2.

1: THE STRUCTURE
OF THE ATOM
 CHAPTER OVERVIEW

2.1: The Structure of the Atom

Unit 2 Objectives

By the end of this unit, you will be able to:

State the postulates of Dalton’s atomic theory
Use postulates of Dalton’s atomic theory to explain the laws of definite and multiple proportions
Outline milestones in the development of modern atomic theory
Summarize and interpret the results of the experiments of Thomson, Millikan, and Rutherford
Describe the three subatomic particles that compose atoms
Define isotopes and give examples for several elements
Write and interpret symbols that depict the atomic number, mass number, and charge of an atom or ion
Define the atomic mass unit and average atomic mass
Calculate average atomic mass and isotopic abundance

2.1.0: A History of Atomic Theory

2.1.1: A History of Atomic Theory (Problems)
2.1.2: The Structure of the Atom and How We Represent It
2.1.3: The Structure of the Atom and How We Represent It (Problems)
2.1.4: Calculating Atomic Masses
2.1.5: Calculating Atomic Masses (Problems)

2.1: The Structure of the Atom is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

1
2.1.0: A History of Atomic Theory
Learning Objectives
By the end of this section, you will be able to:
State the postulates of Dalton’s atomic theory
Use postulates of Dalton’s atomic theory to explain the laws of definite and multiple proportions
Outline milestones in the development of modern atomic theory
Summarize and interpret the results of the experiments of Thomson, Millikan, and Rutherford

A Video Introduction to Atomic Theory through the Nineteenth Century From Crash Course Chemistry

The Creation of Chemistry - The Fundam…

Fundam…

Video 2.1.0.1 : Lavoisier's discovery of The Law of Conservation of Matter led to the Laws of Definite and Multiple
Proportions and eventually Dalton's Atomic Theory.

Atomic Theory through the Nineteenth Century

The earliest recorded discussion of the basic structure of matter comes from ancient Greek philosophers, the scientists of their day. In
the fifth century BC, Leucippus and Democritus argued that all matter was composed of small, finite particles that they called atomos, a
term derived from the Greek word for “indivisible.” They thought of atoms as moving particles that differed in shape and size, and
which could join together. Later, Aristotle and others came to the conclusion that matter consisted of various combinations of the four
“elements”—fire, earth, air, and water—and could be infinitely divided. Interestingly, these philosophers thought about atoms and
“elements” as philosophical concepts, but apparently never considered performing experiments to test their ideas.
The Aristotelian view of the composition of matter held sway for over two thousand years, until English schoolteacher John Dalton
helped to revolutionize chemistry with his hypothesis that the behavior of matter could be explained using an atomic theory. First
published in 1807, many of Dalton’s hypotheses about the microscopic features of matter are still valid in modern atomic theory. Here
are the postulates of Dalton’s atomic theory.
1. Matter is composed of exceedingly small particles called atoms. An atom is the smallest unit of an element that can participate in a
chemical change.
2. An element consists of only one type of atom, which has a mass that is characteristic of the element and is the same for all atoms of
that element (Figure 2.1.0.1). A macroscopic sample of an element contains an incredibly large number of atoms, all of which have
identical chemical properties.
3. Atoms of one element differ in properties from atoms of all other elements.
4. A compound consists of atoms of two or more elements combined in a small, whole-number ratio. In a given compound, the
numbers of atoms of each of its elements are always present in the same ratio (Figure 2.1.0.2).
5. Atoms are neither created nor destroyed during a chemical change, but are instead rearranged to yield substances that are different
from those present before the change (Figure 2.1.0.3).

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 ×
Figure 2.1.0.1 : A pre-1982 copper penny (left) contains approximately 3 1022 copper atoms (several dozen are represented as
brown spheres at the right), each of which has the same chemical properties. (credit: modification of work by “slgckgc”/Flickr)

Figure 2.1.0.2 : Copper(II) oxide, a powdery, black compound, results from the combination of two types of atoms—copper (brown
spheres) and oxygen (red spheres)—in a 1:1 ratio. (credit: modification of work by “Chemicalinterest”/Wikimedia Commons)

Figure 2.1.0.3 : When the elements copper (a shiny, red-brown solid, shown here as brown spheres) and oxygen (a clear and colorless
gas, shown here as red spheres) react, their atoms rearrange to form a compound containing copper and oxygen (a powdery, black
solid). (credit copper: modification of work by http://images-of-elements.com/copper.php).
Dalton’s atomic theory provides a microscopic explanation of the many macroscopic properties of matter that you’ve learned about. For
example, if an element such as copper consists of only one kind of atom, then it cannot be broken down into simpler substances, that is,
into substances composed of fewer types of atoms. And if atoms are neither created nor destroyed during a chemical change, then the
total mass of matter present when matter changes from one type to another will remain constant (the law of conservation of matter (or
mass)).

Want to learn more about the Law of Conservation of Mass?

2.1.0.2 https://chem.libretexts.org/@go/page/210621
 The law of conservation of mass - Todd R…
R…

Video 2.1.0.2 : "We are made of star stuff" - Carl Sagan.

Example 2.1.0.1: Testing Dalton’s Atomic Theory

In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another
element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by
these symbols violate any of the ideas of Dalton’s atomic theory? If so, which one?

Solution
The starting materials consist of two green spheres and two purple spheres. The products consist of only one green sphere and one
purple sphere. This violates Dalton’s postulate that atoms are neither created nor destroyed during a chemical change, but are
merely redistributed. (In this case, atoms appear to have been destroyed.)

Exercise 2.1.0.1

In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another
element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by
these symbols violate any of the ideas of Dalton’s atomic theory? If so, which one

Answer
The starting materials consist of four green spheres and two purple spheres. The products consist of four green spheres and two
purple spheres. This does not violate any of Dalton’s postulates: Atoms are neither created nor destroyed, but are redistributed
in small, whole-number ratios.

Dalton knew of the experiments of French chemist Joseph Proust, who demonstrated that all samples of a pure compound contain the
same elements in the same proportion by mass. This statement is known as the law of definite proportions or the law of constant
composition. The suggestion that the numbers of atoms of the elements in a given compound always exist in the same ratio is consistent
with these observations. For example, when different samples of isooctane (a component of gasoline and one of the standards used in
the octane rating system) are analyzed, they are found to have a carbon-to-hydrogen mass ratio of 5.33:1, as shown in Table 2.1.0.1.
Table 2.1.0.1 : Constant Composition of Isooctane
Sample Carbon Hydrogen Mass Ratio

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 Sample Carbon Hydrogen Mass Ratio
14.82 g carbon 5.33 g carbon
A 14.82 g 2.78 g =
2.78 g hydrogen 1.00 g hydroge
22.33 g carbon 5.33 g carbon
B 22.33 g 4.19 g =
4.19 g hydrogen 1.00 g hydroge
19.40 g carbon 5.33 g carbon
C 19.40 g 3.64 g =
3.63 g hydrogen 1.00 g hydroge

It is worth noting that although all samples of a particular compound have the same mass ratio, the converse is not true in general. That
is, samples that have the same mass ratio are not necessarily the same substance. For example, there are many compounds other than
isooctane that also have a carbon-to-hydrogen mass ratio of 5.33:1.00.
Dalton also used data from Proust, as well as results from his own experiments, to formulate another interesting law. The law of
multiple proportions states that when two elements react to form more than one compound, a fixed mass of one element will react with
masses of the other element in a ratio of small, whole numbers. For example, copper and chlorine can form a green, crystalline solid
with a mass ratio of 0.558 g chlorine to 1 g copper, as well as a brown crystalline solid with a mass ratio of 1.116 g chlorine to 1 g
copper. These ratios by themselves may not seem particularly interesting or informative; however, if we take a ratio of these ratios, we
obtain a useful and possibly surprising result: a small, whole-number ratio.
1.116 g Cl
1 g Cu 2
= (2.1.0.1)
0.558 g Cl 1
1 g Cu

This 2-to-1 ratio means that the brown compound has twice the amount of chlorine per amount of copper as the green compound.
This can be explained by atomic theory if the copper-to-chlorine ratio in the brown compound is 1 copper atom to 2 chlorine atoms, and
the ratio in the green compound is 1 copper atom to 1 chlorine atom. The ratio of chlorine atoms (and thus the ratio of their masses) is
therefore 2 to 1 (Figure 2.1.0.4).

Figure 2.1.0.4 : Compared to the copper chlorine compound in (a), where copper is represented by brown spheres and chlorine by
green spheres, the copper chlorine compound in (b) has twice as many chlorine atoms per copper atom. (credit a: modification of work
by “Benjah-bmm27”/Wikimedia Commons; credit b: modification of work by “Walkerma”/Wikimedia Commons)

Example 2.1.0.2: Laws of Definite and Multiple Proportions

A sample of compound A (a clear, colorless gas) is analyzed and found to contain 4.27 g carbon and 5.69 g oxygen. A sample of
compound B (also a clear, colorless gas) is analyzed and found to contain 5.19 g carbon and 13.84 g oxygen. Are these data an
example of the law of definite proportions, the law of multiple proportions, or neither? What do these data tell you about
substances A and B?
Solution
In compound A, the mass ratio of carbon to oxygen is:
1.33 g O
1gC

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 In compound B, the mass ratio of carbon to oxygen is:
2.67 g O

1 g C

The ratio of these ratios is:

1.33 g O

1 g C 1
=
2.67 g O 2

1 g C

This supports the law of multiple proportions. This means that A and B are different compounds, with A having one-half as much
carbon per amount of oxygen (or twice as much oxygen per amount of carbon) as B. A possible pair of compounds that would fit
this relationship would be A = CO2 and B = CO.

Exercise 2.1.0.2

A sample of compound X (a clear, colorless, combustible liquid with a noticeable odor) is analyzed and found to contain 14.13 g
carbon and 2.96 g hydrogen. A sample of compound Y (a clear, colorless, combustible liquid with a noticeable odor that is slightly
different from X’s odor) is analyzed and found to contain 19.91 g carbon and 3.34 g hydrogen. Are these data an example of the
law of definite proportions, the law of multiple proportions, or neither? What do these data tell you about substances X and Y?

Answer
14.13 g C
In compound X, the mass ratio of carbon to hydrogen is .
2.96 g H

19.91 g C
In compound Y, the mass ratio of carbon to oxygen is .
3.34 g H

The ratio of these ratios is

14.13 g C

2.96 g H 4.77 g C/g H 4

= = 0.800 = .
19.91 g C 5.96 g C/g H 5

3.34 g H

This small, whole-number ratio supports the law of multiple proportions. This means that X and Y are different compounds.

In the two centuries since Dalton developed his ideas, scientists have made significant progress in furthering our understanding of
atomic theory. Much of this came from the results of several seminal experiments that revealed the details of the internal structure of
atoms. Here, we will discuss some of those key developments, with an emphasis on application of the scientific method, as well as
understanding how the experimental evidence was analyzed. While the historical persons and dates behind these experiments can be
quite interesting, it is most important to understand the concepts resulting from their work.

Atomic Theory after the Nineteenth Century

If matter were composed of atoms, what were atoms composed of? Were they the smallest particles, or was there something smaller? In
the late 1800s, a number of scientists interested in questions like these investigated the electrical discharges that could be produced in
low-pressure gases, with the most significant discovery made by English physicist J. J. Thomson using a cathode ray tube. This
apparatus consisted of a sealed glass tube from which almost all the air had been removed; the tube contained two metal electrodes.
When high voltage was applied across the electrodes, a visible beam called a cathode ray appeared between them. This beam was
deflected toward the positive charge and away from the negative charge, and was produced in the same way with identical properties
when different metals were used for the electrodes. In similar experiments, the ray was simultaneously deflected by an applied
magnetic field, and measurements of the extent of deflection and the magnetic field strength allowed Thomson to calculate the charge-
to-mass ratio of the cathode ray particles. The results of these measurements indicated that these particles were much lighter than atoms
(Figure 2.1.0.1).

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 Figure 2.1.0.5 : (a) J. J. Thomson produced a visible beam in a cathode ray tube. (b) This is an early cathode ray tube, invented in
1897 by Ferdinand Braun. (c) In the cathode ray, the beam (shown in yellow) comes from the cathode and is accelerated past the anode
toward a fluorescent scale at the end of the tube. Simultaneous deflections by applied electric and magnetic fields permitted Thomson to
calculate the mass-to-charge ratio of the particles composing the cathode ray. (credit a: modification of work by Nobel Foundation;
credit b: modification of work by Eugen Nesper; credit c: modification of work by “Kurzon”/Wikimedia Commons).
Based on his observations, here is what Thomson proposed and why: The particles are attracted by positive (+) charges and repelled by
negative (−) charges, so they must be negatively charged (like charges repel and unlike charges attract); they are less massive than
atoms and indistinguishable, regardless of the source material, so they must be fundamental, subatomic constituents of all atoms.
Although controversial at the time, Thomson’s idea was gradually accepted, and his cathode ray particle is what we now call an
electron, a negatively charged, subatomic particle with a mass more than one thousand-times less that of an atom. The term “electron”
was coined in 1891 by Irish physicist George Stoney, from “electric ion.”
In 1909, more information about the electron was uncovered by American physicist Robert A. Millikan via his “oil drop” experiments.
Millikan created microscopic oil droplets, which could be electrically charged by friction as they formed or by using X-rays. These
droplets initially fell due to gravity, but their downward progress could be slowed or even reversed by an electric field lower in the
apparatus. By adjusting the electric field strength and making careful measurements and appropriate calculations, Millikan was able to
determine the charge on individual drops (Figure 2.1.0.2).

Figure 2.1.0.6 : Millikan’s experiment measured the charge of individual oil drops. The tabulated data are examples of a few possible
values.
Looking at the charge data that Millikan gathered, you may have recognized that the charge of an oil droplet is always a multiple of a
specific charge, 1.6 × 10−19 C. Millikan concluded that this value must therefore be a fundamental charge—the charge of a single
electron—with his measured charges due to an excess of one electron (1 times 1.6 × 10−19 C), two electrons (2 times 1.6 × 10−19 C),
three electrons (3 times 1.6 × 10−19 C), and so on, on a given oil droplet. Since the charge of an electron was now known due to

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Millikan’s research, and the charge-to-mass ratio was already known due to Thomson’s research (1.759 × 1011 C/kg), it only required a
simple calculation to determine the mass of the electron as well.
1 kg
Mass of electron = 1.602 × 10−19 C × = 9.107 × 10−31 kg (2.3.1)
11
1.759 × 10 C

Scientists had now established that the atom was not indivisible as Dalton had believed, and due to the work of Thomson, Millikan, and
others, the charge and mass of the negative, subatomic particles—the electrons—were known. However, the positively charged part of
an atom was not yet well understood. In 1904, Thomson proposed the “plum pudding” model of atoms, which described a positively
charged mass with an equal amount of negative charge in the form of electrons embedded in it, since all atoms are electrically neutral.
A competing model had been proposed in 1903 by Hantaro Nagaoka, who postulated a Saturn-like atom, consisting of a positively
charged sphere surrounded by a halo of electrons (Figure 2.1.0.3).

Figure 2.1.0.7 : (a) Thomson suggested that atoms resembled plum pudding, an English dessert consisting of moist cake with
embedded raisins (“plums”). (b) Nagaoka proposed that atoms resembled the planet Saturn, with a ring of electrons surrounding a
positive “planet.” (credit a: modification of work by “Man vyi”/Wikimedia Commons; credit b: modification of work by
“NASA”/Wikimedia Commons).
The next major development in understanding the atom came from Ernest Rutherford, a physicist from New Zealand who largely spent
his scientific career in Canada and England. He performed a series of experiments using a beam of high-speed, positively charged alpha
particles (α particles) that were produced by the radioactive decay of radium; α particles consist of two protons and two neutrons (you
will learn more about radioactive decay in the chapter on nuclear chemistry). Rutherford and his colleagues Hans Geiger (later famous
for the Geiger counter) and Ernest Marsden aimed a beam of α particles, the source of which was embedded in a lead block to absorb
most of the radiation, at a very thin piece of gold foil and examined the resultant scattering of the α particles using a luminescent screen
that glowed briefly where hit by an α particle.
What did they discover? Most particles passed right through the foil without being deflected at all. However, some were diverted
slightly, and a very small number were deflected almost straight back toward the source (Figure 2.1.0.4). Rutherford described finding
these results: “It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired
a 15-inch shell at a piece of tissue paper and it came back and hit you”1 (p. 68).

Figure 2.1.0.8 : Geiger and Rutherford fired α particles at a piece of gold foil and detected where those particles went, as shown in
this schematic diagram of their experiment. Most of the particles passed straight through the foil, but a few were deflected slightly and
a very small number were significantly deflected.
Here is what Rutherford deduced: Because most of the fast-moving α particles passed through the gold atoms undeflected, they must
have traveled through essentially empty space inside the atom. Alpha particles are positively charged, so deflections arose when they
encountered another positive charge (like charges repel each other). Since like charges repel one another, the few positively charged α
particles that changed paths abruptly must have hit, or closely approached, another body that also had a highly concentrated, positive

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charge. Since the deflections occurred a small fraction of the time, this charge only occupied a small amount of the space in the gold
foil. Analyzing a series of such experiments in detail, Rutherford drew two conclusions:
1. The volume occupied by an atom must consist of a large amount of empty space.
2. A small, relatively heavy, positively charged body, the nucleus, must be at the center of each atom.
This analysis led Rutherford to propose a model in which an atom consists of a very small, positively charged nucleus, in which most of
the mass of the atom is concentrated, surrounded by the negatively charged electrons, so that the atom is electrically neutral (Figure
2.1.0.5).

Figure 2.1.0.9 : The α particles are deflected only when they collide with or pass close to the much heavier, positively charged gold
nucleus. Because the nucleus is very small compared to the size of an atom, very few α particles are deflected. Most pass through the
relatively large region occupied by electrons, which are too light to deflect the rapidly moving particles.
After many more experiments, Rutherford also discovered that the nuclei of other elements contain the hydrogen nucleus as a “building
block,” and he named this more fundamental particle the proton, the positively charged, subatomic particle found in the nucleus. With
one addition, which you will learn next, this nuclear model of the atom, proposed over a century ago, is still used today.

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 Rutherford Scatterin

Plum Pudding At

Rutherford Atom

Another important finding was the discovery of isotopes. During the early 1900s, scientists identified several substances that appeared
to be new elements, isolating them from radioactive ores. For example, a “new element” produced by the radioactive decay of thorium
was initially given the name mesothorium. However, a more detailed analysis showed that mesothorium was chemically identical to
radium (another decay product), despite having a different atomic mass. This result, along with similar findings for other elements, led
the English chemist Frederick Soddy to realize that an element could have types of atoms with different masses that were chemically
indistinguishable. These different types are called isotopes—atoms of the same element that differ in mass. Soddy was awarded the
Nobel Prize in Chemistry in 1921 for this discovery.
One puzzle remained: The nucleus was known to contain almost all of the mass of an atom, with the number of protons only providing
half, or less, of that mass. Different proposals were made to explain what constituted the remaining mass, including the existence of
neutral particles in the nucleus. As you might expect, detecting uncharged particles is very challenging, and it was not until 1932 that
James Chadwick found evidence of neutrons, uncharged, subatomic particles with a mass approximately the same as that of protons.
The existence of the neutron also explained isotopes: They differ in mass because they have different numbers of neutrons, but they are
chemically identical because they have the same number of protons. This will be explained in more detail later in this unit.

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 The Nucleus: Crash Course Chemistry #1

Video 2.1.0.2 : An Introduction to Subatomic Particles

Summary

The 2,400-year search for the atom - Ther…

Ther…

Video 2.1.0.3 : A summary of discoveries in atomic theory.

The History of Atomic Chemistry: Crash C…

C…

Video 2.1.0.4 : A different summary of discoveries in atomic theory.

The ancient Greeks proposed that matter consists of extremely small particles called atoms. Dalton postulated that each element has a
characteristic type of atom that differs in properties from atoms of all other elements, and that atoms of different elements can combine
in fixed, small, whole-number ratios to form compounds. Samples of a particular compound all have the same elemental proportions by
mass. When two elements form different compounds, a given mass of one element will combine with masses of the other element in a
small, whole-number ratio. During any chemical change, atoms are neither created nor destroyed.

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Although no one has actually seen the inside of an atom, experiments have demonstrated much about atomic structure. Thomson’s
cathode ray tube showed that atoms contain small, negatively charged particles called electrons. Millikan discovered that there is a
fundamental electric charge—the charge of an electron. Rutherford’s gold foil experiment showed that atoms have a small, dense,
positively charged nucleus; the positively charged particles within the nucleus are called protons. Chadwick discovered that the nucleus
also contains neutral particles called neutrons. Soddy demonstrated that atoms of the same element can differ in mass; these are called
isotopes.

Footnotes
1. Ernest Rutherford, “The Development of the Theory of Atomic Structure,” ed. J. A. Ratcliffe, in Background to Modern Science,
eds. Joseph Needham and Walter Pagel, (Cambridge, UK: Cambridge University Press, 1938), 61–74. Accessed September 22,
2014, https://ia600508.us.archive.org/3/it...e032734mbp.pdf.

Glossary
Dalton’s atomic theory
set of postulates that established the fundamental properties of atoms

law of constant composition

(also, law of definite proportions) all samples of a pure compound contain the same elements in the same proportions by mass

law of multiple proportions

when two elements react to form more than one compound, a fixed mass of one element will react with masses of the other element
in a ratio of small whole numbers

law of definite proportions

(also, law of constant composition) all samples of a pure compound contain the same elements in the same proportions by mass

alpha particle (α particle)

positively charged particle consisting of two protons and two neutrons

electron
negatively charged, subatomic particle of relatively low mass located outside the nucleus

isotopes
atoms that contain the same number of protons but different numbers of neutrons

neutron
uncharged, subatomic particle located in the nucleus

proton
positively charged, subatomic particle located in the nucleus

nucleus
massive, positively charged center of an atom made up of protons and neutrons

Contributors and Attributions

Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen
F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative
Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110).
Adelaide Clark, Oregon Institute of Technology
Crash Course Chemistry: Crash Course is a division of Complexly and videos are free to stream for educational purposes.
TED-Ed’s commitment to creating lessons worth sharing is an extension of TED’s mission of spreading great ideas. Within TED-
Ed’s growing library of TED-Ed animations, you will find carefully curated educational videos, many of which represent
collaborations between talented educators and animators nominated through the TED-Ed website.

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2.1.1: A History of Atomic Theory (Problems)
PROBLEM 2.1.1.1

In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of
another element. If the spheres of different elements touch, they are part of a single unit of a compound. The following
chemical change represented by these spheres may violate one of the ideas of Dalton's atomic theory. Which one?

Answer
The starting materials consist of one green sphere and two purple spheres. The products consist of two green spheres and
two purple spheres. This violates Dalton’s postulate that that atoms are not created during a chemical change, but are
merely redistributed.

PROBLEM 2.1.1.2

Which postulate of Dalton's theory is consistent with the following observation concerning the weights of reactants and
products?
When 100 grams of solid calcium carbonate is heated, 44 g of CO2 and 56 g of CaO are produced.

Answer
Atoms are neither created nor destroyed during a chemical change, but are instead rearranged to yield substances that are
different from those present before the change (Based on the Law of Conservation of Mass).

PROBLEM 2.1.1.3

Samples of compound X, Y, and Z are analyzed, with results shown here. Do these data provide example(s) of the law of
definite proportions, the law of multiple proportions, neither, or both? What do these data tell you about compounds X, Y, and
Z?

Compound Description Mass of Carbon Mass of Hydrogen

clear, colorless, liquid with

X 1.776 g 0.148 g
strong odor

clear, colorless, liquid with

Y 1.974 g 0.329 g
strong odor

clear, colorless, liquid with

Z 7.812 g 0.651 g
strong odor

Answer
X+Z are similar compounds (same ratios of C and H), aligning with the Law of Definite Proportions
X+Y and Y+Z are different compounds (differing ratios of C and H), aligning with the Law of Multiple Proportions

Click here to see a video of the solution.

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 Problem 2.1.3

PROBLEM 2.1.1.4

How are electrons and protons similar? How are they different?

Answer
Electrons and protons are both charged subatomic particles.
Protons are much larger than electrons (contributing more mass to the overall atom).
Changing the number of protons changes the identity of the atom, which changing the number of electrons changes the
charge.

PROBLEM 2.1.1.5

How are protons and neutrons similar? How are they different?

Answer
Protons and neutrons are both located in the nucleus of the atom.
Protons and neutrons both contribute to the overall mass of the atom.
Protons carry a charge while neutrons are neutral.

PROBLEM 2.1.1.6

Predict and test the behavior of α particles fired at a “plum pudding” model atom.
(a) Predict the paths taken by α particles that are fired at atoms with a Thomson’s plum pudding model structure. Explain
why you expect the α particles to take these paths.
(b) If α particles of higher energy than those in (a) are fired at plum pudding atoms, predict how their paths will differ
from the lower-energy α particle paths. Explain your reasoning.
(c) Now test your predictions from (a) and (b).

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 Rutherford Scatter

Plum Pudding

Rutherford Atom

Select the “Plum Pudding Atom” tab above. Set “Alpha Particles Energy” to “min,” and select “show traces.” Click on the gun
to start firing α particles. Does this match your prediction from (a)? If not, explain why the actual path would be that shown in
the simulation. Hit the pause button, or “Reset All.” Set “Alpha Particles Energy” to “max,” and start firing α particles. Does
this match your prediction from (b)? If not, explain the effect of increased energy on the actual paths as shown in the
simulation.

Answer a
The plum pudding model indicates that the positive charge is spread uniformly throughout the atom, so we expect the α
particles to (perhaps) be slowed somewhat by the positive-positive repulsion, but to follow straight-line paths (i.e., not to be
deflected) as they pass through the atoms.

Answer b
Higher-energy α particles will be traveling faster (and perhaps slowed less) and will also follow straight-line paths through
the atoms.

Answer c

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 The α particles followed straight-line paths through the plum pudding atom. There was no apparent slowing of the α
particles as they passed through the atoms.

PROBLEM 2.1.1.7

Predict and test the behavior of α particles fired at a Rutherford atom model.
(a) Predict the paths taken by α particles that are fired at atoms with a Rutherford atom model structure. Explain why you
expect the α particles to take these paths.
(b) If α particles of higher energy than those in (a) are fired at Rutherford atoms, predict how their paths will differ from
the lower-energy α particle paths. Explain your reasoning.
(c) Predict how the paths taken by the α particles will differ if they are fired at Rutherford atoms of elements other than
gold. What factor do you expect to cause this difference in paths, and why?
(d) Now test your predictions from (a), (b), and (c).

Rutherford Scatter

Plum Pudding

Rutherford Atom

2.1.1.4 https://chem.libretexts.org/@go/page/210622
 Select the “Rutherford Atom” tab above. Due to the scale of the simulation, it is best to start with a small nucleus, so
select “20” for both protons and neutrons, “min” for energy, show traces, and then start firing α particles. Does this
match your prediction from (a)? If not, explain why the actual path would be that shown in the simulation. Pause or
reset, set energy to “max,” and start firing α particles. Does this match your prediction from (b)? If not, explain the effect
of increased energy on the actual path as shown in the simulation. Pause or reset, select “40” for both protons and
neutrons, “min” for energy, show traces, and fire away. Does this match your prediction from (c)? If not, explain why the
actual path would be that shown in the simulation. Repeat this with larger numbers of protons and neutrons. What
generalization can you make regarding the type of atom and effect on the path of α particles? Be clear and specific.

Answer a
The Rutherford atom has a small, positively charged nucleus, so most α particles will pass through empty space far from
the nucleus and be undeflected. Those α particles that pass near the nucleus will be deflected from their paths due to
positive-positive repulsion. The more directly toward the nucleus the α particles are headed, the larger the deflection angle
will be.

Answer b
Higher-energy α particles that pass near the nucleus will still undergo deflection, but the faster they travel, the less the
expected angle of deflection

Answer c
If the nucleus is smaller, the positive charge is smaller and the expected deflections are smaller—both in terms of how
closely the α particles pass by the nucleus undeflected and the angle of deflection. If the nucleus is larger, the positive
charge is larger and the expected deflections are larger—more α particles will be deflected, and the deflection angles will be
larger.

Answer d
The paths followed by the α particles match the predictions from (a), (b), and (c)

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Contributors and Attributions

Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley
(Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed
under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-
2bd...a7ac8df6@9.110).
Adelaide Clark, Oregon Institute of Technology

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2.1.1: A History of Atomic Theory (Problems) is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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2.1.2: The Structure of the Atom and How We Represent It
Learning Objectives
Describe the three subatomic particles that compose atoms
Define isotopes and give examples for several elements
Write and interpret symbols that depict the atomic number, mass number, and charge of an atom or ion

A Primer on Subatomic Particles and Atoms

What Is An Atom - Part 1 | Properties of …

Video 2.1.2.1 : A Primer on What an Atom is from Fuse School Chemistry.

The development of modern atomic theory revealed much about the inner structure of atoms. It was learned that an atom contains a
very small nucleus composed of positively charged protons and uncharged neutrons, surrounded by a much larger volume of space
containing negatively charged electrons. The nucleus contains the majority of an atom’s mass because protons and neutrons are
much heavier than electrons, whereas electrons occupy almost all of an atom’s volume. The diameter of an atom is on the order of
10−10 m, whereas the diameter of the nucleus is roughly 10−15 m—about 100,000 times smaller. For a perspective about their
relative sizes, consider this: If the nucleus were the size of a blueberry, the atom would be about the size of a football stadium
(Figure 2.1.2.1).

Figure 2.1.2.1 : If an atom could be expanded to the size of a football stadium, the nucleus would be the size of a single
blueberry. (credit middle: modification of work by “babyknight”/Wikimedia Commons; credit right: modification of work by
Paxson Woelber).
Atoms—and the protons, neutrons, and electrons that compose them—are extremely small. For example, a carbon atom weighs
less than 2 × 10−23 g, and an electron has a charge of less than 2 × 10−19 C (coulomb). When describing the properties of tiny
objects such as atoms, we use appropriately small units of measure, such as the atomic mass unit (amu) and the fundamental unit of
charge (e). The amu was originally defined based on hydrogen, the lightest element, then later in terms of oxygen. Since 1961, it
has been defined with regard to the most abundant isotope of carbon, atoms of which are assigned masses of exactly 12 amu. (This
isotope is known as “carbon-12” as will be discussed later in this module.) Thus, one amu is exactly 1/12 of the mass of one
carbon-12 atom: 1 amu = 1.6605 × 10−24 g. (The Dalton (Da) and the unified atomic mass unit (u) are alternative units that are

2.1.2.1 https://chem.libretexts.org/@go/page/210623
equivalent to the amu.) The fundamental unit of charge (also called the elementary charge) equals the magnitude of the charge of an
electron (e) with e = 1.602 × 10−19 C.
A proton has a mass of 1.0073 amu and a charge of 1+. A neutron is a slightly heavier particle with a mass 1.0087 amu and a
charge of zero; as its name suggests, it is neutral. The electron has a charge of 1− and is a much lighter particle with a mass of
about 0.00055 amu (it would take about 1800 electrons to equal the mass of one proton. The properties of these fundamental
particles are summarized in Table 2.1.2.1. (An observant student might notice that the sum of an atom’s subatomic particles does
not equal the atom’s actual mass: The total mass of six protons, six neutrons, and six electrons is 12.0993 amu, slightly larger than
the 12.00 amu of an actual carbon-12 atom. This “missing” mass is known as the mass defect, and you will learn about it in the
chapter on nuclear chemistry.)
Table 2.1.2.1 : Properties of Subatomic Particles
Name Location Charge (C) Unit Charge Mass (amu) Mass (g)

electron outside nucleus −1.602 × 10

−19
1− 0.00055 0.00091 × 10
−24

proton nucleus 1.602 × 10

−19
1+ 1.00727 1.67262 × 10
−24

neutron nucleus 0 0 1.00866 1.67493 × 10

−24

The number of protons in the nucleus of an atom is its atomic number (Z). This is the defining trait of an element: Its value
determines the identity of the atom. For example, any atom that contains six protons is the element carbon and has the atomic
number 6, regardless of how many neutrons or electrons it may have. A neutral atom must contain the same number of positive and
negative charges, so the number of protons equals the number of electrons. Therefore, the atomic number also indicates the number
of electrons in an atom. The total number of protons and neutrons in an atom is called its mass number (A). The number of
neutrons is therefore the difference between the mass number and the atomic number: A – Z = number of neutrons.
atomic number (Z) = number of protons
mass number (A) = number of protons + number of neutrons
A−Z = number of neutrons

Atoms are electrically neutral if they contain the same number of positively charged protons and negatively charged electrons.
When the numbers of these subatomic particles are not equal, the atom is electrically charged and is called an ion. The charge of an
atom is defined as follows:
Atomic charge = number of protons − number of electrons
As will be discussed in more detail later in this chapter, atoms (and molecules) typically acquire charge by gaining or losing
electrons. An atom that gains one or more electrons will exhibit a negative charge and is called an anion. Positively charged atoms
called cations are formed when an atom loses one or more electrons. For example, a neutral sodium atom (Z = 11) has 11 electrons.
If this atom loses one electron, it will become a cation with a 1+ charge (11 − 10 = 1+). A neutral oxygen atom (Z = 8) has eight
electrons, and if it gains two electrons it will become an anion with a 2− charge (8 − 10 = 2−).

Example 2.1.2.1: Composition of an Atom

Iodine is an essential trace element in our diet; it is needed to produce thyroid hormone. Insufficient iodine in the diet can lead
to the development of a goiter, an enlargement of the thyroid gland (Figure 2.1.2.2).

2.1.2.2 https://chem.libretexts.org/@go/page/210623
 Figure 2.1.2.2 : (a) Insufficient iodine in the diet can cause an enlargement of the thyroid gland called a goiter. (b) The
addition of small amounts of iodine to salt, which prevents the formation of goiters, has helped eliminate this concern in the
US where salt consumption is high. (credit a: modification of work by “Almazi”/Wikimedia Commons; credit b: modification
of work by Mike Mozart)
The addition of small amounts of iodine to table salt (iodized salt) has essentially eliminated this health concern in the United
States, but as much as 40% of the world’s population is still at risk of iodine deficiency. The iodine atoms are added as anions,
and each has a 1− charge and a mass number of 127. Determine the numbers of protons, neutrons, and electrons in one of these
iodine anions.
Solution
The atomic number of iodine (53) tells us that a neutral iodine atom contains 53 protons in its nucleus and 53 electrons outside
its nucleus. Because the sum of the numbers of protons and neutrons equals the mass number, 127, the number of neutrons is
74 (127 − 53 = 74). Since the iodine is added as a 1− anion, the number of electrons is 54 [53 – (1–) = 54].

Exercise 2.1.2.1

An ion of platinum has a mass number of 195 and contains 74 electrons. How many protons and neutrons does it contain, and
what is its charge?

Answer
78 protons; 117 neutrons; charge is 4+

Chemical Symbols
A chemical symbol is an abbreviation that we use to indicate an element or an atom of an element. For example, the symbol for
mercury is Hg (Figure 2.1.2.3). We use the same symbol to indicate one atom of mercury (microscopic domain) or to label a
container of many atoms of the element mercury (macroscopic domain).

2.1.2.3 https://chem.libretexts.org/@go/page/210623
Figure 2.1.2.3 : The symbol Hg represents the element mercury regardless of the amount; it could represent one atom of mercury
or a large amount of mercury. from Wikipedia (user: Materialscientist).
The symbols for several common elements and their atoms are listed in Table 2.1.2.2. Some symbols are derived from the common
name of the element; others are abbreviations of the name in another language. Symbols have one or two letters, for example, H for
hydrogen and Cl for chlorine. To avoid confusion with other notations, only the first letter of a symbol is capitalized. For example,
Co is the symbol for the element cobalt, but CO is the notation for the compound carbon monoxide, which contains atoms of the
elements carbon (C) and oxygen (O). All known elements and their symbols are in the periodic table.
Table 2.1.2.2 : Some Common Elements and Their Symbols
Element Symbol Element Symbol

aluminum Al iron Fe (from ferrum)

bromine Br lead Pb (from plumbum)

calcium Ca magnesium Mg

carbon C mercury Hg (from hydrargyrum)

chlorine Cl nitrogen N

chromium Cr oxygen O

cobalt Co potassium K (from kalium)

copper Cu (from cuprum) silicon Si

fluorine F silver Ag (from argentum)

gold Au (from aurum) sodium Na (from natrium)

helium He sulfur S

hydrogen H tin Sn (from stannum)

iodine I zinc Zn

Traditionally, the discoverer (or discoverers) of a new element names the element. However, until the name is recognized by the
International Union of Pure and Applied Chemistry (IUPAC), the recommended name of the new element is based on the Latin
word(s) for its atomic number. For example, element 106 was called unnilhexium (Unh), element 107 was called unnilseptium
(Uns), and element 108 was called unniloctium (Uno) for several years. These elements are now named after scientists or locations;
for example, element 106 is now known as seaborgium (Sg) in honor of Glenn Seaborg, a Nobel Prize winner who was active in
the discovery of several heavy elements.

2.1.2.4 https://chem.libretexts.org/@go/page/210623
 IUPAC
Visit this site to learn more about IUPAC, the International Union of Pure and Applied Chemistry, and explore its periodic
table.

Contributors and Attributions

Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley
(Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed
under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-
2bd...a7ac8df6@9.110).
Adelaide Clark, Oregon Institute of Technology
Fuse School, Open Educational Resource free of charge, under a Creative Commons License: Attribution-NonCommercial CC
BY-NC (View License Deed: https://creativecommons.org/licenses/by-nc/4.0/)

Feedback
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Found a typo and want extra credit? Click here.

2.1.2: The Structure of the Atom and How We Represent It is shared under a not declared license and was authored, remixed, and/or curated by
LibreTexts.

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2.1.3: The Structure of the Atom and How We Represent It (Problems)
PROBLEM 2.1.3.1

In what way are isotopes of a given element always different? In what way(s) are they always the same?

Answer
They always have different masses due to different numbers of neutrons.
They always have the same number of protons (which determines the identity).

PROBLEM 2.1.3.2

Write the symbol for each of the following ions:

(a) the ion with a 1+ charge, atomic number 55, and mass number 133
(b) the ion with 54 electrons, 53 protons, and 74 neutrons
(c) the ion with atomic number 15, mass number 31, and a 3− charge
(d) the ion with 24 electrons, 30 neutrons, and a 3+ charge

Answer a
133
Cs
+
55

Answer b
127
I
-
53

Answer c
31 3-
P
15

Answer d
57
Co
3+
27

Click here for a video of the solution.

PROBLEM 2.1.3.3

Write the symbol for each of the following ions:

(a) the ion with a 3+ charge, 28 electrons, and a mass number of 71

2.1.3.1 https://chem.libretexts.org/@go/page/210624
 (b) the ion with 36 electrons, 35 protons, and 45 neutrons
(c) the ion with 86 electrons, 142 neutrons, and a 4+ charge
(d) the ion with a 2+ charge, atomic number 38, and mass number 87

Answer a
71 3+
31 Ga

Answer b
80 -
35 Br

Answer c
232 4+
90 Th

Answer d
87 2+
38 Sr

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Contributors and Attributions

Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley
(Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed
under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-
2bd...a7ac8df6@9.110).
Adelaide Clark, Oregon Institute of Technology

Feedback
Think one of the answers above is wrong? Let us know here.

2.1.3: The Structure of the Atom and How We Represent It (Problems) is shared under a not declared license and was authored, remixed, and/or
curated by LibreTexts.

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2.1.4: Calculating Atomic Masses
Learning Objectives
Define the atomic mass unit and average atomic mass
Calculate average atomic mass and isotopic abundance
Define the amount unit mole and the related quantity Avogadro’s number
Explain the relation between mass, moles, and numbers of atoms or molecules, and perform calculations deriving these
quantities from one another

What Is An Atom - Part 2 - Isotopes | Pro…

Pro…

Video 2.1.4.1 : A review of counting subatomic particles and a preview of isotopes and relative atomic mass.

Isotopes
The symbol for a specific isotope of any element is written by placing the mass number as a superscript to the left of the element
symbol (Figure 2.1.4.4). The atomic number is sometimes written as a subscript preceding the symbol, but since this number
defines the element’s identity, as does its symbol, it is often omitted. For example, magnesium exists as a mixture of three isotopes,
each with an atomic number of 12 and with mass numbers of 24, 25, and 26, respectively. These isotopes can be identified as 24Mg,
25
Mg, and 26Mg. These isotope symbols are read as “element, mass number” and can be symbolized consistent with this reading.
For instance, 24Mg is read as “magnesium 24,” and can be written as “magnesium-24” or “Mg-24.” 25Mg is read as “magnesium
25,” and can be written as “magnesium-25” or “Mg-25.” All magnesium atoms have 12 protons in their nucleus. They differ only
because a 24Mg atom has 12 neutrons in its nucleus, a 25Mg atom has 13 neutrons, and a 26Mg has 14 neutrons.

Figure 2.1.4.1 : The symbol for an atom indicates the element via its usual two-letter symbol, the mass number as a left
superscript, the atomic number as a left subscript (sometimes omitted), and the charge as a right superscript.
Information about the naturally occurring isotopes of elements with atomic numbers 1 through 10 is given in Table 2.1.4.2. Note
that in addition to standard names and symbols, the isotopes of hydrogen are often referred to using common names and
accompanying symbols. Hydrogen-2, symbolized 2H, is also called deuterium and sometimes symbolized D. Hydrogen-3,
symbolized 3H, is also called tritium and sometimes symbolized T.
Table 2.1.4.1 : Nuclear Compositions of Atoms of the Very Light Elements
Number of Number of % Natural
Element Symbol Atomic Number Mass (amu)
Protons Neutrons Abundance

2.1.4.1 https://chem.libretexts.org/@go/page/210625
 Number of Number of % Natural
Element Symbol Atomic Number Mass (amu)
Protons Neutrons Abundance
1
1H
1 1 0 1.0078 99.989
(protium)
2
1H
hydrogen 1 1 1 2.0141 0.0115
(deuterium)
3
1H
1 1 2 3.01605 — (trace)
(tritium)
3
2 He 2 2 1 3.01603 0.00013
helium
4
2 He 2 2 2 4.0026 100
6
3 Li 3 3 3 6.0151 7.59
lithium
7
3 Li 3 3 4 7.0160 92.41
9
beryllium 4 Be 4 4 5 9.0122 100
10
5B 5 5 5 10.0129 19.9
boron
11
5B 5 5 6 11.0093 80.1
12
6C 6 6 6 12.0000 98.89
13
carbon 6C 6 6 7 13.0034 1.11
14
6C 6 6 8 14.0032 — (trace)
14
7N 7 7 7 14.0031 99.63
nitrogen
15
7N 7 7 8 15.0001 0.37
16
8O 8 8 8 15.9949 99.757
17
oxygen 8O 8 8 9 16.9991 0.038
18
8O 8 8 10 17.9992 0.205
19
fluorine 9F 9 9 10 18.9984 100
20
10 Ne 10 10 10 19.9924 90.48
21
neon 10 Ne 10 10 11 20.9938 0.27
22
10 Ne 10 10 12 21.9914 9.25

2.1.4.2 https://chem.libretexts.org/@go/page/210625
 Build an Atom

Symbol

Atom

Use this Build an Atom simulator to build atoms of the first 10 elements, see which isotopes exist, check nuclear stability, and gain
experience with isotope symbols.

Atomic Mass
Because each proton and each neutron contribute approximately one amu to the mass of an atom, and each electron contributes far
less, the atomic mass of a single atom is approximately equal to its mass number (a whole number). However, the average masses
of atoms of most elements are not whole numbers because most elements exist naturally as mixtures of two or more isotopes.
The mass of an element shown in a periodic table or listed in a table of atomic masses is a weighted, average mass of all the
isotopes present in a naturally occurring sample of that element. This is equal to the sum of each individual isotope’s mass
multiplied by its fractional abundance.

average mass = ∑ i
(fractional abundance × isotopic mass)i (2.1.4.1)

For example, the element boron is composed of two isotopes: About 19.9% of all boron atoms are 10B with a mass of 10.0129 amu,
and the remaining 80.1% are 11B with a mass of 11.0093 amu. The average atomic mass for boron is calculated to be:

2.1.4.3 https://chem.libretexts.org/@go/page/210625
 boron average mass = (0.199 × 10.0129 amu) + (0.801 × 11.0093 amu)

= 1.99 amu + 8.82 amu

= 10.81 amu

It is important to understand that no single boron atom weighs exactly 10.8 amu; 10.8 amu is the average mass of all boron atoms,
and individual boron atoms weigh either approximately 10 amu or 11 amu.

Example 2.1.4.1: Calculation of Average Atomic Mass

A meteorite found in central Indiana contains traces of the noble gas neon picked up from the solar wind during the meteorite’s
trip through the solar system. Analysis of a sample of the gas showed that it consisted of 91.84% 20Ne (mass 19.9924 amu),
0.47% 21Ne (mass 20.9940 amu), and 7.69% 22Ne (mass 21.9914 amu). What is the average mass of the neon in the solar
wind?
Solution
average mass = (0.9184 × 19.9924 amu) + (0.0047 × 20.9940 amu) + (0.0769 × 21.9914 amu)

= (18.36 + 0.099 + 1.69) amu

= 20.15 amu

The average mass of a neon atom in the solar wind is 20.15 amu. (The average mass of a terrestrial neon atom is 20.1796 amu.
This result demonstrates that we may find slight differences in the natural abundance of isotopes, depending on their origin.)

Exercise 2.1.4.1

A sample of magnesium is found to contain 78.70% of 24Mg atoms (mass 23.98 amu), 10.13% of 25
Mg atoms (mass 24.99
amu), and 11.17% of 26Mg atoms (mass 25.98 amu). Calculate the average mass of a Mg atom.

Answer
24.31 amu

We can also do variations of this type of calculation, as shown in the next example.

Example 2.1.4.2: Calculation of Percent Abundance

Naturally occurring chlorine consists of 35Cl (mass 34.96885 amu) and 37Cl (mass 36.96590 amu), with an average mass of
35.453 amu. What is the percent composition of Cl in terms of these two isotopes?
Solution
The average mass of chlorine is the fraction that is 35Cl times the mass of 35Cl plus the fraction that is 37
Cl times the mass of
37
Cl.
35 35 37 37
average mass = (f raction of Cl × mass of Cl) + (f raction of Cl × mass of Cl) (2.1.4.2)

If we let x represent the fraction that is 35Cl, then the fraction that is 37Cl is represented by 1.00 − x.
(The fraction that is 35Cl + the fraction that is 37
Cl must add up to 1, so the fraction of 37
Cl must equal 1.00 − the fraction of
35
Cl.)
Substituting this into the average mass equation, we have:
35.453 amu =( x × 34.96885 amu) + [(1.00 − x) × 36.96590 amu]
35.453 = 34.96885 x + 36.96590 − 36.96590x
1.99705 x = 1.513
1.513
x = = 0.7576
1.99705

35
So solving yields: x = 0.7576, which means that 1.00 − 0.7576 = 0.2424. Therefore, chlorine consists of 75.76% Cl and
24.24% 37Cl.

2.1.4.4 https://chem.libretexts.org/@go/page/210625
Exercise 2.1.4.2

Naturally occurring copper consists of 63Cu (mass 62.9296 amu) and 65Cu (mass 64.9278 amu), with an average mass of
63.546 amu. What is the percent composition of Cu in terms of these two isotopes?

Answer
69.15% Cu-63 and 30.85% Cu-65

Isotopes and Atomic

Mixtures

Isotopes

Use this simulator to make mixtures of the main isotopes of the first 18 elements, gain experience with average atomic mass, and
check naturally occurring isotope ratios.

2.1.4.5 https://chem.libretexts.org/@go/page/210625
 Figure 2.1.4.2 : Analysis of zirconium in a mass spectrometer produces a mass spectrum with peaks showing the different
isotopes of Zr.
The occurrence and natural abundances of isotopes can be experimentally determined using an instrument called a mass
spectrometer. Mass spectrometry (MS) is widely used in chemistry, forensics, medicine, environmental science, and many other
fields to analyze and help identify the substances in a sample of material. In a typical mass spectrometer (Figure 2.1.4.5), the
sample is vaporized and exposed to a high-energy electron beam that causes the sample’s atoms (or molecules) to become
electrically charged, typically by losing one or more electrons. These cations then pass through a (variable) electric or magnetic
field that deflects each cation’s path to an extent that depends on both its mass and charge (similar to how the path of a large steel
ball bearing rolling past a magnet is deflected to a lesser extent that that of a small steel BB). The ions are detected, and a plot of
the relative number of ions generated versus their mass-to-charge ratios (a mass spectrum) is made. The height of each vertical
feature or peak in a mass spectrum is proportional to the fraction of cations with the specified mass-to-charge ratio. Since its initial
use during the development of modern atomic theory, MS has evolved to become a powerful tool for chemical analysis in a wide
range of applications.

Mass Spectrometry MS

Video 2.1.4.2 : Watch this video from the Royal Society for Chemistry for a brief description of the rudiments of mass
spectrometry.

The Mole
The identity of a substance is defined not only by the types of atoms or ions it contains, but by the quantity of each type of atom or
ion. For example, water, H2O, and hydrogen peroxide, H2O2, are alike in that their respective molecules are composed of hydrogen
and oxygen atoms. However, because a hydrogen peroxide molecule contains two oxygen atoms, as opposed to the water molecule,
which has only one, the two substances exhibit very different properties. Today, we possess sophisticated instruments that allow the
direct measurement of these defining microscopic traits; however, the same traits were originally derived from the measurement of
macroscopic properties (the masses and volumes of bulk quantities of matter) using relatively simple tools (balances and

2.1.4.6 https://chem.libretexts.org/@go/page/210625
volumetric glassware). This experimental approach required the introduction of a new unit for amount of substances, the mole,
which remains indispensable in modern chemical science.
The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of
atoms or molecules in a bulk sample of matter. A mole is defined as the amount of substance containing the same number of
discrete entities (such as atoms, molecules, and ions) as the number of atoms in a sample of pure 12C weighing exactly 12 g. One
Latin connotation for the word “mole” is “large mass” or “bulk,” which is consistent with its use as the name for this unit. The
mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental
property, number of atoms, molecules, and so forth.
The number of entities composing a mole has been experimentally determined to be 6.02214179 × 1023, a fundamental constant
named Avogadro’s number (NA) or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly
reported with an explicit unit of “per mole,” a conveniently rounded version being 6.022 × 1023 /mol.

What Is Avogadro's Number - The Mole |…

|…

Video 2.1.4.3 : What is Avogadro's Number?

Consistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other
element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically
different. The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in
units of grams per mole (g/mol) (Figure 2.1.4.3).

Figure 2.1.4.3 : Each sample contains 6.022 × 1023 atoms —1.00 mol of atoms. From left to right (top row): 65.4 g zinc, 12.0 g
carbon, 24.3 g magnesium, and 63.5 g copper. From left to right (bottom row): 32.1 g sulfur, 28.1 g silicon, 207 g lead, and 118.7 g
tin. (credit: modification of work by Mark Ott).

2.1.4.7 https://chem.libretexts.org/@go/page/210625
Because the definitions of both the mole and the atomic mass unit are based on the same reference substance, 12C, the molar mass
of any substance is numerically equivalent to its atomic or formula weight in amu. Per the amu definition, a single 12C atom weighs
12 amu (its atomic mass is 12 amu). The former definition of the mole was that a mole was 12 g of 12C contains 1 mole of 12C
atoms (its molar mass is 12 g/mol). This relationship holds for all elements, since their atomic masses are measured relative to that
of the amu-reference substance, 12C. Extending this principle, the molar mass of a compound in grams is likewise numerically
equivalent to its formula mass in amu (Figure 2.1.4.4). On May 20, 2019 the definition was permanently changed to Avogadro's
number: a mole is 6.02214179 × 1023 of any object, from atoms to apples.1

Figure 2.1.4.4 : Each sample contains 6.022 × 1023 molecules or formula units—1.00 mol of the compound or element. Clock-
wise from the upper left: 130.2 g of C8H17OH (1-octanol, formula mass 130.2 amu), 454.4 g of HgI2 (mercury(II) iodide, formula
mass 454.4 amu), 32.0 g of CH3OH (methanol, formula mass 32.0 amu) and 256.5 g of S8 (sulfur, formula mass 256.5 amu).
(credit: Sahar Atwa).
Table 2.1.4.2 : Mass of one mole of elements
Element Average Atomic Mass (amu) Molar Mass (g/mol) Atoms/Mole

C 12.01 12.01 6.022 × 10

23

H 1.008 1.008 6.022 × 10

23

O 16.00 16.00 6.022 × 10

23

Na 22.99 22.99 6.022 × 10

23

Cl 33.45 35.45 6.022 × 10

23

While atomic mass and molar mass are numerically equivalent, keep in mind that they are vastly different in terms of scale, as
represented by the vast difference in the magnitudes of their respective units (amu versus g). To appreciate the enormity of the
mole, consider a small drop of water after a rainfall. Although this represents just a tiny fraction of 1 mole of water (~18 g), it
contains more water molecules than can be clearly imagined. If the molecules were distributed equally among the roughly seven
billion people on earth, each person would receive more than 100 billion molecules.

How big is a mole? (Not the animal, the …

2.1.4.8 https://chem.libretexts.org/@go/page/210625
Video 2.1.4.4 : The mole is used in chemistry to represent 6.022 × 1023 of something, but it can be difficult to conceptualize such
a large number. Watch this video and then complete the “Think” questions that follow. Explore more about the mole by reviewing
the information under “Dig Deeper.”
The relationships between formula mass, the mole, and Avogadro’s number can be applied to compute various quantities that
describe the composition of substances and compounds. For example, if we know the mass and chemical composition of a
substance, we can determine the number of moles and calculate number of atoms or molecules in the sample. Likewise, if we know
the number of moles of a substance, we can derive the number of atoms or molecules and calculate the substance’s mass.

Example 2.1.4.3: Deriving Moles from Grams for an Element

According to nutritional guidelines from the US Department of Agriculture, the estimated average requirement for dietary
potassium is 4.7 g. What is the estimated average requirement of potassium in moles?
Solution
The mass of K is provided, and the corresponding amount of K in moles is requested. Referring to the periodic table, the
atomic mass of K is 39.10 amu, and so its molar mass is 39.10 g/mol. The given mass of K (4.7 g) is a bit more than one-tenth
the molar mass (39.10 g), so a reasonable “ballpark” estimate of the number of moles would be slightly greater than 0.1 mol.
The molar amount of a substance may be calculated by dividing its mass (g) by its molar mass (g/mol):

The factor-label method supports this mathematical approach since the unit “g” cancels and the answer has units of “mol:”

4.7 g K ( mol K

39.10 g
) = 0.12 mol K

The calculated magnitude (0.12 mol K) is consistent with our ballpark expectation, since it is a bit greater than 0.1 mol.

Exercise 2.1.4.3: Beryllium

Beryllium is a light metal used to fabricate transparent X-ray windows for medical imaging instruments. How many moles of
Be are in a thin-foil window weighing 3.24 g?

Answer
0.360 mol

Example 2.1.4.4: Deriving Grams from Moles for an Element

A liter of air contains 9.2 × 10−4 mol argon. What is the mass of Ar in a liter of air?
Solution
The molar amount of Ar is provided and must be used to derive the corresponding mass in grams. Since the amount of Ar is less
than 1 mole, the mass will be less than the mass of 1 mole of Ar, approximately 40 g. The molar amount in question is
approximately one-one thousandth (~10−3) of a mole, and so the corresponding mass should be roughly one-one thousandth of
the molar mass (~0.04 g):

In this case, logic dictates (and the factor-label method supports) multiplying the provided amount (mol) by the molar mass
(g/mol):

2.1.4.9 https://chem.libretexts.org/@go/page/210625
 9.2 × 10−4 mol Ar ( 39.95 g
) = 0.037 g Ar
mol Ar

The result is in agreement with our expectations, around 0.04 g Ar.

Exercise 2.1.4.4

What is the mass of 2.561 mol of gold?

Answer
504.4 g

Example 2.1.4.6: Deriving Number of Atoms from Mass for an Element

Copper is commonly used to fabricate electrical wire (Figure 2.1.4.6). How many copper atoms are in 5.00 g of copper wire?

Figure 2.1.4.6 : Copper wire is composed of many, many atoms of Cu. (credit: Emilian Robert Vicol)
Solution
The number of Cu atoms in the wire may be conveniently derived from its mass by a two-step computation: first calculating
the molar amount of Cu, and then using Avogadro’s number (NA) to convert this molar amount to number of Cu atoms:

Considering that the provided sample mass (5.00 g) is a little less than one-tenth the mass of 1 mole of Cu (~64 g), a
reasonable estimate for the number of atoms in the sample would be on the order of one-tenth NA, or approximately 1022 Cu
atoms. Carrying out the two-step computation yields:
23

5.00 g Cu ( 63.55
mol Cu
g
) ( 6.022 × 10 atoms
) = 4.74 × 10
22
atoms of copper (2.1.4.3)
mol

The factor-label method yields the desired cancellation of units, and the computed result is on the order of 1022 as expected.

Exercise 2.1.4.6

A prospector panning for gold in a river collects 15.00 g of pure gold. How many Au atoms are in this quantity of gold?

Answer
4.586 × 1022 Au atoms

2.1.4.10 https://chem.libretexts.org/@go/page/210625
 Example 2.1.4.7: Deriving Moles from Grams for a Compound

Our bodies synthesize protein from amino acids. One of these amino acids is glycine, which has the molecular formula
C2H5O2N. How many moles of glycine molecules are contained in 28.35 g of glycine?
Solution
We can derive the number of moles of a compound from its mass following the same procedure we used for an element in
Example 2.1.4.6:

The molar mass of glycine is required for this calculation, and it is computed in the same fashion as its molecular mass. One
mole of glycine, C2H5O2N, contains 2 moles of carbon, 5 moles of hydrogen, 2 moles of oxygen, and 1 mole of nitrogen:

The provided mass of glycine (~28 g) is a bit more than one-third the molar mass (~75 g/mol), so we would expect the
computed result to be a bit greater than one-third of a mole (~0.33 mol). Dividing the compound’s mass by its molar mass
yields:

28.35 g glycine ( mol glycine

75.07 g
) = 0.378 mol glycine

This result is consistent with our rough estimate.

Exercise 2.1.4.7

How many moles of sucrose, C12 H22 O11 , are in a 25-g sample of sucrose?

Answer
0.073 mol

Example 2.1.4.8: Deriving Grams from Moles for a Compound

Vitamin C is a covalent compound with the molecular formula C6H8O6. The recommended daily dietary allowance of vitamin C
for children aged 4–8 years is 1.42 × 10−4 mol. What is the mass of this allowance in grams?
Solution
As for elements, the mass of a compound can be derived from its molar amount as shown:

The molar mass for this compound is computed to be 176.124 g/mol. The given number of moles is a very small fraction of a
mole (~10−4 or one-ten thousandth); therefore, we would expect the corresponding mass to be about one-ten thousandth of the

2.1.4.11 https://chem.libretexts.org/@go/page/210625
molar mass (~0.02 g). Performing the calculation, we get:

1.42 × 10−4 mol vitamin C ( 176.124 g

) = 0.0250 g vitamin C
mol vitamin C

This is consistent with the anticipated result.

Exercise 2.1.4.8

What is the mass of 0.443 mol of hydrazine, N2 H4 ?

Answer
14.2 g

Example 2.1.4.9: Deriving the Number of Molecules from the Compound Mass

A packet of an artificial sweetener contains 40.0 mg of saccharin (C7H5NO3S), which has the structural formula:

Given that saccharin has a molar mass of 183.18 g/mol, how many saccharin molecules are in a 40.0-mg (0.0400-g) sample of
saccharin? How many carbon atoms are in the same sample?
Solution
The number of molecules in a given mass of compound is computed by first deriving the number of moles, as demonstrated in
Example 2.1.4.8, and then multiplying by Avogadro’s number:

Using the provided mass and molar mass for saccharin yields:
23

0.0400 g C7 H5 NO3 S ( 183.18

mol C H NO S
7

g C H NO S
5
) ( 6.022 × 10
3
C7 H5 NO3 S molecules
) (2.1.4.4)
7 5 3 1 mol C7 H5 NO3 S
20
= 1.31 × 10 C7 H5 NO3 S molecules

The compound’s formula shows that each molecule contains seven carbon atoms, and so the number of C atoms in the provided
sample is:

1.31 × 10
20
C7 H5 NO3 S molecules ( 1 C H7NO
C atoms
S molecule
) = 9.20 × 10 21
C atoms
7 5 3

Exercise 2.1.4.9

How many C4 H10 molecules are contained in 9.213 g of this compound? How many hydrogen atoms?

Answer

2.1.4.12 https://chem.libretexts.org/@go/page/210625
 9.545 × 10
22
molecules CH
4 10

9.545 × 10
23
atoms H

How To Use Moles - Part 1 | Chemical C…

C…

Video 2.1.4.5 : A preview of some of the uses we will have for moles in upcoming units

Summary

The Nucleus: Crash Course Chemistry #1

Video 2.1.4.6 : Watch this video for a review of relative atomic mass and isotopes.
An atom consists of a small, positively charged nucleus surrounded by electrons. The nucleus contains protons and neutrons; its
diameter is about 100,000 times smaller than that of the atom. The mass of one atom is usually expressed in atomic mass units
(amu), which is referred to as the atomic mass. An amu is defined as exactly 1/12 of the mass of a carbon-12 atom and is equal to
1.6605 × 10−24 g.
Protons are relatively heavy particles with a charge of 1+ and a mass of 1.0073 amu. Neutrons are relatively heavy particles with no
charge and a mass of 1.0087 amu. Electrons are light particles with a charge of 1− and a mass of 0.00055 amu. The number of
protons in the nucleus is called the atomic number (Z) and is the property that defines an atom’s elemental identity. The sum of the
numbers of protons and neutrons in the nucleus is called the mass number and, expressed in amu, is approximately equal to the
mass of the atom. An atom is neutral when it contains equal numbers of electrons and protons.
Isotopes of an element are atoms with the same atomic number but different mass numbers; isotopes of an element, therefore, differ
from each other only in the number of neutrons within the nucleus. When a naturally occurring element is composed of several
isotopes, the atomic mass of the element represents the average of the masses of the isotopes involved. A chemical symbol
identifies the atoms in a substance using symbols, which are one-, two-, or three-letter abbreviations for the atoms.

2.1.4.13 https://chem.libretexts.org/@go/page/210625
Looking Beyond

Just How Small is an Atom?

Video 2.1.4.7 : Remember our exploration into the size of an atom last week? This video goes deeper into investigating the size of
the subatomic particles we just discussed.

Footnotes
1. Read more about the redefinition of SI units including the kilogram here (Laura Howe, CE&N, Nov. 16, 2018).

Key Equations
average mass = ∑ i
(fractional abundance × isotopic mass)i

Glossary
anion
negatively charged atom or molecule (contains more electrons than protons)

atomic mass
average mass of atoms of an element, expressed in amu

atomic mass unit (amu)

1
(also, unified atomic mass unit, u, or Dalton, Da) unit of mass equal to of the mass of a 12C atom
12

atomic number (Z)

number of protons in the nucleus of an atom

cation
positively charged atom or molecule (contains fewer electrons than protons)

chemical symbol
one-, two-, or three-letter abbreviation used to represent an element or its atoms

Dalton (Da)
alternative unit equivalent to the atomic mass unit

fundamental unit of charge

(also called the elementary charge) equals the magnitude of the charge of an electron (e) with e = 1.602 × 10−19 C

ion
electrically charged atom or molecule (contains unequal numbers of protons and electrons)

mass number (A)

2.1.4.14 https://chem.libretexts.org/@go/page/210625
 sum of the numbers of neutrons and protons in the nucleus of an atom
mole
amount of substance containing the same number of atoms, molecules, ions, or other entities as the number of atoms in exactly
12 grams of 12C
molar mass
mass in grams of 1 mole of a substance
unified atomic mass unit (u)
alternative unit equivalent to the atomic mass unit

Contributors and Attributions

Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley
(Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed
under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-
2bd...a7ac8df6@9.110).

Adelaide Clark, Oregon Institute of Technology

Fuse School, Open Educational Resource free of charge, under a Creative Commons License: Attribution-NonCommercial CC
BY-NC (View License Deed: https://creativecommons.org/licenses/by-nc/4.0/)
Crash Course Chemistry, Crash Course is a division of Complexly and videos are free to stream for educational purposes.
TED-Ed’s commitment to creating lessons worth sharing is an extension of TED’s mission of spreading great ideas. Within
TED-Ed’s growing library of TED-Ed animations, you will find carefully curated educational videos, many of which represent
collaborations between talented educators and animators nominated through the TED-Ed website.

Feedback
Have feedback to give about this text? Click here.
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2.1.4: Calculating Atomic Masses is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

2.1.4.15 https://chem.libretexts.org/@go/page/210625
2.1.5: Calculating Atomic Masses (Problems)
PROBLEM 2.1.5.1

Determine the number of protons, neutrons, and electrons in the following isotopes that are used in medical diagnoses:
(a) atomic number 9, mass number 18, charge of 1−
(b) atomic number 43, mass number 99, charge of 7+
(c) atomic number 53, atomic mass number 131, charge of 1−
(d) atomic number 81, atomic mass number 201, charge of 1+
(e) Name the elements in parts (a), (b), (c), and (d)

Answer a
p: 9; n: 9; e: 10

Answer b
p: 43; n: 56; e: 36

Answer c
p: 53; n: 78; e: 54

Answer d
p: 81; n: 120; e: 80

Answer e
a - F; b - Tc; c - I; d - Tl

Click here to see a video of the solution.

**Please note that there is a misspoken element name in the video - Tl is Thallium, not Tellerium.

PROBLEM 2.1.5.2

Give the number of protons, electrons, and neutrons in neutral atoms of each of the following isotopes:
(a) 105 B
(b) 199
80
Hg

(c) 63
29
Cu

(d) 136 C

2.1.5.1 https://chem.libretexts.org/@go/page/210626
(e) 77
34
Se

Answer a
p&e: 5; n: 5

Answer b
p&e: 80; n: 119

Answer c
p&e: 29; n: 34

Answer d
p&e: 6; n: 7

Answer e
p&e: 34; n: 43

PROBLEM 2.1.5.3

An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance
with 20.99 amu, and 8.82% abundance with 21.99 amu. Calculate the average atomic mass of this element.

Answer
20.16 amu

Click here to see a video of the solution.

Problem 2.3.3

PROBLEM 2.1.5.4

Average atomic masses listed by IUPAC are based on a study of experimental results. Bromine has two isotopes, 79Br and 81Br,
whose masses (78.9183 and 80.9163 amu) and abundances (50.69% and 49.31%) were determined in earlier experiments.
Calculate the average atomic mass of Br based on these experiments. How does this compare to the value given on the periodic
table?

Answer
79.90 amu; this matches the value on the periodic table

2.1.5.2 https://chem.libretexts.org/@go/page/210626
PROBLEM 2.1.5.5

The 18O:16O abundance ratio in some meteorites is greater than that used to calculate the average atomic mass of oxygen on
earth. Is the average atomic mass of an oxygen atom in these meteorites greater than, less than, or equal to a terrestrial oxygen
atom?

Answer
18
Greater, since the contribution to the average atomic mass of O is greater, that will raise the average atomic mass in
meteorites compared to on earth.

PROBLEM 2.1.5.6

Compare 1 mole of H2, 1 mole of O2, and 1 mole of F2.

(a) Which has the largest number of molecules? Explain why.
(b) Which has the greatest mass? Explain why.

Answer a
1 mole is always 6.022 x 1023 molecules. They have the same number of molecules.

Answer b
F2; it has the highest molar mass.

PROBLEM 2.1.5.7

Which contains the greatest mass of oxygen: 0.75 mol of ethanol (C2H5OH), 0.60 mol of formic acid (HCO2H), or 1.0 mol of
water (H2O)? Explain why.

Answer
Formic acid. Its formula has twice as many oxygen atoms as the other two compounds (one each). Therefore, 0.60 mol of
formic acid would be equivalent to 1.20 mol of a compound containing a single oxygen atom.

PROBLEM 2.1.5.8

Determine the mass of each of the following:

(a) 0.0146 mol KOH
(b) 10.2 mol ethane, C2H6
(c) 1.6 × 10−3 mol Na2 SO4
(d) 6.854 × 103 mol glucose, C6 H12 O6
(e) 2.86 mol Co(NH3)6Cl3

Answer a
0.819 g

Answer b
307 g

Answer c
0.23 g

Answer d
1.235 × 106 g (1235 kg)

2.1.5.3 https://chem.libretexts.org/@go/page/210626
 Answer e
765 g

PROBLEM 2.1.5.9

Which of the following represents the least number of molecules?

a. 20.0 g of H2O (18.02 g/mol)
b. 77.0 g of CH4 (16.06 g/mol)
c. 68.0 g of CaH2 (42.09 g/mol)
d. 100.0 g of N2O (44.02 g/mol)
e. 84.0 g of HF (20.01 g/mol)

Answer
20.0 g of H2O represents the smallest number of moles, meaning the least number of molecules present. Since 1 mole =
6.022 × 1023 molecules (or atoms) regardless of identity, the least number of moles will equal the least number of
molecules.

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Let your professors know here.
***Please know that you are helping future students - videos will be made in time for next term's class.

Contributors and Attributions

Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley
(Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed
under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-
2bd...a7ac8df6@9.110).
Adelaide Clark, Oregon Institute of Technology

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Think one of the answers above is wrong? Let us know here.

2.1.5: Calculating Atomic Masses (Problems) is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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