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TAFE: TPC CHEMISTRY A
MODULE: NSWTSCN308A
INVESTIGATE THE CHEMICAL NATURE
OF MATTER
Author:
Robyn Krajniak (1995)
Revised:
Dan Solomon (2013)
Graphics:
Martin Kent
1
TABLE OF CONTENTS
1.
2.
3.
4.
INTRODUCTION – Matter Surrounds Us!
6
CLASSIFICATION SYSTEM 1: STATES OF MATTER
7
1.1
Models of matter
Written Exercise: Which state?
Activity: Making models
Written Exercise: States of matter
7
8
8
10
1.2
Changes of State
11
MODELLING THE SMALLEST PARTICLES OF MATTER
12
2.1
What particles compose the substances around us?
12
2.2
Modelling Atomic Structure
Written Exercise: Timeline
14
14
2.3
Isotopes
17
2.4
Quantifying Subatomic Particles in Atoms
18
2.5
How are Electrons organised around the Nucleus?
Written Exercise: Drawing structural models of elements
Written Exercise: The first 20 elements of the periodic table
Activity: Elemental puns
18
20
21
22
CLASSIFICATION SYSTEM 2:
ELEMENTS, COMPOUNDS AND MIXTURES
23
3.1
Classifying Everyday Substances
Written Exercise: Classifying substances
23
24
3.2
Physical and Chemical Properties
Written Exercise: Physical and chemical changes
Experiment – Making observations
Experiment – Physical versus chemical change
25
26
26
29
3.3
Mixtures and laboratory separation techniques
Experiment – Separation techniques 1
Experiment – Separation techniques 2
30
33
36
ELEMENTS AND THE PERIODIC TABLE
39
4.1
Mapping Elements on the Periodic Table
39
4.2
Distribution of the Elements
41
2
5.
6.
4.3
How Atoms can be arranged in Everyday Elements
Research Task – Allotropes
42
44
4.4
Classification of the Elements
Experiment – Properties of elements
44
44
CHEMICAL BONDING
50
5.1
What makes an Atom Chemically Stable?
50
5.2
Loss and Gain of Electrons – The Ionic Bond
51
5.3
Sharing Electrons – The Covalent Bond
Written Exercise: Lewis electron dot formulae
52
52
5.4
Which Chemical Bond Will Form?
Written Exercise: Type of chemical bond
53
53
5.5
The Metallic Bond
53
5.6
Summary of Bonding
Written Exercise: Formulae and chemical bonding
54
55
5.7
Compound Chemical Formulae
Written Exercise: Interpreting chemical formulae
56
58
5.8
Valency
58
5.9
Chemical Formulae and Naming
Written Exercise: Writing chemical formulae
Written Exercise: Formulae and naming
Activity: Modelling molecules and compounds
60
61
62
64
PROPERTIES OF COMPOUNDS
66
6.1
Comparing properties of elements and compounds
66
6.2
Classification of Compounds
67
6.3
Properties of Compounds
Written Exercise: Solubility
Experiment – Physical properties of compounds
Written Exercise: Hydrogen chloride conductor?
Written Exercise: Compound properties in summary
Experiment – 3 white powders
68
70
71
72
72
73
3
7.
8.
CHEMICAL REACTIONS
75
7.1
Energy and Chemical Processes
Written Exercise: Chemical reactions (introduction)
75
76
7.2
Writing Chemical Equations - Chemical Shorthand
Activity: Modelling a simple chemical reaction
Written Exercise: Balancing chemical reactions
Written Exercise: Expressing chemical reactions
78
79
81
86
7.3
Ionic Equations
Written Exercise: Expressing ionic equations
Experiment – Solubility, precipitation and ionic equations
88
89
90
QUANTITIES AND CHEMICAL PROCESSES
92
8.1
An Introduction to Quantitative Chemistry
92
8.2
The Mole Concept
Written Exercise: Mole calculations
94
96
8.3
Theoretical Yield
97
8.4
Percentage Yield
98
8.5
Limiting Reagents
Written Exercise: Limiting reagent and theoretical yield
Yield Experiment 1 – Magnesium Oxide
Yield Experiment 2 – Decomposition of a carbonate compound
Yield Experiment 3 – Zinc Iodide
99
101
102
103
105
8.6
Moles, Gas Volumes and Yields
Written Exercise: Theoretical yield and gas volumes
107
109
8.7
A quick note on Kelvin
Yield Experiment 4 – Gas volumes
111
112
8.8
Percentage Composition
Written Exercises: Calculating percentage composition
114
114
8.9
Empirical and Molecular Formulae
Written Exercises: Percentage composition, empirical
and molecular formulae
Experiment – Which mercury chloride?
115
Calculating Concentration
120
8.10
4
116
117
9.
8.11
Molarity
Written Exercises: Calculating molarity
Written Exercises: Solution calculations
121
122
124
8.12
Solution Preparation and Quantitative
Dilution in the Laboratory
Written Exercise: Dilution scenario calculations
Experiment – Solution preparation and serial dilution
125
126
128
CHEMICAL ANALYSIS
130
9.1
Qualitative Vs Quantitative Chemical Analysis
130
9.2
Gravimetric Analysis
Practice Questions
Experiment – Sulfate content of plant fertiliser
131
133
134
9.3
Volumetric Analysis
Practice Questions
Experiment 1 – Practice titration and HCl analysis
Experiment 2 – HCl standardisation
Experiment 3 – NaOH standardisation
Experiment 4 – Vinegar analysis
Experiment 5 – Miscellaneous titration
136
141
142
144
145
146
147
9.4
Chromatography
149
9.5
Electrophoresis
151
9.6
Colorimetry
Experiment – Colorimetry
152
154
9.7
Spectroscopy
Experiment 1 – Flame tests
Written Exercises: Flame Emission
Experiment 2 – Introduction to Flame Emission
Experiment 3 – Sodium and potassium in beer
Overview: Infra-red Spectroscopy
155
157
160
162
164
168
ANSWERS TO WRITTEN QUESTIONS
169
References
178
5
INTRODUCTION – Matter Surrounds Us !
Chemistry involves the study of matter. All of the different materials around us are made
of matter. In this module you will learn how to answer questions such as:
-
Why do different materials have different characteristics, ie - properties?
In what ways is sugar different to salt?
How are metals different to plastics or ceramics?
-
How can a knowledge of the properties of a substance help you to select the best
material for a particular purpose?
What would be the best material to choose for making a coffee mug or a saucepan?
-
How can you explain the properties of a substance by understanding its structure?
Why are some substances brittle while others are flexible?
What makes some substances able to conduct electricity?
-
How can you predict what will happen if several substances are reacted together?
What happens when a substance burns?
Why do some substances dissolve in water while other substances don’t?
What is Matter?
Matter is any material of substance. It has mass and takes up space (volume).
Look around you. In any room there are many examples of different types of matter:
flooring materials
paint
metal fittings
probably several plastic items
water-perhaps in a cup or sink
air
Is there anything that is not matter?
Yes, energy is not matter. All forms of energy (heat, light sound etc) can affect matter but
energy has no mass or volume.
6
1.
CLASSIFICATION SYSTEM ONE: STATES OF MATTER
One way of classifying substances is by grouping them according to whether they are
solid, liquid or gas. These are often called the states of matter.
In order to understand the different way solids (eg rock), liquids (eg water) and gases (eg
air) behave, we set up models which we can use to visualise what is happening deep in
the structure of matter. The following models show matter consisting of tiny particles
which are in constant motion - even the particles in a solid piece of rock are constantly
moving!
1.1
Models of matter
In a solid, the particles are just vibrating in
set positions. They are packed very close
together and attract each other strongly.
In a liquid, the particles are further apart.
They can slide past one another as they
move, colliding with each other and the
walls of the container when they reach the
edge. The particles are attracted to each
other, but not as strongly as in the solid
state.
In a gas, the particles fly as far apart as
the container walls will allow. The particles
are only very weakly attracted to each
other, and they collide with each other and
the walls of the container.
7
Written Exercise:
Which State?
Many substances are easy to classify as solid, liquid or gas, others can be difficult,
particularly if they contain more than one state e.g. an inflated balloon! Place the following
substances in their correct category:
sand
lemonade
fog
plasticine
natural gas
balsamic vinegar
steel
mercury
timber
paint
toothpaste
shaving foam
milk
jelly
Add ten more examples including some which may be difficult to classify.
Solid
Activity:
Liquid
Gas
Difficult to classify? Why?
Making models
Making models of solids, liquids, gases using marbles or sand in a transparent container.
Models help us to understand the behaviour and properties of solids, liquids and gases.
The particles of a solid could be represented by gently moving the container so that the
objects vibrate (rock backwards and forwards) without changing position.
8
The particles of a liquid could be represented by tilting the container so that the objects
slide past one another.
The particles of a gas could be represented by shaking the container so that the objects
are in constant motion, flying around colliding with all parts of the container.
Manipulate these models to demonstrate the following properties. Then explain in words
why each is true.
(i)
Only solids can have a fixed shape.
(ii)
Only gases can exert a pressure in all directions at the same time.
(iii)
It would be very difficult for a solid to diffuse.
(iv) Increasing the temperature of a substance involves giving the particles more energy
(greater movement). Use your models to demonstrate the effect of increasing
temperature and corresponding changes of state.
9
Written Exercise:
States of matter
What are some properties of solids, liquids and gases?
1.
Try to complete of the following table. Have a sample of each state of matter in front
of you so that you can refer to it!
Property
Solid
Liquid
Gas
Shape
(3-dimensional
conformation)
Fixed
(doesn’t change)
Takes the shape of
the part of the
container it fills
Takes the shape of
the container
Volume
(the amount of
space taken up)
Compressible?
(can it be made to
occupy a smaller
volume?)
Diffuses?
(can it spread out
by itself?)
2.
Use the particle models to explain these properties by completing the following
descriptions.
In a solid the particles are packed closely together. The particles can vibrate but they
cannot move around. Therefore the solid has a fixed shape and volume (when at
constant temperature). Because the particles are very close together and in fixed
positions, solids are virtually incompressible. A solid cannot easily diffuse.
Complete similar explanations for liquids and gases.
(a) In a liquid the particles ..............
(b) In a gas the particles ................
10
Optional Research:
Other States of Matter
Are there any other states of matter? You might want to research some information on
plasmas as another state.
Ref: C1 pp6-7, 261-263
1.2
CC1 pp13-15
Changes of State
When a solid is heated, the particles vibrate faster until their energy overcomes the forces
(bonds) holding the particles in position. The particles break loose and begin to slide past
one another - the solid melts to form a liquid. Similarly heat can provide the energy for
particles in a liquid to escape from one another and a gas is formed as the liquid
evaporates. Some substances, such as iodine, change directly from a solid to a gas when
they are heated - this change is called sublimation.
Change
Name
solid
liquid
melting, fusion
liquid
solid
freezing, solidification
liquid
gas
boiling, evaporation
gas
liquid
condensing
solid
gas
sublimation
gas
solid
deposition
Ref: CC1 pp14-15
Sublimation
Melting
SOLID
Boiling
LIQUID
Freezing
GAS
Condensation
Deposition
11
2.
MODELLING THE SMALLEST PARTICLES OF MATTER
2.1
What particles compose the substances around us?
When you use the term particle to explain the different properties of a solid, liquid or gas,
you are referring to the simplest, most basic unit of that substance. If you took a sample
of that substance and started to divide it into smaller and smaller portions eventually you
would reach this basic unit that still has the properties of you substance. The type of basic
unit (particle) will depend on how the atoms are arranged in your substance.
For example the simplest unit in a sheet of aluminium foil is the aluminium atom. The
sheet of foil consists of many billions of atoms arranged in a network structure.
Molecules
Molecules are formed in many substances when small groups of atoms join together with
a unique structural conformation. The atoms can be all the same or there can be several
different types. A chemical formula gives information about the types and numbers of
atoms present in each of these molecules.
Examples:
The simplest unit in the
substance water is the water molecule,
H2O which is a cluster of two hydrogen
atoms and one oxygen atom.
Water is therefore composed of 2 different
elements, classifying it also as a compound.
In oxygen gas, O2
there are two oxygen
atoms per molecule.
12
Molecules in simple 3-D
Chemical molecules are discrete structures representing everyday substances in their
most simple form (ie; atomic ratio). You will use a molecular model kit a bit later to
represent a variety of different chemical structures.
The lines or “sticks” between individual atoms represent . . . . . . . . . . . . . . . . . . . . . . . . . . .
Water, H2O
Ref: CC1 pp16-22
Ammonia, NH3
Carbon Tetrachloride, CCl4
C1 pp9-13
Other substances are made up of a large network of ions. Ions are atoms or groups of
atoms which carry a positive or negative electric charge. Common table salt (which has
the chemical name of sodium chloride) is made of a large network of sodium ions, Na+
and chloride ions, Cl -
13
2.2
Modelling Atomic Structure
In order to be able to distinguish these different sorts of particles you must understand
some fundamental ideas about the structure of an atom. Since atoms are too small to be
seen even with the best microscopes, scientists have developed models of atoms to
represent them. No model can be exactly like an atom in all respects and so models are
being constantly revised as further research reveals more about atoms. In order to
explain the properties you are observing, use the simplest possible model.
Some models of the atom
A Basic Timeline of historical development of models of the atom
Written Exercise:
Timeline
Add a least one more name and date to the Timeline above.
Optional Research:
An atomic model
You may like to research the experimental discoveries that led to the changing of one of
the models of the atom.
14
A simple Bohr Model of the Atom
This model of the atom will be of particular use to you in this module. Atoms are units
composed of much smaller subatomic particles: protons, neutrons and electrons.
Electrons:
orbiting about the nucleus
of the atom
Nucleus:
A dense mass of protons
and neutrons at the centre
of the atom
KEY:
+
Proton (mass = 1 a.m.u.)
Neutron (mass = 1 a.m.u.)
–
Electron (mass = 0.00054 a.m.u)
These subatomic particles are measured in arbitrary units called atomic mass units,
shortened to a.m.u. The periodic table lists the atomic weights of all elements in these
units, as individual atoms are far too small and light to weigh individually.
Inside the nucleus are found protons (positive charge) and neutrons (neutral charge).
Almost all the mass of an atom is in the nucleus with an overall positive charge.
Negatively charged electrons orbit the nucleus and define the volume of the atom.
This would indicate that atoms are, by far, mostly empty space! A simple calculation
based on relative atomic masses above will reveal the nucleus as containing around
99.9% by mass of an entire atom!
15
Data Table of Subatomic Particles:
Subatomic
Particle
Charge
Relative mass
(a.m.u.)
Position
Electron: e –
–1
0.00054
orbiting the nucleus
+1
1
nucleus
0
1
nucleus
Proton: p
+
Neutron: n
The electrons form a negatively charged cloud around the nucleus.
Atoms are neutral, so that the number of positively charged protons must equal the
number to negatively charges electrons.
No of protons (p+) = No of electrons (e –)
When electrons are lost an atom becomes a positively charged ion or cation.
When electrons are gained by an atom it becomes a negatively charged ion or anion.
16
2.3
Isotopes
Any element is identified by the number of protons in its nucleus, defined as its Atomic
Number. The number of neutrons within an individual element can vary without changing
its identity, hence the concept of isotopes.
Different isotopes of a given element have the same number of protons, but
different numbers of neutrons in the nucleus.
Case Study 1: Carbon (Atomic Number 6)
All atoms of Carbon will be identified by Atomic Number 6, having this number of protons
within its nucleus. The nuclei of most carbon atoms contain also 6 neutrons giving a
combines mass number of 12. But statistically, a very small proportion of carbon atoms in
nature have 8 neutrons, thus giving a combined mass number of 14.
These are two different isotopes of carbon which can be distinguished by their different
masses; C-12 and C-14. C-14 is unstable and radioactive. Radioactivity is a feature of
many different elemental isotopes. The residual intensity of C-14 radioactivity is used for
carbon dating of ancient fossils.
So why do many elements appearing on the Periodic Table have mass number values
listed with decimal places? Fractions of 1 a.m.u. do not occur in nature!
Case Study 2: Chlorine (Atomic Number 17)
Chlorine is listed with an atomic mass of 35.45. All Chlorine atoms contain 17 protons in
their nucleus. From the atomic mass above it could be concluded that a chlorine nucleus
contains 18.45 neutrons; but there is no such thing as 0.45 of a neutron.
An atomic mass, as appears on the periodic table is an average mass of all isotopes
occurring in nature. This may have decimal places
A mass number is a round figure, summing the protons + neutrons of a single atom.
Statistically in nature, chlorine atoms are composed of 2 isotopes:
Chlorine-35 (having 18 neutrons) accounts for around 76% of chlorine atoms in nature.
Chlorine-37 (having 20 neutrons) accounts for around 24% of chlorine atoms in nature.
 Now, given these statistics of two chlorine isotopes, calculate its atomic mass . . .
17
2.4
Quantifying Subatomic Particles in Atoms
The Periodic Table of the Elements (see also page 40) is a complete database of
known atomic structures. Atoms are listed in order of atomic number, being the number
of protons contained within the nucleus of the atom. An Element is a substance
composed of a single specific atom. The Periodic Table lists over 100 different elements
providing information on their composition of protons, neutrons and electrons.
FEATURES OF ELEMENTS IN THE PERIODIC TABLE:
1. Chemical Symbol:
Shortened form of name.
2. Atomic Number:
Number of Protons in the Nucleus.
3. Mass Number:
Combined mass of Protons and Neutrons.
20
Atomic Number
Ca
Atomic Mass
Chemical Symbol
40.08
Calcium
Chemical Name of Element
DETERMINING ATOMIC STRUCTURE
Number of Protons = Atomic Number
Number of Electrons = Number of Protons, ie Atomic Number
Number of Neutrons = Atomic Mass (rounded) – Number of Protons
2.5
How are Electrons organised around the Nucleus?
Protons and neutrons make up the almost the entire mass of each atom and are located
in the nucleus. In our model of the atom, the electrons are arranged in orbits or shells
around the nucleus and equal the number of protons in a neutral atom. However,
because they may be a long way from the nucleus, outer electrons may be readily
removed or added and then an ion is formed.
Electrons which are in the same shell are about the same distance from the nucleus and
they have about the same energy. Shells are sometimes referred to as energy levels.
More complicated models of the atoms refer to subshells, suborbitals, standing waves and
so on. Shells are numbered from the nucleus.
18
Electron shells (or orbits) around the nucleus
The maximum number of electrons able to “fit” in any given orbital follows what is referred
to as the 2n2 rule, where n is the shell number
Maximum number of electrons, 2n2
Shell number (n)
1
2
2
8
3
18
4
32
And so on . . . . . . . . . . . . .
......................
Electrons fill the inner shells in order from the nucleus and ever outward.
In the third shell, however, eight electrons are placed, then the next two move to the fourth
shell before the final 10 are placed in the third shell. This shows a shortfall in this
particular model that is solved by using move complicated models of the atom that include
subshells.
If we consider only the first twenty elements, ie maximum of 20 electrons, then the
order of filling the shells follows a somewhat different set of rules:
1st shell
2
nd
rd
th
3
4
can take up to
2 electrons
shell
can take up to
8 electrons
shell
can take up to
8 electrons
shell
can take remaining
8 electrons
When drawing the electrons of a particular atom, the shells are depicted as concentric
circles around the nucleus, and the electrons as dashes, dots or “e” symbols.
For example, a hydrogen
atom with 1 electron
19
Written Exercise: Draw structural models of the following elements:
Aluminium
Beryllium
Calcium
Nitrogen
20
THE FIRST 20 ELEMENTS OF THE PERIODIC TABLE
Use a periodic table and your knowledge of electron shell structures to complete the
following table. Some information has been completed for you.
Element
Hydrogen
Symbol
Atomic
Number
Mass
Number
Neutron
Number
7
5
Beryllium
5
Boron
C
2:4
7
Nitrogen
316
Oxygen
10
Fluorine
10
Neon
Na
2:8:1
12
Magnesium
2+
27
Aluminium
14
Silicon
15
Phosphorous
2:8:6
Sulfur
1-
Chlorine
Argon
Potassium
Calcium
Charge of
Ion
 valency
2
Lithium
Sodium
Electron
Configuration
H
Helium
Carbon
Electron
Number
Ar
19
40
21
Activity:
Elemental puns!
Are you a creative thinker? Do you like puns? Even if you do not, you might enjoy
groaning at these elemental puns. Each of the blanks can be completed with the name of
an element. So caesium your pencil and try your luck!
1.
0.5 times holmium:
=
………………………
2.
An officer of the constabulary:
A
..............................................
3.
Funds from your mother’s sister:
...............................................
4.
A comical prisoner:
A
...............................................
5.
What you do with dead cats:
You
...............................................
6.
A driller’s motto:
...............................................
7.
What a good doctor can do for a patient:
8.
When you have a hole in the boat:
You
...............................................
9.
How you mended your clothes:
You
...............................................
10.
What are you doing with that person’s hair?
I’m
11.
A nocturnal trojan:
A
...............................................
12.
A trojan who smokes dope:
A
...............................................
13.
A man who tried to murder his wife with a razorblade only managed to give her:
...............................................
.........................................
……………………………………..
14.
……………………………………………..
Mind your own:
Ref: CC1 pp30-34
22
3.
3.1
CLASSIFICATION SYSTEM TWO:
ELEMENTS, COMPOUNDS AND MIXTURES.
Classifying Everyday Substances
The matter around us can be classified into pure and impure substances.
Pure substances have a definite composition - they are always the same no matter
where the sample comes from.
Common table salt is and example of a pure substance since it is always NaCl (sodium
chloride) whether the sample comes from Australia or anywhere else on earth. The ratio
of sodium ions to chloride ions is 1:1 and it melts at 801 oC.
Water is an example of a pure substance since it is always H2O - it contains twice as
many hydrogen atoms as oxygen atoms. It melts at 0 oC and boils at 100oC at
atmospheric pressure.
Impure substances are normally called mixtures since they have no definite composition
and the proportion of components will vary between samples. They have no fixed melting
and boiling points. Most of the materials surrounding us are mixtures!
Air contains oxygen, nitrogen, carbon dioxide, water vapour and other gases. Can you
give examples of how air samples may vary?
Seawater contains water, dissolved salts and dissolved gases. How do samples of
seawater vary?
Matter
Composed of atoms.
Has mass, volume
Impure
Pure
- Atoms are in fixed proportions
- Atoms are in variable proportions
- Has a definite composition
- Has variable composition
- Has fixed melting and boiling points
- Has no fixed melting or boiling points
Elements
Atoms are all the same
Cannot be decomposed
Compounds
Mixtures
2 or more different
atoms in fixed
proportions
2 or more different
atoms in variable
proportions
Can be
decomposed into
simpler substances
Can be physically separated
into pure substances
23
Pure substances are further divided into those which cannot be decomposed (broken
down into simpler substances) and those which can be decomposed. Elements are our
most basic substances - they are made of one type of atom only and therefore cannot be
decomposed. Compounds are made of at least two different types of atoms and therefore
can be decomposed - although sometimes with great difficulty!
Examples :
Hydrogen, H2 ; oxygen, O2 and ozone, O3 are all elements. Their formulae
show that they each contain only one type of atom.
Water, H2O is a compound formed from hydrogen atoms and oxygen atoms
when they have chemically bonded together. The formula shows that two
different atoms are present in water and that their ratio is fixed – this defines
the chemical composition of water.
Written Exercise:
1.
CLASSIFYING SUBSTANCES
Classify the following list of substances below by placing them into their
corresponding column. Add further examples to each column if you can.
An element is pure substance composed of only one type of atom.
A compound is a pure substance composed of 2 or more different atoms.
A mixture is a substance that contains 2 or more pure substances mixed together.
SUBSTANCES:
Elements
Compounds
Mixtures
Water, Air, Copper, Soil,
Nitrogen, Gold, Ethanol,
Orange Juice, Petrol, Glass,
Vinegar, Brass, Sweat,
Iodine, Carbon Dioxide.
2.
Which laboratory methods would you use to separate the following mixtures?
(a)
Charcoal and salt
(b)
Water and oil
(c)
Alcohol and water
(d)
Sugar from a sugar solution .................................................................................
(e)
Water from muddy river water ...................................................................................
...................................................................................
...................................................................................
...................................................................................
Differences in physical or chemical properties decide how we classify matter.
24
3.2
Physical and Chemical Properties
Physical changes are those which affect only the physical form or state of a substance.
They do not affect the chemical composition of a substance. A physical property of a
substance can usually be observed without a change in chemical composition taking
place. Physical changes can easily be reversed in many instances.
Examples:
Physical Change
Physical Property
The boiling of water
Boiling point
The dissolving of salt in water
Solubility of salt
In a physical change, the elements and / or compounds remain chemically the same!
Chemical changes are those which cause the chemical composition of the substance to
be altered, that is a new substance is formed. A chemical property can be observed as
a substance chemically reacts with another, or breaks down (decomposes), forming new
substance(s). Chemical changes are often difficult, even impossible, to reverse.
Examples:
Chemical Change
Chemical Property
Combustion of fuel
Flammability
Converting water to hydrogen and oxygen
Ease of decomposition
In a chemical change new substances are formed, but the number of atoms remains the
same - they have just been rearranged !
25
Written Exercise:
Physical and Chemical Changes
Classify the following as chemical (C) or physical (P) changes and determine which
property is concerned.
1.
Water freezing to form ice.
P
Melting point (freezing point)
2.
An iron bar rusting in the air.
.............
...............................................
3.
Petrol combusting in a car engine.
.............
...............................................
4.
Moulding plastic into a new shape.
.............
...............................................
5.
Setting off some fireworks.
.............
...............................................
Experiment – MAKING OBSERVATIONS
QUALITATIVE VERSUS QUANTITATIVE OBSERVATION:
Qualitative refers to any observation identifying any characteristic or process which
cannot easily be expressed as an amount or quantity.
Quantitative refers to any observation which can, in theory be easily measured and
therefore be expressed as a number or quantity, for example:
 Mass, Volume, Concentration, Temperature etc……
PHYSICAL AND CHEMICAL PROPERTIES:
A Physical Property is any characteristic of a pure substance which:
A Chemical Property is any characteristic of a pure substance which:
PART A: Observing a burning candle
Aim – to make as many observations as possible on a burning candle.
** See over page for task and results page
26
Task: Light a candle and begin immediately listing your observations in table 1 below.
Classify each observation according to the criteria listed above.
Extinguish the candle and continue making observations in table 2 according to
the same criteria.
TABLE 1 – BURNING CANDLE
Observation
Physical or
Chemical
Change?
Physical or
Chemical Property
Qualitative or
Quantitative?
TABLE 2 – EXTINGUISHED CANDLE
Observation
Physical or
Chemical
Change?
Physical or
Chemical Property
27
Qualitative or
Quantitative?
PART B: Observing and interpreting everyday phenomena
Aim: To interpret the science behind everyday phenomena by observation
Equipment:
-
Ice (Large Beaker)
Bunsen burner
Forks / coin demo
Volumetric flask + dye solution
-
Match / pin demo
Coke / Diet Coke
Bucket
Method:
-
Observe each station carefully.
Record your observations and any other required explanation in the table below.
Results:
STATION 1:
STATION 2:
Matchstick and Bunsen arrangement
Warm coloured water in a sealed container
Observation: ……………
Observation (s) ………………..
Interpretation:
Hint - Identify at least one process you
think might be occurring
Interpretation: ………………
…………………….
STATION 3:
STATION 4:
Arrangement of forks, coin and a beaker
Regular and diet soft drink cans in water
Observation: ……………
Observation: …………………..
Interpretation: ………………
Interpretation: ………………….
STATION 5: Ice
Observation: The ice is turning from a ____________ state into a ______________ state.
Interpretation: ………………
Why do you think this is happening? …………………….
28
Experiment: PHYSICAL VERSUS CHEMICAL CHANGE
Aim: To distinguish a physical from a chemical change based on observation.
Task: (Teacher demonstration)
-
You will be required to record observations at each station set up and identify each as
either a physical or chemical change.
Summarise the process(es) taking place at each station, expressing each in terms of
starting and finishing substances. Expressing physical states as subscripts next to
each substance will assist in this regard.
! Risk Assessment
You will be required to complete a 5 min risk assessment form, that that outlines any
form of risk encountered at each station.
Results:
Station 1:
BOILING WATER
Station 2:
ELECTROLYSIS OF WATER: VOLTAMETER APPARATUS
Station 3:
HEATING IODINE IN A TEST TUBE.
Station 4:
HEATING SUGAR IN A TEST TUBE
Station 5:
A BUNSEN BURNER FLAME
Station 6:
SALT IN WATER
Station 7:
EVAPORATING A SALT SOLUTION
Station 8:
SODIUM METAL IN WATER
Station 9:
MAGNESIUM METAL IN HYDROCHLORIC ACID
Station 10: HEATING MAGNESIUM METAL IN A BUNSEN FLAME
Station 11: ADDITION OF 2 CLEAR / COLOURLESS SOLUTIONS:
 POTASSIUM IODIDE AND LEAD NITRATE
Station 12: ADDITION OF SODIUM CARBONATE POWDER TO A DILUTE
HYDROCHLORIC ACID SOLUTION
Discussion:
Identify one station you felt with confidence demonstrated a physical change. Why
was this obvious?
Identify one station you felt with confidence demonstrated a chemical change. Why
was this obvious?
Was there a station you found difficult to classify as a physical / chemical change?
Did any station demonstrate both a physical and chemical change?
Emphasise the difference between physical and chemical change by defining both.
29
3.3
Mixtures and laboratory separation techniques
Most substances in nature are not found pure but as mixtures. Mixtures can be separated
into pure substances using methods based on differences in their physical properties.
Examples:
A magnet can be used to separate iron (and some other substances) from other solids.
physical property?
differences in magnetic properties
Solids and liquids can be separated by decanting, filtering or centrifuging.
physical property?
differences in density or particle size
Two immiscible liquids can be separated using a separating funnel.
physical property?
differences in density
Example - Separating a mixture of chalk powder and salt
If you were given a mixture of chalk powder and salt to separate you could use differences
in their physical properties to prepare a plan - flowchart - of the steps needed.
Mixture of chalk powder and salt
Add water and stir
Filter to separate liquid and undissolved solid
Chalk powder (wet)
allow to dry
Salt solution
evaporate water by heating
Chalk powder
Salt
In this separation we were using differences in solubility of the two solids in water.
However the water also had to be separated by evaporation.
Some industrial examples of separating mixtures:
Gold panning
Oil refining
Sugar refining
Separating rutile etc from beach sands
Recovering salt from seawater
Recovering pure water from contaminated water
30
Laboratory techniques for separating mixtures
METHOD
(property differences)
Decanting
(density)
Separating funnel
MIXTURE TYPE
Used to separate
undissolved
sediment from a liquid
or
immiscible liquids.
used to separate
immiscible liquids
(density)
Filtration
 Gravity and Vacuum
techniques
used to separate an
undissolved solid from a
liquid.
(particle size)
31
DIAGRAM
METHOD
(property differences)
Centrifuge
(density)
Evaporation /
Crystallisation
Distillation
MIXTURE TYPE
DIAGRAM
used to separate
undissolved substances
(some liquid) of differing
densities
used to separate a
dissolved solid from a
solvent (liquid)
used to separate mixtures of liquids
(boiling point)
Chromatography
used to
separate miscible
(partition – differences in liquids
attraction of sample
matrix for mobile phase
Vs stationary phase)
Note: Solutions are mixtures - solutions will be considered in detail in a later module!
32
Experiment – SEPARATION TECHNIQUES 1
Aim: To devise a method of separating a known sand / salt mixture, and express the
percentage of salt in the mixture.
Equipment:
Method:

Accurately weigh and record the mass of approximately 8 g of sand and 2 g of salt into
a small beaker. Stir gently with a spatula until mixture is homogeneous in composition.

Now devise a series of steps to achieve complete separation of the sand and salt
components below.
Flow Chart:
33
Results:
TRUE VALUES: what mass of sand and salt did you weigh out?
Mass: sand = ________________________ g
Mass: salt = ________________________ g
Total mass of mixture = _________________________ g
% Salt =
MEASURED VALUE: what mass of salt did you separate and weigh back?
SALT
Mass: Clean Dry Container (g)
Mass: Container + Dry Salt (g)
Mass: Salt (g)
% Salt in mixture
=
Mass Salt obtained
Total Mixture Mass
x 100
HOW CLOSE WAS YOUR MEASURED RESULT TO THE TRUE RESULT?
-
Calculate the % error based on the percentage of salt you calculated after the
separation, as follows. A result of 3% or less is considered acceptable at this level.
Step 1:
ABSOLUTE ERROR = Difference between TRUE and MEASURED result
Step 2:
RELATIVE ERROR (%)
= ABSOLUTE ERROR
TRUE VALUE
34
x 100
SALT
Absolute Error =
Relative Error =
Questions:
1. What is meant by the term quantitative transfer? Give an example of one step in
this task that used this technique.
2. Identify one physical property of both the sand and salt that allowed for their
separation to take place.
3. Identify any difficulties or errors you experienced whilst carrying this separation out.
35
Experiment – SEPARATION TECHNIQUES 2
Aim:
To devise and undertake a method for the separation and recovery of a mixture containing
sand, salt and sawdust.
Equipment:
Method:

Accurately weigh and record the mass of an unknown mixture containing sand, salt
and sawdust.

Devise a series of steps to achieve complete separation of the sand and salt
components below.
36
Results:
SAMPLE MASS:
_________________________ g
i) SAND COMPONENT
Description
Quantity / Result
% Sand =
ii) SALT COMPONENT
Description
Quantity / Result
% Salt =
iii) SAWDUST COMPONENT
Description
Quantity / Result
% Sawdust =
37
Discussion:
Compare your results with the true values as supplied by the lab staff.
- Optional: Calculation of corresponding Relative Errors for each separated
substance will assist you in making a more informed comparison.
Identify how individual separations were possible based on your knowledge of physical
properties as they apply to each separated component.
What sources of error can you attribute to this separation task?
Hindsight: can you suggest any improvements to your experimental set up or
technique that might increase your experimental success?
Conclusion:
What were your findings?
How successful overall was this exercise?
38
4.
Elements and the Periodic Table
Elements are pure substances that contain only one type of atom. Elements cannot be
decomposed to give simpler substances. A listing of elements can be obtained from
Chemical Data books or many other texts, however a very useful arrangement of the
elements can be found in a Periodic Table which lists the elements in order of atomic
number.
4.1
Mapping Elements on The Periodic Table
See Periodic Table over page and notice where the various types of elements are placed:
(a)
(b)
(c)
(d)
metals and non-metals
solids, liquids and gases at room temperature
natural and man-made elements
radioactive elements
Periods
Use a key to mark them on you copy of the Periodic Table.
Groups
The horizontal rows on the Periodic Table are called periods. The first period contains
just two elements, hydrogen and helium. (Remember that the first shell of electrons could
hold 2 electrons). Periods 2 and 3 contain eight elements - the end of each period marks
the filling of an electron shell. Periods 4 and 5 contain 18 elements. A section of period 6
is removed and printed below the bottom of the table. This section is named the
lanthanides. Similarly a section of period 7 is also printed below the main table. It is
called the actinides. (The removal of both sections is for ease of printing, so that the table
did not become too wide).
The vertical columns of the table are called groups. They are families of elements with
closely related characteristics. The main group elements are numbered with Roman
numerals:
Group I is headed by Lithium (Li) – referred to as the Alkali Metals.
Group II is headed by Beryllium (Be) – referred to as the Alkaline Earths.
Then skip a few across the central “valley” of transition elements to arrive at . . .
Group III headed by Boron (B).
Group IV headed by Carbon (C).
Group V headed by Nitrogen (N).
Group VI headed by Oxygen (O)
Group VII headed by Fluorine (F) – referred to as the Halogens.
Group VIII headed by Neon (Ne) – referred to as the Noble Gases.
The central block of elements constituting the rectangular “valley” as described above are
referred to as the Transition Metals.
39
THE PERIODIC TABLE OF THE ELEMENTS
1
H
1.008
Hydrogen
KEY
3
Li
6.941
Lithium
4
Be
9.012
Beryllium
11
Na
22.99
Sodium
12
Mg
24.31
Magnesium
19
K
39.10
20
Ca
40.08
Calcium
21
Sc
44.96
Scandium
22
Ti
47.87
Titanium
23
V
50.94
Vanadium
Rubidium
38
Sr
87.62
Strontium
39
Y
88.91
Yttrium
40
Zr
91.22
Zirconium
41
Nb
92.91
Niobium
55
Cs
132.9
Cesium
56
Ba
137.3
Barium
57–71
Lanthanoids
72
Hf
178.5
Hafnium
73
Ta
180.9
Tantalum
87
Fr
[223.0]
Francium
88
Ra
[226.0]
Radium
89–103
Actinoids
Potassium
37
Rb
85.47
2
He
4.003
Helium
Lanthanoids
Actinoids
Atomic Number
Relative Atomic Mass
57
La
138.9
Lanthanum
89
Ac
[227.0]
Actinium
24
25
Cr
Mn
52.00
54.94
Chromium Manganese
26
Fe
55.85
Iron
43
44
Tc
Ru
[97.91]
101.1
Molybdenum Technetium Ruthenium
42
Mo
95.94
74
W
183.8
Tungsten
105
106
Db
Sg
[262.1]
[166.1]
Rutherfordium Dubnium Seaborgium
104
Rf
[261.1]
58
Ce
140.1
Cerium
13
Al
26.98
Aluminium
75
Re
186.2
Rhenium
76
Os
190.2
Osmium
107
Bh
[264.1]
Bohrium
108
Hs
[277]
Hassium
60
61
62
Nd
Pm
Sm
144.1
[144.9]
150.4
Praseodymium Neodymium Promethium Samarium
59
Pr
140.9
90
91
92
Th
Pa
U
232.0
231.0
238.0
Thorium Protactinium Uranium
Symbol of element
Name of element
5
B
10.81
Boron
6
C
12.01
Carbon
7
N
14.01
Nitrogen
8
O
16.00
Oxygen
9
F
Fluorine
19.00
10
Ne
20.18
Neon
13
Al
26.98
Aluminium
14
Si
28.09
Silicon
15
P
30.97
Phosphorus
16
S
32.07
Sulphur
17
Cl
35.45
Chlorine
18
Ar
39.95
Argon
27
Co
58.93
Cobalt
28
Ni
58.69
Nickel
29
Cu
63.55
Copper
30
Zn
65.41
Zinc
31
Ga
69.72
Gallium
32
Ge
72.64
Germanium
33
As
74.92
Arsenic
34
Se
78.96
Selenium
35
Br
79.90
Bromine
36
Kr
83.80
Krypton
45
Rh
102.9
Rhodium
46
Pd
106.4
Palladium
47
Ag
107.9
Silver
48
Cd
112.4
Cadmium
49
In
114.8
Indium
50
Sn
118.7
Tin
51
Sb
121.8
Antimony
52
Te
127.6
Tellurium
53
I
126.9
Iodine
54
Xe
131.3
Xenon
77
Ir
192.2
Iridium
78
Pt
195.1
Platinum
79
Au
197.0
Gold
80
Hg
200.6
Mercury
81
Tl
204.4
Thallium
82
Pb
207.2
Lead
83
Bi
209.0
Bismuth
84
Po
[209.0]
Polonium
85
At
[210.0]
Astatine
86
Rn
[222.0]
Radon
68
Er
167.3
Erbium
69
Tm
168.9
Thulium
70
Yb
173.0
Ytterbium
71
Lu
175.0
Lutetium
109
110
111
112
Mt
Ds
Rg
Cn
[268]
[271]
[272]
[285]
Meitnerium Darmstadtium Roentgenium Copernicium
63
64
Eu
Gd
152.0
157.1
Europium Gadolinium
93
94
95
Np
Pu
Am
[237.0]
[244.1]
[243.1]
Neptunium Plutonium Americium
96
Cm
[247.1]
Curium
65
Tb
158.9
Terbium
66
67
Dy
Ho
162.5
164.9
Dysprosium Holmium
97
98
99
Bk
Cf
Es
[247.1]
[251.1]
[252.1]
Berkelium Californium Einsteinium
100
Fm
[257.1]
Fermium
For some elements, the relative atomic mass of their most common radioactive isotope is shown in brackets.
40
102
103
No
Lr
[259.1]
[262.1]
Mendelevium Nobelium Lawrencium
101
Md
[258.1]
4.2
Distribution of the elements
Approximately 109 elements are known (90 are natural to earth, the rest are man-made).
Throughout the universe, hydrogen (83%) and helium (16%) are by far the most abundant.
On the earth’s crust and atmosphere the distribution by mass of elements is:
In the human body,
the distribution of
elements is different
again . . .
41
Occurrence of the elements
Only a few elements actually exist in a pure form on earth; most are in the form of
compounds.
Activity:
Earth Elements
What are some of the elements that exist uncombined on earth? Two are gold (as
nuggets etc) and oxygen (as gas in the atmosphere). List three others.
1.
....................................................................
2.
....................................................................
3.
...................................................................
Why do the others only exist naturally in compounds? Because most other elements are
too chemically unstable to exist by themselves. They instead react with substances in the
natural environment such as water, oxygen and carbon dioxide, forming various
compounds which have much greater chemical stability.
4.3
How atoms can be arranged in everyday elements
Some elements exist as single free atoms. They are called MONATOMIC elements and
the most common examples are in the group at the far right of the Periodic Table called
the rare or noble gases: He, Ne, Ar, Xe, Rn. These gaseous atoms are extremely
unreactive (ie; chemically stable). They have a full outer shell of electrons.
The other gaseous elements form molecules of two or more atoms.
DIATOMIC elements form molecules composed of 2 like atoms.
Examples are oxygen, O2 ; hydrogen, H2 ; nitrogen N2 ; fluorine, F2 and chlorine, Cl2.
Bromine, Br2 is a liquid at room temperature and iodine, I2 is a solid at room temperature.
POLYATOMIC elements have 3 or more atoms in their molecules. Examples are
phosphorus, P4 and sulphur, S8 although they can also exist in molecules of differing
numbers of atoms, or allotropes.
Metal elements have many atoms grouped together in a regular structure called a crystal
lattice. They do not form separate “molecules”, nor are they free atoms. Some nonmetals such as carbon also have a crystal lattice structure.
42
Allotropes
Some elements can exist in a number of different pure forms, where different numbers of
atoms exist in a molecule, or the atoms are joined in different ways. Examples include
Oxygen which can exist as O2 (oxygen gas) or as O3 (ozone). Carbon can exist as
graphite, diamond or bucky balls! These examples and some others are illustrated below:
OXYGEN
O3
O2
Ozone
C60
Oxygen gas
Graphite
Diamond
CARBON
Buckminsterfullerene
SULFUR
S6
S8
S4
43
Research: Allotropes
You may wish to research one or more of these allotropes, or some others of your
choosing. This would be suitable for including in your portfolio.
Ref: CC1 pp23-25 C1 pp13-16
4.4
Classification of the elements
Experiment – Properties of elements
This practical activity must be done in a science laboratory under the supervision of a
chemistry teacher. The sections marked as demonstrations must only be done by the
teacher with the student observing and recording carefully. Normal laboratory safety
precautions will need to be observed at all times.
In this practical exercise you will move to various workstations. Make and record careful
observations as you proceed. Do not remove samples from their containers.
1. Complete the following table by observing samples of elements on display. All the
elements are contained in glass sample tubes. Some may be stored in a liquid for
safety reasons.
Element Name / Symbol
Colour and Appearance
44
Metal or Non Metal?
2. A range of different elements are available for density calculation:
Lead (Pb), Iron (Fe), Carbon (C), Aluminium (Al), Copper (Cu), Zinc (Zn),
Sulfur (S) and Iodine (I2)
Calculate the density of each element ** via the following simple method:
**
Calculate the density of Sulfur and Iodine (teacher demonstration) in a clean, dry
25 mL measuring cylinder without water
Record the mass of each element.
Fill a measuring cylinder of appropriate size with a volume of tap water enough
to submerge the entire object. Record this volume.
Submerge the object in the water and record the new volume of water.
Calculate the actual volume of water displaced by the object.
Calculate the density of the object, where: Density = Mass (g)
Volume (mL)
Consult an SI Chemical Data book and record the literature density values of
each element: (ρ) g / cm3
Element
Vol 1 (mL)
Meas. Cyl
Vol 2 (mL)
Meas. Cyl
Sulfur, S
N/A
N/A
Iodine, I2
N/A
N/A
ID
Mass (g)
45
Element
Volume
(mL)
Density
(g / mL)
Literature
Density
(ρ) g / cm3
3. (Teacher Demonstration). The Melting Point of available elements will be determined
qualitatively using a bunsen burner on the blue flame setting.
Use the temperature probe to measure the temperature of the blue bunsen
flame. Record this value.
Observe the ease (or otherwise) to which each element melts. Rate each
melting point using word descriptions: very low  low  medium 
medium-high  high  very high.
Consult an SI Chemical Data book and record the literature melting point values
of each element.
Bunsen flame temperature: _________________________
Element ID
Observed Melting Point
46
Literature Melting Point
o
C
4. The Electrical Conductivity of each element can be easily tested by touching live
electrical probes onto the side of each.
Predict the electrical conductivity (good  fair  poor) of each element.
Observe the reading of an ammeter connected in series within the circuit.
Record these observations using similar descriptions.
Element ID
Predicted Electrical
Conductivity
Observed Electrical
Conductivity
Describe on what basis you are making your predictions of electrical conductivity:
47
5. (Teacher Demonstration). Observe the malleability / ductility of available elemental
samples. Malleability can be observed by hammering or attempting to bend each
sample. Ductility is best observed when sodium metal is placed in a laboratory device
called a Sodium Press.
Briefly describe the meaning of the terms:
Malleable:
……………………………………………………………………..
……………………………………………………………………………………………………..
Ductile:
……………………………………………………………………..
……………………………………………………………………………………………………..
Element ID
Malleable / Ductile ?
Yes / No
What class of elements generally exhibit characteristics of malleability / ductility?
………………………………………………………………………………………………..
Ref: Metals C1 pp23-26 Non metals C1 pp51-56 Metals CC1 pp49-60
48
6. (Teacher Demonstration). Elements can be made to undergo physical and chemical
changes under certain conditions. Complete the table below after observing each
scenario.
Scenario
Observations
Zinc + Hydrochloric Acid
Copper + Hydrochloric Acid
Magnesium + Hydrochloric
Acid
Iron + Hydrochloric Acid
Sodium + Water, Δ
Sulfur heated with a bunsen
burner in a gas jar
Addition of Potassium
Iodide, KI to a solution
containing Sodium
Hypochlorite bleach, NaClO
Explain how Physical Changes are different from Chemical Changes .
49
5.
CHEMICAL BONDING
All particles are held together by chemical bonds of some kind.
Water molecules are made up to two hydrogen atoms and one oxygen atom always. The
three atoms are held together by chemical bonds, which are strong and therefore not
easily broken.
How then do ice, liquid water and steam differ? They all contain water molecules.
Gas particles move far away from each other. Liquid particles are closer together. Solid
particles are very close together, held in a fixed position.
The main reason for the difference is that bonds can occur between molecules, called
intermolecular forces, and they increase in strength as the molecules get closer together.
These bonds are much weaker in strength than chemical bonds.
Chemical bonds form between atoms to form stable partnerships, since most atoms are
not stable alone. Chemical bonds exist between the hydrogen and oxygen atoms in the
stable water molecule. Chemical bonds involve the outer electron shells of atoms.
5.1
What makes an Atom Chemically Stable?
The elements in the group at the right-hand end of the periodic table - He, Ne, Ar, Kr, Xe
are called the rare, noble or inert gases. They are inert because they are extremely
unreactive. This means that they don’t form compounds except under rare circumstances.
They are extremely stable as monatomic elements - as single, separate atoms.
Electron
configuration :
(2)
(2,8)
(2,8,8)
In each of these atoms all electron shells present are full. The outer shell of electrons is
called the valence shell.
Chemical stability is obtained when an atom has a complete (full) outer shell of
electrons.
50
5.2
Loss and Gain of Electrons – The Ionic Bond
Sodium and chlorine are extremely reactive as pure elements, but the compound they
form, sodium chloride, is extremely stable. The sodium atom has one electron in its
valence shell, while the chlorine atom has seven electrons in its valence shell (which is
one short of a full shell). A transfer of this one electron from the sodium atom to the
chlorine atom results in the formation of more stable particles, now charged ions, which
have a full outer electron shells. The sodium atom loses an electron to become a positive
sodium ion, while the chlorine atom gains the electron to become a negative chloride ion.
Both the sodium ion and the chloride ion have full outer shells of electrons. Since only the
outer electron shells are involved in chemical bonding between atoms, only the outer shell
of electrons is often shown as an Electron Dot Formula. This concept is also referred to
as a Lewis Structure
Sodium atom 
Sodium ion + e -
Chlorine atom + e -  Chlorine ion
51
5.3
Sharing Electrons – The Covalent Bond
The other way for an atom to fill its outer shell of electrons is to share electrons with other
atoms. When sharing electrons each atom involved contributes an equal number of
electrons - one, two or three - to form a single, double or triple covalent bond. A pair of
shared electrons is represented as a dash ( – ) in a structural formula, ie a single bond.
Some elements have their atoms covalently bonded in simple molecules (such as oxygen)
or even giant networks (such as carbon) to form stable units.
The following common elements have stable diatomic molecules: H2, N2, O2, F2, Cl2 Br2, I2.
Written Exercise:
Lewis Electron Dot Formulae
Use Lewis Structures to illustrate covalent bonds for the following molecular elements:
H2
O2
N2
F2
Structural formula:
H–H
Structural formula:
O=O
Simple molecular compounds contain covalent bonds within their molecules.
Complete the following examples:
H2O
NH3
CH4
Structural formula:
CO2
HCl
52
HCN
5.4
Which Chemical Bond Will Form?
When atoms of one element need to gain electrons to become stable, and atoms of
another element need to lose electrons to become stable then an ionic bond will form.
Ionic bonds usually form between metals and non-metals. Metals will not bond chemically
with other metals to form compounds but they can form mixtures called alloys.
When the atoms of both elements need to gain electrons to become stable then a
covalent bond is formed. This situation arises when compounds form between nonmetals.
Exception: The ammonium ion NH4+ is a positive ion containing non-metals, however it
bonds ionically with any negative ion.
Written Exercise: Type of chemical bond?
Identify the nature of chemical bond that will form between the following elements.
Mg and Cl
...............................
K and S
...............................
C and Br
...............................
O and O
...............................
H and Cl
...............................
Na and Ca
...............................
5.5
take care !
The Metallic Bond
A Model of a Metallic Lattice
Metal atoms generally have few electron shells (1  3) occupying outer valence shells. In
theory these electrons need to be lost in order for the metal to form a stable ion. However
in the pure element there are no atoms to gain them since all are identical in electron
configuration (arrangement).
53
The outer electrons are freed from the metal atoms and form a negative cloud which is the
glue that holds the resultant positive ions (ie nucleus + inner electron orbitals) together in
a regular lattice structure.
The freedom of these electrons to move (even in the solid state) is the reason why metals
are good conductors of electricity. It also helps to explain properties of malleability and
ductility as the free electrons can flow to bind a distortion in the lattice.
5.6
Summary of Bonding
Atoms are held together in compounds by Chemical Bonds. Chemical bonds result from
the sharing or transfer of valence electrons between pairs of atoms. Bonded atoms attain
the stable electron configuration of a noble gas. The noble gases themselves exist as
isolated atoms because that is their most stable condition. The transfer of one or more
electrons between atoms produce positively and negatively charged ions: cations and
anions.
The attraction between a cation and an anion is an ionic bond. A substance with ionic
bonds is an ionic compound. Nearly all ionic compounds are crystalline solids at room
temperature, having high melting points. These solids consist of positive and negative
ions packed in an orderly arrangement. The total positive charge is balanced by the total
negative charge, and therefore the ionic compound (also called a salt) is electrically
neutral.
When atoms share electrons to gain stable electron configuration of a noble gas, they
form covalent bonds. A shared pair of electrons constitutes a single covalent bond.
Sometimes two or three pairs of electrons are shared to give double and triple covalent
bonds.
Metals consist of positively charged ions packed together and surrounded by a sea of their
valence electrons. This arrangement constitutes the metallic bond. The valence
accounts for the excellent electrical conductivity in metals and helps explain why metals
are malleable and ductile.
Ref:
CC1 pp48-51; 63-67
CC1 pp82-85
C1 pp23-25; 50-58
C1 pp69-70
54
Written Exercises:
1.
Write electron dot structures for each of the following elements.
(a)
2.
Cl
5.
6.
8.
(c)
Al
(d)
Li
(e)
C
(b) Al
(c) Na
(d) Li
(e) Ba
(f) Mg
(b) S
(c) N
(d) Cl
(e) P
(f) O
Write the formula for the stable ion formed from each of the following elements.
(a) aluminium
(b) lithium
(c) barium
(d) potassium
(e) calcium
(f) strontium
(g) bromine
(h) oxygen
(i) arsenic
(j) nitrogen
(k) selenium
(l) sulfur
Which of the following pairs of elements are likely to form ionic compounds?
(a) magnesium and bromine
(b) chlorine an iodine
(c) potassium and helium
(d) nitrogen and sulfur
(e) lithium and fluorine
(f) carbon and oxygen
Draw Lewis Structures of the compounds formed from these pairs of ions.
(a) K+ ; S2-
7.
S
How many electrons must be gained by each of the following atoms to attain a
noble gas configuration (full electron shell)?
(a) I
4.
(b)
How many electrons must be lost by each of the following atoms to form an ion?
(a) Ca
3.
Formulae and Chemical Bonding
(b) Ca2+ ; O2-
(c) Al3+ ; Cl-
Write formulae for the ions in the following compounds.
(a) KCl
(b) NaNO3
(c) BaSO4
(d) MgBr2
(e) KOH
(f) Li2CO3
Draw Lewis electron structures of the following. Classify each as ionic or covalent.
(a) H2O
(b) Na2O
(c) CS2
(d) CaS
(e) NH3
(f) SO2
(g) Al2O3
55
5.7
Compound Chemical Formulae
A chemical formula represents one molecule or formula unit of an element or compound.
A chemical formula can feature:
FEATURE
DEFINITION
EXAMPLES
Symbols
identify element(s) present.
C
Subscripts
i) indicate number ratios of
elements or polyatomic formula
units
CO2 or C6H12O6
ii) states of matter
Mg
NaCl or CaO etc
O2 (g) H2O (l) AgCl (s) or HCl (aq)
Superscripts
indicate charge of cation or anion
Coefficient
Numbers
represent mole quantity ratios in
balanced chemical formulae and
equations
Na
+
or
CO3
2-
CuSO4.3H2O
N2 (g) + 3H2 (g)  2NH3 (g)
Brackets
express multiple polyatomic ion
units in a chemical formula.
Al (NO3)3 (aq)
Roman
Numerals
clarify valency of transition metal
elements in word formulae
Iron III Oxide,
Prefix Names
indicate number ratios of non
metal elements in covalent
molecular compounds
Dinitrogen Pentoxide
Fe2O3
N2O5
A molecule is a discreet group of atoms which are bonded together in a fixed ratio
defined by chemical formula.
Examples:
Elements
Compounds
H2 - Hydrogen
H2O - Water
P4 - Phosphorus
C6H12O6 - Glucose (a sugar)
S8 - Sulfur
NH3 - Ammonia
56
A formula unit represents the simplest ratio of atoms in a substance which may be a
giant lattice, ie network structure.
Examples: Each of these substances is present as a giant lattice structure.
Substance
Type
Formula
Diamond
element
C
Sodium Chloride
compound (ionic)
NaCl
Silica
compound
(network covalent)
SiO2
Ions (charged particles) use superscripts to show the electrical charge.
Examples:
Mg2+ means a magnesium ion having a charge of +2.
CO3 2- means one carbon atom and three oxygen atoms forming an ion (called a
carbonate ion) with a charge of -2.
Further Examples:
1.
2 atoms of
hydrogen
H2SO4
1 atom of
sulfur
2.
4 atoms of
oxygen
Ca(OH)2
2 groups of OH
1 atom of
calcium
2 atoms of
oxygen
57
2 atoms of
hydrogen
Written Exercise – Interpreting chemical formulae:
1.
(NH4)2SO4
( ......... groups
of .................... )
............ atoms of
............. atoms of
.............. atoms of ............ atoms of
..........................
............................
............................. ...........................
2.
CH4 represents ......................................................................................................
3.
Sn3(PO4)4 represents ..............................................................................................
4.
CuSO4.5H2O represents..........................................................................................
5.
Cl- represents ........................................................................................................
6.
OH- represents ...........................................................................................................................
5.8
Valency
Valency is a numerical measure of combining power in determining correct compound
formula. Cations have positive valency, Anions negative. Atoms combine together
according to their values of valency to form stable compounds, commonly of neutral
charge, but not always. . .
Where an element can have more than one valency, it is given as a Roman Numeral in
the name of the compound. This commonly occurs with transition elements.
58
VALENCY TABLE
Polyatomic Ions – contain 2 or more elements making up their charged arrangement.
CATIONS
1+
2+
3+
4+/-
Name
Formula
Name
Formula
Name
Formula
Name
Formula
Hydrogen
H+
Calcium
Ca2+
Iron (III)
Fe3+
Carbon
C 4+/-
Lithium
Li+
Magnesium
Mg2+
Aluminium
Al3+
Silicon
Si 4+/-
Sodium
Na+
Copper II
Cu2+
Potassium
K+
Cobalt II
Co2+
* * Carbon and Silicon
Ammonium
NH4+
Zinc II
Zn2+
do not form ions,
Silver I
Ag+
Iron II
Fe2+
rather covalent bonds
Copper I
Cu+
Lead II
Pb2+
ANIONS
1Name
2Formula
–
Name
3Formula
2–
Name
Formula
Nitride
N3–
Fluoride
F
Oxide
O
Chloride
Cl–
Sulfide
S2–
Phosphide
P3–
Bromide
Br–
Carbonate
CO32–
Phosphate
PO43–
Iodide
I–
Sulfate
SO42–
Hydroxide
OH–
Sulfite
SO32–
Ethanoate
CH3COO–
Oxalate
C2O42–
Permanganate
MnO4–
Chromate
CrO42-
Hydrogen
carbonate
HCO3–
Dichromate
Cr2O72-
Nitrate
NO3–
Nitrite
NO2–
Cyanide
CN–
59
5.9
Chemical formulae and naming
The valency of an atom or polyatomic ion can be viewed as a bonding position which must
be filled to form a stable compound. In order to find the formula of a compound you must
recognise the parts or constituents from the name. Since the name is in two parts this
recognition is easy as long as you realise that the ending of the name of the second atom
is often changed to -ide.
Oxygen becomes oxide; sulfur becomes sulfide; chlorine becomes chloride etc.
Na
Na
will join the ratio 1:1
Cl
Cl
 Sodium’s valency is 1+
 Chloride’s valency is 1-
2.
Formula is Na1Cl1 or
NaCl (1 is automatically assumed)
Aluminium nitrate is a compound made from aluminium and nitrate polyatomic ion
NO3
Al
NO3
will join in the ratio 1:3
Al
NO3
NO3
Aluminium has a valency of 3+
Nitrate has a valency of 1-
Formula will be Al(NO3)3
3.
Lead IV oxide is a compound made from lead (valency 4) and oxygen.
Pb
O
will join in the ratio 1:2
O
Pb
O
Lead has a valency of 4+
Oxygen has a valency of 2-
The formula will be PbO2
60
Cross Multiply Method
Step 1.
Write the symbols which represent each part of the name.
Step 2.
Write the valencies as a superscript to each part.
Step 3.
Cancel by dividing the valencies by any common factor
Step 4.
Cross over the numbers to form subscripts, using brackets where necessary.
Examples:
1+
1.
Na
2.
3.
Sodium Chloride
Aluminium Nitrate
Lead (IV) Oxide
Written Exercises:
1Cl
3+
1-
Al
NO3
Formula is Na1Cl1 or NaCl
Formula is Al1(NO3)3 or Al(NO3)3
4+
2-
Pb
O
2 1 4
2
Formula is Pb1O2 or PbO2
Write chemical formulae for the following compounds.
(a) Hydrogen bromide ………………………
(g) Ammonium nitride ………………………
(b) Calcium chloride ………………………..
(h) Iron (III) hydroxide ………………………
(c) Zinc (II) sulfate ………………………..
(i)
Potassium sulfite ………………………..
(d) Magnesium oxide ………………………
(j)
Nickel (II) carbonate …………………….
(e) Lead (IV) sulfide ………………………
(k) Silver (I) phosphate ……………………..
(f) Barium phosphide ………………………
(l)
61
Tin (II) nitrate ………………………….
Prefix Numbering
When two non-metals combine to form a compound, they can often form several different
compounds depending on the prevailing conditions. For example sulfur and oxygen can
form SO2 or SO3. These compounds are named so that the number of each atom present
is indicated by a prefix.
SO2 is sulfur dioxide
SO3 is sulfur trioxide
The prefixes used are :
mono
1
hexa
6
di
2
hepta
7
tri
3
octa
8
tetra
4
nona
9
penta
5
deca
10
The prefix mono is not used for the first atom, but is used for the second.
For example, CO is carbon monoxide - notice also that the final “o” from the prefix name
is dropped so that it is easier to pronounce.
Written Exercises:
1.
Formulae and Naming (also termed “Nomenclature”)
Write formulae for the following compounds:
(a) potassium nitrate ……………….
(b) sodium carbonate …………………………
(c) cobalt (II) sulfate ………………
(d) ammonium carbonate …………………….
(e) calcium phosphate ………………
(f) disulfur dichloride .....……………………….
(g) nitrogen triodide .........................
(h) iodine heptafluoride .................................
(i) silicon dioxide ............................
(j) selenium dibromide ...................................
2.
Name the following compounds:
(a) SiF4 .............................................
(b) CaSO4 …………………………………….…
(c) FeO …………………………………
(d) CuCl ………………………………………….
62
(e) AgCl ………………………………..
(f) NaHCO3 ……………………………………..
(g) NO .................................................. (h) P2O5 ............................................................
(i) PCl3 ................................................. (j) CS2 ...............................................................
3.
Complete the following table:
Formula
Name
Name
MgO
sodium oxide
CaS
calcium carbonate
CO
ammonia
CO2
aluminium hydroxide
Csl
dinitrogen tetroxide
CaBr2
ammonium sulfate
H2O
aluminium oxide
NO2
copper (I) bromide
N2O4
chromium (III) fluoride
AsCl3
manganese (IV) oxide
AgNO3
mercury (II) sulfide
KBr
oxygen difluoride
Ca(OH)2
iron (II) chloride
CoCl2
iron (III) iodide
HI
copper (II) sulfate
SnO2
carbon tetrachloride
CCl4
strontium chloride
SnCl2
potassium nitrate
BaF2
calcium phosphate
63
Formula
Activity – MODELLING MOLECULES AND COMPOUNDS
Aim: To classify and construct models to represent some common substances.
Task:
1. Correspond a chemical formula to each chemical name.
2. Classify each substance according to chemical bond type.
3. Draw a structural representation of each molecular model.
CHEMICAL NAME /
FORMULA
CLASSIFICATION
(Ionic / Covalent)
Chlorine gas
………….…………….
Oxygen gas
……………………………
Nitrogen gas
……………………………
Ammonia
……………………………
Sodium Chloride
……………………………
Lithium Oxide
……………………………
64
STRUCTURE
Sodium Hydroxide
……………………………
Carbon Dioxide
……………………………
Potassium Sulfate
……………………………
Methane
……………………………
Ethanol
……… C2H5OH ……
Ethanoic Acid
……… CH3COOH ……
Question – Consider and write down some of the advantages and disadvantages of using
molecular model kits to understand the nature of chemical substances.
65
6.
Properties of Compounds
6.1
Comparing properties of elements and compounds
Compounds are formed when elements combine together in fixed proportions. The
compound formed will often have properties that are very different from its constituent
elements.
The formation of the compound will involve an absorption or a release of
energy.
Compounds can be decomposed back into its constituent elements, sometimes
with great difficulty.
Example:
When a spark is introduced into a mixture of hydrogen gas and oxygen gas an explosion
occurs and droplets of liquid are formed:
Hydrogen (g)
(element)
+
Oxygen (g)
(element)
Water (I)
(compound)
+
Energy
Comparison of some properties - elements and a compound
Property
Physical
Chemical
Hydrogen
Oxygen
Water
Gas at room temp
- little attraction
between
molecules, low
density.
Gas at room temp
- little attraction
between
molecules, low
density
Liquid at room
temp - attraction
between molecules
evident. Density
approx 1.0 g / mL.
Good solvent for
many substances.
Reacts vigorously
with many
substances.
Necessary for
combustion
burning.
Does not support
combustion.
Burns explosively
in air.
66
6.2
Classification of Compounds
Compounds can be classified in several different ways. The following classifications rely
on differences in both chemical and physical properties.
1.
Organic / Inorganic
Compounds
Inorganic
Organic
Compounds composed of elements
other than carbon and hydrogen.
Exceptions include oxides of carbon:
2CO, CO2, CO3 , HCO3 , cyanides,
CN- and strong acids, HX.
Composed predominately of the
elements carbon and hydrogen. Can
contain small quantities of oxygen,
sulfur, nitrogen, phosphorous and
halogen elements. Found in living
things, (plants & animals), foods, fossil
fuels, fibres, plastics etc . . .
Many organic compounds undergo charring when they are heated. They turn black,
showing a carbon residue. CARE! Always heat small samples in a well ventilated area.
2.
Ionic / Simple Molecular / Network Covalent
Compounds
Ionic
Contains charged
Particles  Ions.
Cation is a metal ion
or ammonium, NH4+
Network arranged.
Covalent
contain non-metals only
Simple Molecular
Small, discrete
groups of atoms
67
Network Covalent
Giant, covalently
bonded lattice
6.3
Properties of Compounds
Melting Point
A substance has a high melting point if it requires lots of energy (heat) to break apart the
forces holding the individual particles together.
Ionic compounds (eg NaCl, common salt) usually have high melting points, 900oC +.
This means that the forces holding the ions together in ionic solids are strong. Ionic bonds
are strong.
Simple molecular compounds (eg H2O water) have low melting points, 0oC. This
means that the forces holding the molecules together in simple molecular solids are weak.
Intermolecular bonds are weak.
Network covalent solids (eg SiO2 silica or quartz) have very high melting points,
1500oC +. This means that the forces holding the atoms together in network covalent
solids are very strong. Covalent bonds are very strong.
Solubility
A substance will dissolve in a liquid if it has similar properties to that liquid.
“Like dissolves like . . .”
Polar inorganic liquids such as water dissolve many ionic solids, like Sodium Chloride,
NaCl but not Silver Chloride, AgCl (!! Solubility rules apply).
Non-polar liquids such as petrol and kerosene will dissolve some simple covalent
molecular solids. Ionic compounds will NOT dissolve in organic solvents.
Polar organic liquids such as alcohols, (ethanol is a good example) are miscible with a
diverse range of organic and inorganic liquids, like petrol and water respectively.
Ionic compounds will commonly dissolve in water:
This means that the water molecules are attracted strongly to the positive and negative
ions such that the energy holding the ions in the crystal lattice is overcome. Cations and
anions randomly move about in aqueous solution (aq). There are still many examples of
insoluble ionic compounds, and therefore there is a general guide that can be referred to
when determining solubility of such compounds.
See Solubility Rules table over page
68
SOLUBILITY RULES
Use this table as a guide to determine the solubility of ionic compounds in water
SOLUBILITY RATING
ION
EXCEPTIONS
Lithium, Li+
Completely
Soluble
Sodium, Na+
None
Potassium, K +
Ammonium, NH4+
Nitrate, NO3Ethanoate, CH3COO –
Mostly
Soluble
Iodide, I -
Insoluble Exceptions:
-
Silver I, Ag+
Bromide, Br -
Lead II, Pb2+
Chloride, Cl
Mercury II, Hg2+
Mostly
Soluble
Sulfate, SO42-
Insoluble Exceptions:
Calcium, Ca2+
Barium, Ba2+
Lead II, Pb2+
Mostly
Insoluble
Hydroxide, OH
-
Soluble Exceptions:
Carbonate, CO32-
Lithium, Li+
Phosphate, PO43-
Sodium, Na+
Potassium, K +
Ammonium, NH4+
69
Written Exercise:
Identify each compound below as either Precipitate or Soluble according to the solubility
rules.
Mercury II Nitrate: _______________
Copper II Hydroxide: _______________
Magnesium Sulfate: ________________
Calcium Carbonate: _______________
Ammonium Phosphate: ______________
Lead II Ethanoate: _________________
Electrical Conductivity
A compound will conduct electricity if it possesses charged particles which are free to
move, ie mobile.
Solutions of ionic compounds are good conductors because they contain ions
which can move in the liquid.
Solid ionic compounds are poor conductors because their ions are not free to
move out of the solid lattice.
Molten (liquid) ionic compounds are good conductors because their ions can
move.
Solutions of molecular compounds are poor conductors because they contain
no ions.
Solid molecular compounds are poor conductors because they contain no ions.
Molten (liquid) molecular compounds are poor conductors because they
contain no ions.
70
Experiment – PHYSICAL PROPERTIES OF COMPOUNDS
Complete the table below carefully. Reference information where necessary from the SI Chemical Data manual.
Test compounds available using appropriate laboratory apparatus.
Compound
Name
Compound
Classification
Compound
Type
Include a
Chemical
Formula
Organic /
Inorganic
Ionic,
Covalent,
Network etc..
Appearance
Melting Point
OBSERVED
 High, Med,
Low, Char.
71
Literature
Value
Solubility
Water
Other
 Name it
Conductivity
Solution
Molten
State
Written Exercise:
Hydrogen chloride conductor?
Hydrogen chloride is a simple molecular compound (gas) and a non-conductor. It is very
soluble in water, forming a solution which is an excellent electrical conductor. Can you
explain this?
...............................................................................................................................................
...............................................................................................................................................
Ref: CC1 pp92-93 C1 pp37-45
Written Exercise:
COMPOUND PROPERTIES IN SUMMARY
Compound Type
COVALENT
MOLECULAR
IONIC
CHEMICAL
EXAMPLE(S)
CHEMICAL
BOND TYPE
SOLID, LIQUID & GAS
PHYSICAL STATE
@ Room Temp
MELTING POINT
(Low, High, Very
High)
WATER
SOLUBILITY
(High, Low, Nil)
SEE SOLUBILITY
RULES
LOW - MEDIUM
ELECTRICAL
CONDUCTIVITY
- Solid State
ELECTRICAL
CONDUCTIVITY
- Molten State
ELECTRICAL
CONDUCTIVITY
- Aqueous Sol’n
NO
Hydrogen salt exceptions
72
COVALENT
NETWORK
Experiment: 3 WHITE POWDERS
Background:
3 different unknown compounds resembling white powders will be issued for positive
identification. You will perform a range of physical tests to classify each compound based
on its unique set of physical properties.
You will then confirm the identity of each white substance based on a simple chemical
diagnostic test. Glucose, Sodium Chloride and Starch are the chemical substances you
must positively identify.
Perform all chemical / physical tests in boiling tubes
PART A – CHEMICAL DIAGNOSTIC TESTS
Complete the following table, classifying the type of chemical compound and providing
results / observations of diagnostic tests:
Glucose
Sodium Chloride
COMPOUND
CLASSIFICATION
? Ionic
? Covalent Molecular
? Covalent Network
? Organic / Inorganic
IODINE
(Observation)
SILVER NITRATE
(Observation)
BENEDICTS
REAGENT
+ HEAT
(Observation)
73
Starch
PART B – RESULTS OBSERVATION TABLE
UNKNOWN
SUBSTANCE
UNKNOWN:
UNKNOWN:
UNKNOWN:
_______
_______
_______
PHYSICAL
PROPERTY
SOLUBILITY
(soluble / insoluble)
CONDUCTIVITY of
aqueous solution
(yes / no)
MELTING POINT
(low / high)
CHAR
(yes / no)
PART C – SUBSTANCE ID
CLASSIFICATION:
? Ionic
? Covalent Molecular
? Covalent Network
CLASSIFICATION:
? Organic
? Inorganic
SUBSTANCE ID
? Glucose
? Sodium Chloride
? Starch
DIAGNOSTIC TEST
(Confirmed / Incorrect)
74
7.
CHEMICAL REACTIONS
Chemical reactions occur when a change leads to the formation of at least one NEW
substance. The starting substances in a reaction are called reactants. The substances
formed during a reaction are called products. During a chemical reaction, atoms are
rearranged and energy is absorbed or released in the process.
A reaction has occurred when one or more of the following can be observed:
evolution of a gas (not the vapour of a reactant)
formation of a precipitate (a solid formed from two solutions)
a permanent colour change
a noticeable temperature change (heat) or other energies released/absorbed
disappearance of a solid that is known not to be soluble.
A chemical equation is a shorthand way of describing the chemical change that has
occurred.
Reactants
Products
In this module we will use word equations to describe the reactions observed. This
concept will then be extended to include balanced chemical equations and use them to
calculate quantities of reactants and products involved.
Chemical reactions are occurring inside us and around us everyday.
Our body’s metabolism involves hundreds of thousands of different chemical reactions
for the production of energy and growth.
Combustion is an important chemical reaction that is also used for energy production
(light, heat) and transport amongst other uses.
Photosynthesis is an essential process for the existence of life on earth. Essential
sugars and oxygen are produced when carbon dioxide and water react in the presence
of sunlight within the green leaves of plants.
Most metal objects have been produced through industrial chemical processes. They
would otherwise remain existing mostly as an Earth based compounds distributed in
rich crustal deposits called ores.
We can look more closely at specific chemical changes:
7.1
Energy and Chemical Processes
When a spark ignites a mixture of hydrogen gas and oxygen gas, an explosion occurs.
A large amount of energy is released. The product formed is water.
Hydrogen (g)
+
Oxygen (g)
Water (l)
+
energy
The subscripts used in this equation indicate the states of reactants and products.
(s) = solid
(l) = liquid
(g) = gas
(aq) = aqueous (dissolved in water)
75
Chemical reactions which release energy are called exothermic, hence “Energy is lost to
the surrounding environment”. Reactions which absorb energy are called endothermic,
hence “energy is absorbed from the surrounding environment”.
Photosynthesis is an example of an endothermic chemical process. Plants absorb energy
in the form of sunlight to change carbon dioxide gas and water to complex organic
molecules such as glucose, releasing oxygen gas. This process is endothermic.
Carbon +
Dioxide (g)
Water(l)
+
Energy
Glucose (aq)
+
Oxygen (g)
Many chemical reactions require some energy to start the reaction. However endothermic
reactions require energy to be available continuously, so that the reaction can continue.
Ref: C1 pp70-71
Written Exercise & Research:
Chemical Reactions
1.
Find two more examples of both exothermic and endothermic reactions.
2.
Do each of the following involve chemical reactions? State what you would
observe in each case to justify this conclusion.
(a)
A piece of wood burns leaving grey ashes
(b)
A piece of sodium metal disappears after fizzing as it zips and releases
sparks over the surface of the water.
(c)
A tablespoon of salt dissolves when mixed in water.
(d)
A headache tablet fizzes as it dissolves in the water.
Practical Activities at Home:
The following chemical reactions can be carried out safely in the home or in the laboratory.
As a safety precaution you should always have long hair tied back, your eyes protected by
safety glasses and your feet fully protected by shoes. Protective clothing such as
laboratory coats are worn in laboratories together with other safety wear as required.
Always use small quantities of reactants and carry out the reactions in open glass
containers.
1. The acid in vinegar is ethanoic (acetic) acid, CH3COOH. Bicarb of soda or sodium
hydrogencarbonate, NaHCO3 is used in cooking, usually because it will release
carbon dioxide gas to help foods to rise. Sodium hydrogencarbonate is also found in
baking powder. When ethanoic acid is added to sodium hydrogencarbonate, the
result is the formation of carbon dioxide, water and a salt called sodium ethanoate,
NaCH3COO.
76
Observations :
You may like to include a simple labelled diagram
Word Equation :
………………………………………………………………………………………………………
Conclusions: Summarise your evidence to show that a chemical reaction occurred.
You may like to substitute other acids in the home for the vinegar and test their reactions
with bicarb of soda or baking powder. Try carbonic acid, H2CO3 in soft drink or citric acid
in lemon juice.
Other carbonates like calcium carbonate in limestone and egg shells can be substituted
for sodium hydrogencarbonate.
2. When two different metals dip into a conducting solution (electrolyte), electrical energy
can be produced. Your metals could be copper and iron or any available. Make sure
the surfaces are clean by rubbing with emery paper. The electrolyte (solution) could
be an acid such as lemon juice, or salty water. A small torch light bulb or a multimeter
can detect the electric current produced.
Electrodes
Connecting wires
Electrolyte
Light Globe
If you used copper and iron, the iron metal will slowly dissolve, forming Fe 2+ and release
electrons (electric current). You may notice a fizzing at the surface of the copper as
hydrogen gas is produced.
Word Equation:
Iron(s) + Acid(aq)
Iron +
Ion (aq)
Is this reaction exothermic or endothermic?
Report on your results.
77
Hydrogen + Energy
7.2
Writing Chemical Equations - Chemical Shorthand
A chemical equation is a shorthand method of communicating information about reactants
and products, quantities involved and sometimes how much energy is needed or released.
It accounts for all the atoms involved in the re-arrangement of bonds.
Reactants
(Ingredients)
(form)
Products
(new substances)
During a chemical reaction, bonds are broken in the reactants and new bonds are formed
making the products. The atoms are re-arranged during a reaction, but the numbers
of the different types of atoms remain constant.
The law of Conservation of Matter (or mass) states that during a chemical reaction the
total mass of reactants is the same as the total mass of products formed. During a
chemical reaction, matter is neither created nor destroyed, it is only changed from
one form into another.
Law of Conservation of Mass (Matter)
Total
Mass of reactants
=
Total
Mass of products
Note: The total number of atoms of each element is unchanged!
Example 1: When a spark ignites a mixture of hydrogen gas and oxygen gas, an
explosion occurs - a large amount of energy is released. The product formed is water.
Chemical reactions which release energy are called exothermic.
Word
Equation :
Symbol
Equation:
(unbalanced)
Hydrogen(g) +
H2(g)
+
Oxygen(g)
Water(l) + energy
O2(g)
H2O(l)
+ energy
The correct chemical formula is written under each reactant and product. The extra
subscripts indicate the state of that substance.
The equation above is said to be unbalanced because the numbers of various atoms
represented on the reactants’ side (left of the arrow) is not the same as those on the
products’ side (right of the arrow). This would mean that mass is not conserved!
78
Reactants ( H2 + O2 )
Products ( H2O )
H=2
H=2
O=2
O=1
The numbers of oxygen atoms are not balanced!
We must adjust the numbers of molecules of each type present to reach a balance! This
involves placing numbers where required in front of each formula. The number will multiply
each atom in the formula. This was referred to previously as a coefficient number. A “2”
in front of the H2O product will provide the required 2 oxygen atoms on the right.
H2(g)
+
O2(g)
2 H2O(l)
+
energy
Reactants ( H2 + O2 )
Products ( 2 H2O )
H=2
H=4
O=2
O=2
Whilst oxygen has now been balanced, Hydrogen is now unbalanced!
In order to balance the hydrogen atoms it is necessary to place a “2” in front of the H2 on
the left. This provides 4 hydrogen atoms on the left also.
Balanced Equation:
2 H2(g) +
O2(g)
2 H2O(l) +
energy
Reactants ( 2 H2 + O2 )
Products ( 2 H2O )
H=4
H=4
O=2
O=2
Diagram:
Activity: Use your atomic models to show the re-arrangement of atoms which takes
place in this reaction. Notice that you will need to use the numbers of molecules as
indicated in the equation, ie; 2 x hydrogen molecules for each oxygen molecule.
79
Example 2. When an emergency flare lights up, magnesium (or aluminium) metal
combines with oxygen gas releasing a bright light. This is also an exothermic reaction. The
substance formed is a metal oxide.
Word
equation:
Magnesium(s)
Symbols:
Mg(s)
+
Oxygen(g)
+
O2(g)
Magnesium + energy
oxide(s)
MgO(s)
+
energy
Reactants ( Mg + O2 )
Products ( MgO )
Mg = 1
Mg = 1
O=2
O=1
The numbers of oxygen atoms are not balanced!
Balance the oxygen atoms by placing a coefficient “2” in front of MgO. This will also
produce 2 magnesium atoms.
Mg(s)
+
O2(g )
2MgO(s)
+
energy
Reactants ( Mg + O2 )
Products ( 2 MgO )
Mg = 1
Mg = 2
O=2
O=2
Then balance the magnesium atoms by placing a coefficient “2” in front of Mg.
Balanced
equation:
2 Mg(s)
+
O2(g)
2 MgO(s)
+
energy
Reactants ( 2 Mg + O2 )
Products ( 2 MgO )
Mg = 2
Mg = 2
O=2
O=2
Ref: C1 pp70-71
80
Written Exercises
1.
Balance the following chemical equations.
(a)
CH4(g) +
O2(g)
CO2(g)
+
H2O(l)
(b)
Na(s) +
H2O(l )
NaOH(aq)
+
H2(g)
(c)
Zn(s)
+
HCl(aq )
(d)
P
+
O2
(e)
NH3
+
H2SO4
(f)
CuO +
(g)
H2O2
H2O
+
O2
(h)
H2CO3
H2O
+
CO2
(i)
Fe
+
O2
Fe2O3
(j)
C8H18 +
O2
CO2
ZnCl2(aq)
+
H2(g)
P2O5
(NH4)2SO4
HCl
CuCl2
+
+
H2O
H2O
2.
You may have earlier experimented at home the reaction between ethanoic (acetic)
acid, CH3COOH with sodium hydrogencarbonate, NaHCO3 to form carbon dioxide, water
and a salt called sodium ethanoate (acetate), NaCH3COO. Write a word and balanced
equation for this reaction.
Word equation:
………………………………………………………………………………………………………..
Balanced equation:
………………………………………………………………………………………………………..
81
EXPRESSING BALANCED CHEMICAL REACTIONS:
A reaction occurs because atoms are REARRANGED. Atoms are not formed or
destroyed. Mass of Reactants and products remains the SAME.
1. Identify Reactants of Chemical Reaction, for example;
Sodium Hydroxide + Sulfuric Acid  ?
2. Try to identify the Nature of reactants. Then correspond with a typical reaction
type, ie
Sodium Hydroxide + Sulfuric Acid  ?
BASE
+
ACID
 SALT + WATER
3. Write down a WORD EQUATION:
Sodium Hydroxide + Sulfuric Acid  Sodium Sulfate + Water
4. Write Symbols for compounds
NaOH + H2SO4  NaSO4 + H2O.
5. Balance atoms in compounds according to Valency:
1+
2-
NaOH + H2SO4  Na2SO4 + H2O.
6. Reactant Atoms = Product Atoms
Balance atoms either side of arrow.
-
Balance Na (2NaOH)
2NaOH + H2SO4  Na2SO4 + H2O.
-
Balance O (2H2O)
2NaOH + H2SO4  Na2SO4 + 2H2O.
Elements in equation now BALANCED.
82
Some Typical Chemical Reactions
-
Atoms can be rearranged in a variety of ways….
-
Write a word and balanced chemical equation of each reaction type listed below in the
corresponding space at the right.
REACTION TYPE
EXAMPLES
ACID + BASE  SALT + WATER
NEUTRALISATION
An ACID mixed with a BASE
(Alkali) produces a Neutral
Salt and Water.
ACID + METAL
An ACID added to a METAL
will produce a Metal Salt and
Hydrogen Gas, H2
ACID + METAL  SALT + HYDROGEN
ACID + CARBONATE  SALT + CARBON DIOXIDE + WATER
ACID + CARBONATE
An ACID added to a Metal
CARBONATE compound will
produce a Salt, Carbon
Dioxide, CO2 and Water,
H2O.
ORGANIC FUEL + OXYGEN  CARBON DIOXIDE + WATER
ORGANIC COMBUSTION
Organic Fuels BURNT in the
presence of OXYGEN
produce Carbon Dioxide and
Water **
** assuming complete combustion.
83
REACTION TYPE
EXAMPLES
ELEMENT + OXYGEN  ELEMENT OXIDE
INORGANIC COMBUSTION
Some elements will readily
burn in the presence of
oxygen to form an oxide
compound.
PRECIPITATION
Two SOLUBLE Compounds
may React and Rearrange to
produce an INSOLUBLE
(solid) Compound, ie
Precipitate.
AB (aq) + CD (aq)  AD (s) + BC (aq)
A  B+C+...
DECOMPOSITION
A single compound can
break down to form 2 or
more smaller compounds or
elements.
HYDROLYSIS
Reaction of any chemical
with WATER
84
TESTING FOR GASES
-
Hydrogen, Carbon Dioxide and Oxygen gases can be identified via simple diagnostic
chemical tests.
TEST TYPE
DESCRIPTION
POP TEST
A diagnostic test for
Hydrogen gas.
SPLINT TEST
A diagnostic test for Carbon
Dioxide gas.
LIMEWATER TEST
A second diagnostic test for
Carbon Dioxide gas.
GLOWING SPLINT TEST
A diagnostic test for Oxygen
gas.
85
CHEMICAL EQUATIONS – Exercise 1.
Carefully read the following chemical reaction scenarios. Firstly decide which
substances are reactants and which are products, then:
i) Write down a word equation of reactants and products.
ii) Write a balanced chemical equation.
1. Aqueous Magnesium Chloride and Hydrogen gas are produced when Magnesium
ribbon is placed into a solution of Hydrochloric acid.
2. Aluminium metal reacts with Nitric acid to produce an aqueous solution of Aluminium
Nitrate and Hydrogen gas.
3. Iron metal rusts in the presence of Oxygen to produce the compound Iron III Oxide.
4. Lithium metal, a reactive group 1 alkali metal, reacts violently when placed in water to
produce a strongly alkaline solution of Lithium Hydroxide and Hydrogen gas.
5. Solid Silver I Carbonate decomposes when heated to produce solid Silver Oxide and
Carbon Dioxide
6. Solid Silver I Chloride and an aqueous solution of Sodium Nitrate are produced when
solutions of Silver I Nitrate and Sodium Chloride are combined.
7. A neutral solution of Sodium Sulfate and water are produced at equivalence point
when Sulfuric Acid is fully neutralised by the addition of Sodium Hydroxide.
86
8. A bright yellow precipitate of Lead II Iodide and aqueous solution of Potassium Nitrate
are produced when aqueous solutions of Lead Nitrate and Potassium Iodide are
combined.
9. When Hexane, C6H14 (a colourless liquid hydrocarbon) combusts in a plentiful supply
of oxygen, the compounds of carbon dioxide and water are formed. Heat is given off
indicating that the process is exothermic.
CHEMICAL EQUATIONS - Exercise 2
Use your knowledge of Chemical Formulae, Valencies and Reactions to
complete the following Exercises. Complete the WORD and BALANCED
CHEMICAL EQUATIONS below.
1. Magnesium + Nitric Acid 
2. Copper + Sulfuric Acid 
3. Potassium + Water 
4. Zinc (II) Carbonate + Nitric Acid 
5. Copper II Carbonate (s) + HEAT 
6. Aluminium + Sulfuric Acid 
7. Sodium Hydroxide (aq) + Copper II Sulfate (aq) 
8. Sodium Hydroxide + Sulfuric Acid 
87
7.3
Ionic Equations
When solutions are involved in a reaction, only some of the ions present are usually
involved. Other ions may be present, but they are still in the solution at the end of the
reaction, unchanged by the chemical process. These ions are called spectator ions and
are best left out of the balanced equation. When spectator ions are left out of an equation
an ionic equation results. Ionic equations are the best representation of the chemical
reaction, since they show only those species which have undergone a change.
Example 1. When a solution of lead (II) nitrate is mixed with a solution of potassium
iodide, a brightly coloured precipitate forms. This precipitate is solid lead(II) iodide. Write
an ionic equation for this chemical change.
General Equation (balanced):
Pb(NO3)2(aq)
+
2KI(aq)
PbI2(s)
+ 2KNO3(aq)
Aqueous compounds split into their respective ions as they are dissolved in water. We can
expand out the general equation as follows:
Expanded Equation
Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq)
PbI2(s) + 2K+(aq) + 2NO3-(aq)
Note: PbI2 (s) on the right does not form ions, as it is in the solid state.
In this “expanded” form, the equation shows the “whole picture” of the chemical process.
If we look a bit closer at the reactants and products however, we can see that some of the
chemical components have remained unchanged in the process. Can you see this ???
If you can, you will have identified 2NO3-(aq) and 2K+(aq) as remaining identical, hence
unchanged on the left and right side of the equation. They are merely “spectator ions”
but can be mistaken as being part of the chemical process when referring back to the
balanced general equation. If they are deleted from the left and right of the equation, we
will arrive at the ionic equation, ie:
Ionic Equation:
Pb2+(aq)
+
2I-(aq)
PbI2(s)
Example 2. When magnesium metal reacts with dilute hydrochloric acid, a solution of
magnesium chloride remains and hydrogen gas is evolved. Write a balanced, general
chemical equation and an ionic equation for this chemical change.
Magnesium(s) +
hydrochloric
acid (aq)
magnesium +
chloride (aq)
hydrogen (g)
MgCl2(aq)
H2(g)
General Equation (balanced):
Mg(s)
+
2HCl(aq)
88
+
Expanded Equation
Mg(s)
+
+
(aq)
2H
+
-
2Cl (aq)
2+
(aq)
Mg
-
+
2Cl (aq) +
H2(g)
Note: Mg (s) on the left and H2 (g) on the right do not form ions. They are in the solid and
gaseous state respectively.
The spectator ions in this case are the chloride ions, which are present in the dilute
hydrochloric acid as Cl-(aq) and remain unchanged on the right as a component of aqueous
magnesium chloride. When Cl-(aq) are deleted from the left and right of the expanded
equation, we arrive at the net ionic equation:
Ionic Equation:
Mg(s) +
2H+(aq)
Mg2+(aq)
+
H2(g)
An Ionic Equation should also appear as a balanced equation. If not, a mistake has been
made and you will need to go back and check where something went wrong.
Ionic equations must also be balanced in terms of their overall charge on the left and
right. This can be done as a simple calculation
Reactants ( Mg(s) + 2H +(aq) )
Products (Mg2+(aq) + H2 (g) )
2 x 1+ = 2+
1 x 2 + = 2+
Total charge on left
hand side of equation
= 2+
Total charge on right
hand side of equation
= 2+
You will find it useful to recognise certain common types of chemical reactions so
that you can correctly predict the products which will form. Ionic equations will be
used where appropriate in the examples following. More work on ionic equations
will be done in another module!
Written Exercises: Write out the net ionic equation of each full equation below.
1.
AgNO3 (aq) + NaCl (aq)  AgCl (s) + NaNO3 (aq)
2.
Na2CO3 (aq) + H2SO4 (aq)  Na2SO4 (aq) + CO2 (g) + H2O (l)
89
Experiment: SOLUBILITY, PRECIPITATION REACTIONS AND IONIC EQUATIONS
Aim:
To observe precipitation reactions and express ionic equations
Method:
Combine pairs of aqueous compounds using a spotting tray as per the table below. Report “yes” or “no” to indicate
formation of a precipitate. Record precipitate colour if observed. Try naming each precipitate compound formed.
Results:
Copper II
Sulfate
Barium
Chloride
Lead II
Nitrate
Calcium
Chloride
Potassium Iodide
Sodium Hydroxide
Sodium Carbonate
Silver I Nitrate
Precipitate?
Precipitate?
Precipitate?
Precipitate?
Colour:
Colour:
Colour:
Colour:
Name:
Name:
Name:
Name:
Precipitate?
Precipitate?
Precipitate?
Precipitate?
Colour:
Colour:
Colour:
Colour:
Name:
Name:
Name:
Name:
Precipitate?
Precipitate?
Precipitate?
Precipitate?
Colour:
Colour:
Colour:
Colour:
Name:
Name:
Name:
Name:
Precipitate?
Precipitate?
Precipitate?
Precipitate?
Colour:
Colour:
Colour:
Colour:
Name:
Name:
Name:
Name:
Written Exercise: Choose one precipitate formed from each row above. Express each process as a word equation, balanced chemical
equation and finally, an ionic equation. See over page for working space and structure:
90
1. Copper II Sulfate
Word Equation:
Copper II Sulfate + ________________  ________________ + _______________
Balanced Chemical Equation:
CuSO4 (aq) + ________________ 
_________________ + _________________
Ionic Equation:
…………………………………………………………………………………………………………
2. Barium Chloride
Word Equation:
Barium Chloride + ________________  ________________ + _______________
Balanced Chemical Equation:
BaCl2 (aq) + _________________ 
__________________ + ________________
Ionic Equation:
…………………………………………………………………………………………………………
3. Lead II Nitrate
Word Equation:
Lead II Nitrate + ________________  ________________ + _______________
Balanced Chemical Equation:
Pb(NO3)2 (aq) + ________________ 
_________________ + _______________
Ionic Equation:
…………………………………………………………………………………………………………
4. Calcium Chloride
Word Equation:
Calcium Chloride + ________________  ________________ + _______________
Balanced Chemical Equation:
CaCl2 (aq) + _________________ 
__________________ + __________________
Ionic Equation:
…………………………………………………………………………………………………………
91
8.
QUANTITIES AND CHEMICAL PROCESSES
How can you determine how much of each substance is needed to make your product from a
chemical reaction? How can you work out the amount of a chemical substance present in a
product from the supermarket?
In this module you will learn how to work out quantities from balanced chemical equations.
You will later encounter practical techniques which are used to analyse samples for particular
chemicals.
8.1
An introduction to Quantitative Chemistry
Read the following 2 scenarios below. Have a go answering each of the questions.
Scenario 1:
If we combined 10 g of Silver Nitrate and 10 g of Sodium Chloride (both in aqueous state), a
chemical reaction occurs and a white solid powder of Silver Chloride forms.
If we evaporate the water after completion of the reaction, what mass of material will
be obtained?
Write down a corresponding chemical equation.
Name which chemical substance(s) would make up this remaining material. What
issues must we consider?
Scenario 2:
If we burn 5 g of magnesium in an open porcelain dish, a bright white flame is observed and
8.3 g of magnesium oxide (a white powder) is obtained.
Write down a corresponding chemical equation.
What mass of oxygen gas was consumed?
Are equal masses of reactants always consumed in a chemical reaction?
Which reactant in this case will there be an excess of? (ie; Excess Reagent)
Which reactant would run out first, (ie; Limiting Reagent)?
Ref: BFK p70
92
All chemical reactions occur in simple ratios of particles. The quantities required are given
by the balancing numbers in the chemical equation.
Examples:
1. Electricity can be used to decompose water molecules into its elements of hydrogen and
oxygen.
RATIO:
2H2O(l)
2H2(g)
2 molecules
of water
2 molecules
of hydrogen
1 molecule
of oxygen
2
2
1
:
+
:
O2(g)
Two molecules of water decompose to form two molecules of hydrogen and one molecule of
oxygen. Of course we do not work with single molecules - they are much too small! However,
the simplest whole number ratio of molecules involved in this process will always be
2 x H2O
:
2 x H2
:
1 x O2
2. Sodium and chlorine react together violently to produce sodium chloride.
2Na(s)
RATIO:
+
Cl2(g)
2NaCl(s)
2 atoms
of sodium
1 molecule
of chlorine
2 units of
sodium chloride
2
1
2
:
:
Two atoms of sodium react with 1 molecule of chlorine to produce 2 formula units of sodium
chloride.
2 x Na : 1 x Cl2 :
2 x NaCl
Again, we cannot see molecules and so making sure that we have the right quantities for a
chemical reaction could be difficult!
Simplest whole number ratios are referred to as coefficient numbers in a balanced
chemical equation. These numbers represent mole quantities, a unit of scientific
measurement which can be reported with accuracy in the laboratory.
93
8.2
The Mole Concept
We need to “scale up” our quantities from the molecular level to an amount which we can
see and measure in the laboratory. The quantity of particles chosen is called the mole
which is approximately 6.02 x 1023 particles of any pure substance (ie; element or
compound). Scientific notation is expressed in this number as this is a gigantic quantity!
The Periodic Table measures the mass of each element in atomic mass units (a.m.u.)
One mole of any substance contains 6.02 x 1023 particles and has a mass measured in
grams equal to its formula weight (a.m.u.)
One mole of water, H2O contains 6.02 x 1023 molecules. The mass of 1 mole of water is
equal to its formula weight (FW), expressed in grams;
FW (Hydrogen) = 1.008 a.m.u.
FW (Oxygen) = 16.00 a.m.u.
(2 x 1.008) + 16.00 = 18.016 g.
Likewise, one mole of hydrogen (gas) contains 6.02 x 1023 molecules of hydrogen and has
a mass of 2 x 1.008 = 2.016 g.
And - one mole of oxygen (gas) contains 6.02 x 1023 molecules of oxygen and has a mass
of 2 x 16.00 = 32.00 g.
We could, if we wanted, determine the ratio, by mass of the substances involved in the
decomposition of water.
2H2O(l)
2H2(g)
+
O2(g)
2 molecules
of water
2 molecules
of hydrogen
1 molecule
of oxygen
2 moles
of water
2 moles
of hydrogen
1 mole
of oxygen
2 x 18.016
= 36.032 g
2 x 2.016
= 4.032 g
1 x 32.00
= 32.00 g
If we started with 1 mole (ie 36.032 g) of water, we would obtain 4.032 g of hydrogen and
32.00 g of oxygen. What mass of hydrogen and oxygen would we obtain if we started with
1.00 g of water?
H2O (l)
H2 (g)
O2 (g)
36.032
36.032
4.032
36.032
32.00
36.032
1.00 g
0.11 g
94
+
0.89 g
If we started with 1.00 g of water we would obtain 0.11 g of hydrogen and 0.89 g of oxygen.
Notice that the balancing numbers do not give us masses directly, that is we cannot say that
2 g of water will give 2 g of hydrogen and 1 g of oxygen!
When we use the mole concept we are taking into account that the various atoms have
different masses, and the balancing numbers give us the right ratios of atoms involved.
It makes life much easier when interpreting number coefficients of a balanced
chemical equation as mole quantities, because that’s exactly what they are !
Take special note that the total mass of reactants equals the total mass of products.
No atoms are lost or gained, as understood from our work on writing and balancing chemical
equations. Atoms are just re-arranged in a chemical reaction.
We could continue our calculations to account for any mass of water, or we could calculate
the quantity of water required to produce a particular amount of hydrogen or oxygen. The
mole ratios of reactants to products are determined from the balanced chemical equation
and a knowledge of the mole concept.
We can determine the number of moles present in a sample of a substance by using one or
more of the following formulae:
No of moles =
No of particles
6.02 x 1023
This mole equation will not be immediately useful as it deals with a known number of
chemical particles, which in reality is mostly unrealistic. You may, however be asked to
calculate a mole quantity given such data.
Mole Equation 1.
No of moles =
Mass (g)
Formula Weight
This equation is useful in the real world, as mass and Formula Weight are measurable
quantities in the real world. Memorise it !!!
Shorthand:
n = m
FW
95
Written Exercises: Mole calculations
1. Calculate the Formula Weight, (FW) for each of the following substances.
(a) Sodium Chloride, NaCl
(b) Ammonium Sulfate, (NH4)2SO4
(c) Glucose, C6H12O6
(d) Potassium Hydrogen Phthalate, KC8H5O4
2. Calculate the number of moles of each substance
(a) 58 g of NaCl
(b) 13.2 g of (NH4)2SO4
(c) 90 g of C6H12O6
(d) 5.0 g of KC8H5O4
3. What mass of each substance is present?
(a) 2.5 moles of NaCl
(b) 0.5 moles of (NH4)2SO4
(c) 4 moles of C6H12O6
(d) 0.015 moles of KC8H5O4
Ref: CC1 p205
C1 pp170-173; 213-218
DB pp20-22
96
8.3
Theoretical Yield (also Chemical Stoichiometry)
A Theoretical Yield is the calculated prediction of an amount of substance produced
(product) in a chemical reaction. It is a very useful application of the mole concept, and is an
integral part of the chemical industry. This concept can alternatively be used to calculate the
amount of chemical substance required (reactant) in a chemical process.
Scenario 1: Predicting the mass of a product obtained in a chemical reaction.
Calculate the mass of Zinc Chloride, ZnCl2 produced when 5.0g of Zinc metal is
dissolved in an excess of Hydrochloric Acid, HCl.
Step 1.
Calculate moles of Zinc metal reacted
Step 2.
Write down a balanced chemical reaction
Step 3.
Express the reaction ratio between the reactant Zn and product ZnCl2
chemicals of interest.
________
:
________
Step 4.
Calculate the number of moles of ZnCl2 produced: apply the reaction ratio
(ie mole ratio) to the calculated number of moles in Step 1.
Step 5.
Calculate the Theoretical Yield mass of ZnCl2 using the mole equation
(rearranged..)
97
Scenario 2: Calculating the required mass of a reactant chemical to produce a desired
known quantity of a given product.
Iron metal, Fe is produced from one of its ore compounds Iron III Oxide, Fe2O3 by
heating in the presence of excess Carbon Monoxide, CO gas. Carbon dioxide is given
off as a by-product.
- Calculate the mass of Iron III oxide, Fe2O3 required to produce 500 g of Iron metal.
Step 1.
Calculate the moles of desired Iron metal.
Step 2.
Write down a balanced chemical reaction.
Step 3.
Express the reaction ratio between the reactant / product chemicals of
interest.
Fe2O3
Fe
_______ : _______
Step 4.
Apply reaction ratio to calculate the moles of Iron III Oxide (Fe 2O3) required.
Step 5.
Convert moles of Iron III Oxide (Fe2O3) into a mass (via mole equation
again…)
8.4
Percentage Yield
In reality, it is very unlikely that a 100% yield of product is obtained. Experimental errors,
transfer losses and incomplete reaction often tend to contribute to a lower than expected
yield.
(Note: yields greater than 100% indicate that an error in calculation, or perhaps moisture in
the final weighed product might have been encountered).
% Yield =
obtained yield
theoretical yield
x 100
This percentage calculation is quite easy when comparing your obtained yield with the
theoretical (ie expected yield).
98
Scenario 2 Revisited: Calculating a percentage yield
The required mass of Iron III oxide, Fe2O3 calculated in Scenario 2 previously was carefully
weighed out, and chemically converted into Iron via the same chemical process. A mass of
462.2 g of Iron was recovered.
i)
Calculate the % Yield of this process
ii)
Suggest reasons for this obtained %
8.5
Limiting Reagents
As we have seen, it is possible to calculate the exact quantities of reactants that would in
theory react chemically right down to the very last molecule. This, in reality does not occur –
there is always an excess of one substance (quite possibly verrry verrry small or large
depending on the situation). The chemical substance totally consumed in a chemical process
is referred to as the Limiting Reagent – it is the substance that determines when the
reaction will cease and the amount of product that will be obtained.
As an example – the pop test is a chemical reaction between Hydrogen Gas, H2 and Oxygen
Gas, O2. Water, H2O is the resulting product. The amount of water produced in this case is
limited by the volume (and hence mass) of hydrogen gas. Atmospheric oxygen is in plentiful
supply (21%) and is therefore not particularly likely to be entirely consumed…… Oxygen is
referred to here as the Excess Reagent.
Identifying the Limiting Reagent is yet another application of the mole concept. The mole
quantities of each chemical reactant present is applied to the reaction ratio according to the
balanced chemical equation. A decision can be made from here via simple calculation as to
which one is in limited supply, (ie; Limiting Reagent identification)
Scenario 1 Revisited:
10.0g of Zn (s) is chemically reacted with 10.0g of HCl (aq).
PART A: Identify the Limiting Reagent
Step 1.
Calculate moles Zn
Step 2.
Calculate moles HCl
99
Zn
HCl
Step 3.
Express reaction ratio: _______ : _______ (from Step 2, Scenario 1)
Step 4.
Apply reaction ratio to moles of Zn (Step 1.) to calculate the actual mole
requirement of HCl
Moles HCl Required:
___________________________
Moles HCl Available:
___________________________ (from Step 2)
Step 5.
Are there sufficient moles of HCl available ??
-
If yes, Zn will be the Limiting Reagent (it will be consumed first)
-
If no, then HCl will be the Limiting Reagent. There is insufficient HCl to fully
react with the mass of Zn available, hence HCl will be consumed first.
Limiting Reagent ID: ____________________________________
PART B: Calculate the Theoretical Yield of ZnCl2 produced in this scenario.
- The amount of ZnCl2 is governed by the available amount of Limiting Reagent.
Step 1.
Identify the limiting reagent
(as per Step 5, Part A):
_____________________
Step 2.
Calculate moles of Limiting Reagent
(as per Step 2, Part A): ______________________
Step 3.
Express the reaction ratio between the
limiting reagent and the product ZnCl2
Step 4.
L.R.
ZnCl2
_______ : _______
Apply this reaction ratio in calculating the moles of ZnCl2 produced.
Moles ZnCl2 = _____________________
Step 5.
Calculate the Theoretical Yield mass of ZnCl2 (using the mole equation…)
100
PART C: Calculate the mass of Excess Reagent remaining unreacted.
Step 1.
Identify the excess reagent: _________________________
Step 2.
Calculate the moles of excess reagent available (as per Step 1, Part A):
Step 3.
Calculate moles of limiting reagent (as per Step 2, Part A):
Step 4.
Express the reaction ratio between the excess and limiting reagent:
Zn
Step 5.
:
HCl
=
_______ : _______
Apply reaction ratio and calculate the moles of excess reagent consumed
- Moles Zn consumed in reaction = _________________________
Step 6.
Convert moles of excess reagent into mass of excess reagent consumed.
Step 7.
Calculate the mass (g) of excess reagent remaining.
Written Exercise: Hydrazine, N2H4 is used as a rocket fuel. It reacts with oxygen
to form nitrogen gas and water vapour.
(a)
Write a balanced chemical equation for this reaction.
(b)
1 kg of hydrazine and 1 kg of oxygen are available on the rocket. Identify the
Limiting Reagent in this scenario.
(c)
What mass of water vapour is produced during the flight?
101
Yield Experiment 1 – MAGNESIUM OXIDE
Aim: To predict a theoretical yield when magnesium metal is combusted.
Method:
(record all results in the table below)
Obtain and record accurately the mass of a clean empty crucible + lid.
Accurately weigh approximately 0.2g of Mg metal into your crucible. Record the combined
mass.
Heat crucible and contents with a bunsen flame. Use tongs to lift crucible lid periodically
and observe until all Mg has combusted.
Allow 10 minutes to cool. Reweigh Crucible and combusted contents.
Results:
Mass: empty crucible + lid (g)
Mass: crucible + lid + Mg (g)
Calculate mass: Mg metal (g)
Mass: crucible + lid + MgO (g)
Calculate mass: MgO (g)
Calculations (Theoretical Yield):
1.
Moles Mg:
Moles = Mass / FW
………………………………………………………………………………………………………
2.
Write a word and balanced chemical equation for the combustion of Mg (s).
i)
Word Equation:
………………………………………………………………………………………………………
ii)
Balanced Chemical Equation:
………………………………………………………………………………………………………
Mg
MgO
3.
Reaction Ratio between Mg (s) and MgO (s)
= _____ :
4.
Moles MgO = ………………………………………..
5.
Mass MgO (g) (theoretical yield): Mass = Moles x FW (MgO)
______
…………………………………………………………………………………………………
Calculations; Percentage Yield:
6.
Calculate this % based on the mass of MgO obtained experimentally.
102
Yield Experiment 2 – DECOMPOSITION OF A CARBONATE COMPOUND
Background:
Decomposition Reactions involve the break down of large compounds into 2 or more smaller
components. Heat is a large factor of many decompositions. Bicarb Soda, a carbonate used
in baking, will decompose into simpler substances – most notably Carbon Dioxide. This is
what causes bread to rise.
Procedure:
Obtain and record accurately the mass of a clean empty porcelain crucible + lid.
Accurately weigh approximately 1.5g of Copper Carbonate Powder into the crucible.
Record the combined mass of crucible, lid and Copper Carbonate powder.
Set up a heatproof mat, Bunsen burner, tripod and pipeclay triangle.
Carefully heat the Carbonate substance with a blue Bunsen flame for between 5 and
10 minutes. Use tongs to occasionally lift the crucible lid and observe the change in
colour of the compound.
Allow 10 minutes for the crucible and lid to cool. Reweigh and record the new mass.
Results:
Mass: Crucible + Lid (g)
Mass: Crucible + Lid + CuCO3 (g)
Mass: CuCO3 (g)
After heating with Bunsen Burner:
Mass: Crucible + Lid + CuO (g)
Mass: CuO (g)
Calculations
1.
Theoretical Yield:
Moles CuCO3 :
Moles = Mass ÷ FW
………………………………………………………………………………………………………
103
2.
Write a word and balanced chemical equation for the decomposition of CuCO3 (s)
i)
Word
Equation: ……………………………………………………………………………………
ii)
Chemical
Equation: ……………………………………………………………………..……………
CuCO3
3.
Reaction (mole) Ratio between CuCO3 and CuO =
4.
Moles CuO = ………………………………………………………………..
5.
Mass CuO (g)
Mass = moles x FW (CuO)
______ :
CuO
______
(this value is the theoretical yield)
………………………………………………………………………………………………………
CALCULATIONS
6.
Percentage Yield:
Calculate the Percentage Yield of CuO, based on the mass you obtained
experimentally.
% Yield = Obtained Yield
x 100
Theoretical Yield
Questions:
1.
Atoms are neither created nor destroyed in chemical reactions – rather they are
rearranged. How can we therefore account for the loss of mass in the reaction?
2.
Explain why it is not necessary to periodically remove the crucible lid to ensure
completion of the chemical reaction.
3.
Yield percentages of over 100% are commonly encountered in this task. Explain why.
104
Yield Experiment 3 – ZINC II IODIDE
Background:
Zinc metal, Zn reacts with iodine, I2 to produce the compound zinc II iodide, ZnI2.
You will react an accurately known mass of iodine crystals with excess zinc metal, so that all
the iodine can react but some zinc will remain. Ethanol will be used as a solvent for the
iodine to speed up the reaction. Zinc iodide is highly hygroscopic - it will absorb water from
the air.
! Risk Assessment:
Iodine vapour is mildly corrosive and an irritant. Iodine will stain skin and clothing easily. The
reaction between zinc and iodine is exothermic and iodine vaporises easily.
Method:
1.
2.
Collect approximately 2g of zinc pieces and record their mass accurately.
Obtain a sample tube and lid. Tare sample tube and accurately weigh approximately
1g of iodine. Record this value.
3.
Carefully add the zinc pieces to the tube containing the iodine.
4.
Collect about 20 mL of ethanol and add it to the tube a few drops at a time while
gently swirling the contents of the tube. The ethanol should be added very slowly
and the tube should become very warm between additions. If the ethanol is added too
quickly the reaction will become very slow.
5.
Continue to add the ethanol and swirl the contents until all reaction has ceased. The
brown colour of dissolved iodine should have almost disappeared. Allow the tube to
stand covered until all colour has been lost.
6.
Carefully weigh a clean, dry evaporating basin. Decant the liquid from the sample tube
into the basin, retaining unreacted zinc in the sample tube.
7.
Wash the zinc and the insides of the tube with several small quantities of ethanol. Add
washings to the evaporating basin.
8.
Evaporate the ethanol from the evaporating basin using an infra-red lamp if
possible. Otherwise use an electric hotplate on a low setting.
9.
Allow the dry evaporating basin to cool in a desiccator and weigh the basin
and
contents quickly.
From your measurements you should now be able to calculate the percentage yield of zinc
iodide.
Results:
Mass: Zinc =
__________________ g
Mass: Iodine =
__________________ g
Mass: Clean, dry evaporating basin = _________________________ g
Mass: Evaporating basin + dry Zinc Iodide = _______________________ g
Mass: Zinc Iodide = _________________________ g
105
Calculations:
1.
Calculate to show that the Iodine was the Limiting Reagent
2.
Calculate the Theoretical Yield of Zinc Iodide
3.
Calculate the Percentage Yield of Zinc Iodide
106
8.6
Moles, Gas Volumes and Yields
It is often inconvenient to have to weigh a gas in order to determine the quantity present.
Why? Try weighing an inflated balloon in order to determine the mass of gas inside! Various
problems such as buoyancy effects and the mass of the container become apparent. When
handling gases, it is preferable to measure volume.
Of course we must use set conditions of temperature and pressure in order to compare
quantities, as differences in both temperature and pressure change volume!
There are 2 conditions of Molar Volume chosen by scientists universally. Each is a measure
of the volume of 1 mole of any gas. They are as follows:
Standard Temperature and Pressure, S.T.P.
Temperature
Pressure
Molar Volume
0oC (273K)
100 kPa
22.71 L
Standard Laboratory Conditions, S.L.C.
Temperature
Pressure
Molar Volume
25oC (298K)
100 kPa
24.79 L
100kPa, for interests sake is very close to that of atmospheric air pressure
It has been found that 1 mole of any gas will occupy the corresponding volume according to
the respective conditions outlined in the tables above. We can use this information to
therefore easily determine the number of moles present.
No of moles = volume (L)
molar volume (L)
Example:
Carbon disulfide, CS2 is an important industrial solvent. It is prepared by the reaction of coke
(almost pure carbon) with sulfur dioxide gas. What mass of coke and volume of sulfur dioxide
(at 25oC and 100 kPa) would be required to produce 1 kg of carbon disulfide?
Equation:
5C(s) +
2SO2 (g)
Mole ratio:
5 moles :
2 moles
CS2 (l) +
:
1 mole
107
4CO (g)
:
4 moles
i)
Calculate required mole quantities:
The balancing numbers tell us that if we wanted 1 mole of carbon disulfide we would require
5 moles of carbon (coke) and 2 moles of sulfur dioxide.
How many moles of carbon disulfide do we require? Using the formula:
No of moles =
(of CS2)
Mass (g)
FW
=
1000 g
( 1 kg = 1000 g)
12.01 + (2 x 32.07)
=
13.1
Therefore we must produce 13.1 moles of carbon disulfide. For each mole of carbon
disulfide we need 5 moles of coke.
No of moles (of C)
=
5 x 13.1
= 65.5
For each mole of carbon disulfide we need 2 moles of sulfur dioxide.
No of moles (of SO2) =
ii)
2 x 13.1
= 26.2
Calculate corresponding mass (C) and volume (SO2) quantities
The question requires answers as mass (C) and volume (SO2) and so we will need to use the
corresponding mole equations, covered previously to calculate these.
1. Calculating Mass of Carbon, C
Moles C = 65.5
Mass C = ?
FW (C) = 12.01
No of moles =
Mass (g)
Formula Weight
 rearrange to make “Mass” the subject of the formula
Mass (g) = No of moles x FW
= 65.5 x 12.01
= 787 g
(rounded)
See over page for calculation of SO2 volume
108
2. Calculating Volume of SO2 (g)
Moles SO2 = 26.2
Volume SO2 = ?
Molar Volume (SLC) = 24.79 L
No of moles =
Volume (L)
24.79
 rearrange to make “Volume” the subject of the formula
Volume (L) = No of moles x 24.79
= 26.2 x 24.79
= 650 L
(rounded)
If the gases present were not under S.L.C. (or S.T.P.) conditions, you would need to use
another mathematical formula:
PxV = nxRxT
where;
P = Gas pressure (kPa)
V = Volume of gas (L)
n = No of moles of gas present
R = the Gas Constant
T = Temperature (Kelvin)
Ref: C1 pp276-279
Written Exercise: Theoretical Yield and Gas Volumes
1. Limestone and shells are made mostly of calcium carbonate. When they are heated to
produce “quicklime” or calcium oxide, carbon dioxide gas is released. What volume of
carbon dioxide (measured at 25oC and 100 kPa) is released in the production of 112 kg
of quicklime?
(i)
Write a balanced equation for the reaction, and hence indicate mole ratios.
(ii)
How many moles of quicklime are produced?
(iii)
How many moles of carbon dioxide are produced?
(iv)
What volume of carbon dioxide is produced?
109
2.
When hydrogen gas, H2 is needed in the laboratory it is commonly produced by the
reaction of dilute hydrochloric acid on zinc metal. What volume of hydrogen gas
o
could be produced at 25 C and 100 kPa, when 13.0 g of zinc is reacted with
excess hydrochloric acid.
(i)
Write a balanced equation for the reaction, and hence indicate mole ratios.
(ii)
How many moles of zinc metal are used?
(iii)
How many moles of hydrogen will be produced?
(iv)
What volume of hydrogen will be produced?
3. Silver metal can be reclaimed from wastes containing silver chloride by reacting them
with hydrogen gas. What volume of hydrogen (measured at 25 oC and
100 kPa) would be needed to change 28.7 g of AgCl to silver metal?
AgCl(s) +
H2(g)
Ag(s)
(i)
Balance the equation above.
(ii)
Write the mole ratio under the balanced equation.
(iii)
How many moles of AgCl need to be reacted?
(iv)
How many moles of H2 will be required?
(v)
What is this volume of H2?
+
HCl(g)
4. Acetylene, C2H2 is an important fuel which is used in welding. We can generate
acetylene in the laboratory by reacting calcium carbide, CaC2 with water. Calculate the
volume of acetylene released at 25oC and 100 kPa if 0.450 mol of calcium carbide is
reacted with excess water.
CaC2(s)
+
H2O(l)
Ca(OH)2(aq) + C2H2(g)
See question structure over page
110
(i) Balance the equation.
(ii) Write the mole ratio under the balanced equation.
(iii) How many moles of acetylene will one mole of CaC2 produce?
(iv) How many moles of acetylene will 0.450 moles of CaC2 produce?
(v) What is this volume of acetylene?
Another bad joke:
Did you hear about the cat which accidently consumed a meal of calcium carbide after falling
pregnant? It later had acetylene kittens!
8.7
A Quick note on Kelvin . . .
Temperature is measured on a scale of degrees Celsius, (oC) as we are all familiar with.
Many calculations encountered in chemistry (and physics) often require values of
temperature to be measured not in oC, rather in Kelvin, (K).
Where does temperature start and finish? Cold temperatures could be measured in the order
of –10 oC in a very cold environment, or maybe even –150 oC when some gases liquefy.
Air temperature is around 25 oC, a volcano might be 1000 oC  5000 oC, whereas the core
temperature of a star might be in the order of 1,000,000 oC or more!
In short, temperature has no maximum value, but does have a minimum value, referred to
as Absolute Zero. Absolute Zero has been scientifically calculated to be –273.16 oC. This is
normally rounded to –273 oC. It is not possible to achieve a temperature colder than this!
Kelvin is referred to as an absolute measure of temperature, and therefore has a minimum
value of 0 Kelvin, or 0 K. This is equivalent to the Celsius value identified above as
–273 oC.
Hence: 0 K = –273 oC
Any calculation requiring conversion of oC  K is simply a matter of adding 273 to the
value of oC.
Kelvin = oC + 273
111
Yield Experiment 4 – GAS VOLUMES
Background:
Predicting mass yields in chemical processes via mole based quantities is by now a familiar
concept. Gases produced in chemical reactions have until now been accounted for as a
mass loss. However there is a directly proportional relationship between the amount of any
gas produced and its volume. The molar volume of any gas is identical whilst temperature
and pressure remain unchanged. Most laboratories operate in an environment close to
Standard Laboratory Conditions (S.L.C.)
Prework:
Part A – Carbon Dioxide Volume Yield:
Accurately weigh approximately 0.2 g of Calcium Carbonate and record in the table over
page. Based on the mass you weighed out, calculate the volume yield of Carbon Dioxide
expected at S.L.C.
Part B – Hydrogen Gas Volume Yield
Accurately weigh approximately 0.05 g of Aluminium metal and record in the table over page.
Based on the mass you weighed out, calculate the volume yield of Hydrogen gas expected at
S.L.C.
PROCEDURE:
Perform the general procedure below separately for both Parts A and B.
Set up equipment required for transferring and collecting the gas produced into an
inverted 100 mL measuring cylinder containing tap water.
Record the initial volume of air in the inverted measuring cylinder.
112
Measure around 20 mL of 5.0 M HCl. Transfer into the conical flask and immediately
attach bored stopper.
Ensure the gas displaced is successfully transferred into the inverted measuring
cylinder.
Record the final volume of gas.
Report the volume of gas displaced, and % Yield based on this result.
RESULTS:
PART A:
Calcium Carbonate + HCl
PART B:
Aluminium + HCl
Mass weighed (g)
Initial Volume in Measuring
Cylinder (mL)
Final Volume in Measuring
Cylinder (mL)
Gas Volume Displaced (mL)
CALCULATIONS
% Yield = Obtained Yield
x 100
Theoretical Yield
Percentage Yield:
Calculate the Percentage Yield of obtained gas, based on the volume you obtained
experimentally.
PART A: Calcium Carbonate + HCl
PART B: Aluminium + HCl
QUESTIONS:
1.
Identify possible errors in the design and execution of this experiment.
2.
The gas collected in the measuring is a mixture of displaced air and carbon
dioxide. Does this contribute errors in your volume measurements? Explain.
3.
Summarise the conditions of S.T.P. and S.L.C. Identify what both conditions stand
for and their respective conditions of pressure and temperature.
113
8.8
Percentage Composition
Any chemical compound has a set chemical formula and therefore set percentages of
elements by mass in it. If we can determine the percentage composition by mass of a
compound in a laboratory, we can often identify it.
A chemical formula makes it possible to calculate the percentage total of each element.
% element
in compound
=
total mass component of element
formula weight of compound
x 100
Consider water as an example:
H2O contains two elements, hydrogen and oxygen. From the formula, there are twice as
many hydrogen atoms as oxygen atoms. However, hydrogen atoms are much lighter than
oxygen atoms, so the fraction of the mass due to oxygen will presumably be much greater
than one third!
Formula weight
H2O =
% hydrogen =
% oxygen
(2 x 1.008) +
(1 x 16.00)
2 x 1.008
18.016
x
100
x
100
=
11.2 %
=
1 x 16.00
18.016
=
88.8 %
=
18.016
Of course, we did not need to calculate % oxygen once we knew % hydrogen! Since oxygen
is the only remaining element in water:
% oxygen
Written exercise:
=
100% – 11.2%
= 88.8%
Percentage composition
1. Calculate the percentage composition by mass of each element in the following
compounds.
(a) CO2
(b) H2SO4
(c) C6H12O6
114
2. Zinc and iodine react to form a compound, zinc iodide, ZnI2. Calculate the percentage by
mass of each element in zinc iodide.
3. Mercury reacts to form two different chlorides; mercury (I) chloride, Hg2Cl2 and mercury
(II) chloride HgCl2. Mercury (II) chloride is more toxic although both must be handled with
care. Calculate the percentage composition by mass of Hg and Cl in each compound.
8.9
Empirical and Molecular Formulae
The Empirical Formula of a compound is the simplest whole number ratio of elements
composing a compound. The Molecular Formula is the true chemical formula of the
compound.
Octane is an organic compound having a molecular formula of C8H18. Its empirical formula
would be C4H9.
Propane is another organic compound having a molecular formula of C 3H8. Its empirical
formula is also C3H8 as it cannot be simplified any further.
You can calculate the molecular formula for a compound if you are given its elemental
percentage composition and its formula weight. Consider the following:
Example:
A compound in white wine has been analysed to give the following percentage composition.
C = 52.3%
H = 13.1%
C
O = 34.6%
:
H
:
O
Ratio by mass
52.3
13.1
34.6
Ratio by number
of atoms
52.3
12.01
13.1
1.008
34.6
16.00
4.36
13.1
2.16
4.36
2.16
13.1
2.16
2.16
2.16
2
:
6
:
(
atomic mass)
(
smallest number ratio)
1
Following these steps will calculate the empirical (simplest ratio) formula, ie C2H6O.
115
Its formula weight has been is known to be approximately 46. What is its molecular formula?
The empirical formula weight, C2H6O
=
=
(2 x 12.01) + (6 x 1.008) + 16.00
46.068
This is very close to the given formula mass and therefore the molecular formula is the same
as the empirical formula C2H6O. If the molecular FW was larger than the empirical formula
weight, then a multiplying factor would apply. If the molecular FW was say 92, which is 2x the
empirical FW, then this factor would apply to the empirical formula. It would hence be
C4H12O2.
Written Exercises:
Percentage Composition, Empirical and Molecular Formulae
1.
The percentage composition of an organic compound is 48.6% carbon, 8.10%
hydrogen and 43.2% oxygen. What is its empirical formula?
2.
A compound is analysed to give the following results:
H = 5.00%
F = 95.00%
Its formula weight was determined to be 40.0. Find its empirical and molecular
formula.
3.
The mass of phosphorus in an 18.3 g sample of a compound containing only
phosphorus and oxygen is 10.3 g. The formula weight of the compound is 220.
Find the empirical and molecular formula of the compound.
Ref: C1 pp211-213;
CC1 p202
116
Experiment:
Which Mercury Chloride?
Task:
You will be given a sample powder (approximately 3g) in a specimen tube. Your task is to
identify your given compound as mercury (I) chloride, Hg2Cl2 or mercury (II) chloride, HgCl2.
Both of these compounds are white powders. Use the percentage composition calculated in
the previous exercise to help you make your decision.
You will be able to determine the percentage of mercury in your compound by determining
the mass of the white powder (the compound) and then decomposing it to obtain the mercury
metal which can be weighed also.
% Hg =
mass of mercury metal
mass of compound
x
100
Decomposition of the mercury chloride is achieved by heating the substance with a mixture
of sodium hypophosphite and hydrochloric acid until the element mercury is produced. Do
not attempt to write a balanced equation for this reaction at this stage.
! Risk Assessment:
All compounds of mercury should be considered to be highly poisonous.
Don't breathe in the fumes from the solution – carry experiment out in a fume cupboard
and make sure that the mercury is covered by solution during heating.
Wash your hands thoroughly at the completion of the exercise and make sure all spillages
are cleaned up!
All mercury residues must be disposed of in the labelled waste collection containers.
Chemicals required:
Unknown sample tube containing approximately 3g of mercury(I) or mercury(II) chloride.
Sodium hypophosphite, 2g in 5 mL H2O + conc HCl 3 mL
methylated spirit 5 mL
Procedure:
1.
2.
3.
4.
5.
6.
7.
8.
9.
Obtain a sample tube containing a mercury chloride unknown and record the
identification code on the tube.
Weigh and record the mass of the sample tube, lid and contents.
Obtain a clean, dry evaporating basin. Label with your name. Weigh and record its
mass.
Carefully add the mercury chloride to the evaporating basin.
Collect 20 mL of distilled water and add the water to the evaporating basin. Re-weigh
the empty sample tube.
Add the sodium hypophosphite / hydrochloric acid solution to the contents of the
evaporating basin.
Place over a steam bath in a fume cupboard and stir to mix the reactants.
! Care - liquid is corrosive.
Continue heating and stirring until mercury collects in large silvery globules.
! Avoid inhaling vapour while heating.
When all the mercury globules have coalesced, remove the basin from the steam bath
and decant the colourless liquid.
117
10.
11.
12.
13.
Wash the contents of the basin twice with water and once with small amounts of
methylated spirits. Collect all washings in the waste container provided.
Dry the mercury by pulling pieces of filter paper through it as demonstrated.
Re-weigh the basin containing the dry mercury.
Return the mercury to the class collection bottle.
Results
MERCURY COMPOUND A
MERCURY COMPOUND B
ID Code:
Mass: Sample Tube +
compound
Mass: Sample Tube
(empty)
Mass: Mercury
Compound (g)
Mass: Evaporating Basin
(empty)
Mass: Evaporating Basin
+ Mercury
Mass: Mercury (g)
Mass: Chloride (g)
Calculations
MERCURY COMPOUND A
% Mercury =
% Chloride =
Mole Ratio Mercury =
Mole Ratio Chloride =
Empirical Compound
Formula
Molecular Compound
Formula
118
MERCURY COMPOUND B
Questions
1.
Why is it unlikely that your experimental composition was exactly the same as the
calculated values from the formula?
2.
Is your mercury chloride a covalent or ionic compound? Why?
3.
This exercise began with a mercury chloride, and resulted in mercury metal.
Why did we not need to measure the amount of chlorine produced?
4.
Name two (2) safety procedures which you needed to observe while completing this
experiment.
5.
Mercury is classified as a heavy metal and the term “heavy metal residues” often
appears in pollution reports. You may wish to find out about pollution by mercury
compounds, where they are found and how they are detected.
119
8.10
Calculating Concentration
Concentration is a measure of the quantity of one substance within another. Concentration
can be measured as a function of all 3 states of matter: solids, liquids and gases.
Liquids and gases are normally expressed as volume based concentrations. A liquid
containing one or more dissolved substances becomes a mixture by definition. A liquid
mixture is more commonly referred to as a solution.
A common example of a solution: seawater
Seawater, simplistically speaking contains salt dissolved in water, ie;
Seawater
=
Water
+
Solution
=
Solvent +
Salt
Solute
The Solvent is the main liquid component. The solute is normally the “ingredient of interest”
dissolved within the solvent and may be in the form of a solid, liquid or gas. As soon as any
substance becomes dissolved in a liquid, it is recognised as being aqueous, (aq) in its
physical form.
Solution concentration is more conventionally measured as the amount of solute dissolved
within a set volume of solution, (mL, 100 mL, or L)
The solute component is often expressed as a mass (g, mg, or ug) but can also be
measured as a volume where a liquid solute, eg: ethanol applies.
Mass based concentrations can be calculated using the following general equation;
Concentration
=
Amount of Solute
Volume of Solvent
Examples: Calculate the concentration of substances in the scenarios below.
1.
4.8 grams of salt dissolved in 2 litres of water
Concentration
2.
=
4.8 g
2L
= 2.4 g / L
456 milligrams of lead dissolved in 80 litres of water
Concentration
=
456 mg
80 L
= 5.7 mg / L
120
3.
18 mL of alcohol dissolved in 375 mL of beer
Concentration
=
18 mL
375 mL
= 0.048 mL / mL
This concentration might not sound overly meaningful, or even confusing as it is a figure
comparing mL with mL . . . .
Normally alcohol concentration is reported as a volume based percentage (%).
In simple terms, if there is 0.048 mL of alcohol in 1 mL of beer, then there would be 100 x as
much alcohol in 100 mL of beer . . . makes sense ?
This is essentially a unit conversion, but in short if you multiply the calculated concentration
above by 100, then a more meaningful alcohol concentration of
4.8 mL / 100 mL will be arrived at. This could also be reported as 4.8% v/v.
In conclusion, you will notice that the units of concentration corresponding with the
calculated number are determined by the individual measurement units of the solute
and solvent.
8.11
Molarity
A term not associated whatsoever with Morality!
We saw in 8.10 that solution concentration can be calculated and expressed in a variety of
different units of measurement. In one example, concentration was calculated from a mass
based solute (grams) and volume based solvent (measured in Litres), ie:
Concentration
(g / L)
=
Mass (g)
Volume (L)
We have also explored the mole concept, whereby a given mass (g) of pure substance can
be expressed as a quantity in moles.
It follows logic that if convert a mass (g) of solute instead into moles, then it would modify our
units of concentration, ie;
Concentration
(moles / L)
=
Moles
Volume (L)
This measurement of concentration is more conventionally referred to as a MOLARITY, and
is calculated using the following equation:
Mole Equation 2.
Molarity
=
Moles
Volume (L)
NOTE ! ! Volume is always measured in litres (L)
Shorthand:
M = n
V (L)
121
We can see that Molarity is a measurement of concentration in units of Moles / L. These
units can be further shortened to the symbol M, meaning Molar.
Either way, Molarity, Moles / L, Molar and M are all different representations of exactly
the same measurement!
To further clarify, a 1M concentration of any chemical solution will contain 1 mole of chemical
per 1 litre of solution. For example, a 1M solution of Sodium Hydroxide, NaOH would contain
1 mole of NaOH per litre of solution. This 1 mole of NaOH (FW 40.08 a.m.u.) per litre could
be expressed back to a mass based concentration of 40.08 g/L.
Practice Examples:
Use mole equation 2 to calculate the molarity of solution in the following examples:
i)
6 moles in 2 L
ii)
0.5 moles in 200 mL
iii)
0.0025 moles in 25 mL
The examples above were fairly straight forward, except for the requirement to convert units
of mL  L where identified.
The following 3 examples deal with a specific mass of compound dissolved in a given volume
of solution. Each mass will firstly need to be converted to moles (mole equation 1), before
calculating the molarity of solution.
iv)
58.5g of NaCl in 2L of
v)
4.0 g of NaOH in 200 mL
vi)
0.5 g of Na2CO3 in 50 mL
The six practice examples above all used the mole equation 2 in its basic form to calculate
the concentration of solution. However, another important application of this equation is to
assist us in preparing a solution of known concentration. For this to happen, a laboratory
technician will need to know 2 things about the solution to be prepared:
1.
2.
its desired Concentration (M), and
the Volume required
The task therefore becomes calculating the (3.) quantity of solute that will need to be
weighed and dissolved in solution.
122
These 3 pieces of data will also fit into the mole equation 2:
Molarity
=
moles ?
Volume (L)
The values of Molarity and Volume are known, making the value of moles as our unknown
quantity in this scenario. It will be necessary to rearrange this equation so that moles
becomes the subject of the formula, ie;
moles = Molarity x Volume (L)
This intermediate calculation gives us a quantity of moles that are required to be weighed
and dissolved, however we cannot weigh mole quantities of any substance!
We therefore refer back to our original mole equation 1 to convert our intermediate
calculation of moles into a mass, ie;
Moles =
mass (g) ?
FW
You will become familiar with solution preparation scenarios as requiring a 2 step calculation.
Practice examples help!
Example:
Explain how to make 500 mL of a 0.10 M NaCl solution.
Volume = 500 mL = 0.5 L
Concentration = 0.10 M
No of moles of NaCl =
=
0.10 x 0.5
0.050
Mass of NaCl
No of moles x FW
0.050
x 58.44
2.9 g
=
=
=
Therefore to make the desired solution we would need to accurately weigh 2.9 g of NaCl,
add water until the volume reaches 500 mL and mix well to dissolve. Normally a 500 mL
volumetric flask would be used for this procedure.
123
Written Exercises:
Solution calculations
1. Explain how to make the following solutions.
(a) 250 mL of a 0.100 M AgNO3 solution.
(b) 100 mL of a 0.33 M K2CO3 solution.
2. What is the resulting molarity of the following solutions?
(a) 9.20 g of C2H6O in 500 mL of solution.
(b) 1.06 g of LiCl in 10.0 L of solution.
(c) 1.31 g of Ba(NO3)2 in 250 mL of solution.
3. What mass of solute is contained in each of the following solutions?
(a) 250 mL of 0.010 M Na2CrO4 solution.
(b) 1000 L of 1.0 x 10-4 M AuCl3 solution.
(c) 1.0 mL of 18 M H2SO4.
4. What volume of a 0.200 M solution contains these masses of solute?
(a) 8.40 g of NaHCO3
(b) 5.95 g of KBr
(c) 1.17 g of NaCl
BONUS QUESTION:
Carbon dioxide reacts with limewater, Ca(OH)2(aq) as follows:
Ca(OH)2(aq) +
CO2(g)
CaCO3(s) + H2O(l)
100 mL of a 0.10 M Ca(OH)2 solution was used as a test for carbon dioxide. Assuming an
excess of CO2 gas is passed through the solution, what is the maximum mass of CaCO3
precipitate which could form?
124
8.12
Solution Preparation and Quantitative Dilution in the Laboratory
Written calculation and explanation of solution preparation in the classroom is one thing,
however achieving this skill in a laboratory practical sense is another thing altogether.
Knowledge of appropriate laboratory glassware, equipment and technique is another thing
completely!
Quantitative (ie accurate) dilution of an existing solution is another important role of a
laboratory chemist. Some methods of chemical analysis can require solutions to be
extremely diluted down, sometimes to the extent of being diluted down to one thousandth of
their original concentration or less! This cannot be achieved in a single dilution step alone, so
the technique of serial dilution is employed.
An upcoming experiment will get you to put into practice the theoretical and practical
concepts of solution preparation and dilution. So far, the calculation concepts of solution
preparation have been covered – so lets look at what quantitative dilution is all about!
The process of diluting a solution achieves accuracy through the careful use of quantitative
glassware including pipettes and volumetric flasks.
The extent of a single dilution can typically be done accurately in the range of 1:2 down to
1:100. The dilution equation relates the mathematical relationship between the original and
diluted solution:
Concentration 1 x Volume 1 = Concentration 2 x Volume 2
OR:
C1.V1 = C2.V2
This equation has a simple message – adding water to an existing solution (C1.V1) gives you
a higher volume of solution (V2) at a lower concentration (C2). The amount of solute in the
original solution in reality remains the same.
A dilution factor is a number multiplier that expresses the magnitude to which the solution
has been diluted. For example, if you quantitatively add 50.0 mL of a solution to a 100.0 mL
volumetric flask and fill to volume with distilled water, then the new solution is ½ the original
concentration. It is therefore twice as dilute, otherwise reported as 2x.
Scenario:
A student wished to make 100 mL of a potassium chloride solution of 5 g/L.
1. Calculate the mass of KCl required.
Hint:
125
Concentration = Mass (g)
Volume (L)
2. The student then pipetted 5 mL of this solution into a second 100 mL volumetric flask and
made it up to volume.
a)
What was the dilution factor applied here?
b)
Calculate the new concentration of the solution.
3. The student was required to dilute the original solution (5 g/L) down to a concentration of
1 g/L
a)
What dilution factor is required here?
b)
Briefly outline a means of achieving this dilution (ie; what glassware will you
use?)
4. A further 1:20 dilution of the 1 g/L solution in 3. above was performed.
a)
Briefly outline a means of achieving this dilution.
b)
Calculate:
Written Exercises:
1.
i)
the new solution concentration.
ii)
the overall dilution factor applied
Dilution Scenario Calculations
Calculate the dilution factor involved in the following scenarios:
i)
5 mL of solution is diluted to 50 mL
ii)
20 mL of solution is diluted to 500 mL
iii) 15 mL of solution is diluted to 250 mL
126
2.
Calculate the concentration of the diluted solution in the following:
i)
25 mL of 2.5 g/100 mL is diluted to 100 mL
ii)
15 mL of 2.5 mole/L is diluted to 2.5 L
iii)
5 mL of 400 mg/L is diluted to 200 mL
3.
Calculate the volume of original solution required to perform the following dilutions:
i)
How much 2000 mg/L solution is required to make 500 mL of 100 mg/L?
ii)
How much 25 %v/v solution is required to make 250 mL of 5 %v/v?
iii) How much 0.05 M solution is required to make 200 mL of 0.01 M?
Ref: DB pp27-29
127
Experiment – Solution Preparation and Serial Dilution
A: Sodium Chloride
Prework:
In the table below, briefly outline a method of solution preparation.
Test by Conductivity
Solution
Method
Your Sample
Comparison
Standard
50 g / L
NaCl
Next perform a series of 1:10 dilutions from the initial solution.
Serial
Dilutions
Aliquot
Volume
Volumetric
Flask Size
Diluted NaCl
concentration
NaCl
Test by Conductivity
Your
Sample
Comparison
Standard
Dilution 1
Dilution 2
Dilution 3
B: Potassium Permanganate
Prework:
In the table below, briefly outline a method of solution preparation.
Test by Absorbance
Prepare a
Solution
Method
Your Sample
1000 mg / L
KMnO4
Next perform a series of specified dilutions from the initial solution (over page)
128
Comparison
Standard
Serial
Dilutions
Aliquot
Volume
Volumetric
Flask Size
Diluted
KMnO4
Concentration
KMnO4
Test by Absorbance
Your Sample
Comparison
Standard
Dilution 1
1:10
Dilution
1:5
Dilution
1:4
Dilution
1:2
C: Hydrochloric Acid
Perform a series of 1:10 dilutions on an existing HCl solution.
The concentration of the original HCl solution is:
________________ M
Serial
Dilutions
Aliquot
Volume
Volumetric
Flask Size
Diluted
Concentration
Original HCl
Solution
N/A
N/A
N/A
Dilution 1
Dilution 2
Dilution 3
Dilution 4
129
pH
Yours
N/A
Comparison
Standard
9.
CHEMICAL ANALYSIS
Chemical analysis is a necessary requirement of many industrial processes, particularly
in satisfying government regulations. It is necessary for chemists to be able to identify the
composition of raw materials, manufactured products and substances which are
discharged as wastes. An integral component of a typical chemist’s role is in Quality
Control.
9.1
Qualitative Vs Quantitative Chemical Analysis
When determining the composition of a substance we need to find out:
 What chemicals are present? This identification process is called qualitative
analysis. Sometimes this is all that is required, but often we also need to know:
 how much of each chemical is present. This measurement process is called
quantitative analysis.
In the mercury chloride laboratory task you determined whether your unknown sample
contained mercury (I) or mercury (II) chloride. The methods you used involved quantitative
analysis since you measured masses of reactants and products in order to analyse your
sample.
Consider the following examples:
A jar of peanut butter has listed as ingredients: peanuts, vegetable oils, sugar salt.
This represents a qualitative analysis only. Normally ingredients are listed in order of
decreasing mass but no actual quantities have been given.
On a packet of pain relief capsules the following ingredients are listed:
Paracetamol:
500 mg ;
Codeine phosphate:
8 mg
This represents a quantitative analysis. Note that there may be other ingredients also
which are unlisted, since the manufacturers are only obliged to list active ingredients, not
fillers, binders etc.
Written Exercise / Research
1. Examine the labels of other containers for examples of qualitative and quantitative
analysis. List three further examples of each.
2. Care must be taken with some ingredients to make sure that their concentrations stay
within set limits. For example, the allowable pesticide residue limits in foods is carefully
controlled. Find another example and find out how this product is analysed.
What methods can we use in chemical analysis?
Chemists employ many methods in order to analyse the millions of different substances
recognised today. We will look at some examples from which you may be able to choose
for your major investigation in this module - some involve simple laboratory apparatus
130
while others will involve sophisticated instruments. You will need to adapt any procedure
given to make use of the resources available to you. Consult a chemistry teacher once
you have made your choice. You should practice as many techniques as possible.
Consult references to find out about other techniques.
9.2
Gravimetric Analysis
Gravimetric analysis involves the separation and weighing of a substance obtained from
a sample. This form of analysis separates the substance of interest from the sample
based on its physical or chemical properties. The concentration of separated substance
is calculated and reported as a percentage by mass.
Often there are three main steps to this procedure:
1. Obtain and weigh sample mass
2. Employ means of physical or chemical separation of the analyte
3. Reweigh sample (or separated analyte component)
Gravimetric analysis in its most simple form could be the analysis of the moisture content
of say, a food substance like cereal. A suitable sample container needs to be chosen as it
must not change mass during the analysis.
An accurately weighed mass of cereal is placed into the container and heated in air oven
at around 120 oC for a minimum time period to evaporate the entire moisture content. The
container is removed, cooled and reweighed. The difference in mass (confidently
assuming it became lighter !) is calculated and expressed as a percentage of the sample
mass.
All calculations in moisture determinations are purely mass based, but there are further
examples of gravimetric analysis involving mole based calculations . . .
If we were to determine the salt content of a food sample we must first dissolve the food
in water. Any undissolved substance will need to be removed using a separation method
such as filtration. The food solution can be analysed for salt, as NaCl by adding an excess
of AgNO3 solution. This precipitates out the entire chloride ion, Cl- (aq) content as AgCl (s)
according to the following chemical equation:
and then filtering and weighing the precipitate which forms:
Ag+(aq)
+
Cl-(aq)
AgCl(s)
The precipitate is separated by filtration. After the precipitate is dried, it is then weighed
and its mass recorded.
The number of moles of AgCl formed can be calculated. From the equation, this is the
same as the number of moles of Cl- in the food solution.
This is therefore equal to the number of moles of salt, NaCl in the food sample analysed.
We can convert this number of moles to a mass and express it as a percentage of the
mass of the original sample.
131
Example:
A 25.0 mL sample of soup was analysed for its salt content, as NaCl. The sample was
filtered to remove any suspended solids. An excess of Silver nitrate solution was added to
the filtrate to precipitate out the available salt component. After standing for a few minutes
a further few drops of silver nitrate were added to make sure that precipitation was
complete. The precipitate was collected by filtration, then dried and weighed. A mass of
0.23g of precipitate was obtained. Find the percentage (g / 100 mL) of salt in the soup.
No of moles of AgCl produced
Chemical equation:
+
(aq)
Ag
+
From the equation above, moles Cl-
=
mass of AgCl
FW
=
0.23
143.4
=
1.60 x 10-3
-
Cl (aq)
AgCl(s)
=
moles of AgCl
=
1.60 x 10-3
We are assuming that all the chloride ions present came from dissolved salt - NaCl.
This is a reasonable assumption in most food samples.
Therefore, moles of NaCl in the soup sample
=
mass of NaCl in the soup sample =
% of NaCl in soup sample
1.60 x 10-3
moles x FW
=
1.60 x 10-3 x 58.45
=
0.0938
=
0.0938 x 100
25.0
=
0.38 % w/v
You should note the following important points:
solid samples need to be dissolved so that the substance to be detected is in solution.
Any undissolved material should be removed by filtration.
the precipitate formed needs to be a highly insoluble substance, so that the proportion
of soluble ions remaining in solution for analysis is negligible. Check the table of
solubility rules, under the chapter “properties of compounds” – your teacher will assist
you with this!
the filtered precipitate needs to be washed carefully with small volumes of water to
remove other soluble substances present in the solution.
the precipitate should be dried in an oven at 110 oC until constant mass is obtained
and allowed to cool before weighing.
132
Practice Questions:
1. A student analysed a popular breakfast cereal for its moisture content . The standard
procedure was followed and results were listed in the student's logbook. Determine the
moisture content:
Empty container mass:
25.3403g
Mass container + cereal:
29.3489g
Mass container + cereal (2hr @ 110 oC):
29.2897g
2. The Limestone, CaCO3 content of a rock sample was analysed by dissolving 8.3917g
of rock into a solution of HCl. The CO2 evolved was collected in a special sodaasbestos tube. It gained in mass by 2.4918g. Calculate the %w/w of limestone in the
sample.
BONUS QUESTION
3. Aluminium can be analysed by precipitation with the organic compound
8-hydroxyquinoline, C9H6NO. A 0.3271g of a bauxite ore was analysed for its
aluminium content. A mass of 1.4827g of the aluminium precipitate, Al(C9H6NO)3 was
recovered. Determine the percentage of Aluminium in the ore.
133
Experiment – Sulfate Content of Plant Fertilizer
Background:
Sulfate is an important nutrient for plant growth. It is added to many
fertilizer materials, present in soluble forms, most notably as
Ammonium Sulfate. This method aims to dissolve the sulfate
component of a fertilizer sample into a slightly acidified aqueous
solution. The available Sulfate, SO4 2- (aq) will be precipitated, filtered and weighed as
BaSO4 (s) by addition of a soluble Barium Chloride solution, BaCl2 (aq)
Method:
1. Obtain a 600 mL beaker, place onto a 3 decimal place balance and tare.
2. Accurately weigh approximately 0.75g of fertiliser and record this mass into the
results table below.
3. Dissolve the weighed material in around 100 mL of distilled water.
4. Add around 5 mL of the supplied 2M HCl and heat to boiling on a hotplate.
5. Remove from heat. Filter solution through #4 (fast) filter paper. Ensure quantitative
transfer and rinse final residue in filter paper with distilled water.
6. Place filtered solution onto a hotplate and heat until just boiling.
7. Remove again from heat and add 25 mL of 5% BaCl2.
8. Cover beaker with a watchglass and transfer solution onto a hotplate (! low heat).
9. Add a further 2 mL of the BaCl2 solution after 10 min.
10. Continue to digest your solution on low heat for a further 20 min.
11. Obtain and record the mass of a sintered glass crucible in the results table.
12. Remove from hotplate and cool in an ice bath for 10 min.
13. Set up vacuum filtration apparatus and filter your solution through the pre-weighed
sintered glass crucible. Wash remaining residue with a small volume of distilled
water.
14. Place into an air oven to dry and leave until next
class to weigh back.
15. Remove from oven. Allow 10min to cool and reweigh.
Results:
Mass (g): Fertiliser
Mass (g): Sintered Glass crucible
Mass (g): Sintered Glass Crucible + BaSO4 (s)
Mass (g): BaSO4 (s)
134
Calculations:
Moles BaSO4 (s)
Balanced Chemical Equation (precipitation process):
Reaction Ratio BaSO4 : SO4 2-
=
_________ : __________
Moles SO4 2Mass SO4 2-
% SO4 2-
% SO4
2-
=
2-
Mass SO4
Sample Mass
x 100
Questions:
1 a)
Why is it important to initially filter and remove any undissolved fertilizer in step 5 ?
1 b)
Why would this undesirable solid residue be washed with distilled water?
2)
Explain the purpose of cooling the precipitate prior to filtering in step 12.
3)
Explain the importance of washing the solid BaSO4 compound with distilled water
prior to drying in step 13
4)
Why was the measured volume of distilled water (ie 100 mL) instructed as
approximate and not accurate?
135
9.3
Volumetric Analysis
Unknown concentrations of soluble chemicals in aqueous solution can be determined by a
technique referred to as Volumetric Analysis, or Titration. A carefully measured volume of
the unknown solution is transferred to a conical flask by pipette. This volume is referred
to as an aliquot.
A second chemical solution is chosen that is known to chemically react with the unknown
solution in a predictable manner, via a balanced chemical reaction. This solution is must
have a known concentration as it is referred to as a standard solution. A burette is
filled with the standard solution where it is then slowly delivered into the aliquot of
unknown solution, allowing the chemical reaction to progressively “consume” the chemical
analyte within the unknown solution aliquot.
This method of analysis continues to deliver standard solution into the unknown solution
aliquot (in the conical flask) until all of the chemical analyte in the unknown solution is
consumed. At this point the chemical analysis is complete, and addition of further solution
is ceased. The volume of standard solution delivered to the unknown solution by burette is
carefully measured and recorded. This is referred to as an endpoint volume.
Diagram of basic
Titration apparatus:
Bulb Pipette
Apparatus
The fundamental basis of all volumetric analyses is the chemical reaction. It is essential
that a balanced chemical equation be expressed as part of the calculation procedure.
Generic Equation:
HX(aq) +
1
YOH(aq)
:
XY (aq) + H2O (l)
1
In this example the reaction ratio, and hence mole ratio is 1:1, however different chemical
reactants may have mole ratios like 1:2, 1:3, 3:2 and so on. This ratio is an integral part
of the calculation procedure in determining the unknown solution concentration.
136
Volumetric analysis (titration) is commonly employed to determine the concentrations of
dilute acids and bases. As covered previously, acids and bases react together
predictably in what is referred to as a neutralisation reaction, ie;
Acid + Base  Salt + Water
Avoid potential contamination – rinse your glassware correctly prior to use:
Burettes require a small sacrificial rinse with the solution it will be filled with.
Pipettes require a small sacrificial rinse with the solution it will deliver.
Conical flasks require a rinse with distilled water only.
Volumetric Flasks (used sometimes for titration) require a rinse with distilled water only.
The general steps involved in a titration, in summary are:
A measured pipette volume (aliquot) of the unknown solution is added to a conical
flask.
A few drops of a pH sensitive indicator are normally added to the conical flask to
produce a noticeable colour change at endpoint.
The burette is filled with the standard solution and volume adjusted to the top volume
reading of 0 mL.
The standard solution is carefully delivered to the flask. As the colour change becomes
more apparent, the delivery rate from the burette is slowed from a steady stream to
smaller “squirts” and ultimately dropwise until a permanent colour change is observed.
Cease delivering standard solution from burette. Record burette volume. This is the
volume of standard solution that was required to consume the last of the analyte in the
standard solution.
There are 4 measurable quantities recorded in a single titration analysis. 3 of them are
known quantities and 1 is unknown initially. Place a tick in the boxes below which
correspond with known quantities. Place a cross in the box corresponding with the value
to be calculated (unknown).
Burette
Conical flask
 Standard solution:
 Unknown solution
aliquot
Volume
Volume
Concentration
Concentration
137
The titration technique requires considerable practice in order to obtain reliable results.
Volumetric analysis is commonly performed in triplicate. It is desirable that your 3 titration
volume results are within 0.2 mL  0.3 mL. This can be difficult to achieve initially!
Titrations are used in many situations such as when we want to find the proportion of
particular pollutants present in industrial waste water, or the proportion of vitamin C in
commercially produced fruit juices.
There are a variety of other chemical reaction types that can be used for the technique of
volumetric analysis including:
 Reduction – Oxidation (Redox: Covered in TPC Chemistry B)
 Complexometric , and
 Precipitation
(TAFE: Cert IV / Diploma in Laboratory Techniques)
The following examples use acid / base reactions although many other types of chemical
reactions can be used in volumetric analysis.
Example 1
A 20.0 mL aliquot of hydrochloric acid, HCl was titrated to endpoint with 15.5 mL of a
standardised 0.112 M solution of sodium carbonate, Na2CO3. Determine the concentration
of the HCl.
moles of Na2CO3 titrated =
Concentration x Volume (L)
=
0.112
=
Equation:
Mole ratio:
2HCl(aq)
2
+
:
1.74 x 10
x 0.0155
-3
Na2CO3(aq)
2NaCl(aq)
+ H2O(l) + CO2(g)
1
The mole ratio above indicates there will be twice as many moles of HCl than there will be
of Na2CO3.
Therefore, moles of HCl reacted
Concentration of HCl
138
=
2 x 1.74 x 10-3
=
3.47 x 10-3
=
moles HCl
Volume (L)
=
3.47 x 10
0.02 L
=
0.174 M
-3
(ie; moles / L)
Example 2
How much ammonia (a base) is in ‘Cloudy Ammonia’ ?
A 25.0 mL aliquot of ‘Cloudy Ammonia’ was diluted to 250.0 mL in a volumetric flask.
A 25.0 mL aliquot of this diluted solution was transferred by pipette into a conical flask.
A few drops of methyl orange indicator were added (a yellow solution results).
A standardised 0.099 M HCl solution was delivered by burette with constant swirling of the
flask. The titration was completed when the colour of the solution just changed to a
permanent tinge of apricot-pink.
The volume of HCl added was noted. The procedure was repeated three times until
consistent results were obtained.
Results:
Titration No
Volume HCl added (mL)
1
28.90
2
27.60
3
27.60
4
27.65
Titration 1 result is ignored as it is not consistent with the following titrations. The results
from titrations 2, 3 and 4 are used to find an average titre.
Average titre = 27.60 + 27.60 + 27.65
3
moles of HCl titrated
Equation:
Mole ratio:
Therefore:
HCl(aq) +
1
:
=
=
=
= 27.62 mL
concentration x volume (L)
0.0991
x 0.02762 L
-3
2.74 x 10
NH3(aq)
NH4Cl(aq)
1
moles of NH3
= 2.74 x 10-3 (since reaction ratio is 1:1)
These 2.74 x 10-3 moles of NH3 are contained in the 25.0 mL titrated.
Concentration of NH3 in the diluted solution
=
moles NH3
Volume (L)
=
2.74 x 10-3
0.025 L
=
0.110 M **
** The original sample had been diluted by 10x in the 250 mL volumetric flask.
Therefore, the concentration of NH3 in the ‘Cloudy Ammonia’
= 0.110 M x 10
= 1.10 M
139
TITRATION CALCULATION STEPS
Remember - a titration is always a chemical reaction between 2 reacting solutions:
A Solution of KNOWN concentration (Standard Reagent) or quantity (Primary
Standard) is fully titrated with your solution of UNKNOWN concentration.
REMEMBER MOLE EQUATIONS:
Equation 1. Moles = Mass (g)
FW
Equation 2. Molarity = Moles
Volume (L)
1. You must firstly read the scenario carefully and identify your KNOWN reagent and
UNKNOWN reagent.
2. Calculate moles of Known Reagent
3. Write down a BALANCED CHEMICAL EQUATION between the 2 reagents:
Acid + Base  Salt + H2O
Acid + Carbonate  Salt + CO2 + H2O
4. Determine and write down the REACTION RATIO between your Known and Unknown
reagent.
5. Calculate MOLES of your Unknown Reagent:

Multiply or Divide your Moles of KNOWN reagent (step 1) by the
Reaction Ratio (step 3).
6. Calculate the MOLARITY of your Unknown Reagent:
MOLARITY = Moles Unknown Reagent
Volume (L) Unknown Reagent
140
Practice Questions:
1.
Determine the concentration of the unknown solution given the following data:
a)
25.0 mL of NaOH is titrated to endpoint with 18.3 mL of 0.1050M HCl.
b)
10.0 mL of H2SO4 is titrated to endpoint with 24.8 mL of 0.0978M NaOH.
2.
A solution of approximately 0.1M HCl was standardised via sodium carbonate,
Na2CO3 as the primary standard. A mass of 0.1472 g of the primary standard was
weighed and required 23.7 mL of the HCl to reach endpoint. Calculate its
concentration.
3.
A sample of vinegar, thought to contain around 4% Ethanoic Acid, CH3COOH was
analysed by titration with a standardised 0.1082M NaOH solution. A 10 mL aliquot of
the vinegar was diluted to 100 mL in a volumetric flask. Triplicate 25 mL aliquots of
this diluted solution were titrated to endpoint by an average of 18.3 mL of the
standardised NaOH. Determine the Ethanoic Acid content (%w/v) of the vinegar.
141
Experiment 1 – PRACTICE TITRATION AND HCl ANALYSIS
Part A:
Indicator Colour
Record colours of the following indicators as they apply to a typical acid – base
titration.
Screened Methyl
Orange
Methyl Orange
Phenolphthalein
Acid (HCl)
Base (NaOH)
Endpoint
Part B:
Practice Titration
! Be sure to rinse all glassware appropriately prior to use . . .
Obtain triplicate 25.0 mL volumes of HCl (0.1M) by pipette and place into 3 conical
flasks. Add 3-5 drops of phenolphthalein indicator.
Fill your burette with with NaOH (0.1M) to the 0mL mark. (don’t forget sacrificial
rinsings where required….)
Titrate your solutions to endpoint placing emphasis on the precision of your
results. Aim to titrate each set of 3 analyses within 0.2 mL  0.3 mL.
Repeat titration steps above using screened methyl orange indicator and if time,
methyl orange indicator.
RESULTS
NaOH concentration: ________________ M
NaOH Titration Volume (mL)
HCl Aliquot
Volume (mL)
1. Phenolphthalein
2. Screened MO
25.0
25.0
25.0
AVERAGE
142
3. MO
Part C:
HCl Analysis – A Titration Calculation
Choose the best set of triplicate titration results from Part B above. This would be
the indicator which achieved the smallest titration volume range.
Use the average titration value of this indicator to calculate the molarity of the
hydrochloric acid, following steps 1 – 5 below:
Calculations:
1. Moles NaOH Titrated: = Molarity x Volume (L)
2. Chemical Equation: HCl + NaOH  _______________ + _______________
3. Reaction Ratio:
____ : ____
4. Moles HCl =
5. Molarity HCl: = Moles
Volume(L)
QUESTIONS
1. Summarise the rinsing procedure(s) for the following pieces of laboratory
equipment:
Burette
Conical Flask
Pipette
2. How is endpoint different to equivalence point when performing volumetric analysis
(ie titration)?
143
Experiment 2 – HCl STANDARDISATION
Aim: To standardise a solution of hydrochloric acid using the primary standard Na2CO3.
Prework:
Calculate the mass of Sodium Carbonate, Na2CO3 required to react with 25 mL of an
approximately 0.1M HCl solution.
Method:
HCl Standardisation
Weigh accurately, approximately the mass of Na2CO3 calculated above into 3 conical
flasks. ! Label 1 – 3.
Dissolve in a small volume of water. Add 5 drops of the methyl orange indicator.
Fill your burette with the HCl solution to be standardised.
Titrate your standard Na2CO3 solution with the HCl until endpoint is reached.
Titrate your standard Na2CO3 solution with the HCl until endpoint is reached. Record
this volume and repeat for flasks 2 and 3
Results:
Mass Na2CO3 (g)
HCl Volume (mL)
Calculations: Perform separately for each titration result.
i)
Calculate moles of sodium carbonate
ii)
Write a balanced chemical reaction, and hence determine the reaction ratio.
iii)
Calculate moles of HCl
iv)
Calculate the molarity of HCl
Questions:
1. Why are titration volumes between 15 – 30 mL the most desirable for titrations?
2. Why is it OK to average the titration volumes performed on triplicate analyses of
solution aliquots, but NOT OK for 3 separate masses of Sodium Carbonate carefully
weighed and dissolved within 3 separate conical flasks?
144
Experiment 3 – NaOH STANDARDISATION
Aim: To standardise a solution of sodium hydroxide using the primary standard KC8H5O4
(Potassium Hydrogen Phthalate, KHP)
Prework:
Calculate the mass of Potassium Hydrogen Phthalate, KC8H5O4 required to react with
25 mL of an approximately 0.1M NaOH solution.
Method:
Weigh accurately, approximately the mass of KC8H5O4 calculated above into 3 conical
flasks. ! Label 1 – 3.
Dissolve in a small volume of water. Add 5 drops of the phenolphthalein indicator.
Fill your burette with the NaOH solution to be standardised.
Titrate your standard KC8H5O4 solution with the NaOH until endpoint is reached.
Record this volume and repeat for flasks 2 and 3
Results:
Mass KC8H5O4 (g)
NaOH Volume (mL)
Calculations: Perform separately for each titration result.
i)
Calculate moles of KHP
ii)
Write a balanced chemical reaction, and hence determine the reaction ratio.
iii)
Calculate moles of NaOH
iv)
Calculate the molarity of NaOH
145
Experiment 4 – VINEGAR ANALYSIS
Aim: To determine the concentration of Ethanoic Acid, CH3COOH in Vinegar.
Method:
1. Perform a 10x dilution on a vinegar sample by transferring a 25.0 mL aliquot of vinegar
sample into a 250.0 mL volumetric flask and making up to volume.
2. Transfer 25.0 mL aliquots of the diluted vinegar in triplicate by pipette into 3 conical
flasks.
3. Add 3 – 5 drops of phenolphthalein indicator to each flask and titrate with standardised
sodium hydroxide, NaOH reagent.
Results Table:
NaOH = ____________________ M
Aliquot volume (mL)
Volume NaOH (mL)
25.0
25.0
25.0
Volume NaOH (Average)
Calculate the concentration of the vinegar in %w/v….
1. Moles NaOH titrated =
2. Chemical Reaction (and hence reaction ratio):
3. Moles CH3COOH present in 25.0 mL (diluted) aliquot =
4. Molarity CH3COOH (diluted) =
5. Dilution Factor =
6. Molarity CH3COOH (undiluted solution) =
7. % w/v, (ie g/100 mL vinegar) = ……………..
i)
Moles / L =
ii) g / L =
iii) g /100 mL =
146
Experiment 5 – MISCELLANEOUS TITRATION: RESULTS PROFORMA
1. Sample ID:
4. Total sample volume prepared:
_________________________________
________________________________
2. Analyte ID:
5. Titrant ID:
_________________________________
________________________________
3. Sample amount taken:
6. Titrant molarity:
_________________________________
________________________________
Aliquot (mL)
Titrant Volume (mL)
1.
2.
3.
Average Titrant Volume
1. Calculation of Analyte Concentration (M)
i) Moles of titrant:
ii) Balanced chemical reaction and reaction ratio:
iii) Moles of analyte in sample solution:
Calculations continued over page
147
iv) Molarity of sample solution:
v) Dilution factor (if applicable)
2. Report your analyte concentration as a percentage, ie;
Liquid samples as % w/v, (ie g/100 mL)
i)
Moles / L =
ii)
g/L=
iii)
g /100 mL = _______________
OR
Solid samples % w/w, (ie g/100 g)
i)
Moles / L =
ii)
g/L=
iii)
g / total sample volume (volumetric flask)
iv)
g / sample amount taken
v)
% w/w = ________________
148
9.4
Chromatography
Chromatography is a procedure for separating mixtures containing several dissolved
substances. Each dissolved substance has a difference in attraction for the solvent,
referred to as the mobile phase, versus the blotting paper, referred to as the stationary
phase.
By comparing the distance moved by one substance in the mixture with that of a known
substance, we can identify the presence of that substance in the mixture.
You can separate some food colourings or inks using this method. Concentrated dots of
colour are placed along one edge of a piece of blotting or other absorbent paper as
shown. In this example, three coloured inks A, B and C and a known substance S are
spotted. The paper is suspended in a suitable solvent in a covered glass container for
several hours.
At the beginning
`
cover
covered glass
container
blotting paper
spots are small and concentrated
pencil line marking
origin
solvent level is below the
spots on the paper
A B C S
Known reference substance (S) can be spotted next to the substances to be tested.
After several hours
----------------
solvent front
distance
solvent front
has moved
known reference substance
(S)
A B C S
149
Substance A contains substance S and at least one other substance (since this solvent
may not have separated all the substances present). B and C do not contain S because
there are no spots at this height on the chromatogram.
The components of the inks can be identified by the distance they travel through the paper
compared to the distance that the solvent front has travelled. This is expressed as an R f
value for each component.
Rf = distance component spot has moved from the origin
distance solvent front has moved from the origin
solvent front
component P
component Q
Distance moved by Q
Distance moved by P
Distance moved by solvent
=
=
=
5.2 cm
11.8 cm
15.0 cm
Rf for component P =
11.8
15.0
=
0.79
Rf for component Q =
5.2
=
0.35
The solvent that is best for a particular separation is found by experimentation. You might
try various mixtures of water, alcohol (colourless eg vodka, white rum), white vinegar or
ammonia solutions.
Other absorbent materials such cotton or even a stick of chalk can be used for the
stationary phase of the chromatogram. Thin layer chromatography makes use of a fine
layer of silica or aluminium oxide on a glass plate. The absorbent material is packed in to
a glass column for column chromatography and a solvent is dripped through after the
sample is placed on top.
Practical Activity / research:
1.
2.
3.
Chromatography
Use chromatography to separate some pigments of your choice. In your report
comment on the effectiveness of your procedure and suggest improvements.
Research instrumental methods such as high performance liquid
chromatography and gas-liquid chromatography. Observe demonstrations of
these methods if possible.
Chromatography is used extensively in industry. Use some modern sources to
investigate one application.
Ref: CC1 pp172-174
C2 pp13-14
CC2 pp25-29; 58-61
150
9.5
Electrophoresis
Anything which conducts an electric current must contain mobile electric charge carriers.
Conduction can arise in two different ways: by movement of electrons or by movement of
ions. In metal wires, only electrons move through the rigid crystal lattice but this is not the
case in solutions where ions move. Solutions which contain ions and therefore conduct
electricity are called electrolytic solutions while any substance which produces ions in a
solution is an electrolyte.
Electrolytic solutions conduct because they contain charge-carriers (ions) which are
absent from solutions of non-electrolytes and virtually absent from water itself. The
electrically charged ions in the compound sodium chloride, Na+ & Cl- will conduct an
electric current if they are free to move. The ions can move if the compound is molten
(liquid) or dissolved in water.
Electrophoresis uses this movement of ions to separate them in a process which is, in
some ways is similar to chromatography. The positive ions are attracted to the wire and
metal clip attached to the negative terminal of the electricity source. The negative ions are
attracted to the oppositely charged positive electrode. Smaller ions can move faster than
larger ones and so a separation and identification can take place. If the ions are not
coloured they can be made visible by spraying with a suitable dye or other means.
Diagram
151
9.6
Colorimetry
The intensity of a coloured solution is directly related to its concentration. The more
intense the colour, the more concentrated the solution. A copper(II) sulfate solution of
unknown concentration is mid-blue. We can estimate its concentration by comparing its
colour intensity with other copper sulfate solutions of known concentrations. This analysis
technique is called colorimetry.
The colour intensity of solutions can be more reliably and accurately compared using an
instrument called a colorimeter than using our eyes. The instrument measures the
intensity of light which passes through the solution of unknown concentration. A series of
measurements is first carried out on the standard solutions and a calibration graph is
drawn. The concentration of the unknown solution can then be read from the graph.
Colourless substances need to be converted first to coloured compounds before simple
colorimetry can be used. Phosphates are converted to a blue complex by reacting them
with ammonium molybdate. In this way colorimetry can be used to calculate the amount of
phosphates in detergent samples.
Example: The concentration of a solution of dye orange-X needed to be determined.
Standard solutions of orange-X were made up using distilled water to the concentrations
shown in the table. A blank (plain distilled water) was measured for absorbance in a
colorimeter as was each of the standards.
Solution concentration (mg / L)
Absorption
0 (blank)
0.00
0.10
0.16
0.20
0.30
0.30
0.45
0.40
0.61
0.50
0.76
Unknown
0.37
A calibration graph has been plotted. Your task is to interpolate from the graph a
corresponding for the unknown solution.
See over page
152
Orange - X Analysis by Colorimetry
0.8
0.7
Absorbance
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
Concentration of Orange - X (mg / L)
From the calibration graph the concentration of the orange-X solution was
determined to be ................ mg / L.
Ref: CC2 pp38-39
C2 pp49-50
153
0.6
Experiment – COLORIMETRY
Aim:
To analyse the relationship between colour intensity and concentration of an analyte in
aqueous solution.
! Graph Paper required as part of this exercise
Method:
1. Devise a simple method for the preparation of a series of standard copper sulfate,
CuSO4 solutions from a 1.0 M stock solution:
2. Analyse standard solutions through the colorimeter. Record each standard
absorbance value in the results table.
3. Record the absorbance of your unknown(s).
4. Make a calibration graph from your CuSO4 standards
5. Determine the concentration of your unknown solutions.
Results:
Standard Solution
Method of Preparation
Absorbance
0M
0.2 M
0.4 M
0.6 M
0.8 M
Unknown Solution
Unknown Solution Concentration: _________________________
Questions:
1. Consider the accuracy of this method. Suggest ways in which it could be improved.
2. Discuss the use of a calibration graph as a mathematical model in determining the
concentration of your unknown.
154
9.7
Spectroscopy
Chemicals are capable of emitting or absorbing energy in the form of visible light,
ultraviolet and infrared radiation. By measuring the amount and types of energy absorbed
or emitted we have a method of identifying the chemical (qualitative analysis) and in some
cases determining the amount present (quantitative analysis).
Visible and U.V. spectra
An electron cannot move in an orbit of just any radius - there are in fact very few orbits
whose radius values are “legal”, and an electron must be in one or another of these. If an
electron in a “legal” orbit near the nucleus absorbs enough energy it can move to an orbit
which is further away. The electron is now said to be in an “excited” state and will usually
fall back fairly promptly into an empty “legal” orbit of lower energy closer to the nucleus.
When this happens the energy the electron no longer carries is released as a pulse of
radiant energy such as light.
This process can only be studied properly in a collection of isolated atoms, for in
polyatomic molecules or in condensed phases (solids or liquids) adjacent atoms interfere
with the free movement of outer electrons. Ways of obtaining isolated atoms include:
1)
2)
vaporising a substance at high temperatures (eg in a bunsen flame);
subjecting a gas at low pressure to an electric discharge (eg in a fluorescent tube).
Both processes result in a collection of isolated gaseous atoms in excited states which will
release light or other radiant energies as they revert to less energetic states.
Common applications of the principle are the red neon advertising signs and both mercury
(blue) and sodium (orange-yellow) streetlights.
Each element produces energies of a characteristic pattern which makes this
phenomenon a powerful tool in chemical analysis.
The instrument for examining the wavelengths of these energies is a spectroscope. An
incandescent liquid (eg molten iron) or solid (eg the tungsten filament of an electric light
globe) when viewed through a spectroscope shows a “continuous” spectrum, ie; a rainbow
of colours with no gaps, ranging from red to violet. Incandescent gaseous atoms however
produce light at only certain sharply-defined positions in the spectrum (only at specific
wavelengths), with darkness in between. Through the spectroscope discrete bright
coloured lines appear across a dark background. It is the number and positions (colours)
of these lines which constitute the unique “fingerprint” of each element called its emission
spectrum.
Continuous spectrum
155
Sodium spectrum
Hydrogen spectrum
It is also possible to analyse which frequencies of light from the continuous spectrum are
absorbed by a particular element - an absorption spectrum. The spectrum appears as the
continuous spectrum with black lines across it corresponding to those frequencies which
would be emitted by the same element (ie. a reverse of the emission spectrum).
Flame Emission Spectrophotometry
Many metal compounds at the high temperature of a bunsen flame or similar produce
excited metal atoms whose emission spectra are analytically useful. The technique is very
simple; the only apparatus required is:
· a burner which can be adjusted to give an almost non-luminous flame
· a platinum wire holder
or
· a plastic squeeze bottle with a very fine nozzle from which a fine spray of a solution of
the compound may be delivered.
Either method will result in a coloured flame characteristic of a particular element. Not all
metals emit energy in the visible light region of the electromagnetic spectrum. Magnesium,
for example, does not. However for those that do, the colour may be used to identify the
metal either directly by observation or by spectroscopic analysis. Non-metals do not emit
in this way. Each metal gives off its own unique spectrum even if other metals are present
in the specimen as well. The spectrum produced will contain all the lines from each
individual metal’s spectrum.
Potassium alone
Sodium alone
If the colour of one metal masks that of another because of overwhelming brightness,
spectroscopic examination will still reveal the presence of the fainter one. Sodium, for
example, generally outshines all others, and a mixed sodium-potassium specimen will give
a flame colour which looks yellow like the sodium alone to the naked eye.
156
Spectroscopically however the flame appears like this:
Mixture (sodium-potassium)
Flame spectroscopy is the most convenient way of detecting the various “alkali metals”
(Group I) in their compounds. They have very similar properties and often occur naturally
together.
Written exercises: Visible spectra
1.
Why does salty water boiling over in a saucepan turn a gas flame yellow?
2.
How can hydrogen with only one electron produce multiple lines in its spectrum?
Why are some lines brighter than others?
3.
The following spectra are provided so that you can answer the questions below.
(a)
(b)
Identify the metallic elements in the unknown specimen.
Which alkali metal is most abundant in the atmosphere of the sun?
Experiment 1 – FLAME TESTS
The colour of light emitted when different metal ions are sprayed into a flame are
characteristic for each metal element, as it appears on the Periodic Table.
PART A - METAL ION ID
Metal Ion
Colour
157
PART B - UNKNOWN METAL ION IDENTIFICATION
Try matching an element to the colour of the unknown metal ion upon being aspirated into
the flame.
Unknown Metal Ion ID
Colour
Metal Ion
PART C - COLOUR INTENSITY AND METAL ION CONCENTRATION
1. Prepare a solution containing an approximately 10% concentration of Sodium
Chloride.
2. Aspirate this through the flame.
3. Dilute 10 mL of this solution up to 100 mL with distilled water.
4. Aspirate the diluted solution through the flame.
5. Repeat steps 1 – 4 another 5 times.
Summarise your observations of this task below.
Ref: CC1 p27 (flame tests)
C2 pp10-13
CC2 pp29; 383
158
Analysis of Sodium and Potassium by Flame Emission Spectrophotometry
Potassium (K+) and Sodium (Na+) are soluble metal ions commonly found in natural water
bodies. The ocean is a salt water environment. This salt is composed predominantly of
table salt, known chemically as Sodium Chloride (NaCl).
Sodium ions (Na+) exist in concentrations of around 1.2%, or 1.2 g/100 mL
Potassium (K+) ions exist at lower concentration - around 0.4%, or 0.4 g/100mL
We have seen in previous lessons how various metal ions release different colours of light
when placed into a hot gas flame. This is referred to as Light Emission.
A relationship exists between the concentration of soluble metal ion and colour intensity.
As concentration increases, flame colour emitted becomes brighter, or more INTENSE.
Flame colour intensity therefore relates directly to metal ion concentration.
An instrument called a FLAME EMISSION SPECTROPHOTOMETER is designed to
aspirate (or spray) small volumes of solution into a gas flame. The instrument has a
detector designed to measure the intensity of coloured light being emitted by the metal
ion.
If we can prepare a series of different known concentrations of a given metal ion (eg
Sodium or Potassium), we will see that the colour intensity of light emitted changes
according to the relationship described above.
If we make a graph of metal ion concentration versus colour intensity (vertical axis), we will
see on graph paper that colour intensity increases as soluble metal ion concentration
increases (see graph).
An unknown concentration of the same metal ion will give a particular flame intensity when
aspirated through the flame photometer. We can use our graph to convert this intensity to
a concentration.
Flame Emission Spectrophotometers are designed to analyse for the elements Sodium
(Na) and Potassium (K). The instrument will accurately analyse concentrations of these
elements between 1mg/L and 100 mg/L.
Outside of these concentrations the light emitted is too weak or too strong to analyse
accurately. Samples which are too concentrated must be accurately diluted until the
Sodium / or Potassium concentration is lowered back within this range.
159
Graph of Metal Ion Concentration versus
Colour Intensity
Colour Intensity
120
100
80
60
40
20
0
0
20
40
60
80
100
Concentration (mg / L)
Written Exercises:
1. What is the chemical name of the salt found predominantly in sea water?
2. Name 2 soluble metal ions commonly found in sea water.
3. What do the following symbols stand for:
Na+
______________________
mL:
K+
______________________
g/100mL: ___________________________
g:
______________________
g/L:
__________________________
mg:
______________________
mg/L:
__________________________
L:
______________________
__________________________
4. What is the name of the instrument used to analyse for potassium and sodium ion
concentration?
160
5. A small volume of solution is aspirated into the photometer flame to measure analyte
concentration. What is meant by the term "aspirate"?
6. When metal ions release different colours of light in a gas flame, this referred to as
Fl_________________
E___________________.
7. What is the purpose of the detector within the photometer machine?
8. According to the Flame Emission principle, what is light intensity dependant on?
9. How might we be able to still analyse a sample for both Na+ and K+ when their
concentrations are too high to be measured?
161
Experiment 2 – INTRODUCTION TO FLAME EMISSION
Background:
We can see the relationship between metal ion concentration and colour intensity through
the method of Flame Emission. A series of known concentrations (or standards) of the
sodium metal ion will be aspirated through a Flame Emission Spectrophotometer for this
purpose.
This experiment aims to develop the following skills:
1. Preparation of standard solutions through serial dilution.
2. Tabulating and graphing results.
3. Determining the concentration of an unknown solution.
Method:
-
Complete the information below summarising how you will prepare your standard
solutions.
Run your prepared standards and unknown sample through the spectrophotometer.
Record the intensity readings of each solution in the table at right of page.
Results:
Sodium (Na+) ion Stock Solution: ___________________ mg/L
Intermediate solution ___________________ mg/L
Preparation:
__________ mL (Pipette) 
__________ mL (Volumetric flask)
STANDARD SOLUTION PREPARATION
Intermediate (mL)
- Pipette
Total Volume
- Volumetric
Flask
Standard 1
0 mg/L
Standard 2
20 mg/L
Standard 3
40 mg/L
Standard 4
60 mg/L
Unknown
162
ANALYSIS:
FLAME EMISSION
SPECTROPHOTOMETER
Sodium (Na+)
Intensity
Analysis:
1. Construct a calibration graph of Concentration (axis
) versus
+
Colour Intensity (axis ) for your Sodium (Na ) standard solutions.
PASTE CALIBRATION GRAPH HERE
2. From your graph, describe how the colour intensity changes with sodium
concentration.
3. Use your graph to calculate the concentration of Na+ in the unknown solution.
163
Experiment 3 – SODIUM AND POTASSIUM IN BEER
Background:
We have seen there is a relationship between metal ion concentration and colour intensity
through the method of Flame Emission. The concentration of Sodium (Na+) and
Potassium (K+) ions in beer can be determined via a similar method.
Method:
-
Complete the information below summarising how you will prepare your standard Na +
and K+ solutions.
De-gas a beer sample by pouring from one beaker to another several times.
Prepare and aspirate your prepared standards through the spectrophotometer and
record your results in the table below.
Quantitatively dilute beer sample until Na+ / K+ intensities fall within the range of
standard concentrations.
Aspirate diluted beer sample and record this diluted Na + / K+ intensity.
Draw a calibration graph for both Na+ and K+
+
+
Calculate the K and Na in your original beer sample.
Results:
PART A:
Sodium
Na+ Stock Solution: ___________________ mg/L
STANDARD SOLUTION PREPARATION
Na+
Stock Na+ (mL)
- Pipette
Total Volume
- Volumetric
Flask
Standard 1
0 mg/L
Standard 2
10 mg/L
Standard 3
20 mg/L
Standard 4
30 mg/L
DILUTED BEER
164
ANALYSIS:
FLAME EMISSION
SPECTROPHOTOMETER
Sodium (Na+)
Intensity
Dilution Factor - Beer:
Beer Volume PIPETTE
= ____________ mL
PART B:
Total Volume VOL FLASK
= ____________ mL
DILUTION FACTOR
x __________
Potassium
K+ ion Stock Solution: ___________________ mg/L
STANDARD SOLUTION PREPARATION
+
K
Stock K+ (mL)
- Pipette
Total Volume
- Volumetric
Flask
ANALYSIS:
FLAME EMISSION
SPECTROPHOTOMETER
Potassium (K+)
Intensity
Standard 1
0 mg/L
Standard 2
20 mg/L
Standard 3
40 mg/L
Standard 4
60 mg/L
DILUTED BEER
Dilution Factor - Beer:
Beer Volume PIPETTE
= ____________ mL
Total Volume VOL FLASK
= ____________ mL
Analysis - Over Page
165
DILUTION FACTOR
x __________
Analysis:
PART A - Sodium
Construct a calibration graph of Concentration (axis
) versus
Colour Intensity (axis ) for your Sodium (Na+) standard solutions.
PASTE CALIBRATION GRAPH HERE
Calculations: Sodium Ion Concentration:
1. Na+ intensity (diluted beer) = _________________
2. Na+ concentration (diluted beer) = ________________ mg/L
3. Dilution Factor (beer) =
x ______________
Na+ concentration (beer sample)
___________________ mg/L
166
Analysis:
PART B - Potassium
Construct a calibration graph of Concentration (axis
) versus
+
Colour Intensity (axis ) for your Potassium (K ) standard solutions.
PASTE CALIBRATION GRAPH HERE
Calculations: Potassium Ion Concentration:
1. K+ intensity (diluted beer) = _________________
2. K+ concentration (diluted beer) = ________________ mg/L
3. Dilution Factor (beer) =
x ______________
K+ concentration (beer sample)
___________________ mg/L
167
Infra-Red (I.R.) Spectroscopy
Other radiations such as infra-red can also be absorbed or emitted by chemicals in a
spectrum which allows them to be identified and even measured. You may be able to
observe infra-red spectroscopy being used. It is particularly useful in identifying covalently
bonded molecules and it is the type of bond which absorbs the energy enabling it to be
identified. X-rays are another radiation which is often used in this identification process.
Figure 1: An infrared spectrum of the compound butanoic acid
Figure 2: A general infrared spectrum. Fingerprint region is unique to each compound
168
SOLUTIONS TO WRITTEN EXERCISES
Page 8
Solid
sand
plasticine
steel
timber
Liquid
lemonade
mercury
milk
Gas
natural gas
Difficult to classify? Why?
fog
balsamic vinegar
paint
toothpaste
shaving foam
jelly
Page 10
Property
Shape
Solid
fixed (own) shape
Gas
takes the shape of the
container
fixed at constant
temperature
Liquid
takes the shape of the
part of the container it
fills
fixed at constant
temperature
Volume
(the amount of space
taken up)
Compressible?
(can it squash into a
smaller volume?)
no
no
yes
Diffuses?
(can it spread out by
itself?)
no
yes
yes
fills whatever
container it is in and
has its volume
(b) In a liquid the particles can slide past one another but they are still very close. Their sliding movement
means that they will diffuse until they meet the walls of the container and the liquid will take part of its shape.
The particles are still very close and so the volume of the liquid is fixed at constant temperature and it is
incompressible.
(c) In a gas the particles can fly apart until they collide with the walls of their container. Therefore the gas
takes the shape and volume of the container. The particles have a lot of space in between them and
therefore a gas is easily compressible. Gases diffuse quickly because of the flying movement of their
particles.
Page 21
Element
Hydrogen
Helium
Lithium
Beryllium
Boron
Carbon
Nitrogen
Oxygen
Fluorine
Neon
Sodium
Magnesium
Aluminium
Symbol
Atomic
Number
Mass
Number
Neutron
Number
Electron
Number
Electron
Configuration
H
He
Li
Be
B
C
N
O
F
Ne
Na
Mg
Al
1
2
3
4
5
6
7
8
9
10
11
12
13
1
4
7
9
11
12
14
16
19
20
23
24
27
0
2
4
5
6
4
7
8
10
10
12
12
14
1
2
3
4
5
6
7
8
9
10
11
12
13
1
2
2:1
2:2
2:3
2:4
2:5
2:6
2:7
2:8
2:8:1
2:8:2
2:8:3
169
Charge of
Ion
 valency
1+
0
1+
2+
3+
4 +/3210
1+
2+
3+
Silicon
Phosphorous
Sulfur
Chlorine
Argon
Potassium
Calcium
Page 22
Si
P
S
Cl
Ar
K
Ca
14
15
16
17
18
19
20
28
31
32
35
40
39
40
14
16
16
18
22
20
20
14
15
16
17
18
19
20
2:8:4
2:8:5
2:8:6
2:8:7
2:8:8
2:8:8:1
2:8:8:2
Elemental puns
1.
0.5 times holmium:
=
......hafnium.................
2.
An officer of the constabulary:
A
.....copper...............................
3.
Funds from your mother’s sister:
4.
A comical prisoner:
A
......silicon...............................
5.
What you do with dead cats:
You
.......barium.............................
6.
A driller’s motto:
.......boron..................
7.
What a good doctor can do for a patient:
.......helium.......or curium........
8.
When you have a hole in the boat:
You
.......zinc..................................
9.
How you mended your clothes:
You
........sodium............................
10.
What are you doing with that person’s hair?
I’m
........platinum........................
11.
A nocturnal trojan:
A
........nitrogen..........................
12.
A trojan who smokes dope:
A
........hydrogen........................
13.
A man who tried to murder his wife with a razorblade only managed to give her: ….arsenic…
14.
Mind your own:
Page 24
1.
4+/3210
1+
2+
Elements
Copper
Nitrogen
Gold
Iodine
......antimony...........................
….bismuth…….
Compounds
Water
Ethanol
Vinegar or 
Carbon Dioxide
Mixtures
Air
Soil
Orange Juice
Petrol
Glass
Vinegar
Brass
Sweat
2.
(a)
Charcoal and salt
...........dissolution and filtration...................
(b)
Water and oil
......separation funnel or distillation................
(c)
Alcohol and water
..........distillation........................................
(d)
Sugar from a sugar solution
..............evaporation or distillation ...........
(e)
Water from muddy river water
.............filtration or distillation......................
Page 26
Classify the following as chemical (C) or physical (P) changes and determine which property is concerned.
2.
An iron bar rusting in the air.
........C.....
...........Corrosion / reaction with oxygen....
3.
Petrol combusting in a car engine.
........C.....
...........Combustion / reaction with oxygen..
170
4.
Moulding plastic into a new shape.
.......P......
...........Malleable...................
5.
Setting off some fireworks.
.......C......
...........Combustion. / reaction with oxygen..
Page 53
Mg, Cl
....ionic..........................
O, O
....covalent....................
K, S
....ionic..........................
H, Cl
....covalent....................
C, Br
....covalent..................
Na, Ca
....metallic…....
Pages 55
..
:S:
c)
: Al .
a)
..
: Cl :
b)
‘
Ca - 2 b)
Al - 3
c)
3)
a)
I-1
S-2
c)
4)
(a) aluminium
Al
(d) potassium
K
(g) bromine
Br
(j) nitrogen
N
1)
a)
2)
5)
b)
3+
d)
Li .
e)
:C:
Na - 1 d)
Li - 1
e)
Ba - 2 f)
Mg - 2
N-3
Cl - 1
e)
P-3
O-2
d)
+
(b) lithium
Li
+
(e) calcium
Ca
-
(h) oxygen
O
(k) selenium
Se
(a) magnesium and bromine
yes
(b) chlorine an iodine
no
(c) potassium and helium
no
(d) nitrogen and sulfur
no
(e) lithium and fluorine
yes
(f) carbon and oxygen
no
+
2-
3-
2+
2-
2-
3+
Sr
(i) arsenic
As
(l) sulphur
S
-
7)
(a) KCl
K and Cl
(d) MgBr2
Mg
(a) H2O
covalent
(b) Na2O
ionic
(c) CS2
covalent
(d) CaS
ionic
(e) NH3
covalent
(f) SO2
covalent
(g) Al2O3
ionic
2+
and Br
-
(c) Al
(f) strontium
K2S
-
CaO
Ba
(a) K ; S
+
; O
2-
+
-
(b) NaNO3
Na and NO3
(e) KOH
K and OH
+
-
2+
c) barium
6)
8)
(b) Ca
2+
f)
; Cl
AlCl3
2+
Ba
and SO4
(f) Li2CO3
Li and CO3
+
2-
(NH4)2SO4
of ........ NH4.... )
......2.... atoms of
......8.... atoms of
.......1..... atom of
.....4..... atoms of
.......Nitrogen.....
.........Hydrogen....
.............Sulfur.....
......Oxygen.......
CH4 represents .........Carbon x 1 and Hydrogen x 4.......................
171
2-
(c) BaSO4
( ...2.... groups
2.
3-
2-
Page 58
1.
2+
3.
Sn3(PO4)4 represents .....Tin x 3, Phosphorus x 4 and Oxygen x 16.....
4.
CuSO4.5H2O represents.....Cu x 1, Sulfur x 1, Oxygen x 9, and Hydrogen x 10....
5.
Cl
-
represents ....... Chloride anion......................................................................
-
6.
OH represents ...Hydrogen x 1 and Oxygen x 1. Net charge of -1 (or: hydroxide polyatomic ion)...
Page 61
Written Exercises:
Write formulae for the following compounds.
(a) Hydrogen bromide ……HBr…………
(g) Ammonium nitride ………(NH4)3N……
(b) Calcium chloride ……CaCl2………..
(h) Iron (III) hydroxide ……Fe(OH)3……….
(c) Zinc (II) sulfate …………ZnSO4………
(i)
Potassium sulfite ………K2SO3……..
(d) Magnesium oxide ……MgO………
(j)
Nickel (II) carbonate ……NiCO3….
(e) Lead(IV) sulphide ……PbS2………
(k) Silver (I) phosphate ……Ag3PO4……..
(f)
Barium phosphide …Ba3P2…………
(l)
Tin (II) nitrate ………Sn(NO3)2…….
Pages 62-63
Formulae and Naming (also termed “Nomenclature”)
Written Exercises:
1.
Write formulae for the following:
(a) potassium nitrate …KNO3….
(b) sodium carbonate …..Na2CO3……
(c) cobalt II sulfate …..CoSO4….
(d) ammonium carbonate …..(NH4)2CO3
(e) calcium phosphate ....Ca3(PO4)2...
(f) disulfur dichloride .....S2Cl2......
(g) nitrogen triodide ....NI3..........
(h) iodine heptafluoride .........IF7............
(i) silicon dioxide ........SiO2..........
(j) selenium dibromide ....SeBr2.................
2.
Name the following compounds:
(a) SiF4 ......Silicon tetrafluoride.........
(b) CaSO4 ......Calcium sulfate.........
(c) FeO ......Iron II oxide.........
(d) CuCl ......Copper I chloride.........
(e) AgCl......Silver I chloride.........
(f) NaHCO3 ......Sodium hydrogen carbonate.........
(g) NO .......Nitrogen monoxide....
(h) P2O5 ......Diphosphorous pentoxide.........
(i) PCl3 ......Phosphorous trichloride....
(j) CS2 ..........Carbon disulfide...........
3.
Formula
Name
Name
Formula
MgO
magnesium oxide
sodium oxide
Na2O
CaS
calcium sulfide
calcium carbonate
CaCO3
CO
carbon monoxide
ammonia
NH3
CO2
carbon dioxide
aluminium hydroxide
Al(OH)3
172
Csl
caesium iodide
dinitrogen tetroxide
N2O4
CaBr2
calcium bromide
ammonium sulfate
(NH4)2SO4
H2O
dihydrogen monoxide
aluminium oxide
Al2O3
NO2
nitrogen dioxide
copper (I) bromide
CuBr
N2O4
dinitrogen tetroxide
chromium (III) fluoride
CrF3
AsCl3
arsenic trichloride
manganese (IV) oxide
MnO2
AgNO3
Silver I nitrate
mercury (II) sulfide
HgS
KBr
potassium bromide
oxygen difluoride
OF2
Ca(OH)2
calcium hydroxide
iron (II) chloride
FeCl2
CoCl2
cobalt (II) chloride
iron (III) iodide
Fel3
HI
hydrogen iodide
copper (II) sulfate
CuSO4
SnO2
tin (IV) oxide
carbon tetrafluoride
CF4
CCl4
carbon tetrachloride
strontium chloride
SrCl2
SnCl2
tin (II) chloride
potassium nitrate
KNO3
BaF2
barium fluoride
calcium phosphate
Ca3(PO4)2
Page 72
Written Exercise:
Hydrogen chloride conductor?
.......As a solid or gas hydrogen chloride assumes a covalent molecular form. It is unique as having the
properties of a strong acid, releasing hydrogen ions and also chloride ions when dissolved in water. In this
form it is an excellent conductor of electricity....
Page 72
Compound Type
CHEMICAL
EXAMPLE(S)
CHEMICAL BOND
TYPE
PHYSICAL STATE
@ Room Temp
MELTING POINT
(Low, High, Very High)
WATER SOLUBILITY
(High, Low, Nil)
COMPOUND PROPERTIES IN SUMMARY
IONIC
COVALENT MOLECULAR
COVALENT
NETWORK
Sodium Chloride
Carbon Dioxide
Silica
Ionic
Covalent
Covalent
Solid
SOLID, LIQUID & GAS
Solid
High
Low
Very High
SEE SOLUBILITY RULES
LOW - MEDIUM
Insoluble
173
No
No
No
Yes
No
No
ELECTRICAL
CONDUCTIVITY
- Solid State
ELECTRICAL
CONDUCTIVITY
- Molten State
Yes
ELECTRICAL
CONDUCTIVITY
- Aqueous Sol’n
NO
Hydrogen salt exceptions
Page 81
1.
2.
(a)
CH4(g)
+
2O2(g)
(b)
2Na(s)
+
(c)
Zn(s)
(d)

CO2(g) +
2H2O(l)
2H2O(l) 
2NaOH(aq) +
H2(g)
+
2HCl(aq ) 
ZnCl2(aq) +
H2(g)
4P
+
5O2
(e)
2NH3
+
H2SO4 
(f)
CuO
+
2HCl

CuCl2
(g)
2H2O2

2H2O
+ O2
(h)
H2CO3

H2O
+ CO2
(i)
4Fe
3O2

2Fe2O3
(j)
2C8H18 +
25O2

16CO2 + 18H2O
+
Sodium hydrogen +
carbonate
NaHCO3
+

2P2O5
(NH4)2SO4
+
ethanoic  Sodium +
acid
ethanoate
H2O
water
CH3COOH  NaCH3COO + H2O
+
carbon
dioxide
+ CO2
Page 86-87
1.
Magnesium + hydrochloric acid  magnesium chloride + hydrogen gas
Mg (s) + 2HCl (aq)  MgCl2 (aq) + H2 (g)
2.
Aluminium + nitric acid  aluminium nitrate + hydrogen gas
2Al (s) + 6HNO3(aq)  2Al(NO3)3 (aq) + 3H2 (g)
3.
Iron + oxygen  iron III oxide
4Fe (s) + 3O2 (g)  2Fe2O3 (s)
4.
Lithium + water  lithium hydroxide + hydrogen gas
2Li (s) + 2H2O (l)  2LiOH (aq) + H2 (g)
5.
Silver I carbonate  silver oxide + carbon dioxide
Ag2CO3 (aq)  Ag2O (s) + CO2 (g)
6.
Silver I nitrate + sodium chloride  silver chloride + sodium nitrate
AgNO3 (aq) + NaCl (aq)  AgCl (s) + NaNO3 (aq)
7.
Sulfuric acid + sodium hydroxide  sodium sulfate + water
H2SO4 (aq) + 2NaOH (aq)  Na2SO4 (aq) + 2H2O (l)
174
No
8.
Lead II nitrate + potassium iodide  lead II iodide + potassium nitrate
Pb(NO3)2 (aq) + 2KI (aq)  PbI2 (s) + 2KNO3 (aq)
9.
Hexane + oxygen  carbon dioxide + water (+ energy)
C6H14 (l) + 9½O2 (g)  6CO2 (g) + 7H2O (l)
or
2C6H14 (l) + 19O2 (g)  12CO2 (g) + 14H2O (l)
Page 87
1.
Magnesium + Nitric Acid  Magnesium nitrate + hydrogen gas
Mg (s) + 2HNO3 (aq)  Mg(NO3)2 (aq) + H2 (g)
2.
Copper + Sulfuric Acid  Copper II sulfate + hydrogen gas
Cu (aq) + H2SO4 (aq)  CuSO4 (aq) + H2 (aq)
3.
Potassium + Water  Potassium hydroxide + hydrogen gas
2K (s) + 2H2O (l)  2KOH (aq) + H2 (g)
4.
Zinc II Carbonate + Nitric Acid  Zinc II nitrate + carbon dioxide + water
ZnCO3 (aq) + 2HNO3 (aq)  Zn(NO3)2 (aq) + CO2 (g) + H2O (l)
5.
Copper II Carbonate (s) + HEAT  copper II oxide + carbon dioxide
CuCO3 (s)  CuO (s) + CO2 (g)
6.
Aluminium + Sulfuric Acid  Aluminium sulfate + hydrogen gas
2Al (s) + 3H2SO4 (aq)  Al2(SO4)3 (aq) + 3H2 (g)
7.
Sodium Hydroxide + Copper II Sulfate  Copper II hydroxide + sodium sulfate
2NaOH (aq) + CuSO4 (aq)  Cu(OH)2 (aq) + Na2SO4 (aq)
8.
Sodium Hydroxide + Sulfuric Acid  Sodium sulfate + water
2NaOH (aq) + H2SO4 (aq)  Na2SO4 (aq) + 2H2O (l)
Page 89
+
-
1.
Ag
2.
2CO3 (aq)
(aq)
+ Cl
(aq)
+ 2H
 AgCl (s)
+
(aq)
 CO2 (g) + H2O (l)
Page 96
1.
a) 58.44
b) 132.15
c) 180.16
d) 204.22
2.
a) 0.99 mol
b) 0.10 mol
c) 0.50 mol
d) 0.024 mol
3.
a) 146.1 g
b) 66.08
c) 720.64 g
d) 3.06 g
Page 101
a)
N2H4 (g) + O2 (g)  N2 (g) + 2H2O (g)
b)
L.R. = Hydrazine (just!)
c)
1124 g
Pages 109-111
1.
i)
ii)
iii)
iv)
CaCO3 (s)  CaO (s) + CO2 (g)
1997.1 mol
1997.1 mol
49 508 L
175
2.
i)
ii)
iii)
iv)
Zn (s) + 2HCl (aq)  ZnCl2 (aq) + H2 (g)
0.20 mol
0.20 mol
5.0 L
3.
i)
ii)
iii)
iv)
v)
2AgCl (s) + H2 (g)  2 Ag (s) + 2HCl (aq)
2
:
1
:
2 :
2
0.20 mol
0.10 mol
2.48 L
4.
i)
ii)
iii)
iv)
v)
CaC2 (s) + 2H2O (l)  Ca(OH)2 (aq) + C2H2 (g)
1
:
2 :
1
:
1
1
0.450 mol
11.16 L
Pages 114-115
1.
a) C = 27.3% ; O = 72.7%
b) H = 2.1% ; S = 32.7% ; O = 65.2%
c) C = 40.0% ; H = 6.7% ; O = 53.3%
2.
Zn = 20.5% ; I = 79.5%
3.
Hg2Cl2: Hg = 85.0% ; Cl = 15.0%
HgCl2: Hg = 73.9% ; Cl = 26.1%
Page 116
1.
C3H6O2
2.
Emp Formula: HF ; Molec Formula: H2F2
3.
Emp Formula: P2O3 ; Molec Formula: P4O6
Page 122
i) 3M
iv) 0.5M
ii) 2.5M
v) 0.5M
iii) 0.1M
vi) 0.09M
Page 124
1.
a) Dissolve 4.248 g of AgNO3. Make up to 250.0 mL
b) Dissolve 4.561 g of K2CO3. Make up to 100.0 mL
2.
a) 0.40M
b) 0.0025M
c) 0.02M
3.
a) 0.405 g
b) 30.335 g
c) 1.766 g
4.
a) 500 mL
b) 250 mL
c) 100 mL
BONUS QUESTION:
1.001 g
Pages 126-127
1.
i) 10x
ii) 25x
2.
i) 0.63 g/100 mL
3.
i) 25 mL
iii) 16.67x
ii) 0.015M
ii) 50 mL
iii) 10 mg/L
iii) 40 mL
176
Page 133
1.
1.5% w/w
2.
67.5% w/w
3.
26.6% w/w
1.b)
0.1213M
2.
0.1172M
Page 141
1.a)
0.0769M
3.
4.76% w/v
Page 160-161
1.
sodium chloride, NaCl
2.
sodium Na
+
3.
potassium K
+
magnesium Mg
2+
calcium Ca
2+
What do the following symbols stand for:
+
Na
+
K
g:
mg:
L:
__sodium ion____
__potassium ion_____
____grams__________
___milligrams_____
____litres_________
mL:
_______millilitres________
g/100mL: ___grams per 100 millitres___
g/L:
_____grams per litre_____
mg/L:
____milligrams per litre_____
4.
flame emission spectrophotometer
5.
to spray a finely dispersed mist of solution
6.
flame emission
7.
detector measures light intensity emitted
8.
analyte concentration
9.
quantitative dilution of the sample
177
REFERENCE LIST
Some of these references are referred to in the module text by abbreviation:
CC1
James et al (1991)
Chemical Connections. Book One. Jacaranda ISBN O 7016 2750 6
C1
Elvins et al. (1995)
Chemistry One. Materials: Chemistry in Everyday Life. Second Edition
Heinemann. ISBN 0 85859 708 X
Elvins et al. (19xx). Chemistry One. Teachers’ Resource Book. Heinemann. ISBN 0 85859 xxx x
C2
Commons et al. (1994) Chemistry Two. Chemistry and the Marketplace; Energy and Matter.
Heinemann. ISBN 0 85859 755 1
Commons et al. (1995). Chemistry Two. Teachers’ Resource Book.
Heinemann. ISBN 085859 758 6
CC2
James et al. (1991) Chemical Connections. Book One. Jacaranda. ISBN 0 7016 2750 6
James et al. (1992) Chemical Connections. Book Two. Jacaranda. ISBN 0 7016 3062 0
Selinger, B. (1994). Chemistry in the Marketplace. Fourth Edition. Harcourt Brace & Co.
ISBN 0 7295 0334 8
Elvins et al. (1995). Chemistry One. Materials; Chemistry in Everyday Life. Second Edition. Heinemann.
ISBN 0 85859 708 X
Commons et al. (1994). Chemical Connections. Book Two. Chemistry and the Marketplace; Energy and
Matter. Heinemann. ISBN 0 85859 755 1
DB
Barker, D. (2001). 6849AA Calibration and Data Handling
BFK
Barker, D. Fullick, G and Krajniak, E. (2003). Chemistry For Technicians. Harcourt Brace & Co.
ISBN 0 7295 3291 7
Laboratory Manuals
Deretic, G. & Ware, G. (1995). Senior Chemistry Practical Manual. Heinemann. ISBN 0 85859 789 6.
Burke, S. (1992). Chemistry Works. Science Press. ISBN 0 85583 170 7
Further References
Smith, A. and Dwyer, C. (1991). Key Chemistry. Book One. Materials and Everyday Life. ISBN 0 522 844502
Smith, A. et al. (1992). Key Chemistry. Book Two. Energy, Matter and the Marketplace. ISBN 0 522 84461 8
178