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CHEMISTRY
FOR TECHNICIANS
BARKER FULLICK KRAJNIAK
Publishing Editor: David Barker
Cover design: Harcourt Brace Australia
1st edition copyright ©1996
by Harcourt Brace & Company, Australia
This edition copyright ©2003
by David Barker, Graeme Fullick & Edward Krajniak
Hunter Institute
Locked Bag 45
Hunter Region Mail Centre 2310
This publication is copyright. Except as expressly provided in
the Copyright Act 1968, no part of this publication may be
reproduced by any means (including electronic, mechanical,
microcopying, photocopying, recording or otherwise) without
prior written permission from the authors.
National Library of Australia Cataloguing-in-Publication
Data
Barker, David J.
Chemistry for technicians
Bibliography
Includes index.
ISBN 0 7295 3291 7.
1. Chemistry
I. Fullick, G.
II. Krajniak, Edward Richard.
III. Title
540
Printed in Australia by WHO Presentation Services 02 4969 4044
Distributed by Newcastle College Bookshop 02 4923 7389
ii
CONTENTS
How to use this book
vii
Chapter 1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
What Is Chemistry?
Chemistry and science
Classification systems for matter
Classification by properties
Classification by composition
Solutions: The most important mixture
Physical and chemical changes
Elements
What you need to be able to do
Terms and definitions
1
1
2
2
5
10
11
13
14
15
Chapter 2
2.1
2.2
2.3
2.4
2.5
2.6
2.7
Atoms: The Building Blocks of Matter
Evolution of the atomic concept
The solar system model of the atom
How atoms differ
Atomic weights
Can atoms be changed?
The electron shell model
Another look at the formation of ions
What you need to be able to do
Terms and definitions
16
16
17
19
22
24
25
27
28
29
Chapter 3
3.1
3.2
3.3
3.4
Organising the Elements
A brief look at some important elements
Some similarities appear
A first look at the periodic table
The periodic table and electron configurations
What you need to be able to do
Terms and definitions
30
30
34
36
38
38
39
Chapter 4
4.1
4.2
4.3
4.4
4.5
4.6
Names and Formulae
Chemical formulae
Types of compounds
Binary molecular compounds
Ionic compounds
Common acids and bases
Formula weights
What you need to be able to do
Terms and definitions
40
40
42
43
45
49
49
50
51
iii
Chapter 5
5.1
5.2
5.3
5.4
5.5
Properties and Structure
A survey of the properties of pure substances
Ionic substances
Metallic substances
Non-metallic non-ionic substances
What type of bond will form?
What you need to be able to do
Terms and definitions
52
52
56
59
60
67
68
69
Chapter 6
6.1
6.2
6.3
6.4
6.5
Chemical Reactions I
What is a chemical reaction?
How can we tell that a chemical reaction has occurred?
Describing chemical reactions
Common types of reactions
Mass relationships in chemical reactions
What you need to be able to do
Terms and definitions
70
70
71
72
77
79
82
83
Chapter 7
7.1
7.2
7.3
7.4
The Periodic Table
Why the periodic table is still used
Sections of the periodic table
Trends in the main group elements
Trends in other sections of the table
What you need to be able to do
Terms and definitions
84
84
85
86
89
90
91
Chapter 8
8.1
8.2
8.3
8.4
8.5
8.6
Names And Formulae Revisited
Hydrated compounds
Some more ions
Empirical, molecular and structural formulae
What are organic compounds?
Functional groups
An introduction to naming organic compounds
What you need to be able to do
Terms and definitions
92
92
93
94
95
96
97
99
99
Chapter 9
9.1
9.2
9.3
9.4
9.5
9.6
How Atoms Bond
Why atoms bond
Electron dot diagrams
Formation of ionic bonds
Formation of covalent bonds
Electronegativity
Intermolecular forces
What you need to be able to do
Terms and definitions
iv
100
100
101
101
102
104
106
110
110
Chapter 10
10.1
10.2
10.3
10.4
Preparing Solutions
Concentration units
How much solute do I need?
Dilution of solutions
Interconversion of concentration units
What you need to be able to do
Terms and definitions
111
111
113
116
120
121
122
Chapter 11
11.1
11.2
11.3
11.4
11.5
11.6
Chemical Reactions II
Ionic equations
Reactions of acids
An introduction to titrations
The equilibrium process
Factors affecting equilibrium
Industrial chemistry: Application of chemical equilibria
What you need to be able to do
Terms and definitions
123
123
124
126
128
131
133
134
135
Glossary
136
Answers to Practice Questions
140
Index
150
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vi
HOW TO USE THIS BOOK
FOR ALL READERS
You might wonder why yet another chemistry textbook has been released onto a very
crowded market. You might say that little in the basic principles of chemistry has
changed in the past 50 years. Can there really be a new way of expressing the same ideas
that have been expressed in thousands of chemistry books in that time?
The answer is, of course, no, though a qualified no. This book has been written with
a very specific purpose in mind: to meet the requirements of a wide range of disciplines
in vocational science courses: chemical analysis, materials testing, biology and
pathology, health sciences, food technology and geology among them.
This book assumes no previous knowledge of chemistry assumed and does not
extend beyond the boundaries of the basic chemical knowledge required across these
various areas. This makes it different to other texts in the area, which assume previous
knowledge or cover much more ground.
FOR STUDENTS
It is hoped that you will find this book easier to read, since it has been the authors’
intention to use a style that is conversational where possible. The book attempts to
explain as many of the basic concepts as possible in simple terms, with relevant
examples, but limitations of space do not allow it to be a ‘teach-yourself’ manual in the
true sense of the word. It is hoped that you will understand much of these basic chemical
principles, but nothing replaces the assistance of someone who has studied the area in
some detail, that is, your teacher or mentor.
Each chapter follows a consistent pattern, which is described below.
Purpose
A simple statement of the content of the chapter, which reads like an aim in a high school
science experiment.
There is no assumed prior knowledge at all. However, some of the later chapters require
that you have done earlier ones: at the beginning of Chapters 7, 8, 9 and 11, there is a
recommendation that you revise the related earlier chapters.
This symbol indicates that there is an important term (or terms) highlighted
in bold type in the paragraph. Also in the paragraph which will be the
meaning of that term (its definition) which be in italics. These terms are
also found in the Glossary (Appendix 1) at the end of the book.
EXAMPLE
Fully worked and explained questions which demonstrate the principles discussed in the
paragraph(s) above.
vii
The PRACTICAL WORK is intended to make the theory easier to understand.
The exercises are relatively simple, and should work successfully, but the stepby-step instructions are not, in some cases, detailed enough to allow a learner to
correctly perform the task. This is deliberate.
If you are using this book to teach yourself something about the basics of
chemistry, you should not attempt to perform any of the practical work
described alone. Chemicals are potentially hazardous, and should be used only
under the supervision of someone who knows about their correct use.
The LIBRARY WORK is intended to extend some of the basics in the book and
to allow you to discover for yourself, aspects that space does not allow us to
provide. Your teacher may use the topics for assignments.
PRACTICE QUESTIONS
You are not expected to understand all the theory without some practice, so most
sections will have practice questions to help you. The answers to all questions are
found in Appendix 2 at the end of the book.
WHAT YOU NEED TO BE ABLE TO DO
•
A list of the key parts of the chapter that you must be able to do before
attempting assessment tasks.
TERMS AND DEFINITIONS
Because many of the basic skills of chemistry revolve around learning the jargon, this
section helps reinforce the meaning of the important terms highlighted throughout the
chapter by the
symbol.
FOR TEACHERS
The authors recognise that there are many ways of skinning this particular cat — the
delivery of the fundamentals of chemical theory — and that the order of material as
presented in this book may not fit your vision. If you have decided to recommend this
book to your students, then some jumping around in the sequence may be necessary to
match your preferred approach. We hope, however, that the material is presented in a
suitable way, and meets the requirements for students and teachers in all related areas of
laboratory work.
Limitations of space and cost do not allow glossy illustrations or comprehensive
examples across the range of disciplines which are addressed by courses that incorporate
these modules. It is, therefore, necessary that you, as the teacher, provide the
contextualisation to meet the needs of your students, be they technicians in a chemistry,
pathology, metallurgical or food laboratory.
In the same way, the practical work and library work are suggestions about possible
exercises that could be employed to assist learners.
However, they are not
comprehensive, nor prescriptive, and it is anticipated that teachers will expand the areas
that suit their classes.
viii
The practical work serves to illustrate the theory, and if this book is being used in a
classroom environment, it is anticipated that the practical work would be done
immediately following the related theory. We believe this to be a better arrangement
than large exercises at the end of a whole topic. The instructions are not exhaustive, and
inexperienced learners will require assistance in undertaking the exercises.
The library work presents some suggestions for assignment work, but is by no
means comprehensive.
ACKNOWLEDGMENTS
I would like to thank my colleagues, Graeme Fullick and Ed Krajniak, for their advice,
expertise and friendship. My wife, Elaine, has my heartfelt thanks for her support,
encouragement and, above all, patience. Thanks to my cats for keeping me company
during long hours spent at the computer, and their occasional additions to the text when
they stepped on the keyboard. Finally, thanks to Beethoven and Bach for keeping me
sane through it all.
David Barker
Principal author
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x
1
WHAT IS CHEMISTRY?
PURPOSE
To discuss the purposes of chemistry, and to introduce the ways that chemistry seeks to
explain the behaviour of matter.
1.1
CHEMISTRY AND SCIENCE
Science is basically the attempt to explain the behaviour of everything that
we see, smell, hear etc. The different areas of science — physics,
physics, chemistry, biology, geology and others — look at different parts of
different parts of this ‘everything’. Chemistry deals with the properties of
properties of the materials (matter) that make up the Earth, the living
organisms on it, the atmosphere we live in, and the universe around us.
Chemistry crosses many boundaries of science: it is required as background knowledge
for the food sciences, geology, materials technology, geography, biology and other health
sciences, environmental science, and the list goes on.
Within chemistry, there are a range of speciality areas (including analytical,
physical, organic and inorganic), and a range of reasons for studying it: a simple background from school science is important for general knowledge and day-to-day life, while
senior high school chemistry leads into further education areas involving chemistry;
university chemistry relies a lot on theory and research; and finally technician-level
chemistry uses a practical approach to chemistry to help explain the reasons behind the
testing of materials of whatever nature. This book is aimed at the third type of chemistry:
it will give you the basic language and ideas of chemistry in an applied and practical
way. Figure 1.1 shows some of these relationships.
Food
Science
SCIENCE
Environmental
Science
Biology
CHEMISTRY
• technician
• university
• school
Geology
Medicine
Health
Science
Figure 1.1 The importance of chemistry
Geography
Material Science
Chemistry for Technicians
By studying the properties of ‘matter’, chemists are able to develop new materials, make
predictions on the behaviour of this material, monitor and modify the behaviour of
existing materials, among many other activities. Consider one typical day. Every single
activity will involve us using matter; matter which has been tested, developed or studied
by a chemist somewhere: food, air, water, petrol, chairs, cars, basketballs, every single
thing.
Certainly, the work of chemists is not all positive: new materials developed by
chemists, such as DDT, CFCs and the lead added to petrol, have proven to be significant
hazards to life after many years of apparent benefit. Now chemists study ways of
measuring, controlling and destroying these materials.
How do chemists study the properties of matter? It is done by observations and
measurements, by classification of that matter into groupings of like materials, and by
prediction on the basis of those classifications.
PRACTICE QUESTION
1. Consider the process of making a cup of coffee. List the ways in which chemists
have been involved in allowing you to drink that coffee safely and with pleasure.
1.2
CLASSIFICATION SYSTEMS FOR MATTER
The number of different types of matter is almost infinite; to study all of them in fine
detail would be almost impossible, and a waste of time. Chemists classify matter by its
behaviour and properties, and in doing so, simplify the process of understanding and
prediction. If you read a description of matter which indicates that it is a solid, nonmetallic molecular compound, then (by the end of this textbook at least) you will have a
good idea of its properties in general.
There are many different classification systems used, some linked together, others
separate, some quite general and easily understood by non-scientific people, others very
specialised and of use only to those studying in that particular area. At this point in your
studies in chemistry, we can only look at the general; more specific knowledge will come
in time as you continue to study. Figure 1.2 illustrates the relationships of the
components of matter and how they can be classified; some of the terms will be defined
in the text below, some will already be familiar, and others will have to wait for later
chapters.
EXAMPLES
To help illustrate the classification systems, we will use a number of examples, from a
range of the fields addressed in this book: sugar, blood, stainless steel, sea water and
helium gas.
1.3
CLASSIFICATION BY PROPERTIES
1.3.1 Physical State: Solid/Liquid/Gas
The most easily recognised distinction of all is based on how the substance behaves at
room temperature (25°C). To further classify matter in this general area, the melting
point and boiling point of the material can be measured.
2
Chapter 1
What Is Chemistry?
The melting (or freezing) point is the temperature at which the matter
turns from solid to liquid (or the reverse). Gases and liquids have melting
points below room temperature.
•
•
•
MATTER
Element
solid/liquid/gas
metal/non-metal
acid/base/neutral
Compound
Atom
•
•
•
•
Mixture
•
•
organic
inorganic
ionic
molecular
homogeneous
heterogeneous
Figure 1.2 The classification of matter
The boiling point is the temperature at which a liquid turns into a gas.
Gases have boiling points below room temperature. Table 1.1 summarises
the melting and boiling point ranges for the three phases.
Table 1.1
Melting and boiling points
State of Matter
Melting point
Boiling point
Solid
greater than 25°C
greater than 25°C
Liquid
less than 25°C
greater than 25°C
Gas
less than 25°C
less than 25°C
EXAMPLE
What physical state is a substance with a melting point of 5°C and a boiling point of 178°C
at (a) room temperature (b) 0°C and (c) 200°C?
(a)
(b)
(c)
At room temperature (25°C), the melting point has been exceeded but not the boiling
point, so the substance is liquid.
At 0°C, the substance is below its melting point, and therefore is a solid.
The substance is a gas since its boiling point has been exceeded.
Classify the example substances by their physical state.
Sugar
solid
Blood
liquid
Stainless steel
solid
Sea water
liquid
Helium
gas
3
Chemistry for Technicians
PRACTICAL WORK
1. With the assistance of your teacher, set up a distillation apparatus.
2.
Perform the distillation of samples of pure water, sea water and ethanol.
Record the temperature at 30-second intervals once boiling has
commenced.
Q.
Plot the temperature/time graphs for each liquid. Comment on how they
differ. Determine the boiling point.
PRACTICE QUESTION
2. What are the characteristics of solids, liquids and gases that help you to classify
matter in this way?
1.3.2 Metal/Non-metal
The most recognisable sign of a metal is its shiny appearance, but this is not entirely
reliable, since very fine powdered metals look like soot. More usefully, metals can be
distinguished by their ability, in the solid state, to allow heat and electricity to pass
through them (they are conductors as solids, unlike the vast majority of non-metals), and
their density (metals are generally much heavier than non-metals).
EXAMPLE
Classify the example substances by their metallic nature.
Sugar
non-metal
Blood
non-metal
Stainless steel
metal
Sea water
non-metal
Helium
non-metal
1.3.3 Natural/Man-made
Many substances freely used these days are not available from natural sources, but this
distinction is not at all useful for chemists, because it tells us little or nothing about the
properties of the substance. Many natural substances can be man-made and samples from
each source are absolutely identical.
Furthermore, many substances that we use have been obtained from nature, but
processed (e.g. purified) by man before use. When all is said and done, nothing on Earth
is new when we look at the basic components of matter — atoms (see below): everything
is just being rearranged. Some of the atoms that are today swimming around in our
bloodstrea, were yesterday part of a plant or animal, and before that, part of the air
around us or the soil beneath us.
4
Chapter 1
What Is Chemistry?
EXAMPLE
Classify the example substances by their origin.
Sugar
occurs in this form naturally, but is purified by man
Blood
natural
Stainless steel
man-made
Sea water
natural
Helium
occurs in this form naturally, but is purified by man
1.3.4 Acidic/Basic/Neutral
Classification of matter by these properties can be done by measurements in a laboratory,
or by simple observations which have been known for many centuries. Acids and bases
have tastes that are distinctive: acids are sour (e.g. lemon juice) and bases bitter. Taste is
not, however, a laboratory method for classification of substances!
The behaviour of certain vegetable extracts (e.g. litmus) with acids and bases is
widely used to classify substances. These change colour depending on the nature of the
matter. Litmus is known as an acid-base indicator. Acids turn litmus red, bases (also
known as alkalies) turn it blue and neutral substances do not affect it at all. There are a
number of other indicators used, some of which can distinguish the level of acidity or
basicity. An example is universal indicator, which has at least four colour variations.
PRACTICAL WORK
Measure the acidity of a series of common substances, such as milk, lemon
juice, washing powder, shampoo, soap, salt water. Use either test strips or
universal indicator. Your teacher will explain the meaning of the results.
Q.
On the basis of your results, comment on the statement ‘Few, if any, food
items are alkaline’.
EXAMPLE
Classify the example substances by their acidity.
Sugar
neutral
Blood
slightly alkaline
Stainless steel
neutral
Seawater
slightly alkaline
Helium
neutral
So far, the classifications have been on the basis of the measurable properties of the
matter. The following divisions of matter are determined by the composition of the
matter — the substances that are combined to form the matter. The two systems are
basically independent of each other, for reasons which will be explained after we look at
these classifications.
1.4
CLASSIFICATION BY COMPOSITION
1.4.1 Pure/Mixture
This is one of the most important classifications. A mixture is a substance containing
more than one different type of pure substance. Mixtures can vary in their properties, not
only because different substances are involved, but because of different proportions of
the same substance. However, the substances making up the mixtures will not vary in
their properties at all when in the pure form. For example, pure iron rusts quite rapidly,
while stainless steel, a mixture of about 80% iron, and 20% chromium and nickel (and a
few other metals) rusts very slowly.
5
Chemistry for Technicians
How do we recognise whether matter is pure or a mixture? If it does not
look the same throughout — it is not uniform — then it is certainly a
certainly a mixture, and is called a heterogeneous mixture. An example is
An example is concrete, which has rocks, pebbles, cement and sand
sand combined to form the overall material. Slight differences in the
proportions of each change the physical properties of the concrete.
If the matter is uniform, it can still be a mixture. Sea water, once the
‘floaties’ are removed, is uniform, but is a mixture of water and sodium
and sodium chloride (common salt) and a large range of other substances. It
substances. It is called a homogeneous mixture. If we drink enough sea
enough sea water (which has about 35 grams of salt per litre), we will die of
dehydration. A saline drip solution, used in hospitals to maintain liquid
levels in patients unable to drink and eat, contains nine grams of salt per
litre. Obviously, these two mixtures are quite different in some of their
properties. Figure 1.3 shows the difference between homogeneous and
heterogeneous mixtures.
(a)
(b)
Figure 1.3 Types of mixtures: (a) homogeneous and (b) heterogeneous
Mixtures can be identified by the ease with which their components can be separated. A
mixture of solid and liquid can be separated by filtration (see Figure 1.4(a)). Compounds
dissolved in a liquid can be recovered by boiling off the liquid (evaporation), as in
Figure 1.4(b). The ink in the pen that you are writing with can be separated into its
individual dye components by putting a mark on a piece of paper, and allowing liquid to
pass over it (known as chromatography) (see Figure 1.4(c)). Each of these processes is
relatively simple and requires little energy input. If the separated components are
recombined, we get exactly the same matter as before the separation. By subjecting
matter to one or more of these processes, we can determine whether it is pure.
Mixtures tend to have melting and boiling points that are dependent on the
proportions in the mixture, but also are not a single temperature, but range over 5–10°C.
Pure substances cannot be separated further without changing their form. For
example, water can be broken down by passing electrical current through it, but in the
process hydrogen gas and oxygen gas (not gaseous water which is steam) are formed.
These are not the same as the water — it is definitely not a good idea to try to drink
hydrogen gas!
PRACTICAL WORK
1.
Obtain a small quantity of samples made from (i) mixing iron and sulfur and
(ii) heating iron and sulfur.
2.
Place the samples on separate pieces of filter paper, and examine each
using a magnifying lens. Note whether each is homogeneous or
heterogeneous.
3.
Pass a magnet under each piece of filter paper. Check again with the
magnifying lens.
6
Chapter 1
What Is Chemistry?
4.
Note the results obtained from other demonstrations in the laboratory, such
as paper chromatography of food dyes.
5.
Perform the distillation of a 1:1 mixture of water and ethanol.
Q.
Explain your observations for each test in terms of the purity of the matter
being examined.
(a)
Solids retained by filter paper
Liquid passes through
(b)
Mixture of solid and
liquid
Liquid
turns
into vapour
Heat
Solid remains
(c)
Spots corresponding to individual
substances in dye
Original spot of ink
Figure 1.4 Methods of separating mixtures: (a) filtration (b) evaporation (c) chromatography
EXAMPLE
Classify the example substances as pure or mixture (homogeneous or heterogeneous).
Sugar
pure
Blood
heterogeneous mixture
Stainless steel
homogeneous mixture
Sea water
homogeneous mixture (unless there are lots of solids floating
around in it!!)
Helium
pure
7
Chemistry for Technicians
PRACTICE QUESTION
3. How would you separate a mixture of sand and salt so that you end up with
containers of the individual solids?
1.4.2
Atom/Molecule
All matter is composed of atoms, the smallest individual units, the building
blocks of matter. There are different types of atoms — 107 of them — and
joined and mixed together, these atoms give us the vast range of matter we
see (and don’t see) around us.
In many types of matter, atoms join together in a small (or not so small)
group, known as a molecule. The oxygen that we breathe is one of the
simplest molecules: it has two atoms joined together. The plastics that we
use contain huge molecules with thousands of atoms joined together.
EXAMPLE
Classify the pure example substances as atoms or molecules.
At this stage of your chemical knowledge, you can’t be expected to answer this one.
However, to maintain continuity for the examples, the answers are provided.
Sugar
molecule
Helium
atom
Molecules with two atoms are called diatomic, and triatomic if they have three atoms.
These atoms do not have to be of the same element. More than three atoms in a molecule
earns it the name polyatomic (though in some cases, prefixes for four (tetra), five
(penta), and six (hexa) can be used).
EXAMPLE
List some diatomic, triatomic and polyatomic molecules.
Diatomic molecules — oxygen gas, nitrogen gas, carbon monoxide
Triatomic molecules — ozone gas, carbon dioxide, water
Polyatomic molecules — ammonia (one nitrogen and three hydrogens), butane cigarette
lighter fuel (four carbons and 10 hydrogens)
1.4.3
Element/Allotrope/Compound
Pure substances will be one of two chemical types — element or
compound. An element is a substance containing only one type of atom.
For example, oxygen gas is an element, as is 24 carat (pure) gold.
Some elements exist in a number of different forms, where the atoms are
arranged in different ways. These are known as allotropes. For example,
For example, ozone, the substance which shields the Earth from the sun’s
from the sun’s radiation, is an allotrope of oxygen. The oxygen that we
oxygen that we breathe has two atoms of oxygen in a molecule, while ozone
has three. Another important group of allotropes are soot, graphite and
diamond — all different forms of the element carbon.
8
Chapter 1
What Is Chemistry?
Compounds have atoms of more than one element joined together in a
fixed and unchanging ratio. Water has two hydrogen atoms and one oxygen
atom in each molecule, regardless of whether it is in the form of steam,
liquid water or ice.
A different ratio would mean a different substance. For example, bleaching solution
contains the compound, hydrogen peroxide, which has hydrogen and oxygen atoms, like
water, but has two oxygens with two hydrogen atoms. The difference between the two
compounds is shown in Figure 1.5.
Water
Figure 1.5
Hydrogen peroxide
Differences in compounds (oxygen atoms are white circles, hydrogen atoms black)
EXAMPLE
Classify the pure substances among the examples as elements or compounds.
Sugar
compound
Helium
element
There are a number of other very important classification systems used by chemists, such
as ionic/molecular and organic/inorganic, but at this stage, it is too difficult to describe
them without more knowledge. We will study them later in this textbook.
PRACTICE QUESTIONS
4. Classifying the following substances, using all the systems described in Sections
1.3 and 1.4: (a) glass, (b) unpolluted air, (c) a 24-carat gold ring, (d) milk,
(e) salt, (f) muddy water.
5.
Two clear, colourless liquids (A and B) are placed in evaporating basins and
heated until no liquid remains in either. In the basin that contained liquid A,
nothing remains, while liquid B has left a white powder. What conclusions can
be made about the purity of each liquid?
6.
Are the following statements TRUE or FALSE?
(a) A mixture can usually be separated into its components by simple
procedures.
(b) Two elements when combined in a mixture have different properties to
when they are combined in a compound.
(c) Most naturally occurring substances are mixtures.
7.
Figure 1.6 (overleaf) shows containers with substances composed of one or two
different types of atoms — black and white — illustrating the terms element,
compound, homogeneous mixture and heterogeneous mixture. Match the
containers with the terms.
9
Chemistry for Technicians
A
B
Figure 1.6
1.5
C
D
Practice Question 7
SOLUTIONS: THE MOST IMPORTANT MIXTURE
Solutions are homogeneous mixtures and, in most cases, the physical state is liquid. Why
are solutions so important? Seventy per cent of the Earth’s surface is covered in a
solution, most of our bodily functions work using solutions, most chemical analyses to
check the composition of matter rely on making the matter into a solution, and so on.
There are a number of key terms that help to describe a solution. Basically,
a solution is one or more substances dissolved in a liquid. The substances
The substances dissolved are called solutes, and the liquid is called the
called the solvent. Thus, salt is the solute, and water the solvent in sea
solvent in sea water. Solutions with high levels of solute are concentrated,
those with little solute are dilute.
The concentration of a solution is the amount of a solute per volume of
solution. The numerical value of concentration is an extremely important
one: it defines some of the properties of the solution. The earlier example of
salt in blood serum is significant. Saline drip solution being made for hospitals must be
checked for its concentration; if it is not the right value (nine grams of salt per litre of
solution), then it will be harmful if used.
EXAMPLE
Of the following solutions, which is the most concentrated? Which is the most dilute?
A. 20 g of salt in 500 mL of solution
B. 50 g of salt in 2.5 L of solution
C. 5 g of salt in 10 mL of solution
D. 20 g of salt in 100 mL of solution
The ratio of mass to volume is all important in defining concentration. If we calculate the
grams per mL for each of these solutions, we can answer the question.
A. 20 ÷ 500 = 0.04 g/mL
C. 5 ÷ 10 = 0.5 g/mL
B. 50 ÷ 2500 = 0.02 g/mL
D. 20 ÷ 100 = 0.2 g/mL
Thus, solution C is the most concentrated, even though it has the least mass of salt.
Likewise, solution B, with the largest mass, is the most dilute.
There is a limit to the amount of a particular solute that can be dissolved in a given
volume of solution. The degree to which a substance will dissolve is called its solubility.
This might be a value (e.g. the solubility of salt in water is 360 g/L) or it might be a
relative term (e.g. the solubility of sugar in water is high).
10
Chapter 1
What Is Chemistry?
When two liquids are mixed together and dissolve, they are said to be
miscible. For example, alcohol and water are miscible, but petrol and water
petrol and water are not — they are immiscible. You may be wondering
wondering which liquid is the solvent and which is the solute in a mixture
in a mixture of water and alcohol. Good question! The answer is the solvent
is the one which is in greater proportion in the mixture. Thus in beer, which
is about 5% alcohol, water is the solvent and alcohol one of the solutes. In a
very strong spirit, such as Bundaberg™ rum, which is more than 50%
alcohol, it could be said that water becomes the solute.
PRACTICE QUESTIONS
8. What is the solvent and give TWO examples of solutes in a cup of black tea with
one teaspoon of sugar.
9.
Arrange the following salt water solutions by order of increasing concentration.
Which is the most dilute?
A. 10 g of salt and 2 L of water
B. 10 g of salt and 200 mL of water
C. 5 g of salt and 500 mL of water
D. 5 g of salt and 20 mL of water
10. Complete the blanks in the following.
When salt is dissolved in water, we call the salt the ____(a)____ and the water
the ___(b)___. The overall combination of the two is called a ____(c)____.
When liquid alcohol is mixed with water, the two liquids are said to be
___(d)___.
1.6
PHYSICAL AND CHEMICAL CHANGES
Earlier, we distinguished mixtures from pure substances by using simple
processes that separated the components of a mixture. These processes did
processes did not change the chemical composition of the matter, only its
matter, only its physical form. These are called physical changes. Thus,
changes. Thus, salt water when boiled and the vapour collected was still
composed of salt (left behind in the container) and water (in the vapour
form or liquefied by cooling). The processes are also reversible. If the salt
and water were recombined, the result would be identical to the original salt
water solution.
If matter undergoes a transformation which alters its chemical form (i.e.
different substances are produced) then a chemical change occurs. When
occurs. When sugar in grape juice ferments, alcohol and carbon dioxide gas
carbon dioxide gas result. The sugar is consumed. The change is not easily
change is not easily reversible: you cannot take a mixture of alcohol and
carbon dioxide and hope to turn it into sugar.
Chemical changes are often indicated by:
•
the production of heat
•
evolution of gas bubbles
•
colour changes
•
the formation of solids from solutions (called precipitates).
11
Chemistry for Technicians
EXAMPLE
What are some physical and chemical changes that could be performed on sea water?
Physical changes don’t affect the chemical form of the components of the matter. We
could obtain pure water by evaporation and cooling. The salt could be crystallised to form
pure table salt.
A chemical change affects the chemical form: hydrogen and oxygen gases could be
formed by passing electricity through the sea water; sodium metal and chlorine gas could
be formed by melting the pure salt and passing electricity through it.
PRACTICAL WORK
1. Place a small amount of naphthalene (half a spatula full) in a test tube, and
note its smell and appearance.
Using a test tube holder, hold the test tube in the water bath until the
solid melts. Note its appearance.
Remove the test tube from the water bath, and stand it in a beaker
until it cools and the solid reforms. Note its odour and appearance again.
2.
Gently heat a small amount of ammonium chloride in a test tube. Record
your observations.
3.
Pour about 100 mL of water into a 250 mL beaker. Add a few drops of
phenolphthalein indicator to the water. Measure the temperature of the
water.
Place one piece of clean sodium in the beaker, and observe. Measure
the temperature of the water after reaction has ceased.
A beaker containing sodium hydroxide solution and phenolphthalein will
be available for observation.
4.
Place a small amount of copper sulfate in an evaporating basin, and add hot
water 1 mL at a time until the copper sulfate dissolves. Note the colour of
the solution. Evaporate the water by heating on a water bath. Note the
appearance of the residue.
5.
PERFORM THIS IN A FUME HOOD. Place a piece of copper metal in a
beaker, and very carefully add 5 mL of concentrated nitric acid. Heat gently
until a reaction begins.
Q1. Comment on the changes undergone during the heating of naphthalene and
ammonium chloride. Are they chemical or physical changes?
Q2. Give TWO pieces of evidence to support the theory that the combination of
water and sodium metal is a chemical change. Describe ONE further test or
procedure that would help confirm that it was a chemical change.
Q3. Classify the two processes involving copper or copper sulfate as physical or
chemical changes and explain your answers.
12
Chapter 1
What Is Chemistry?
PRACTICE QUESTIONS
11. Classify the following processes as physical or chemical changes:
(a) dissolving sugar in water
(b) wine going ‘off’ to form vinegar
(c) forming ice in the freezer
(d) steel corroding to form rust.
12. Which of the following is NOT a physical change?
(a) lighting a match
(b) heating water
(c) melting butter
(d) releasing compressed air from a gas cylinder.
13. What distinguishing feature of physical changes allowed you to answer Q11 and
Q12?
1.7
ELEMENTS
There are at least 107 known elements, 91 of which are naturally occurring. The others
are man-made, and only occur very briefly in nuclear reactors. The atoms of each
element differ in their size, weight and internal composition. We will look more at the
nature of atoms and how they differ between elements in the next chapter, and at the
properties of some important elements in Chapter 3.
It should be made very clear at this point the difference in the usage of the terms
atom and element. Many books tend to use the latter term in two ways: firstly, atoms of a
particular type and secondly, the pure substance itself. In this book, the term element will
be used to describe the pure substance only. When the term atom is used, it can refer to
its occurrence in elements, compounds or mixtures.
EXAMPLE
Reword the following sentences: Water contains two types of elements — hydrogen and
oxygen. Water can be made by combining the elements hydrogen and oxygen.
The elements hydrogen and oxygen are both reactive gases. The sentence should read
‘Water contains two types of atoms — hydrogen and oxygen. Water can be made by
combining the elements hydrogen and oxygen.’
Each element (or type of atom) has a name, but also a unique symbol of
generally one or two letters which simplifies matters when it comes to
comes to referring to the atom in molecules and compounds (the most
(the most recently discovered man-made elements have been given threegiven three-letter symbols, but we won’t worry about them). A number of
different approaches to giving the element a symbol have been used. These
include:
• the first letter in the name (e.g. carbon, C)
• the first two letters in the name (e.g. cobalt, Co)
• the first and third or further letters in the name (e.g. chlorine, Cl;
californium, Cf)
• abbreviation of an earlier name in another language (e.g. gold Au,
from the Latin name aurum; tungsten W, from the German name
wolfram).
13
Chemistry for Technicians
Table 1.2 (overleaf) lists some of the names and symbols which you should
start to remember, since they form the ‘alphabet’ which is used to write
chemical ‘words’, ‘sentences’ and ‘stories’. The entire list of known
elements is tabulated in a special way (known as the periodic table), and is
included on the inside front cover of this book. We will look at the periodic
table, what the numbers mean and why it is the shape it is in coming
chapters.
PRACTICE QUESTION
14. Look at the periodic table on the inside front cover of your book. Find more
examples of element symbols of the four basic types described above.
Table 1.2 Some common elements and their symbols
Element
Symbol
Element
Symbol
Hydrogen
H
Chlorine
Cl
Carbon
C
Potassium
K
Nitrogen
N
Calcium
Ca
Oxygen
O
Iron
Fe
Sodium
Na
Copper
Cu
Magnesium
Mg
Zinc
Zn
S
Lead
Pb
Sulfur
WHAT YOU NEED TO BE ABLE TO DO
•
•
•
•
•
•
•
•
Explain how chemistry attempts to explain our observations.
List and apply important classification systems for matter.
Distinguish between heterogeneous and homogeneous mixtures.
Define important terminology related to solutions.
Define important terminology related to pure substances.
Distinguish between chemical and physical changes.
Explain the need for a system for element symbols.
Recall common elements and their symbols.
14
Chapter 1
What Is Chemistry?
TERMS AND DEFINITIONS
During this chapter, you have met a number of new terms, listed below. Match them with
their correct definitions.
Terms
A science
C
matter
E
heterogeneous mixture
G molecule
I
allotrope
K solution
M solvent
O physical change
B
D
F
H
J
L
N
P
chemistry
homogeneous mixture
atom
element
compound
solute
miscible
chemical change
Definitions
1
the building blocks of matter
2
matter containing more than one pure substance, but of uniform appearance
3
a different physical form of an element
4
a homogeneous mixture, normally a liquid
5
the study of the behaviour of our surroundings
6
the substance dissolved in the liquid
7
groups of atoms linked together in a fixed way
8
the liquid that forms the solution
9
all the ‘stuff’ around us
10 two liquids that are soluble in each other
11 a substance containing one type of atom
12 a process where different substances are produced
13 a process where the physical, but not the chemical, form of the matter is altered
14 matter containing more than one pure substance, with varying appearance
throughout
15 a pure substance containing more than one type of atom
16 the study of the behaviour of matter
15
ATOMS: THE BUILDING
BLOCKS OF MATTER
2
PURPOSE
To describe the structure of atoms and ions.
The story so far: if chemistry is about studying the properties of matter, then it needs to
explain why there is so much variation.
2.1
EVOLUTION OF THE ATOMIC CONCEPT
Two-and-a-half thousand years ago, the Greeks developed an interest in explaining — in
essentially scientific terms — the world around them. Their approach was based not on
experiment, but on ideas. They proposed a theory of matter, which attempted to explain
why water was different to air and the earth, and so on. The basis of the theory was that
all matter was composed of four ‘elements’ — earth, wind, water, fire — in varying
mixtures. While the elements themselves have been redefined (we know of 107), the
basic idea is still current.
The Greeks disagreed on how matter was made up: most thought that it was
continuous, but some thought it was made up of very small indivisible particles. What is
the difference? Imagine a cube of butter. Cut it in half, then halve one of the portions,
and keep doing so, all the way down to the invisible. Some of the Greeks believed that
this process could be done forever, that there was no smallest entity — matter was
continuous.
Others believed there would come a point at which an entity was obtained
that could not be divided any further, which they called ‘atomos’, from which our word
atom derives. All this was done without any sophisticated equipment, in fact no
equipment at all, since the Greeks did not believe in experimental measurement, but
simply observation and thought. That they came so close to the truth (or at least the
twentieth-century scientific view of the truth) is quite remarkable.
Between the time of the ancient Greeks, and the beginning of the nineteenth
century, interest in the sciences faded and grew. The picture of matter and the atom
changed little though much experimental work was done, particularly from the
seventeenth century onwards. Finally, an English schoolteacher, John Dalton, proposed a
theory based on all this work which, in summary, states the following:
•
•
•
•
•
•
all matter is made up of extremely small particles called atoms
elements contain one type of atom
all atoms of a given element are identical, but atoms of different elements are
different
atoms of different elements have different properties
atoms of one element cannot be changed, nor are they created or destroyed in
reactions
compounds are made up of atoms of different elements, and in a fixed proportion.
16
Chapter 2
Atoms: The Building Blocks of Matter
Dalton was very close to the truth: some of his definitions should be familiar from
Chapter 1 (e.g. elements, compounds). However, he was wrong in one respect. His
picture of the atom was that of a billiard ball — hard, uniform and indivisible. More
experiments in the next hundred or so years were to change that picture substan-tially.
These discoveries and the changes to the picture of the atom are summarised in Figure
2.1.
Dalton’s billiard ball
Thomson discovers particle
contained within atom — 1897
Thomson’s plum pudding model
Rutherford discovers that atoms are mostly empty
space with dense centre — 1911
Bohr’s solar system model
Figure 2.1 Evolution of the picture of the atom
LIBRARY WORK
Thomson, Millikan, Rutherford and Chadwick were key players in the
development of the picture of the atom. They each performed an experiment
which was critical in furthering the atomic theory. Look up reference sources in
the library to see what they did, and what the results meant to the picture of the
atom.
2.2
THE SOLAR SYSTEM MODEL OF THE ATOM
The key discoveries in changing the picture of the atom were that it:
•
contained smaller components (Thomson called them electrons), and
•
was mostly empty space, with a dense, positively charged centre (Rutherford).
17
Chemistry for Technicians
These discoveries led a Danish physicist, Niels Bohr, to propose a picture
of the atom, which was essentially a very small-scale version of the solar
of the solar system: a central massive body, called the nucleus (like the
(like the sun) with the very small electrons circling around it (like the
(like the planets). The solar system model of the atom is shown in Figure
2.2.
Orbiting electrons
Nucleus
Figure 2.2 The solar system model of the atom
The electrons form the ‘outer skin’ of the atom, hiding away the positively charged
mass in the centre. Therefore, it is perhaps not surprising that the electrons are more
likely to be involved in changes to the atom. We will look at this in more detail in
Section 2.5.
In the next 20 years, two more components of the atom were discovered. In 1919,
Ernest Rutherford found that the positive charge that balanced the negative charge of the
electrons was possessed by individual units, which he called protons. Protons were
found in the nucleus, and were much heavier than electrons.
Finally in 1932, James Chadwick found that the mass of the atom was not entirely
due to the heavy protons, but that another particle, of similar mass but with no charge,
also existed in the nucleus. He called these particles neutrons. The three components of
the atom were called subatomic particles. Table 2.1 compares the properties of the
proton, electron and neutron.
Table 2.1
Properties of the major subatomic particles
Particle
Charge
Mass (g)
Relative mass
Position in atom
Proton
+1
1.67 x 10–24
1
nucleus
Electron
–1
9.11 x 10–28
0.00054
orbiting
Neutron
0
1.67 x 10–24
1
nucleus
In the age of nuclear research, other subatomic particles have been found, some of
which are the components of the major subatomic particles. However, these do not affect
our understanding of the atom in the study of chemistry.
Further development of the solar system model stated that electrons circled, like
planets, at fixed distances from the nucleus, and that certain maximum numbers of
electrons could exist in a given orbit (unlike planets). This will be further examined in
Section 2.6.
The Bohr model is a good approximation of the structure of the atom; it has,
however, been modified to comply with more sophisticated experiments that have been
conducted since Bohr’s time, but for explaining simple chemical behaviour, the solar
system model is entirely adequate.
18
Chapter 2
Atoms: The Building Blocks of Matter
PRACTICE QUESTIONS
1. What feature of the solar system model of the atom makes it very different from
the ‘plum pudding’ model?
2.
Calculate the percentage of the total atom’s mass possessed by the nucleus, in a
carbon atom which contains six of each subatomic particle.
2.3
HOW ATOMS DIFFER
Why is hydrogen different to carbon? To gold? To any of the other 104 elements known
at this moment in time? The simplest answer is that each element is composed of
different types of atoms. How do the atoms of hydrogen differ to those of all the
elements? The answer is in the building blocks of the atoms themselves: the sub-atomic
particles.
Hydrogen is the simplest element of all. Its atoms contain one proton and one
electron. No other element has only one proton. Table 2.2 lists the four simplest
elements, and the breakdown of the number of each of the subatomic particles to be
found in atoms of those elements.
Table 2.2
Element
Sub-atomic particles in the first four elements
Protons
Electrons
Neutrons
Hydrogen
1
1
0
Helium
2
2
2
Lithium
3
3
4
Beryllium
4
4
5
If you look at the figures in the above table, a number of observations might
emerge:
•
•
•
there are the same number of protons and electrons in each atom
the number of protons (and electrons) are increasing in a sequence
the number of neutrons are increasing as well, but not in an orderly fashion.
If we were to continue the table to cover all the known elements, the number of protons
and electrons would continue to grow one at a time. This is the distinguishing feature of
elements at the atomic level: the number of protons in the atom. No two elements have
atoms with the same number of protons.
For atoms of the pure element, the same statement can be made about electron
numbers. However, the number of electrons in an atom changes when the element
combines to form compounds, but the number of protons do not. In Section 2.5, we will
examine the effect of changing subatomic particle numbers on the nature of an atom.
Atoms are neutral: they have no overall charge. The size of the charge on protons
and electrons has been measured to be identical, but opposite in sign. To obtain a neutral
atom, the number of electrons in an atom must be equal to the number of protons.
19
Chemistry for Technicians
So, the number of protons is an important defining aspect of atoms and
elements. It is given the special name atomic number (sometimes
(sometimes abbreviated as A). Hydrogen has one proton and thus, atomic
thus, atomic number one. If you look at the periodic table on the inside
the inside front cover of this book, the atomic number for each element is
given in the top left-hand corner of the box. It is like the catalogue number
or the address of the element.
HOW CAN WE COUNT THE SUBATOMIC PARTICLES WHEN WE CAN’T SEE THEM?
The picture of the atom — with various numbers of protons, electrons and neutrons — is
just that: a picture. It could be totally and utterly wrong. Without being able to see the
subatomic particles, it is the best guess scientists can make at this time.
The negative charge can be ‘peeled’ away from the atom, one unit at a time. This is
measurable. So, this unit has been called the electron. The negative charge held by the
electrons must be balanced by the same amount of positive charge.
The question arises: is this positive charge in single units of equal size but opposite sign
to the electrons, or is it one large block? The best justification for single units is that the
negative charge is that way. It is much simpler to picture the mass and charge carriers of
the atom as individual units. The electrons seem to be, why not the others?
The simple hydrogen atom becomes the reference point: it has one unit of positive charge
and one unit of negative charge. Different atoms simply have extra amounts of these, plus
extra mass units from the neutronsIt is possible that one day someone will find an atom
smaller or simpler than hydrogen. If that happens, you may tip this textbook into the
garbage, because all of the chemistry fundamentals will have to be re-written.
You will notice that the atomic numbers increase in an unbroken sequence until
107. What does this mean? Element 108 has yet to be discovered! Before you begin
writing your Nobel Prize speech, you should know that all the elements after number 94
have only been produced in incredibly complex and unbelievably expensive equipment
by causing other atoms to have high speed, head-on collisions.
Chadwick’s discovery of the neutron was, in part, brought about by the puzzling
discoveries of ‘heavy forms’ of many elements. For example, 1% of hydrogen atoms are
twice as heavy as the other 99%, but are chemically identical. Until his discovery, the
mass of the atom was thought to be contributed only by the protons, but a hydrogen atom
could not be twice as heavy by having an extra proton, because then it would have the
same atomic number as helium. The discovery of the neutron (of equal mass to the
proton, but of no charge) solved this mystery.
The two types of hydrogen with the different masses are called isotopes:
atoms of the same element with differing numbers of neutrons. Isotopes
Isotopes differ in their mass, which is essentially determined by the
by the combined number of protons and neutrons. This number is called the
number is called the mass number (sometimes abbreviated as Z — don’t
ask why!). The three naturally occurring isotopes of hydrogen have mass
numbers 1 (one proton), 2 (one proton and one neutron) and 3 (one proton
and two neutrons), as shown in Figure 2.3. Mass numbers are not unique:
for example, there are seven elements with isotopes having a mass number
of 69.
20
Chapter 2
Atoms: The Building Blocks of Matter
Most elements have a number of naturally occurring isotopes, hydrogen being an
example already discussed. Tin is an extreme example: it has 10 naturally occurring
isotopes. Sodium, on the other hand, has only one isotope found in nature: its mass
number is 23. Sodium isotopes with every other mass number from 19 to 31 are known,
but were all manufactured in nuclear reactors and are very unstable. Unstable isotopes
have too many or too few neutrons in the nucleus, when compared to the number of
protons.
Because isotopes of a particular element have the same name, some means of
distinguishing them is needed. The mass number is written after the name of element, for
example hydrogen-1, hydrogen-2, carbon-14, uranium-235. When symbols are used, the
mass number is superscripted before the symbol, for example 1H, 2H, 14C, 238U.
Figure 2.3
Representation of the isotopes of hydrogen (protons are white circles, neutrons grey
and electrons black)
The basic relationships between the numbers of subatomic particles and the atomic
and mass numbers in neutral atoms are summarised in Equation 2.1a–c.
Atomic number
Mass number
Number of neutrons
=
=
Number of protons
Number of electrons
Eqn 2.1a
=
Number of protons
+ number of neutrons
Eqn 2.1b
Mass number – atomic number
Eqn 2.1c
=
EXAMPLES
1.
How many of each subatomic particle are in neutral atoms of helium-4,
uranium-235?
23Na
and
To answer these questions, you need to use the equations above, and the periodic table
on the inside front cover (which gives the atomic number in the top left corner of the
element box).
Helium-4: its atomic number is 2, so it has 2 protons and 2 electrons. Its mass number
is 4, so it must have 2 neutrons also.
23Na: sodium has the atomic number 11, so it has 11 protons and 11 electrons. Its mass
number is 23, so it must have 23 – 11 = 12 neutrons also.
Uranium-235: its atomic number is 92, so it has 92 protons and 92 electrons. Its mass
number is 235, so it must have 235 – 92 = 143 neutrons also.
2.
What is the name, symbol, atomic number and mass number of an element with
neutral atoms containing 43 protons, 43 electrons and 56 neutrons. Give the number
of subatomic particles in another isotope of this element.
Its atomic number is 43 — equal to the number of protons (or electrons).
Referring to the periodic table, the element is technetium (Tc).
The mass number of this isotope is 99 (43 + 56).
A different isotope has more or less neutrons, but the same number of protons and
electrons, for example 57 neutrons, 43 protons, 43 electrons.
21
Chemistry for Technicians
PRACTICE QUESTIONS
3. Explain why mass numbers are not unique, but atomic numbers are.
4. Rewrite the following paragraph, correcting any errors. The atomic number of
carbon is 6.
A neutral atom of carbon contains six protons, seven electrons and six neutrons.
The protons are found circling around a group of electrons in the centre. The
neutrons can be found anywhere
5.
Give the number of protons, electrons and neutrons for an isotope of carbon.
6.
Complete Table 2.3 for neutral atoms.
Table 2.3 Question 6
Atomic number
(A)
Mass
number (Z)
Protons (p)
(a)
6
(b)
22
26
29
36
(f)
210
(g)
197
2.4
12
11
(e)
(h)
Neutrons (n)
7
11
(c)
(d)
Electrons (e)
48
128
79
92
140
ATOMIC WEIGHTS
We can no more weigh individual atoms than we can count them individually. We can
make relative measurements, where equal numbers of different elements are weighed,
and a relative weight established. This was how the chemists of the nineteenth century
measured their ‘atomic’ and ‘compound’ weights. Hydrogen atoms were assigned the
relative mass of 1, and everything else was based on that. This can only be done because
a volume of any gas at a set pressure contains the same number of molecules (or atoms).
The atom now used as the standard is carbon-12, which is given the mass of exactly 12
(and the unit amu — atomic mass unit).
Carbon-12 has 12 mass-carrying subatomic particles (six protons and six neutrons).
You might then expect that all atoms would have whole number (integer) masses. This is
more or less the case at the atomic level, but when measurements are done on real
samples of the elements, few have relative masses that are whole numbers. For example,
chlorine has a mass (relative to carbon-12) of 35.45.
Why? Does this mean chlorine has 0.45 of a neutron in its structure? Of course, the
answer is no. The non-whole number mass is due to the fact that chlorine has two natural
isotopes — mass numbers 35 and 37 — and the 35.45 mass is a reflection of the average
mass of a large collection of these atoms. In fact, 76 of every 100 chlorine atoms have a
mass of 35, and the other 24 have a mass number of 37. The result is an average of 35.45,
as illustrated in Figure 2.4.
22
Chapter 2
Atoms: The Building Blocks of Matter
The atomic weight of an element is defined as the average mass of the
naturally occurring isotopes of that element. Man-made isotopes that don’t
occur in nature can’t be incorporated into atomic weights since their
proportion is not fixed.
EXAMPLE
Why does the element bromine have an atomic weight of 80 when its stable isotopes
have mass numbers of 79 and 81?
Bromine’s natural isotopes occur in almost exactly equal proportions: Br-79 50.69% and
Br-81 49.31%. When you calculate the average mass, it equals almost exactly 80.
Average mass
= (50.69% x 79) + (49.31% x 81)
= 79.99 amu
A box containing 100
chlorine atoms — 76 are
Cl-35 and 24 are Cl-37
Total mass on ‘atomic
balance’
3548 amu
Average mass = 35.48 amu
Figure 2.4 ‘Weighing’ 100 chlorine atoms
LIBRARY WORK
Your teacher will assign you one element to look up in a reference book, which
lists all the known isotopes, and the proportions of the natural isotopes. You will
be required to find the following information:
•
•
•
•
•
atomic number of the element
mass numbers of all known isotopes
for each isotope, indicate whether it is stable or unstable
numbers of subatomic particles for two isotopes
atomic weight.
PRACTICE QUESTION
7. Calculate the atomic weight of iron given the proportion of natural isotopes as
follows: 54 (5.8%), 56 (91.72%), 57 (2.2%) and 58 (0.28%).
23
Chemistry for Technicians
2.5
CAN ATOMS BE CHANGED?
The defining aspect of a given atom is the number of each of the three subatomic
particles of which it is composed. Is it possible to change the number of any of these
particles? If so, what is the result?
2.5.1 Protons
The number of protons defines the element of which the atom is a ‘member’. If the
number of protons could be changed, a different element would result. Such an outcome
could be highly desirable. For example, removing three protons from an atom of lead
(atomic number 82) gives us an atom of gold (atomic number 79). However, there would
be three electrons too many, yielding a non-neutral atom, and there may also be too many
neutrons. The atom may not be stable, and would break up into smaller bits: our gold
would not last!
Transformations of the nucleus are very difficult to achieve and almost impossible
to control. You don’t need to consider them in your studies of simple chemistry.
2.5.2 Neutrons
Neutron changes fit into the same category as those affecting the number of protons.
Theoretically, it is possible to add neutrons to an atom, but the equipment is highly
complex, and the outcome of the process is unpredictable. Again, don’t worry about this.
2.5.3 Electrons
As mentioned earlier, the electrons are able, under normal conditions, to be removed
from or added to the atom because they form the ‘skin’ of the atom. The presence of
electricity, other chemicals, heat or radiation may cause a change in the number of
electrons in an atom. What happens when such an event occurs?
EXAMPLE
What happens to an atom of hydrogen-1 if it (a) gains and (b) loses an electron?
Hydrogen-1 has one proton and one electron. If it gains an electron, it has one more
negative charge than positive. If it lost its electron, it would have only a positive charge. In
both cases, the resulting atom would not be neutral. The table below illustrates each case.
Case
Neutral atom
Gain electron
Lose electron
Protons
1
1
1
Electrons
1
2
0
Overall charge
0
–1
+1
By definition, atoms and molecules are neutral, that is they have no charge. Can a
non-neutral atom exist? The answer is yes, but not in all cases.
Chemical species (species is a general term which may refer to any type of
chemical substance) that ‘look’ like atoms do exist with a positive or
positive or negative charge. They are called ions. A single positive or
positive or negative ion cannot exist by itself without another of the
of the opposite charge. To make the distinction between atoms and ions
clear: atoms are neutral, ions are charged. Ions may be species containing
one atom (monatomic ion) or a number of atoms (like charged molecules,
known as polyatomic ions).
24
Chapter 2
Atoms: The Building Blocks of Matter
Ions may be classified by the sign of their charge. Positive ions are called cations,
while negative ions are called anions. Cations result when an atom loses one or more
more electrons (more protons than electrons), while anions result from an atom
gaining one or more electrons (more electrons than protons).
Ions can exist in charges ranging from 3– to 4+: the most common tend to be 1+,
1–, 2+ and 2–. The conventional way for writing the charge of an ion is as shown here:
the value of the charge followed by the sign.
EXAMPLE
How do the ions of sodium (1+) and sulfur (2–) occur?
Sodium would have to lose one electron, leaving it with 11 protons and 10 electrons.
Sulfur would have to gain two electrons, leaving it with 16 protons and 18 electrons.
PRACTICE QUESTIONS
8. Explain why the 1+ sodium ion cannot be formed by a sodium atom gaining an
extra proton.
9.
What happens to an atom of fluorine when it loses one electron?
10. A neutral zinc atom contains 30 protons, 30 electrons and 34 neutrons. What
happens when the atom changes to the zinc 2+ ion?
2.6
THE ELECTRON SHELL MODEL
One significant limitation of Bohr’s model of the atom was that the position of the
electrons in their orbits was not clear. Did all electrons occur in the same orbit? Or did
each have its own orbit? Or maybe, there was no fixed orbit to which a given electron
belonged.
Scientists continued to study the atom, and one of the most important experiments
was the measurement of the ionisation energy for all the known elements: the amount of
energy required to remove one electron from an atom of the element. The data for the
elements with atomic numbers 1–40 are shown in Figure 2.5.
The most important observation from the data in Figure 2.5 was that there seemed
to be elements that were difficult to ionise (marked with arrows), immediately followed
by elements that were ionised relatively easily. The extra electron in the latter type of
element clearly did not belong in the same pattern as those preceding it. These results
eliminated some of the possibilities mentioned above.
The most logical conclusion that could be drawn from the data was that each orbit
had a certain number of ‘holes’ in it to contain electrons. Once these holes were filled,
the next electron had to go to another orbit, most likely one further out from the nucleus.
Thus was the electron shell theory developed. The atom begins to appear as an onion,
with its concentric skins corresponding to the shells.
25
Chemistry for Technicians
2
10
18
36
1
6
11
16
21
26
31
36
41
Atomic number
Figure 2.5 Ionisation energies for elements 1–40
The data also allowed the prediction of the maximum number of
electrons per shell. From the peaks in the ionisation energies (elements 2,
(elements 2, 10, 18 and 36) of the first 40 elements, it was clear that the
clear that the first four shells could contain 2 (first shell), 8, 8 and 18
and 18 electrons as a maximum. Further work extended the picture to the
next two shells, which could contain 18 and 32 electrons. You will see in
the next chapter that these numbers were particularly significant in terms of
the periodic table, which already existed at the time.
You are now able to draw a more detailed picture of the atom, locating the electrons
not simply in a ‘general’ orbit somewhere around the nucleus, but in particular shells.
The rule for assigning electrons to the shells is that the first shell is filled before an
electron is added to the next one.
In this chapter, you will deal only with the first 20 elements, which involve the first
three shells and two electrons in the fourth shell. Figure 2.6 shows the electronic
arrangements for a few of the elements.
The information about electron arrangements, as shown in Figure 2.6, can
be summarised using a shorthand version, known as the electron
electron configuration, which is a numerical representation of the picture.
of the picture. The electron configuration for an element (or ion) is simply
ion) is simply the number of electrons in each shell, separated by a full stop
and reading from the first shell outwards. The electron configurations for
some elements are given in Table 2.4.
PRACTICE QUESTION
11. Draw the electron shell arrangement for calcium.
26
Chapter 2
Atoms: The Building Blocks of Matter
H
He
Li
Na
Figure 2.6 Electron arrangements in shells
Table 2.4
Element
Electron configurations for some elements
Atomic number
Electron configuration
Hydrogen
1
1
Helium
2
2
Lithium
3
2.1
Carbon
6
2.4
Sodium
11
2.8.1
Silicon
14
2.8.4
Chlorine
17
2.8.7
Calcium
20
2.8.8.2
PRACTICE QUESTION
12. Write the electron configurations for (a) oxygen, (b) sulfur and (c) potassium.
2.7 ANOTHER LOOK AT THE FORMATION OF IONS
The electron shell model was supported by observations about the charge on monatomic
ions. We can use it to help determine and remember their charge. The example of sodium
and chlorine is useful to consider: two of the most unstable elements react to form one of
the most stable compounds. Why? Table 2.5 shows the change in electron configurations
from atom to ion.
Table 2.5 Changes to electron configurations in ion formation
Element
Configuration as atom
Configuration as ion
Sodium
2.8.1
2.8
Chlorine
2.8.7
2.8.8
27
Chemistry for Technicians
What difference between atom and ion is observed? The last shell of each
atom containing an electron has been affected, but not the shells already
shells already filled. The last shell is called the valence or outer shell. In
outer shell. In sodium, its third shell containing one electron, is its valence
is its valence shell. Chlorine’s third shell is also its outer shell, containing
seven electrons. These electrons are known as valence electrons.
In each ion, you will notice that the outer shell is full. Sodium has achieved this by
losing its valence shell electron, exposing the full second shell. Chlorine, on the other
hand, has filled its third shell with an extra electron.
This is what atoms try to do when forming ions: attain an electron
configuration with a full outer shell. This is sometimes known as the octet
known as the octet rule, since for the first 20 elements, eight is the number
is the number of the electrons in the full outer shell. This can be achieved
be achieved by filling up (gaining electrons) or emptying (losing) the outer
electron shell.
Sodium could have gained seven electrons to become a 7– ion (2.8.8), and chlorine lost
seven to become a 7+ ion (2.8). However, this does not happen: the atoms take the
shortest route to the full outer shell.
EXAMPLE
Predict the electron configuration and charge of the aluminium ion (atomic number 13).
The electron configuration for aluminium will be 2.8.3. The shortest route to a full outer
electron shell for aluminium is to lose the three electrons in its third shell. Thus, it
becomes 2.8 and has a 3+ charge.
PRACTICE QUESTIONS
13. How many valence electrons do atoms of (a) lithium (b) fluorine and (c) neon
have?
14. What ions would be formed by (a) oxygen (b) sulfur and (c) potassium?
15. Explain why the octet rule is a misleading name when applied to elements 1–5.
WHAT YOU NEED TO BE ABLE TO DO
•
•
•
•
•
•
•
•
•
•
•
Describe the development of knowledge of the structure of the atom.
Describe the structure of the atom.
Distinguish between the three sub-atomic particles.
Define the terms atomic number, mass number, isotope and atomic weight.
Determine these values for a given atom.
Define the terms ion, monatomic ion, polyatomic ion, cation and anion.
Explain how ions form.
Explain what is meant by the term electron shell.
Write electron configurations for elements up to atomic number 20.
Explain why ions form.
Predict their charge based on the electron configuration of the neutral atom.
28
Chapter 2
Atoms: The Building Blocks of Matter
TERMS AND DEFINITIONS
During this chapter, you have met a number of new terms, listed below. Match them with
their correct definitions.
Terms
A proton
B electron
C neutron
D atomic number
E isotope
F mass number
G atomic weight
H ion
I
anion
J
cation
K monatomic ion
L nucleus
M valence shell
Definitions
1
the number of protons in an atom
2
a negatively charged ion
3
a positively charged subatomic particle
4
atoms with the same number of protons, but different numbers of neutrons
5
a charged atom or molecule
6
an ion containing a single atom
7
the centre of the atom, which contains most of the mass
8
a negatively charged subatomic particle
9
a positively charged ion
10 the combined total of protons and neutrons
11 the outermost shell containing electrons
12 a subatomic particle with no charge
13 the average mass of naturally occurring atoms of an element
29
3
ORGANISING THE ELEMENTS
PURPOSE
To describe the properties of a number of important elements, and to identify some
trends.
The story so far: matter is made up of different sorts of atoms, but can we learn anything
about the different types?
3.1
A BRIEF LOOK AT SOME IMPORTANT ELEMENTS
Even before Dalton proposed his theory of matter, chemists endeavoured to produce
‘pure’ matter — that which could not be broken down any further, that is elements. The
great majority of the elements known today were produced in pure form before Bohr
proposed his theory of the atom’s structure. Figure 3.1 shows the rise in the number of
known elements over the last two centuries or so.
120
Known elements
100
80
60
40
20
0
1700
1750
1800
1850
1900
1950
2000
Year
Figure 3.1 Discovery of the elements
How do the known elements fit some of the important property classifications
described in Chapter 1? Figure 3.2 illustrates some of these.
30
Chapter 3
Organising the Elements
(b)
(a)
Gas
Liquid
Metal
10%
2%
77%
Solid
88%
Metalloid
Non-metal
7%
16%
(c)
Man-made
14%
Natural
86%
Figure 3.2 Classification of the elements by (a) physical state (b) metallic character (c) origin
As more and more elements were discovered, chemists wondered whether there
might be any trends in the properties of the elements. Atomic weight could be measured
with reasonable accuracy in the 1800s, so if the elements were arranged in order of
increasing weight, could any trend be discerned? Three possibilities emerged:
•
•
•
there would a gradual change in properties, or
there would be a repeating pattern, or
there would be no trend at all — they would all be vastly different.
The following descriptions introduce you to some of the first 20 elements (by atomic
weight).
3.1.1 Hydrogen (Atomic Number 1)
Hydrogen is the most common element in the universe; it accounts for more than 90% of
the atoms and 75% of the mass of the universe. In outer space, it occurs uncombined. On
Earth, apart from a very low concentration of the element as a diatomic gas in the
atmosphere, hydrogen only exists in the form of compounds, due to its reactivity. Most
hydrogen atoms are found in water molecules, but they are also an important component
of organic substances. Commercially, the element hydrogen is formed by the
decomposition of water. It is used in the production of ammonia, and may also become
an important energy source in the future. In ionic form, hydrogen usually exists in the 1+
state, but it can also form a 1– ion.
31
Chemistry for Technicians
3.1.2 Helium (Atomic Number 2)
The second most common element in the universe, helium is, however, a very rare
element on Earth. This is due to its limited reactivity: it is stable as a single atom and
forms very few compounds. This is the main reason why it was among the last elements
of the first 20 (by atomic weight) to be discovered (first isolated in 1895). It is found
mixed with crude oil and trapped in rocks, these being the sources of the gas for
commercial purposes.
3.1.3 Carbon (Atomic Number 6)
In elemental form, carbon is widespread across the Earth and through the universe. It has
been known since prehistoric times, and exists in three forms: diamond, graphite and
amorphous (better known as soot or charcoal). Diamond is one of the hardest substances
known, while graphite is one of the softest. Carbon forms millions of compounds, and is
the basis for all organic substances, which are the basis of life. It does not form simple
monatomic ions.
3.1.4 Nitrogen (Atomic Number 7)
First isolated in pure form in 1772, gaseous nitrogen is 78% of the atmosphere, occurring
as a diatomic molecule. It is generally unreactive, but will combine with oxygen under
conditions of extreme heat. The possibility of the atmosphere catching ‘fire’ was a risk
that was considered when the first atomic bomb test was carried out. The reaction does
occur in a car engine, where the emission of nitrogen–oxygen compounds is considered a
pollution problem. These compounds of nitrogen and oxygen are quite toxic and reactive,
and can form nitric acid when mixed with water. Ammonia is one of the most important
industrial compounds, and can be formed from nitrogen and hydrogen. Nitrogen forms a
3– ion, but it is very unstable.
3.1.5 Oxygen (Atomic Number 8)
Oxygen is the substance which our body requires when we breathe. It is a diatomic gas,
and makes up about 21% of the atmosphere. It was purified two centuries ago, and used
in many experiments to study other elements (by looking at the way they reacted with
oxygen). By mass, it is the most abundant element on earth, being found in the oceans,
air, organisms and the Earth’s crust. It is quite reactive, being one of the major
substances involved in rusting of iron metals. It exists in a second form, ozone, which has
three oxygen atoms per molecule. You will, no doubt, have heard about the ozone layer
50 kilometres up in the atmosphere, protecting us from harmful ultraviolet solar
radiation, yet ozone is harmful when it is formed at ground level in smogs in large cities.
Oxygen forms 2– ions.
3.1.6 Fluorine (Atomic Number 9)
Fluorine is the most reactive of all the elements, and was first isolated in pure form in
1886 as a diatomic gas. It will react with many substances, and so doesn’t exist in
elemental form in nature. It forms many very stable compounds, a good case being the
infamous CFCs which cause problems in the ozone layer because of their stability in the
lower atmosphere. Teflon plastic is a carbon-fluorine compound of great heat resistance
and stability. Flourine is the smallest of the group of elements called the halogens.
Fluorine readily forms a 1– ion.
3.1.7 Sodium (Atomic Number 11)
Sodium has been known in compounds for centuries, but was first isolated as the pure
metal in 1807. It is an unusually soft metal, being able to be cut with a knife. It reacts
vigorously with water and many other substances. Its compounds are widely used in
industry, commerce and the home. Table salt and soap are two of the best known
examples of sodium compounds. In ionic form, it is important for good health, being
required in the transmission of nerve signals through the body. It is, however, a cause of
high blood pressure, when eaten too much. It is most stable when present as a 1+ ion.
32
Chapter 3
Organising the Elements
3.1.8 Magnesium (Atomic Number 12)
Like sodium, its compounds have been known for centuries, but magnesium metal was
only isolated in 1808 by the same scientist who isolated sodium: Sir Humphrey Davy. It
is the eighth most abundant element in the Earth’s crust, and the second most common
cation in the oceans. It only exists in nature in compound form, because of its reactivity.
The metallic element is of low density and high strength, and is used in aeroplane
construction because of these properties. It also burns well, making it useful in fireworks
and emergency flares. The green pigment in plants, chlorophyll, is a magnesium
compound. Magnesium forms a 2+ ion in compounds.
3.1.9 Aluminium (Atomic Number 13)
Aluminium is the most industrially important metal among the first 20 elements. It was
first purified in the early 1800s. However, one of its most important compounds, alum
(from which its name derives), was used by the ancient Romans in medicine and in
dyeing clothes. Aluminium is too soft and weak for construction purposes, but when
mixed with other metals, forms a strong, light alloy. Though it is not as good a conductor
of electricity as copper, it is used for electricity transmission cables, because of cost and
construction advantages. Its oxygen compound occurs widely in ores, and is the major
component in precious gemstones, such as rubies and sapphires. American chemists call
it aluminum (no second i). It forms 3+ ions.
3.1.10 Chlorine (Atomic Number 17)
Chlorine, first isolated in 1774, has many features in common with fluorine: it is a
reactive diatomic gas, only found in compounds in nature. It is used to produce safe
drinking water and in the production of paper. Carbon-chlorine compounds are used as
industrial solvents and in the plastics industry. Its fumes are an irritant to our respiratory
system. In the ionic form, chlorine has a 1– charge.
3.1.11 Potassium (Atomic Number 19)
Potassium was another element first produced in pure form by Davy, in 1807. It is very
similar to sodium in many ways, and is even more reactive. Its compounds are not as
widely used as those of sodium, but some find use in fertilisers, because of the
importance of potassium in plant growth. Potassium is also involved in the functioning of
the nervous system. It forms a 1+ ion, like sodium.
LIBRARY WORK
Eleven of the first 20 elements are described above. Use reference sources to
find out similar information on the other nine. You might find out about some of
the heavier elements, both well known and rare.
PRACTICAL WORK
1. Make a note of the appearance of the following elements: magnesium,
aluminium, phosphorus (CAUTION: this must not be handled outside a fume
hood) and sulfur.
2.
Burn a small amount of each element, as directed by your teacher. Note the
physical state of the product.
3.
Place a small amount of the oxygen compounds of Na, Mg, Al, Si and P in
test tubes and add about 5 mL of water to each.
Measure the acidity of the solutions using universal indicator solution or
test strip.
33
Chemistry for Technicians
4.
For those substances which do not dissolve, try their solubility in
concentrated acid and base. If a compound dissolves in acid, then it is
basic; if it dissolves in the base, it is an acid. If it fails to dissolve at all, then
it is neutral.
5.
Look up the density, atomic weight and melting point of Na, Mg, Al, Si, P, S,
Cl and Ar in a reference book.
Q1. List the elements in order of increasing atomic weight. What trends are can
identified for:
• physical state
• metallic character
• acidity of the oxygen compounds.
Q2. On the basis of your answer to Q1, can it be said that the elements undergo
a gradual change in properties as atomic weight increases?
PRACTICE QUESTION
1. Of the elements described in Section 3.1, group those which seem to have
similar properties.
3.2
SOME SIMILARITIES APPEAR
It should be apparent from your work so far that the gradual change option in the
properties of elements is certainly not observed. It may also have occurred to you that
there are:
•
•
clusters of gases and solids, metals and non-metals,
elements with some common properties, for example, fluorine and chlorine are both
reactive diatomic gases.
In the 1860s, several chemists reported the observation of a number of groups of
elements with similar properties. One of the chemists who brought these observations to
the notice of the scientific world called them triads, because most of the groups had
three elements. These triads included lithium/sodium/potassium and chlorine/bromine/
iodine. The properties that were used to link the elements in the triads included:
•
•
•
•
•
metallic character
molecular form if non-metal
reactivity
nature of oxygen and chlorine compounds
charge of ions
EXAMPLE
What properties do the elements lithium, sodium and potassium have in common?
•
•
•
•
•
•
•
light grey in colour
metallic
soft and low density (for metals)
very reactive
form 1+ ions
oxygen compounds have two atoms of the element for every one oxygen atom
oxygen compounds are alkaline.
34
Chapter 3
Organising the Elements
LIBRARY WORK
Look up the properties of chlorine, bromine and iodine. What do they have in
common?
About the same time as these observations were being reported, a breakthrough was
made in the measurement of reasonably accurate atomic weights for the known elements,
which numbered around 60. When one chemist arranged the elements in order of
increasing atomic weight, a grid appeared, with the members of the triads lining up under
one another. Figure 3.3 shows such an arrangement of some of the elements with the
triads in columns, and atomic weight increasing across a row.
H
1
Li
7
Be
9
B
11
C
12
N
14
O
16
Mg
24
Al
27
Si
28
P
31
S
32
Cl
35.5
K
39
Ca
40
As
75
Se
79
Br
80
Rb
85
Sr
88
Te
127
I
127
Cs
133
Ba
137
Cu
63
Zn
65
Ag
108
Cd
112
Figure 3.3
Na
23
Grid arrangement of the elements showing the triads (in columns) and atomic weights
(of the time)
As more elements were discovered, they were slotted into positions in the element
grid. There were a few that didn’t quite seem to fit, but the overall pattern was
sufficiently convincing that most scientists agreed that the grid was a breakthrough in
organising the elements, and in summarising their properties. Some of the discrepancies
were later found to be caused by incorrect atomic weights, but the main problem was
solved when atomic numbers were employed.
PRACTICE QUESTION
2. Fluorine was discovered 20 years after the first element grids appeared. It was
known in compound form for many years, but could not be purified until the
1880s. It was expected to belong with chlorine, bromine and iodine. Predict the
following properties of fluorine: metallic character, charge of ion, reactivity,
number of atoms in molecules of element, number of atoms in oxygen
compound, acidity of oxygen compound.
35
Chemistry for Technicians
3.3
A FIRST LOOK AT THE PERIODIC TABLE
The arrangement of the elements in Figure 3.3 is the basic structure of the most
significant development in chemistry of the nineteenth century: the periodic table. It was
proposed by two chemists, a Russian, Dmitri Mendeleev, and a German, Lothar Meyer,
working independently of each other in the 1860s.
Mendeleev has been given most of the credit for the periodic table. This was in part
due to luck (Meyer’s article in a scientific magazine was delayed because he changed
jobs), and partly because Mendeleev was vigorous in his attempts to convince the
scientific world of the value of his periodic table. He made a number of predictions about
undiscovered elements and was prepared to move some elements around to fit the pattern
better, even though this disrupted the atomic weight sequence.
The periodic table now contains 107 elements, arranged in rows (or
periods, numbered 1 to 7 down the table) of increasing atomic number, and
atomic number, and columns (or groups or families, labelled 1A to 8A and
1A to 8A and 1B to 8B) which contain up to six elements of similar
similar properties. There are no obvious flaws — elements that don’t really
belong in their position, according to atomic number.
A periodic table has been included on the inside cover of this book for easy
reference. Its shape is more complex than that first mapped out by Mendeleev, but the
basic principles remain.
PRACTICE QUESTIONS
3.
(a) List all the elements in one group of the periodic table. Give the code
number for that group.
(b) List all the elements in one row of the periodic table. Give the number of
that row.
4.
Which elements are out of place in Figure 3.3 (Mendeleev’s first periodic table)
by comparison with the modern table?
The groups headed 1A to 8A contain the first 20 elements, and are those on which
we will concentrate in this book. The chemistry of the other groups is more complex, and
is briefly outlined in Part B of this book. Some of the main groups have commonly used
names, which are shown in Table 3.1.
Table 3.1
Group
Common names for main groups
Common name
1A
Alkali metals
2A
Alkaline earth metals
7A
Halogens
8A
Inert (or noble gases)
36
Chapter 3
Organising the Elements
3.3.1 What Can We Use the Periodic Table for?
At this level, we can use the periodic table for two main purposes:
•
as a summary of atomic weights and numbers, symbols and names for the elements,
as shown in Figure 3.4, and
to help in our knowledge of the properties of elements, since elements in the groups
share many properties in common, as summarised in Table 3.2.
•
69
Tm
Thulium
168.93
Atomic number
Symbol
Name
Atomic weight
Figure 3.4 Element information contained in the periodic table
Note that some elements have their atomic weight in brackets. These elements have no
stable or naturally occurring isotopes, so the figure given is the mass number of the least
unstable isotope.
PRACTICE QUESTION
5. Find the required information about the following elements:
(a) the symbol for platinum
(b) the name of the element with the symbol Hf
(c) the atomic number of gold
(d) the atomic weight of iodine.
Table 3.2
Characteristic properties of groups 2A to 8A
Group
Characteristic properties
2A
Form 2+ ions; metals; quite reactive, form alkaline compounds with oxygen, form ionic
compounds with groups 5A–7A
3A
Form 3+ ions; metals; moderately reactive; oxygen compounds can be acidic or
alkaline; form ionic compounds with groups 5A–7A
4A
Do not form monatomic ions; form compounds with four hydrogen atoms; solids
5A
Form 3– ions; form ionic compounds with groups 1A–3A but are very reactive; form
non-ionic compounds with groups 5A–7A; form compounds with three hydrogen atoms;
solids except nitrogen; oxygen compounds are acidic
6A
Form 2– ions; form stable ionic compounds with groups 1A–3A; form compounds with
two hydrogen atoms; low melting point solids except oxygen; oxygen compounds are
acidic
7A
Diatomic molecules; form 1– ions; form compounds with one hydrogen atom; highly (F)
to fairly (I) reactive; form stable ionic compounds with groups 1A–3A; form non-ionic
compounds with groups 5A–7A
8A
Monatomic gases; unreactive
37
Chemistry for Technicians
This is only a very brief introduction to the periodic table. The module Chemical
Principles covers a great deal more on the subject, and is addressed in Chapter 7 this
book.
PRACTICE QUESTIONS
6. Water (the hydrogen compound of oxygen) has two hydrogen atoms for every
one oxygen. What do you think the case would be for the other elements in the
same group as oxygen?
7. From the following descriptions, link each element to a group in the periodic
table.
Element A is a monatomic gas, which is unreactive.
Element B is a light moderately reactive metal, which forms 2+ ions.
Element C is a diatomic molecule, liquid and highly reactive at room
temperature. It forms 1– ions.
3.4
THE PERIODIC TABLE AND ELECTRON CONFIGURATIONS
If you look back to Table 2.4, you will notice that lithium and sodium, which are in the
same group in the periodic table, have a common feature with their electron
configurations. They both have one outer shell electron. Practice Question 12 in Chapter
2 showed the same occurs for the third member of the group, potassium. Likewise,
oxygen and sulfur, which are in another group, each have six electrons in their outer
shell. This holds for the groups containing the elements we have looked at (the first 20):
elements in a group have the same number of electrons in their outer shell.
This is not a coincidence, since the electron shell theory was, in fact, developed
from the shape of the periodic table. It was considered that the fact that lithium, sodium
and potassium each lost one electron to form their ions meant that the electrons must be
organised so that a single electron was ‘exposed’ in these elements, and that the other
electrons were somehow ‘protected’. The same story applied to other groups, so the shell
theory was developed.
PRACTICE QUESTION
8. For the eight groups that contain the first 20 elements, construct a table that
links the group to the number of electrons in the outer shell.
WHAT YOU NEED TO BE ABLE TO DO
•
•
•
•
•
Define the terms group and period.
Explain how the periodic table arrangement of the elements was developed.
Recognise characteristic properties of groups of elements.
Describe changes in properties of elements across the third row of the table.
Relate positions in the periodic table to electron configuration.
38
Chapter 3
Organising the Elements
TERMS AND DEFINITIONS
During this chapter, you have met a number of new terms, listed below. Match them with
their correct definitions.
Terms
A triad
C period
B periodic table
D group
Definitions
1
a column of elements in the periodic table
2
a group of three elements with similar properties
3
a row of elements in the periodic table
4
an arrangement of the known elements by ascending atomic number and similar
properties
39
4
NAMES AND FORMULAE
PURPOSE
To develop skills in the writing of formulae and names for simple compounds.
The story so far: 107 elements in various combinations make a lot of compounds —
there had better be some simple way of labelling them.
Just as we need a language in everyday life to communicate with others, chemists
need a language to talk about the various substances that are used. Part of the chemist’s
language is a system that gives a name to every different pure substance (element or
compound), and also a formula which tells us the number and type of each atom in the
substance.
The most important aspect of the system is that it is unique in both directions, that
is, each compound has a unique name, and from that name, we can write the formula.
That means we don’t need to remember both the name and formula of a compound — the
system allows us to work out one from the other.
4.1
CHEMICAL FORMULAE
What distinguishes one chemical substance from another? The answer is
the type and number of atoms in it (and also the way they are joined
joined together, but more of that in later chapters). A chemical formula
chemical formula lists precisely that information, so that we can rapidly
can rapidly know something of its nature very easily by looking at its
formula.
A chemical formula is written in a consistent way so that anyone with chemical
knowledge can interpret its meaning. The formula of the element hydrogen — a
diatomic molecule - is shown in Figure 4.1.
Symbol for
element hydrogen
H2
Subscript after symbol
indicates that two hydrogen
atoms are present
Figure 4.1 Formula of hydrogen gas
When we write the formula for a compound, we write the symbol for each element
present. Figure 4.2 shows the formula for water. The order of writing the symbols varies
depending on what type of compound we are looking at. The idea of different types of
compounds and the rules associated with writing names and formulae will be covered in
the next section.
40
Chapter 4
Names and Formulae
H2O
Symbol for
element oxygen
No subscript after symbol
indicates that one oxygen
atom is present
Figure 4.2 Formula of water
Thus, the important components of a formula of a molecule are:
•
•
the symbols for each element involved, and
a numerical subscript after the symbol indicating the number of atoms of that
element (where no subscript is written, one atom is present).
In the previous chapter, you were introduced to the idea of ions: substances like
atoms or molecules, but with a positive or negative charge. The formulae of ions not only
need to include types and numbers of atoms, but also the size and sign of the charge.
Figure 4.3 shows the formula of a simple monatomic ion, hydrogen, and a polyatomic
ion, sulfate, which contains one sulfur atom and four oxygen atoms, and has a total
charge of 2–.
+
H
A superscript indicates
the size and sign
of the charge
on the ion
SO42-
Figure 4.3 Formulae of hydrogen and sulfate ions
Substances containing ions are a very important class of compounds, and will be one of
the types of compounds whose names are covered in this chapter. There are many ions,
positive and negative, and Table 4.1 below lists the most common, with which you
should become familiar.
PRACTICE QUESTIONS
1. Explain what is meant by the formulae:
(a) O3
(b) SO2
(c) C2O42–
(d) VO2+
(e) KMnO4
(f) C9H12Br3N4O6.
2.
Write a formula for:
(a) an ion which contains one iron atom, four chlorine atoms with a single
negative charge
(b) a compound with three carbon atoms, four hydrogens, two oxygens and a
nitrogen
(c) an ion which contains one vanadium atom, two oxygen atoms and a double
positive charge.
41
Chemistry for Technicians
Table 4.1
Names and formulae of common ions
Cations
Name
Formula
Name
Formula
Name
Formula
Hydrogen
H+
Calcium
Ca2+
Iron (III)
Fe3+
Sodium
Na+
Magnesium
Mg2+
Aluminium
Al3+
Potassium
K+
Copper
Cu2+
Ammonium
NH4+
Zinc
Zn2+
Lead
Pb2+
Anions
Name
Formula
Name
Formula
Name
Formula
Fluoride
F–
Oxide
O2–
Phosphate
PO43–
Chloride
Cl–
Carbonate
CO32–
Bromide
Br–
Sulfide
S2–
Iodide
I–
Sulfate
SO42–
Hydroxide
OH–
Sulfite
SO32–
Hydrogen
carbonate
HCO3–
Nitrate
NO3–
4.2
TYPES OF COMPOUNDS
There are a range of different classes of compounds, a division based on the way that
their atoms are connected, the type of elements and their properties. For each different
class of compounds, there is a different naming system, unfortunately. You will need to
be familiar with two basic types:
•
ionic — compounds containing cations and anions, for example, sodium chloride,
NaCl, and
•
binary molecular — where the atoms are held together differently, for example
water, H2O, and only two types of elements are involved.
If you are given a formula, you need to be able to come up with a name, and vice versa.
How can you tell ionic and binary molecular compounds apart at this stage of your
studies? Figure 4.4 shows you how. (Once your knowledge of the properties of different
substances increases, you will be able to recognise ionic and molecular compounds
without the need for this flowchart, but this takes time.)
EXAMPLE
Use the flowchart in Figure 4.4 to determine the type of the following compounds: nitrogen
dioxide, magnesium sulfate, NH4NO3, CO2.
Compound
nitrogen dioxide
magnesium sulfate
NH4NO3
CO2
Sequence of answers from
flowchart
Type of compound
name Õ no Õ
name Õ yes Õ
formula Õ yes Õ
formula Õ no Õ no Õ
binary molecular
ionic
ionic
binary molecular
42
Chapter 4
Names and Formulae
Are you looking at
a name or a formula?
Formula
Name
Does the compound have
more than two elements?
Does the first part of the name
appear in your list of cations?
No
Yes
Yes
Does the first
element appear in your
list of cations?
Yes
IONIC
IONIC
No
BINARY
MOLECULAR
No
BINARY MOLECULAR
Figure 4.4 Flowchart for identifying binary molecular and ionic compounds
PRACTICE QUESTION
3. What types of compound (ionic or binary molecular) are the following:
(a) nitrogen tribromide, (b) CuI2, (c) sodium nitrate, (d) OBr2.
4.3
BINARY MOLECULAR COMPOUNDS
As with all rules and systems, there are exceptions to make life difficult. Water is a prime
example: no one calls it by its systematic name, dihydrogen monoxide. Ammonia (NH3)
is similar.
4.3.1 Names from Formulae
The basic rule for writing a name from a formula is that the first element in the
formula is named first, the second element second with the ending -ide replacing
part of the element’s name.
EXAMPLE
What are the names for chlorine, nitrogen and oxygen when occurring as the second part
of a name?
Chlorine becomes chloride.
Nitrogen becomes nitride.
Oxygen becomes oxide.
43
Chemistry for Technicians
PRACTICE QUESTION
4. What are the names for fluorine, bromine, iodine and sulfur when modified to
form the second word in a binary molecular name?
EXAMPLES
1.
What is the name of the compound corresponding to the formula CO?
The name would be carbon oxide.
2.
What name would correspond to the formula CO2?
It would seem (without any other information) that the name would again be carbon
oxide.
The examples above illustrate a problem. At this point, our naming system has
given the same name to two different compounds. This is not desirable. The difference
between the two compounds is the number of atoms of oxygen. We need to indicate how
many atoms of each element are present in the molecules. Prefixes before the name are
used for this purpose, and are given in Table 4.2.
Table 4.2
Prefixes used to indicate numbers of atoms (letters in brackets may
not always appear in the name in order to improve pronunciation)
1
mon(o)
2
di
3
tri
4
tetr(a)
5
pent(a)
6
hex(a)
7
hept(a)
8
oct(a)
EXAMPLE
Give the name corresponding to the following formulae: CO, CO2, NO, SO3, IBr
Formula
Name
CO
carbon monoxide
CO2
carbon dioxide
NO
nitrogen monoxide
SO3
sulfur trioxide
IBr
iodine monobromide
Prefixes are used sometimes when one atom is present, but not always, which is
rather confusing. You will never be wrong to always use mono! In general, if the first
element has only one atom in the compound, the prefix can be left off, but should always
be added to the second element.
44
Chapter 4
Names and Formulae
PRACTICE QUESTIONS
5. Name the following molecular compounds:
(a) PCl5
(b) NO2
(c) SF6
(d) P2O5
(e) CS2.
6. Water and hydrogen peroxide (H2O2) are the common acceptable names for these
compounds. What are their systematic names?
4.3.2 Formulae from Names
Writing the formula for a molecular compound from its name is very easy. The name
tells you what elements and how many atoms of each are present.
EXAMPLE
Give the formulae corresponding to the following names of molecular compounds:
dinitrogen tetroxide, dihydrogen sulfide, carbon tetrachloride.
Name
Formula
dinitrogen tetroxide
N2O4
dihydrogen sulfide
H2S
carbon tetrachloride
CCl4
PRACTICE QUESTION
7. Write the formula for the following binary molecular compounds.
(a) silicon dioxide
(b) oxygen diiodide
(c) boron trifluoride.
4.4
IONIC COMPOUNDS
Ionic compounds are composed of positive (cations) and negative (anions) ions. You
cannot have a bottle of positive ions, or of negative ions. In any ionic compound, there
will always be the number of cations and anions so that the overall charge of the
compound is zero, that is, the positive charge cancels out the negative charge.
EXAMPLE
In the compound aluminium chloride, what is the ratio of cation to anion?
Aluminium ion has a 3+ charge, while chloride ion is 1–. For every one 3+ ion, there must
three 1– ions to balance the charge. Therefore, the ratio is 1:3.
45
Chemistry for Technicians
PRACTICE QUESTIONS
8. Construct a table with the columns headed A+, B2+ and C3+ and the rows labelled
D–, E2– and F3–. Fill the grid positions representing the intersection of a column
and a row with the numbers of each cation and anion that would make up a
neutral ionic compound. For example, the compound from B2+ and D– would
have 1 cation (total charge 2+) for every 2 anions (2 × 1– = 2–).
9. Which of the following correctly describes the compound containing Zn2+ and
Cl– ions?
(a) one zinc and one chlorine
(b) two zincs and one chlorine
(c) one zinc and two chlorines
(d) no definite number.
4.4.1 Names from Formulae
The basic rule for writing a name from a formula is cation name first, and anion name
second. Thus, you need to be able to recognise the ions that make up the compound with
that formula. To do this, you will need Table 4.1 to help you until you become familiar
with the common ions. There are no prefixes in ionic names, so once you have
recognised the ions, name them in the correct order, and it’s done!
EXAMPLE
What are the names of the compounds corresponding to the following formulae: NaCl,
NaNO3?
NaCl — the two elements are sodium and chlorine; chlorine in ionic form is called
chloride, and its name is second. Thus, the name is sodium chloride.
NaNO3 — sodium is recognisable; the anion is a polyatomic ion called nitrate (looking
back at Table 4.1). The name is sodium nitrate.
If there is more than one of a particular polyatomic ion, you must put brackets
around it to separate its formula from the subscripted number, indicating how many of
that ion are in the compound. Figure 4.5 shows an example, and the exact meaning of
the formula.
Ca(NO3)2
The brackets enclosing the
nitrate ion mean that the
subscript number 2 outside the
brackets refers to the entire ion;
this compound has 1 Ca, 2 N
and 6 O atoms
Figure 4.5 Formula involving brackets
EXAMPLE
If brackets weren’t used in Figure 4.5, what would the formula mean?
The formula would become CaNO32, indicating that the compound contains one calcium,
one nitrogen and 32 oxygen atoms.
46
Chapter 4
Names and Formulae
PRACTICE QUESTION
10. How many atoms of each element are present in the following compounds?
(a) (NH4)2CO3
(b) Cu(NO3)2.
EXAMPLE
Give the names for the following compounds: KBr, CaCl2, NH4NO3, Al2(SO4)3, FeCl2,
FeCl3.
Formula
Cation
Anion
Name
KBr
K
Br
potassium bromide
CaCl2
Ca
Cl
calcium chloride
NH4NO3
NH4
NO3
ammonium nitrate
Al2(SO4)3
Al
SO4
aluminium sulfate
FeCl2
Fe
Cl
iron chloride
FeCl3
Fe
Cl
iron chloride
What do you notice about the last two examples above? They have different
formulae, but the same name. According to the system, this can’t be correct: different
formulae must have different names. The problem arises because iron and copper ions
exist as two different charges each: Fe2+ and Fe3+, Cu+ and Cu2+. Compounds involving
these ions must be named to indicate the charge of the cation, and this is done with
roman numerals in brackets after the cation name, that is, iron (II), iron (III), copper (I)
and copper (II).
How can we tell from the formula which charged ion is involved? Look at the
number and charge of anions in the compound. Work out how much negative charge is
involved. There must be an equal amount of positive charge, and this will be evenly
divided if there is more than one cation.
EXAMPLE
What is the name corresponding to the formula Fe2(CO3)3?
There are three carbonate ions, each of 2– charge. Therefore, the total charge is 6–.
The two iron cations must have a total charge of 6+, and thus are 3+ each.
The name is iron (III) carbonate.
PRACTICE QUESTION
11. Name the following ionic compounds.
(a) NH4Cl
(b) Na2CO3
(c) CuSO4
(d) FeO
(e) Mg3(PO4)2
(f) Fe2S3
47
Chemistry for Technicians
4.4.2 Formulae from Names
The cation is always named before the anion, and this carries through to the formula. To
determine the formula, write down the formula for each ion, including charge, in the
correct order.
If the ions have the same numerical charge, then the formula is one of each, for
example calcium oxide, which contains Ca2+ and O2– ions. Since each ion has a charge of
2, the formula is CaO.
If the charges on the ions are not the same value, then you can either use simple
maths to determine how many of each ion occurs in the compound or use the ‘cross-over’
technique, as in Figure 4.6.
C m+
The numerical value of the
charge on the cation is the
number of anions in the
compound, and vice versa.
A n–
CnAm
Figure 4.6 The ‘cross-over’ technique for determining ionic formulae
EXAMPLES
1.
What is the formula for aluminium oxide?
Aluminium is Al3+ and oxide is O2–.
The charge of 3 on the
aluminium tells us that
there must be three oxide
ions
Al3+
O2-
The charge of 2 on the oxide tells
us that there must be two
aluminium ions
Al2O3
2.
Give the formulae for the following ionic compounds: calcium iodide, ammonium
hydroxide, potassium carbonate, calcium sulfate, zinc nitrate.
Name
Cation
Anion
Formula
calcium iodide
Ca2+
I–
CaI2
ammonium hydroxide
NH4+
OH–
NH4OH
potassium carbonate
K+
CO32–
K2CO3
calcium sulfate
Ca2+
SO42–
CaSO4
zinc nitrate
Zn2+
NO3–
Zn(NO3)2
48
Chapter 4
Names and Formulae
PRACTICE QUESTIONS
12. Write the formulae for the following ionic compounds.
(a) ammonium sulfate
(b) magnesium bromide
(c) iron (III) oxide
(d) copper (II) hydrogen carbonate
(e) zinc sulfite.
13. Given the formulae for some unfamiliar ions below, work out the correct
formulae for the following compounds:
(a) vanadyl vanadate
(b) thorium permanganate
(c) thallium dichromate.
Vanadyl
VO2+
Dichromate
Cr2O72–
4+
Thorium
Th
Vanadate
VO43–
3+
Thallium
Tl
Permanganate MnO4–
14. Give the names corresponding to the following formulae:
(a) SiCl4
(b) Fe(NO3)3
(c) Ca(HCO3)2
(d) N2O.
15. Give the formulae corresponding to the following names:
(a) carbon tetrafluoride
(b) sulfur trioxide
(c) aluminium hydroxide
(d) zinc carbonate
(e) aluminium phosphate.
4.5
COMMON ACIDS AND BASES
All the names above refer to compounds — pure substances. There are a few mixtures —
important acid or base solutions in water — which are so commonly used that they have
names. You will need to remember them, both name and formula. Table 4.3 lists these
substances.
Table 4.3
Names and formulae for common acids and bases
Name
Formula
Hydrochloric acid
HCl
Name
Nitric acid
HNO3
HClO4
Sulfuric acid
H2SO4
Perchloric acid
Phosphoric acid
H3PO4
Ammonium hydroxide
4.6
Formula
NH3
FORMULA WEIGHTS
You have met the terms atomic weight and formula. If we combine them,
we get the term formula weight (FW), which is the average total weight of
total weight of the atoms combined in a given compound, and is simply
is simply calculated as the sum of the individual atomic weights of each
weights of each atom. You will need to refer to the periodic table on the
inside cover for the atomic weights.
49
Chemistry for Technicians
How is the formula weight calculated? Firstly, count how many atoms of each
element are present, then multiply the atomic weight for each element by the number of
atoms of that element, and then add all these together.
EXAMPLE
Calculate the formula weight of the following substances: KI, SO3, Al2(SO4)3.
KI has one potassium atom (AW 39.10) and one iodine atom (126.91). Its formula weight
is 39.10 + 126.91 = 166.01.
SO3 has three oxygen atoms and one sulfur atom. Therefore, its atomic weight is 16.00 +
16.00 + 16.00 + 32.06 = 80.06.
Al2(SO4)3 is more complicated, so to be safe, a table can be used to avoid missing any
atoms. There are three sulfate ions, which means three sulfur atoms and 12 oxygen
atoms (three ions each with four oxygen atoms).
Element
Number of atoms
AW
Total weight
Al
2
26.98
2 x 26.98 = 53.96
S
3x1= 3
32.06
3 x 32.06 = 96.18
O
3 x 4 = 12
16.00
12 x 16
= 192.00
TOTAL
PRACTICE QUESTIONS
16. Calculate the formula weight of each compound in Q5.
17. Calculate the formula weight of each compound in Q11.
WHAT YOU NEED TO BE ABLE TO DO
•
•
•
•
•
•
Interpret chemical formulae.
Distinguish between ionic and molecular compounds.
Write names from formulae for ionic and molecular compounds.
Write formulae from names for ionic and molecular compounds.
Recall the names and formulae for common acid solutions.
Calculate formula weights.
50
342.14
Chapter 4
Names and Formulae
TERMS AND DEFINITIONS
During this chapter, you have met a number of new terms, listed below. Match them with
their correct definitions.
Terms
A
chemical formula
C
formula weight
B
systematic name
Definitions
1
the sum of the atomic weights of all atoms in a molecule
2
a means of summarising the type and number of elements in a substance
3
a unique identifier for every known compound
51
5
PROPERTIES AND STRUCTURE
PURPOSE
To examine the properties of a range of pure substances and the structures causing the
differences.
The story so far: the foundations are in place, so let’s join some of the building blocks
together. The variety is almost endless, but are there some common elements (pun
intended)?
5.1
A SURVEY OF THE PROPERTIES OF PURE SUBSTANCES
There are millions of pure substances (most of them are compounds), but some
classification by properties can simplify what would otherwise be an extremely
confusing area. A question then presents itself: why should different compounds have
similar properties? Or perhaps a better question is: why do atoms hang around in
‘groups’ (molecules/compounds) in the first place? Water, for example, is not simply a
collection of atoms, two-thirds of them being hydrogen and the other third being oxygen.
The answer to the second question is that they are held together by bonds.
A bond is a connection between two atoms, brought about by changes in
changes in the electron arrangements of the atoms involved (little more will
(little more will be said about the part that electrons play in bonding — this
bonding — this is left to the module Chemical Principles in Chapter 9).
Bonds form to increase the stability of the collection of atoms (similarly no
more will be said on this topic until Chapter 9).
A bond may be strong or weak, depending on what is causing it to happen; it may
require lots of energy to break it (a strong bond), or it may fall apart readily (a weak
bond). Without bonding, atoms and molecules would drift apart from each other.
Bonds can be broken and the atoms they hold together released from the grip. This
requires energy, which may be of a variety of forms, including heat, radiation and
electricity. The stronger a bond, the more energy is required to break it. Some bonds are
so weak that a temperature of –269°C is enough to break them; others are not ‘cracked’
without the temperature being raised to many thousands of degrees.
To link the measurable properties of substances with the bonding, a number
bonding, a number of substances will be compared. They have been chosen
been chosen to illustrate the wide range of structures and bonding possible
bonding possible in pure substances. A structure is the way that the atoms
that the atoms are linked to each other. It is possible to have two
compounds with the same number and types of atoms, but with
substantially different properties, because of the different ways that the
atoms are bonded.
52
Chapter 5
The 12 substances we wish to study are:
•
ammonium chloride
•
ethanol (alcohol)
•
hydrogen chloride
•
naphthalene (in mothballs)
•
sodium
•
sulfur
Properties and Structure
•
•
•
•
•
•
diamond
helium
iron
oxygen
sodium chloride
water
Using some of the simple classifications outlined in Chapter 1, these substances can be
divided up by the characteristics listed below.
5.1.1 Physical State
Solid (7)
sulfur, diamond, iron, sodium, sodium chloride, ammonium
chloride, naphthalene
Liquid (2)
water, ethanol
Gas (3)
helium, hydrogen chloride, oxygen
5.1.2 Metallic Character
Metal (2)
iron, sodium
Non-metal (10)
sulfur, diamond, sodium chloride, ammonium chloride, water,
ethanol, helium, hydrogen chloride, naphthalene, oxygen
5.1.3 Type of Pure Substance
Element (6)
sulfur, diamond, iron, sodium, helium, oxygen
Compound (6)
— Ionic (2)
sodium chloride, ammonium chloride
— Molecular (4)
hydrogen chloride, water, ethanol, naphthalene
Further classification will be needed to identify some of the differences in the way that
these 2 compounds are bonded together. These other methods involve measurement of
certain physical properties, including melting and boiling points, density, solubility and
electrical conductivity.
5.1.4 Melting and Boiling Points
To an extent, these substances have already been separated on the basis of these two
measurements, that is, by their physical state, since the distinction between solid, liquid
and gas is due to the relationship of a substance’s melting and boiling points to room
temperature. However, it is useful to look at the actual values, because within a particular
physical state group (particularly solids), some important variations occur. Table 5.1 lists
the melting and boiling points for the 12 substances.
What do these results say? Firstly, the trends displayed by melting and boiling
points are very similar. Secondly, it is clear that there is a wide range within the
classification of solid. It is not simply a division between metal and non-metal, since
metallic sodium has a lower melting point than non-metallic sulfur, while non-metallic
diamond has a much higher melting point that metallic iron. The ionic compounds are
mid-range, while the molecular compounds are lower.
53
Chemistry for Technicians
Table 5.1
Melting and boiling points for survey substances
Substance
Melting point (°C)
Boiling point (°C)
n/a 1
–269
Helium
Oxygen
–218
–183
Hydrogen chloride
–115
–85
Ethanol
–117
78
0
100
Naphthalene
81
218
Sodium
98
883
113–120
445
Water
Sulfur
Ammonium chloride
340 2
Sodium chloride
801
1413
1535
2750
approx. 3600
not known
Iron
Diamond
n/a
Notes
1.
Helium is the only known substance which cannot be solidified under normal conditions.
2.
Ammonium chloride sublimes at this temperature: that is, it passes directly from solid to
gas, without passing through the liquid state. It, therefore, has no boiling point.
5.1.5 Density
When looking at densities, it is essential that the comparison is done on a level playing
field: all in the same physical state. This is not as easy as it seems, given the variety of
behaviour undergone by the substances during changes of state. The solid state is the
most useful choice, but this, of course, eliminates helium. The density of solid hydrogen
chloride was not available, and that for ethanol is only an estimate. Table 5.2 lists the
densities of the substances as solids.
The density of a substance is affected by two factors: the atomic weight of the
atoms, and how close the atoms are to each other. This makes interpretation of the data
somewhat difficult. It is obvious that the density of iron is substantially higher than the
other substances, and this cannot be due entirely to the higher atomic weight of iron
atoms. It is a characteristic of most metals, sodium being one of the exceptions.
Diamond, despite the relatively low atomic weight of carbon, has a high density for a
non-metal, which indicates that its atoms must be quite close together.
Table 5.2
Densities of survey substances
Density range (g/cm3)
Substances
0.8–1.0
Naphthalene, water, sodium
1.0–1.5
Oxygen, ammonium chloride
1.5–2.5
Sulfur, sodium chloride
2.5–5.0
Diamond
> 5.0
Iron
54
Chapter 5
Properties and Structure
5.1.6 Solubility
Two vastly different solvents are chosen to test the survey substances for solubility:
water and petrol. Unfortunately, it is a bit difficult to explain here why they are different,
so unfortunately, you will have to accept that they are. Table 5.3 lists the solubilities.
Table 5.3
Solubility of survey substances
Substance
Water
Petrol
high
nil
nil
nil
Ethanol
high
nil
Helium
nil
nil
high
nil
Iron
nil
nil
Naphthalene
nil
high
Oxygen
slight
nil
Sodium
nil
nil
high
nil
Sulfur
nil
medium
Water
n/a
nil
Ammonium chloride
Diamond
Hydrogen chloride
Sodium chloride
The variety in behaviour among the non-metallic substances is significant. It would
appear that naphthalene and sulfur have some common property that allows them to
dissolve only in petrol. The ionic substances share a solubility in water with others that
are not ionic.
5.1.7 Electrical Conductivity
This is the ability to conduct electrical current, and has been found to occur in some
solids, some liquids and in the aqueous solutions of some substances. This being the
case, this survey will consider each physical state.
Solid: only the metals (iron and sodium) conduct electricity in the solid state. This is a
characteristic of metals.
Liquid: again the metals are conductive, but also molten sodium chloride.
Aqueous solution: only those substances soluble in water can be considered; water itself
is non-conductive, and so is an aqueous solution of ethanol. The three chloride
compounds are conductive when in aqueous solution.
The ability of a substance to conduct electricity has been linked to the presence in
the substance of charged species which are capable of movement. You are aware that
ionic compounds have positive and negative components, but there is an obvious
difference between solid and liquid sodium chloride in terms of the ability of the ions to
move.
55
Chemistry for Technicians
5.1.8 Overall Conclusions
•
metals clearly have similar structures
•
non-metals have a wide variety of structures, but within that classification, ionic
substances appear to have similarities
•
there must be differences in structure between the non-metallic non-ionic
substances to explain the wide variation in properties
•
the distinction between element and compound is irrelevant when considering
properties and structure.
Having looked in some depth at the range of properties that these relatively simple
substances display, it’s time to introduce you to the different types of structures and
bonds. In doing so, most of the properties can be explained.
5.2
IONIC SUBSTANCES
From Chapters 2 and 4, you are familiar with the idea that some compounds contain
positively and negatively charged species called ions. They form when electrons are lost
or gained by atoms. These compounds, not surprisingly, are called ionic compounds. By
the way, there are no ionic elements: no element is able to form both anion and cation.
Positively and negatively charged objects (regardless of size) attract each
other. This is known as electrostatic attraction. Similarly charged objects
charged objects repel each other: electrostatic repulsion. The cations and
cations and anions in ionic compounds are held together by the attractive
the attractive force between positive and negative: this is known as an ionic
bond.
In the solid state, a large three-dimensional structure of alternating positive and
negative ions is produced, as shown in Figure 5.1. It is often known as the crystal lattice.
The arrangement of cations and anions in the crystal lattice maximises positive–negative
contact and minimises positive–positive and negative–negative repulsion (electrostatic
attraction and repulsion is affected substantially by distance of separation). The regular
geometric shape of the lattice is what causes the crystal shape in table salt, as well as
many other substances.
Na+
Cl–
Figure 5.1
Ions in crystal lattice of NaCl
56
Chapter 5
Properties and Structure
If the ratio of cations to anions in the compound was other than 1:1 (e.g. CaCl2),
then numbers of ions of each type in the above picture would vary (one Ca2+ for every
two Cl–), but the basic idea that positives sit next to negatives is retained.
In an ionic compound in the solid state, the concept of a molecule as being a single
unit of fixed atomic composition is not really valid, since who can tell which cation
belongs to which anion. If you look carefully at Figure 5.1, and in particular at the anion
in the very centre of the cube, you should notice that every cation (14 of them) in that
cube is adjacent to that anion. But the formula of sodium chloride isn’t Na14Cl! If another
cube was drawn with the positions of the ions reversed, the same could be said about the
central cation.
So, individual NaCl molecules don’t exist in the solid state. This is a feature of
all ionic compounds, and is important in affecting the overall properties of the
compound. Ionic bonds are relatively strong, and every bond in NaCl is as strong as any
other. This is significant in making the melting point of ionic substances relatively high.
At room temperature, all ionic compounds are solid.
The density is moderately high, because of the regular arrangement, but the ions
can’t get too close, because of the repulsions. The ions are not able to move because of
the bonds holding them in place: no electrical conductivity can occur.
The only movement that occurs in a crystal lattice is the vibration of the
ions: they shake but they can’t break free. When an ionic compound is
compound is heated, the amount of vibration increases. Eventually, the ions
Eventually, the ions shake themselves free of their shackles, and begin to
and begin to move around randomly. This is the melting point of the
substance: the temperature at which the individual species have enough
energy to break free of the bonds that hold them in the solid form.
The amount of heat energy required to melt ionic compounds is great, because the
bonds are quite strong, and there are many of them. The crystal lattice is destroyed by the
heat, but enough electrostatic attraction remains to keep the ions from completely
separating. They ‘swim’ around in a random fashion, now able to conduct electricity.
When an ionic compound dissolves in water, its conductivity indicates that the ions
are free to move, as in the liquid (molten) form. In this case, water has the remarkable
ability to break up the lattice, even without the addition of heat. This is shown in Figure
5.2. So, still no molecules of NaCl will exist. In general, other solvents are not able to
penetrate the lattice and therefore can’t dissolve ionic compounds.
Cation
Anion
Figure 5.2
Ionic compounds in aqueous solution
57
Chemistry for Technicians
So in the solid state, in liquid form and in solution, no individual molecules
of an ionic compound occur. What does this say about the formula of an
ionic compound, for example, NaCl? It is the simplest ratio of cation to
anion.
In the gas phase, the behaviour of ionic compounds is relatively complex and will
not be dealt with. It is, after all, an extreme condition, requiring very high temperatures.
EXAMPLE
Which of the following statements about Al2(CO3)3 are true?
(a)
(b)
(c)
(d)
There are two aluminium ions for every three carbonate ions in pure aluminium
carbonate.
There are three aluminium ions for every two carbonate ions in pure aluminium
carbonate.
When aluminium carbonate melts, molecules break free of the crystal lattice and
move around freely.
When aluminium carbonate melts, aluminium, carbon and oxygen ions are
produced.
The formula indicates two aluminium and three carbonates by the subscripts so (a) is true
and (b) is false.
(c) is false, because molten ionic compounds release the ions to move freely.
(d) is ‘half true’ and ‘half false’ because the aluminium ions are released, but the C and O
atoms remain as carbonate ions.
PRACTICAL WORK
You are required to test the following properties of the ionic compounds
potassium chloride, sodium nitrate, calcium carbonate, lead sulfate.
•
•
•
•
melting (or boiling) point
solubility in water and trichloromethane (or similar organic solvent)
conductivity in aqueous solution
density.
Melting point
Estimate the melting point using the following steps:
1.
2.
Place a small amount on the end of a metal spatula
Heat it in a hot Bunsen flame — if it melts, then its melting point is less than
1000°C; if it doesn’t, then the melting point is greater than 1000°C.
Solubility and conductivity in water
3. Place a ‘spatula-full’ or 2 Pasteur pipette volumes of the compound into a
150 mL beaker, and add 50–75 mL of water.
4. Stir well and observe whether all, some or none of the compound dissolves.
5. Measure the resistance of the solution, as demonstrated by your teacher.
6. Calculate the inverse value of the resistance — this is the conductivity.
Conductivity of melt
A demonstration of the conductivity of a molten ionic compound should be
organised, if possible. It should show not only the conductivity when the
compound is molten, but also the loss of conductivity after the heat is removed.
58
Chapter 5
Properties and Structure
Density
7. Grind a small amount of the solid into a fine powder.
8. Pour enough to occupy 2 mL in a pre-weighed measuring cylinder, and
record the mass. Tap the base of the cylinder to settle the solid.
9. Calculate the density (mass ÷ volume).
Q1. Tabulate your results.
Q2. Are there any results which do not fit with the picture of ionic compounds so
far? Suggest what might cause the variation.
Q3. Explain why ionic compounds are conductive in water, but are not
conductive in the solid form.
PRACTICE QUESTIONS
1. When ammonium chloride is dissolved in water, it forms ammonium ions and
chloride ions. What does this indicate about the structure of the compound?
2.
5.3
Summarise the important physical properties of ionic compounds and explain
them in terms of what you know about their structure.
METALLIC SUBSTANCES
Metals and ionic compounds have a few properties in common: for example the ability to
conduct electricity in the molten form. However, metals can also conduct when solid.
Clearly, this indicates a different structure. The ability to conduct does mean that a metal
must have charged species in its structure, and furthermore, those charged species must
have the ability to move, even in the solid state.
The presence of charged species suggests electrostatic forces, as in ionic crystals. A
pure metal, such as iron, is an element, not a compound, and there is no such thing as an
ionic element. There are no metallic compounds. Where do the charged species come
from?
The concept of ion formation was first introduced in Chapter 2. In Section 2.7, it
was noted that elements, such as sodium, got rid of electrons in the formation of ions.
This is in fact typical of all metals: the desire to rid themselves of extra atoms in the
outer shell.
In metallic substances (pure or mixtures), the atoms ‘cheat’: they ‘dump’ the
unwanted electrons from the outer shell. These electrons ‘float’ around the resulting
cations. This is known as the metallic bond and is another form of electrostatic
attraction, as shown in Figure 5.3. It is the negative charge of the electrons that forms the
‘glue’ that holds the positively charged metal ions together.
Metal ion
Electron
Figure 5.3 An atomic view of the metallic structure
59
Chemistry for Technicians
The cations adopt crystal structures similar to those in ionic compounds, but the ‘anions’
— electrons — are very small and so the distance between cations is less than in ionic
compounds. This means that the density of metals, in general, is high.
WHY IS MERCURY A LIQUID?
Of the 95 known metallic elements, only mercury and gallium are liquids. Gallium
is near a group of elements in the periodic table that have properties of both
metals and non-metals, so its odd behaviour can possibly be excused.
The elements in mercury’s group and row are quite definitely solids. So
what is special about mercury? No one knows! It just is. One of the mysteries of
chemistry.
The ‘free’ electrons are quite able to move under the influence of an electrical field,
so the solid metal is conductive. In molten form, the crystal structure breaks down as
before, but the electrons are still free to move.
Metals do not dissolve in any solvent, because to do so would mean that the
individual metal atoms would have to reclaim their ‘lost’ electrons and move separately
through the liquid. Note that some metals (e.g. sodium) undergo a chemical change with
water. The metal could not be recovered from the solution by simply boiling off the
water.
PRACTICAL WORK
Estimate the melting point and measure the density of the following metals:
tin, lead, copper, bronze (an alloy of copper and tin).
If possible, a demonstration of the electrical conductivity of solid and molten
metal should be organised.
Q1. Suggest reasons for variation in the properties of the metals studied.
Q2. Explain why metals, but not ionic compounds, conduct electricity in the solid
state.
PRACTICE QUESTIONS
3. If metals could dissolve in water, do you think the solution would conduct
electricity? Explain your answer.
4.
Summarise the important physical properties of metals and explain them in terms
of what you know about their structure.
5.
How would the picture of a metal alloy (a mixture of metallic elements) differ
from that in Figure 5.3?
5.4
NON-METALLIC NON-IONIC SUBSTANCES
Now it is time to try to make sense of the vast range of properties exhibited by the
remaining substances: melting points ranging from 200 degrees below zero to almost
4000 degrees, widely varying solubility characteristics and significant differences in
density. Can all these substances have similar structures? The answer is yes and no. Yes,
because the bond holding the atoms together in all of the substances (except helium) is
the same, and no, because the structures are quite individual.
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Chapter 5
Properties and Structure
Firstly, let’s get the structure of helium clear. It doesn’t have one: well, not if structure is
defined as how the atoms are linked together. Helium (and the other elements in its
group) do not form molecules. They are quite stable as single atoms (monatomic), and
have very little attraction of any kind to other atoms. This explains their very low melting
and boiling points (compared to other species of similar formula weights). Figure 5.4
(overleaf) shows the ‘structure’ of helium in the gas phase.
Figure 5.4 Random ‘structure’ of gaseous helium
Figure 5.4 presents your first picture of a gas: individual species (atoms, in
this case, but usually molecules) arranged and moving randomly. There is
randomly. There is little or no connection between the atoms in helium (or
in helium (or the molecules in other gases). Gaseous compounds (at room
compounds (at room temperature) tend to be those with low formula
weights.
The other apparently ‘odd’ substance in the set of survey substances is diamond.
Its melting and boiling points are the ‘opposite’ of helium’s.
EXAMPLE
What properties of diamond indicate that its atoms are held together by a bond that is not
ionic or metallic? Comparatively, how strong is this bond?
Diamond is one of the allotropes of the element carbon, so it cannot be an ionic
substance, since these are all compounds. It is crystalline, but does not conduct electricity
in the solid state. This means that there are no charged species present in diamond’s
structure.
Its bonds must be very strong, since the amount of heat required to melt it is very
great. It also does not dissolve in water (or any other solvent), which suggests that its
bonds cannot be disturbed by this means either.
This ‘new’ bond that is present in diamond is called a covalent bond. In electron terms,
each atom in the bond contributes an electron to the bond in a shared arrangement. In
In this way, no charges are produced, because the electrons are still retained by the
original atoms. The result is a bond that is stronger than the electrostatic attraction
produced in ionic and metallic substances. Covalent bonds are the most important type
of intramolecular bonds: bonds linking atoms within molecules.
The structure of diamond is a crystal lattice, where the carbon atoms are covalently
bonded: each atom is bonded to four others in a three-dimensional arrangement, as
shown in Figure 5.5.
61
Chemistry for Technicians
The structure of diamond is known as a covalent network: a very large
three-dimensional covalently bonded structure, where all the bonds in the
substance are covalent bonds. Silicon dioxide, which is the main
component of sand, is an example of a compound featuring this structure.
Covalent networks involve huge numbers of the atoms in the substance: the tiniest grain
of sand would contain 1020 atoms! Such a large collection of atoms will inevitably mean
that the substance is a solid, since each particle of the substance is one continuous
network of atoms. Therefore, like ionic compounds, the concept of a molecule is not
appropriate
to
covalent networks.
Figure 5.5
Partial lattice structure of diamond
The great strength of covalent bonds means that a covalent network has an
extremely high melting point, because large amounts of heat are required to break the
very large number of covalent bonds. As a covalent network melts, its structure is
destroyed, and when it re-solidifies, its structure may be completely different. If you
melted a diamond, you may get graphite or soot back when it cooled: an expensive
experiment indeed!
Diamond is relatively dense, despite the low atomic weight of carbon, because of
the compactness of the network lattice. It does not conduct electricity because there are
no charged species present. Covalent networks do not dissolve in water, because the
covalent bonds are too strong to be overcome by the dissolving power of water.
LIBRARY WORK
Look up the structure and properties of graphite, another allotrope of carbon, in a
reference book. Draw a diagram illustrating its structure. Explain how the
structure differs from diamond, and why this affects the properties.
How do the properties of oxygen differ from those of the other gas,
helium? The most significant difference in physical properties is that
is that oxygen can be cooled sufficiently to solidify. However, this is partly
this is partly a factor of its increased weight. Experiment has shown that
shown that oxygen exists as a true molecule: two atoms linked by a covalent
bond. This is known as a diatomic molecule. In the gas phase, these two
atoms move separately to other molecules. Figure 5.6 shows the structure of
gaseous oxygen.
62
Chapter 5
Properties and Structure
Figure 5.6 Gaseous oxygen molecules
Since diamond and oxygen both contain covalent bonds, you might wonder why
their properties vary so greatly. In diamond, every atom was joined to four others in a
continuous structure. In oxygen gas, each atom is strongly bonded to only one other. The
individual molecules are at best weakly attracted to each other. Therefore, oxygen is a
gas. Because of the covalent bond, oxygen molecules have no ability to conduct
electricity, regardless of form. To distinguish oxygen and other true molecules from
covalent networks, the term covalent molecule is used for classifying. The remaining
survey substances are all covalent molecules.
Hydrogen chloride is also a diatomic molecule: the fundamental difference
between it and oxygen is that HCl is a compound, where the two atoms are of different
elements. It is a gas for the same reason that oxygen is. Figure 5.7 shows the structure of
gaseous HCl.
Oxygen and hydrogen chloride exhibit two important differences in properties: HCl
is very soluble in water, and its solution conducts electricity. The explanation of the
solubility of HCl is beyond the scope of this module (it is covered in the next part of the
book, if you wish to read Chapter 9).
Figure 5.7 Gaseous hydrogen chloride molecules
EXAMPLE
Based on existing knowledge, how can the conductivity of HCl solution be explained?
Conductivity in water is the property demonstrated by compounds that produce ions in
solution. The conclusion that must be made about HCl in water is that for some reason,
the covalent bond is broken, and the ions H+ and Cl– form.
Water’s properties can be explained, at least at the basic level, by its covalent
molecular structure. With three atoms in the molecule, water’s structure has two possible
arrangements, but only one actually exists, as shown in Figure 5.8. The covalent bonds
are represented by solid lines: this is the conventional way of depicting these bonds in
drawings of structures.
H
H
O
H
(a)
Figure 5.8
O
(b)
Possible structures of water — (b) is the correct structure
63
H
Chemistry for Technicians
Covalent molecules undergo easily explained changes between the three physical
states: solid, liquid and gas. No change in the intramolecular bonding occurs as the
molecule undergoes a change of physical state. Liquid water and its other forms, ice
and steam, each have the structure show in Figure 5.8(b).
The difference in the form of the three states is the proximity of the
molecules to each other and the presence of weak forces of attraction
between the molecules, known as intermolecular forces. These forces
increase in strength as the molecules get close to each other.
In a solid, the molecules are close together, and are held that way because
the intermolecular forces are strong enough to keep the molecules in place.
molecules in place. More often than not, the molecules arrange themselves
arrange themselves into regular patterns, like the crystal lattice of an ionic
lattice of an ionic compound. As heat is added, the molecules vibrate, and at
their melting point, are moving enough to break free of some of the
intermolecular forces. The liquid state allows some freedom of movement,
but complete freedom is only attained in the gaseous state when enough
heat has been added to break all intermolecular forces. These changes are
illustrated in Figure 5.9.
The strength of intermolecular forces also depends on the molecule itself. The
molecules of some substances, for example water, have a much stronger attraction for
each other, than others, for example, helium. The causes of intermolecular forces and
differences in their strengths are not part of this module. They are dealt with in Chapter 9
of this book, as part of the module Chemical Principles.
Solid
Liquid
Gas
Increasing temperature
Figure 5.9
Changes in covalent molecules between physical states (intermolecular forces are
indicated by dotted lines)
EXAMPLE
Water has a formula weight lower than that of both oxygen and hydrogen chloride, yet its
boiling point is much higher. Why?
Substances with lower formula weights tend to have lower boiling points, unless the
intermolecular forces are strong. This means that water must have much stronger
intermolecular forces than oxygen or HCl.
64
Chapter 5
Properties and Structure
PRACTICE QUESTIONS
6. What property of HCl indicates that it is not ionic?
7. The boiling point of argon is –186°C.
(a) Compare this to those of oxygen and hydrogen chloride, substances that
have similar formula weights to argon.
(b) Given argon’s position in the periodic table, suggest what structure it is
likely to have.
(c) Compare the relative strengths of the intermolecular forces in the three
compounds.
8. What conclusion can be made about the formula of a covalent molecular
substance?
The remaining three covalent molecular compounds — ethanol, naphthalene and
sulfur — have the structures shown in Figure 5.10. They exhibit more complex
structures, but obey the general rules of their class. The structures shown in Figure 5.10
are often known as structural formulae, since they give more information about the
bonding arrangements than a simple formula.
The structure of sulfur shown in Figure 5.10 is one of its allotropes. Naphthalene is
an organic compound: a covalent molecular substance involving carbon, and capable of
complex bonding arrangements as seen in the structure of naphthalene. The double lines
denote double covalent bonds, where two sets of electrons form bonds between the same
atoms.
H
H
H—C—C—O—H
H
H
H
H
H
C
C
C
C
H
C
C
H
(a)
Figure 5.10
C
C
C
C
H
H
S S
S
S
S
S
S S
H
(b)
(c)
Structures of (a) ethanol (b) naphthalene and (c) sulfur
PRACTICAL WORK
You are required to test the properties (as before) of two unknown substances to
determine their class of substance from the following categories: ionic, covalent
molecule, covalent network, metallic.
Melting (or boiling) point
If the compound is solid, then estimate the melting point using the following
steps:
1.
2.
3.
Place a small amount on the end of a metal spatula.
Heat the compound on a water bath by resting the tip of the spatula against
the metal lid of the bath — if the compound melts then the melting point is
less than 100°C.
If it doesn’t melt, repeat the process on a hotplate — if it melts, the melting
point is 100–400°C.
65
Chemistry for Technicians
4.
If it still doesn’t melt, heat it in a hot Bunsen flame — if it melts, then its
melting point is 400–1000°C; if it doesn’t, then the melting point is greater
than 1000°C.
If the compound is a liquid, heat a few mL of it in a test tube on the water
bath to determine whether its boiling point is greater or less than 100°C
Q1. Explain why ionic compounds are conductive in water, but the great majority
of covalent compounds are not.
Q2. Using your results from earlier work, explain the order of melting points
observed for potassium chloride, methanol and silica, in terms of the
bonding.
Q3. Identify the bonding type in each unknown sample, and explain your
reasons.
LIBRARY WORK
Using a reference book, look up the melting point, density and solubility in water
and an organic solvent for all the chloride compounds of the elements of the third
row of the periodic table. Identify each as ionic, molecular covalent or covalent
network.
PRACTICE QUESTIONS
9. What change in bonding type occurs for the compounds of elements in a row of
the periodic table from left to right?
10. Identify the type of substances from the properties below.
Melting point
1750
121
850
3850
A
B
C
D
Density
2.1
1.8
8.0
2.2
Solubility in water
medium
low
nil
nil
11. Explain this apparent contradiction: Covalent bonds are the strongest, but
covalent molecules are the easiest to melt.
5.4.1 Bonding in Polyatomic Ions
The atoms in ions such as ammonium, sulfate and nitrate, are held together by covalent
bonds, but are charged because electrons have been added to or lost from one or more of
the atoms. Thus, ammonium chloride has ionic bonds between the NH4+ and Cl–, and
covalent bonds between the N and H atoms, as shown in Figure 5.11.
H
Cl–
N
H
H
+
H
Cl–
N
H
H
Figure 5.11
+
H
H
Bonding in ammonium chloride
66
Chapter 5
Properties and Structure
When ammonium chloride dissolves in water, the ammonium ion stays intact because of
the strength of the covalent bonds holding it together. An ‘atomic level’ snapshot of a
solution of ammonium chloride in water would look like Figure 5.12 (overleaf).
H—O—H
H—O—H
Cl–
Figure 5.12
Cl–
H
|
O
|
H
NH4+
H—O—H
H—O—H
NH4+
H
|
O
|
H
Representation of ammonium chloride in water
PRACTICE QUESTION
12. Draw a representation of aqueous solutions of (a) ethanol and (b) sodium nitrate,
using Figure 5.12 as a model.
5.5
WHAT TYPE OF BOND WILL FORM?
In this topic, four types of bonds have been discussed: ionic, metallic, covalent and
intermolecular. These bonds occur because of the different types of atoms involved.
There are other types of bonds between atoms, but these are less common, and are
beyond the scope of this module.
Some generalisations can be made about the type of bond that atoms from different
groups of the periodic table will form. A simplified scheme which uses Tables 5.4 and
5.5 has been devised to predict bonding types. It is not perfect because it can only
generalise: there are exceptions to almost everything in science. Only those elements
which are in groups 1A to 8A are covered by this scheme. It also excludes intermolecular
bonds.
Table 5.4
Classification of groups 1A–8A according to bonding rules
Class 1
Class 2
Class 3
Class 4
1A, 2A, 3A
4A
5A, 6A, 7A, H
8A
Table 5.5
Rules for predicting bond types
Bond type
Atom 1
Atom 2
Ionic
Class 1
Class 3
Metallic
Class 1
Class 1
Covalent
Class 2 or 3
Class 2 or 3
Note:
No bond is formed by elements in class 4, or where atoms
from classes 1 and 2 are mixed.
67
Chemistry for Technicians
EXAMPLES
Predict the type of bond formed between the following pairs of elements:
(a) carbon and chlorine
(b) potassium and sulfur
(c) sodium and potassium
(d) calcium and carbon
(e) helium and silicon.
Class 1 = C1, Class 2 = C2, Class 3 = C3, Class 4 = C4.
Element 1
Element 2
Type of bond
Carbon (C2)
Chlorine (C3)
covalent
Potassium (C1)
Sulfur (C3)
ionic
Sodium (C1)
Potassium (C1)
metallic
Calcium (C1)
Carbon (C2)
none
Helium (C4)
Silicon (C2)
none
PRACTICE QUESTION
13. Predict the bond type formed between
(a) lithium and oxygen
(b) hydrogen and sulfur
(c) calcium and aluminium
(d) silicon and sodium
(e) neon and calcium.
WHAT YOU NEED TO BE ABLE TO DO
•
•
•
•
•
•
•
List the variety of forms of structures that pure substances can adopt.
Summarise the properties associated with these structure types.
Define the term bond.
Distinguish between intramolecular and intermolecular bonding.
Draw diagrams showing the structure of substances containing ionic, covalent
and metallic bonds.
Predict the type of bond formed between atoms.
Distinguish between the structures associated with three states of matter.
68
Chapter 5
Properties and Structure
TERMS AND DEFINITIONS
During this chapter, you have met a number of new terms, listed below. Match them with
their correct definitions.
Terms
A bond
C
ionic bond
E
ionic formula
G gas
I
covalent network
K intramolecular bond
B
D
F
H
J
L
structure
melting point
metallic bond
covalent bond
diatomic molecule
intermolecular force
Definitions
1
a bond between two atoms in different molecules
2
attraction between positive and negative species
3
a link between two atoms
4
the temperature at which the individual species in a substance break free of the
bonds that hold them together in a solid
5
two atoms linked by a covalent bond, forming a separate molecule
6
a large three-dimensional structure of covalent bonds
7
a physical state where the species are able to move randomly without interaction
with others
8
the way that atoms are joined to each other
9
the simplest ratio of numbers of cation and anion
10 electrostatic attraction between free electrons and cations
11 a bond between two atoms in the same molecule
12 the sharing of two electrons between a pair of atoms to increase the number of outer
shell electrons
69
6
CHEMICAL REACTIONS I
PURPOSE
To study and describe simple chemical reactions.
6.1 WHAT IS A CHEMICAL REACTION?
A chemical change occurs when substances rearrange their atoms between
the molecules to give different chemical substances. This is also known as a
also known as a chemical reaction. The starting materials in a chemical
a chemical reaction are known as the reactants, and the new substances
substances produced by the chemical reaction are called the products.
Chemical reactions occur because the combination of reactant substances is not
stable under the conditions of temperature, concentration and sometimes pressure. Only
some combinations of substances cause a reaction, for example, iron will rust in the
presence of water and oxygen, but gold will not. Also, reaction may only occur if the
conditions are right, even though the mixture of reactants can lead to a reaction. Dilute
nitric acid will not react with copper, but concentrated nitric acid will.
In a chemical reaction, no atoms are destroyed nor are any created. This is
known as the law of conservation of mass, and can also be stated as the
stated as the total mass of material does not change during a chemical
chemical reaction. While we cannot see the atoms, we can measure the
measure the masses of the starting materials, and those left over, and these
will be equal.
Chemical reactions are not always 100% efficient, that is, all reactants are not
necessarily consumed. This may be because the reaction does not completely occur (it
stops at some point short of 100%), or because one reactant has been completely used up
before the others. However, the law of conservation of mass still holds. The total amount
of material does not change.
EXAMPLES
1.
If we mix two aqueous solutions, one containing 10 g of sodium iodide, the other 10
g of lead nitrate in water, a yellow solid (lead iodide) forms. If we remove the water
by evaporation, what mass of material will be obtained?
There will be 20 g of material remaining, some of which will be the products and some will
be leftover reactant.
2.
If we burn 5 g of magnesium in a sealed container which has been filled with
oxygen gas, a white powder (magnesium oxide) weighing 8.3 g is obtained. How
much oxygen gas has been used in the reaction?
If all the magnesium has been burnt, then it contributes 5 g of the 8.3 g of magnesium
oxide. The other 3.3 g must have come from the oxygen.
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Chapter 6
Chemical Reactions I
PRACTICAL WORK
1. Weigh about 5 g of sodium iodide into a 250 mL beaker, recording the exact
mass.
2.
Weigh about 5 g of lead nitrate into a 250 mL beaker, recording the exact
mass.
3.
Add 50 mL of water to each beaker, and stir until each solid has completely
dissolved.
4.
Pour one solution into the other, stir well and heat on a water bath for 10
minutes.
5.
Cool in an ice bath, and filter off the yellow solid as directed by your teacher.
6.
Collect the yellow solid and dry between filter papers in a 110°C oven.
7.
Evaporate the filtered solution in a pre-weighed evaporating basin on a
water bath. Once most of the water has gone, dry in a 110°C oven.
8.
Weigh the dry yellow powder and evaporating basin.
9.
Calculate the mass of residue in the basin.
Q.
Compare the mass of starting material to the total mass of material obtained
after the reaction. Is the law of conservation of mass obeyed? Suggest
some causes of experimental error.
PRACTICE QUESTIONS
1. 20 g of magnesium is burnt in a sealed vessel containing 75 g of oxygen. All the
magnesium reacts. The product is weighed, and is found to weigh 33.2 g. How
much oxygen gas remains in the vessel?
2.
10 g of hydrogen and 10 g of oxygen are reacted to form water in a sealed
vessel. 12 g of water is produced by the reaction. What total mass of unreacted
hydrogen and oxygen remains in the vessel?
3.
10 g of sodium and 20 g of chlorine gas are mixed in a sealed vessel. Sodium
chloride is produced. What is the mass of the contents of the vessel?
4.
For each of the chemical substances named in Q1–3, determine whether it is a
reactant or a product.
6.2
HOW CAN WE TELL THAT A CHEMICAL REACTION HAS
OCCURRED?
In simple terms, a chemical reaction occurs when a mixture of substances is not stable
under the existing conditions of temperature and pressure. Only certain mixtures will
react. Some require continual heating to allow the reaction to occur, others will provide
their own heat after an initial input of heat (e.g. burning petrol). How do we know that
we are not simply heating or vaporising the substances, causing a physical change but
with no chemical reaction?
The definition of a chemical reaction tells us that new substances are formed. This
can be determined by simple observations or chemical analysis. Chemical analysis is
beyond the scope of this subject.
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Chemistry for Technicians
Some of the observations that can be made which suggest a chemical reaction has
occurred are:
•
colour changes
•
formation of precipitate (solid)
•
changes in acidity
•
temperature changes
•
formation of gases (but not simply from boiling of a liquid).
PRACTICAL WORK
Do not worry about the M value for the various solutions used in these exercises.
They are given to assist your teacher, and are not significant in your results.
1.
Into three separate beakers, measure 10 mL of the following: 0.01 M
KMnO4, 0.5 M H2SO4, 0.05 M iron (II) sulfate. Note the colour of each
solution. Measure the temperature of one of the solutions. Combine the
three solutions and observe the final colour. Measure the temperature of
this solution.
2.
Pour 25 mL of water into a beaker and measure its temperature and acidity.
Drop one small piece of sodium into the water. Observe the reaction. Once
all the sodium has dissolved, measure the temperature and acidity.
3.
Pour 25 mL of 1 M HCl and 25 mL of 0.5 M Na2CO3 into separate beakers.
Measure the temperature of one of the solutions. Combine the solutions,
observe any changes and measure the temperature after reaction has
ceased.
4.
Weigh 1.32 g of Na2CO3 into a 150 mL beaker. Add 25 mL of 0.5 M HCl.
Observe any changes and measure the temperature after reaction has
ceased.
5.
Measure 10 mL of 0.1 M BaCl2 and 10 mL of 0.1 M Na2SO4 into separate
beakers. Measure the temperature of one of the solutions. Combine the
solutions, observe any changes and measure the final temperature.
Q1. What observable signs of a chemical reaction did you detect in each of the
above reactions?
Q2. Did you observe any differences between reactions 3 and 4? Can you
account for them?
6.3
DESCRIBING CHEMICAL REACTIONS
Chemical reactions are the foundation of chemistry in the real world — in
making the materials we use, in checking the purity of our foods and so on.
foods and so on. To simplify the relationship between reactants and
reactants and products, a chemical equation is written. This is a ‘sentence’
is a ‘sentence’ using chemical formulae or names which describes what and
how much reacts to form what and how much.
An equation normally involves formulae, but will be initially written in terms of the
names of the substances.
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Chapter 6
Chemical Reactions I
EXAMPLE
Write the word equations for the reactions in the examples in Section 6.1, based on the
information provided.
magnesium + oxygen → magnesium oxide
lead nitrate + sodium iodide → lead iodide
Note the use of the + and → symbols. They are used to separate the names
(and formulae) of the various substances involved. The + separates the
separates the reactants (or products) substances from each other, while the
other, while the → separates the reactants on the left hand side of the
side of the arrow from the products on the right hand side.
To see better what is happening in the reaction, the formulae corresponding to the
names are used. This allows us to:
•
•
•
see what is happening to the ions or atoms
confirm the conservation of mass
determine the ratio of the number of reactant and product molecules involved, since
this is fixed for each given reaction.
EXAMPLE
Write the formula equations for the reactions of (i) hydrogen and oxygen to form water
and (ii) lead nitrate with sodium iodide.
(i)
(ii)
H2 + O2 → H2O
Pb(NO3)2 + NaI → PbI2
Is the law of conservation of mass (atoms) obeyed by these equations? At the
moment, no! In (i), an oxygen on the reactant side has gone missing on the product side,
and in (ii), elements (Na, N, O) are missing from the products.
This is because we have not finished the process of writing the chemical
equation. It must be balanced, so that there is the same number of atoms of
number of atoms of each element on both sides of the arrow. This means
This means that the identity of all the reactants and products must be
must be known. In example (ii), this is clearly another product.
6.3.1 Balancing Chemical Equations
The actual substances involved in a reaction are fixed for that reaction. If we know the
identity of each reactant and product, then any difference in the number of atoms
between reactants and products cannot be corrected by changing the substances. It can
only be done by changing the number of molecules of one or more of the substances
involved.
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Chemistry for Technicians
EXAMPLE
What can and can’t be done to balance the hydrogen/oxygen equation above?
The equation is short one oxygen on the product side. We can’t balance it by:
•
changing the formula of oxygen gas to O (not the right substance)
•
changing the formula of water to H2O2 (not the right substance)
•
adding an oxygen atom as an extra product (you’ll have to take our word that water
is the only product!).
We have to balance it by:
•
adding more molecules of the existing substances until the atoms are equal on each
side.
Some equations are easier than others to balance; some are already balanced. The scheme
suggested below for balancing equations is not compulsory, but is at least a systematic
way of approaching the job. You may prefer a different way, or to short-cut this method
after you become practised at balancing equations.
6.3.2 A Suggested Method for Balancing Equations
1. Make sure that all substances involved in the reaction are identified.
2. Write the equation in formulae first, making sure that the formulae are correct.
3. Construct a grid listing the elements involved for both sides of the equation.
4. Fill the grid with the number of atoms of each element on both sides.
5. Look at which elements are not balanced, and how that imbalance could be
corrected one element at a time.
6. For the element to be balanced, identify which substance contains that element and
on which side of the equation, extra atoms need to be added. This will normally
mean adding an extra molecule(s) of a given substance, which may mean that other
elements will be affected, for example, when trying to balance hydrogens, you add
an extra water molecule to one side, so not only do you add two hydrogens but an
extra oxygen.
7. Fix the grid on both sides with the new numbers of atoms. Don’t forget to change
all elements involved in the species you have added.
8. Return to step 4 until the equation is balanced.
EXAMPLES
1.
H2 + O 2
H
2
O
2
H2O
→
→
H
2
O
1
The absence of a number in front of a substance is taken to mean 1.
Oxygen is unbalanced: it needs one extra atom on the product side. If we add
another water to the product side, this will add the required O and create a balance.
However, it will add two hydrogens to the right-hand side as well.
H2 + O 2
→
H
2
→
O
2
2H2O
H
4
O
2
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Chapter 6
Chemical Reactions I
Now there are too many hydrogens on the product side. If we add one more hydrogen
molecule, that will fix that.
2H2 + O2
→
H
4
→
O
2
2H2O
H
4
O
2
The equation is balanced.
2.
The lead nitrate/sodium iodide reaction involves a seemingly more complex process,
since there are five different elements.
NaI
+
Pb(NO3)2
→
NaNO3
+
PbI2
Na
1
I
1
Pb
1
N
2
O
6
→
Na
1
I
2
Pb
1
N
1
O
3
Only the lead and sodium are balanced at the moment. Which element to try to correct
first? It doesn’t really matter, so we will work from left first, choosing I. One more I is
needed on the reactant side; this means one extra NaI.
2NaI
Na
2
+
I
2
Pb(NO3)2
Pb
1
N
2
→
O
6
→
NaNO3
Na
1
+
I
2
Pb
1
PbI2
N
1
O
3
Now the sodiums are unbalanced; another is required on the product side, from an extra
NaNO3.
2NaI
Na
2
+
I
2
Pb(NO3)2
Pb
1
N
2
→
O
6
→
2NaNO3
Na
2
+
I
2
Pb
1
PbI2
N
2
O
6
The equation is balanced.
PRACTICE QUESTION
5. Balance the following equations.
(a) AlCl3 + H2O → Al2O3 + HCl
(b) C2H4 + O2 → CO2 + H2O
(c) Fe2O3 + C → Fe + CO2
(d) Zn + HCl → ZnCl2 + H2
(e) AgNO3 + H2SO4 → Ag2SO4 + HNO3
(f) Na2O + H2O → NaOH
(g) CH4 + Br2 → CBr4 + HBr.
6.3.3 What Do the Numbers in Front of the Formulae Mean?
A balanced equation tells us what is happening between the atoms/molecules that react
together and form. The numbers in front of each reactant or product give the simplest
ratio of substances involved in the reaction, and are called coefficients. As an example,
let’s look at the reaction of hydrogen and oxygen gases to form water. The balanced
equation is:
2H2 +
1O2
→
2H2O
The numbers in front of the formulae tell us that for every oxygen molecule, two
hydrogen molecules must be available to react, and that two water molecules would
form. This is depicted in Figure 6.1.
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Chemistry for Technicians
EXAMPLES
1.
How many oxygen molecules would be needed to react with 250 hydrogen
molecules?
Since twice as many hydrogen molecules are required in the reaction,
250 ÷ 2 = 125 oxygen molecules would be required.
2.
How many water molecules would form if 2000 hydrogen molecules were combined
with 1000 oxygen molecules?
The same number of molecules of water form as there are hydrogen molecules that
react. Therefore, 2000 water molecules would form.
Reactions occur by ratio of numbers, not masses. But we can’t see atoms or
molecules to count them, so how can we tell whether we have the right amount of each to
fit the equation? We will answer that question in Section 6.5.
Figure 6.1 Reaction of hydrogen and oxygen gases to form water (black circles are hydrogen,
white are oxygen)
PRACTICE QUESTIONS
6. What would happen to the final mixture if the following changes were made to
the reactant mixture in Figure 6.1? Assume any further possible reaction would
occur.
(a) five more hydrogen molecules and two oxygen molecules
(b) two more oxygen molecules
(c) a water molecule
7. Explain why the balanced equation for the reaction of hydrogen and oxygen
doesn’t mean that 20 g of hydrogen reacts with 10 g of oxygen to form 20 g of
water.
6.3.4 Indications of Physical State of Reaction Components in Equations
Knowing the physical state of the reactants and products of a reaction is important for
studying or using that reaction. The four important physical states that typically occur are
solid, liquid, gas and aqueous (water) solution. These are indicated in an equation by
subscripted abbreviation in brackets after the compound formula, i.e. (s), (l), (g) and (aq).
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Chapter 6
Chemical Reactions I
EXAMPLE
Sodium carbonate powder is mixed with hydrochloric acid, resulting in the formation of
carbon dioxide gas and a solution of sodium chloride. Write a full balanced equation for
this reaction, showing physical states of the substances.
Na2CO3 (s) + 2HCl (aq) → CO2 (g) + 2NaCl (aq) + H2O (l)
PRACTICE QUESTION
8. Write balanced equations to describe the following reactions, including physical
state indicators:
(a) calcium carbonate reacts with sulfuric acid to give water, calcium sulfate
precipitate, and carbon dioxide gas
(b) hydrogen gas and nitrogen gas form ammonia gas
(c) molten potassium bromide is converted to potassium metal and bromine
liquid.
6.4
COMMON TYPES OF REACTIONS
There are reactions which produce similar results with a range of different possible
reactants. Some of the important reaction types are described below. Some of these
classifications overlap, that is, one reaction could be classified as both a neutralisation
and a gas formation reaction.
6.4.1 Combustion
Reaction with oxygen gas under high temperatures is called combustion; we might also
call it burning, though sometimes the familiar flame is not produced. Metals and nonmetals alike can react with oxygen. Oxides are the compounds produced by combustion;
this includes carbon dioxide (from carbon compounds) and water (from hydrogencontaining compounds). An example is the combustion of natural gas (which is
composed mainly of the carbon compound known as methane).
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)
6.4.2 Decomposition
This occurs when a compound is broken down into its elements. Water can be
decomposed into hydrogen and oxygen gas by electrical current.
2H2O (l) → 2H2 (g) + O2 (g)
6.4.3 Combination
This is the opposite of decomposition, where elements combine to form a compound. An
example is the reaction of sodium and chlorine to form sodium chloride.
2Na (s) + Cl2 (g) → 2NaCl (s)
6.4.4 Neutralisation
Acids are compounds that produce H+ ions in water, bases produce OH– in water, even
though the actual substance may not contain these ions itself. When a base reacts with an
acid, it is in fact these ions that react to form water, and is known as neutralisation. The
reaction is indicated by a change in solution acidity. An example is the reaction of
ethanoic acid (found in in vinegar) with sodium hydroxide.
C2H4O2 (aq) + NaOH (aq) → NaC2H3O2 (aq) + H2O (l)
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Chemistry for Technicians
6.4.5 Precipitation
Precipitation reactions only occur in solution, and involve the mixing of two solutions
and the production of a solid. An example is the formation of the scum when soap
(NaC18H35O2) is used in hard water, which contains calcium chloride.
CaCl2 (aq) + 2NaC18H35O2 (aq) → Ca(C18H35O2)2 (s) + 2NaCl(aq)
6.4.6 Gas formation
This classification is applied to reactions where a gas is formed from liquid or solid
reactants. It does not include reactions where all species are in the gas phase. An example
is the reaction of calcium carbonate (limestone) and hydrochloric acid.
CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l)
6.4.7 Reversible reactions
One of the characteristics of a chemical reaction is the difficulty in reversing the process.
This is not strictly true, since many reactions can be turned around, so that the reactants
are reformed.
One of the features of these reversible reactions is that they do not proceed
to completion in either direction; they reach a point of equilibrium, after
equilibrium, after which no change in amount of reactant and product
and product occurs. A change in conditions, for example, temperature,
temperature, causes the equilibrium position to shift, giving more products
or more reactants.
An example is the reaction of two nitrogen dioxide molecules to form a ‘double
molecule’ or dimer, dinitrogen tetroxide. Reversible reactions are indicated by doubleended arrows.
2NO2 (g) ↔ N2O4 (g)
PRACTICAL WORK
1. Observe the change in colour in a closed system containing nitrogen dioxide
(red-brown) and dinitrogen tetroxide (colourless) as it is heated and cooled.
2.
Obtain a solution of red iron (III) thiocyanate complex, and pour three equal
portions into test tubes. Under certain conditions, it decolourises, due to the
decomposition of the compound.
Heat a test tube of the solution and observe the colour change. Allow
to return to room temperature, and compare with a portion of the solution
that has not been used.
Cool a test tube of the solution in an salt-chilled ice bath and observe
any colour changes. Allow to return to room temperature, and compare with
a portion of the solution that has not been used.
Q1. Do your observations suggest these reactions are reversible? Explain.
Q2. Classify the reactions performed in earlier laboratory work in this chapter by
the types described in this section.
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Chapter 6
Chemical Reactions I
PRACTICE QUESTION
9. Classify the following reactions.
(a) Na2CO3 (aq) + Pb(NO3)2 (aq) → NaNO3 (aq) + PbCO3 (s)
(b) Cl2 (g) + H2 (g) → HCl (g)
(c) HgO (s) → Hg (l) + O2 (g)
(d) HCl (aq) + NaOH (s) → NaCl (aq) + H2O (l)
(e) C2H4 (g) + O2 (g) → CO2 (g) + H2O (g)
6.5
MASS RELATIONSHIPS IN CHEMICAL REACTIONS
You now realise that the number of molecules (or atoms) is the important factor in the
way chemical reactions occur. If we want to use chemical reactions to make things, to
analyse materials etc, we must be able to ‘count’ the reactants and/or products. But atoms
and molecules are very, very, very, very (you get the idea) small. How small? If you
added up all the human beings in all of time and then multiplied that number by itself,
you might get towards the number of molecules in the smallest speck of dust! Atoms
have radii around 10 nm, or one hundred-thousandth of a millimetre. Counting something
that small would be somewhat of a challenge. Only very recently has technology
developed to the point of making atoms visible, and even then, only under special
circumstances.
So, we need to have a way of relating the number of atoms (or molecules) to
something we can readily measure — mass. Further, this relationship should be simple
and the same for all chemical substances. It is called the mole.
6.5.1
The Mole Concept
Simply put, the mole is the number of atoms or molecules of a given
substance that has a mass equal to the formula weight in grams. One mole
of anything contains the same number, just like a dozen. In this case, the
number is very large.
The actual number of a given substance (atom, molecule, ion etc) that
makes up a mole is 6.02 x 1023, and is called Avogadro’s number. This has
number. This has been measured by experiment. Thus, 6.02 x 1023
1023 molecules of water weighs 18.02 g, while the same number of
number of molecules of oxygen gas weighs
32.00 g. In practical terms, the actual value of Avogadro’s number is not
useful, since we still work in mass.
Equation 6.1 links the three important terms — number of moles, mass and formula
weight — together.
Number of moles = Mass (in grams) ÷ Formula Weight (FW)
Eqn 6.1
EXAMPLES
1.
Calculate the number of moles of molecules contained by 45 g of water.
FW of water is 18.02 g/mole.
Moles = 45/18.02 = 2.50.
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Chemistry for Technicians
2.
How many atoms of Cu are contained in 1 mg?
Convert the mass of copper to grams (divide by 1000).
FW copper = 63.54.
Moles = 1 x 10–3/63.54 = 1.57 x 10–5.
Number of atoms = moles x Avogadro’s number = 1.57 x 10–5 x 6.02 x 1023 =
9.48 x 1018.
3.
What mass of sodium chloride is contained by 5 x 10–3 moles?
Mass = moles x FW = 5 x 10–3 x 58.44 = 0.29 g.
PRACTICE QUESTION
10. Complete Table 6.1.
Table 6.1
(a)
(b)
(c)
(d)
(e)
Question 10
Substance
Mass
Sodium chloride
Water
Iron
Sugar (FW 180)
Nitrate ion
50 mg
1 kg
Number of moles
1.25 x 10–4
3.33
0.1575 g
6.5.2 The Significance of the Mole in Chemical Reactions
From balancing equations, we know that the coefficients tell us the ratio of the number of
species involved. Thus, if we use our measurable quantity of numbers, the mole, then we
can make some predictions about reactions. Using the simple example of the formation
of water 2H2 + O2 → 2H2O, it is valid to say that two moles of hydrogen react with one
mole of oxygen to give two moles of water, and that any fraction of moles can equally
be used in the same ratio.
EXAMPLES
1.
How many moles of oxygen would be required to react with 0.05 moles of hydrogen?
Since for every two hydrogen molecules, there is one oxygen involved in the
reaction, only half of 0.05 moles of oxygen — 0.025 moles — will be required.
2.
How many moles of hydrogen and oxygen would be required to make three moles of
water?
To make any number of moles of water requires the same number of hydrogen and
half that of oxygen. Therefore to make three moles of water would require three
moles of hydrogen and 1.5 moles of water.
But we can’t measure out a given number of moles, only a certain mass, so these
mole relationships in equations need to be converted to mass to be immediately useful.
Using Equation 6.1 above, we can convert our mole relationships into masses. This gives
the mass ratio for each species involved.
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Chapter 6
Chemical Reactions I
EXAMPLES
1.
2.
Using the number of moles to equal the coefficients in the water equation, calculate
the masses of each substance involved.
2 moles H2
+ 1 mole O2
FW
2.02
32.00
18.02
Mass
4.04 g
32.00 g
36.04 g
→
2 moles H2O
What mass of hydrogen would be required to react with 320 g of oxygen?
320 g of oxygen is 10 times the mass of oxygen in the example above. We should
need 10 times the mass of hydrogen, which is 40.4 g.
3.
What masses of hydrogen and oxygen should be combined to form 0.36 g of water?
0.36 g of water is 1/100 of the mass of water in the example above. Therefore, we
need only 1/100 of the masses of hydrogen and oxygen: 0.04 g and 0.32 g
respectively.
In this module, we will only deal with ‘easy’ amounts of masses or moles, like 10
times or half, but the principle is exactly the same if 0.0181 moles of hydrogen or 690 g
of water were involved.
PRACTICAL WORK
1. Calculate the masses of 0.005 moles of sodium sulfate and barium chloride
(to work out the formula of this compound, find barium in the periodic table
and consider what elements that you do know are in the same group).
2.
Check these masses with your teacher.
3.
Weigh these exact masses into separate 250 mL beakers.
4.
Dissolve each solid in 50 mL of water.
5.
Combine the solutions and heat for 10 minutes on a water bath.
6.
Cool the solution in an ice bath, and filter through a pre-weighed crucible as
shown by your teacher.
7.
Dry the precipitate in a 110°C oven.
8.
Re-weigh the crucible and determine the mass of precipitate.
Q1. Write a balanced equation to describe this reaction — the precipitate is
barium sulfate.
Q2. Calculate (a) the number of moles and (b) the mass of barium sulfate
expected.
Q3. Compare your result with expected value. Account for any differences.
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Chemistry for Technicians
PRACTICE QUESTIONS
11. In the zinc/HCl reaction in Q5(d):
(a) what is the maximum number of moles of zinc that can react with a
solution containing two moles of HCl? What mass of zinc is this?
(b) how many moles of HCl would be required to make 0.5 moles of zinc
chloride?
12. To analyse the amount of sulfate in a water sample, the sulfate is precipitated by
reaction with barium ions to form barium sulfate. If 0.5 moles of BaSO4 was
produced, (a) how many moles and (b) what mass of sulfate was present?
13. Using the water formation reaction and mass relationships on page 81, calculate:
(a) what mass of oxygen would be required to react with 20 g of hydrogen?
(b) what masses of hydrogen and oxygen would be required to make 3.6 kg of
water?
14. Write a balanced equation for the reaction of sodium chloride and silver nitrate
to form silver chloride and sodium nitrate.
(a) Calculate the mass relationships.
(b) What mass of each reactant would be required to form 14.3 g of silver
chloride?
WHAT YOU NEED TO BE ABLE TO DO
•
•
•
•
•
Define the terms chemical reaction, reactant and product.
Explain why chemical reactions occur.
Write full balanced equations.
Interconvert moles, mass and formula weight.
Determine simple mole and mass relationships for chemical reactions.
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Chapter 6
Chemical Reactions I
TERMS AND DEFINITIONS
In this chapter, you have met a number of new terms, listed below. Match them with their
correct definitions.
Terms
A chemical reaction
C
product
E
chemical equation
G coefficients
I
mole
B
D
F
H
J
reactant
law of conservation of mass
balanced equation
equilibrium
Avogadro’s number
Definitions
1
a new substance formed in a reaction
2
a process where the chemical form of the substances involved is changed
3
an equation where there are equal numbers of each type of atom on both sides
4
the number of individual species in one mole
5
a substance initially present before a reaction commences
6
a reaction which does not go to completion, but reaches a point where no further
change in reactant and product concentration occurs
7
numbers in front of formulae which indicate the ratio of numbers of molecules
involved in a reaction
8
a ‘sentence’ written in chemical formulae describing a reaction
9
the number of atoms, molecules or ions that comprise the formula weight of the
substance in grams
10
no matter is created or destroyed in a reaction
83
7
THE PERIODIC TABLE
PURPOSE
To describe the layout and features of the periodic table, and to study trends in related
elements.
It is recommended that you revise Chapter 3 before progressing with this chapter.
7.1
WHY THE PERIODIC TABLE IS STILL USED
Chapter 3 described the essential points in the development of the periodic table: the
grouping of elements of similar properties and the ordering of all the elements by atomic
weight (at first) then atomic number.
If Mendeleev’s table had been no more than a listing of elements, it would not have
been successful. He was able to convince the scientific world of the value of his periodic
table because he used the repeating patterns to predict the existence and properties of
unknown elements. Their existence was indicated by the fact that there were ‘holes’ in
the table, left blank because the next element by atomic weight did not fit into that
column, but into the next one. The properties were predicted from those of the elements
above and below the missing element in a column. Such a case is shown in Figure 7.1.
B
C
N
Al
Si
P
Zn
?
?
As
Cd
In
Sn
Sb
Figure 7.1 The case of the missing elements
A well-documented example is the element germanium, which was unknown in
Mendeleev’s time. There was a space in the column under silicon. By doing nothing
more sophisticated that averaging the atomic weights, melting points and other
measurements and assuming the same chemical form for the oxide, Mendeleev made a
prediction for this element, which he called eka-silicon, as shown in Table 7.1. When an
element fitting the space was discovered, and its properties found to match those of
Mendeleev’s prediction, the future of the periodic table was assured.
84
Chapter 7
Table 7.1
The Periodic Table
Comparison of predicted and actual properties of germanium
Property
Prediction
Actual
grey
light grey
Atomic weight
72
72.59
Density
5.5
5.35
GeO2
GeO2
5.7
5.70
Appearance
Formula of oxide
Density of oxygen compound
Mendeleev also claimed that certain experimental measurements had to be wrong. An
example is the case of beryllium, which at the time of the first publication of the periodic
table had been given an atomic weight of 14. This placed it in a group containing
nitrogen and phosphorus, with which it shared no similarities. Mendeleev believed
beryllium belonged with magnesium, because of the similarities in their oxides, and so
stated that the formula weight of beryllium must be less than 10. He was proven correct.
MENDELEEV WASN’T ALWAYS CORRECT!
Another apparently wrong atomic weight — tellurium 127 — caused a problem
with the sequence because it and iodine were in the wrong groups. One of
Mendeleev’s believers wasted many years trying to purify tellurium so that its
atomic weight could be corrected to about 125. He never succeeded because
tellurium does weigh more than iodine!
LIBRARY WORK
Look up the atomic weight, melting point and density of aluminium and indium
and the formulae of the oxides in a reference book. Use this information to
predict the properties of eka-aluminium, which was eventually called gallium.
PRACTICE QUESTIONS
1.
Justify Mendeleev’s assertion that beryllium should belong with magnesium,
rather than nitrogen.
2.
Find all the examples of elements that do not fit the sequence of atomic weights
originally used to arrange the elements in the periodic table. Explain why this
occurs.
7.2
SECTIONS OF THE PERIODIC TABLE
The terms group (column) and period (row) are often used to describe a set of elements,
for example, the column with Li at the top is called Group 1A; Na through to Ar is called
the third period. Figure 7.2 shows the main ‘regions’ of the periodic table.
The position of hydrogen is ‘floating’ because it has some properties in common
with both groups 1A (it forms 1+ ions) and 7A (it forms diatomic molecules and covalent
bonds).
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Chemistry for Technicians
The main-group elements were those from which the first idea of periodic
behaviour in the elements was developed.
H
Main-group
elements
(Groups
1A to 2A)
Main-group elements
(Groups 3A to 8A)
Transition metals
(Group B elements)
Lanthanides
Actinides
Figure 7.2 Regions of the modern periodic table
The transition metals are so-called because they form a transition between the two
blocks of main-group elements. Their chemical properties are unlike those of the main
group metals (1A–3A).
The lanthanides and actinides fit in atomic number order after element 57
(lanthanum) and element 89 (actinium), respectively. They are placed separately to make
the table more compact. They are all metals: the lanthanides are often called the rare
earth metals, because they have been isolated from rocks and are extremely uncommon.
PRACTICE QUESTION
3. To what region of the periodic table does each of the following elements belong:
(a) rhenium (b) indium (c) uranium (d) praseodymium?
7.3
TRENDS IN THE MAIN-GROUP ELEMENTS
The periodic table was developed on the basis of similar properties. As more elements
were added to it, some trends in properties were observed, in groups or periods. The
properties were those used to build the groups in the first place:
•
•
•
•
metallic character
density
melting and boiling points
reactivity and properties of simple compounds (chlorides and oxides).
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Chapter 7
The Periodic Table
These trends varied from one section of the table to another, and are of most use to the
main-group elements. It means that we can know something about all these elements
without looking their properties up in a book.
One of the earliest observations was that the metals were concentrated on
the left-hand side of the table, and the non-metals on the right. In the
In the middle were some elements with properties of each class: these were
class: these were called the metalloids. Figure 7.3 shows the dividing line
dividing line between metals and non-metals, and the metalloid elements.
B
C
N
O
F
Al
Si
P
S
Cl
Ga
Ge
As
Se
Br
In
Sn
Sb
Te
I
Tl
Pb
Bi
Po
At
← Metals
Non-metals →
Figure 7.3 The dividing line between metals and non-metals (metalloid elements are shaded)
LIBRARY WORK
1. Look up a reference book to find properties of metalloid elements that
clearly show the dual nature of this class of elements (e.g. appearance,
conductivity, melting point). The CRC Handbook of Physics and Chemistry
(published by CRC Press) is a good source of general information on the
elements.
2. Your teacher will assign you a group and a period of the periodic table.
Using the reference books available, find the following information about the
elements in the set you are assigned: physical state, metal/non-metal,
electrical conductivity, melting and boiling points, density, acidity of oxide,
ionic character of chloride.
What trends do you observe down the group/across the row for each
property? Do they correspond with those in Table 7.2?
PRACTICAL WORK
If you haven’t done the practical work in Section 3.1 of Part A of this textbook,
which looks at the properties of row 3 elements, it would be suitable to perform it
at this point.
Other trends in properties are outlined in Table 7.2. Those relating to measured
quantities (e.g. melting point) are not totally universal — there are minor exceptions in
most cases. However, the overall trends allow chemists to make predictions and to
maintain a working knowledge of the general properties of most of the elements without
the need for too much memorising.
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Chemistry for Technicians
Figure 7.4 shows the trends in ionisation energy (the energy required to form a 1+
ion) across the main group elements in periods 2–6. Of the properties of the elements that
have been studied, the ionisation energies probably best show the consistent patterns
evident in the periodic table.
Table 7.2
Trends in the periodic table
Down a group
Across a period (L to R)
increases (except 4A)
increase to 4A then decrease
Metallic character
increase
decrease
Density
increase
decrease
Atomic size
increase
decrease
Acidity of oxide
decrease
increase
Ionic nature of
chloride
increase
decrease
Reactivity
increase
1A and 7A maximum,
4A and 8A minimum
Melting point/boiling
point
Period 2
Period 3
Period 4
Period 5
Period 6
1A
2A
3A
4A
5A
6A
7A
8A
Group
Figure 7.4 Trends in ionisation energy
Some of these trends in properties can be explained by aspects of bonding, others
by the increasing atomic weight of the elements, and others by effects which are beyond
the scope of this module.
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Chapter 7
The Periodic Table
EXAMPLE
Why does the metallic character of elements increase down a group, but decrease across
a period?
The metallic bond is formed by elements wanting to lose electrons. As we go down a
group, the valence (outer shell) electrons are further away from the nucleus, and,
therefore, the electrostatic attraction gets less. Therefore, the atoms are even more able
to lose their electrons.
Across a period, the elements at Group 4A begin to need to gain electrons to
become stable, which is the opposite of metallic character.
PRACTICE QUESTIONS
4. Where in the periodic table (left/right/centre, top/bottom) would you expect to
find elements fitting the following descriptions?
(a) a shiny grey solid with the ability to form ions and covalent bonds, a
limited ability to conduct electricity
(b) a very reactive, low density metal, forms an alkaline ionic oxide
(c) a moderately reactive, purple solid, non-conductive
(d) a reactive gas
5.
Francium and astatine were the last two main group elements to be discovered,
since they are very unstable and extremely rare, if they exist at all on Earth.
(a) plot a graph of melting point against table period for the elements above
them in each group.
(b) use this graph to predict their melting points given the data below.
Period
2
3
4
5
6
Group 1A
181
98
64
39
29
Group 7A
–220
–101
–7
114
—
6.
What properties would you expect the element polonium to have, compared to
the other elements in the same period and group?
7.
Explain why:
(a) the size of the atom gets larger down a group, but smaller across a period
(b) the metallic character increases down a group, but decreases across a period
(c) elements with the highest melting points are found in the centre of a period.
7.4
TRENDS IN OTHER SECTIONS OF THE TABLE
You might have gained the impression that the main-group elements were the only
important ones, given the amount of ‘airplay’ they have had. Not so! Look at the
transition metals: how many commercially important metals can you find? Just about all
of them — aluminium is probably the only metal that is produced in large quantities by
the metals industry that isn’t a transition metal. Likewise, the lanthanides and actinides
contain some significant (though less well known) members.
The reason that we have concentrated on the main-group elements is that their
chemistry is much simpler and predictable. What do we mean by this? A main-group
metal, such as sodium or aluminium, forms one cation as predicted by the octet rule. No
covalent bonds are possible, and the ionic salts are almost always colourless. Their
physical properties vary depending on the group.
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Chemistry for Technicians
A transition metal will normally form at least two different cations, and have a
number of other ways of becoming stable. Transition metal compounds often feature
bonds other than ionic, and are coloured more often than not. The means by which
transition elements achieve stability cannot be explained at this stage in your knowledge,
another reason to leave them alone.
The 10 groups that form the transition metals do have similarities within the groups,
but vary much less from one group to another than do the main group elements. The
physical properties do exhibit some common trends: for example, the copper/silver/gold
group show the best conductivity properties of each period, while the peak density occurs
in the cobalt/rhodium/iridium group.
EXAMPLE
What properties does the transition metal, manganese, exhibit?
It forms 2+ and 3+ ions, an oxide with the formula MnO2 (this is not an example of Mn
occurring as a 4+ ion — the bonding is different), and joins with oxygen in the anion
permanganate, MnO4–, which forms intensely purple-coloured ionic compounds.
The 14 lanthanide (often called rare earth) elements are remarkably similar in
their physical and chemical properties. They all form 3+ ions, and some also form 2+ or
4+ ions. Their melting points vary from 800–1600°C (compare the range of 29–725°C
for the main-group metals in period 6), and the densities from 6.8 to 9.85 g/cm3
(1.9–11.9 g/cm3 for main-group period 6). Again, explanations are not possible at this
stage of your learning.
A similar story applies to the actinides, which are the equivalent of the lanthanides
in the next period. Of these elements, only numbers 90–94 are naturally occurring; the
rest are man-made in nuclear reactors.
PRACTICE QUESTION
8. List characteristics of (a) the transition metals and (b) the lanthanides that make
them different to the main-group metals.
LIBRARY WORK
Choose an element from each row of the transition metal block, the lanthanides
and actinides. Find out as much as you can about the chemical and physical
properties of these elements, some compounds that they form and their
commercial or industrial importance.
WHAT YOU NEED TO BE ABLE TO DO
•
•
•
•
Explain why the periodic table idea was successful.
Name the regions of the periodic table.
Predict the properties of main group elements.
Describe trends in properties across periods and down groups in the periodic
table.
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Chapter 7
The Periodic Table
TERMS AND DEFINITIONS
During this chapter, you have met a number of new terms, listed below. Match them with
their correct definitions.
Terms
A main-group element
C
lanthanide element
B
D
transition metal
actinide element
Definitions
1. an element in periods 4–6 in the block between the main-group elements
2. an element in the groups 1A–8A
3. an element in the separate block of the table with atomic numbers 90–103
4. an element in the separate block of the table with atomic numbers 58–71
91
NAMES AND
FORMULAE REVISITED
8
PURPOSE
To extend the rules for naming inorganic compounds and to introduce the class of
organic compounds and their naming.
It is recommended that you revise Chapter 4 before progressing with this chapter.
8.1
HYDRATED COMPOUNDS
In your laboratory work so far, you may have noticed on the labels of some
ionic compounds, the inclusion of water molecules in the formula, for
formula, for example, CuSO4.5H2O. Such compounds are called hydrated:
called hydrated: they contain water molecules as part of their crystal
crystal structure, with the anions and cations. They are as much a part of
the formula as the ions themselves, and must be included when calculating
the formula weight.
The ionic compound is not wet as we know it, but generally formed of regular
crystals. The water cannot be removed by heating at 100°C, but generally require up to
300°C, which is further evidence that it is more than just moisture. An ionic compound
that can exist as hydrated, but is not, is called anhydrous.
Naming these compounds is a matter of naming the ionic part as normal then
attaching a separate word on the end to indicate the presence of water molecules, using a
prefix for the number and the suffix hydrate for water. The prefixes are the same as
those used in molecular compounds (see Table 4.2).
In a formula, the water molecules are written after the ionic formula with a full stop
separating the two parts of the compound.
EXAMPLES
1.
What name corresponds to CuSO4.5H2O?
Copper (II) sulfate pentahydrate
2.
What is the formula of iron (III) nitrate nonahydrate?
Fe(NO3)3.9H2O
PRACTICE QUESTIONS
1. Name the following compounds:
(a) MgSO4.7H2O
(b) Ba(OH)2.8H2O.
2. Write the formula corresponding to the following names.
(a) iron (II) fluoride tetrahydrate
(b) calcium sulfate dihydrate.
92
Chapter 8
8.2
Names and Formulae Revisited
SOME MORE IONS
You should be familiar with the names and formulae of the common ions listed in
Chapter 4. It is now time to add some more to the list. These are given in Table 8.1. What
makes some of these unusual is that they are anions that contain metals.
Table 8.1
Names and formulae of some more ions
Name
Formula
2–
Name
Formula
Nitrite
NO2–
CN–
Chromate
CrO4
Dichromate
Cr2O72–
Cyanide
Permanganate
MnO4–
Perchlorate
ClO4–
PRACTICE QUESTIONS
3. Name the following compounds:
(a) KClO4
(b) Na2CrO4.
4.
Write the formula corresponding to the following names:
(b) zinc dichromate
(c) sodium permanganate.
8.2.1 Some Comments About the Names of Ions
By now, you may have noticed some common features about the confusing names that
exist for anions of sulfur and nitrogen. Each has a common series of endings: -ide, -ite
and -ate. If you look at each series of ions, the pattern becomes apparent: -ide means no
oxygen, -ate means the most oxygens and -ite actually means one less than -ate.
Unfortunately, after this ‘system’ was developed (it is also used for ions of
chlorine, bromine and iodine), other ions with more oxygen than the -ate ion (or less than
-ite) were found, so the prefix per- was brought into use to mean ‘more than -ate’, and
hypo- to mean less than -ite. If this is confusing, don’t worry — just remember the names
of the ions that have been listed here and in Chapter 4.
One other naming aspect for ions refers to the old way of distinguishing between
cations of different charge of the same element, such as Fe2+ and Fe3+. The endings -ous
and -ic were used to indicate the lower and higher charge, respectively. So, iron (II) and
iron (III) used to be called ferrous and ferric. This is not part of the new system, but is
still commonly used, for example, on bottle labels of chemicals.
EXAMPLES
1.
How does the naming system work for anions of chlorine?
You have met chloride (Cl–) and perchlorate (ClO4–). These are the extremes in numbers
of oxygens per chlorine in the sequence. Other oxychlorine ions are:
hypochlorite ClO–
chlorite ClO2–
chlorate ClO3–
2.
What were the old names for copper (I) and copper (II)?
Cuprous and cupric, respectively.
Note:
The aspects of anion and cation naming discussed in this example are not
required knowledge, but have been included for your general information.
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Chemistry for Technicians
8.3
EMPIRICAL, MOLECULAR AND STRUCTURAL FORMULAE
Up to now, we have used the term formula to refer to the number of atoms of each
element existing in a compound. This is the simplest definition of a formula, but it
doesn’t tell us everything we may wish to know about the compound.
The formula of molecular compounds reflects the number of atoms in the molecule.
However, when we look at ionic compounds, the formula is only the simplest ratio of the
number of atoms of each type. The term ‘ionic molecule’ doesn’t apply in the solid or
liquid form.
To distinguish between the information that each formula provides, two
different names have been given:
•
empirical formula, which is the simplest ratio of atom numbers (it
can be applied to any type of compound), and
•
molecular formula, which is the actual number of atoms in a
molecule (and can’t be applied to ionic compounds in solid or liquid
form).
EXAMPLE
What is the empirical and molecular formula of dinitrogen tetroxide?
The name tells us that there are two nitrogen atoms and four oxygens. Thus, the
molecular formula is N2H4, while the empirical formula is NH2.
The empirical formula in ionic compounds, where the ions are polyatomic, could be
simplified entirely to numbers of atoms, but this would not be very informative. For
example, the ‘true’ empirical formula for copper (II) nitrate would be CuN2O6. Such a
formula masks the fact that nitrate is involved.
The molecular formula doesn’t tell us what atoms are bonded to each other,
and what type of bonds are involved: this is known as the compound’s
compound’s structure. The structural formula spells out in detail the
detail the intramolecular bonding, that is, which atoms are connected to
connected to each other.
EXAMPLE
Draw the structural formula for N2H4.
H
H
N
H
N
This would often be abbreviated
as NH2–NH2 to save space
H
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Chapter 8
Names and Formulae Revisited
PRACTICE QUESTIONS
5. Explain why NaCl is an empirical formula, not a molecular formula.
6.
An important laboratory chemical is the ionic compound sodium oxalate. The
oxalate ion has the formula C2O42–.
(a) What is the ratio of sodium and oxalate ions in the compounds?
(b) What is the empirical formula for the compound?
(c) Explain why this would not be a suitable description of the compound.
7.
Draw the structural formula for (a) water and (b) boron trifluoride.
8.4
WHAT ARE ORGANIC COMPOUNDS?
The simple definition of an organic compound is one that contains carbon. This is
perhaps a bit simple, because salts of carbonate and hydrogen carbonate are not organic,
but contain carbon. So our definition becomes a combination of the two: a compound
containing carbon atoms, other than carbonate and hydrogen carbonate salts.
Carbon has the unusual ability to form covalent bonds with itself and other nonmetal elements in many configurations. This means that organic compounds are the most
common class of compound on Earth. There are millions of organic compounds, and
the number continues to grow. Many are naturally occurring, but more and more are
man-made. Figure 8.1 shows some of the possible structures formed by carbon
compounds.
Br
Non-carbon atom
C
Ring
C
C
C
C
C C
Double bond
C
Chain
C
C
Figure 8.1
Branched chain
C
Structural features occurring in carbon compounds
Carbon atoms always form four covalent bonds. In Figure 8.1 the remaining bonds are to
hydrogen atoms, and have been omitted for the sake of simplicity. This is very common
in the drawing of organic compounds.
Other elements involved in organic compounds are hydrogen (in almost all
compounds), oxygen, nitrogen, the halogens, phosphorus and sulfur.
Most of the compounds whose names you would recognise from media reports and
from everyday use are organic compounds or mixtures. Table 8.2 lists some of those
compounds, sub-dividing them as naturally occurring or man-made.
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Chemistry for Technicians
Table 8.2
Some well-known organic substances
Man-made
Natural
Polythene, DDT, heroin, aspirin
Alcohol, penicillin, strychnine
Paracetamol, TNT, nylon
Sugar, protein, MSG, LSD
The names given in general use to these substances are normally not systematic:
they are common names, often commercial brand names, used by non-scientific people to
refer to these substances. Sometimes they are abbreviated for ease of use, for example,
MSG from monosodium glutamate. The problem with common names is that they do not
give a direct indication of the structure of the substance, so that you need to know both
the name and structure of the substance. If you remember, the most important aspect of
the systematic naming rules is that if you know the rules, you can work out the name
from the structure and vice versa. Common, non-systematic names do not allow this.
PRACTICE QUESTIONS
8. Re-draw the carbon compound in Figure 8.1 with all hydrogen atoms shown.
9.
Which of the following formulae would not be that of an organic compound?
(a) NaC2H3O2
(b) C4H5ClO3
(c) C50H102
8.5
FUNCTIONAL GROUPS
One of the features of organic compounds is that common groups of atoms occur in many
different substances, and give these substances certain predictable physical and chemical
properties. Ionic compounds are similar: the nitrate ion, which occurs with many
different cations, gives those compounds certain properties. In organic compounds, these
atom groupings are called functional groups. They form parts of molecules: they are the
building blocks. Table 8.3 lists some of the important functional groups with which you
should become familiar.
Table 8.3
Common functional groups (hydrogens attached to carbons have been omitted)
Name
Structure
Name
Structure
C
Alkane
C
C
Alkanol
Alkene
C
C
Carbonyl
C
O
Alkyne
C
C
Alkanoic acid
C
O
OH
OH
Alkyl halide
C
X
Amine
(X = F, Cl, Br, I)
96
C
N
Chapter 8
Names and Formulae Revisited
EXAMPLE
What functional groups are in the compound shown in Figure 8.1?
There are lots of alkane groups (carbon–carbon single bond), an alkene group
(carbon–carbon double bond), and there is a alkyl halide group (carbon–halogen covalent
bond).
Functional groups make the study of organic compounds much easier: the presence of an
alkanol group tells us something about the properties of that compound. It doesn’t matter
whether the compound has a few carbons or lots of them, the alkanol group will behave
chemically and physically in the same way.
PRACTICE QUESTION
10. What functional groups are contained in the compounds in Figure 8.2?
Carbon–hydrogen bonds are not included in structures for simplicity.
(a)
(b)
C
C
C
C
C
(c)
C
C
C
C
C
OH
C
C
C
C
O
(d)
(e)
C
C
C
C
(f)
OH
C
O
C
H
C
O
(g)
(h)
C
C
C
C
C
C
(i)
C
C
C
C
NH2
Br
C
C
C
C
Figure 8.2 Question 10
8.6
AN INTRODUCTION TO NAMING ORGANIC COMPOUNDS
It should be obvious that organic compounds are molecular compounds. However, they
are not named in the same way that we used in Chapter 4. They have their own system of
naming, which is quite different. This section is only intended to give the very basics of
that system. Further detail is covered in the module Qualitative Organic Chemistry,
which may not be part of the course in which you are enrolled.
We will limit ourselves to straight-chain compounds containing one of the following functional groups: alkanes, alkenes, alkynes, alkanols, carbonyl (in two different
forms) and alkanoic acids. The naming system for these simple organic compounds is
based on a two-part single word name using the following structural features:
•
the number of carbons in the chain (indicated in the first half of the name and
listed in Table 8.4), and
•
the type of functional group (indicated by the second half of the name and listed
in Table 8.5).
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Chemistry for Technicians
Figure 8.3 shows the basic structure of the name.
Name start
defined by length of chain
Name ending
defined by functional group
Figure 8.3 Basic structure of name of organic compounds (straight chain only)
Table 8.4
Chain length names
Carbons
Name start
Carbons
Name start
1
meth-
5
pent-
2
eth-
6
hex-
3
prop-
7
hept-
4
but-
8
oct-
Table 8.5
Functional group endings
Functional group
Name ending
Functional group
Name ending
Alkane
-ane
Alkanol
-anol
Alkene
-ene
Carbonyl
-anal (end of chain)
-anone (not at end)
Alkyne
-yne
Alkanoic acid
EXAMPLES
Name the following compounds.
(a) an alkane with four carbons
(b) an alkanoic acid with two carbons
(c) CH3CH2CH2CH=CH2
(d)
H
H
(a)
(b)
(c)
(d)
(e)
C
H
H
C
H
H
C
H
H
C
H
-anoic acid
(e)
H
C
H
H
C
O
H
H
C
H
H
C
O
C
H
H
but- (four carbons) and -ane (alkane) = butane
eth- (two carbons) and -anoic acid (alkanoic acid) = ethanoic acid
pent- (five carbons) and -ene (alkene) = pentene
hex- (six carbons) and -anal (carbonyl on end of chain) = hexanal
prop- (three carbons) and -anone (carbonyl not on end of chain) = propanone
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Chapter 8
Names and Formulae Revisited
There is a problem with Example (c): another compound which could also be called
pentene, but which is a different compound. It has the double bond between the next two
carbons — CH3CH2CH=CH2CH3. Since they are different compounds, they must have
different names. The compound in Example (c) is called 1-pentene (the 1 indicates that
the double bond is on the end carbon) and the other is called 2-pentene (the double bond
is on the carbon second from the end). This extra part of the naming system will not be
discussed any further here.
PRACTICE QUESTION
11. Name each of the compounds in Q10(a)–(g). Which of these names requires
numbering to make them unique?
WHAT YOU NEED TO BE ABLE TO DO
•
•
•
•
•
•
Explain the meaning of the terms hydrated and anhydrous.
Name hydrated ionic compounds.
Distinguish between empirical, molecular and structural formulae.
Define the terms organic compound and functional group.
Recognise common functional groups.
Name simple organic compounds.
TERMS AND DEFINITIONS
During this chapter, you have met a number of new terms, listed below. Match them with
their correct definitions.
Terms
A hydrated
B anhydrous
C
empirical formula
D molecular formula
E
structural formula
F
organic compound
G functional group
Definitions
1
a compound, which could be hydrated, which has had the water molecules removed
2
an expanded formula or diagram showing the bonding arrangements for the atoms
3
the simplest ratio of numbers of elements (or ion) in a compound
4
a compound that contains a fixed number of water molecules in its structure
5
the actual number of atoms of each element in a compound
6
a grouping of atoms that occurs in many different organic compounds, and defines
their properties
7
any compound containing carbon, other than salts of carbonate and hydrogen
carbonate
99
9
HOW ATOMS BOND
PURPOSE
To describe the atomic changes occurring during bond formation, and to outline the
range of intermolecular bonds.
It is recommended that you revise Chapter 5 before progressing with this chapter.
9.1
WHY ATOMS BOND
The three main types of interatomic bonds — ionic, covalent and metallic — form as the
atoms try to attain a full outer shell of electrons by losing or gaining the required
number. In doing so, the atoms attain the same electron configuration as the nearest
Group 8A inert gas element. The chemical inertness of the noble gases (Group 8A) is
well known: why should they react when they have the stable electron configuration?
EXAMPLE
Which inert gases have the same electron configurations as sodium and chlorine when
they react to form sodium chloride?
Sodium has a configuration of 2.8.1. To become stable, it must lose the outer one, giving
a configuration of 2.8, which is the same as that for neon. Chlorine has a configuration of
2.8.7, and needs to fill its third shell by gaining an electron to become 2.8.8, which is the
same as that of argon.
The number of electrons that an element needs to lose or gain to achieve
the full outer shell is called its valency; thus sodium and chlorine both have
valencies of one.
Sometimes, the valency is further distinguished by a positive or negative sign, when
referring to ions. Table 9.1 summarises the changes required to achieve stability for
elements in the main-group elements.
PRACTICE QUESTIONS
1 What are the requirements for stability for (a) hydrogen, (b) calcium, (c) sulfur
and (d) neon?
2
What are the valencies of the elements in Q1?
100
Chapter 9
Table 9.1
How Atoms Bond
Requirements for periodic table groups to achieve full outer shells
Group
Number of outer shell
electrons
Change in electrons to achieve
full outer shell
1A
1
lose one
2A
2
lose two
3A
3
lose three
4A
4
lose or gain four *
5A
5
gain three
6A
6
gain two
7A
7
gain one
8A
8
no change
NOTE: the elements in Group 4A differ from top to bottom in the group in terms of the way that
they achieve stability. Carbon silicon and germanium attempt to gain rather than lose the four
electrons needed while tin and lead are more likely to lose rather than gain.
9.2
ELECTRON DOT DIAGRAMS
In the next two sections we will look at how the two main types of chemical
bonds (ionic and covalent) form. To help do this, we will use electron dot
electron dot diagrams to make the process simpler and clearer. An
clearer. An electron dot diagram of an element shows only its outer shell
outer shell electrons as dots or crosses around the symbol, as shown in
Figure 9.1. The first four electrons are indicated as separate dots (top,
bottom, left and right), and the remaining four form pairs.
••
Ca •
•
O
•
••
•
Figure 9.1 Electron dot diagrams for calcium and oxygen
PRACTICE QUESTION
3. Draw the electron dot diagrams for (a) aluminium, (b) sulfur and (c) magnesium.
9.3
FORMATION OF IONIC BONDS
In Chapter 5, we looked at the structure of ionic compounds. Here we will look at how
the ions form the neutral atoms. You are aware that an ionic compound must have cations
and anions with a balance of charges.
In Chapter 2, we saw that ions form by the loss or gain or electrons. Ionic
compounds form when one element loses electrons and another element gains them, as
shown in Figure 9.2. Sodium’s one outer shell electron is transferred to fill the one space
in chlorine. Sodium’s full second shell is now its outer shell.
101
Chemistry for Technicians
••
Na •
•
Cl
••
+
•
•
Na
••
•
•
Cl– ••
••
Na loses its one
electron to Cl
Figure 9.2 Formation of ions in NaCl
EXAMPLE
How does calcium chloride form?
Calcium has two outer shell electrons to lose, while each chlorine needs to gain only one.
Therefore, two chlorines must be present for every calcium.
••
•
•
Cl •
••
••
•
Ca •
•
Cl
•
•
••
Ionic compounds with polyatomic ions form in a similar way, but the picture is
somewhat more complicated and won’t be dealt with here.
PRACTICE QUESTION
4. Using electron dot diagrams, show how the ionic compounds (a) magnesium
oxide, (b) sodium oxide and (c) aluminium sulfide form.
9.4
FORMATION OF COVALENT BONDS
As described in Chapter 5, elements needing to gain electrons to become stable can form
covalent bonds, where the pair of atoms in a bond share a pair of electrons. In most
covalent bonds, each atom contributes one of its outer shell electrons. In doing so, each
‘sees’ another electron, without actually gaining it. No charge results because the atom
has a half-share in two electrons, which is equal to the one electron it owned previously.
Such an example is shown in Figure 9.3 with the formation of HCl.
••
H×
•
••
electrons to be
shared in bond
Figure 9.3
Cl
•
•
••
•
×
H Cl
••
•
•
H ‘sees’ 2 electrons in the
bond
Formation of a covalent bond between H and Cl
102
Cl ‘sees’ 2 electrons in the
bond and 6 others in its outer
shell — a total of 8
Chapter 9
How Atoms Bond
Such a bond is called a single covalent bond. Double and triple covalent
bonds (known as multiple bonds) can also occur, where two and three
and three pairs of electrons are shared, as in carbon dioxide and nitrogen,
and nitrogen, respectively. These are shown in Figure 9.4. Double and
Double and triple bonds are shown in structural formulae with double and
triple lines.
••
O
••
Figure 9.4
••
•×
•×
C
ו
ו
×
×
O
N
×××
•••
N ••
••
O=C=O
N≡N
(a)
(b)
Covalent bonds in (a) carbon dioxide and (b) nitrogen (shorthand version also shown)
PRACTICE QUESTIONS
5. Show how two covalent bonds between O and H are formed in water.
6.
Show how (a) O2 (b) NH3 and (c) CH4 form.
If you look at the O, N and Cl atoms in the above figures, you will notice
that not all the eight electrons in the outer shell are involved in covalent
in covalent bonds: chlorine has six, oxygen four and nitrogen two valence
two valence electrons that are not part of the covalent bonds. These are
These are known as non-bonded electrons, or lone pair electrons. They
are significant in the formation of intermolecular forces and other types of
chemical bonds not covered in this module.
PRACTICE QUESTIONS
7. Explain why carbon atoms in covalent compounds do not have any non-bonded
electrons.
8.
Explain why non-bonded electrons always occur in pairs.
Each covalent bond formed adds one extra electron to the outer shell of each atom.
Elements needing to gain more than one electron may form all single bonds or
combinations of single and multiple bonds. Table 9.2 lists the options for covalent bonds
for some common elements.
PRACTICE QUESTIONS
9. Explain what is wrong with the following structures.
(a) N=O—H
(b) H=C=O
(c) Cl=C—Cl
10. Construct a table for the three bonding types (ionic, covalent and metallic),
which indicates whether these bonds allow electron loss or gain).
103
Chemistry for Technicians
Table 9.2
Covalent bonding by common elements
Number of bonds
Elements
Types of bonds
1
H, F, Cl, Br, I
one single
2
O, S
one double
two single
one triple
3
N
one double and one single
three single
one triple and one single
4
C
two double
one double and two single
four single
9.5
ELECTRONEGATIVITY
You have seen the different ‘appetite’ of elements for electrons to make
them stable: metals, such as sodium, want to get rid of them, the halogens
the halogens desperately need them. A measure of the need of elements for
elements for electrons is called electronegativity. Elements with a need to
with a need to gain electrons to become stable have higher electronegativity
values than those wishing to lose electrons.
If we concentrate on only the main-group elements, the general trend is for an
increase in electronegativity from left to right (but not including the inert gases), but
decreasing down a group. Thus, fluorine is the most electronegative element, and
caesium the least. Table 9.3 lists electronegativity values for some of the main-group
elements.
The actual electronegativity values are not important to us; what is significant is the
relative values for two elements in a bond. The greater the difference, the more unevenly
shared the electrons, and vice versa. The extreme is an ionic bond, where one atom is
very hungry for electrons and the other is a willing supplier. However, this isn’t telling us
anything we didn’t know already: metals form ionic bonds with non-metals.
Table 9.3
Electronegativity values (EN) for selected elements
Element
EN
Element
EN
Element
EN
Hydrogen
2.20
Oxygen
3.44
Phosphorus
2.19
Lithium
0.98
Fluorine
3.98
Sulfur
2.58
Boron
2.04
Sodium
0.93
Chlorine
3.16
Carbon
2.55
Magnesium
1.31
Bromine
2.96
Nitrogen
3.04
Silicon
1.90
Iodine
2.66
104
Chapter 9
How Atoms Bond
What electronegativity values tell us is that all covalent bonds are not the same.
Until now, we have described a covalent bond as each atom putting one electron into the
bond, and sharing equally in the pair. A simple picture would be an inflated balloon with
the atoms at either end. However, if one of the atoms has a stronger pull on the electrons
(because of a higher electronegativity), the balloon gets ‘squeezed’ so that this atom has
more than its share. Figure 9.5 illustrates this point.
(a)
(b)
Atoms of similar EN
with an even share of
the electron 'balloon'
Atoms of different EN
with an uneven share of
the electron 'balloon'
Figure 9.5 The balloon analogy for covalent bonds
How does an uneven share of the electrons affect the atoms? To consider this, we have to
view the electrons in the bond, not as individual units, but as contributing to a volume
(i.e. the balloon) containing a 2– charge. That volume can be divided in any way. In
Figure 9.5(a), the even sharing of the electrons means that each atom has a 1– charge as
its share, which is equal to the 1– charge it put into the bond.
In Figure 9.5(b), the atom on the right (which has the higher EN) has more
of the volume and therefore, slightly more than half the 2– charge. It has
charge. It has gained a small amount of negative charge in the formation of
formation of the bond, while the other atom has lost some, and therefore,
and therefore, gains a slight positive charge. This is known as a polar
covalent bond, because of the slight + and – charges or poles in the bond.
The other bond (in (a)) is known as a non-polar covalent bond. In general,
a difference of at least 1 in EN values indicates a polar bond.
It should be made very clear that the + and – charges in polar covalent bonds are
not full charges, like they are in ionic bonds. The distribution of the electrons is not even
between the two atoms; atom A might have the ‘equivalent’ of 1.1 electrons and atom B
the other 0.9 — though it must be said that these fractions of electrons are not real, but
just a way of picturing what is happening. A polar covalent bond in a structural formula
will often have the symbols δ+ and δ– to clearly distinguish it from an ionic bond (δ is
frequently used in science to symbolise ‘a small amount of’).
105
Chemistry for Technicians
EXAMPLE
What type of covalent bonds would form between the following pairs of atoms?
(a)
carbon and hydrogen
(b)
oxygen and hydrogen
(c)
oxygen and oxygen
(a)
(b)
(c)
Carbon has an EN of 2.55 and hydrogen 2.2: a difference of 0.35. This is relatively
small, and the bond is generally considered to be non-polar.
Oxygen has an EN of 3.44: now the difference is 1.24. This is quite significant, and
the bond is polar.
Bonds between atoms of the same element must be non-polar, because the EN
values are the same.
PRACTICE QUESTIONS
11. Which of the following covalent bonds is considered polar?
(a) C=C (b) C=O (c) C—N (d) N—O (e) C—F
12. Which of the elements listed in Table 9.3 would form a polar covalent bond with
hydrogen?
9.6 INTERMOLECULAR FORCES
Intermolecular forces (or bonds) are those between molecules. They affect the physical
properties of the substance: melting and boiling points, density, solubility, etc. The
stronger the intermolecular forces are, the higher these values will be because the
molecules are held together more strongly.
Ionic and covalent network compounds do not really have individual molecules.
The forces that hold the atoms together also hold the ‘molecules’ together. This gives
these types of substances very high melting and boiling points, relatively high densities
(though not as high as metals) and in the case of networks, very limited solubility.
In this section, the effect of intermolecular forces on the physical properties of
‘true’ molecular compounds — covalent molecular compounds — will be described. In
this type of compound, there are the strong covalent bonds within the molecules, holding
the atoms together, and relatively weak forces between the molecules, as shown in Figure
9.6.
Weak intermolecular force
Covalent intramolecular bonds
Figure 9.6 Intra- and intermolecular bonds in covalent molecules
106
Chapter 9
How Atoms Bond
What causes these intermolecular forces in covalent molecules? In the previous
section, we saw that certain covalent bonds developed small positive and negative
charges. The most significant intermolecular forces are simply caused by attraction
between atoms in different molecules with opposite δ+ and δ– charges. The greater the
difference in EN values between atoms in a covalent bond, the greater the charge buildup and the stronger the intermolecular force as a result. Figure 9.7 shows the
intermolecular forces in water.
The forces between the slightly positive atoms (e.g. hydrogen) and slightly
negative atoms (e.g. oxygen) are called dipole attraction. Because, in
Because, in water, they are particularly strong, and involve hydrogen as
hydrogen as one of the atoms, they are given a special name — hydrogen
— hydrogen bonds. They occur in compounds where hydrogen is attached
to a strongly electronegative element, that is F, Cl, N and O. Hydrogen
bonds are the strongest of the intermolecular forces.
δ+ δ+
H δ− H
O
Hδ+
O δ−
H δ+
Hδ+
O δ−
H δ+
Figure 9.7 Intermolecular forces in water
Dipole attractions between other polar covalent bonds are not as strong as hydrogen
bonds. Figure 9.8 shows dipole attractions between molecules of methanal.
H δ+
C O
δ−
H
H δ+
C O
δ−
H
Figure 9.8 Dipole attraction between methanal molecules
Molecules without polar covalent bonds exhibit very weak intermolecular
forces, known as dispersion forces. It is somewhat difficult to explain why
any attraction should occur between molecules, such as methane, and we
won’t attempt to do so here.
PRACTICE QUESTION
13. What type of intermolecular forces would occur in (a) HCl and (b) ethane?
107
Chemistry for Technicians
9.6.1 Effect on Melting and Boiling Points
What effect do the different types of intermolecular forces have on the physical
properties of covalent molecules? Because even hydrogen bonds are much, much weaker
than ionic or covalent bonds, it doesn’t take much energy (heat) to ‘unstick the glue’
between the molecules. Therefore, the melting and boiling points of covalent molecules
are much lower than ionic, metallic or covalent network substances.
Within the class of covalent molecules, those with hydrogen bonds will have the
highest melting and boiling points and densities. One other point that needs to be made
when comparing physical properties is that the formula weight of a compound should be
taken into account. The heavier the molecule, the denser it will be and the more heat is
required to melt and boil it.
EXAMPLES
1.
Explain why water has a boiling point almost 300°C higher than oxygen gas.
Water has hydrogen bonds which are much stronger. The bond in oxygen gas
(O=O) is non-polar, because both atoms have the same electronegativity. It requires
much more heat to break the hydrogen bonds in water than the weak dispersion
forces in oxygen, even though oxygen is heaver (32 compared to 18).
2.
Arrange the following organic molecules in order of increasing boiling point:
methanol, ethanol, ethane, methanal.
The formula weights for methanol (32), ethane (28) and methanal (30) are similar,
but ethanol (46) is higher.
H
H
C
δ−
O
H
H
H
Methanol
δ+
C δ− H
δ+
C
O
H
H
H
H
Ethanol
H
H
C
H
H δ+ H
C
H
C
H
H
Ethane
O
δ−
Methanal
The structural formulae above for the four compounds shows that only ethane has
no polar bonds. It, therefore, must have the lowest boiling point. Of the others, the
polar bond in methanol and ethanol is the OH bond, which allows for hydrogen
bonding. The C=O bond in methanal allows dipole attraction, but is weaker than that
in the alkanols. Methanal will have the second lowest boiling point. Methanol will be
next highest, and ethanol the highest because its intermolecular forces are as strong
as methanol and it is heavier.
Order (with actual b.p. in °C given): ethane (–89), methanal (–21), methanol (61),
ethanol (78).
PRACTICE QUESTIONS
14. Water and ammonia both have hydrogen bonds between the molecules. Why
does water have a higher boiling point?
15. How would you expect the boiling point of HF to compare with that of water and
ammonia?
16. Why does peroxide (H2O2) have a higher boiling point than water?
17. Explain the boiling point trend in the simple organic compounds, propanol
(97°C), propanal (–21°C) and butane (–89°C).
108
Chapter 9
How Atoms Bond
LIBRARY WORK
Look up the boiling points of the hydrogen compounds of Groups 6A and 7A.
Explain the trends within a group and between compounds of elements in the
same row.
9.6.2
Effect On Solubility
The golden rule for predicting the solubility of compounds in solvents is
‘like dissolves like’: compounds dissolve in solvents with similar types of
similar types of bonds. Thus, compounds with polar or ionic bonds dissolve
bonds dissolve in solvents with polar bonds (there are no ionic liquids),
liquids), while compounds without polar bonds dissolve in non-polar
solvents. Table 9.4 classifies some important solvents by their polarity.
Table 9.4
Classification of solvents by polarity
Polarity
Common solvents
Very high
water
Moderate
methanol, ethanol, propanone (acetone)
Low (or non-polar)
hexane, trichloromethane (chloroform)
EXAMPLES
1.
Explain why sodium chloride and ethanol both dissolve in water, but ethane doesn’t.
Water contains polar covalent bonds, which will have an attraction for the ions in
sodium chloride, and the polar OH bond in ethanol. Ethane, without polar bonds is
not attracted to the water.
2.
Predict the most suitable solvent (from water, ethanol and hexane) for calcium
chloride, methanal and butane.
Calcium chloride (ionic): water
Methanal (polar covalent): water or ethanol
Butane (non-polar): hexane
PRACTICE QUESTIONS
18. Explain why butanal is more soluble in hexane than butanol.
19. Which would be more soluble in water — butanol or butanoic acid?
20. What would be a suitable solvent for naphthalene (see Figure 5.10 for its
structure)?
109
Chemistry for Technicians
WHAT YOU NEED TO BE ABLE TO DO
•
•
•
•
•
•
•
•
Predict the requirement for stability of an element based on its position in the
periodic table.
Draw electron dot diagrams for the first 20 elements.
Describe how ionic and covalent bonds form.
Indicate the number of covalent bonds formed by a given element.
Define the terms electronegativity and polar covalent bond.
Describe trends in electronegativity in the periodic table.
Describe intermolecular forces between covalent molecule.s
Explain differences in physical properties of covalent molecules.
TERMS AND DEFINITIONS
During this chapter, you have met a number of new terms, listed below. Match them with
their correct definitions.
Terms
A valency
C
double covalent bond
E
polar covalent bond
G hydrogen bond
I
dispersion force
K ‘like dissolves like’
B
D
F
H
J
electron dot diagram
electronegativity
non-polar covalent bond
dipole attraction
non-bonded electrons
Definitions
1
four electrons shared between a pair of atoms
2
very weak intermolecular forces between non-polar bonds
3
valence electrons not involved in covalent bonding
4
a covalent bond where each atom’s electronegativity is similar
5
the number of electrons that an atom needs to lose or gain to achieve stability
6
attraction between polar covalent bonds, where the positive and negative centres in
different molecules are weakly linked
7
a shorthand way of representing the number of valence electrons in an atom
8
the solubility rule which says that compounds will dissolve in solvents of similar
polarity
9
a covalent bond where one atom’s electronegativity is considerably greater than the
other
10 a force between a hydrogen atom attached to an electronegative element and that
element in another molecule
11 a measure of an element’s need for electrons
110
10
PREPARING SOLUTIONS
PURPOSE
To perform calculations associated with solution preparation, to prepare those solutions
and check their concentrations.
10.1 CONCENTRATION UNITS
Many chemical processes occur in solution. How then can the amount of reactant added
be controlled if it is in the form of a solution? If 10 g or 0.1 moles of reactant is required,
how much solution should be used? The answer is by a knowledge of the concentration
of the solution and the unit in which the concentration is expressed.
The solution concentration, is a numerical measure of the amount of
solute per volume of solution. The great majority of concentration units,
concentration units, and there are lots of them, have a common form: they
common form: they are a ratio of the amount unit (e.g. moles, grams, mL)
grams, mL) to the volume unit (e.g. litres). The unit that expresses the
concentration is a reflection of this ratio (e.g. mole/L, g/100 mL, mg/mL).
It is important at this stage to make absolutely clear that the unit is a ratio, and
therefore, completely independent of the volume of the solution. A cup of sea water has
the same concentration as the ocean that it came from. The mass of salt in the cup is
much, much less than in the ocean, but so is the volume, and so the concentration ratio is
the same. This works because solutions are homogeneous. Figure 10.1 further illustrates
this point.
10 g
10 L
Figure 10.1
1g
1L
1g
1L
1g
1L
1g
1L
1g
1L
1g
1L
1g
1L
1g
1L
1g
1L
1g
1L
10 grams in 10 litres or 10 separate one-litre volumes each containing one gram.
Are they the same? Yes!
In this module, we will concentrate on the most important units (listed in Table
10.1) and the calculations associated with preparing solutions. The actual practical skills
of solution preparation are dealt with more thoroughly in the module Chemical
Laboratory Techniques.
111
Chemistry for Technicians
Table 10.1 Concentration units discussed in this chapter
Name
Abbreviation
Meaning
Molarity
M or mole/L
the number of moles of solute per litre of solution
Grams per litre
g/L
Grams per 100
millilitres
g/100 mL
or %w/v
the mass of solute in grams per 100 mL of
solution
mg/L
the mass of solute in milligrams per litre of
solution
Milligrams per litre
Grams per 100
grams
Millilitres per 100
millilitres
%w/w *
%v/v
the mass of solute in grams per litre of solution
the mass of solute in grams per 100 grams of
solution
the volume of liquid solute in mL per 100 mL of
solution
NOTE: the %w/w unit is more frequently used for reporting concentrations of analyte (see Section 11.3) in
solid materials, where the definition is ‘grams of analyte per 100 g of sample’. It is really only used for
solutions of concentrated acids, for example, sulfuric acid, 98%w/w.
Concentration units involving mass are not affected by the nature of the solute: one
litre of 1 g/L NaCl solution and one litre of 1 g/L ethanol both contain one gram of
solute. If the unit involves moles, then the solute identity is important. One litre of 1 M
NaCl solution contains 58.45 grams of solute, while one litre of 1 M alcohol solution
contains only 46 g of solute. Concentration units, regardless of type, are not affected by
the type of solvent used.
10.1.1 Why Are There So Many Different Units?
The first reason is that the amount of solute can be measured in either mass, volume or
number of moles, meaning that there are three classes of units for solution
concentrations. Secondly, within each class, a number of units are used so that dilute and
concentrated solutions can be described without using very small or very large numbers.
The general rule is that units with milli- or micro-amounts are used for dilute
solutions. For example, the unit mg/L would be better than g/L to express the
concentration of low levels of mercury in a river, because milligrams are smaller units
than grams (1 mg/L is the same as 0.001 g/L). We will look at how to convert from one
unit to another in Section 10.4.
PRACTICE QUESTIONS
1. Explain why the statement ‘The concentration of ethanol (alcohol) in a glass of
beer is the same as in the keg that the beer came from’ is true only if the beer is
homogeneous.
2.
Figure 10.1 divides 10 L of 1 g/L solution into 10 individual one-litre portions.
Divide the 10 L into 100 portions. What volume (in L) is each portion? What
mass (in g) is in each portion? What is the concentration (in g/L) in each portion?
112
Chapter 10
Preparing Solutions
10.2 HOW MUCH SOLUTE DO I NEED?
Preparing solutions by dissolving a certain amount of solute in a certain volume of
solution is one of the common tasks performed by a laboratory technician. It is vital that
the solution is of the concentration expected, otherwise many other tasks using the
solution will be affected. Some simple calculations have to be done before a solution of a
specified concentration can be made.
The calculations in this section assume that the solute is pure. If it is in solution
already, then different calculations are required. These will be covered in the next
section. Having determined that the solute is pure, we then need to consider the amount
unit in the concentration unit. Is it mass, moles or volume?
In the majority of commonly used solution concentration units (%w/w is the main
exception), the concentration is equal to the number of amount units divided by the
number of volume units, as shown in Equation 10.1. If you remember this, your solution
making will be much easier.
Concentration =
number of amount units

number of volume units
Eqn 10.1
10.2.1 Mass units
Equations 10.2a–c summarise the relationships between concentration, mass and volume.
Concentration
= mass ÷ volume
Eqn 10.2a
Mass
= concentration x volume
Eqn 10.2b
Volume
= mass ÷ concentration
Eqn 10.2c
To use any of these equations, you must know two of the three pieces of information.
If you think you know only one, then you are in trouble!
EXAMPLES
1.
What concentration in g/L is obtained when 1.5 g of solute is dissolved in 300 mL of
solution?
Always convert the mass and volume units of the data you are given to those
specified in the concentration unit before proceeding!
300 mL
concentration
2.
=
=
=
=
=
(300 ÷1000) L
0.3 L
mass ÷ volume
1.5 g ÷ 0.3 L
5 g/L
Calculate the mass required to make 2L of 200 mg/L solution.
mass
= concentration x volume
= 200 x 2
= 400 mg
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Chemistry for Technicians
3.
You are required to dissolve 100 mg of solute in sufficient solution to give a
concentration of 0.05 g/100 mL. What volume is required?
100 mg
4.
=
=
volume
=
=
=
=
What concentration
solution?
(100 ÷ 1000) g
0.1 g
mass ÷ concentration
0.1 ÷ 0.05
2 x 100 mL
200 mL
(in %w/w) is obtained by dissolving 3 g of solute in 500 g of
%w/w is the major exception to the amount/volume rule. %w/w means grams of
solute per 100 g of solution, and can be calculated from the equation below.
concentration
= mass of solute x 100 ÷ mass of solution
= 3 x 100 ÷ 500
= 0.6 %w/w
PRACTICE QUESTIONS
3. Calculate the mass of solute required to prepare the following solutions:
(a) 100 mL of 0.1 g/L
(b) 2.5 L of 200 mg/L
(c) 50 mL of 4 g/100 mL
(d) 250 g of 0.5 %w/w.
4.
Calculate the concentration of the following solutions in the unit given:
(a) 2 g in 100 mL (g/L)
(b) 0.015 g in 500 mL (g/100 mL)
(c) 0.015 g in 500 mL (mg/L)
(d) 1.25 g in 75 g (%w/w).
5.
What volume of solution should be prepared in the following?
(a) 50 g dissolved to make 5 g/L
(b) 100 mg dissolved to make 0.2 g/100 mL
(c) 0.2 g dissolved to make 1000 mg/L.
10.2.2 Molarity
The basic relationships are the same between concentration, amount of solute and
volume as those above, with moles instead of mass. However, we know that we can’t
measure the number of moles directly. We need to use the formula weight to convert
moles to mass before we can measure anything out. Equation 10.3 is the equivalent of
10.2a but using molarity as the concentration unit. The volume unit is litres. You can
work out for yourself the equations where mass and volume are the subject of the
equations.
Molarity
= mass ÷ (volume x formula weight)
114
Eqn 10.3
Chapter 10
Preparing Solutions
EXAMPLES
1.
What molarity is obtained when 1.5 g of NaCl is dissolved in 300 mL of solution?
We need to calculate the formula weight (FW) of NaCl before proceeding: 22.99 +
35.45 = 58.44 g.
300 mL
= (300 ÷1000) L
= 0.3 L
molarity
= mass ÷ (volume x FW)
= 1.5 g ÷ (0.3 L x 58.44)
= 0.086 M
2.
Calculate the mass required to make 2L of 3 M NaCl solution.
mass
3.
= molarity x FW x volume
= 3 x 58.44 x 2
= 351 g
You are required to dissolve 58.44 mg of NaCl in sufficient solution to give a
concentration of 0.05 M? What volume is required?
58.44 mg
volume
=
=
=
=
=
(58.44 ÷ 1000) g
0.05844 g
mass ÷ (molarity x FW)
0.05844 ÷ (0.05 x 58.44)
0.02 L
PRACTICE QUESTIONS
6. Calculate the molarity of the following solutions:
(a) 0.2 moles in 4 litres
(b) 0.075 moles in 250 mL
(c) 0.106 g of sodium carbonate in 500 mL.
7.
How many moles and what mass are contained in the following volumes of the
solutions below?
(a) 1.5 L of 0.015 M HCl
(b) 300 mL of 0.3 M NaNO3
(c) 250 L of 0.0001 M CuSO4.5H2O.
10.2.3 Volume Units
The only common concentration unit for liquid solutes is mL/100 mL or %v/v. We can
use a simplified version of the equation, as shown in Equation 10.4. By the way, this
equation can be applied to g/100 mL (or %w/v), replacing mL of solute with grams of
solute.
Concentration = 100 x mL of solute ÷ mL of solution
Eqn 10.4
EXAMPLES
1.
What concentration of ethanol is obtained when 50 mL is dissolved in 500 mL of
solution?
Concentration = 100 x 50 ÷ 500 = 10 mL/100 mL or 10%v/v.
2.
How much ethanol is required to make 2.5 L of 3%v/v solution?
Convert the solution volume to mL. 2.5 L = 2500 mL.
Volume of solute
= concentration x mL of solution ÷ 100
= 3 x 2500 ÷ 100
= 75 mL
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Chemistry for Technicians
PRACTICE QUESTIONS
8. What volume of liquid solute is required to prepare the following solutions?
(a) 5 L of 10%v/v
(b) 300 mL of 0.1 %v/v.
9.
What is the concentration in %v/v of the following solutions?
(a) 20 mL in 1.5 L of solution
(b) 5 mL in 500 mL of solution.
It should be emphasised that you don’t have to remember all these equations. All
you should remember is the one basic idea about concentration, shown in Equation 10.1,
and apply it to whatever situation is required.
PRACTICAL WORK
1. Prepare 100 mL of 5 g/L CuSO4.5H2O solution. Measure its colour intensity
(absorbance) as shown by your teacher. Check your value against the
standard value.
2.
Prepare 25 mL of 1 M NaOH solution as shown by your teacher.
Add 25 mL of 1 M HCl to your NaOH solution. Measure the acidity with
a test strip. It should be close to 7.
Prepare 50 mL of 5%v/v ethanol solution. Check your work by
measuring its density or refractive index, as shown by your teacher.
Compare your result with that from the literature or a solution of known 5%
concentration.
10.3 DILUTION OF SOLUTIONS
Many important chemicals are supplied in solution form, particularly acids and bases.
Normally, this is done because the solute is difficult to contain in pure form (e.g. a gas or
reactive liquid). Table 10.2 lists the concentrations of the concentrated solutions of some
common laboratory chemicals.
Table 10.2 Concentrations of concentrated laboratory chemicals
Chemical
Molarity
g/L
Hydrochloric acid
11.6
425
Sulfuric acid
18.3
1800
Nitric acid
15.6
980
Ammonia
14.8
250
If we require a solution of a particular concentration, prepared from the
concentrated solutions, then we must perform a dilution, where we
where we measure out a certain amount of the concentrated solution, and
solution, and add enough solvent to it to make the required lower
lower concentration solution.
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Chapter 10
Preparing Solutions
Equation 10.5 is an important one to remember. It applies to all concentration units. The
only point to remember is that the units for concentration and volume are be the same on
each side of the equation.
C1V1 = C2V2
Eqn 10.5
where C1 and V1 are the concentration and volume of the concentrated
solution, and C2 and V2 refer to the dilute solution.
When a solution is diluted accurately, using volumetric glassware such as
pipettes and volumetric flasks, the ratio of C1/C2 (or (V2/V1) is called the
dilution factor. It is used in calculations, where the solution has been
diluted to be analysed, and then a correction must be made for this dilution.
DILUTION FACTORS — CONFUSION REIGNS
Depending on who you listen to, or where you work, you will hear dilutions referred to in
different ways. The variations are mostly just in the wording, but there is one important
area, where the method of performing the dilution is quite different.
Let’s look at the terms first. For a 10 mL (initial volume) to 100 mL (final volume) dilution,
the following means of describing it are all equivalent:
•
•
•
•
•
10 in 100
10 to 100
1 in 10
1 to 10
10 times.
Are you confused yet? Well, the best/worst is yet to come. In the biology/pathology areas,
dilutions are often done in a different way. Rather than measure the final volume, the
volume of diluting solvent is measured instead. So in our example, 10 mL (initial solution)
and 90 mL of solvent would be combined. This is referred to as:
•
•
10 is to 90
1 is to 9.
The reason for this is the small volumes frequently required in these areas: volumetric
flasks don’t exist below 10 mL.
EXAMPLES
1.
10 mL of 500 mg/L solution is diluted to 250 mL. What is the dilution factor? What is
the final concentration?
The dilution factor is 250 ÷ 10 = 25.
To determine the final concentration (C2), C1 = 500 mg/L, V1 = 10 mL and
V2 = 250 mL.
500 x 10 = C2 x 250
Therefore, C2 = 20 mg/L.
2.
How much 5 g/L solution is required to prepare 2 L of 0.1 g/L solution?
C1 = 5 g/L, V1 = ?, C2 = 0.1 g/L, V2 = 2 L.
5 x V1 = 0.1 x 2
Therefore, V1 = 0.04 L = 40 mL.
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Chemistry for Technicians
3.
How would you perform the following dilutions to obtain 100 mL of the diluted
solution? (a) 1 to 10 (b) 1 is to 4 (c) 2 times
(a) In 1 to 10, the numbers refer to the ratio of initial to final volumes. If 100 mL of
diluted solution is required, then 10 mL of initial solution is required.
(b) Is to refers to the dilution method where the first number is the solution to be
diluted and the second number the volume of solvent. If 100 mL of total solution
is required, then 20 mL of initial solution and 80 mL of solvent are required.
(c) A 2 times dilution refers to the dilution factor. The final volume is twice the initial:
50 mL is required.
4.
A solution is diluted 1 in 10, and the concentration of this solution analysed to be 23
mg/L. What is the concentration of the original solution?
‘1 in 10’ means the final volume was ten times the initial volume, so the dilution
factor is 10. The initial concentration is ten times higher than the final: 230 mg/L.
PRACTICE QUESTIONS
10. What is the concentration of the diluted solution in the following?
(a) 10 mL of 0.5 M is diluted to 250 mL
(b) 5 mL of 1.25 g/100 mL is diluted to 100 mL
(c) 25 mL of 1000 mg/L is diluted to 500 mL.
11. What volume of concentrated solution would be required to prepare the
following?
(a) 1 L of 100 mg/L from 500 mg/L
(b) 100 mL of 0.2 g/L from 2 g/L
(c) 300 mL of 1.5 g/100 mL from 10 g/100 mL.
12. What dilution factors are available using the glassware in Table 10.3 if:
(a) a 100 mL volumetric flask is used to contain the diluted solution
(b) a 25 mL pipette is used to measure out the concentrated solution.
13. A solution is diluted from 10 mL to 100 mL. The diluted solution is analysed and
found to have a concentration of 50 mg/L. What is the concentration of the
original solution?
14. Describe how you perform the following dilutions:
(a) a 1 to 5 dilution to obtain 50 mL of diluted solution
(b) a 10 times dilution to obtain 250 mL of diluted solution
(c) a 1 is to 9 dilution, where 0.5 mL of initial solution is to be diluted.
Sometimes it is not practical to prepare a solution by dilution in one step,
because the dilution factor is too great. For example, a dilution factor of
dilution factor of 1000 would mean 1 mL of concentrated solution to 1 L of
solution to 1 L of dilute solution. Small volumes are not very accurate, and
accurate, and large volumes may be inconvenient — you might only want
50 mL of the diluted solution, so making 1 L is wasteful. Serial dilution
means a series of dilution steps, where the first diluted solution is further
diluted to make a second, and so on if necessary.
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Chapter 10
Preparing Solutions
In serial dilution, the dilution factors are multiplied together. Thus, our 1000 times
dilution could be broken up into a 20 times step followed by a 50 times dilution.
There are, of course, many ways of achieving this overall dilution. Where exactness
is not required, and measuring cylinder and beakers are used, the possibilities are
endless. However, if the dilution must be exact (quantitative), then the options are
limited by the range of volumetric glassware available. The dilution factors available
from these pieces of glassware are the ratio of the flask volume to the pipette volume.
Table 10.3 lists the common pipette and volumetric flask volumes as a reminder.
Table 10.3
Common volumetric glassware
Pipette volumes (mL)
Volumetric flask (mL)
5, 10, 15, 20, 25, 50
50, 100, 200, 250, 500, 1000
EXAMPLES
1.
5 mL of 10 g/L is diluted to 100 mL, and 10 mL of this solution further diluted to 500
mL. What is the concentration of the final solution?
First dilution. C1 = 10 g/L, V1 = 5 mL, C2 = ?, V2 = 100 mL.
10 x 5 = C2 x 100
Therefore, the first diluted solution has a concentration of 0.5 g/L.
Second dilution. C1 = 0.5 g/L, V1 = 10 mL, C2 = ?, V2 = 500 mL.
0.5 x 10 = C2 x 500
Therefore, the final solution concentration = 0.01 g/L.
An alternative method for this calculation uses the dilution factors (DF).
DF (step 1) = 100 ÷ 5 = 20. DF (step 2) = 500 ÷ 10 = 50.
Overall DF = 20 x 50 = 1000
Final concentration = initial concentration ÷ DF = 10 ÷ 1000 = 0.01 g/L.
You can use either method.
2.
How could 100 mL of 0.5 mg/L solution be prepared from 1000 mg/L solution?
The dilution factor is 1000 ÷ 0.5 = 2000.
The final volume of the second solution is required to be 100 mL, so the greatest
second dilution we can achieve is 5 mL to 100 mL, a DF of 20.
This leaves us with a remaining DF of 100 (2000 ÷ 20 = 100). This could be
achieved in two 10 to 100 steps, or a 5 to 500 step. The single step is quicker and
probably more accurate (less chance of error, since the glassware use is less).
So, our suggested method is:
Step 1: 5 mL of 1000 mg/L diluted to a final volume of 500 mg/L.
Step 2: 5 mL of this solution diluted to 100 mL.
PRACTICE QUESTION
15. How would you prepare (a) 2 L of 0.1 M HCl and (b) 500 mL of 1 M H2SO4
from the concentrated solutions specified in Table 10.2? The final concentrations
are only approximate (±10%). Use sensible volumes — multiples of 5 mL —
which are not necessarily pipette volumes.
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Chemistry for Technicians
PRACTICAL WORK
1.
You are required to prepare 100 mL of 0.01 M HCl and 100 mL of 0.001 M
HCl solutions from provided 0.1 M HCl. Calculate how you will prepare
these solutions by one-step dilutions.
2.
Check your calculations with your teacher, and prepare the solutions.
3.
Use serial dilution to prepare 100 mL of 0.0001 M solution without using a
pipette smaller than 5 mL. Check with your teacher, and perform the
dilution.
4.
Measure the pH of each solution, as shown by your teacher.
10.4 INTERCONVERSION OF CONCENTRATION UNITS
In some circumstances, it is necessary to change the unit of concentration from that used
in a procedure to that required for a final report. The reasons for this include the final
answer being too small or large a number to be conveniently understood (numbers
between 0.1 and 100 are more easily comprehended), or the unit itself not being the one
required by standard practice (e.g. alcohol contents in drinks must be %v/v on the label).
To convert the concentration from one unit to another requires a knowledge of the
metric system, and possibly some simple mass relationships (i.e. moles = mass ÷ FW,
density = mass ÷ volume). The process is most simply done by treating the concentration
unit as a fraction, and dealing with the amount unit first and the volume unit next. This is
best done by a series of examples.
EXAMPLES
Convert a concentration of 1 M ethanol to: (a) g/L; (b) g/100 mL; (c) mg/L; (d) %v/v. The
formula weight of ethanol is 46, and the density is 0.8 g/mL.
(a)
The amount unit must be converted, but the volume unit is the same in both so can
be ignored. How do we convert moles to mass? Multiply by the FW.
mole
x 46
g
1  → 46 
L
L
Having converted from moles to mass, we will use the answer from (a) in the following
calculations.
(b)
Converting from g/L to g/100 mL means correcting the volume unit, and ignoring the
mass unit. How do we get from 1L to 100 mL? Divide by 10.
g
g
46  → 4.6 
L
÷ 10
100 mL
(c)
Converting from g/L to mg/L means correcting the mass unit, and ignoring the
volume unit. How do we get from grams to milligrams? Multiply by 1000.
g x 1000
mg
46  → 46 000 
L
L
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Chapter 10
(d)
Preparing Solutions
Converting from g/L to mL/100 mL means converting amount and volume units. How
do we convert grams of ethanol to volume of ethanol? Using the density equation —
volume = mass ÷ density.
g
÷ 0.8
mL
46  → 57.5 
L
L
How do we convert from L to 100 mL? As we did in (b) — divide by 10.
mL
mL
57.5  → 5.75 L
÷ 10
100 mL
Which of the units in the example above is the most suitable? The mg/L is a large
number, and therefore not normally used. Molarity means nothing to anyone outside
chemistry and biochemistry, so it would not be used in general reporting, so the
remaining values 46 g/L, 4.6 g/100 mL and 5.75 %v/v are the most suitable here. As
mentioned above, ethanol levels in drinks are reported as %v/v.
PRACTICE QUESTION
16. Perform the following unit conversions:
(a) 0.1 M HCl to g/100 mL
(b) 5 g/L of KBr to mole/L
(c) 0.001 g/L to mg/L
(d) 50 000 mg/L to g/100 mL
(e) 50 000 mg/L to %v/v (density = 0.9 g/mL).
WHAT YOU NEED TO BE ABLE TO DO
•
•
•
•
•
•
Describe the basic structure of a concentration unit.
List the common solution concentration units.
Explain why a number of units are widely used.
Perform calculations associated with solution concentration.s
Explain the meaning of the terms serial dilution and dilution factor.
Interconvert concentration units.
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Chemistry for Technicians
TERMS AND DEFINITIONS
During this chapter, you have met a number of new terms, listed below. Match them with
their correct definitions.
Terms
A solution concentration
C
serial dilution
E
1 to 5 dilution
B
D
F
dilution
dilution factor
1 is to 5 dilution
Definitions
1
the amount of solute per unit volume of solution
2
the ratio of initial to final concentration (or final to initial volume) in a dilution step
3
1 volume unit of initial solution is diluted to a final volume of 5 units
4
addition of solvent to a volume of existing solution (usually with known initial and
final volumes)
5
1 volume unit of initial solution is mixed with 5 units of solvent
6
where a solution is diluted a number of times, each diluted solution becoming the
starting solution for the next step
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11
CHEMICAL REACTIONS II
PURPOSE
To describe some significant chemical reactions, and to introduce the concept of
chemical equilibrium.
It is recommended that you revise Chapter 6 before progressing with this chapter.
11.1 IONIC EQUATIONS
In many reactions involving solutions of ionic substances, some of the ions
do not actually participate in the reaction. Such ions are called spectator
called spectator ions. They have to be there to balance charges, since we
charges, since we can’t have a beaker of positive (or negative) ions alone.
ions alone. They can be left out of the balanced equation, so that the actual
reaction is clearer. These equations are known as ionic equations.
An example is the reaction of an acid (e.g. hydrochloric acid) with a base (e.g.
sodium hydroxide). If the NaOH is in solid form, then the full equation must be written,
as below in Equation 11.1a.
HCl (aq) + NaOH (s) → NaCl (s) + H2O (l)
Eqn 11.1a
If this reaction was carried out using solutions of both species, then the sodium and
chloride ions do no more than ‘swim’ around. Remember that ionic species in solution
have the individual ions already free of each other. The full equation is shown in
Equation 11.1b.
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
Eqn 11.1b
If the ionic species in Equation 11.1b are separated, as they exist in solution, then
Equation 11.1c results,
H+ (aq) + Cl– (aq) + Na+ (aq) + OH– (aq) → H2O (l) + Na+ (aq) + Cl– (aq)
Eqn 11.1c
Clearly, the sodium and chloride ions are in exactly the same form on both sides of
the equation, and are making the equation more complicated than is necessary for some
purposes. This can be simplified to the ionic equation in 11.1d.
H+ (aq) + OH– (aq) → H2O (l)
Eqn 11.1d
You must be aware, however, that ionic equations can oversimplify the reaction
description. The above example (11.1d) tells you nothing about the actual nature of the
reactants — only that an acid–base reaction occurred. This ionic equation would be
exactly the same, regardless of the acid solution or the nature of the cation in the
hydroxide solution.
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Chemistry for Technicians
PRACTICE QUESTION
1. Write the ionic equations for the following:
(a) Pb(NO3)2 (aq) + NaI (aq) → PbI2 (s) + 2NaNO3 (aq)
(b) 2Fe (s) + 6HCl (aq) → 2FeCl3 (aq) + 3H2 (g)
(c) H2SO4 (aq) + 2KOH (aq) → K2SO4 (aq) + H2O (l)
11.2 REACTIONS OF ACIDS
The meaning of the term acid has changed over history as understanding of
chemistry has increased. In this section, we will simply treat an acid as any
an acid as any species that possesses at least one hydrogen atom that is
atom that is able to break free of the molecule to form a H+ ion. The
ion. The remainder of the molecule, therefore, becomes an anion, as shown
in Equation 11.2 (where HA is any acid, and A– is the anion). Table 11.1
shows this for four common acids.
HA → H+ + A–
Eqn 11.2
Table 11.1 Ionisation of common acids
Acid
Molecular form
Ionised form
Hydrochloric
H–Cl
H+ Cl–
H
SO4
Sulfuric
2H+
SO42–
H
Nitric
H+ NO3–
H–NO3
O
Ethanoic
H
O
C
C
H
H
H+
CH3COO–
H
Note: Only the hydrogen attached to the O in ethanoic acid forms an H+ ion.
11.2.1 Hydroxides
The reaction of acids with hydroxides is one of a number that fits under that category of
neutralisation discussed in Chapter 6. The products are an ionic salt and water. The
ionic salt is composed of the cation from the hydroxide and the anion from the acid.
e.g. HCl + NaOH → NaCl + H2O
11.2.2 Metal Oxides
This is a very similar reaction to acid/hydroxide, because metal oxides, if soluble in
water, form the metal hydroxide. The products are an ionic salt and water.
e.g. 2HNO3 + CaO → Ca(NO3)2 + H2O
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Chapter 11
Chemical Reactions II
11.2.3 Metals
Acids are able to convert many metals into their cations and so form ionic salts with the
anion from the acid. Hydrogen gas is the other product. Some metals, such as gold and
copper, do not react with acids in this way.
e.g. 2Fe + 6HCl → 2FeCl3 + 3H2
11.2.4 Carbonate or hydrogen carbonate salts
These salts also form ionic salts and water when reacted with acid, but release carbon
dioxide gas.
e.g. Na2CO3 + 2HCl → 2NaCl + CO2 + H2O
KHCO3 + CH3CH2COOH → 2KCH3CH2COO + CO2 + H2O
11.2.5 Sulfite salts
The reaction of sulfite with acid produces sulfur dioxide gas, together with salt and
water.
e.g. MgSO3 + 2HNO3 → 2NaNO3 + SO2 + H2O
11.2.6 Sulfide salts
The well-known, but highly toxic, ‘rotten egg’ gas, which is dihydrogen sulfide is
produced, together with salt by the reaction of a sulfide salt with acid.
e.g. CuS + 2HCl → CuCl2 + H2S
11.2.7 Other salts
There is no reaction between acids and salts of sulfate, nitrate or the halides.
PRACTICAL WORK
Part I
1. Test the reaction of the species listed below with 1 M HCl. Two ‘spatulafulls’ of the solid and about 5 mL of 1 M HCl are sufficient.
2.
Before you commence, predict what gas (if any) is likely to be produced by
each reaction, so that you can be ready. You will need to test for gas
evolution at the very start of the reaction, otherwise most of the gas will
escape, and the test will not work.
3.
Determine whether heat has been produced by feeling the base of the test
tube.
4.
After you have tested for gases and heat, add more of each test substance
until no further reaction occurs, or the substance will not dissolve. Test the
acidity of the reaction mixtures and of the HCl using a pH test strip.
Test substances: NaCl, NaOH, Na2CO3, NaHCO3, Na2SO3, CaO, Mg, Na2S (this
must be done in a fume hood).
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Chemistry for Technicians
Test procedures for gases evolved
Hydrogen: will extinguish a small flame with a ‘pop’.
Carbon dioxide: limewater (a solution of Ca(OH)2) turns milky; hold a capillary
containing some limewater in the mouth of the test tube.
Sulfur dioxide: turns dichromate green; use the capillary method.
Dihydrogen sulfide (remember to test this in the fume hood): turns lead
ethanoate solution black — soak a strip of filter paper in the lead solution (WEAR
GLOVES!) and hold in the mouth of the test tube.
Q1. Write a full balanced equation for each reaction.
Q2. Write an ionic equation for each reaction.
Q3. Explain how the acidity test results indicate whether a reaction has occurred
in the cases where no gas is produced.
Part II
1. In separate test tubes place about 2 mL of 0.5 M CH3COOH and 0.5 M NH3.
Note the smell of each.
2.
Add 5 mL of 1 M NaOH to the ethanoic acid, and 5 mL of 1 M HCl to the
ammonia. Record your observations regarding the smell of the solutions.
Q1. Write a full balanced equation for each reaction.
Q2. Explain your observations about the change in smell of the solutions after
reaction.
PRACTICE QUESTION
2. Predict the products of the following reactions:
(a) HNO3 + NaOH
(b) H2SO4 + CaO
(c) HCl + Na2SO4
(d) HNO3 + Mg
(e) HCOOH + CaCO3
(f) HCl + ZnS
(g) HCl + CaSO3.
11.3 AN INTRODUCTION TO TITRATIONS
A titration is a chemical reaction used to determine (or analyse) the
amount of one of the reactants present. It achieves this by the controlled
controlled addition of known amounts of one of the other reactants (known
reactants (known as the standard) until the species being analysed (known
analysed (known as the analyte) is all used up. The balanced equation is
then used, together with the amount of the known reactant added, to
calculate the amount of the reactant being analysed.
•
•
•
The key points of a titration are:
the accurate addition of the standard reactant (generally involves the use of a
burette)
the concentration of the standard is known accurately (in molarity)
the point at which the analyte reactant is used up must be able to be detected.
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Chapter 11
Chemical Reactions II
These points will be covered in practical and theoretical detail in other modules —
Chemical Laboratory Techniques and Non–Instrumental Analysis.
EXAMPLE
What mass of NaOH is in 25 mL of a solution if it takes 15.6 mL of 0.105 M HCl to react
with it all?
The balanced equation for this reaction is NaOH + HCl → NaCl + H2O.
This means that for every mole of NaOH present, the same number of moles of HCl must
be added.
Number of moles of HCl
= molarity x volume of HCl (in L)
= 0.105 x (15.6/1000)
= 1.64 x 10–3 moles
Moles of NaOH present
= moles of HCl added
= 1.64 x 10–3 moles
Mass of NaOH
= moles x FW
= 1.638 x 10–3 x 40
= 0.0655 g
PRACTICAL WORK
1. Calculate how much pure sodium hydroxide would be required to prepare
100 mL of 0.1 M solution. Check your answer with your teacher.
2.
Prepare the NaOH solution as per step 1.
3.
Obtain about 100 mL of standard 0.1 M HCl. Rinse and fill a burette with it.
4.
Pipette 25 mL of your NaOH solution into each of three 250 mL conical
flasks. Add 3–4 drops of phenolphthalein indicator to each flask.
5.
Add HCl from the burette carefully until the purple indicator colour just
disappears. Note the volume. Repeat for each flask.
Q1. Average the titration volumes. Calculate the molarity of NaOH in your
solution.
Q2. Suggest reasons for differences between the expected NaOH concentration
(as prepared) and that found by titration.
Titrations most frequently use acid–base neutralisation reactions, but others
including precipitations are also important. The reactants in titrations must be in solution
form.
PRACTICE QUESTIONS
3. Identify the standard and analyte in the titration reactions in Q4.
4.
Calculate the moles and mass of analyte reactant, given the following
information for titrations:
(a) 20.3 mL of 0.0567 M NaOH is needed to react with a solution of HNO3
(b) 11.8 mL of 0.0536 M Na2CO3 reacts with a solution of sulfuric acid
(c) 84.9 mL of 0.102 M H2SO4 reacts with 20 mL of a solution of KOH
(d) 25.7 mL of 0.0946 M HCl reacts with an ammonia solution (1 mole of HCl
reacts with 1 mole of ammonia).
127
Chemistry for Technicians
5.
6.
A vinegar sample is analysed for its ethanoic acid content by titration with
standard sodium hydroxide solution. A 5 mL sample of the vinegar reacts with
15.3 mL of 0.252 M NaOH. The reaction between NaOH and ethanoic acid is
1:1.
(a) How many moles of CH3COOH are present?
(b) What is the molarity of the ethanoic acid in the vinegar?
(c) What is the g/100 mL concentration of ethanoic acid in the vinegar?
A washing soda sample is analysed by titration for its Na2CO3 content. A 0.2387
g sample of the powder was dissolved in water, then titrated with 16.6 mL of
0.0527 M H2SO4.
(a) What mass of sodium carbonate was present in the sample?
(b) What is the percentage of sodium carbonate in the washing soda?
11.4 THE EQUILIBRIUM PROCESS
In Chapter 6, you were introduced to the idea that not all reactions go completely from
reactants to products, and that all chemical changes are not irreversible. Reactions that do
not go to completion are said to reach an equilibrium position, a point at which the
concentrations of reactants and products reach a steady, unchanging level.
Exactly why this should be is too hard to explain here, but part of the reason is that
the reaction can be reversed, that is the products can undergo the reverse process and reform the reactants. Therefore, the overall process is a competition between the forward
(reactants → products) reaction and the reverse (products → reactants) reaction.
Equilibrium is reached when this competition becomes equal.
This leads to the question: since the reaction is reversible, how can we say which
are the reactants and which are the products? The answer is that it depends on how you
define the process. In the equation, the substances on the left-hand side are the reactants,
and those on the right-hand side, the products.
PRACTICAL WORK
Silver ions form precipitates with ethanoate ion (white) and iodide ion (yellow).
1.
Combine 25 mL solutions of 0.2 M silver nitrate and 0.2 M sodium
ethanoate.
2.
Allow the precipitate of silver ethanoate to settle, then decant some of the
liquid into a test tube.
3.
Add a few drops of 0.1 M sodium iodide, and observe any changes.
4.
Measure about 1 g of silver ethanoate into a 100 mL beaker, and add 50 mL
of 0.1 M sodium nitrate.
5.
Stir well for a few minutes, then allow the remaining solid to settle.
6.
Decant some of the liquid into a test tube, and repeat step 3.
Q1. Write a balanced equation for the reaction.
Q2. How do your observations from step 3 and 6 indicate that the reaction is
reversible and an equilibrium reaction?
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Chapter 11
Chemical Reactions II
When equilibrium is reached, the concentrations of the various substances
stop changing. However, the forward and reverse reactions do not stop:
do not stop: they continue, but cancel each other out, because they are
they are each occurring to the same extent. The equilibrium process is said
process is said to be dynamic: at the molecular level, a lot is happening, but
to no overall effect.
A DEMONSTRATION OF DYNAMIC EQUILIBRIUM
We can use the physical process of sublimation, where a solid vaporises into
a gas without passing through the liquid form first. Iodine is such a solid. Set
up a large glass jar with a tight sealing lid, with enough iodine crystals to cover
the base. Place it near a window. Over a week or two, iodine crystals will form
on the side of the jar. If you make a note of a few crystals in particular, and
watch them over time, some will grow and other shrinks. This must be due to
the forward and reverse processes in the equilibrium I2 (s) ↔ I2 (g) continuing to
occur.
How does the reaction proceed as it moves towards equilibrium? Does it get there
instantly or take some time? Does its speed (how many reactions are occurring per
second) vary or is it constant? Figure 11.1 illustrates the way that concentrations of the
species involved in a simple reaction change over time.
Concentration
Equilibrium reached
Concentration of reactant
Concentration of product
Time
Figure 11.1
Progress of a reaction towards equilibrium
You can see that the reaction does not occur instantly, nor does it proceed at a
constant speed. The point at which equilibrium is reached is when the concentrations of
A and B level off. The time taken to reach equilibrium is dependent on the reaction itself
and the conditions under which it is carried out. This is examined further in the next
section. Until something is done to alter the reaction mixture (e.g. heat or adding some
more reactant), nothing will change.
129
Chemistry for Technicians
PRACTICAL WORK
1. Your teacher will assign you an alkanol (C1–C4 alkanols are suitable); the
reaction of this and ethanoic acid will be studied as an exercise in the
equilibrium process.
2. Using formula weights and densities from the literature, calculate the
volumes corresponding to 0.35 moles of each reactant. Check these with
your teacher.
3.
Combine the liquids in a two- or three-necked 100 (or 250) mL roundbottom flask. Carefully add 1 mL of concentrated sulfuric acid.
4.
Set up a reflux apparatus in a fume hood, as demonstrated by your teacher.
Do not begin heating at this stage.
5.
Take a 0.5 mL sample (by pipette) of the reaction mixture (carefully removing the stopper from the second neck) and run it into a conical flask
containing about 50 mL of ice water. Add a few drops of phenolphthalein.
When time allows, titrate this solution with standardised 0.1 M NaOH.
6.
Turn on the heating and bring to the boil as quickly as possible. Begin
timing when the first bubbles appear. Reduce the heat to a steady boiling
rate.
7.
Take 0.5 mL sample after 10, 20, 30, 40 and 60 minutes, and treat as in
step 5.
8.
Leave the apparatus running. Further samples after 2 and 12 hours should
be collected (the heating can be stopped overnight for safety reasons and
restarted the next day — the times are for refluxing only). Each sample
should be treated as in step 5. Storage in the refrigerator for titration in the
next session is acceptable.
Q1. Plot a graph of titration volume versus time, drawing a smooth curve
through the data points.
Q2. How does your graph correspond to (a) others in the class (b) Figure 11.1?
Q3. Draw a representation of the change in alkanol and product concentrations
over the same time.
Q4. How could we prove that this reaction had reached equilibrium, rather than
having just stopped when one of the reactants was used up?
Q5. What does the shape of the graph suggest about the speed at which the
reaction occurs during the time period studied?
PRACTICE QUESTION
7. Which, if any, of the following statements about equilibrium processes are true?
(a) At equilibrium, nothing happens.
(b) At equilibrium, the concentrations of all substances are equal.
(c) At equilibrium, the number of forward and reverse reactions are equal.
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Chapter 11
Chemical Reactions II
11.5 FACTORS AFFECTING EQUILIBRIUM
When we talk about a system at equilibrium, we are assuming that it is a closed system:
nothing is happening to change the contents. It has reached a point of stability that is set
for a given reaction at a fixed temperature.
The equilibrium position is defined by the ratio of the concentrations of the
various species involved and is known as the equilibrium constant, K. A
constant, K. A simplified expression for the equilibrium constant is given
constant is given in Equation 11.3. However, it should be noted that this is
noted that this is not the full equation, since an extra complication is not
necessary at this stage.
concentration of products
K = 
concentration of reactants
Eqn 11.3
The value of the equilibrium constant is an indication of the extent to which the
reaction has proceeded from reactants to products. The further the reaction proceeds to
products, the higher their concentration (and the lower that of the reactants) and,
therefore, the value of K is greater.
EXAMPLE
Reaction A has an equilibrium constant value of 3490, while K for reaction B is 0.21. In
which reaction are more products formed?
Since K is a ratio of product concentration over reactant concentration, higher values of K
are obtained when more products are formed. Therefore, reaction A yields more
products.
In general, reactions with values of K between 0.001 and 1000 are considered to
have measurable quantities of all species involved. Reactions with K less than 0.001 have
almost no product formation, while those with K greater than 1000 are almost 100%
complete.
The equilibrium constant is not altered by:
•
•
•
the initial concentrations of the species involved (because it is a ratio)
the direction from which equilibrium is reached, that is whether only the reactants
were present initially, or only the ‘product’, or a mixture (because the forward or
reverse reaction will dominate until the set position is reached)
the addition or removal of substances from the system (the system will return to
equilibrium with different concentrations of each substance).
Figure 11.2 (overleaf) shows what happens to a system at equilibrium (point )
when more reactant is added.
What is Figure 11.2 telling us about the changes that have occurred after extra
reactant was added (point )? A new equilibrium has been reached (point ) after some,
but not all, of the extra reactant has been converted into product. At the molecular level,
the forward reaction has dominated over the reverse reaction for a short period of time,
until the correct equilibrium position was reached.
131
Chemistry for Technicians
EXAMPLE
Calculate the value of K at points 1–3. What do the results mean?
Point
Concentration of reactant
Concentration of product
K
0.59
0.66
0.65
0.40
0.41
0.43
0.68
0.62
0.68
The value of K has not changed between points 1 and 3 because the system is at
equilibrium. The concentrations are different, but the ratio is unchanged!
At point 2, it is not equilibrium, as can be seen by the changes in concentrations, and
the different value of K.
1
Reactant
0.59
0.40
Product
0
Concentration
Figure 11.2
The effects of change on a system at equilibrium
The simplest statement that can be made about what happens to an
equilibrium system when something happens to alter the system, is that the
system reorganises itself to partly reverse those changes. This is known as
Le Chatelier’s principle.
When concentration or pressure changes occur, the system returns to the same value
of equilibrium constant as before. If a change in temperature occurs, the system will
again attempt to reverse that change (e.g. if the system is cooled, reaction will occur to
release heat). However, the value of the equilibrium constant will change, because it is
temperature dependent.
132
Chapter 11
Chemical Reactions II
PRACTICAL WORK
If you haven’t performed the practical exercise in Section 6.4 on page 78, it
would be useful to do so now.
The anion chromate, which is yellow in colour, when subjected to acidic
conditions, reacts to form dichromate, which is orange. The reverse happens
when dichromate is subjected to alkaline conditions.
1.
Obtain 0.001 M solutions of potassium chromate in 0.1 M NaOH and
potassium dichromate in 0.5 M H2SO4. Measure the colour spectrum of
each solution, as shown by your teacher, between 300 and 500 nm.
2.
Adjust the acidity of each solution to neutral (pH 7) using concentrated
H2SO4 for chromate, and 5 M NaOH for the dichromate.
3.
Measure the colour spectrum of each solution.
4.
Continue adding acid to the chromate solution, and alkali to the dichromate
until their colours do not appear to change any more.
Q1. What do your results suggest about the equilibrium position as approached
from each direction?
Q2. What do your results suggests about the reversibility of the reaction?
PRACTICE QUESTIONS
8. Explain why each of the following statements is incorrect.
(a) When an equilibrium system is changed, it ‘repairs’ itself so that it returns
to exactly the same state as before.
(b) The equilibrium constant for a particular reaction will be different
depending on the concentrations initially present.
9. Draw a representation of the changes to the equilibrium system in Figure 11.2 if:
(a) some of the product is removed
(b) some of the reactant is removed.
10. At 25°C, a reaction has a value of K of 35, while at 100°C, K equals 492. Which
temperature is better for maximising the amount of product?
11.6 INDUSTRIAL CHEMISTRY: APPLICATION OF CHEMICAL
EQUILIBRIA
Many industrial processes, involved with the production of important substances, are
equilibrium reactions. This means that the yield of product (how much reactant is
actually converted into product) is affected by the equilibrium constant, and will never be
100%. Much work is put into the design of the process to maximise the yield by
knowledge of how equilibrium processes work. In this section, you will examine one
very significant process: the production of ammonia.
11.6.1 The Haber Process
Ammonia is one of the most important industrial chemicals, mainly because of its use in
the production of fertilisers, such as ammonium nitrate, ammonium sulfate and
ammonium phosphate. The Haber process converts nitrogen and hydrogen gases to
ammonia gas, as shown in the equation below.
3H2 (g) + N2 (g) ↔ 2NH3 (g)
133
Chemistry for Technicians
Under normal atmospheric conditions, the reaction basically doesn’t occur (K is
very small). The reaction, in fact, needs a catalyst: an added substance which allows the
reaction to occur. With the presence of a catalyst, it has been found that the reaction is
most efficient when subject to extremely high pressures — several hundred times
atmospheric pressure — but that increasing the temperature in fact causes a decrease in
ammonia production. Figure 11.3 (overleaf) shows the effect of temperature and pressure
on the process.
At 200°C, it is evident that an almost 100% conversion of reactants is obtained at
pressures above 750 times atmospheric pressure. In practice, most production systems
use significantly lower pressures (around 200 atmospheres) because of the difficulties of
containing large amounts of materials at the high pressures. This, of course, reduces the
yield of ammonia. The temperature used is around 500°C, which also limits the
percentage conversion, but speeds up the process.
According to Figure 11.3, the percentage of ammonia obtained under these
conditions is only 20%. While this seems low, it is the choice of the engineers who
design the plant to maximise the rate of producing the ammonia and to minimise costs
and safety concerns in the building and operation of the plant.
LIBRARY WORK
You should study at least one other important equilibrium process, preferably
one that is related to your area of interest or employment.
% Ammonia
100
200°C
80
300°C
60
400°C
500°C
40
600°C
700°C
20
0
0
200
400
600
800
Pressure (atm)
Figure 11.3
Effect of temperature and pressure on the Haber process
WHAT YOU NEED TO BE ABLE TO DO
•
•
•
•
•
•
•
•
Write ionic equations.
Predict the products of various reactions of acids.
Outline the titration process.
Perform simple titration calculations.
Define the terms dynamic equilibrium and reversible reaction.
Describe the features of an equilibrium reaction.
Outline factors affecting equilibrium reactions.
Describe some industrially significant equilibrium processes.
134
1000
Chapter 11
Chemical Reactions II
TERMS AND DEFINITIONS
During this chapter, you have met a number of new terms, listed below. Match them with
their correct definitions.
Terms
A spectator ions
C
acid
E
standard
G dynamic equilibrium
I
Le Chatelier’s principle
B
D
F
H
ionic equation
titration
analyte
equilibrium constant
Definitions
1
a chemical reaction used to measure the amount of one reactant
2
a numerical value representing the ratio of the concentrations of products to
reactants
3
one of the reactants in a titration, whose amount is exactly known
4
a substance that releases H+ in solution
5
a system at equilibrium, when changed, will attempt to partly reverse the effects of
that change
6
the reactant in a titration whose amount is being measured
7
an equation where the spectator ions are omitted
8
a process where reactions at the molecular level are still proceeding, but there is no
overall change to the levels of the substances involved
9
ions in a reaction which are not involved in the reaction
135
GLOSSARY
acid — a substance that releases H+ in solution
actinide element — an element in the separate block of the table with atomic numbers
90–103
allotrope — a different physical form of an element
analyte — the reactant in a titration whose amount is being measured
anhydrous — a compound, which could be hydrated, which has had the water molecules
removed
anion — a negatively charged ion
atom — the building blocks of matter
atomic number — the number of protons in an atom
atomic weight — the average mass of naturally occurring atoms of an element
Avogadro’s number — the number of individual species in one mole
balanced equation — an equation where there are equal numbers of each type of atom
on both sides
bond — a link between two atoms
cation — a positively charged ion
chemical change — a process where different substances are produced
chemical equation — a ‘sentence’ written in chemical formulae describing a reaction
chemical formula — a means of summarising the type and number of elements in a
substance
chemical reaction — a process where the chemical form of the substances involved is
changed
chemistry — the study of the behaviour of matter
coefficients — numbers in front of formulae which indicate the ratio of numbers of
molecules involved in a reaction
compound — a pure substance containing more than one type of atom
covalent bond — the sharing of two electrons between a pair of atoms to increase the
number of outer shell electrons
covalent network — a large three-dimensional structure of covalent bonds
diatomic molecule — two atoms linked by a covalent bond, forming a separate molecule
dilution — addition of solvent to a volume of existing solution (usually with known
initial and final volumes)
dilution factor — the ratio of initial to final concentration (or final to initial volume) in
a dilution step
dipole attraction —attraction between polar covalent bonds, where the positive and
negative centres in different molecules are weakly linked
dispersion force — very weak intermolecular forces between non-polar bonds
136
Chapter 11
Chemical Reactions II
double covalent bond — four electrons shared between a pair of atoms
dynamic equilibrium — a process where reactions at the molecular level are still
proceeding, but there is no overall change to the levels of the substances involved
electron — a negatively charged subatomic particle
electron dot diagram — a shorthand way of representing the number of valence
electrons in an atom
electron shell — a region surrounding the nucleus which contains electrons
electronegativity — a measure of an element’s need for electrons
element — a substance containing one type of atom
empirical formula — the simplest ratio of numbers of elements (or ions) in a compound
equilibrium — a reaction which does not go to completion, but reaches a point where no
further change in reactant and product concentration occurs
equilibrium constant — a numerical value representing the ratio of the concentrations
of products to reactants
formula weight — the sum of the atomic weights of all atoms in a molecule
functional group — a grouping of atoms that occurs in many different organic
compounds, and defines their properties
gas — a physical state where the species are able to move randomly without interaction
with others
group — a column of elements in the periodic table
heterogeneous mixture — matter containing more than one pure substance, with
varying appearance throughout
homogeneous mixture — matter containing more than one pure substance, but of
uniform appearance
hydrated — a compound that contains a fixed number of water molecules in its structure
hydrogen bond — a force between a hydrogen atom attached to an electronegative
element and that element in another molecule
intermolecular force — a bond between two atoms in different molecules
intramolecular bond — a bond between two atoms in the same molecule
ion — a charged atom or molecule
ionic bond — attraction between positive and negative species
ionic equation — an equation where the spectator ions are omitted
ionic formula — the simplest ratio of numbers of cations and anions
isotope — atoms with the same number of protons, but different number of neutrons
lanthanide element — an element in the separate block of the table with atomic
numbers 58–71
law of conservation of mass — no matter is created or destroyed in a reaction
137
Chemistry for Technicians
Le Chatelier’s principle — a system at equilibrium, when changed, will attempt to
partly reverse the effects of that change
‘like dissolves like’ — the solubility rule which says that compounds will dissolve in
solvents of similar polarity
main-group element — an element in the groups 1A–8A
mass number — the combined total of protons and neutrons
matter — all the ‘stuff’ around us
melting point — the temperature at which the individual species in a substance break
free of the bonds that hold them together in a solid
metallic bond — electrostatic attraction between free electrons and cations
miscible — two liquids that are soluble in each other
mole — the number of atoms, molecules or ions that comprise the formula weight of the
substance in grams
molecular formula — the actual number of atoms of each element in a compound
molecule — groups of atoms linked together in a fixed way
monatomic ion — an ion containing a single atom
neutron —a sub-atomic particle with no charge
non-bonded electrons — valence electrons not involved in covalent bonding
non-polar covalent bond — a covalent bond in which each atom’s electro-negativity is
similar
nucleus — the centre of the atom, which contains most of the mass
organic compound — any compound containing carbon, other than salts of carbonate
and hydrogen carbonate
period a row of elements in the periodic table
periodic table — an arrangement of the known elements by ascending atomic number
and similar properties
physical change — a process where the physical, but not chemical, form of the matter is
altered
polar covalent bond — a covalent bond in which one atom’s electro-negativity is
considerably greater than the other
product — a new substance formed in a reaction
proton — a positively charged subatomic particle
reactant — a substance initially present before a reaction commences
science — the study of the behaviour of our surroundings
serial dilution — where a solution is diluted a number of times, each diluted solution
becoming the starting solution for the next step
solute — the substance dissolved in the solution
solution — a homogeneous mixture, normally a liquid
solution concentration — the amount of solute per unit volume of solution
solvent — the liquid that forms the solution
138
Chapter 11
Chemical Reactions II
species — a general term which may refer to any type of chemical substance
spectator ion — an ion in a reaction which is not involved in the reaction, though it is
part of one of the reactant compounds
standard — one of the reactants in a titration, whose amount is exactly known
structural formula — an expanded formula or diagram showing the bonding
arrangements for the atoms
structure — the way that atoms are joined to each other
systematic name — a unique identifier for every known compound
titration — a chemical reaction used to measure the amount of one reactant
transition metal — an element in periods 4–6 in the block between the main-group
elements
triad — a group of three elements with similar properties
valence shell — the outermost shell containing electrons
valency — the number of electrons that an atom needs to lose or gain to achieve stability
139
ANSWERS TO PRACTICE QUESTIONS
CHAPTER 1
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
Water: purification, testing.
Coffee/milk/sugar: processing, testing.
Kettle materials: production. testing.
Electricity generation and supply: production and testing of materials.
You can probably think of some more.
Solids have a fixed volume and shape. Liquids have a fixed volume, but their shape
changes to fit the container. Gases have neither fixed shape nor volume — they are
determined by the container.
Add water to the mixture and stir until the salt dissolves. Pass the mixture through
filter paper — the sand is retained and can be dried and separated. The solution
containing the salt must be boiled to remove all the water, leaving the salt behind.
(a) solid, non-metal, mostly manmade, neutral, mixture, molecules, compounds
(b) gas, non-metal, natural, neutral, mixture, mostly molecules with some atoms,
mostly elements with some compounds
(c) solid, metal, natural, neutral, pure, atom, element
(d) liquid, natural, acid, homogeneous mixture, molecules, compounds
(e) solid, natural, neutral, pure, molecule, compound
(f) liquid, natural, neutral, heterogenous mixture, molecules, compounds.
Liquid B is definitely a mixture since evaporation of the liquid substance has left a
solid substance. Liquid A could be pure (e.g. water), or it could be a mixture of two
liquids (e.g. water and alcohol).
All statements are true.
A — element; B - compound; C — homogenous mixture; D — heterogenous
mixture.
Solvent — water; Solutes — caffeine, tannin, sugar (tea leaves are not a solute).
Concentrations in g/mL: A (0.005, most dilute), C (0.01), B (0.05), D (0.25).
(a) solute, (b) solvent, (c) solution, (d) miscible.
(a) physical, (b) chemical, (c) physical, (d) chemical.
(a) is a chemical change.
Physical changes are easily reversed.
From the symbols starting with the letter P: phosphorus P, polonium Po, platinum
Pt, lead Pb.
CHAPTER 2
1.
2.
3.
4.
Much of the atom in the solar system model is empty space.
99.97%.
Elements can vary greatly in the number of neutrons, so an atom with one less
proton than another may have one more neutron, making the combined total equal,
e.g. carbon-14 (6 p, 8 n) and nitrogen-14 (7 p, 7 n).
Corrections in bold. ‘A neutral atom of carbon contains six protons, six electrons
and six neutrons. The electrons are found circling around a group of protons in the
centre. The neutrons can be found in the nucleus.’
140
Appendix 2
Answers To Practice Questions
5.
Carbon-13: 6 p, 6 e, 7 n. The number of protons and electrons is identical for all
neutral atoms of carbon; only the number of neutrons varies.
6. (a) A = 6, Z = 13, e = 6
(b) A = 11, Z = 23, p = 11
(c) A = 11, e = 11, n =11
(d) Z = 55, p = 26, e = 26
(e) A = 36, Z = 84, p = 36.
(f) A = 82, p = 82, e = 82.
(g) A = 79, p = 79, n = 118
(h) Z = 232, p = 92, e = 92.
7. 55.91. This is not exactly the real atomic weight; one reason is that the mass
numbers are not the exact masses of the atoms.
8. If sodium gained another proton, the atom would gain a 1+ charge, but would have
12 protons, and therefore, be a magnesium atom.
9. It would become positively charged (1+), since it would have 9 protons and 8
electrons.
10. It loses two electrons.
11.
12.
13.
14.
15.
(a) 2.6 (b) 2.8.6 (c) 2.8.8.1
(a) 1 (b) 7 (c) 8.
(a) 2– (b) 2– (c) 1+.
These elements don’t attempt to have eight electrons in their valence shell when
stable. A full first shell has only two electrons.
CHAPTER 3
1.
2.
3.
4.
5.
6.
7.
Na and K, F and Cl.
non-metal, 1–, very reactive, two, two, acid.
(a) Group 1A — Li, Na, K, Rb, Cs, Fr.
(b) Row (period) 2 — Li, Be, B, C, N, O, F, Ne.
Cu and Ag, and Zn and Cd are no longer grouped with the same elements, but do
belong in groups together.
(a) Pt (b) hafnium (c) 79
(d) 126.9.
The same.
A — 8A, B — 2A, C — 7A.
141
Chemistry for Technicians
8.
The group number is also the number of electrons in the outer shell, e.g. 1A — 1,
7A — 7.
Group Electrons
1A
1
2A
2
3A
3
4A
4
Group Electrons
5A
5
6A
6
7A
7
8A
8
CHAPTER 4
1.
2.
3.
4.
5.
6.
7.
8.
(a) a molecule with three oxygen atoms
(b) a compound with one sulfur and two oxygen atoms
(c) an ion with two carbon and four oxygen atoms and with 2– charge
(d) an ion with one vanadium and two oxygen atoms with a 1+ charge
(e) a compound with one potassium, one manganese and four oxygen atoms
(f) a compound with nine carbon, 12 hydrogen, three bromine, four nitrogen and
six oxygen atoms.
(a) FeCl4– (b) C3H4NO2 (c) VO22+.
(a) molecular (b) ionic (c) ionic (d) molecular.
fluoride, bromide, iodide, sulfide.
(a) phosphorus pentachloride
(b)
nitrogen dioxide
(c) sulfur hexafluoride
(d)
diphosphorus pentoxide
(e) carbon disulfide.
Dihydrogen monoxide, dihydrogen dioxide.
(a) SiO2 (b) OI2 (c) BF3.
cation : anion.
D–
E2–
F3–
A+
1:1
2:1
3:1
B2+
1:2
1:1
3:2
C3+
1:3
2:3
1:1
9. (c).
10. (a) 2 N, 8 H, 1 C, 3 O.
(b) 1 Cu, 2 N, 6 O.
11. (a) ammonium chloride
(c) copper (II) sulfate
(e) magnesium phosphate
12. (a) (NH4)2SO4
(c) Fe2O3
(e) ZnSO3.
13. (a) (VO)3(VO4)2
(c) Tl2(Cr2O7)3.
14. (a) silicon tetrachloride
(c) calcium hydrogen carbonate
15. (a) CF4
(c) Al(OH)3
(e) AlPO4.
16. (a) 208.22
(c) 146.06
(e) 76.13.
17. (a) 53.50
(c) 159.60
(e) 262.87
142
(b)
(d)
(f)
(b)
(d)
sodium carbonate
iron (II) oxide
iron (III) sulfide.
MgBr2
Cu(HCO3)2
(b)
Th(MnO4)4
(b)
(d)
(b)
(d)
iron (III) nitrate
dinitrogen monoxide.
SO3
ZnCO3
(b)
(d)
46.01
141.94
(b)
(d)
(f)
105.99
71.85
207.88.
Appendix 2
Answers To Practice Questions
CHAPTER 5
1.
The bonds holding the atoms in the ammonium ions must be stronger than ionic
bonds.
2. Solid — it takes a lot of heat to break ionic bonds.
Crystalline and dense — due to the regular arrangement of ions.
Electrical conductivity — ions are free to move in solution or in melt, but not in
solid state.
Solubility in water but not in organic solvents — beyond your knowledge at this
time, but it is worthwhile knowing that water is a good solvent for substances
containing charges.
3. Yes, if the ions and electrons remained separated, but not if the electrons returned to
the atoms.
4. Solid — it takes a lot of heat to break metallic bonds.
Crystalline and dense — due to the regular arrangement of atoms.
Electrical conductivity — electrons are free to move in the solid or liquid state.
Solubility in water — won’t dissolve because the metallic bond would have to break
and lose the stability.
5. Still a crystal lattice of ions surrounded by electrons, but the ions would be different
sizes.
6. It is a gas.
7. (a) Argon has the lowest, HCl the highest.
(b) Monatomic.
(c) HCl must have the strongest
forces, argon the weakest, based on the amount
of heat required to boil the substances.
8. It is the actual number of atoms of each type in the molecule.
9. Ionic to covalent, from left to right.
10. A — ionic, B — covalent molecular, C — metal,
D — covalent network.
11. The bonds within a covalent molecule are covalent and will not break easily.
However, the forces between the molecules are relatively weak, and can be broken
with a small amount of heat.
12. (a)
H—O—H
C2H6O
H
|
O
|
H
C2H6O
(b)
H—O—H
Na+
NO3–
H—O—H
H
|
O
|
H
13. (a) ionic (b) covalent (c) metallic
(d) none (e) none.
143
Chemistry for Technicians
CHAPTER 6
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
61.8 g
8g
30 g
1 — Reactants: magnesium, oxygen.
2 — Reactants: hydrogen, oxygen; product: water.
3 — Reactants: sodium, chlorine; product: sodium chloride.
(a) 2AlCl3 + 3H2O → Al2O3 + 6HCl
(b) C2H4 + 3O2 → 2CO2 + 2H2O
(c) 2Fe2O3 + 3C → 4Fe + 3CO2
(d) Zn + 2HCl → ZnCl2 + H2
(e) 2AgNO3 + H2SO4 → Ag2SO4 + 2HNO3
(f) Na2O + H2O → 2NaOH
(g) CH4 + 4Br2 → CBr4 + 4HBr.
(a) 2 extra H2O, 1 extra H2
(b) 2 extra O2
(c) 1 extra H2O.
The 2:1 coefficients in the equation refer to the number of molecules of each
reactant, not their mass.
(a) CaCO3 (s) + H2SO4 (aq) →
CaSO4 (s) + H2O (l) + CO2 (g)
(b) 3H2 (g) + N2 (g) → 2NH3 (g)
(c) 2KBr (l) → 2K (s) + Br2 (l) .
(a) precipitation
(b)
combination
(c) gas formation, decomposition
(d)
neutralisation
(e) combustion.
(a) 8.56 x 10–4 moles
(b)
55.5 moles
(c) 6.98 mg
(d)
599 g
(e) 2.54 x 10–3 moles.
(a) 1 mole, 65.38 g
(b)
1 mole.
(a) 0 5 moles
(b)
48 g
(a) 158.7 g
(b)
403 g H2, 3.2 kg O2.
AgNO3 + NaCl → NaNO3 + AgCl
(a) AgNO3 169.91 g, NaCl 58.44 g, NaNO3 85.0 g, AgCl 143.35g
(b) 16.9 g AgNO3, 5.83 g NaCl.
CHAPTER 7
1.
2.
3.
4.
5.
Beryllium and magnesium are both low-density reactive metals, forming 2+ ions.
Nitrogen is a gas, which forms 3– ions and covalent bonds.
Ar/K, Co/Ni, Te/I, Th/Pa, U/Np. In these cases, the first element has an average
number of neutrons in its isotopes that increases its mass past that of the second
element.
(a) transition metals
(b) main group
(c) actinides
(d) lanthanides.
(a) centre (b) left (c) bottom right
(d) top right.
(b) Fr 20–25°C, At 220–230°C
(estimated from graph of
melting point
US period).
144
Appendix 2
6.
7.
8.
Answers To Practice Questions
Given that it is on the metallic side of the dividing line, it should a fair conductor of
electricity, have a metallic appearance, form ions and covalent bonds and have a low
melting point for metals.
(a) Down a group, the atom grows in size because there is an extra electron shell
per row, but
across a period, the greater
charge in the nucleus will pull
the electrons closer.
(b) Metals need to lose electrons to become stable, which becomes less likely
across the table, where elements require extra electrons. Down a group, the
outer electrons become farther away from the positive nucleus and are more
easily lost.
(c) The elements in the centre of a row tend to form covalent
networks.
(a) Transition metals are denser, more varied in properties, can form ions with
different charges and also special covalent bonds.
(b) Lanthanide metals are less
reactive than main group metals
and
more uniform in their
properties.
CHAPTER 8
1.
2.
3.
4.
5.
6.
7.
(a) magnesium sulfate heptahydrate
(b) barium hydroxide octahydrate.
(a) FeF2.4H2O
(b) CaSO4.2H2O.
(a) potassium perchlorate
(b) sodium chromate.
(a) ZnCr2O7
(b) NaMnO4.
Ionic compounds do not form separate molecules under normal conditions, so a
molecular formula is not possible, only a ratio, i.e. empirical formula.
(a) 2 Na+ to 1 oxalate.
(b) NaCO2
(c) This formula ‘hides’ the poly-atomic structure of the oxalate ion.
(a) Water is not a linear molecule — the angle between the two bonds is less than
180° (you weren’t expected to know this, so H—O—H is OK).
H
O
H
(b) The three fluorines are equivalent, so they must each connect to the boron.
F
B
F
F
8.
Br
H
H
H
C
C
C
H
C
C
C
H
H
CH3
H
CH2CH 2CH 3
H
C
CH3
145
Chemistry for Technicians
9. (a).
10. All contain alkane groups.
(a) alkene
(c) alkanone
(e) alkanal
(g) alkyne
(i) alkyl halide.
11. (a) hexene *
(c) pentanone *
(e) ethanal
(g) butyne *
Those marked with * require
(b)
(d)
(f)
(h)
alkanol
alkanoic acid
alkane
amine
(b)
(d)
(f)
propanol *
butanoic acid
propane
numbering.
CHAPTER 9
1.
2.
3.
(a) lose one or gain 1
(b) lose 2
(c) gain 2
(d) no change.
(a) 1 (b) 2 (c) 2 (d) 0.
(a)
(b)
(c)
••
• •
••
•
4.
Mg •
S
Al •
•
•
(a) Magnesium loses both electrons
to oxygen
••
• •
••
Mg •
O
•
••
•
•
2+
Mg
• 2–
•
O
••
(b) Two sodium atoms each lose one electron to oxygen
(c) Two aluminium atoms each lose
three electrons; three sulfur
each gain two electrons.
5.
••
•
×
•
H O
× H
••
6.
(a)
××
••
•×
•×
O O
••
××
146
atoms
Appendix 2
6.
Answers To Practice Questions
(b )
••
H
N• ×
ו
•
×
H
H
7.
8.
9.
(c) 4 H atoms each share an electron with C.
Carbon needs to gain four electrons. To do this, it must participate in four covalent
bonds, and, therefore, must share all of its four.
The number of electrons in a full shell is always even, as is the number of electrons
in a covalent bond. Therefore, the number of any non-bonded electrons in the shell
will also be even.
(a) 3 bonds to O
(b) 2 bonds to H
(c) 2 bonds to Cl, 3 bonds to C
10.
Covalent
Ionic
Metallic
Lose
no
yes
no
Gain
yes
yes
yes
11. (e); normally (b) would also be considered polar.
12. O, F, Cl, Br, I; normally N–H would also be considered polar.
13. (a) H-bonds
(b) dispersion forces.
14. Because the O–H bond is more polar than the N–H bond, the H–bonds in water are
stronger, giving it a higher boiling point.
15. Higher (for the same reason as in Q14).
16. It has a higher formula weight.
17. Propanol has the polar
O–H bond which allows
H-bonding, propanal can undergo weaker dipole attraction, while butane has only
very weak dispersion forces.
18. Butanal is less polar than butanol, and therefore is closer in polarity to hexane.
19. Butanoic acid — more H-bonding sites.
20. A non-polar solvent, such as hexane.
CHAPTER 10
1.
2.
3.
4.
5.
6.
7.
If a solution is homogeneous, then the concentration — the ratio of amount to
volume — is the same throughout the solution. Any sample taken from that solution
will have the same concentration as the whole.
0.1 L, 0.1 g, 1 g/L.
(a) 0.01 g
(b)
0.5 g
(c) 2 g
(d)
1.25 g.
(a) 0.2 g/L
(b)
0.003 g/100 mL
(c) 30 mg/L
(d)
1.67 %w/w.
(a) 10 L
(b)
50 mL
(c) 200 mL.
(a) 0.05 M
(b)
0.3 M
(c) 0.002 M.
(a) 0.0225 moles, 0.82 g
(b)
0.09 moles, 7.65 g
(c) 0.025 moles, 6.24 g.
147
Chemistry for Technicians
8. (a) 500 Ml
(b)
0.3 mL.
9. (a) 1.33 %v/v
(b)
1%v/v.
10. (a) 0.02 M
(b)
0.0625 g/100 mL
(c) 50 mg/L.
11. (a) 200 mL
(b)
10 mL
(c) 45 mL.
12. (a) 2, 4, 5, 6.67, 10, 20
(b)
2, 4, 8, 10, 20, 40.
13. 500 mg/L.
14. (a) 10 mL of solution made to a final volume of 50 mL.
(b) 25 mL of solution made up to a final volume of 250 mL.
(c) 0.5 mL of solution added to 4.5 mL of solvent.
15. Other answers are possible.
(a) 20 mL of concentrated acid diluted to 100 mL. 85 mL of this solution to 2L.
(b) 30 mL of concentrated acid to 500 mL of solution.
16. (a) 0.365 g/100 mL
(b)
0.042 M
(c) 1 mg/L
(d)
5 g/100 mL (%w/v)
(e) 5.56 mL/100 mL (% v/v).
CHAPTER 11.
1.
1.
2.
(a) Pb2+ (aq) + 2I– (aq) → PbI2 (s)
(b) 2Fe (s) + 6H+ (aq)
→ 2Fe3+ (aq) + 3H2 (g)
+
–
(c) H (aq) + OH (aq) → H2O (l) (note that this ionic equation simplifies matters too
much if the reaction ratio is needed!).
(a) sodium nitrate and water
(b) calcium sulfate and water
(c) no reaction
(d) magnesium nitrate and hydrogen gas
(e) Ca(HCOO)2 + CO2 + H2O
(f) zinc chloride and dihydrogen sulfide
(g) calcium chloride, water and sulfur dioxide.
3.
(a)
(b)
(c)
(d)
4.
5.
6.
7.
8.
Standard
NaOH
Na2CO3
H2SO4
HCl
Analyte
HNO3
H2SO4
KOH
NH3
(a) 1.15 x 10–3 moles, 72.5 mg
(b) 6.32 x 10–4 moles, 62.0 mg
(c) 1.73 x 10–2 moles, 972 mg
(d) 2.43 x 10–3 moles, 41.4 mg.
(a) 3.86 x 10–3 moles
(b) 0.771 M
(c) 4.63 g/100 mL.
(a) 92.7 mg
(b) 38.8 %w/w.
(c).
(a) The system does not return to the same state, where all the concentrations are the
same as before. Concentrations will be different, but the equilibrium constant is the
same as before
(except where the temperature is changed, where a new constant
is reached as well).
148
Appendix 2
Answers To Practice Questions
8.
(b) At the same temperature, the equilibrium constant will always be the same,
regardless of the initial concentrations.
9. (a) The concentration of product drops as a result of its removal, then begins to
increase slowly to another equilibrium point, which will be less than before removal.
The concentration of reactant will slowly decrease until equilibrium is reached,
since it is being consumed to replace some of the lost product.
(b) The concentration of reactant drops as a result of its removal, then begins to
increase slowly to another equilibrium point, which will be less than before removal.
The concentration of product will slowly decrease until equilibrium is reached,
since it is being consumed to replace some of the lost reactant.
10. 100°C
.
149
INDEX
A
chemical change, see chemical reaction
chemical reactions, 11, 70–83, 123–135
products, 70
reactants, 70
recognition of, 11–12, 71–72
reversibility of, 78
types of, 77–78
chlorine, properties, 34
chromatography, 7, 8
combination reaction, 77
combustion, 77
compound, 9
naming,
binary molecular, 44–46
hydrated, 92
ionic, 46–50
organic, 97–98
types of, 43–44
concentration, solution, 111
calculation of,
dilution, see dilution
g/L, 113–114
g/100 g (%w/w), 114
g/100 mL (%w/v), 113–114
mg/L, 113–114
mL/100 mL (%v/v), 115
molarity, 114–115
units, 112
interconversion of 119–120
covalent molecule, 63
covalent network, 61–62
acid-base indicator, 5
acids, 5, 124
common, 50
reactions of, 124–126
actinide elements,
position in periodic table, 86
trends in properties, 90
allotropes, 8
aluminium, properties, 34
anhydrous compounds, 92
ammonium chloride,
properties, 53–55
structure, 66–67
ammonium hydroxide, 50
analyte, 126
atom, 8
changes to structure, 24–25
composition, 19–23
theories of structure, 17–19
atomic number, 20–22, 38
atomic weight, 23–24, 38
standard for, 23
Avagadro’s number, 79
B
Bohr, 18–19
boiling point, 3, 53–54
effect on by intermolecular forces, 107
of mixtures, 6
bond, 52, 67–68
covalent, 61
by common elements, 104
formation of, 102–103
multiple, 103
polarity, 105
sharing of electrons in, 104–105
hydrogen, 106
intermolecular, 64, 105–108
effect on physical properties,
107–108
types of, 107
intramolecular, 61, 63, 106
ionic, 56
formation of, 101–102
metallic, 60
D
Dalton, 17–18
decomposition, 77
density, 54
diamond,
properties, 53–55, 63
structure, 61–62
dilution, solutions, 116
calculation of, 116–119
serial, 118–119
dilution factor, 116–117
dipole attraction, 107–108
dispersion forces, 108
E
C
electron, 18
changes to atomic structure, 24–25
configuration, 27–28, 39, 100
non-bonded, 103
properties, 19
carbon,
allotropes of, 8
atomic weight standard, 23
properties, 33
150
Appendix 2
Answers To Practice Questions
hydrogen chloride,
properties, 53–55, 63
structure, 63, 64
shell theory, 27
valence, 28, 100
electron dot diagrams, 101
electronegativity, 104
values for common elements, 104
elements, 8
arrangement of, 35–37
discovery of, 31
properties of,
by group, 38
selected, 32–34
trends in,
symbols, 13–14
equations,
balancing, 73–75
ionic, 123–124
meaning of coefficients, 75
physical state indicators, 76
symbols, 73
writing, 72–75
chemical equilibrium, 78, 128–133
changes occurring in reaction, 129,
constant, 131–132
dynamic, 129
factors affecting, 130–132
Le Chatelier’s principle, 132
ethanol,
properties of, 53–55
structure of, 65
I
ion,
formation of, 25, 28–29
names of common, 43, 93
spectator, 123
types of, 25
ionic compounds,
properties, 56–58
structure 56–57
ionisation energy, 26, 88
iron,
properties, 53–55
isotope, 21–22
L
lanthanide elements,
position in periodic table, 86
trends in properties, 90
law of conservation of mass, 70
liquid, 64
indication of in equation, 76
M
magnesium, properties, 34
main group elements,
position in periodic table, 86
trends in properties, 86–88
mass number, 21–22
matter, 1
classification of, 2–9
by acidity, 5
by chemical form, 8–9
by composition, 5–9
by metallic state, 4
by physical state, 2–4
by properties, 2–5
by purity, 5–7
by source, 4
melting point, 3, 53–54
effect on by intermolecular forces, 107
of mixtures, 6
Mendeleev, 36, 37, 85
mercury, 60
metal, 4, 87, 88
properties, 59–60
structure, 59
metalloids, 87
Meyer, 36, 37
miscibility, 11
mixture, 5–7
heterogeneous, 5
homogeneous, 6
separation of, 6–7
mole, 79
calculations, 79
F
filtration, 7
fluorine, properties, 33
formula weight, 50
formulae,
empirical, 94
ions, 42
molecular, 94
molecules, 41, 47
structural, 65, 94
functional groups, 96
G
gas, 2–3, 61, 64
formation in reaction, 78
indication of in equation, 76
germanium, properties, 87
groups (of elements),
common names, 37
numbering, 37
H
Haber process, 133–134
helium,
properties, 32, 53–55
structure, 60–61
hydrated compounds, 92
hydrochloric acid, 50
hydrogen, properties, 32
151
Chemistry for Technicians
significance in reactions, 80–81
molecule, 8
types of, 8
R
Rutherford, 18, 19
S
N
octet rule, 28
organic compound, 65, 95
functional groups, 96
naming, 97–98
oxygen,
allotropes of, 8
properties, 33, 53–55, 63
structure, 62, 64
sodium,
properties, 33, 35, 53–55, 60
sodium chloride,
properties, 53–55
structure, 56–57
solid, 64
indication of in equation, 76
solubility,
effect on by intermolecular forces, 108
solution,
aqueous, 76
concentration, see concentration
indication of in equation, 76
preparation of, 111–122
terminology, 10
standard, 126
structure, 52
sulfur,
properties, 53–55
structure, 65
sulfuric acid, 50
P
T
perchloric acid, 50
periodic table, 37
as a store of information, 37–38
earliest form, 36
predictions using, 84–85
sections of, 85–86
phosphoric acid, 50
physical change, 11
pipette volumes, 119
potassium, properties, 34, 35
practical work,
acidity measurement, 5
boiling point measurement, 3
chemical equilibria, 127, 128, 130–131
observation of chemical reactions, 72
identification of substances, 65–66
law of conservation of mass, 71
mass/mole relationships, 81
physical and chemical changes, 12
preparation of solutions, 117, 120
properties of ionic compounds, 58
properties of metals, 60
properties of Row 3 elements, 34–35
purity of substances, 8
reactions of acids, 125–126
reversibility of reactions, 78, 132
titrations, 127
precipitation, 77
proton, 19
changes to atomic structure, 24–25
properties, 19
Thomson, 18
titration, 126
calculations, 127
transition metals,
position in periodic table, 86
trends in properties, 89–90
naphthalene,
properties, 53–55
structure, 65
neutralisation, 77
neutron, 19
changes to atomic structure, 25
properties, 19
nitric acid, 50
nitrogen, properties, 33
nucleus, 19
O
V
volumetric flask volumes, 119
W
water,
properties, 53–55
structure, 63, 64
152