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William Collins’ dream of knowledge for all began with the publication of his first book in 1819. A self-educated mill worker, he not only enriched millions of lives, but also founded a flourishing publishing house. Today, staying true to this spirit, Collins books are packed with inspiration, innovation and practical expertise. They place you at the centre of a world of possibility and give you exactly what you need to explore it. Collins. Freedom to teach HarperCollins Publishers The News Building 1 London Bridge Street London SE1 9GF
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10 9 8 7 6 5 4 3 2 1 © HarperCollinsPublishers 2016 ISBN 978-0-00-818078-2 Collins® is a registered trademark of HarperCollinsPublishers Limited www.collins.co.uk A catalogue record for this book is available from the British Library Thanks to John Bentham and Graham Curtis for their contributions to the previous editions. Commissioned by Gillian Lindsey Edited by Alexander Rutherford Project managed by Maheswari PonSaravanan at Jouve Development by Tim Jackson Copyedited and proofread by Janette Schubert Typeset by Jouve India Private Limited Original design by Newgen Imaging Cover design by Angela English Printed by CPI Group (UK) Ltd, Croydon, CR0 4YY Cover image © Shutterstock/isaravut All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission in writing of the Publisher. This book is sold subject to the conditions that it shall not, by way of trade or otherwise, be lent, re-sold, hired out or otherwise circulated without the Publisher’s prior consent in any form of binding or cover other than that in which it is published and without a similar condition including this condition being imposed on the subsequent purchaser. HarperCollins does not warrant that www.collins.co.uk or any other website mentioned in this title will be provided uninterrupted, that any website will be error free, that defects will be corrected, or that the website or the server that makes it available are free of viruses or bugs. For full terms and conditions please refer to the site terms provided on the website.
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Contents 3.1
Physical chemistry 3.1.1 Atomic structure 3.1.2 Amount of substance 3.1.3 Bonding 3.1.4 Energetics 3.1.6 Chemical equilibria, Le Chatelier’s principle and Kc 3.1.7 Oxidation, reduction and redox equations
4 4 13 21 33 39 49
3.2
Inorganic chemistry 3.2.1 Periodicity 3.2.2 Group 2, the alkaline earth metals 3.2.3 Group 7(17), the halogens
55 55 57 60
Examination preparation Practical and mathematical skills Practice exam-style questions Answers
66 69 80
Glossary
85
Index
89
Notes
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3.1
Physical chemistry
3.1.1 Atomic structure
3.1.1.1 Fundamental particles Evolution of atomic structure over time Ideas about atoms and their internal structure have evolved over thousands of years, with our understanding accelerating in the last 200 years. Element: Aristotle (c350 bce) proposed that earthly matter was made up of four elements (earth, air, wind, fire). Atom: Democritus (c400 bce) proposed that matter was made up of indivisible particles. These became known as atoms from the Greek ‘atomos’, meaning ‘cannot be divided’. Dalton, in his Atomic Theory (1803), resurrected the idea of atoms as being indivisible and indestructible. Atoms have different masses, and combine to form what we now call compounds. Nucleus: When studying the penetrating effects of emissions from a radioactive source, Rutherford (1911) deduced that almost all of the mass of an atom is concentrated in its tiny centre. Electron: Experiments on electrical discharge tubes led to the discovery of cathode rays, and then to Thomson’s discovery of the electron (1897).
Table 1. and bonding by assuming that an atom is made from the particles shown in Students of chemistry can develop all the necessary explanations for structure detailed and sophisticated knowledge. amongst others, Bohr, de Broglie, and Schrödinger, has led to our current more did not cease with the discovery of the neutron, and that subsequent work by, It should be appreciated that our understanding of the structure of the atom neutrons do exist. lack of charge made its proof harder. Chadwick (1932) finally proved that proposed at about the same time as the discovery of the proton, the neutron’s Neutron: Although the presence of neutral particles in the nuclei of atoms was of an atom. particles called protons were responsible for the positive charge on the nucleus Proton: Rutherford, with Moseley, developed the idea that positive charged cathode rays, and then to Thomson’s discovery of the electron (1897). Electron: Experiments on electrical discharge tubes led to the discovery of concentrated in its tiny centre. source, Rutherford (1911) deduced that almost all of the mass of an atom is Nucleus: When studying the penetrating effects of emissions from a radioactive combine to form what we now call compounds. atoms as being indivisible and indestructible. Atoms have different masses, and ‘cannot be divided’. Dalton, in his Atomic Theory (1803), resurrected the idea of particles. These became known as atoms from the Greek ‘atomos’, meaning Atom: Democritus (c400 bce) proposed that matter was made up of indivisible elements (earth, air, wind, fire). Element: Aristotle (c350 bce) proposed that earthly matter was made up of four years, with our understanding accelerating in the last 200 years. Ideas about atoms and their internal structure have evolved over thousands of
Evolution of atomic structure over time
3.1.1.1 Fundamental particles
Proton: Rutherford, with Moseley, developed the idea that positively charged particles called protons were responsible for the positive charge on the nucleus of an atom. Neutron: Although the presence of neutral particles in the nuclei of atoms was proposed at about the same time as the discovery of the proton, the neutron’s lack of charge made its proof harder. Chadwick (1932) finally proved that neutrons do exist.
Notes The SI unit of charge is the coulomb (C). SI is the abbreviation used for the international system of units of measurement (Système International d’Unités).
Table 1 Properties of fundamental particles
It should be appreciated that our understanding of the structure of the atom did not cease with the discovery of the neutron, and that subsequent work by, amongst others, Bohr, de Broglie, and Schrödinger, has led to our current more detailed and sophisticated knowledge. Students of chemistry can develop all the necessary explanations for structure and bonding by assuming that an atom is made from the particles shown in Table 1. Particle
Mass/kg
Charge/C
Relative charge
1
1
1.673 10
1.602 10
neutron
1.675 10
–27
0
1
electron
9.109 10 –31
1.602 10 –19
5.45 10–4
proton
–19
Relative mass
–27
0 –1
The mass of an electron is so small in comparison with the mass of either a proton or a neutron that its relative mass is often taken to be zero.
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Protons, neutrons and electrons An atom consists of electrons surrounding a small, heavy nucleus that contains protons and neutrons (except for the hydrogen atom, 1H, which has only one proton and no neutrons in the nucleus).
Notes The atomic radius of a hydrogen atom is about 10 000 times the radius of the nucleus.
3.1.1.2 Mass number and isotopes Definition The mass number, A, of an atom is the total number of protons and neutrons in the nucleus of one atom of the element.
Definition The atomic (proton) number, Z, of an atom is the number of protons in the nucleus of an atom.
An atom is neutral; it has no overall charge. The charge on a proton is equal but opposite to the charge on an electron. Therefore the atomic number must also be equal to the number of electrons in a neutral atom. In an element, each atom has the same atomic number, the same number of protons and the same number of electrons. Isotopes of the same element consist of atoms with the same atomic number but different numbers of neutrons; their mass numbers are therefore different. The notation for an isotope gives the mass number and the atomic number: A mass number → 12 C or in general Z X atomic number → 6
Some isotopes of hydrogen are
1 2 3 1H 1H 1H
Some isotopes of chlorine are
35 37 17Cl 17Cl
The number of neutrons in the nucleus of an isotope can be calculated as follows: mass number number of protons number of neutrons s o number of neutrons mass number − atomic number e.g. for 126C, the number of neutrons 12 − 6 6 for 136C, the number of neutrons 13 − 6 7 The chemical properties of isotopes are almost identical, because isotopes have the same number of protons and electrons. Chemical properties are dictated by the number and the arrangement of electrons. The only differences between isotopes are in physical properties, such as rates of diffusion, which depend on the mass of the particles, or in nuclear properties such as radioactivity and the ability to absorb neutrons. Different isotopes of the same element also have slightly different boiling points.
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Principles of a time of flight mass spectrometer Mass spectrometry is a powerful instrumental method of analysis. It can be used to find the mass and abundance of each isotope in an element, allowing its relative atomic mass to be determined
l
help to identify molecules by determining their relative molecular mass.
l
Applications include the testing of athletes’ blood or urine samples for the illegal use of performance-enhancing substances, and in space research to analyse rock and atmosphere samples. A common form of mass spectrometry is time of flight mass spectrometry (Fig 1), in which atoms or molecules are ionised to form positive 1 ions. The ions are then accelerated to a point where they all have the same kinetic energy. The time taken to travel a further fixed distance is used to find the mass of each ion in the sample. Fig 1 A time of flight mass spectrometer (simplified)
Ionisation region (where ions form)
Drift region (no accelerating field: heavy ions drift more slowly, light ions drift faster)
Detection (ions arrive with different times of flight, each producing a current)
Acceleration region time of flight is measured (electric field gets all ions between these points to the same kinetic energy)
Ionisation The neutral atoms or molecules in the sample must be turned into positive ions. Two methods to achieve this are electron impact ionisation and electrospray ionisation. Electron impact ionisation (also known as electron ionisation) The sample being analysed is vaporised and high-energy electrons are fired at it. The high-energy electrons come from an ‘electron gun’. The electron gun has a hot wire filament with a current running though it; the heated wire emits electrons. These electrons are accelerated by attraction to a positively charged electrode. Such high-energy electrons usually knock off only one electron from each atom or molecule in the sample forming a 1 ion. M(g) e− → M(g) 2e− (also written as M(g) → M(g) e−)
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The 1 ions are then attracted towards a negative electric plate and therefore accelerate to gain the same kinetic energy regardless of mass. This technique is used for elements and inorganic or organic molecular substances with low molecular mass. When molecules are ionised in this way, in addition to forming a molecular ion, they can break down into smaller fragments some of which are also detected in the mass spectrum. Electrospray ionisation The sample (X) is dissolved in a volatile solvent (e.g. water or methanol) and injected through a thin hypodermic needle to give a fine mist (aerosol). The needle tip is attached to the positive terminal of a high voltage power supply. The particles are ionised by gaining a proton (H) from the solvent as they leave the needle producing XH ions (ions with a single positive charge and a mass of Mr 1).
Essential Notes All the 1 ions have the same kinetic energy.
Notes The study of fragments is not required within this specification.
Essential Notes Particles gain a proton from the solvent so X becomes XH.
X H → XH The solvent evaporates away and the XH ions are attracted towards the negative plate in the same way as the X ions were in electron impact ionisation. This technique is used for many substances with high molecular mass, including many biological molecules such as proteins. This procedure is known as a ‘soft’ ionisation technique (that is, low energy) and fragmentation rarely takes place. This outcome is an advantage because many large organic molecules subjected to electron impact ionisation fragment so readily that a molecular ion is not formed, so that the relative molecular mass cannot be determined.
Acceleration The positive ions are accelerated using an electric field so that they all have the same kinetic energy, Ek. E k 1 mv 2 2
Ek is the kinetic energy of particle (J) m is the mass of the particle (kg)
Essential Notes Soft ionisation impedes the fragmentation of the protonated molecular ion XH.
Notes You will be given this equation if you are expected to use it in an exam.
v is the speed of the particle (m s−1) So, on rearranging, the speed of each particle is given by: v
2Ek m
Given that all the particles have the same kinetic energy, the speed of each particle depends on its mass. Lighter particles have greater speed, and heavier particles have a lesser speed.
Ion drift The positive ions travel through a hole in the negatively charged plate into a tube. The time of flight of each particle through this flight tube depends on its speed (t d/v) and therefore on its mass.
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Essential Notes The mass of the charged particle determines its speed and hence its time of flight.
The time of flight along the flight tube is given by the following expression where d is the length of the tube: t d
m 2Ek
t is the time of flight (s) Ek is the kinetic energy of particle (J) m is the mass of the particle (kg) d is the length of flight tube (m)
This equation shows that the time of flight is proportional to the square root of the mass of the ions. Therefore lighter ions travel fast and reach the detector in less time and the heavier particles travel more slowly and take longer to reach the detector. For example, ions of the three isotopes of magnesium (24Mg, 25Mg, 26Mg) will travel at different speeds through the flight tube and separate, with the lightest ion (24Mg) reaching the detector first. Example Typical time of flight calculation: 26
Mg ion has relative mass 26
The actual mass of the ion 26/L g 26 10−3/L kg, where L the Avogadro constant, 6.022 1023 mol−1 Therefore m 26 10−3/6.022 1023 4.35 10−26 kg So if Ek 2.175 10−16 J, and d 0.6 m Then t 0.6
4.35 × 10−26 2 × 2.175 × 10−16
0.6 10−5 s 6 10−6 s
Detection
Essential Notes In mass spectrometry, m/z is known as the mass-to-charge ratio, where m is the relative molecular, atomic or fragment mass and z is the charge on the ion. Because the ionisation process is designed to produce only 1 ions, m/z values in a printout are a direct measure of relative mass.
At the end of the drift tube, the positive ions strike a negatively charged electric plate. When they hit this detector plate, the positive ions are neutralised by gaining electrons from the plate. This process generates a flow of electrons and hence an electric current that is then amplified to produce a signal on a computer. The relative intensity of the peak in the resulting mass spectrum produced by an ion with a particular m/z value (mass-to-charge ratio) is proportional to the magnitude of the amplified current. This current is proportional to the number of ions hitting the plate. Therefore the current and hence the peak height give a measure of the abundance of the ion. A typical printout from a sample of chlorine in a mass spectrometer that uses electron impact ionisation is shown in Fig 2. Molecules ionised using electron impact ionisation give rise to a peak with a maximum value of m/z Mr.
l
Molecules ionised using electrospray ionisation give rise to a peak with a maximum value of m/z (Mr 1).
l
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Relative intensity
Fig 2 The mass spectrum of chlorine
Essential Notes
20
30
40
m/z
50
60
The mass spectrometer is sensitive enough to distinguish clearly between isotopes.
70
The peak at m/z 35 represents the 35Cl ion. The ratio of peak heights at m/z 35 and 37 is 3 : 1. The peak heights for the Cl2 ions are in the ratio 9 : 6 : 1. This ratio represents the proportions of Cl2 ions with m/z 70, 72 and 74, respectively.
Notes For simplicity, the atomic number of isotopes is often omitted in symbols such as 35 Cl.
The peaks in the mass spectrum can be assigned as in Table 2. m/z
35
37
70
Ion from
35
37
35
Cl
Cl
Cl —
72 74 35
Cl
37
Cl — 35Cl
37
Cl —
37
Cl
The relative atomic mass (Ar) of an isotopic mixture can be calculated by using information from a mass spectrum. The spectrum above shows that the relative proportions of isotopes in that sample of chlorine are: 35
Cl :
so
3 4
37
Cl
3 : 1
of the sample is 35Cl and
so Ar
3 4
35
1 4
1 4
of the sample is 37Cl
37 35.5
The effective relative atomic mass of an element is determined by calculating the weighted mean of the individual relative atomic masses of the isotopes. The abundance of the different isotopes is found from a mass spectrum. The relative molecular mass (Mr) of a substance can also be determined from its mass spectrum. When a sample, X, is introduced into a mass spectrometer the peaks near the maximum value of m/z correspond to molecular ions, X, made up from the various isotopic mixtures. The relative molecular mass of the substance can be calculated from these peaks.
Table 2 Assignment of peaks in the mass spectrum of chlorine
Essential Notes A mass spectrometer can be used to identify elements from the m/z values of their isotopes. Elements that have an extra–terrestrial origin (e.g. those found in meteorites) often have a different ratio of isotopes compared with the same element on planet Earth.
Notes Isotopic masses are quoted as integer values but these are not the exact mass values.
3.1.1.3 Electron configuration Electron arrangement in atoms and ions It was originally considered that the maximum number of electrons that could be accommodated in the outside layer of an atom was eight and that these electrons occupied a circular or spherical orbit. The elements from helium to krypton were thought to be inert because their outer orbits were full. This model of atomic structure has been replaced by one that is able to account for observed facts, such as the reactions of fluorine with these gases which are now referred to as ‘noble’ rather than ‘inert’. In this later model, the electrons in atoms are arranged into main (or principal) energy levels, which are numbered. Level 1 contains electrons which are closest to the nucleus. Within levels there are sublevels designated s, p, d, f. The maximum number of sub-levels is different for each level and is shown in Table 3.
Essential Notes For definitions of the terms Ar and Mr see this book, section 3.1.2.1.
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Each sub-level consists of orbitals. Each orbital can hold a maximum of two electrons which have opposite spin. The number of orbitals and the maximum number of electrons which can be accommodated in each sub-level are shown in Table 4. Table 3 Types of sub-level in each electron energy level (shell) Table 4 Number of orbitals and maximum number of electrons in each sub-level
Notes Main (principal) energy levels for electrons are sometimes referred to as shells.
Main (principal) level
1
2
3
4
Sub-levels in that level
s
s, p
s, p, d
s, p, d, f
p
d
f
Number of orbitals in sub-level 1
3
5
7
Maximum number of electrons 2
6
10
14
Sub-level
Owing to differences in shielding from the nucleus, different sub-levels within a level have slightly different energies. A typical energy-level diagram for an atom is shown in Fig 3.
Fig 3 A typical energy-level diagram
Essential Notes If electrons are regarded as clouds rather than as particles, the electron cloud has a characteristic shape for each type of orbital.
Essential Notes An s orbital is an electron cloud with spherical symmetry.
The diagrams of p orbitals are shown superimposed on 3 dimensional x, y, z axes. Each p orbital has twin lobes.
s
level 4 level 3
— (4s) — (3s)
level 2
— (2s)
level 1
— (1s)
— — — (4p) — — — (3p)
— — — — — (3d)
— — — (2p)
each orbital can hold up to 2 electrons
Level 2 sub-levels all lie below level 3 sub-levels but, in atoms, sub-level 3d is higher in energy than sub-level 4s. The electron configurations of atoms of elements can be deduced from this diagram. Typical configurations are: Li
1s22s1
F
1s22s22p5
l l
Fe 1s22s22p63s23p64s23d6
l
The electron configurations of ions can be deduced from the configuration of the neutral atom by adding or removing electrons. A complication is that, for transition-metal ions, the 3d sub-level is lower in energy than the 4s, so that the 4s electrons are removed first: Li 1s2
l
F – 1s22s22p6
l
Fe3 1s22s22p63s23p63d5
l
px orbital
py orbital
First ionisation energy and electron arrangements Definition
pz orbital (d and f orbitals have more complicated shapes.)
The first ionisation energy of an element is defined as the enthalpy change for the removal of one mole of electrons from one mole of atoms of the element in the gas phase: X(g) → X(g) e–
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First ionisation energy/kJ mol
−1
The first ionisation energies of the Group 2 metals (Be–Ba) vary as shown in Fig 4. Fig 4 First ionisation energies of the Group 2 elements
900 800 700 600
Notes
500
Electron configurations are sometimes abbreviated by giving only the electrons beyond the previous noble gas. For example
400 300 200 100
Fe
0 Be
Mg
Ca Group 2 elements
Sr
3
Fe
Ba
There is a successive decrease in first ionisation energy from beryllium to barium. Magnesium has a lower first ionisation energy than beryllium because its outer electron is in a 3s sub-level rather than a 2s sub-level. The 3s sub-level is higher in energy than the 2s sub-level. The 3s electron is further from the nucleus and is more shielded from the nucleus by inner electrons. Thus, the 3s electron is more easily removed. This trend in ionisation energies is evidence for the electrons of atoms being organised in levels. A similar decrease in ionisation energy occurs down each group in the Periodic Table.
First ionisation energy/kJ mol
−1
The first ionisation energies of the elements from neon to potassium vary as shown in Fig 5.
[Ar] 4s23d6 [Ar] 3d5
For neutral atoms the 3d sub-level is higher in energy than the 4s. For ions, this is no longer true. The 3d sub-level becomes lower in energy than the 4s.
Notes Note that the decrease in ionisation energy down the group is small in comparison with the absolute magnitude of the ionisation energies.
2500 2000 1500 Fig 5 First ionisation energies of the elements neon to potassium
1000 500 0 Ne
Notes Na
Mg Al Si P S Cl Elements from neon to potassium
Ar
K
There is a general increase in ionisation energy across Period 3 (sodium to argon). Across the period from Na (11 protons) to Ar (18 protons) the nuclear charge in each element increases. As a result, the electrons are attracted more strongly to the nucleus and it takes more energy to remove one from the atom. There is a fall in ionisation energy from magnesium to aluminium because the outer electron in Al (configuration 1s 22s 22p63s 23p1) is in a p sub-level. The p sub-level electron is higher in energy than the outer electron in Mg (1s 22s 22p63s 2) which is in an s sub-level.
The first ionisation energy of neon is greater than that of sodium because the outermost electron in sodium is in a main level which is further from the nucleus and more shielded. This big difference in ionisation energy from neon to sodium is strong evidence for the existence of main (principal) electron energy levels.
The fall in ionisation energy from phosphorus to sulfur can be explained by considering their electronic arrangements (see Fig 6). 11
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Essential Notes Electrons have a property called spin. A spinning electron can be represented by an arrow. Electrons can only spin in one of two directions and are shown by an up or a down arrow, i.e. ↑ or ↓. The arrows represent the magnetic field produced by the spinning electron.
The 3p electrons in phosphorus are unpaired. If there are several empty sublevels all of the same energy, electrons will organise themselves so that they remain unpaired and occupy as many sub-levels as possible. In sulfur the fourth 3p electron is paired. There is some repulsion between paired electrons in the same sub-level, which increases their energy. Therefore it is easier to remove one of these paired 3p electrons from sulfur than it is to remove an unpaired 3p electron from phosphorus. phosphorus:
This variation across Period 3 is regarded as evidence for the existence of electronic sub-levels.
sulfur:
(3p) (3s)
(2p)
Fig 6 Energy levels for phosphorus and sulfur
Notes
(3p) (3s)
(2p)
(2s)
(2s)
(1s)
(1s)
Subsequent ionisation energies and their relationship to electron shells The second and third ionisation energies of an element X are the enthalpy changes for the reactions: X(g) → X2(g) e−
Definition
X(g) → X2(g) e–
Fig 7 Successive ionisation energies for aluminium
1
Successive ionisation energies can provide a very useful guide to the number of electrons in the outside shell (electron energy level) of an element. For example, the successive ionisation energies for aluminium vary as Fig 7 shows. Successive ionisation energy/kJ mol
Second ionisation energy is the enthalpy change for the removal of one mole of electrons from one mole of unipositive ions in the gas phase:
X2(g) → X3(g) e−
25000 20000 15000 10000 5000 0
1
2
3 4 5 Number of electrons removed
6
7
There is a big jump after removal of the third electron because the next electron must be removed from an inner shell. The graph in Fig 7 shows that aluminium has three electrons in its outside shell and is therefore in Group 3.
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Successive ionisation energy/kJ mol
1
Example In which group in the Periodic Table is this element to be found? Answer The large jump after the fifth electron shows that this element is in Group 5.
1 2 3 4 5 6 7 Number of electrons removed
3.1.2 Amount of substance
3.1.2.1 Relative atomic mass and relative molecular mass The isotope 12C is the standard for relative mass: X (g) → X (g) from e phosphorus. unpaired 3p electron remove paired 3p electrons from sulfur than it is to remove an X (g)one → of X these (g) e in the same sub-level, which increases their energy. Therefore it is easier to changes the reactions: fourth 3pfor electron is paired. There is some repulsion between paired electrons The second and third energies of an element X are In thesulfur enthalpy remain unpaired and ionisation occupy as many sub-levels as possible. the to electron shells levels all of the same energy, electrons will organise themselves so that they Subsequent ionisation energies and their relationship The 3p electrons in phosphorus are unpaired. If there are several empty sub-
relative atomic mass (Ar)
average mass per atom of an element 1 12
mass of one atom of 12C
average mass of a molecule
relative molecular mass (Mr)
1 12
mass of one atom of 12C
For compounds that are not molecules, relative formula mass is used: relative formula mass (Mr)
Notes The average mass must be used to allow for the presence of isotopes.
average mass of an ‘entity’ 1 12
mass of one atom of 12C
3.1.2.2 The mole and the Avogadro constant The Avogadro constant is a quantity and is given the symbol L: L 6.022 1023 mol−1 The name mole is given to the amount of substance: 1 mol of particles/entities is 6.022 1023 particles/entities For example: 1 mol of lithium atoms has a mass of 1.152 10−23 g 1 mol of lithium atoms contains 6.022 10−23 lithium atoms therefore, 1 mol of lithium atoms has a total mass of (1.152 10−23) 6.022 1023 6.937 g 1 mol of 12C has a mass of precisely 12.000 g because 12C is the standard.
Essential Notes An ‘entity’ is a ‘formula unit’.
Essential Notes The symbol for the unit of the mole is mol. The SI units of the Avogadro constant are mol−1. It is often helpful to think of this as the number of particles that make up one mole of these particles.
Essential Notes The mass of one atom of an element is very tiny. The mass of L atoms of an element is a recognisable number of grams. 13
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1 mol of CH4 molecules has a mass of approximately 16.0 g. It is calculated by adding up the individual values of the relative atomic masses:
Notes
1 C 12.0
This degree of accuracy – to one decimal place – is sufficient for most chemical purposes.
4 H 4 1.0 4.0 total mass 12.0 4.0 16.0 This mass is approximate for two reasons: the relative atomic mass of 11H is 1.0078
l
carbon and hydrogen occur naturally as isotopic mixtures.
l
Different isotopes of carbon and hydrogen, such as 13C and 2H, occur naturally, so a few of the methane molecules have a mass which is greater than the mass of a 12C1H4 molecule. A chemical equation usually implies quantities in moles. For example: 2O2 → CO2 2H2O CH4 1 mol 2 mol 1 mol 2 mol of methane of oxygen of carbon of water molecules molecules dioxide molecules molecules The mole can be applied to electrons, atoms, molecules, ions, formulas and equations. The concentration of a solution is a quantitative expression with units of mol dm−3: amount in moles of solute concentration volume of solution in dm3
Notes Problems with units? Think of a metre rule:
A 1.0 mol dm−3 solution contains 1 mol of a substance which has been dissolved in enough water to make 1 dm3 of solution. For example:
1 m 10 dm 100 cm 1 m3 103 dm3 (100)3 cm3
106 cm3
a 1.0 mol dm−3 solution of Na2SO4 contains: (23.0 2) 32.0 (16.0 4) 142.0 g of Na2SO4 in 1 dm3 of solution Three of the most useful methods for calculating the amount in moles of a substance are as follows: 1 For a known mass of substance: mass mass mass in g amount in moles Mr mass of 1 mol Mr expressed in g 2 For solutes in a solution, if the volume of the solution is known: amount in moles of solute volume of solution in dm3 concentration
Essential Notes This equation is discussed in more depth in the next section.
volume of solution in cm3 1000
concentration
3 For gases: pV amount in moles RT
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3.1.2.3 The ideal gas equation An ideal gas obeys the assumptions of the kinetic theory of gases. According to this theory, ideal gas particles (molecules or free atoms) are treated as hard spheres of negligible size which move with rapid random motion and experience no intermolecular forces. The ideal gas equation is: pV nRT p is the pressure of the gas in Pa V is the volume of the gas in m3
Essential Notes One pascal (Pa) is one newton per square metre (N m−2)
n is the amount in moles of gaseous particles
Essential Notes
R is the gas constant (8.31 J K1 mol1)
100 kPa 100 000 Pa 200 cm3 200 10–6 m3 25°C 298 K
T is the temperature in kelvin (add 273 to the temperature in °C) This equation can be used to find the amount in moles (n) of a gaseous substance. For example, in 200 cm3 of CH4 at 25 °C and 100 kPa the amount in moles of methane is: pV 100 000 200 10−6 n 0.00808 mol RT 8.31 298 8.08 10−3 mol (to three significant figures) If the mass and the amount in moles of a sample are known, it is possible to calculate the relative molecular mass (Mr). The mass of the sample of methane above is 0.129 g. Hence: Mr
mass 0.129 16.0 amount in moles 0.00808
3.1.2.4 Empirical and molecular formula
Notes This is the basis of experiments to determine Mr by measuring the mass of a given volume of gas or vapour at a known temperature and pressure (gas syringe or bulb experiments).
The empirical formula is the formula which represents the simplest ratio of atoms of each element in a compound. The molecular formula gives the actual number of atoms of each element in a molecule (or the amount in moles of each type of atom in 1 mol of the compound).
Calculation of empirical formulas The empirical formula of a compound can be calculated from data which give the percentage composition by mass of each element in the compound.
Calculation of molecular formulas The molecular formula can be deduced from the empirical formula if the relative molecular mass is known. A value for Mr can be determined as shown using the ideal gas equation or from a mass spectrum. If a compound with empirical formula CH2O has Mr 180, the molecular formula can be calculated as shown at the end of the example on the following page.
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AQA Chemistry AS/A-level Year 1
Example A compound containing carbon, hydrogen and oxygen gave, after elemental analysis, the following percentages by mass: C 40% and H 6.7%. The percentage of oxygen is often calculated by difference. In this case, the percentage of oxygen 100 (40 6.7) 53.3%. The empirical formula can be calculated as follows. Assume that there are 100 g of the compound, then the masses of the elements are: C 40 g; H 6.7 g; O 53.3 g The amount in moles of each element is calculated as follows: carbon:
mass 40 6.7 53.3 3.3 hydrogen: 6.7 oxygen: 3.3 Ar 12.0 1.0 16.0
These amounts in moles can be expressed as a simple ratio by dividing through by the smallest number: ratio of moles of C : H : O 3.3 : 6.7 : 3.3
Notes The molecular formula is always a whole number times the empirical formula.
3.3 6.7 3.3 : : 3.3 3.3 3.3 1:2:1 Therefore the empirical formula is CH2O The empirical formula mass of CH2O is 12.0 2.0 16.0 30.0 The ratio of Mr : empirical formula mass 180 : 30.0 6 : 1 Therefore, in comparison with the empirical formula, the molecular formula must contain 6 times the number of atoms. Therefore the molecular formula is 6 CH2O C6H12O6
Essential Notes For ionic equations the charges on the ions must also balance. Thus Fe3 Zn → Fe Zn2 is not a balanced equation but 2Fe3 3Zn → 2Fe 3Zn2 is balanced (6 positve charges on each side).
3.1.2.5 Balanced equations and associated calculations A full equation is a balanced symbol equation with the formulas of all reagents on the left-hand side and the formulas of all the products on the right-hand side. An ionic equation is a simplified version of a balanced symbol equation, showing only the ions which are actively involved in the reaction. Ions which do not take part in the reaction (spectator ions) are eliminated. The resulting ionic equation is no longer specific to a single reaction but is now a generalised expression of the essential chemistry. State symbols are often included in both full and ionic equations. These are letters, in brackets, which are added after a formula to indicate the state the substances are in: (s) solid (l) liquid (g) gas (aq) aqueous solution/dissolved in water
Balancing equations 16
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Balanced equations must have the same number of atoms of each element on the left-hand side and on the right-hand side of the ‘arrow’.
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Physical chemistry
To balance equations, work through these steps. 1 Write the equation, then pick one element and see if the number of atoms of that element is equal on both sides of the arrow. 2 If the equation needs balancing, write the necessary number in front of the appropriate formula or symbol to make that element balance. 3 Move on to each new element and balance it in turn. 4 Check for fractions and multiply them out. Example Consider the reaction between sulfuric acid and sodium hydroxide. The balanced symbol equation is: H2SO4(aq) 2NaOH(aq) → Na2SO4(aq) 2H2O(l) Showing separate ions this becomes: 2H(aq) SO42−(aq) 2Na(aq) 2OH−(aq) → 2Na(aq) SO42−(aq) 2H2O(l) Water is covalent, so is not shown as ions. Sulfate ions and sodium ions, the spectator ions, appear on both sides so are cancelled out to leave: 2H(aq) 2OH−(aq) → 2H2O(l) This simplifies to give: H(aq) OH−(aq) → H2O(l) This ionic equation now represents the reaction of any acid with any hydroxide in solution. Example Consider the unbalanced equation: Al NaOH → Na3AlO3 H2 Taking each element in turn: Al There is one atom (or 1 mol of atoms) on each side, so Al balances. Na There is one Na on the left and three on the right – the equation is unbalanced.
Therefore use 3NaOH and the equation is now Al 3NaOH → Na3AlO3 H2
O There are now three Os on each side, so O balances. H There are three Hs on the left and two on the right. Using 3 H on the right-hand side balances the equation: 2 2 Al 3NaOH → Na3AlO3
3 H 2 2
This equation is balanced, but it is better multiplied by 2 to avoid the fraction 32 : 2Al 6NaOH → 2Na3AlO3 3H2
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AQA Chemistry AS/A-level Year 1
Calculating reacting masses and reacting volumes of gases This skill is again best learned from examples.
Notes In calculations of this type the answer is usually expressed to three significant figures. It may be necessary to carry more precise numbers through the calculation, but the answer should be rounded.
Example Consider the following equation: 2HCl Na2SO3 → 2NaCl H2O SO2 If 1.00 g of Na2SO3 is reacted with an excess of HCl, calculate: (i) the mass of NaCl produced by complete reaction (ii) the volume of SO2 gas produced at 25°C and 100 kPa pressure. Answer Calculations like this almost always involve, as an intermediate step, working out amounts in moles. The answers can be deduced as follows: (i) Mr for Na2SO3 is (2 23.0) 32.0 (3 16.0) 126.0
moles of Na2SO3 used
mass 1.00 0.00794 Mr 126.0
From the equation, moles of NaCl 2 moles of Na2SO3 2 0.00794 0.0159 Mr for NaCl 23.0 35.5 58.5 mass of NaCl moles Mr 0.0159 58.5 0.929 g
Notes Remember 1 m3 1 106 cm3
(ii) From the equation, moles of SO2 moles of Na2SO3 0.00794 nRT 0.00794 8.31 298 volume V 1.97 10−4 m3 197 cm3 p 100 000
1 cm3 1 10−6 m3
Calculating concentrations and volumes of aqueous reagents Work through this example of a typical problem.
Notes The general way to progress through calculations like this is •c alculate the moles of known substance •u se the equation for the reaction and the moles of known substance to state the moles of unknown substance • proceed to the answer.
Example 25.0 cm3 of 0.102 mol dm−3 NaOH are exactly neutralised by a solution of 0.0830 mol dm−3 H2SO4: H2SO4(aq) 2NaOH(aq) → Na2SO4(aq) 2H2O(l) Calculate: (i) the volume of sulfuric acid required for the neutralisation (ii) the concentration of sodium sulfate in the resulting solution. Answer (i) moles of NaOH volume (in dm3) concentration volume (in cm3) concentration 1000 25.0 0.102 1000
0.00255 mol
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Physical chemistry
Notes
From the equation: moles of H2SO4
1 2
Remember
moles of NaOH
concentration
0 .00255 2
Also 1 dm3 1 103 cm3 1 cm3 1 10−3 dm3
0.001275 mol
amount in moles of H2SO4 volume in dm3 concentration
moles of solute volume of solution in dm3
amount in moles therefore volume of H2SO4(aq) concentration 0.001275 0.0830
0.0154 dm
15.4 cm3
Notes 3
(ii) moles of Na2SO4 produced moles of H2SO4 used 0.001275 mol Ignoring the small amount of water produced in the reaction:
The making up of a volumetric solution and the carrying out of an acid– base titration is a required practical activity.
total volume of final solution 25.0 15.4 40.4 cm3 40.4 10−3 dm3 amount in moles concentration of Na2SO4(aq) volume in dm3 0.001275 0.0316 mol dm−3 40.4 10−3
Percentage yield Percentage yield is a practical measure of the efficiency of a reaction. It takes into account reactions that do not go to completion. It can only be calculated from experimental data.
percentage yield
actual mass of product maximum theoretical mass of product
100
Example Consider the following equation for the production of dichloromethane (CH2Cl2): CH4 2Cl2 → CH2Cl2 2HCl In an experiment, 21.3 g of CH2Cl2 were produced when 8.00 g of methane were reacted with an excess of chlorine. 8.00 amount in moles of methane 0.500 16.0 Maximum amount in moles of CH2Cl2 that can be formed from 0.500 mol of CH4 0.500 mol Maximum mass of CH2Cl2 that can be formed amount in moles Mr 0.500 85.0 42.5 g
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AQA Chemistry AS/A-level Year 1
Actual mass of CH2Cl2 formed 21.3 g
yield
actual mass of CH2Cl2 100 maximum theoretical mass of product
21.3 100 50.1% 42.5
This answer suggests that the reaction did not go to completion or that some of the methane was converted into by-products (or a combination of the two).
Percentage atom economy The percentage atom economy is a measure of how much of a desired product in a reaction is formed from the reactants. It is a theoretical quantity calculated from a balanced equation. mass of desired product percentage atom economy 100 total mass of reactants Example Consider the following equation for the production of dichloromethane (CH2Cl2): CH4 2Cl2 → CH2Cl2 2HCl mass of one mole of CH2Cl2 100 percentage atom economy (mass of one mole of CH4 mass of two moles of chlorine)
85.0 100 53.8% (16.0 142.0)
This answer shows that (100 – 53.8) 46.2% of the mass of reactants is converted into a co-product other than the desired product.
Economic, ethical and environmental advantages of high atom economy Unlike percentage yield improvement, which focuses only on the amount of product formed in a reaction, high atom economy, developed from the principles of ‘green’ or sustainable chemistry, focuses on economic, ethical and environmental issues. Adoption of high atom economy processes reduces waste products and so reduces both the cost of hazardous waste treatment and potential damage caused by its release into the environment. In addition, high atom economy synthetic routes are less wasteful of natural resources. If the quest for higher atom economy processes also incorporates the use of less toxic starting materials and safer solvents, greater use of renewable resources, the development of more efficient catalysts, and less energy usage, there are further indirect advantages. One example of a successful increase in atom economy is the synthesis of Ibuprofen. The original manufacturing process had an atom economy of 40% but switching to an alternative synthetic route has increased this to 77%. 20
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Inorganic chemistry
NaX
Observations
Products
Type of reaction
NaF
steamy fumes
HF
acid–base (F2 acting as a base)
NaCl
steamy fumes
HCl
acid–base (Cl2 acting as a base)
NaBr
steamy fumes
HBr
acid–base (Br2 acting as a base)
colourless gas
SO2
redox (reduction product of H2SO4)
brown fumes
Br2
redox (oxidation product of Br2)
NaI
steamy fumes
HI
acid–base (I2 acting as a base)
colourless gas
SO2
redox (reduction product of H2SO4)
yellow solid
S
redox (reduction product of H2SO4)
smell of bad eggs
H2S
redox (reduction product of H2SO4) redox (oxidation product of I2)
black solid, purple I2 fumes
These results indicate that: iodide ions can reduce the sulfur in H2SO4 from oxidation state 16 to 14, as SO2, then to 0, as the element sulfur, and finally to 22, as H2S
l
bromide ions can reduce the sulfur in H2SO4 from oxidation state 16 to 14, as SO2
l
Table 23 The reactions of concentrated sulfuric acid with solid sodium halides
Essential Notes These reactions can be demonstrated in a laboratory fume cupboard but full safety precautions must be taken. Hydrogen fluoride is an extremely dangerous gas and, in the presence of water, will even etch glass.
Notes Deriving equations for the reactions which occur provides valuable revision of redox reactions.
fluoride and chloride cannot reduce the sulfur in H2SO4 under these conditions.
l
The use of silver nitrate solution to identify and distinguish between halide ions Silver fluoride is soluble in water but silver chloride, silver bromide and silver iodide are all insoluble. Silver chloride, bromide and iodide are precipitated when an aqueous solution containing the appropriate halide ion is treated with an aqueous solution of silver nitrate. Dilute nitric acid is added to the solution under test before addition of silver nitrate solution to prevent the formation of other insoluble compounds, such as Ag2CO3. The colours of the three silver salts formed with chloride, bromide and iodide ions, and their different solubilities in aqueous ammonia, can be used as a test for the presence of the halide. These results are summarised in Table 24. Halide
Precipitate
Observation Solubility of precipitate in ammonia solution
F2 none
–
AgCl
white solid
2
Br
AgBr
cream solid sparingly soluble in dilute NH3(aq), soluble in concentrated NH3(aq)
I2
AgI
yellow solid insoluble in concentrated NH3(aq)
soluble in dilute NH3(aq)
These results show that the solubility of the silver halides in ammonia solution decreases in the following order: AgCl . AgBr . AgI
Table 24 Testing for halide ions using AgNO3(aq) and NH3(aq)
–
2
Cl
Notes If you are asked to describe an observation, you must always link a colour to a solution or a solid (precipitate).
Notes The carrying out of simple test-tube reactions to identify halide ions is a required practical activity. 63
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The addition of ammonia solution to the silver halide precipitate formed is used to eliminate any potential confusion caused by the similar colours of these precipitates. The different solubilities of AgCl, AgBr and AgI in ammonia lead to a clear identification of the halide ion originally present.
3.2.3.2 Uses of chlorine and chlorate(I) The products obtained when chlorine reacts with water depend on the conditions used. Under normal laboratory conditions, a very pale green solution is formed, showing the presence of the element chlorine, and an equilibrium is established: Cl2 1 H2O
HCl 1 HClO
This reaction is an example of a disproportionation reaction in which one species, in this case chlorine, is simultaneously both oxidised and reduced: Oxidation state of chlorine:
0 21 11
Cl2 1 H2O
Essential Notes Because this reaction is rather slow, it is best to leave an inverted test tube containing chlorine water in sunlight for several days, after which sufficient oxygen will have been produced to give a positive test with a glowing splint.
HCl 1 HClO
If universal indicator is added to a solution of chlorine water, it first turns red since both the reaction products are acids, i.e. hydrochloric acid, HCl, which is a strong (fully ionised) acid, and chloric(I) acid, HClO, which is a weak (slightly ionised) acid. The red colour then disappears and a colourless solution is left because chloric(I) acid is a very effective bleach. If chlorine is bubbled through water in the presence of bright sunlight, or the green solution of chlorine water is left in bright sunlight, a colourless gas is produced and the green colour, due to chlorine, fades. Tests show that the colourless gas evolved is oxygen. Under these conditions, chlorine oxidises water to oxygen and is itself reduced to chloride ions: 2Cl2 1 2H2O → 4H1 1 4Cl2 1 O2
Water treatment Chlorine and chlorine compounds are used in water treatment. For many years, small quantities of chlorine have been added to drinking water and to swimming pools in order to kill disease-causing bacteria. In drinking water, the major public health hazards are due to bacteria that cause cholera and typhus. In swimming pools, the dangerous bacteria killed by chlorine are often types of E. coli that originate from human waste. The decision about the amount of chlorine to be added to drinking water supplies is a good illustration of how society has to assess the advantages and disadvantages of using chemicals to sterilise water supplies. Too little chlorine is ineffective in killing bacteria, but too much damages the health of consumers. Clearly, the health benefits of carefully controlled chlorination of drinking water outweigh the potential toxic effects. More recently, fluoridation of the water supply to improve dental health has required similar assessment of the advantages and disadvantages. The concentration of chlorine in drinking water is approximately 0.7 mg dm–3. Higher concentrations are used in swimming pools. Great care is taken to ensure that the correct amounts of chlorine are used because chlorine itself is very toxic. In addition, chlorine can react with organic waste material in water to form organochlorine compounds which may be toxic. Although it is well known that failure to chlorinate water results in serious health risks, there is little information to support the claim that the formation of organochlorine compounds in water is a long-term health risk. 64
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Inorganic chemistry
Reaction of chlorine with cold dilute aqueous sodium hydroxide When chlorine reacts with cold water, an equilibrium is established between the reactants and the two acidic products: Cl2 1 H2O
HCl 1 HClO
If water is replaced by cold dilute sodium hydroxide, the effect is to displace the equilibrium to the right as the hydroxide ions react with the acids produced: Cl2 1 2NaOH → NaCl 1 NaClO 1 H2O or Cl2 1 2OH2 → Cl2 1 ClO2 1 H2O This reaction is of great commercial importance because the mixture of sodium chloride and sodium chlorate(I) is used as a bleach.
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