CHEMISTRY

IN CONTEXT
CHEMISTRY

Seventh edition
CHEMISTRY IN CONTEXT for Cambridge International AS & A Level
Authors
for Cambridge International Graham Hill
John Holman
AS & A Level Seventh edition Philippa Gardom Hulme

Chemistry in Context for Cambridge International AS & A Level directly matches the latest
syllabus, supporting comprehensive understanding of key scientific concepts. A stretching
approach focuses on the development of advanced skills, for strong assessment potential.
IN CONTEXT
for Cambridge International
• Fully prepare for exams – comprehensive coverage of the course
• Develop advanced skills – critical thinking and practical skills extend performance
• Progress to the next stage – differentiated extension material eases the transition to
university study
AS & A Level
Empowering every learner to succeed and progress
Seventh edition
Complete Cambridge syllabus match
Comprehensive exam preparation
Reviewed by subject specialists
Embedded critical thinking skills
Progression to the next educational stage

Also available:
978 0 19 839962 9 978 0 19 839959 9

Philippa Gardom Hulme

John Holman
Graham Hill
Graham Hill
John Holman
Philippa Gardom Hulme

Oxford excellence for Cambridge AS & A Level

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 Contents

Introduction viii 3.5 Describing covalent bonds in terms

of overlapping orbitals 48
3.6 Co-ordinate (dative covalent) bonding 50
Chapter 1 Atoms, molecules and equations
3.7 The shapes of simple molecules 51
3.8 Metallic bonding 54
1.1 Atoms and molecules 1
3.9 Polar bonds 55
1.2 Comparing the masses of atoms 2
3.10 Polar and non-polar molecules 56
1.3 Relative atomic masses – the 12C scale 4
3.11 Intermolecular forces 57
1.4 Moles, molar masses and the
3.12 Hydrogen bonding 59
Avogadro constant 5
3.13 Estimating the strength of
1.5 Using mass spectra to calculate
hydrogen bonds in water 61
relative atomic masses 7
3.14 The effect and importance of
1.6 Empirical and molecular formulae 8
hydrogen bonding in water and ice 62
1.7 Predicting and writing chemical equations 10
1.8 Calculations involving formulae End of chapter questions 64
and equations 12
1.9 Determining equations 15 Chapter 4 States of matter
1.10 Ionic equations 16
End of chapter questions 18 4.1 Evidence for moving particles 65
4.2 The kinetic–molecular theory 66
Chapter 2 Atomic and electronic structure 4.3 Changes of state 67
4.4 The gaseous state 69
2.1 Introduction 20 4.5 The kinetic theory and ideal gases 72
2.2 Evidence for atomic structure 20 4.6 Determining the relative
2.3 Sub-atomic particles 24 molecular masses of gases and
2.4 Proton number, nucleon number volatile liquids 74
and isotopes 25 4.7 The solid state 76
2.5 Isotopes, nucleon numbers and 4.8 Evidence for the structure of solids 77
relative isotopic masses 26 4.9 Giant metallic structures 78
2.6 Evidence for the electronic 4.10 The properties and uses of metals 79
structure of atoms 27 4.11 Giant ionic structures 80
2.7 Obtaining ionisation energies from 4.12 The properties of ionic solids 82
emission spectra 28 4.13 Giant molecular (giant covalent)
2.8 Using ionisation energies to predict structures 83
electronic structures – evidence for 4.14 Silicon(iv) oxide, silicates and
shells of electrons 30 ceramics – more giant structures 85
2.9 How are the electrons arranged in 4.15 Simple molecular structures 86
larger atoms? 31 4.16 Fullerenes 87
2.10 Evidence for sub-shells of electrons 33 4.17 Materials as finite resources – recycling 88
2.11 Trends in ionisation energies 36 4.18 Comparing typical solid structures 89
2.12 Electrons and orbitals 38 End of chapter questions 90
End of chapter questions 40
Chapter 5 Chemical energetics
Chapter 3 Chemical bonding and intermolecular
forces 5.1 Energy and energy changes 92
5.2 The ideas and language of
3.1 A theory for chemical bonding 42 thermochemistry 92
3.2 Transfer of electrons – ionic 5.3 Measuring enthalpy changes 94
(electrovalent) bonding 43 5.4 The standard conditions for
3.3 The properties of ionic compounds 45 thermochemical measurements 97
3.4 Sharing electrons – covalent bonding 45 5.5 Standard enthalpy changes 98

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5.6 Measuring standard enthalpy 7.14 Acids, bases and equilibria 153
changes of formation 100 7.15 The Brønsted–Lowry theory of
5.7 Using Hess’ law and enthalpy acids and bases 154
changes of formation to calculate 7.16 Acid–base reactions: competition
the energy changes in reactions 101 for protons 156
5.8 Using standard enthalpy changes End of chapter questions 157
of formation to predict the relative
stabilities of compounds 102
Chapter 8 Rates of reaction
5.9 Predicting whether reactions will occur 104
5.10 Enthalpy changes and bond energies 105
8.1 Introduction 159
5.11 Using bond energies 106
8.2 The concept of reaction rate 159
5.12 Enthalpy changes and ionic bonding 107
8.3 Factors affecting the rate of a reaction 161
5.13 Born–Haber cycles 109
8.4 Investigating the effect of
5.14 Testing the model of ionic bonding 110
concentration changes on the rates
5.15 Enthalpy changes when ionic
of reactions 163
substances dissolve 112
8.5 Explaining the effect of changes in
5.16 Energy for life 114
concentration, pressure and surface
End of chapter questions 116 area on the rates of reactions 165
8.6 Investigating the effect of
Chapter 6 Redox temperature change on the rates of
reactions 165
6.1 Introduction 118 8.7 Explaining the effect of
6.2 Redox processes in terms of temperature change on the rates of
electron transfer 118 reactions 167
6.3 Important types of redox reaction 120 8.8 Explaining the effect of catalysts on
6.4 Electrolysis 121 the rates of reactions 171
6.5 Explaining the electrolysis of mixtures 122 8.9 Interpreting the action of
6.6 Electrolysis in industry 124 catalysts in terms of the Maxwell–
6.7 Oxidation numbers 126 Boltzmann distribution 172
6.8 Explaining redox in terms of 8.10 The importance of transition metals
oxidation numbers 129 and their compounds as catalysts 173
6.9 The advantages and disadvantages End of chapter questions 175
of oxidation numbers 130
End of chapter questions 131 Chapter 9 The periodic table and periodicity

Chapter 7 Equilibria 9.1 Families of similar elements 176

9.2 Mendeléev’s periodic table 177
7.1 Reversible reactions 132 9.3 Modern forms of the periodic table 178
7.2 Equilibria in physical processes 133 9.4 Metals, non-metals and metalloids 182
7.3 Characteristic features of a dynamic 9.5 Periodic properties 183
equilibrium 136 9.6 The periodicity of physical properties 183
7.4 Equilibria in chemical reactions 137 9.7 The periodicity of atomic properties 185
7.5 Studying the equilibrium of a 9.8 The periodicity of chemical properties 190
solute between two immiscible 9.9 Variation in the oxidation numbers
solvents: partition coefficients 139 of the elements in Periods 2 and 3 192
7.6 Solvent extraction 141 9.10 Patterns in the properties of chlorides 194
7.7 The equilibrium constant 141 9.11 Patterns in the properties of oxides 195
7.8 The equilibrium law 143 End of chapter questions 200
7.9 Equilibrium constants in gaseous systems 146
7.10 The effect of concentration changes
on equilibria 147 Chapter 10 Group II – the alkaline-earth metals
7.11 The effect of pressure changes on
equilibria 149 10.1 Introduction 202
7.12 The effect of catalysts on equilibria 150 10.2 Electron structures and
7.13 The effect of temperature changes atomic properties 202
on equilibria 151 10.3 Physical properties 204

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10.4 Chemical properties of the Group II 13.7 Instrumental methods of analysis 260
elements 205 13.8 Spectroscopy 263
10.5 Reactions of the compounds of 13.9 Chromatography 273
Group II metals 206 End of chapter questions 278
10.6 Occurrence of the alkaline-earth metals 209
10.7 Uses of the alkaline-earth metals
Chapter 14 Alkanes
and their compounds 210
End of chapter questions 213 14.1 Crude oil 280
14.2 The composition of crude oil 282
Chapter 11 Group VII – the halogens 14.3 Naming alkanes 283
14.4 Physical properties of alkanes 286
11.1 Introduction 215 14.5 Reaction types and reaction
11.2 Sources of the halogens 215 mechanisms in organic chemistry 288
11.3 Obtaining the halogens 216 14.6 Reactivity of alkanes 292
11.4 Structure and physical properties 14.7 Important reactions of alkanes 294
of the halogens 218 14.8 The environmental impact of
11.5 Chemical reactions of the halogens 220 motor vehicle fuels 296
11.6 The halogens as oxidising agents 222 14.9 The greenhouse effect and global
11.7 The reactions of halogens with alkalis 223 climate change 298
11.8 Reactions of halide ions 224 End of chapter questions 301
11.9 Uses of the halogens and their
compounds 226
Chapter 15 Alkenes
End of chapter questions 229
15.1 Naming alkenes 303
Chapter 12 From nitrogen and sulfur to ammonia 15.2 The nature of the double bond 304
and sulfuric acid 15.3 Cis–trans isomerism 305
15.4 Mechanism of addition to
12.1 The properties and reactions of a double bond 306
nitrogen and sulfur 231 15.5 Important reactions of alkenes 307
12.2 The uses of nitrogen and sulfur 233 15.6 Addition polymerisation 311
12.3 Pollution from the oxides of 15.7 Rubber – a natural addition polymer 314
nitrogen and sulfur 233 End of chapter questions 316
12.4 The hydrides of nitrogen and sulfur 235
12.5 Ammonia as a base 236
Chapter 16 Organic halogen compounds
12.6 The manufacture of ammonia 237
12.7 The modern Haber process 239
16.1 Anaesthetics 318
12.8 From ammonia to nitric acid 241
16.2 Naming halogen compounds 319
12.9 Fertilisers and explosives from
16.3 The nature of the carbon–halogen
nitric acid 242
bond 319
12.10 Problems with the over-use
16.4 Nucleophilic substitution 321
of fertilisers 244
16.5 Important substitution reactions of
12.11 The manufacture of sulfuric acid 245
halogenoalkanes 325
12.12 The importance and uses of
16.6 Elimination reactions 326
sulfuric acid 248
End of chapter questions 328
End of chapter questions 249

Chapter 17 Alcohols
Chapter 13 Introduction to organic chemistry
17.1 Fermentation 330
13.1 Carbon – a unique element 250
17.2 Naming alcohols 331
13.2 Organic chemistry 251
17.3 Alcohols as a homologous series 332
13.3 Functional groups 252
17.4 The amphoteric nature of alcohols 334
13.4 Finding the formulae of organic
17.5 Reactions involving the carbon
compounds 253
skeleton 336
13.5 Writing structural formulae 255
13.6 Isomerism 257 End of chapter questions 340

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 Contents

Chapter 18 Carbonyl compounds Chapter 21 Ionic equilibria in aqueous solution

18.1 The carbonyl group 342 21.1 Introduction 397

18.2 Aldehydes and ketones – nature 21.2 The solubility of sparingly soluble
and naming 343 ionic solids in water 397
18.3 Addition reactions of carbonyl 21.3 Calculating solubility products
compounds 345 and solubilities 399
18.4 Condensation reactions of carbonyl 21.4 Limitations to the solubility
compounds 347 product concept 400
18.5 Oxidation of carbonyl compounds 347 21.5 Using the solubility product concept 400
18.6 Effect of the carbonyl group on 21.6 The strengths of acids and bases 403
neighbouring atoms 349 21.7 The dissociation of water 404
18.7 Sugars – naturally occurring carbonyl 21.8 The pH scale 405
compounds 350 21.9 Measuring the hydrogen ion
End of chapter questions 354 concentration and pH 406
21.10 Dissociation constants of acids and bases 407
21.11 Acid–base indicators 410
Chapter 19 Carboxylic acids and their derivatives
21.12 pH changes during acid–base titrations 412
21.13 Buffer solutions 414
19.1 Carboxylic acids 356
19.2 Some important reactions of End of chapter questions 418
carboxylic acids 358
19.3 Esters 360 Chapter 22 Reaction kinetics
19.4 The carboxyl group and acidity 364
19.5 Acyl chlorides 365 22.1 Introducing reaction kinetics 420
End of chapter questions 368 22.2 Techniques for studying the rates
of reactions 421
Chapter 20 Electrochemistry 22.3 Investigating the effect of
concentration on reaction rate
20.1 What is electrochemistry? 370 more fully – rate equations 423
20.2 Electrochemical cells 370 22.4 Order of reaction and rate equations 425
20.3 Cell potentials 372 22.5 The initial rates method for
20.4 Standard electrode potentials 375 determining rate equations 429
20.5 Measuring standard electrode 22.6 Investigating the reaction between
potentials 376 hydrogen and nitrogen monoxide
20.6 Relative strengths of oxidising and using the initial rates method 430
reducing agents 379 22.7 The effect of temperature change
20.7 The use of standard electrode potentials 380 on reaction rates 431
20.8 Limitations to the predictions from 22.8 The effect of catalysts on reaction rates 433
standard electrode potentials 383 22.9 The importance of reaction rate studies 435
20.9 Calculating the effect of concentration End of chapter questions 439
on standard electrode potential: the
Nernst equation 385
Chapter 23 Entropy and Gibbs free energy
20.10 Commercial cells 387
20.11 Fuel cells 389
23.1 Introduction 441
20.12 Measuring the quantity of
23.2 Chance, probability and the
electricity (electric charge) 391
direction of change 441
20.13 How much electric charge is
23.3 Entropy 442
needed to deposit one mole of
23.4 Entropy changes 444
copper during electrolysis? 391
23.5 Entropy changes and Gibbs free
20.14 Determining the Avogadro
energy changes 446
constant by an electrolytic method 393
20.15 Calculating the amounts of End of chapter questions 450
substances produced during electrolysis 393
End of chapter questions 395

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Chapter 24 Group IV – carbon to lead, non-metal 26.10 Phenol – an aromatic hydroxyl

to metal compound 499
26.11 Substitution reactions of the
24.1 Introducing Group IV 451 aromatic ring in phenol 501
24.2 The occurrence, extraction and End of chapter questions 503
uses of Group IV elements 451
24.3 Variation in the physical properties
of the elements 454 Chapter 27 Organic nitrogen compounds
24.4 Variation in the chemical
properties of the elements 455 27.1 Dyes and the development of the
24.5 General features of the compounds 456 organic chemical industry 505
24.6 Bonding and properties 27.2 Important organic nitrogen compounds 506
of the tetrachlorides 457 27.3 The nature and occurrence of amines 508
24.7 Bonding and properties of the oxides 458 27.4 Making amines 509
27.5 Amines as bases 510
End of chapter questions 460 27.6 Other reactions of phenylamine 512
27.7 Diazonium salts 513
Chapter 25 The transition elements 27.8 Amides 515
27.9 Amino acids, proteins and polymers 516
25.1 Introduction 461 End of chapter questions 517
25.2 Ions of the transition metals 462
25.3 What is a transition element? 463
25.4 Trends in atomic properties across Chapter 28 Polymers
the transition metals 464
25.5 General properties of the first 28.1 The importance of polymers 519
transition series (Ti to Cu) 466 28.2 Polymer properties 520
25.6 Variable oxidation states 467 28.3 Types of polymerisation: addition
25.7 The formation of complex ions 470 and condensation 523
25.8 Naming complex ions 471 28.4 Important condensation polymers:
25.9 The structure of complex ions 472 polyesters and polyamides 524
25.10 Ligand exchange reactions 474 28.5 Recycling polymers 528
25.11 Stability constants 475 End of chapter questions 530
25.12 The importance and use of
complex ions 476 Chapter 29 Proteins and DNA
25.13 Coloured compounds and
coloured ions 479 29.1 Sickle cell disease 531
25.14 Catalytic properties 481 29.2 Amino acids, the building blocks
25.15 Toxic trace metals in the environment 482 for proteins 532
25.16 Fundamental reactions in 29.3 The structure of proteins 535
inorganic chemistry 483 29.4 Enzymes 538
End of chapter questions 484 29.5 Nucleic acids: DNA and RNA 540
29.6 From DNA to proteins 545
Chapter 26 Aromatic hydrocarbons and phenol 29.7 Using DNA 546
End of chapter questions 548
26.1 Aromatic hydrocarbons 486
26.2 The structure of benzene 486 Chapter 30 Organic synthesis
26.3 Naming aromatic compounds 488
26.4 Chemical characteristics of benzene 490 30.1 Introduction 549
26.5 The mechanism of substitution 30.2 Polyfunctional compounds 550
reactions of benzene 491 30.3 Principles of organic synthesis 551
26.6 Important electrophilic 30.4 Working out synthetic routes 553
substitution reactions of benzene 492 30.5 Synthesis and chiral molecules 555
26.7 Other important reactions of benzene 494
26.8 Other arenes 495 End of chapter questions 558
26.9 The position of substitution in
benzene derivatives 498 Index 559

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 Introduction

Chemistry in Context is a well-respected text to which The layout of the book is designed to cover information
students and teachers turn for a clear and authoritative in a clear, accessible way. Its features include:
explanation of post-16 and A Level Chemistry. A
• Accessible language facilitated by the ‘Definitions’
comprehensive student book, it presents chemistry in
and ‘Key Points’ features, which enable you to review
its wider context and emphases its relevancy as a vital
and revise the content of each chapter more easily.
contribution to society, industry and civilisation.
Bold type is used to emphasise the key terms which
As ever, this new edition aims to make your study of you will need to be able to define and understand.
chemistry successful and interesting. It was revised to Definitions for these terms are given in the ‘Definitions’
meet the requirements of Supplementary Level (AS boxes in the margin.
Level) and Advanced Level (A Level) Chemistry, and • Frequent cross referencing to link different topics
is particularly suited to the Cambridge International and provide you with a fully integrated understanding
syllabus. The content also covers the requirements of the of chemistry as a whole.
Caribbean Advanced Proficiency Examination (CAPE)®.
• Step-by-step worked examples helping you apply
The book is divided into 30 chapters: broadly, chapters 1–19 theory into practice
cover the AS Level requirements, while chapters 20–30 • Questions appearing at frequent intervals throughout
cover A Level. the book to encourage you to read more critically and
The text draws on experimental evidence to develop maximise your understanding. They provide a quick
key ideas and establish laws and theories. New ideas are check on how well you have learnt and understood the
presented in the book in a careful step-by- step manner factual content of the section you just finished studying.
to allow you to develop a firm understanding of key • End-of-chapter questions testing the full range of
concepts and ideas. skills expected at AS and A Level, including application
of knowledge, understanding, analysis, synthesis and
Key concepts are essential principles, theories and ideas
evaluation. The questions cover the material in the
that help you to develop a deeper comprehension of
same chapter and give you practice of the type of
chemistry and to make relevant links between different
questions you can expect in an examination.
topics. They are, in effect, the foundations upon which
the whole subject is based. An awareness of key concepts
allows you to see chemistry as an interrelated, coherent,
albeit complex, whole. What’s on the website?

Once you have mastered the key concepts, you will Extra materials specifically written to support your
be able to use them to describe and explain facts and learning, help you prepare for the examination and
processes in detail and with accuracy, to solve problems achieve your best are available at
and tackle related material that is completely new to you. www.oxfordsecondary.com/9780198396185
The key concepts in chemistry cover the following ideas: • Thirty auto-marked interactive activities, one for
each chapter of the student book.
• Atoms and forces determine the physical and chemical
properties of matter, including bonding and reactivity. • A multiple choice quiz in the style of the
Cambridge AS Chemistry examination. Consisting
• Experiments and evidence help chemists to build of 40 questions, the quiz is auto-marked and offers
models and formulate theories that explain the post-submission summative feedback, allowing you
structure and properties of materials. to identify any areas of improvement.
• Patterns in chemical behaviour and reactions • Ten On Your Marks activities consisting of
predict the properties of substances and suggest how questions with sample answers and specific teacher
substances undergoing chemical reactions can give rise feedback. They are structured in the same way as
to new materials. the A2 examination and cover a range of questions
• An understanding of chemical bonding allows types that are used in the examination.
chemists to predict patterns of reactivity. • Answers to in-chapter questions and to end of
• Energy changes during a chemical reaction can be chapter questions, which you can use to check and
used to predict the extent and the rate of that reaction. reinforce your understanding of the work.

viii

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 1 Atoms, molecules and
equations

1.1 Atoms and molecules

The first chemist to use the name ‘atom’ was John Dalton (1766–1844). Dalton In this section you will learn to:
used the word ‘atom’ to mean the smallest particle of an element. He then • State the difference between an
went on to explain how atoms could combine together to form molecules atom and a molecule
which he called ‘compound atoms’.
• Describe the size of an atom
For example, chlorine consists of particles of Cl2 under ordinary conditions, but
at very high temperatures these split up to form particles of Cl. So, molecules of
chlorine are written as Cl2 and atoms of chlorine are written as Cl. DEFINITIONS
An atom is the smallest particle of
an element.
A molecule is a particle containing
two or more atoms joined together
chemically.

Fig 1.1 John Dalton collecting ‘marsh gas’ (mainly methane) from rotting vegetation at the bottom
of a pond. Dalton was born in 1766 in the village of Eaglesfield in Cumbria. He was the son of a
handloom weaver. For most of his life, Dalton lived in Manchester and taught at what was then the
Presbyterian College

QUESTIONS
1 Name one substance in Dalton’s list of elements (Figure 1.2) which we
now know is a compound and not an element.
2 Name an element whose molecules consist of two atoms under normal
conditions, like chlorine. Fig 1.2 Dalton’s symbols for the elements. In
1803, Dalton published his atomic theory. He
3 Name an element that exists as separate single atoms under normal suggested that all matter was composed of
conditions. small particles which he called ‘atoms’. Later,
Dalton went on to suggest symbols for the
4 How many atoms are there in one molecule of sugar (sucrose), atoms of different elements as shown above.
‘Azote’, the second element in Dalton’s list, is
C12H22O11? now called nitrogen
5 How many different types of atom are there in one molecule of sugar?

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 Atoms, molecules and equations

Most atoms have a radius of about 10–10 m, but the unit used in measuring
atomic distances is usually the nanometre (nm).
1 m = 109 nm
10–9 m = 1 nm
∴ 10–10 m = 0.1 nm

Fig 1.3 In the 1990s, two scientists working for So, the radii of atoms are about 0.1 nm.
IBM introduced a little xenon into an evacuated
container containing a small piece of nickel at Atoms, of course, are far too small to be seen even with the most powerful
–269 °C. Some xenon atoms stuck to the surface light microscope. However, scientists have used electron microscopes to pick
of the nickel. Then, using a special instrument out individual atoms (Figure 1.3).
called a ‘scanning tunnelling microscope’, they
moved individual xenon atoms around the nickel
surface to make the IBM logo. Each blue blob is
the image of a single xenon atom

1.2 Comparing the masses of atoms

In this section you will learn to: Individual atoms are far too small to be weighed, but, in 1919, F.W. Aston
• List the stages used in mass invented the mass spectrometer. This gave chemists an accurate method of
spectroscopy analysis comparing the relative masses of atoms and molecules.
• Describe how this analysis can A mass spectrometer separates atoms and molecules according to their mass
tell us about the masses of and also shows the relative numbers of the different atoms and molecules
different atoms or molecules present. Figure 1.4 shows a diagram of a simple mass spectrometer.

Vaporised
sample
Ionisation chamber

+ Accelerating electric field

–
Electron gun Magnetic field

Recorder Heavier particles

Particles of
intermediate mass
Lighter particles

Amplifier

Ion detector

To vacuum pump
Fig 1.4 A diagram of a simple mass spectrometer

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 1.2 Comparing the masses of atoms

Before atoms can be detected in a mass spectrometer, they must be converted

to positive ions in the vapour or gaseous state. Inside a mass spectrometer,
there is a vacuum. This allows particles from the chemical under test to be
studied without interference from particles in the air.
There are five stages in a mass spectrometer: vaporisation, ionisation,
acceleration, deflection and detection.

Vaporisation
Gases, liquids and volatile solids vaporise when injected into the instrument
just before the ionisation chamber. Less volatile solids are preheated to help
them to vaporise.

Ionisation
Vaporised atoms and/or molecules pass into the ionisation chamber. Here they
are bombarded with a beam of high-energy electrons. These knock electrons
off the atoms or molecules in the sample forming positive ions:
e– + X → X+ + e– + e–
high-energy atom in positive electron knocked high-energy
electron sample ion out of X electron retreating

Acceleration
Positive ions, such as X+, are now accelerated by an electric field.
Fig 1.5 Aston’s mass spectrometer.
Deflection A vaporised sample of the element in the glass
The accelerated ions pass into a magnetic field. As the ions pass through the bulb on the left was bombarded by electrons.
magnetic field, they are deflected according to their mass and their charge. The ions produced were then accelerated by
an electric field towards the magnetic field on
the right (produced by the hundreds of coils in
Detection the electromagnet)
If the accelerating electric field and the deflecting magnetic field stay
constant, ions of only one particular mass-to-charge ratio will hit the ion
detector at the end of the instrument. Ions of smaller mass-to-charge ratio
will be deflected too much. Ions of greater mass-to-charge ratio will be NOTE
deflected too little. Particles can only be attracted and
The ion detector is linked through an amplifier to a recorder. As the strength of accelerated into the spectrometer
the magnetic field is slowly increased, ions of increasing mass are detected and if they are positively charged.
a mass spectrum similar to that shown in Figure 1.6 can be printed out.
By first using a reference compound with a known structure and relative
molecular mass, the instrument can print a scale on the mass spectrum.
The relative heights of the peaks in the mass spectrum give a measure of the
relative amounts of the different ions present. (Strictly speaking, it is the areas
(∝ relative abundance)

under the peaks and not the peak heights which give the relative amounts or
Detector current

abundances.)

QUESTIONS
6 How many different ions are detected in the mass spectrum of naturally
occurring magnesium in Figure 1.6?
7 What are the relative masses of these different ions?
8 The relative masses of the atoms which formed these ions are virtually
22 23 24 25 26 27
the same as the relative masses of the ions. Explain why.
Relative atomic mass
9 Estimate the relative proportions of the different ions. Fig 1.6 A mass spectrometer trace for naturally
occurring magnesium

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 Atoms, molecules and equations

1.3 Heading atomic

Relative A masses – the 12C scale

In this section you will learn to: Chemists originally measured atomic masses relative to hydrogen.
• Outline how the relative atomic Hydrogen was chosen initially because it had the smallest atoms and
masses of elements is found these could be assigned a relative atomic mass of 1. At a later date, when
scientists realised that one element could contain atoms of different mass
• Calculate the relative molecular
(isotopes), it became necessary to choose a single isotope as the reference
mass or relative formula mass of
standard for relative atomic masses. Isotopes are studied more fully in
a substance
Section 2.4.
In 1961, the isotope carbon-12 (12C) was chosen as the new standard, because
DEFINITIONS carbon is a solid which is much easier to store and transport than hydrogen
(which is a gas).
Isotopes are atoms of the same
element with different masses. On the 12C scale, atoms of the isotope carbon-12 are assigned a relative
atomic mass, or more correctly a relative isotopic mass, of exactly 12.
Chemists use the relative atomic
So, the relative atomic mass of an element is the average mass of an atom
mass scale to compare the masses
of the element relative to one-twelfth the mass of an atom of the isotope
of different atoms and isotopes.
carbon-12.
Atoms of the isotope carbon-12
are assigned a relative atomic mass average mass of an atom of the element
Relative atomic mass =
of exactly 12. 1
12 × the mass of one atom of carbon-12

The relative atomic mass of an

Using the 12C scale, the relative atomic mass of hydrogen is 1.008 and that of
element is the average mass of one
magnesium is 24.312. So, as the symbol for relative atomic mass is Ar, we can
atom of the element relative to
write:
one-twelfth the mass of one atom
of carbon-12. Ar(H) = 1.008 and Ar(Mg) = 24.312 or simply H = 1.008 and Mg = 24.312
for short.
The relative isotopic mass of an
isotope is the mass of one atom A few other relative atomic masses are listed in Table 1.1. The relative atomic
of the isotope relative to one- masses of all elements are shown in the Periodic Table, at the back of this book.
twelfth the mass of one atom of In calculations using relative atomic masses, you will only be expected to work
carbon-12. to one decimal place.
The relative molecular mass of
Table 1.1 The relative atomic masses of some elements
an element or compound is the
sum of the relative atomic masses Element Symbol Relative
atomic mass
of all the atoms in its molecular
12
formula. Carbon-12 C 12.000
Carbon C 12.011
The relative formula mass of
Chlorine Cl 35.453
a compound is the sum of the
relative atomic masses of all the Copper Cu 63.540
atoms in its formula. Hydrogen H 1.008
Iron Fe 55.847
Magnesium Mg 24.312
Oxygen O 15.999
Sulfur S 32.064

Notice in Table 1.1 that the relative atomic mass of carbon is 12.011. This means
that the average mass of a carbon atom is 12.011, not 12.000. This is because
naturally occurring carbon contains a few atoms of carbon-13 and carbon-14
mixed in with those of carbon-12.

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 1.4 Moles, molar masses and the Avogadro constant

QUESTIONS
Use Table 1.1 to answer the following questions: 11 Which element has atoms approximately twice as
heavy as sulfur atoms?
10 Roughly, how many times heavier are:
a carbon atoms than hydrogen atoms 12 Why do the values of relative atomic masses have
no units?
b magnesium atoms than carbon atoms?

Relative atomic masses can also be used to compare the masses of different QUESTIONS
molecules. These relative masses of molecules are called relative molecular
masses (symbol Mr). 13 What is the relative molecular
mass of:
So, the relative molecular mass of water,
a hydrogen, H2
Mr(H2O) = 2 × Ar(H) + Ar(O) = (2 × 1.0) + 16.0 = 18.0 b carbon disulfide, CS2
and the relative molecular mass of chloromethane, c sulfuric acid, H2SO4?
Mr(CH3Cl) = Ar(C) + 3 × Ar(H) + Ar(Cl) 14 What is the relative formula
= 12.0 + (3 × 1.0) + 35.5 = 50.5 mass of:
a copper(ii) chloride, CuCl2
Metal compounds, such as sodium chloride and copper sulfate, consist of giant
structures containing ions, not molecules. So, it would be wrong to use the b magnesium nitrate,
term ‘relative molecular mass’ for ionic compounds. Instead, chemists use the Mg(NO3)2
term ‘relative formula mass’ for ionic compounds. c hydrated iron(ii) sulfate,
FeSO4·7H2O?
So, the relative formula mass of iron(iii) sulfate,
Mr(Fe2(SO4)3) = 2 × Ar(Fe) + 3 × Ar(S) + 12 × Ar(O)
= (2 × 55.8) + (3 × 32.1) + (12 × 16.0) = 399.9

1.4 Moles, molar masses and the

Heading A
Avogadro constant
Chemists measure the amount of a substance in moles. The word ‘mole’ In this section you will learn to:
comes from a Latin word meaning a heap or pile. One mole of an element or
• Relate mass, moles and molar
a compound has a mass equal to its relative atomic mass, its relative molecular
masses by calculation
mass or its relative formula mass in grams.
• Use the Avogadro constant to
So, one mole of iron, Fe, is 55.8 g, one mole of magnesium, Mg, is 24.3 g. One find the number of particles in
mole of water, H2O, is 18.0 g and one mole of chloromethane, CH3Cl, is 50.5 g. an amount of a substance
Similarly, one mole of iron(iii) sulfate, Fe2(SO4)3, is 399.9 g.
These masses of 1 mole of different substances are usually called molar
DEFINITION
masses and given the symbol M.
Molar mass is the mass of
So, the molar mass of iron, M(Fe) = 55.8 g per mole, usually written as one mole of an element or a
55.8 g mol–1. The molar mass of water, M(H2O) = 18.0 g mol–1 and the molar compound. The unit for molar
mass of iron(iii) sulfate, M(Fe2(SO4)3) = 399.9 g mol–1. masses is g mol–1. The symbol for
molar mass is M. Don’t confuse
this with the symbol Mr for
relative molecular mass or relative
formula mass, which has no units.

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 Atoms, molecules and equations

Amounts in moles
Chemists often count in moles because one mole is the mass of ‘one formula
NOTE
worth’ of a substance. In fact, the mole is the SI unit for amount of substance
SI units: The abbreviation ‘SI’ whereas the kilogram is the SI unit for mass.
from the French term ‘Système
International’ refers to the The amount of a substance is therefore measured in moles, which is usually
internationally accepted system abbreviated to ‘mol’. So,
of standard units for physical 55.8 g of iron contain 1 mol of iron atoms,
quantities.
111.6 g of iron contain 2 mol of iron atoms and
558.0 g of iron contain 10 mol of iron atoms.
KEY POINT
The mole is the basic unit for These simple calculations show that:
mass of substance (g)
measuring amounts of substances. Amount of substance (mol) =
molar mass (g mol–1)
The Avogadro constant
Since one atom of carbon is 12 times as heavy as one atom of hydrogen, it
(a) (b) follows that 12 g of carbon will contain the same number of atoms as 1 g of
hydrogen. In the same way, one atom of oxygen is 16 times as heavy as one
atom of hydrogen, so 16 g of oxygen will also contain the same number of
atoms as 1 g of hydrogen.

(d)
In fact, the relative atomic mass in grams (i.e. 1 mole) of every element
(1 g hydrogen, 12 g carbon, 16 g oxygen, etc.) will contain the same number
of atoms. Experiments show that this number is 6.02 × 1023 mol–1. Written
out in full, this is 602 000 000 000 000 000 000 000 per mole. The number is
(c) (e) usually called the Avogadro constant in honour of the Italian scientist Amedeo
Avogadro, and given the symbol L.
Fig 1.7 One mole samples of (a) carbon,
(b) sulfur, (c) magnesium, (d) copper and L = 6.02 × 1023 mol–1
(e) iron
The Avogadro constant is the number of atoms, molecules or formula units
in one mole of any substance. Therefore, 1 mole of iron (55.8 g) contains
6.02 × 1023 Fe atoms, 1 mole of water (18.0 g) contains 6.02 × 1023 H2O
molecules and 1 mole of iron(iii) sulfate (399.9 g) contains 6.02 × 1023 Fe2(SO4)3
formula units.
1 mole of water contains 6.02 × 1023 H2O molecules, 2 moles of water contain
KEY POINT
2 × 6.02 × 1023 H2O molecules and 10 moles of water contain 10 × 6.02 × 1023
The relative atomic mass in grams H2O molecules. Therefore,
(molar mass) of any element
contains 6.02 × 1023 atoms. Number of atoms,
= amount of substance in moles × Avogadro constant
molecules or formula units
Notice how important it is to specify exactly which particles you mean in
discussing the number of moles of different substances. For example, the
statement ‘one mole of oxygen’ is ambiguous. It could mean one mole of
oxygen atoms (O), i.e. 16.0 g; or it could mean one mole of oxygen molecules
(O2), i.e. 32.0 g. To avoid this ambiguity, it is important to state the formula of
the substance involved, for example, ‘one mole of oxygen, O2’.

DEFINITIONS
The Avogadro constant (6.02 × 1023 mol–1) is the number of atoms in
Fig 1.8 One mole of sodium chloride exactly 12 g of the isotope carbon-12.
(salt), NaCl (58.5 g), and one mole of water,
H2O (18.0 g) A mole is the amount of substance which contains the same number of
particles (atoms, molecules or formula units) as there are atoms in exactly
12 g of 12C (i.e. 6.02 × 1023 particles).
The term ‘mol’ is the symbol for mole. It is not an abbreviation for ‘molecule’
or ‘molecular’.
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 1.5 Using mass spectra to calculate relative atomic masses

QUESTIONS
15 What is the amount in moles of: 17 Using the Avogadro constant, calculate the number of:
a 13.95 g of Fe atoms, a S atoms in 4.125 g of sulfur,
b 7.1 g of Cl2 molecules, b N atoms in 8.0 g of NH4NO3,
c 15.19 g of FeSO4? c SO42– ions in 4 mol of Fe2(SO4)3.
16 What is the mass of: 18 One cubic decimetre of Pellegrino natural mineral
a one atom of carbon-12, water contains 179 mg of calcium ions (Ca2+) and
445 mg of sulfate ions (SO42–). What are:
b 6.02 × 1023 atoms of copper,
a these masses of Ca2+ and SO42– in grams,
c 0.5 mol of SCl2,
b the amount of Ca2+ and SO42– in moles,
d 0.25 mol of SO42– ions?
c the numbers of Ca2+ and SO42– ions?

1.5 Using mass spectra to calculate relative

atomic masses

Look at the mass spectrum of naturally occurring chlorine in Figure 1.9. In this section you will learn to:
This shows that chlorine consists of a mixture of two isotopes with relative isotopic • Read a mass spectrum and
masses of 35 and 37. These isotopes can be labelled 35Cl and 37Cl respectively. calculate relative atomic masses
When chlorine is analysed in a mass spectrometer, the beam of ions separates from it
into two paths producing two peaks in its mass spectrum corresponding to the • Describe that mass spectra can
two isotopes. also determine the mass and
structure of compounds
The heights of the two peaks in Figure 1.9 show that the relative proportions
of 35Cl to 37Cl are 3:1. So, of every four chlorine atoms, three are chlorine-35
and one is chlorine-37.
On average, there are three atoms of 35Cl and one atom of 37Cl in every
four atoms of chlorine. This is ¾ or 75% chlorine-35 and ¼ or 25% chlorine-37.
So the average mass of a chlorine atom on the 12C scale, which is the relative
atomic mass of chlorine is:
Relative abundance

(¾ × 35) + (¼ × 37) = 26.25 + 9.25 = 35.5

Mass spectrometers can also be used to study compounds. When molecules
of the vaporised compound are bombarded by electrons, some molecules lose
just one electron while other molecules break into fragments. Because of the
vacuum inside the mass spectrometer, chemists can study these molecular
fragments which do not exist under normal conditions.
The peak with the highest mass in the spectrum is produced by ionising
molecules without breaking them into fragments. So the mass of a ‘molecular
ion’ gives the relative molecular mass of the compound:
e– + M(g) → M+(g) + e– + e– 33 34 35 36 37 38 39
}

high-energy molecule in ‘molecular electrons Relative isotopic mass

electron sample ion’
Fig 1.9 A mass spectrum for naturally
By carefully studying the data from mass spectra, chemists can deduce: occurring chlorine

• the isotopic composition (abundance) of elements,

• the relative isotopic masses of atoms,

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 Atoms, molecules and equations

• the relative atomic masses of elements,

• the relative molecular masses of compounds.
After identifying the fragments in mass spectra, chemists can also piece
together the fragments and deduce the possible structures of compounds.
This technique is particularly useful in determining the structures of newly
synthesised compounds (Section 13.7).

QUESTION
19 Natural silicon in silicon-containing ores contains 92% silicon-28, 5%
silicon-29 and 3% silicon-30.
a What are the relative isotopic masses of the three silicon isotopes?
b What is the relative atomic mass of silicon?
c Samples of pure silicon obtained from ores, mined in different parts of
the world, have slightly different relative atomic masses. Why is this?

Dating the polar ice-caps

The existence of naturally-occurring 18O in addition to 16O has been used to
calculate the age of the polar ice-caps. Polar ice sheets have formed over
thousands of years as water molecules evaporate from the oceans, move
towards the poles and then fall as snow or hail.
Fig 1.10 In 2015, the United States men’s
4 × 100 meters relay team were stripped of the However, H216O molecules with a lower mass than H218O molecules evaporate
silver medal they won at the London Olympic
Games in 2012, after one of the team’s and diffuse more readily. So the ice collecting in the polar regions contains
members tested positive for performance a higher proportion of H216O than water in the oceans. But, the process
enhancing substances. Tests involving mass of evaporation and diffusion of the water molecules is also temperature-
spectrometry are routinely used by the World
Anti-Doping Association to ensure that dependent and ice which collects at the Poles in winter has a higher proportion
competing athletes are ‘clean’ of H216O than ice collecting in the summer.
So, by measuring the ratio of H216O to H218O continuously through the
depth of the polar ice-caps and counting the annual changes in the ratio
H216O : H218O, the age of the ice-cap can be found. Between 1970 and 1983,
a Russian expedition to the Antarctic collected an ice-core to a depth of 2000
metres. The bottom of this ice-core had formed about 150 000 years ago, but it
was only about half-way down into the ice sheet.

1.6 Empirical and molecular formulae

In this section you will learn to: It is possible to predict the formulae of most compounds, but the only certain
• Calculate the empirical formula way of knowing a formula is by experiment.
of a compound The word ‘empirical’ means ‘from experiment’ or ‘from experience’, so
• Explain the difference between chemists use the term ‘empirical formulae’ to describe formulae which have
the empirical and molecular been calculated from the results of experiments.
formulas of a substance
An experiment to calculate an empirical formula involves three stages:
1 Measure the masses of elements which combine in the compound.
2 Calculate the number of moles of atoms which combine.
3 Calculate the simplest ratio for the atoms which combine.
This gives the empirical formula of the compound.

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 1.6 Empirical and molecular formulae

WORKED EXAMPLE
When 10.00 g of ethene was analysed, it was found to contain 8.57 g of DEFINITIONS
carbon and 1.43 g of hydrogen. What is its formula? An empirical formula shows
Answer the simplest whole number ratio
Carbon Hydrogen for the atoms of each element in a
compound.
1 Masses of combined elements: 8.57 g 1.43 g
A molecular formula shows the
Molar mass of elements: 12.00 g mol–1 1.00 g mol–1 actual number of atoms of each
8.57 g 1.43 g element in one molecule of a
2 Moles of combined atoms:
12.00 g mol–1 1.00 g mol–1 compound.
= 0.714 mol 1.43 mol
0.714 1.43
3 Ratio of combined atoms: =1 =2
0.714 0.714
Therefore, the empirical formula of ethene is CH2.

This formula for ethene shows only the simplest ratio of carbon atoms to
hydrogen atoms. The actual formula showing the correct number of carbon
atoms and hydrogen atoms in one molecule of ethene could be CH2, C2H4, C3H6,
C4H8, etc. because all these formulae give CH2 as the simplest ratio of atoms.
Experiments show that the relative molecular mass of ethene is 28, which
corresponds to an actual formula of C2H4 and not CH2. Formulae, such as C2H4
for ethene, which show the actual number of atoms of each element in one
molecule of a compound are called molecular formulae.
The empirical formulae of some compounds like ethene can be calculated using
combustion data in place of their composition by mass. The combustion data is
obtained by burning the compound in oxygen to produce carbon dioxide and Fig 1.11 This scientist is using a mass
water. From the masses of carbon dioxide and water produced, it is possible to spectrometer to find the relative molecular
mass of a substance in order to confirm its
calculate the masses of carbon and hydrogen combined in the compound. These molecular formula
masses of carbon and hydrogen can then be used to calculate the empirical formula
of the compound. Study the worked example below to see how this is done.

WORKED EXAMPLE
A sample of a compound containing only carbon and hydrogen was burned
completely in oxygen. All the carbon was converted to 3.38 g of carbon
dioxide and all the hydrogen was converted to 0.692 g of water. What are
the masses of carbon and hydrogen combined in the compound?
Answer
One mole of CO2 contains one mole of C.
So, 44 g CO2 contains 12 g of C.
12 3
Fraction of C in CO2 = 44 = 11
3 QUESTION
∴ Mass of C in 3.38 g CO2 = 3.38 × = 0.92 g
11
1 mole of H2O contains 2 moles of H. 20 Use the masses of carbon and
So, 18 g H2O contains 2 g of H. hydrogen from the worked
2 1 example on the left to calculate
Fraction of H in H2O = 18 = 9 the empirical formula of the
1 compound.
∴ Mass of H in 0.692 g H2O = 0.692 × 9 = 0.077 g

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 Atoms, molecules and equations

1.7 Heading A and writing chemical equations

Predicting

In this section you will learn to: When methane in natural gas burns on a hob, it reacts with oxygen in the air
• Outline the stages in writing to form carbon dioxide and water. A word equation for the reaction is:
chemical equations methane + oxygen → carbon dioxide + water
• Balance chemical equations
Word equations like this give the names of the reactants and products, but as
and explain the importance of
chemists we should always aim to write balanced chemical equations using
doing so
symbols and formulae.
There are three key stages in writing chemical equations:
1 Write a word equation:
methane + oxygen → carbon dioxide + water
2 Write symbols for elements and formulae for compounds in the word equation:
CH4 + O2 → CO2 + H2O
Remember that the elements oxygen, hydrogen, nitrogen and halogens exist
as diatomic molecules containing two atoms, so they are written as O2, H2, N2,
Cl2, Br2 and I2 in equations. All other elements are written as single atoms
(e.g. Cu, C, Fe) in equations.
3 Balance the equation by writing numbers in front of the symbols and
formulae in order to have the same number of each kind of atom on both
sides of the equation.
In the equation above in Stage 2, there are 4 hydrogen atoms on the left and
Fig 1.12 Methane (natural gas) burning on only 2 on the right. Therefore, H2O on the right must be doubled (2H2O).
a hob There are now 2 oxygen atoms in O2 on the left, but 4 oxygen atoms in CO2
and 2H2O on the right. So, O2 on the left must be doubled (2O2). The balanced
equation is:
CH4 + 2O2 → CO2 + 2H2O
Remember that formulae must never be altered in balancing an equation. The
KEY POINT formula of methane is always CH4 and never CH2 or C2H4. Similarly, carbon
Note the three stages in writing a dioxide is always CO2 and water is always H2O. Atoms in an equation can
balanced chemical equation: only be balanced by putting a number in front of a symbol or a formula, thus
1 writing a word equation doubling or trebling, etc. the whole formula.
2 writing symbols for elements
and formulae for compounds Normally state symbols are also included, so the final equation is:
3 balancing the equation.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

QUESTION
21 Write balanced chemical equations for the following word equations:
a copper + oxygen → copper(ii) oxide
b potassium + oxygen → potassium oxide
c sodium + water → sodium hydroxide + hydrogen
d magnesium + hydrochloric acid → magnesium chloride + hydrogen
e copper(ii) oxide + sulfuric acid → copper(ii) sulfate + water
f hydrogen + oxygen → water

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 1.7 Predicting and writing chemical equations

A balanced chemical equation tells us

DEFINITION
• the reactants and products A balanced chemical equation
• the formulae of reactants and products describes a chemical reaction
• the relative amounts in moles of the reactants and products. using symbols and formulae as
appropriate for the reactants and
For example, the equation CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) tells us that: products. It has the same number
• Methane reacts with oxygen to form carbon dioxide and water. of each kind of atom on both sides
of the equation.
• The formula of methane is CH4, oxygen is O2, carbon dioxide is CO2 and water
is H2O.
• 1 mole of methane (CH4) reacts with 2 moles of oxygen molecules (2O2) to
produce 1 mole of carbon dioxide (CO2) and 2 moles of water (2H2O).
In addition, we can show the states of reactants and products by putting state
symbols in the equation.
Using relative atomic masses it is also possible to calculate the relative masses of
reactants and products involved in reactions (Section 1.8).
Chemical equations are very useful, but it is important to appreciate their
limitations. An equation cannot tell us: KEY POINT
Chemical equations can tell us the
• the rate of the reaction – whether it is fast or slow, whether it will explode or amounts and states of reactants
take years to react and products, but not how or why
• how or why the reaction takes place. the reactions occur.
For example, when natural gas (methane) burns, the balanced equation for the
reaction is:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
The equation does not show that a spark or a flame is needed to start the
reaction. Nor does it show that once the reaction has started it will continue
as long as both methane and oxygen are available. The equation does not tell
us why the methane reacts with oxygen or how the molecules and atoms
behave in the reaction. Does the oxygen react with methane as O2 molecules

Fig 1.13 The photographs show three stages in the manufacture of copper from copper ore
(copper pyrites). The left-hand photo shows the huge open-cast copper mine at Brigham near
Salt Lake City, USA. The ore is crushed and then reduced in furnaces to produce molten copper
(centre). This can be made into solid bars or rolls of cable (right)

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 Atoms, molecules and equations

or split into separate oxygen atoms which then react? Also how and when do
the carbon and hydrogen atoms in methane break apart? The equation cannot
answer these questions.
Equations can only tell us about the overall chemical change – the amounts
and states of the reactants and products. They cannot tell us anything about
what happens on the pathway between the reactants and the products.

QUESTION
22 a What is the relative formula mass of copper pyrites, CuFeS2?
(Cu = 63.5, Fe = 55.8, S = 32.1)
b How much copper can be obtained from 1 mole of CuFeS2?
c How much copper can be obtained from 1 tonne (1000 kg) of CuFeS2?

1.8 Calculations involving formulae

Heading A
and equations

In this section you will learn to: In industry and research and everywhere that chemists carry out reactions, it
• Calculate the mass of a is useful and sometimes essential to know how much product can be obtained
substance given in a reaction from a particular amount of reactants. Chemists can calculate these quantities
using formulae, equations and relative atomic masses.
• Calculate the volume of a gas
given in a reaction
• Calculate the volume and Calculating the masses of reactants and products
concentration of a solution used There are four stages in calculating the masses of reactants and products in a
in a reaction chemical reaction:
1 Write a balanced equation for the reaction.
2 Note the amounts in moles of relevant reactants and products in the
equation.
3 Calculate the masses of relevant reactants and products using relative atomic
masses.
4 Scale the masses of relevant reactants and products to the required
quantities.

WORKED EXAMPLE
Calculate the mass of lime (calcium oxide, CaO) that can be obtained from
1 tonne (1000 kg) of pure limestone (calcium carbonate, CaCO3).

Answer
1 CaCO3(s) → CaO(s) + CO2(g)
2 1 mole CaCO3 → 1 mole CaO
3 (40.1 + 12.0 + (3 × 16.0)) g CaCO3 → (40.1 + 16.0) g CaO
100.1 g CaCO3 → 56.1 g CaO
56.1
4 1 g CaCO3 → g CaO
100.1
∴ 1000 kg CaCO3 → 56.1 × 1000 kg CaO
100.1
Fig 1.14 Lime kilns are used to decompose
limestone to lime at high temperatures. This
= 560.4 kg of CaO (lime)
shows a traditional lime kiln in India

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 1.8 Calculations involving formulae and equations

QUESTIONS
23 Iron reacts with chlorine to form iron(iii) chloride: 25 Aluminium is manufactured by electrolysis
a How much chlorine reacts with 1.86 g of iron? (electrolytic decomposition) of bauxite, Al2O3.
How much aluminium can be obtained from 1 kg
b What mass of iron(iii) chloride is produced?
of pure bauxite?
24 A typical jumbo jet burns fuel at a rate of 200 kg per
minute. Assuming jet fuel is C11H24, what mass of
carbon dioxide does a jumbo jet produce per minute?

Calculating the volumes of gases

The volume of any gas depends on its temperature and pressure.
Repeated measurements with different gases show that the volume of 1
mole of every gas occupies approximately 24 dm3 (24 000 cm3) under room
conditions (25 °C and atmospheric pressure of 101 kPa). This volume of
one mole of gas is described as the molar volume (symbol Vm). So, Vm =
24 dm3 mol–1 under laboratory conditions at room temperature and pressure.
At standard temperature and pressure (s.t.p.), i.e. 273 K (0 °C) and 101 kPa
pressure, the molar volume is approximately 22.4 dm3.

Clamp

DEFINITION
The molar volume of a gas is
the volume of 1 mole. Under
Graduated 100 cm3 syringe
laboratory conditions, the
molar volume of all gases is
approximately 24 dm3 mol–1
Boiling tube
(24 000 cm3 mol–1). At standard
temperature and pressure (s.t.p.),
273 K and 101 kPa pressure,
Excess, dilute hydrochloric acid the molar volume of all gases is
approximately 22.4 dm3 mol–1.

0.081 g of magnesium

Fig 1.15 Measuring the volume of hydrogen produced when magnesium reacts with
hydrochloric acid

So under laboratory conditions, 1 mol of hydrogen, 1 mol of chlorine and 1 mol

of methane all occupy 24 dm3; 2 mol of each gas will occupy 48 dm3 and 0.1 mol
of each gas will occupy 2.4 dm3, i.e.
volume of gas(dm3)
Amount of gas(mol) =
molar volume(dm3 mol–1)
Figure 1.15 shows the apparatus that can be used to measure the volume of
hydrogen produced when magnesium reacts with dilute hydrochloric acid.
When the 0.081 g of magnesium ribbon is added to the excess acid, a vigorous
reaction occurs and hydrogen is evolved. When all the magnesium has reacted,
effervescence stops and the syringe shows that 80 cm3 of gas have been produced.
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 Atoms, molecules and equations

QUESTIONS So, 0.081 g of Mg react with HCl(aq) to produce 80 cm3 of H2.

0.081 g
26 Calculate the amount in moles Now, 0.081 g of Mg = = 0.0033 mol Mg
24.3 g mol –1
at room temperature and
80 cm3
pressure of: and 80 cm3 of H2 = = 0.0033 mol H2
24 000 cm3 mol–1
a 960 cm3 of chlorine,
∴ 0.0033 mol Mg produces 0.0033 mol H2.
b 144 dm3 of methane,
c 120 cm3 of oxygen. So, 1 mol Mg produces 1 mol H2.

27 Calculate the volumes of Assuming that the other product of the reaction is magnesium chloride, MgCl2,
the following gases at room the overall equation for the reaction is:
temperature and pressure: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
a 5 mol of hydrogen,
b 6.4 g of oxygen. Calculating the concentrations and volumes of solutions
28 a Write an equation for the As many reactions only take place in aqueous solution, it is important for
complete combustion of chemists to know the concentrations of these solutions. Chemists usually
propane, C3H8 (Calor gas) measure the concentration of solutions in moles of solute per cubic decimetre
with oxygen. (mol dm–3) of solution.
b What volume of oxygen 1 decimetre (dm) = 1⁄10 metre = 10 cm
reacts with 2 dm3 of propane and 1 dm3 = (10 cm)3 = 1000 cm3
and what volume of carbon
So, a solution of sodium chloride, NaCl(aq), containing 1.0 mol dm–3 has 1 mole
dioxide is produced?
of sodium chloride (58.5 g NaCl) in 1 dm3 (1000 cm3) of solution. A solution of
(Assume measurements are
NaCl containing 0.2 mol dm–3 contains 11.7 g (0.2 × 58.5 g) in 1 dm3 of solution.
made at room temperature
and pressure.) Notice that these concentrations are expressed as the number of moles in 1 dm3
of solution, not 1 dm3 of solvent. Here is the method which you should use to
prepare 1 dm3 of 1.0 mol dm–3 sodium chloride solution (Figure 1.16).
• Weigh out 1 mol of NaCl (58.5 g).
• Dissolve this in about 500 cm3 of distilled water in a beaker.
• Add this solution to the 1 dm3 volumetric flask plus washings from the beaker.
• Add more distilled water up to the 1 dm3 mark on the neck of the flask.
• Finally, mix the solution thoroughly.
• Can you see that 1 dm3 of 1.0 mol dm–3 NaCl contains less than 1 dm3 of water?
1 dm3 mark on volumetric
flask A solution of sodium chloride containing 1.0 mol dm–3 is usually written as:
[NaCl] = 1.0 mol dm–3 or simply [NaCl] = 1.0 M
Volumetric flask
Notice that square brackets around a formula are used to indicate the
58.5 g NaCl concentration of a substance and that mol dm–3 can be abbreviated to M.
The amounts of solute needed to make solutions of different concentration and
1 dm3 Watch glass volume can be worked out by simple proportion.

Fig 1.16 Preparing 1 dm3 of 1.0 mol dm–3 WORKED EXAMPLE

sodium chloride solution
What mass of solute is needed to make 250 cm3 of a solution containing
0.1 mol dm–3 sodium chloride, NaCl?

Answer
KEY POINT 1 dm3 of 1.0 M NaCl(aq) contains 1 mol NaCl
The concentrations of solutes in 250
250 cm3 of 1.0 M NaCl(aq) contains 1 × mol NaCl
solutions are usually expressed 1000
in moles per cubic decimetre 250
250 cm3 of 0.1 M NaCl(aq) contains 1 × × 0.1 mol NaCl
(mol dm–3). M is sometimes used 1000
as an abbreviation for mol dm–3. = 0.025 mol = 0.025 mol × 58.5 g mol–1 = 14.625 g NaCl

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 1.9 Determining equations

When ionic compounds like NaCl and H2SO4 dissolve, they are fully DEFINITION
dissociated into ions. Ions are positive and negative
For example: H2SO4 (aq) → 2H+(aq) + SO42–(aq) particles formed from atoms by
the loss or gain of one or more
So, if the concentration of H2SO4 is 1 mol dm–3, the concentration of H+ ions electrons.
is 2 mol dm–3 and that of SO42– is 1 mol dm–3.

QUESTIONS
29 What is the concentration in mol dm–3 of a solution containing:
a 22.22 g of calcium chloride, CaCl2, in 500 cm3 of solution,
b 39.63 g of ammonium sulfate, (NH4)2SO4 in 400 cm3 of solution,
c 24.96 g of hydrated copper sulfate, CuSO4·5H2O in 250 cm3 of solution?
30 What mass of solute is required to make:
a 3 dm3 of 3.0 mol dm–3 sodium hydroxide, NaOH,
b 200 cm3 of 0.2 mol dm–3 sulfuric acid,
c 50 cm3 of 2.0 mol dm–3 potassium carbonate, K2CO3?

1.9 Determining equations

There are three key stages in determining the equation for a particular reaction: In this section you will learn to:
1 Find by experiment the masses or volumes of the reactants and products • Describe the stoichiometric
involved in the reaction. relationship between substances
2 Convert these masses or volumes to amounts in moles of the substances in an equation
concerned. • Calculate the mole : mole ratio
3 Calculate the simplest whole number ratios for the amounts in moles of between substances in a reaction
substances involved in the reaction.

WORKED EXAMPLE NOTE

0.27 g of aluminium reacts with 2.40 g of bromine to form aluminium Stoichiometry: The simplest
bromide. What is the equation for the reaction? whole number ratio for the
molar amounts of reactants and
Answer products in a chemical reaction
1 Assuming the formula of aluminium bromide is AlBr3, we know that: is sometimes described as the
stoichiometric relationship
0.27 g Al + 2.40 g Br2 → 2.67 g AlBr3 or the stoichiometry of the
0.27 reaction.
2 Moles of Al reacting = mol = 0.010 mol Al
27.0
2.40
Moles of Br2 reacting = mol = 0.015 mol Br2
2 × 79.9
2.67
Moles of AlBr3 produced = = 0.010 mol AlBr3
27.0 + (3 × 79.9)

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 Atoms, molecules and equations

3 Ratio of moles of substances involved:

Al : Br2 : AlBr3 = 0.010 : 0.015 : 0.010
0.010 0.015 0.010
= : :
0.010 0.010 0.010
= 1 : 1.5 : 1
Therefore, the simplest whole number ratio of Al : Br2 : AlBr3 = 2 : 3 : 2
So, the equation is: 2Al + 3Br2 → 2AlBr3

QUESTIONS
Fig 1.17 A drawing of the original balance 31 3.81 g of copper reacted with oxygen to produce 4.29 g of a copper oxide.
which Dalton used to study the reacting
quantities of different substances Calculate a formula for the copper oxide and then write an equation for
the reaction.
32 25 cm3 of a gas containing only nitrogen and oxygen decomposed to
form 25 cm3 of nitrogen and 50 cm3 of oxygen. All the volumes were
measured at the same temperature and pressure. Write an equation for
the reaction.
33 18 cm3 of 1.0 M H2SO4 just reacted with 24 cm3 of 1.5 M NaOH to form
sodium sulfate and water. Calculate the amounts in moles of sulfuric
acid and sodium hydroxide reacting and write an equation for
the reaction.

1.10 Ionic equations

In this section you will learn to: Many reactions involve ionic compounds. The part played in a reaction by the
• Describe different types of ionic separate ions of these compounds can often be shown more clearly using an
equations ionic equation. Here are five important types of reaction where ionic equations
can be used.
• Show cancellation of spectator
ions to give an ionic equation
1 The reactions of metals with non-metals

NOTE When magnesium reacts with oxygen, the product is magnesium oxide – a
The only substances which contain solid ionic compound, Mg2+O2–(s):
ions are compounds of metals 2Mg(s) + O2(g) → 2Mg2+O2–(s)
with non-metals (salts and bases)
and aqueous acids.
2 The reactions of metals with acids
Earlier in this chapter we studied the reaction between magnesium
and hydrochloric acid, HCl(aq), forming magnesium chloride solution,
MgCl2(aq), and hydrogen.
The HCl(aq) and MgCl2(aq) are solutions of ionic compounds, fully
dissociated into separated ions which are free to move apart. So, we can
write an ionic equation as:
Mg(s) + 2H+(aq) + 2Cl–(aq) → Mg2+(aq) + 2Cl–(aq) + H2(g)

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 1.10 Ionic equations

By cancelling the Cl– spectator ions which appear on both sides of the
equation and take no part in the reaction, we get
Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g)

3 The reactions between acids and bases (neutralisation) DEFINITIONS

When hydrochloric acid reacts with sodium hydroxide solution, the equation Ionic equations show the part
for the reaction is: played by separate ions in a
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) reaction.

However, HCl(aq), NaOH(aq) and NaCl(aq) consist of dissociated ions in Spectator ions are ions which
aqueous solutions, so we can write an ionic equation as: take no part in a reaction.

H+(aq) + Cl–(aq) + Na+(aq) + OH–(aq) → Na+(aq) + Cl–(aq) + H2O(l)

Cancelling the Cl– and Na+ spectator ions on both sides of the equation
gives:
H+(aq) + OH–(aq) → H2O(l)

4 The precipitation of insoluble ionic solids

When solutions of silver nitrate and sodium chloride are mixed, an insoluble
precipitate of white silver chloride forms:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
Ions in the aqueous solutions are fully dissociated, but those in the
precipitate cling together as a solid and we write this in ionic equations as
Ag+Cl–(s). So, the ionic equation for the reaction is:
Ag+(aq) + NO3–(aq) + Na+(aq) + Cl–(aq) → Ag+Cl–(s) + Na+(aq) + NO3–(aq)
By cancelling spectator ions in this case, we get:
Ag+(aq) + Cl–(aq) → Ag+Cl–(s)

5 The reactions at electrodes during electrolysis

When liquids are electrolysed, the reactions which take place at the
electrodes involve ions. These reactions which occur during electrolysis are
discussed in Sections 6.4 to 6.5.

QUESTION
34 Write ionic equations for the following reactions:
a Iron + chlorine → iron(iii) chloride
b Zinc + nitric acid → zinc nitrate + hydrogen
c Sodium + water → sodium hydroxide + hydrogen
d Potassium hydroxide + sulfuric acid → potassium sulfate + water

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