Reaction energy
absorption/release, 224
diagrams, 369–370, 369f, 389
interpretation, 371
usage, 373
Reaction rates, 367–371, 389
activation energy dependence, 370
concentration dependence, 370
concentration/temperature increase, impact, 368, 368f
energy, changes (impact), 368–371
variation, 368f
Reaction(s)
categories, 153f
chemical, 10, 18, 154–155
classification, 153–155
decomposition, 154
double-displacement, 154, 155
enthalpy (ΔHrxn), 225–227, 228, 230
usage, 225–226
equilibrium, 339
forward, 339f
occurrence, 372
half-reactions, 407–411
heat, finding
bomb calorimetry, usage, 223
coffee cup calorimetry, usage, 221–222
metal displacement, 319–320
neutralization, 168, 341–342
oxidation numbers, identification, 400–401
oxidation-reduction, 156
precipitate, production, 165f
products, energy (inclusion), 214
relationship, 184f
reverse, 339f
occurrence, 372
single-displacement, 154, 155
solution, 319–320
synthesis (combination reactions), 154

1225
 theoretical yield (prediction), stoichiometry problems (usage), 196
vessels, usage, 179f, 384
Reactive nonmetals, 65f
Reactor
loop, 487
Reagents
excess, 194
limiting, 193–195
Receptor sites, 136
Red light, wavelength, 88
Redox reaction, see Oxidation-reduction (redox) reaction
Reduced, term (usage), 156
Reducing agent, impact, 399
Reduction, 172, 399–401, 417
half-reaction, 407
occurrence (cathode location), 409
Reduction-oxidation reaction, see Oxidation reduction (redox) reaction
Renewable Fuel Standard (RFS), 209–210
Resonance, 247–249
Resonance structures, 247f, 248, 261, 440
charge distribution, evaluation, 248
indication, double-headed arrow (usage), 248f
phosphate ion representation, 248f
second/third covalent bonds, involvement, 248
Reverse direction, activation energies (determination), 374–375
Reverse reaction, 339f
forward reaction, equivalence, 373f
occurrence, 372
Rhodium (Rh), beta decay, 477f
Ribonucleic acid (RNA), 456, 458
Robusta coffee, 303
Rock formation, 12, 13f
Root name, 130
Roughage (fiber), 450
Rubbing alcohol (isopropanol), 441
Rust
formation, 156f
iron(III) oxide compound, 401f
Ruthenium
catalysts, usage, 459–460, 459f
decay, 477f
Rutherford, Ernest, 68, 73

1226
s sublevel, 95, 109
electrons, presence, 95f
Salicylic acid, presence/usage, 253
Saline solution
hospital IV pack, 306f
isotonic characteristic, 318, 318f
sodium chloride, gram calculation, 306
Salt (ionic compound), 168, 341, 355
bridge, 409f, 410
dissociation, 133f
road deposit, 314f
Salt/water mixture, 6f
Saltwater solution, pure water (separation), 316
Samarium-153 EDTMP, 478
Sample, heating, 9f
Sand/water mixture, 6f
Saturated, term (usage), 305
Scanning tunneling microscope, 237
Scanning tunneling microscopy, usage, 60f
Schou, Mogens, 114f
Science, foundation, 25–36
Scientific law, 15, 18
Scientific method, 14–15, 18
cyclic method, 14f
Scientific notation, 25–27, 48, 345
coefficient, usage, 345f
conversion, 26
involvement, 26
multipliers, usage, 26t
usage, 26–27
Scintillation camera, 482
Scintillation counter, 479, 493
Scrubbers, usage, 344
Self-ionization, 358
reaction, 345, 356
Semiconductors (metalloids), 64
counter, 479, 493
energy measurement, 480f
Semiconductors, silicon (usage), 108f
Semipermeable membrane, 316
water molecules, movement, 316f
Serotonin, amine functional groups, 445f
Sievert (Sv), 480, 493
Signal flare, chemical change, 58

1227
Significant digits (significant measurements, 31, 48
addition/subtraction, 35
multiple calculations, 36
multiplication/division, 35
number, determination, 32–33
usage, 32, 34–35
Silicon
electronic device component, 99
metalloid element, 64f
Silicon dioxide (SiO2), mass calculation, 182
Silver
metal, single-displacement reaction, 412
transition metal, 63f
Silver acetate solution, ions (presence), 165
Silver chloride (AgCl) precipitate, production, 320
Silver ions, chloride ions
combination, 319f
reaction, 320
Silver nitrate
moles, calculation, 310
solution, elemental copper presence (metal displacement), 319f
Single bonds, 436, 457
usage, 241
Single displacement, 153, 172
Single-displacement reaction, 154, 155f, 343
SI units, 28
Skeletal structures, 433–436, 457, 460
conversion, 434–435
cycloalkanes, 438f
usage, 433f
Slaked lime, 145
Small-scale gold mining, mercury contamination, 55–56
Small units, usage, 28
Sodium
alkali metal, 64f
electron configuration, 102, 103
electrons, levels/sublevels, 100
metals, water (reaction), 190
neon, similarity, 118f
reaction, 406f
valence electron, presence, 118f
Sodium acetate, osmotic pressure (contrast), 318
Sodium azide (NaN3), Lewis structure, 244–245

1228
Sodium bicarbonate (NaHCO3) (baking soda)
addition, 322
H+ determination, 349
Sodium chloride (NaCl)
freezing point, 314f
moles, conversion, 185
naming, 125
production, 156f
solution, calcium chloride solution (separation), 318
Sodium hydroxide (NaOH)
base, 336
hydrochloric acid, combination, 196
requirement, volume determination, 352
solution, preparation, 310
usage, 350
Soft drinks
acidic/basic determination, 348
bubbles, formation, 153f
Solar cells, 85, 108, 397
Solar power, 85
Solar roof, 85f
Solids, 8, 18, 272–278
involvement, 379–381
transition, 9f
Solubility
products, 380–381, 389
concentration, determination, 381
determination, 380–381
rules, usage, 165
Soluble, term (usage), 132–134, 137
Solute, 305, 326
moles, 311
Solution(s), 132–134, 137, 305, 326
dilute, preparation, 311
electrolyte, 312–318
hypertonic, 318
hypotonic, 318
isotonic, 318
preparation
concentration knowledge, 310f
molarity knowledge, 310
reactions, 319–320
stoichiometry, 320–323
Solvent, 305, 310, 326

1229
 involvement, 378
Sound waves, travel, 39
Space-filling model, 275f
Space heaters, BTU release, 212f
SpaceX, 85
Spearmint oil, component, 442f
Specific heat, 217–219, 230
capacity, 217, 227
comparison, 218
equation, 217
finding, coffee cup calorimetry (usage), 220
materials ranking, 218t
usage, 218
Spectator ion, 164, 172
Spectral lines, 91
color lines, 91f
Speed, wavelength/frequency (relationship), 88f
Spending, analogy, 211
Spin, 95
orientation, 95
Sponges, scientist collection, 182
Stability, band, 485f
Standard notation, conversion, 26
Standard pressure, 280
Standard temperature and pressure (STP), 285, 285f, 294
Starch, 450
alpha(1→4) bonds, breakdown, 450f
State, 153
Statue (erosion), acid rain (impact), 344f
Stoichiometry
conversions, 192
gas stoichiometry, 291–293
Stoichiometry problems, 188–190, 200
advanced problems, 322–323
solving, 192f
strategies, 191–193
Strength (parameter), 325, 325f
ranking, 404t
Strong acids, 338, 358
types, 338t
Structure, 4–5
representation, 433
Subatomic particles, 66, 68t, 78
ion charge, relationship, 75

1230
 nuclear symbols, 472t
types, 484t
Sublevel(s), 94–97, 109
d, orbitals, 96f
energy levels/electron capacity, relationship, 96t
filling sequence (determination), periodic table (usage), 105f
f, orbitals, 96f
overlap, 100f
p, orbitals, 95–96
s, orbitals, 95
1s, cutaway view, 95f
2s, cutaway view, 95f
Subscripts, 147–148, 172, 220
adjustment, avoidance, 150
coefficients, difference, 148f
Subshells, 95
Substances
amounts, relationship, 189
behavior, 9
composition, 9
forces of attraction, 271
formation, 10
fuel value, 224
ionic, 272
metallic, 272–273
molecular, 273–277
particles, arrangement, 9
potential energy, 12
pure, 6, 7f, 18
usage, examples, 4f
Sucrose, freezing point, 314f
Sugar
dissolving, 305f
impact, absence, 379f
nonionic compounds, dissolving, 133f
table sugar, compound, 449f
Sulfur
−2 ions, formation, 120f
fluorine gas, reaction, 197
oxidation number, identification, 400, 401
oxygen, reaction, 160f
reaction, 10f
Sulfur dichloride, net dipole, 257
Sulfuric acid (H2SO4)

1231
 diprotic acid, 336
neutralization, KOH solution (usage), 352
phenolphthalein, addition, 352
Sulfur oxides (SOx),
toxicity, 159
water (reaction), 344
Sulfur trioxide (SO3), molecules (presence), 187
Sun, light production (reason), 92
Surroundings, 214, 230
system, impact, 215
Synthesis, 172
reactions (combination reactions), 154
Syringol, 444
System, 230
defining, 214
impact, 214–215

T
Table salt, 65, 65f
Table sugar, compound, 449f
Tannins, 335f
Taxol, 1–2
discovery, 14
Taxus brevifolia,, 1
Teflon, 65, 65f
Temperature, 230
equilibrium, relationship, 385
fusion reaction condition, 489
heat energy, relationship, 217–223
impact, 368
kelvin representation, 284
measurement, 45–46
pressure, correlation, 289
standard temperature and pressure (STP), 285
volume-temperature problems (solving), Charles’s law (usage), 283–284
Temple of the Foliated Cross, 145f
Ten, powers, 25t
Terephthalic acid, 448
Tesla battery system, achievement, 86
Tesla Motors, 85
Tetrahedral geometry, 251, 261
three-dimensional volume, 251
Tetrahedron, X-shape, 251f

1232
Thallium-209, beta decay, 475
Theoretical yield, 196–198, 200
finding, 197
Theory, 15, 18
Thermochemistry, 211
Thermodynamics, 211, 230
first law, 216
Thermonuclear weapons, 489
Thiocyanate ion (SCN−), 244
Thomson, J. J., 66–68, 73, 76
Thorium, decay series, 475
Thr-Ala-Cys (tripeptide), drawing, 453
Thymine, 454–455
Tin(IV) ion, 401f
formation, 158
Tin(IV) oxide, ionic compound, 158
Tire, pressure recommendation, 281
Titration, 350, 358
coefficients, inequality, 352
HCl neutralization, 351
long tube, usage, 350f
usage, 352
Tokamak design, 490, 490f
Tornado, pressure areas, 279f
Total dissolved solids (TDS), 325, 325f
Total energy change, measurement, 213
Tour, James, 237
Tranilast, functional groups, 446
Transition elements, 63
Transition metals, 78
electrons, loss, 118
ion formation, 118f
types, 63f
Transition state, 369, 389
Trichloroacetic acid (HC2Cl3O2), weak acid, 339
Trigonal planar, term (usage), 250–251, 250f, 251f, 261
Trinitrotoluene (TNT), benzene ring, 440, 440f
Tripeptide, 452
Triple bonds
carbon–carbon triple bonds, 439, 457
representation, 433
Triprotic acids, 336
Tritium, 489

1233
Tryptophan, connection, 452
Tsunami, impact, 469–470, 492
Tucson Fuel Cell, 414f
Two-amino-acid structure, components, 452f
Tylenol, acetaminophen component, 434f

U
Ultraviolet (UV) region, 88
Uncertainty, 29
principle, 93–94, 109
Unit conversion, 37–42
Units of measurement. See Measurement
Unshared electrons, display, 433
Uranium
decay series, 475, 476f
enrichment, 486–487
fluorine gas, reaction, 195
isotopes, 70
uranium-235 (U-235), nuclear fission, 486f
uranium-238 (U-238), alpha decay, 474
yellowcake uranium, 487
Uranium hexafluoride (UF6), 487
gas centrifuge separation, 487f
Uranium oxide (U3O8), mining, 487
Urea (CH4N2O), covalent compound, 240
U.S. drinking water, contaminants (monitoring), 307t

V
Valence
configurations, 104
electrons, relationship, 117
energy, filling, 100, 104, 109, 117
level, 100, 109, 239f
filling, 117–118
shell, 100
Valence electrons, 99–100, 117
addition, 241, 246
configuration (identification), periodic table (usage), 105f
identification, 101, 240
main-group numbers, 106f
presence, 245
Valence shell electron pair repulsion (VSEPR) model, 250, 259, 261

1234
Vaporization, 8, 18
Villaseñor, Isabel, 145, 145f
Vinegar, acetic acid solution, 443f
Visible light
involvement, 93
production, 92
spectrum portion, 87f
Visible spectrum, 87, 109
ranges, 87f
Volta, Alessandro, 66–67, 397, 413
Volume
mass
conversion, 43–44
relationship, 43–45
measurement
beaker/graduated beaker, usage, 31f
dependence, 40f
units, conversion, 40–42
volume-temperature problems (solving), Charles’s law (usage), 283–284
volume-temperature relationship, extrapolation, 283f

W
Wall, Monroe, 1
Wani, Mansukh, 1
Water (H2O)
acids, dissolving, 335f
addition, 383f
aggregation, 317f
base, 337
behavior, 338
bonding electrons/lone pairs sets, 252f
bonds, polarity, 256
carbon dioxide, reaction, 293
composition, 5f
conductor, 133f
covalent compounds, 272
dipole, 256f
electronic geometry, 252
exhalation, 249
exit, 318f
forces of attraction, 271f
formation, 148, 168, 342
hydrogen bonding, impact, 275f
ion attraction, 133f

1235
 metals
heat loss, 220
reaction, 405–407
reactivity, 406t
monitoring, 307–308
movement, 316f, 384f
net dipole, 258f
O-H bonds, 275
oxygen atom, valence level, 239f
pH, change, 353f
polar covalent bonds, 257f
production, 10f
properties, 274t
purity, testing, 133
salt, addition, 305
self-ionization reaction, 345
specific heat, comparison, 218
splitting, 67f
temperature (increase), heat requirement (kilojoules calculation), 218
U.S. drinking water, contaminants (monitoring), 307t
Waugh, Lauren, 23–24, 47
Wavelength, 87–88, 109
distance, measurement, 87f
energy, relationship, 89
frequency/light energy, relationship, 89
speed/frequency, relationship, 88f
Wave nature, 93–94
Weak acids, 338–339, 358
conjugates (basic characteristic), 341
Weak bases, conjugates (acidic characteristic), 341
Wealth
analogy, 211
transfer, 211f
Weighted average (isotopes), 71
consideration, 71f
Weighted average value, 72
White light, prism (usage), 90f
Whole-number ratio, 59
Width measurements, 37
Work, 211–217, 230
energy transfer, 213
production, 213f
transfer, 211f

1236
X
X-ray crystallography, 60, 60f
Xenon, noble gas, 100

Y
Yaghi, Omar, 295, 295f
Yellowcake uranium, 487
Yew cells, growth, 2

Z
Zeros, significance, 32
Zimbabwe, rock formation, 12, 13f
Zinc
anode, aqueous zinc chloride solution immersion, 411
nitric acid, reaction, 408
reaction, 10f
washers, usage, 67f
Zinc chloride, production, 405f
Zinc metal
aqueous copper(II) chloride, reaction, 192
aqueous copper(II) ions, reaction, 402f
aqueous copper(II) sulfate, reaction, 402
aqueous hydrochloric acid, reaction, 323
hydrochloric acid, reaction, 405f
Zinc sulfide
formation, 10f
radiation, impact, 474

1237
Useful Tables and Figures
Common Unit Conversions

Length Volume

1 m = 3.281 ft 1 L = 1 dm3

1 km = 0.621 mi 1 mL = 1 cm3

1 cm = 0.394 in 1 L = 0.264 gal

Mass or Weight

1 kg = 2.205 lb

Energy Pressure

1 J = 1 kg · m2/s2 1 atm = 760 mm Hg (torr)

1 cal = 4.184 joules 1 atm = 14.70 psi

1,000 cal = 1 kcal = 1 Calorie 1 atm = 101.3 kPa

1 BTU = 1,055 J 1 atm = 1.013 bar

Common Metric Prefixes

Prefix Symbol Meaning

tera- T 1012 1,000,000,000,000

giga- G 109 1,000,000,000

mega- M 106 1,000,000

kilo- k 103 1,000

deci- d 10−1

110
centi- c 10−2

1100

milli- m 10−3

11,000
micro- µ

1238
 10−6

11,000,000
nano- n 10−9

11,000,000,000

pico- p 10−12

11,000,000,000,000

Mathematical Relationships and Constants

Density

d=mv

Temperature

K=

°C+273.15
°F=95°C+32

°C=59(°F−32)

Energy, frequency, and wavelength

c = λν c = 3.00 × 108 m/s

E = hν h = 6.63 × 10−34 J · s

Avogadro’s number

6.02 × 10−23 particles = 1 mole

Energy and temperature

ΔE = q + w q = CΔT
q = msΔT swater = 4.184 J/g · °C

Gas laws

P1V1T1=P2V2T2
PV = nRT R = 0.0821 L · atm/mol · K

Concentration

ppm=mass solutemass mass %=mass of solutemass of

1239
 solution×100%

solution×106
ppb=mass solutemass volume %=volume of solutevolume of
solution×100%

solution×109

Molarity=moles soluteL mass/volume %=mass of solutevolume of

solution×100%

solution

Acids and bases

[H+][OH−] = 1 × 10−14 pH = −log[H+]

The Activity Series

Elemental Color Codes

1240
Solubility Rules

1241
Pauling’s Table of Electronegativities

The Complete Mole Map

Common Ions

1242
1243
1244
目录
Front Matter 2
Cover Page 2
Inside Front Cover 3
Half Title Page 4
Title Page 5
Copyright Page 6
Brief Contents 7
Contents 9
A word from the author 28
Acknowledgments 30
Dedication 36
Chapter One: Foundations 37
Taxol 37
1.1 Chemistry: Part of Everything You Do 40
1.2 Describing Matter 43
Composition and Structure 43
Pure Substances and Mixtures 44
States of Matter 50
Properties and Changes 53
1.3 Energy and Change 59
1.4 The Scientific Method 64
Summary 66
You Can Do This 67
Key Terms 68
Additional Problems 70
Chapter Two: Measurement 78
A Strange Death 78
2.1 Measurement: A Foundation of Good Science 81
Scientific Notation: Working with Very Large and Very
81
Small Numbers
Converting from Scientific Notation to Standard Notation 82
Converting from Standard Notation to Scientific Notation 83

1245
 Calculations Involving Scientific Notation 84
Units of Measurement 85
Describing the Quality of Measurements 89
Describing Precision: Significant Digits 92
Determining the Number of Significant Digits in a
95
Measurement
Working with Exact Numbers 97
Using Significant Digits in Calculations 97
2.2 Unit Conversion 104
Dimensional Analysis 104
Problems Involving Multiple Conversions 107
Converting between Volume Units 109
Converting Units Raised to a Power 111
2.3 Density: Relating Mass to Volume 116
Converting between Mass and Volume 118
Will It Float? 120
2.4 Measuring Temperature 122
Summary 124
A Strange Death 125
Key Terms 127
Additional Problems 128
Chapter Three: Atoms 141
Mercury Contamination from Small-Scale Gold Mining 141
3.1 Atoms: The Essential Building Blocks 145
Uncovering the Atom: From Democritus to Dalton 145
Can We See Atoms? 150
3.2 The Periodic Table of the Elements 154
Regions of the Periodic Table 159
Metals, Nonmetals, and Metalloids 160
Groups (Families) of Elements 162
3.3 Uncovering Atomic Structure 168
The Discovery of Charged Particles 168
The Discovery of the Nucleus 172
3.4 Describing Atoms: Identity and Mass 176
Atomic Number and Mass Number 176

1246
 Average Atomic Mass 179
3.5 Electrons—A Preview 183
The Bohr Model and the Quantum Model 183
The Formation of Ions 183
Summary 188
Mercury: Ancient Treasure, Modern Toxin 190
Key Terms 192
Additional Problems 194
Chapter Four: Light and Electronic Structure 208
Edging toward Solar Energy 208
4.1 The Electromagnetic Spectrum 212
Wavelength and Frequency 213
The Energy of a Photon 216
4.2 Color, Line Spectra, and the Bohr Model 220
Color and Line Spectra 220
The Bohr Model 223
4.3 The Quantum Model and Electron Orbitals 229
The Uncertainty Principle and the Wave Nature of Electrons 229
Energy Levels and Sublevels 231
4.4 Describing Electron Configurations 238
Valence Electrons and the Octet Rule 242
Electron Configurations for Larger Atoms 244
Electron Configurations for Ions 246
4.5 Electron Configuration and the Periodic Table 249
Summary 257
Solar Cells: Converting Light into Electric Current 258
Key Terms 260
Additional Problems 262
Chapter Five: Chemical Bonds and Compounds 275
An Unexpected Combination: Lithium Carbonate and Bipolar
275
Disorder
5.1 Lewis Symbols and the Octet Rule 278
5.2 Ions 280
Cations: Ions with a Positive Charge 280
Naming Cations 282

1247
 Anions: Ions with a Negative Charge 283
Naming Anions 285
Polyatomic Ions 286
Naming Polyatomic Ions 286
A Summary of the Common Ions 288
5.3 Ionic Bonds and Compounds 291
Ionic Bonds and Ionic Lattices 291
Predicting Formulas for Ionic Compounds 292
Naming Ionic Compounds 294
5.4 Covalent Bonding 298
Nonmetal–Nonmetal Bonds 298
Covalent Compounds 300
Naming Covalent Compounds 304
5.5 Distinguishing Ionic and Covalent Compounds 307
5.6 Aqueous Solutions: How Ionic and Covalent Compounds
310
Differ
5.7 Acids—An Introduction 313
Naming Acids 314
Binary Acids 314
Oxyacids 314
Summary 316
Continuing Cade’s Work 317
Key Terms 319
Additional Problems 321
Chapter Six: Chemical Reactions 339
Lost Cities of the Maya 339
6.1 Chemical Equations 342
Balancing Equations 346
Strategies for Balancing Equations 352
Equations with Phase Notations 354
6.2 Classifying Reactions 356
6.3 Reactions between Metals and Nonmetals 361
6.4 Combustion Reactions 366
6.5 Reactions in Aqueous Solution 370
Representing Dissociation: Molecular and Ionic Equations 370

1248
 Solubility Rules and Precipitation Reactions 372
Acid-Base Neutralization Reactions 383
Acids and Bases 383
Neutralization Reactions 384
Summary 388
The Mayan Lime Cycle 390
Key Terms 392
Additional Problems 394
Chapter Seven: Mass Stoichiometry 409
Process Development 409
7.1 Formula Mass and Percent Composition 412
Formula Mass 412
Percent Composition 413
How Chemists Measure Formula Mass and Percent
415
Composition
Mass Spectrometry 416
Elemental Analysis 418
7.2 Connecting Atomic Mass to Large-Scale Mass: The Mole
419
Concept
Avogadro’s Number and the Mole 419
Converting between Grams, Moles, and Particles 421
7.3 The Mole Concept in Balanced Equations 426
Stoichiometry Problems 427
Gram-to-Gram Questions 432
Strategies for Solving Stoichiometry Problems 433
Calculations with Limiting Reagents 436
Finding the Leftovers 439
7.4 Theoretical and Percent Yield 442
Summary 445
Process Development, Continued 446
Key Terms 448
Additional Problems 449
Chapter Eight: Energy 468
The Corn Ethanol Debate 468
8.1 Energy, Work, and Heat 471

1249
 Units of Energy 472
Heat and Work in Chemical Changes 474
Endothermic and Exothermic Changes 475
The Law of Conservation of Energy 479
8.2 Heat Energy and Temperature 482
Specific Heat and Heat Capacity 482
Calorimetry: Measuring Heat Flow 486
Coffee Cup Calorimetry 486
Bomb Calorimetry 492
8.3 Heat Energy and Chemical Reactions 495
Fuel Value 495
Reaction Enthalpy 496
Summary 502
Gasoline or Ethanol—Which Fuel Is Better? 503
Key Terms 505
Additional Problems 507
Chapter Nine: Covalent Bonding and Molecules 521
The Shortest Race 521
9.1 Covalent Molecules 525
Representing Covalent Structures 525
Exceptions to the Octet Rule 528
Drawing Lewis Structures 529
9.2 Molecules and Charge 533
Polyatomic Ions and Formal Charge 533
Drawing Lewis Structures for Polyatomic Ions 537
Choosing the Best Lewis Structure 539
Resonance 540
9.3 Shapes of Molecules 544
9.4 Polar Bonds and Molecules 553
Electronegativity and Polar Covalent Bonds 553
Molecules with Dipoles 557
Identifying Molecules with a Net Dipole 558
How Dipoles Affect Properties—A Preview 561
Summary 562
Building a Nanocar 564

1250
 Key Terms 566
Additional Problems 568
Chapter Ten: Solids, Liquids, and Gases 589
The North Dakota Boom 589
10.1 Interactions between Particles 592
10.2 Solids and Liquids 596
Ionic Substances 596
Metallic Substances 597
Molecular Substances 598
Dipole–Dipole Interactions 598
Hydrogen Bonding 600
London Dispersion Forces 604
Summarizing Intermolecular Forces 606
Covalent Networks and Polymers 608
10.3 Describing Gases 611
Pressure 612
Measuring Pressure 612
10.4 The Gas Laws 617
Boyle’s Law 617
Charles’s Law 618
Using Charles’s Law to Find Absolute Zero 618
Solving Volume–Temperature Problems Using Charles’s
620
Law
The Combined Gas Law 620
Avogadro’s Law 623
The Ideal Gas Law 624
Mixtures of Gases 625
A Molecular View of the Gas Laws 628
10.5 Diffusion and Effusion 632
10.6 Gas Stoichiometry 635
Summary 640
Rethinking Gas Storage 643
Key Terms 645
Additional Problems 647
Chapter Eleven: Solutions 661

1251
 The Perfect Cup of Coffee 661
11.1 Describing Concentration 664
Concentration by Percent 664
Percent by Mass and Volume 664
Mass/Volume Percent 666
Very Dilute Solutions: ppm and ppb 668
Molarity 672
Preparing Solutions of Known Molarity 674
Preparing Dilute Solutions 676
Using Square Brackets to Represent Concentration 678
11.2 Electrolyte Solutions 680
Electrolyte Concentrations 681
Colligative Properties 683
Freezing Point Depression 683
Boiling Point Elevation 685
Osmotic Pressure 686
11.3 Reactions in Solution—A Review and a Preview 694
Precipitation Reactions 694
Acid-Base Neutralization Reactions 695
Metal Displacement Reactions 695
11.4 Solution Stoichiometry 697
Gravimetric Analysis 698
Advanced Stoichiometry Problems 700
Summary 704
Amazing Coffee—The Importance of Concentration 705
Key Terms 707
Additional Problems 709
Chapter Twelve: Acids and Bases 723
Cocaine: Ruin and Recovery 723
12.1 Introduction to Acids and Bases 726
The Arrhenius Definition 726
Polyprotic Acids 730
The Brønsted-Lowry Definition 730
12.2 Acid-Base Equilibrium Reactions 735
Are Conjugate Bases Basic? 739

1252
 12.3 Reactions Involving Acids and Bases 741
Neutralization Reactions 741
Reactions of Acids with Metal 743
Formation of Acids from Nonmetal Oxides 744
12.4 Acid and Base Concentration 748
Concentrations of H+ and OH− in Aqueous Solutions 748
Math Review: Exponential and Scientific Notation 749
Connecting [OH−] and [H+] 750
The pH Scale 750
12.5 Measuring Acid and Base Concentration 757
Determining pH in the Laboratory 757
Acid-Base Titrations 758
Acid-Base Titrations with Different Coefficients 761
12.6 Buffers and Biological pH 764
Summary 769
The War on Drugs, Then and Now 771
Key Terms 773
Additional Problems 775
Chapter Thirteen: Reaction Rates and Equilibrium 793
The Haber-Bosch Process 793
13.1 Reaction Rates 796
How Concentration and Temperature Affect Reaction Rates 798
How Changes in Energy Affect Reaction Rates 798
Describing Energy Changes in Chemistry: Reaction
800
Energy Diagrams
Catalysts 803
13.2 Equilibrium Reactions 806
13.3 Equilibrium Expressions 813
Equilibrium Expressions Involving Solvents 816
Equilibrium Expressions Involving Solids 819
Solubility Products 820
Equilibrium Expressions Involving Gases 823
13.4 Le Chatelier’s Principle 826
Equilibrium and Concentration 826
Equilibrium and Temperature 830

1253
 Equilibrium and Pressure 831
Summary 835
Miracles and Monstrosities: The Brutal Ironies of Fritz Haber 836
Key Terms 838
Additional Problems 839
Chapter Fourteen: Oxidation-Reduction Reactions 860
Volta’s Marvel 860
14.1 Oxidation and Reduction 863
Oxidation Numbers 863
14.2 Types of Redox Reactions 868
Reactions of Metals with Nonmetals 868
Combustion Reactions 868
Metal Displacement Reactions 870
The Activity Series 873
Reactions of Metals with Acid and Water 876
14.3 Half-Reactions and Batteries 881
Half-Reactions 881
Batteries 883
14.4 Balancing Redox Equations 891
14.5 Other Applications of Redox Reactions 894
Electroplating 894
Fuel Cells 896
Summary 898
Charging Ahead: Batteries Today and Tomorrow 900
Key Terms 902
Additional Problems 903
Chapter Fifteen: Organic Chemistry and Biomolecules 923
Forming New Bonds: The Grubbs Catalyst 923
15.1 Organic Chemistry and the Carbon Cycle 927
15.2 Covalent Bonding with Carbon and Other Nonmetals 930
15.3 Drawing Covalent Structures 936
Condensed Structures 936
Skeletal Structures 938
15.4 Major Functional Groups 943
Hydrocarbon Functional Groups 943

1254
 Alkanes and Cycloalkanes 943
Alkenes and Alkynes 948
Aromatic Compounds 949
Oxygen-Containing Functional Groups 951
Alcohols and Ethers 951
Carbonyl Groups 953
A Summary of Oxygen-Containing Groups 956
Nitrogen-Containing Functional Groups 958
15.5 Polymers and Plastics 962
15.6 Biomolecules—An Introduction 966
Carbohydrates 966
Amino Acids and Proteins 969
Amino Acids 969
Peptides and Peptide Bonds 971
DNA 975
Summary 981
How Catalysts Work 984
Key Terms 987
Additional Problems 990
Chapter Sixteen: Nuclear Chemistry 1009
Fukushima 1009
16.1 Nuclear Changes 1012
The Nucleus—A Review 1012
Nuclear Reactions 1014
16.2 Radioactivity 1018
Types of Radioactive Decay 1020
Alpha Decay 1020
Beta Decay 1021
Gamma Decay 1022
Radioactive Decay Series 1022
Half-Life 1024
Health Effects of Radiation Exposure 1027
Measuring Radiation 1028
Common Exposure Levels 1031
Uses of Radioactive Nuclides 1033

1255
 Uses in Medicine 1033
Uses in Geology and Archaeology 1035
16.3 Energy Changes in Nuclear Reactions 1038
Mass Defect, Binding Energy, and Einstein’s Famous
1038
Equation
Nuclide Stability 1040
16.4 Nuclear Power: Fission and Fusion 1043
Fission 1043
Uranium Enrichment 1045
Fission Reactor Design 1048
Waste from Nuclear Fission 1049
Fusion 1049
Replicating Fusion on Earth 1049
Are We There Yet? 1052
Summary 1055
Powering the Future 1057
Key Terms 1059
Additional Problems 1061
Back Matter 1074
Answers to Odd-Numbered Problems 1074
Glossary 1159
Index 1174
A 1174
B 1179
C 1182
D 1190
E 1192
F 1196
G 1199
H 1201
I 1204
J 1206
K 1206
L 1207
M 1209

1256
 N 1213
O 1216
P 1218
Q 1224
R 1224
S 1226
T 1232
U 1234
V 1234
W 1235
X 1237
Y 1237
Z 1237
Useful Tables and Figures 1238
Back Cover 1244

1257