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AS and A Level Chemistry
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Coursebook
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David Sang,
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Richard Woodside
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Lawrie
Ryan and Roger
Norris Chadha

Lawrie Ryan and Roger Norris

Cambridge International AS and A Level

Chemistry
Coursebook
Second Edition



Lawrie Ryan and Roger Norris

Cambridge International AS and A Level

Chemistry
Coursebook
Second Edition


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Contents
How to use this book

vi

Chapter 1: Moles and equations

1

Masses of atoms and molecules
Accurate relative atomic masses
Amount of substance
Mole calculations
Chemical formulae and chemical equations
Solutions and concentration
Calculations involving gas volumes

Chapter 2: Atomic structure

2
3
5
6
10
14
18

24

Elements and atoms
25
25
Inside the atom
Numbers of nucleons
28
Isotopes28
How many protons, neutrons and electrons?
29

Chapter 3: Electrons in atoms
Simple electronic structure
Evidence for electronic structure
Subshells and atomic orbitals
Electronic configurations
Orbitals and the Periodic Table
Patterns in ionisation energies in the
Periodic Table

Chapter 4: Chemical bonding
Types of chemical bonding
Ionic bonding
Covalent bonding
Shapes of molecules
More molecular shapes
Metallic bonding
Intermolecular forces
Hydrogen bonding
Bonding and physical properties

Chapter 5: States of matter
States of matter
The gaseous state
The liquid state
The solid state
Simple molecular lattices
Carbon nanoparticles
Conserving materials

Chapter 6: Enthalpy changes

89

What are enthalpy changes?
90
Standard enthalpy changes
92
94
Measuring enthalpy changes
Hess’s law
97
Enthalpy change of reaction from enthalpy
changes of formation
97
Enthalpy change of formation from enthalpy
changes of combustion
98
Calculating the enthalpy change of hydration of an
anhydrous salt
99

99
Bond energies and enthalpy changes
Calculating enthalpy changes using
bond energies
101

Chapter 7: Redox reactions

106

33
34
37
38
40

What is a redox reaction?
107
Redox and electron transfer
108
Oxidation numbers
109
Redox and oxidation number
110
Naming compounds
111
From name to formula
112
Balancing chemical equations using oxidation
numbers112

41

Chapter 8: Equilibrium116

32

48
49
49
51
55
56
58
60
64
66

72
73
73
77
78
80
82
83

Reversible reactions and equilibrium
117
Changing the position of equilibrium

119
Equilibrium expressions and the equilibrium
constant, Kc123
Equilibria in gas reactions: the equilibrium
constant, Kp127
Equilibria and the chemical industry
129
Acid-base equilibria
130

Chapter 9: Rates of reaction

140

Reaction kinetics
141
The effect of concentration on rate of reaction 143
143
The effect of temperature on rate of reaction
Catalysis144
Enzymes145

Chapter 10: Periodicity
Structure of the Periodic Table
Periodicity of physical properties
Periodicity of chemical properties

148
149
149

154

iii


Oxides of Period 3 elements
Chlorides of Period 3 elements

Chapter 11: Group 2
Physical properties of Group 2 elements
Reactions of Group 2 elements
Thermal decomposition of Group 2 carbonates
and nitrates
Some uses of Group 2 compounds

Chapter 12: Group 17

156
158

163
164
165
168
169

171

Physical properties of Group 17 elements
172

Reactions of Group 17 elements
173
Reactions of the halide ions
175
Disproportionation177
Uses of the halogens and their compounds
178

Chapter 13: Nitrogen and sulfur
Nitrogen gas
Ammonia and ammonium compounds
Uses of ammonia and ammonium compounds
Sulfur and its oxides
Sulfuric acid
iv

Chapter 14: Introduction to organic

chemistry

180
181
182
183
185
185

The homologous series of alcohols
Reactions of the alcohols
Carboxylic acids

226
226
231

Chapter 18: Carbonyl compounds

234

The homologous series of aldehydes and
ketones235
236
Preparation of aldehydes and ketones
Reduction of aldehydes and ketones
237
237
Nucleophilic addition with HCN
Testing for aldehydes and ketones
238
Reactions to form tri-iodomethane
240
241
Infra-red spectroscopy

Chapter P1: Practical skills 1

188

Chapter 15: Hydrocarbons201
202

202
204
207
208
210
211
213

Chapter 16: Halogenoalkanes217
Nucleophilic substitution reactions
218
Mechanism of nucleophilic substitution in
halogenoalkanes220
Elimination reactions
222
Uses of halogenoalkanes
222

246

Review of practical knowledge and
understanding247
Manipulation, measurement and observation 249
Presentation of data and observations
250
Analysis, conclusions and evaluation
251

Chapter 19: Lattice energy

Representing organic molecules
189
Functional groups
192
192
Naming organic compounds
Bonding in organic molecules
193
Structural isomerism
194
Stereoisomerism195
Organic reactions – mechanisms
196
Types of organic reaction
198

The homologous group of alkanes
Sources of the alkanes
Reactions of alkanes
The alkenes
Addition reactions of the alkenes
Oxidation of the alkenes
Addition polymerisation
Tackling questions on addition polymers

Chapter 17: Alcohols, esters and carboxylic
225
acids

Defining lattice energy

Enthalpy change of atomisation and
electron affinity
Born–Haber cycles
Factors affecting the value of lattice energy
Ion polarisation
Enthalpy changes in solution

257
258
258
259
262
263
265

Chapter 20: Electrochemistry273
Redox reactions revisited
274
Electrolysis275
Quantitative electrolysis
276
Electrode potentials
278
Measuring standard electrode potentials
282
O
values284
Using E —
Cells and batteries
293

More about electrolysis
295

Chapter 21: Further aspects of equilibria

303

The ionic product of water, Kw304
pH calculations
305
Weak acids – using the acid dissociation
constant, Ka307
Indicators and acid–base titrations
309
Buffer solutions
313


Equilibrium and solubility
Partition coefficients

316
319

Chapter 22: Reaction kinetics

324

Factors affecting reaction rate
325

Rate of reaction
325
Rate equations
330
Which order of reaction?
332
Calculations involving the rate constant, k334
Deducing order of reaction from raw data
335
Kinetics and reaction mechanisms
338
Catalysis
340

Chapter 23: Entropy and Gibbs free energy 349
Introducing entropy
Chance and spontaneous change
Calculating entropy changes
Entropy and temperature
Entropy, enthalpy changes and free energy
Gibbs free energy
Gibbs free energy calculations

Chapter 24: Transition elements
What is a transition element?
Physical properties of the transition elements
Redox reactions
Ligands and complex formation

Chapter 25: Benzene and its compounds

The benzene ring
Reactions of arenes
Phenol
Reactions of phenol

Chapter 26: Carboxylic acids and their
derivatives
The acidity of carboxylic acids
Oxidation of two carboxylic acids
Acyl chlorides

350
350
354
357
357
358
360

366
367
369
369
371

381
382
384
387
388

393
394
395
396

Chapter 27: Organic nitrogen compounds 400
Amines
Formation of amines
Amino acids
Peptides
Reactions of the amides
Electrophoresis

401
402
404
405
406
407

Chapter 28: Polymerisation
Condensation polymerisation
Synthetic polyamides
Biochemical polymers
The importance of hydrogen bonding in DNA
Polyesters
Designing useful polymers
Degradable polymers
Polymer deductions

Chapter 29: Analytical chemistry
Chromatography
Proton (1H) nuclear magnetic resonance
Carbon-13 NMR spectroscopy
Mass spectrometry

411
412
413
414
418
421
422
425
426

433
434
439
444
446

Chapter 30: Organic synthesis

456

Designing new medicinal drugs

457

Chapter P2: Practical skills 2
Written examination of practical skills
Planning
Analysis, conclusions and evaluation

464
465
465
468
v

Appendix 1: The Periodic Table of the
Elements

473

Appendix 2: Selected standard electrode
 potentials

474

Appendix 3: Qualitative analysis notes

475

Glossary

477

Index

486

Acknowledgements

493

CD-ROM

CD1

Introduction to the examination and changes
to the syllabus
Advice on how to revise for and approach
examinations
Answers to end-of-chapter questions
Recommended resources

CD1
CD4
CD10
CD76


How to use this book
Each chapter begins with a short
list of the facts and concepts that
are explained in it.

There is a short
context at the
beginning of each
chapter, containing
an example of how
the material covered
in the chapter relates
to the ‘real world’.

This book does not contain detailed instructions for doing particular experiments,
but you will find background information about the practical work you need to
do in these boxes. There are also two chapters, P1 and P2, which provide detailed
information about the practical skills you need to develop during the course.

vi

Important equations and
other facts are shown in
highlight boxes.

Questions throughout the text give
you a chance to check that you have
understood the topic you have just
read about. You can find the answers
to these questions on the CD-ROM.

The text and illustrations describe and
explain all of the facts and concepts
that you need to know. The chapters,
and often the content within them

as well, are arranged in the same
sequence as in your syllabus.


How to use this book

Wherever you need to know how to use a formula to carry out a calculation,
there are worked example boses to show you how to do this.

Definitions that are required by the
syllabus are shown in highlight boxes.

Key words are highlighted in the text
when they are first introduced.

You will also find definitions of
these words in the Glossary.

vii

There is a summary of key
points at the end of each
chapter. You might find
this helpful when you are
revising.

Questions at the end of each chapter are more
demanding exam-style questions, some of which
may require use of knowledge from previous
chapters. Answers to these questions can be

found on the CD-ROM.



Chapter 1:
Moles and equations
Learning outcomes
you should be able to:
■■

■■

■■

■■

define and use the terms:
– relative atomic mass, isotopic mass and
formula mass based on the 12C scale
– empirical formula and molecular formula
– the mole in terms of the Avogadro constant
analyse and use mass spectra to calculate the
relative atomic mass of an element
calculate empirical and molecular formulae using
combustion data or composition by mass
write and construct balanced equations

■■

■■

perform calculations, including use of the mole
concept involving:
– reacting masses (from formulae and equations)
– volumes of gases (e.g. in the burning of
hydrocarbons)
– volumes and concentrations of solutions
deduce stoichiometric relationships from
calculations involving reacting masses, volumes of
gases and volumes and concentrations of solutions.

1


Cambridge International AS Level Chemistry

Introduction
For thousands of years, people have heated rocks and
distilled plant juices to extract materials. Over the
past two centuries, chemists have learnt more and
more about how to get materials from rocks, from
the air and the sea, and from plants. They have also
found out the right conditions to allow these materials
to react together to make new substances, such as
dyes, plastics and medicines. When we make a new
substance it is important to mix the reactants in the
correct proportions to ensure that none is wasted. In
order to do this we need to know about the relative
masses of atoms and molecules and how these are
used in chemical calculations.

Figure 1.1  A titration is a method used to find the amount of
a particular substance in a solution.

2

Masses of atoms and molecules
Relative atomic mass, Ar

Atoms of different elements have different masses. When
we perform chemical calculations, we need to know how
heavy one atom is compared with another. The mass of
a single atom is so small that it is impossible to weigh it
directly. To overcome this problem, we have to weigh a lot
of atoms. We then compare this mass with the mass of the
same number of ‘standard’ atoms. Scientists have chosen
to use the isotope carbon-12 as the standard. This has been
given a mass of exactly 12 units. The mass of other atoms is
found by comparing their mass with the mass of carbon-12
atoms. This is called the relative atomic mass, Ar.

We use the average mass of the atom of a particular element
because most elements are mixtures of isotopes. For example,
the exact Ar of hydrogen is 1.0079. This is very close to 1 and
most periodic tables give the Ar of hydrogen as 1.0. However,
some elements in the Periodic Table have values that are not
whole numbers. For example, the Ar for chlorine is 35.5. This
is because chlorine has two isotopes. In a sample of chlorine,
chlorine-35 makes up about three-quarters of the chlorine
atoms and chlorine-37 makes up about a quarter.

Relative isotopic mass

Isotopes are atoms that have the same number of protons
but different numbers of neutrons (see page 28). We represent
the nucleon number (the total number of neutrons plus
protons in an atom) by a number written at the top
left-hand corner of the atom’s symbol, e.g. 20Ne, or by
The relative atomic mass is the weighted average mass of
a number written after the atom’s name or symbol, e.g.
naturally occurring atoms of an element on a scale where
neon-20 or Ne-20.
an atom of carbon-12 has a mass of exactly 12 units.
We use the term relative isotopic mass for the mass
of a particular isotope of an element on a scale where
From this it follows that:
an atom of carbon-12 has a mass of exactly 12 units. For
Ar [element Y]
example, the relative isotopic mass of carbon-13 is 13.00.
average
mass
of
one
atom
of
element
Y
×
12
If we know both the natural abundance of every isotope

 ​
= ​ ____________________________________
    
   
of an element and their isotopic masses, we can calculate
mass of one atom of carbon-12


Chapter 1: Moles and equations

the relative atomic mass of the element very accurately.
To find the necessary data we use an instrument called a
mass spectrometer (see box on mass spectrometry).

Relative molecular mass, Mr

The relative molecular mass of a compound (Mr) is the
relative mass of one molecule of the compound on a
scale where the carbon-12 isotope has a mass of exactly
12 units. We find the relative molecular mass by adding
up the relative atomic masses of all the atoms present in
the molecule.
For example, for methane:
formulaCH4
atoms present
1 × C; 4 × H
(1 × Ar[C]) + (4 × Ar[H])
add Ar values
= (1 × 12.0) + (4 × 1.0)
Mr of methane

= 16.0

Accurate relative atomic masses
Mass spectrometry

Biological
drawing
1.1: 1.2)
A mass spectrometerbox
(Figure
can be used
to measure the mass of each isotope present
in an element. It also compares how much of
each isotope is present – the relative abundance
(isotopic abundance). A simplified diagram of a
mass spectrometer is shown in Figure 1.3. You will
not be expected to know the details of how a mass
spectrometer works, but it is useful to understand
how the results are obtained.

Relative formula mass

For compounds containing ions we use the term relative
formula mass. This is calculated in the same way as for
relative molecular mass. It is also given the same symbol,
Mr. For example, for magnesium hydroxide:

formulaMg(OH)2
ions present

1 × Mg2+; 2 × (OH–)
(1 × Ar[Mg]) + (2 × (Ar[O] + Ar[H]))
add Ar values
Mr of magnesium
hydroxide
= (1 × 24.3) + (2 × (16.0 + 1.0))

= 58.3

3

Figure 1.2  A mass spectrometer is a large and complex
instrument.

question

vaporised sample
positively charged
electrodes accelerate positive ions

1
Use the Periodic Table on page 473 to calculate the
relative formula masses of the following:
acalcium chloride, CaCl2

magnetic field

bcopper(II) sulfate, CuSO4
cammonium sulfate, (NH4)2SO4
dmagnesium nitrate-6-water, Mg(NO3)2.6H2O

Hint: for part d you need to calculate the mass of
water separately and then add it to the Mr of Mg(NO3)2.

heated
filament
produces
high-energy
electrons

ionisation
chamber flight tube

ion
detector

recorder
computer

Figure 1.3  Simplified diagram of a mass spectrometer.


Cambridge International AS Level Chemistry

Determination of Ar from mass spectra

Mass spectrometry (continued)
The atoms of the element in the vaporised sample
are converted into ions. The stream of ions is
brought to a detector after being deflected (bent)
by a strong magnetic field. As the magnetic field is

increased, the ions of heavier and heavier isotopes
are brought to the detector. The detector is
connected to a computer, which displays the
mass spectrum.
The mass spectrum produced shows the relative
abundance (isotopic abundance) on the vertical
axis and the mass to ion charge ratio (m/e) on the
horizontal axis. Figure 1.4 shows a typical mass
spectrum for a sample of lead. Table 1.1 shows
how the data is interpreted.

■■
■■

multiply each isotopic mass by its percentage abundance
add the figures together
divide by 100.

We can use this method to calculate the relative atomic
mass of neon from its mass spectrum, shown in Figure 1.5.
The mass spectrum of neon has three peaks:
20Ne

(90.9%), 21Ne (0.3%) and 22Ne (8.8%).

Ar of neon

(20 × 90.9) + (21.0 × 0.3) + (22 × 8.8)
 
​= 20.2

   
  
= _______________________________
​ 
100

204

205 206 207 208
Mass/charge (m/e) ratio

209

Figure 1.4  The mass spectrum of a sample of lead.

For singly positively charged ions the m/e values
give the nucleon number of the isotopes detected.
In the case of lead, Table 1.1 shows that 52% of the
lead is the isotope with an isotopic mass of 208.
The rest is lead-204 (2%), lead-206 (24%) and lead207 (22%).
Isotopic mass

Relative abundance / %

204

  2

206

 24

207

 22

208

 52

total

100

Table 1.1  The data from Figure 1.4.

60
40
20
0

8.8 %

0

80

0.3 %

1

90.9 %

100

2
Relative abundance / %

Detector current / mA

■■

Note that this answer is given to 3 significant figures,
which is consistent with the data given.

3

4

We can use the data obtained from a mass spectrometer
to calculate the relative atomic mass of an element very
accurately. To calculate the relative atomic mass we follow
this method:

19
20
21
22
Mass/charge (m/e) ratio

23

Figure 1.5  The mass spectrum of neon, Ne.

A high-resolution mass spectrometer can give very
accurate relative isotopic masses. For example 16O = 15.995
and 32S = 31.972. Because of this, chemists can distinguish
between molecules such as SO2 and S2, which appear to
have the same relative molecular mass.


Chapter 1: Moles and equations

question

36.7 %

2
Look at the mass spectrum of germanium, Ge.

10
0




27.4 %

7.6 %

20

7.7 %

30

20.6 %

Abundance / %

40

70
75
Mass/charge (m/e) ratio

80

Figure 1.6  The mass spectrum of germanium.

aWrite the isotopic formula for the heaviest isotope
of germanium.

We often refer to the mass of a mole of substance as its
molar mass (abbreviation M). The units of molar mass
–1
are g mol .
The number of atoms in a mole of atoms is very large:
6.02 × 1023 atoms. This number is called the Avogadro
constant (or Avogadro number). The symbol for the

Avogadro constant is L (the symbol NA may also be used).
The Avogadro constant applies to atoms, molecules, ions
and electrons. So in 1 mole of sodium there are 6.02 × 1023
sodium atoms and in 1 mole of sodium chloride (NaCl) there
are 6.02 × 1023 sodium ions and 6.02 × 1023 chloride ions.
It is important to make clear what type of particles
we are referring to. If we just state ‘moles of chlorine’, it is
not clear whether we are thinking about chlorine atoms
or chlorine molecules. A mole of chlorine molecules, Cl2,
contains 6.02 × 1023 chlorine molecules but twice as many
chlorine atoms, as there are two chlorine atoms in every
chlorine molecule.

bUse the % abundance of each isotope to calculate
the relative atomic mass of germanium.

Amount of substance

5

The mole and the Avogadro constant

The formula of a compound shows us the number of
atoms of each element present in one formula unit or one
molecule of the compound. In water we know that two
atoms of hydrogen (Ar = 1.0) combine with one atom of
oxygen (Ar = 16.0). So the ratio of mass of hydrogen atoms
to oxygen atoms in a water molecule is 2 : 16. No matter
how many molecules of water we have, this ratio will
always be the same. But the mass of even 1000 atoms is

far too small to be weighed. We have to scale up much
more than this to get an amount of substance that is easy
to weigh.
The relative atomic mass or relative molecular mass of
a substance in grams is called a mole of the substance. So a
mole of sodium (Ar = 23.0) weighs 23.0 g. The abbreviation
for a mole is mol. We define the mole in terms of the
standard carbon-12 isotope (see page 28).
One mole of a substance is the amount of that substance
that has the same number of specific particles (atoms,
molecules or ions) as there are atoms in exactly 12 g of the
carbon-12 isotope.

Figure 1.7  Amedeo Avogadro (1776–1856) was an Italian
scientist who first deduced that equal volumes of gases
contain equal numbers of molecules. Although the Avogadro
constant is named after him, it was left to other scientists to
calculate the number of particles in a mole.

Moles and mass

The Système International (SI) base unit for mass is the
kilogram. But this is a rather large mass to use for general
laboratory work in chemistry. So chemists prefer to use
the relative molecular mass or formula mass in grams
(1000 g = 1 kg). You can find the number of moles of a
substance by using the mass of substance and the relative
atomic mass (Ar) or relative molecular mass (Mr).
 mass of substance in grams (g)
number of moles (mol) = ​ __________________________

   
  
 ​
molar mass (g mol–1)


Cambridge International AS Level Chemistry

worked example

worked example

1 How many moles of sodium chloride are present in
117.0 g of sodium chloride, NaCl?

2 What mass of sodium hydroxide, NaOH, is present in
0.25 mol of sodium hydroxide?

( Ar values: Na = 23.0, Cl = 35.5)

( Ar values: H = 1.0, Na= 23.0, O = 16.0)





molar mass of NaCl= 23.0 + 35.5

molar mass of NaOH = 23.0 + 16.0 + 1.0

= 58.5 g mol–1





mass
=​ ___________
   ​ 
molar mass





= 0.25 × 40.0 g



117.0
=​ _____ ​  



= 10.0 g NaOH

= 2.0 mol

number of moles

58.5

= 40.0 g mol–1

mass = number of moles × molar mass

question
4
Use these Ar values: C = 12.0, Fe = 55.8, H = 1.0,
O = 16.0, Na = 23.0.


Calculate the mass of the following:

a0.20 moles of carbon dioxide, CO2
b0.050 moles of sodium carbonate, Na2CO3
c5.00 moles of iron(II) hydroxide, Fe(OH)2

Mole calculations

6

Reacting masses
Figure 1.8  From left to right, one mole of each of copper,
bromine, carbon, mercury and lead.

question
3aUse these Ar values (Fe = 55.8, N = 14.0, O = 16.0,
S = 32.1) to calculate the amount of substance in
moles in each of the following:

i
10.7 g of sulfur atoms

ii
64.2 g of sulfur molecules (S8)

iii
60.45 g of anhydrous iron(III) nitrate, Fe(NO3)3.
bUse the value of the Avogadro constant (6.02 ×
1023 mol–1) to calculate the total number of atoms in
7.10 g of chlorine atoms. (Ar value: Cl = 35.5)

When reacting chemicals together we may need to know
what mass of each reactant to use so that they react exactly
and there is no waste. To calculate this we need to know
the chemical equation. This shows us the ratio of moles
of the reactants and products – the stoichiometry of the
equation. The balanced equation shows this stoichiometry.
For example, in the reaction
Fe2O3  +  3CO

2Fe  +  3CO2

1 mole of iron(III) oxide reacts with 3 moles of carbon
monoxide to form 2 moles of iron and 3 moles of carbon

dioxide. The stoichiometry of the equation is 1 : 3 : 2 : 3.
The large numbers that are included in the equation
(3, 2 and 3) are called stoichiometric numbers.
In order to find the mass of products formed in a
chemical reaction we use:

To find the mass of a substance present in a given number
■■ the mass of the reactants
of moles, you need to rearrange the equation
■■ the molar mass of the reactants
■■ the balanced equation.
of
substance
in
grams
(g)
 mass
   
  
number of moles (mol) = ​ __________________________
 ​
–1
molar mass (g mol )
mass of substance (g)
= number of moles (mol) × molar mass (g mol–1)


Chapter 1: Moles and equations

worked example

4 Iron(III) oxide reacts with carbon monoxide to form
iron and carbon dioxide.
Fe2O3  +  3CO

2Fe  +  3CO2



Calculate the maximum mass of iron produced when
798 g of iron(III) oxide is reduced by excess carbon
monoxide.



(Ar values: Fe = 55.8, O = 16.0)

Step 1 Fe2O3  +  3CO

Figure 1.9  Iron reacting with sulfur to produce iron sulfide.
We can calculate exactly how much iron is needed to react
with sulfur and the mass of the products formed by
knowing the molar mass of each reactant and the balanced
chemical equation.

worked example
3 Magnesium burns in oxygen to form magnesium oxide.
2Mg  +  O2


2MgO

We can calculate the mass of oxygen needed to react
with 1 mole of magnesium. We can calculate the mass
of magnesium oxide formed.

Step 1  Write the balanced equation.
Step 2  Multiply each formula mass in g by the
relevant stoichiometric number in the equation.
2Mg +O2
2 × 24.3 g 1 × 32.0 g
48.6 g 32.0 g



2MgO
2 × (24.3 g  +  16.0 g)
80.6 g



From this calculation we can deduce that:

■■

32.0 g of oxygen are needed to react exactly with
48.6 g of magnesium
80.6 g of magnesium oxide are formed.

■■

If we burn 12.15 g of magnesium (0.5 mol) we get
20.15 g of magnesium oxide. This is because the
stoichiometry of the reaction shows us that for every
mole of magnesium burnt we get the same number of
moles of magnesium oxide.

In this type of calculation we do not always need to know
the molar mass of each of the reactants. If one or more of
the reactants is in excess, we need only know the mass in
grams and the molar mass of the reactant that is not in
excess (the limiting reactant).

2Fe  +  3CO2

Step 2 1 mole iron(III) oxide

2 moles iron

(2 × 55.8) + (3 × 16.0)

2 × 55.8

159.6 
g Fe2O3

111.6 g Fe
111.6
Step 3 798 g​ _____ 

 ​ × 798
159.6

= 558 g Fe


You can see that in step 3, we have simply used ratios
to calculate the amount of iron produced from 798 g of
iron(III) oxide.

question
5aSodium reacts with excess oxygen to form sodium
peroxide, Na2O2.
2Na  +  O2

Na2O2

Calculate the maximum mass of sodium
peroxide formed when 4.60 g of sodium is burnt
in excess oxygen.

(Ar values: Na = 23.0, O = 16.0)

bTin(IV) oxide is reduced to tin by carbon. Carbon
monoxide is also formed.
SnO2  +  2C

Sn  +  2CO

Calculate the mass of carbon that exactly reacts

with 14.0 g of tin(IV) oxide. Give your answer to 3
significant figures.

(Ar values: C = 12.0, O = 16.0, Sn = 118.7)

The stoichiometry of a reaction

We can find the stoichiometry of a reaction if we know the
amounts of each reactant that exactly react together and
the amounts of each product formed.
For example, if we react 4.0 g of hydrogen with 32.0 g of
oxygen we get 36.0 g of water. (Ar values: H = 1.0, O = 16.0)

7


Cambridge International AS Level Chemistry

hydrogen (H2)  +  oxygen (O2)
water (H2O)
32.0
36.0
4.0
_______
   ​​ 
 
  ​​ 
  _____________
  
 

​ ______
2 × 1.0
2 × 16.0
(2 × 1.0) + 16.0
 ​ = 2 mol
= 1 mol
= 2 mol

worked example (continued)

This ratio is the ratio of stoichiometric numbers in the
equation. So the equation is:
2H2  +  O2

2H2O

We can still deduce the stoichiometry of this reaction
even if we do not know the mass of oxygen that reacted.
The ratio of hydrogen to water is 1 : 1. But there is only one
atom of oxygen in a molecule of water – half the amount
in an oxygen molecule. So the mole ratio of oxygen to
water in the equation must be 1 : 2.
Question
6
56.2 g of silicon, Si, reacts exactly with 284.0 g of
chlorine, Cl2, to form 340.2 g of silicon(IV) chloride,
SiCl4. Use this information to calculate the
stoichiometry of the reaction.

(Ar values: Cl = 35.5, Si = 28.1)

8

Significant figures

When we perform chemical calculations it is important
that we give the answer to the number of significant
figures that fits with the data provided. The examples show
the number 526.84 rounded up to varying numbers of
significant figures.
rounded to 4 significant figures = 526.8
rounded to 3 significant figures = 527
rounded to 2 significant figures = 530
When you are writing an answer to a calculation, the
answer should be to the same number of significant figures
as the least number of significant figures in the data.



Note 1  Zeros before a number are not significant
figures. For example, 0.004 is only to 1 significant
figure.



Note 2  After the decimal point, zeros after a number
are significant figures. 0.0040 has 2 significant figures
and 0.004 00 has 3 significant figures.

Note 3  If you are performing a calculation with
several steps, do not round up in between steps.
Round up at the end.

Percentage composition by mass

We can use the formula of a compound and relative atomic
masses to calculate the percentage by mass of a particular
element in a compound.
% by mass
atomic mass × number of moles of particular
      element in a compound      
 ​  × 100
    
   
=  ​  ______________________________________
molar mass of compound


worked example
6 Calculate the percentage by mass of iron in iron(III)
oxide, Fe2O3.
(Ar values: Fe = 55.8, O = 16.0)
2 × 55.8
  
   ​ × 100
% mass of iron = ​ __________________
(2 × 55.8) + (3 × 16.0)

= 69.9 %

worked example
5 How many moles of calcium oxide are there in 2.9 g of
calcium oxide?
(Ar values: Ca = 40.1, O = 16.0)


If you divide 2.9 by 56.1, your calculator shows
0.051 693 …. The least number of significant figures
in the data, however, is 2 (the mass is 2.9 g). So your
answer should be expressed to 2 significant figures,
as 0.052 mol.

Figure 1.10  This iron ore is impure Fe2O3. We can calculate
the mass of iron that can be obtained from Fe2O3 by using
molar masses.


Chapter 1: Moles and equations

Question

worked examples

7
Calculate the percentage by mass of carbon in
ethanol, C2H5OH.

7 Deduce the formula of magnesium oxide.

(Ar values: C = 12.0, H = 1.0, O = 16.0)



Empirical formulae

The empirical formula of a compound is the simplest whole
number ratio of the elements present in one molecule or
formula unit of the compound. The molecular formula of a
compound shows the total number of atoms of each element
present in a molecule.
Table 1.2 shows the empirical and molecular formulae
for a number of compounds.
■■

■■

■■

The formula for an ionic compound is always its
empirical formula.
The empirical formula and molecular formula for simple
inorganic molecules are often the same.
Organic molecules often have different empirical and
molecular formulae.

Compound

Empirical
formula

Molecular
formula

water

H2O

H2O

hydrogen peroxide

HO

H2O2

sulfur dioxide

SO2

SO2

butane

C2H5

C4H10

cyclohexane

CH2

C6H12

Table 1.2  Some empirical and molecular formulae.

Question
8
Write the empirical formula for:



This can be found as follows:
■■ burn a known mass of magnesium (0.486 g) in
excess oxygen
■■ record the mass of magnesium oxide formed
(0.806 g)
■■ calculate the mass of oxygen that has combined
with the magnesium (0.806 – 0.486 g) = 0.320 g
■■ calculate the mole ratio of magnesium to oxygen
■■ ( Ar values: Mg = 24.3, O = 16.0)
0.486 g
moles of Mg = ​ __________
  ​ 
= 0.0200 mol
24.3 g mol–1
0.320 g
moles of oxygen = ​ __________
  ​ 
= 0.0200 mol

16.0 g mol–1
The simplest ratio of magnesium : oxygen is 1 : 1. So the
empirical formula of magnesium oxide is MgO.

8 When 1.55 g of phosphorus is completely combusted
3.55 g of an oxide of phosphorus is produced. Deduce
the empirical formula of this oxide of phosphorus.
(Ar values: O = 16.0, P = 31.0)


P

O

Step 1 note the mass
1.55 g3.55 – 1.55

of each element
= 2.00 g
Step 2 divide by atomic
2.00 g
1.55 g
   ​​ 
 ___________
   ​ 

masses​  __________
31.0 g mol–1 16.0 g mol–11

= 0.05 mol
= 0.125 mol
Step 3 divide by the
0.05
0.125
____
_____

lowest figure​  0.05  ​ = 1​  0.05 ​  = 2.5

Step 4 if needed, obtain
P2O5

the lowest whole

number ratio

to get empirical

formula

ahydrazine, N2H4
boctane, C8H18
cbenzene, C6H6
dammonia, NH3

The empirical formula can be found by determining
the mass of each element present in a sample of the
compound. For some compounds this can be done by
combustion. An organic compound must be very pure in

order to calculate its empirical formula. Chemists often
use gas chromatography to purify compounds before
carrying out formula analysis.

An empirical formula can also be deduced from data that
give the percentage composition by mass of the elements
in a compound.

9


Cambridge International AS Level Chemistry

worked example

worked example (continued)

9 A compound of carbon and hydrogen contains 85.7%
carbon and 14.3% hydrogen by mass. Deduce the
empirical formula of this hydrocarbon.

Step 2  divide the relative molecular mass by
187.8
the empirical formula mass:  _____
​ 
 ​   = 2
93.9

(Ar values: C = 12.0, O = 16.0)

Step 3  multiply the number of atoms in the empirical
formula by the number in step 2:
2 × CH2Br, so molecular formula is C2H4Br2.



C

Step 1 note the % by mass 85.7

H
14.3

85.7
14.3
Step 2 divide by Ar values​ ____  ​  = 7.142​ ____ ​   = 14.3
12.0
1.0




14.3
7.142
Step 3 divide by the lowest _____
 ​    = 1​ _____  ​   = 2
​ 
figure
7.142
7.142

Empirical formula is CH2.

Question
9
The composition by mass of a hydrocarbon is 10%
hydrogen and 90% carbon. Deduce the empirical
formula of this hydrocarbon.

(Ar values: C = 12.0, H = 1.0)
10

Molecular formulae

The molecular formula shows the actual number of each
of the different atoms present in a molecule. The molecular
formula is more useful than the empirical formula. We
use the molecular formula to write balanced equations
and to calculate molar masses. The molecular formula is
always a multiple of the empirical formula. For example,
the molecular formula of ethane, C2H6, is two times the
empirical formula, CH3.
In order to deduce the molecular formula we need
to know:
■■
■■

the relative formula mass of the compound
the empirical formula.

worked example

10 A compound has the empirical formula CH2Br. Its
relative molecular mass is 187.8. Deduce the molecular
formula of this compound.
(Ar values: Br = 79.9, C = 12.0, H = 1.0)

Step 1  find the empirical formula mass:
12.0 + (2 × 1.0) + 79.9 = 93.9

Question
10The empirical formulae and molar masses of three
compounds, A, B and C, are shown in the table below.
Calculate the molecular formula of each of these
compounds.
(Ar values: C = 12.0, Cl = 35.5, H = 1.0)

Compound

Empirical formula

Mr

A

C3H5

 82

B

CCl3

237

C

CH2

112

Chemical formulae and
chemical equations
Deducing the formula

The electronic structure of the individual elements in a
compound determines the formula of a compound (see
page 33). The formula of an ionic compound is determined by
the charges on each of the ions present. The number of positive
charges is balanced by the number of negative charges so that
the total charge on the compound is zero. We can work out the
formula for a compound if we know the charges on the ions.
Figure 1.11 shows the charges on some simple ions related to
the position of the elements in the Periodic Table. The form
of the Periodic Table that we shall be using has 18 groups
because the transition elements are numbered as Groups
3 to 12. So, aluminium is in Group 13 and chlorine is in
Group 17.
For simple metal ions in Groups 1 and 2, the value of
the positive charge is the same as the group number. For
a simple metal ion in Group 13, the value of the positive
charge is 3+. For a simple non-metal ion in Groups 15 to

17, the value of the negative charge is 18 minus the group


Chapter 1: Moles and equations

18

Group
1

2

Li+

Be2+

Na+

Mg2+

K+

Ca2+

Rb+

Sr2+

H+

13

14

15

transition
elements

worked examples

16

17

2–

F–

none

11 Deduce the formula of magnesium chloride.

2–

–

Ions present: Mg2+ and Cl–.

Cl

none




Br–

none

For electrical neutrality, we need two Cl– ions for every
Mg2+ ion. (2 × 1–) + (1 × 2+) = 0

I–

none



So the formula is MgCl2.

O
Al3+

none

S

Ga3+

Figure 1.11  The charge on some simple ions is related to
their position in the Periodic Table.

number. The charge on the ions of transition elements can
vary. For example, iron forms two types of ions, Fe2+ and
Fe3+ (Figure 1.12).

12 Deduce the formula of aluminium oxide.


Ions present: Al3+ and O2–.



For electrical neutrality, we need three O2– ions for
every two Al3+ ions. (3 × 2−) + (2 × 3+) = 0



So the formula is Al2O3.

The formula of a covalent compound is deduced from the
number of electrons needed to achieve the stable electronic
configuration of a noble gas (see page 49). In general,
carbon atoms form four bonds with other atoms, hydrogen
and halogen atoms form one bond and oxygen atoms form
two bonds. So the formula of water, H2O, follows these
rules. The formula for methane is CH4, with each carbon
atom bonding with four hydrogen atoms. However, there
are many exceptions to these rules.

Compounds containing a simple metal ion and nonmetal ion are named by changing the end of the name of
the non-metal element to -ide.
sodium chloride

sodium + chlorine
Figure 1.12  Iron(II) chloride (left) and iron(III) chloride (right).
These two chlorides of iron both contain iron and chlorine,
but they have different formulae.

Ions that contain more than one type of atom are called
compound ions. Some common compound ions that you
should learn are listed in Table 1.3. The formula for an
ionic compound is obtained by balancing the charges of
the ions.
Ion

zinc + sulfur

zinc sulfide

Compound ions containing oxygen are usually called
-ates. For example, the sulfate ion contains sulfur
and oxygen, the phosphate ion contains phosphorus
and oxygen.
Question
11aWrite down the formula of each of the following
compounds:

Formula

ammonium

NH4+


i
magnesium nitrate

carbonate

CO32–


ii
calcium sulfate

hydrogencarbonate

HCO3–

hydroxide

OH–

nitrate

NO3–

phosphate

PO43–

sulfate

SO42–

Table 1.3  The formulae of some common compound ions.


iii
sodium iodide

iv
hydrogen bromide

v
sodium sulfide
bName each of the following compounds:

i
Na3PO4iiiAlCl3

ii
(NH4)2SO4

i vCa(NO3)2

11

Cambridge International AS Level Chemistry

Balancing chemical equations

worked examples (continued)

When chemicals react, atoms cannot be either created
or destroyed. So there must be the same number of each
type of atom on the reactants side of a chemical equation
as there are on the products side. A symbol equation is a
shorthand way of describing a chemical reaction. It shows
the number and type of the atoms in the reactants and the
number and type of atoms in the products. If these are
the same, we say the equation is balanced. Follow these
examples to see how we balance an equation.

Step 3 balance the Fe2O3 + CO
2Fe + CO2
iron

2[Fe] + 1[C] + 2[Fe] + 1[C] +

3[O]1[O]2[O]



Step 4 balance the Fe2O3 + 3CO
2Fe + 3CO2

oxygen 2[Fe] + 3[C] + 2[Fe] + 3[C] +

3[O]3[O]6[O]


worked examples

In step 4 the oxygen in the CO2 comes from two places,
the Fe2O3 and the CO. In order to balance the equation,
the same number of oxygen atoms (3) must come from
the iron oxide as come from the carbon monoxide.

13 Balancing an equation
Step 1  Write down the formulae of all the reactants and
products. For example:
H2

+

O2



Question

H2O

12Write balanced equations for the following reactions.

Step 2  Count the number of atoms of each reactant
and product.

H2

+

O2



aIron reacts with hydrochloric acid to form iron(II)
chloride, FeCl2, and hydrogen.
bAluminium hydroxide, Al(OH)3, decomposes on
heating to form aluminium oxide, Al2O3, and water.

H2O

2[H] 2[O]
2[H] + 1[O]
12



Step 3  Balance one of the atoms by placing a number
in front of one of the reactants or products. In this case
the oxygen atoms on the right-hand side need to be
balanced, so that they are equal in number to those on
the left-hand side. Remember that the number in front
multiplies everything in the formula. For example, 2H2O
has 4 hydrogen atoms and 2 oxygen atoms.

H2

+

O2



2H2O

2[H] 2[O]4[H] + 2[O]
Step 4  Keep balancing in this way, one type of atom at
a time until all the atoms are balanced.
2H2

+

O2



2H2O

cHexane, C6H14, burns in oxygen to form carbon
dioxide and water.

Using state symbols

We sometimes find it useful to specify the physical states
of the reactants and products in a chemical reaction. This
is especially important where chemical equilibrium and

rates of reaction are being discussed (see Chapter 8 and
Chapter 9). We use the following state symbols:
■■
■■

4[H] 2[O]
4[H] + 2[O]

■■



■■

Note that when you balance an equation you must not
change the formulae of any of the reactants
or products.

14 Write a balanced equation for the reaction of iron(III)
oxide with carbon monoxide to form iron and carbon
dioxide.
Step 1formulae Fe2O3 + CO

Fe + CO2

Step 2 count the

number

of atoms

Fe2O3 + CO

Fe + CO2



3[O]
1[O]
2[O]

2[Fe] + 1[C] + 1[Fe] 1[C] +

(s) solid
(l) liquid
(g) gas
(aq) aqueous (a solution in water).

State symbols are written after the formula of each
reactant and product. For example:
ZnCO3(s) + H2SO4(aq)

ZnSO4(aq) + H2O(l) + CO2(g)

Question
13Write balanced equations, including state symbols, for
the following reactions.
aSolid calcium carbonate reacts with aqueous
hydrochloric acid to form water, carbon dioxide
and an aqueous solution of calcium chloride.

Chapter 1: Moles and equations

with charges:

Question (continued)

Zn(s)  +  Cu2+SO42–(aq)

Zn2+SO42–(aq)  +  Cu(s)

bAn aqueous solution of zinc sulfate, ZnSO4,
reacts with an aqueous solution of sodium
hydroxide. The products are a precipitate of zinc
hydroxide, Zn(OH)2, and an aqueous solution of
sodium sulfate.

cancelling spectator ions: Zn(s)  +  Cu2+SO42–(aq)

Zn2+SO42–(aq)  +  Cu(s)

ionic equation:

Zn(s)  +  Cu2+(aq)
Zn2+(aq)  +  Cu(s)

In the ionic equation you will notice that:
■■

■■

there are no sulfate ions – these are the spectator ions as
they have not changed
both the charges and the atoms are balanced.

The next examples show how we can change a full
equation into an ionic equation.
worked examples
15 Writing an ionic equation
Step 1  Write down the full balanced equation.

Figure 1.13  The reaction between calcium carbonate and
hydrochloric acid. The equation for this reaction, with all the
state symbols, is:
CaCO3(s)  +  2HCl(aq)

CaCl2(aq)  +  CO2(g)  +  H2O(l)

Balancing ionic equations

When ionic compounds dissolve in water, the ions
separate from each other. For example:
NaCl(s)  +  aq

Na+(aq)  +  Cl–(aq)

Ionic compounds include salts such as sodium bromide,
magnesium sulfate and ammonium nitrate. Acids and
alkalis also contain ions. For example H+(aq) and Cl–(aq)

ions are present in hydrochloric acid and Na+(aq) and
OH–(aq) ions are present in sodium hydroxide.
Many chemical reactions in aqueous solution involve
ionic compounds. Only some of the ions in solution take
part in these reactions.
The ions that play no part in the reaction are called
spectator ions.
An ionic equation is simpler than a full chemical
equation. It shows only the ions or other particles that are
reacting. Spectator ions are omitted. Compare the full
equation for the reaction of zinc with aqueous copper(II)
sulfate with the ionic equation.
full chemical equation:

Zn(s)  +  CuSO4(aq)
ZnSO4(aq)  +  Cu(s)

Mg(s)  +  2HCl(aq)

MgCl2(aq)  +  H2(g)

Step 2  Write down all the ions present. Any reactant
or product that has a state symbol (s), (l) or (g) or is a
molecule in solution such as chlorine, Cl2(aq), does not
split into ions.


Mg(s)  +  2H+(aq)  +  2Cl–(aq)
Mg2+(aq)  +  2Cl–(aq)  +  H2(g)

Step 3  Cancel the ions that appear on both sides of
the equation (the spectator ions).


Mg(s)  +  2H+(aq)  +  2Cl–(aq)
Mg2+(aq)  +  2Cl–(aq)  +  H2(g)

Step 4  Write down the equation omitting the
spectator ions.


Mg(s)  +  2H+(aq)

Mg2+(aq)  +  H2(g)

16 Write the ionic equation for the reaction of aqueous
chlorine with aqueous potassium bromide. The
products are aqueous bromine and aqueous
potassium chloride.
Step 1  The full balanced equation is:
Cl2(aq)  +  2KBr(aq)

Br2(aq)  +  2KCl(aq)

Step 2  The ions present are:
Cl2(aq)  +  2K+(aq)  +  2Br–(aq)
Br2(aq)  +  2K+(aq)  +  2Cl–(aq)
Step 3  Cancel the spectator ions:
Cl2(aq)  +  2K+(aq)  +  2Br–(aq)
Br2(aq)  +  2K+(aq)  +  2Cl–(aq)

Step 4  Write the final ionic equation:
Cl2(aq)  +  2Br–(aq)

Br2(aq)  +  2Cl–(aq)

13


Cambridge International AS Level Chemistry

compound is dissolved to make 1 dm3 of solution the
concentration is 1 mol dm–3.

Question
14Change these full equations to ionic equations.
a H2SO4(aq) + 2NaOH(aq)


bBr2(aq)  +  2KI(aq)

2H2O(aq) + Na2SO4(aq)
2KBr(aq)  +  I2(aq)

Chemists usually prefer to write ionic equations for
precipitation reactions. A precipitation reaction is a
reaction where two aqueous solutions react to form a solid
– the precipitate. For these reactions the method of writing
the ionic equation can be simplified. All you have to do is:
■■
■■

write the formula of the precipitate as the product
write the ions that go to make up the precipitate as
the reactants.

worked example
17 An aqueous solution of iron(II) sulfate reacts with an
aqueous solution of sodium hydroxide. A precipitate of
iron(II) hydroxide is formed, together with an aqueous
solution of sodium sulfate.
■■

14

Write the full balanced equation:

FeSO4(aq) + 2NaOH(aq)
■■

Fe(OH)2(s) +  Na2SO4(aq)

The ionic equation is:

Fe2+(aq)  +  2OH–(aq)

Fe(OH)2(s)

concentration (mol dm–3)
number of moles of solute (mol)
   

 ​
​     
= ___________________________
volume of solution (dm3)
We use the terms ‘concentrated’ and ‘dilute’ to refer to the
relative amount of solute in the solution. A solution with a
low concentration of solute is a dilute solution. If there is a
high concentration of solute, the solution is concentrated.
When performing calculations involving
concentrations in mol dm–3 you need to:
■■
■■

change mass in grams to moles
change cm3 to dm3 (by dividing the number of cm3 by 1000).

worked example
18 Calculate the concentration in mol dm–3 of sodium
hydroxide, NaOH, if 250 cm3 of a solution contains 2.0 g
of sodium hydroxide.
(Mr value: NaOH = 40.0)

Step 1 Change grams to moles.
2.0

​ ____  ​ = 0.050 mol NaOH
40.0
Step 2 Change cm3 to dm3.
250

250 cm3 = ​ _____  ​  dm3 = 0.25 dm3
1000
Step 3 Calculate concentration.
0.050 (mol)

​ __________
 
 ​ 
= 0.20 mol dm–3
0.25 (dm3)

Question
15Write ionic equations for these precipitation reactions.



aCuSO4(aq)  +  2NaOH(aq)
bPb(NO3)2(aq)  +  2KI(aq)

Cu(OH)2(s)  +  Na2SO4(aq)
PbI2(s)  +  2KNO3(aq)

Solutions and concentration
Calculating the concentration of
a solution

The concentration of a solution is the amount of
solute dissolved in a solvent to make 1 dm3 (one cubic
decimetre) of solution. The solvent is usually water. There
are 1000 cm3 in a cubic decimetre. When 1 mole of a

Figure 1.14  The concentration of chlorine in the water in a
swimming pool must be carefully controlled.


Chapter 1: Moles and equations

We often need to calculate the mass of a substance present
in a solution of known concentration and volume. To do
this we:
■■

Question
16aCalculate the concentration, in mol dm–3, of the
following solutions: (Ar values: C = 12.0, H = 1.0,
Na = 23.0, O = 16.0)

rearrange the concentration equation to:
number of moles (mol)
= concentration (mol dm–3) × volume (dm3)

■■


i
a solution of sodium hydroxide, NaOH,
containing 2.0 g of sodium hydroxide in 50 cm3
of solution

multiply the moles of solute by its molar mass


ii
a solution of ethanoic acid, CH3CO2H, containing
12.0 g of ethanoic acid in 250 cm3 of solution.

mass of solute (g)
= number of moles (mol) × molar mass (g mol–1)

bCalculate the number of moles of solute dissolved
in each of the following:

worked example


i
40 cm3 of aqueous nitric acid of concentration
0.2 mol dm–3

19 Calculate the mass of anhydrous copper(II) sulfate in
55 cm3 of a 0.20 mol dm–3 solution of copper(II) sulfate.


ii
50 cm3 of calcium hydroxide solution of
concentration 0.01 mol dm–3.

(Ar values: Cu = 63.5, O = 16.0, S = 32.1)

Step 1 Change cm3 to dm3.

55
   ​ = 0.055 dm3

= ​ _____
1000

Step 2 moles= concentration (mol dm–3)
×  volume of solution (dm3)


= 0.20 × 0.055 = 0.011 mol

Step 3 mass (g) = moles ×  M


= 0.011 ×  (63.5 + 32.1 + (4 × 16.0))



= 1.8 g (to 2 significant figures)

15

Carrying out a titration
A procedure called a titration is used to determine the
amount of substance present in a solution of unknown
concentration. There are several different kinds of
titration. One of the commonest involves
the exact neutralisation of an alkali by an acid
(Figure 1.15).

If we want to determine the concentration of a solution
of sodium hydroxide we use the following procedure.
■
■

■
■

■
■

Get some of acid of known concentration.
Fill a clean burette with the acid (after having washed
the burette with a little of the acid).
Record the initial burette reading.
Measure a known volume of the alkali into a titration
flask using a graduated (volumetric) pipette.
Add an indicator solution to the alkali in the flask.
Slowly add the acid from the burette to the flask,
swirling the flask all the time until the indicator changes
colour (the end-point).

■

■

■

■

Record the final burette reading. The final reading
minus the initial reading is called the titre. This first
titre is normally known as a ‘rough’ value.
Repeat this process, adding the acid drop by drop
near the end-point.
Repeat again, until you have two titres that are no
more than 0.10 cm3 apart.
Take the average of these two titre values.

Your results should be recorded in a table, looking
like this:
rough

1

2

3

final burette
reading / cm3

37.60

38.65

36.40

34.75

initial burette
reading / cm3

 2.40

 4.00

 1.40

 0.00

titre / cm3

35.20

34.65

35.00

34.75