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CHEMISTRY
What is chemistry?
It is the study of the composition, structure and properties of matter and the reactions by which one form of matter
may be converted into another form.
Why should we study chemistry?
Branches of chemistry
Analytical Chemistry
Analytical chemistry is the study involving the analysis of chemical components of substances. Examples of
areas using analytical chemistry include forensic science, environmental science, and drug testing.
Biochemistry
This is the study of chemical processes in living organisms.
Inorganic chemistry
Inorganic chemistry covers all materials that are not organic and are termed as non-living substances – those
compounds that do not contain carbon hydrogen (C-H) bond.
Organic chemistry
Organic chemistry deals with the study of carbon and the chemicals in living organisms. It is the opposite of
inorganic chemistry.
Physical chemistry
Physical chemistry deals with the study of the physical properties of matter.
It deals with the principles and methodologies of both chemistry and physics and is the study of how
chemical structure impacts physical properties of a substance.
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Matter
Objectives:
 describe the solid, liquid and gaseous states of matter and explain their interconversion in terms of the kinetic
particle theory and of the energy changes involved
 describe and explain evidence for the movement of particles in liquids and gases
 explain everyday effects of diffusion in terms of particles
 state qualitatively the effect of molecular mass on the rate of diffusion and explain the dependence of rate of
diffusion on temperature
Matter is anything that has mass and volume. There are three forms in which matter can exist; these forms are called
the states of matter.



Solid
Liquid
Gas
The small particles which matter is made up of are called the basic units of matter. The basic units of matter are



Atoms
Molecules
Ions
Properties of matter
The properties of the three states of matter can be discussed in terms of shape and volume as in the table below.
Physical state
Solid
Liquid


Gas

Shape
Have fixed shapes
No fixed shape takes the
shape of the container
No fixed shape


Volume
Have fixed volume
Have fixed volume

Have no fixed volume
The table below summarises the movement and arrangement of particles
Property
Arrangement
of particles
Attractive

Solid
Particles are
closely packed

Held by very

Liquid
Particles are
loosely packed


Held by weak

Gas
Particles are far
apart from each
other
Held by very weak
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forces
between the
particles
Motion of the
particles

strong forces of
attraction
in fixed positions
Particles vibrate in
fixed positions
forces of
attraction

Molecules slide
over each other
and move slowly in
random manner
forces

Molecules move
very fast in a
random manner
Changes of state of matter
Matter can change form from one physical state to another physical state. The conditions needed for transformation
of mater into different states are heating and cooling.
The graphs below show the effect of heating and cooling of substances.
Heating
curve
The above figure shows a heating curve of a solid. At point A the state is solid. At point B the solid is
melting, at this stage it is a mixture of solid and liquid. At point C the state is liquid. At point D the liquid is
evaporating, it is a mixture of liquid and gas. At point E the state is gas. Temperature X is the melting point
and the temperature Y is the boiling point of the solid.
Cooling curves
The figure below shows a cooling curve of a gas. At point A the state is gas. At point B the gas is
condensing the state over here is a mixture of gas and liquid, at point C the state is liquid. At point D the
liquid is freezing at this stage there is a mixture of liquid and solid. At point E the state is solid. Temperature
X is the melting point and temperature Y is the boiling point of the gas.
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Processes associated with changes of state of matter
A is melting brought about by heating. A solid is changing to liquid state; this process is accompanied by
absorption of heat.
 Since the process takes in heat energy to break the bonds and cause melting, it is called endothermic.
B is freezing /solidification the process is accompanied with the release of heat energy to the surrounding.
Since heat is released to the surrounding, the process is exothermic.
 C is condensation
 D is evaporation
 E is sublimation
Evaporation and boiling as processes
 Evaporation and boiling both involve the change of state of matter from liquid to gas
 Evaporation is a process when the particles at top of the surface of the liquid get more energy as
compared to the particles at the bottom thus they leave the liquid as vapour.
 Liquids which easily evaporate are referred to as volatile liquids e.g. ethanol, paraffin, methylated
spirit.
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The differences between evaporation and boiling are shown below
EVAPORATION
BOILING

Slow process

Fast process

Occurs only at the top surface of liquid

Occurs throughout the liquid

Occurs at any temperature

Occurs at a fixed temperature

Results in cooling

Results in heating
Kinetic theory of matter
The kinetic theory of matter is a theory that explains the behavior and evidence of matter.
1. Matter is made up of small particles
2. The particles from which matter is made of are in continuous motion
3. The speed of the particles increase as the temperature increases
4. Particles exert strong forces upon each other when they collide
The idea that matter is made up of tiny particles can be explained by diffusion and the Brownian
experiments.
Diffusion
It is the movement of particles from a region of high concentration to a region of low concentration.
Observing diffusion in gases
The diagrams above explain how the movement of particles in gases occurs. The particles of bromine will
completely mix with the air particles by the process of diffusion.
Demonstrating diffusion in liquids
Diffusion in liquids can be explained using a coloured substance such as potassium manganate or potassium
dichromate in a colourless liquid like water. The colour of the small particle spread throughout the liquid
over some time.
 This can only happen if the particles of the small solute are moving to all the parts of the liquid.
Brownian motion
Brownian motion also provides evidence that matter is made of particles which are in a random motion.
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
It explains the movement of particles in fluids ( gases and liquids)

In the experiment above, the smoke particles will be moving in a zigzag (random) motion. This is
because the smoke particles are colliding with the molecules of air which are moving in a random
motion.
Factors that affect the rate of diffusion
Temperature: the higher the temperature the higher the rate of diffusion. This is because; increasing the
temperature increases the speed of the particles.
Concentration: the larger the difference in the concentration of the particles between two points, the higher
the rate of diffusion
Particle size: smaller particles move at high speed than bigger particles. This means that, the smaller the
particles the higher the rate of diffusion.
An experiment to explain how particle size affects the rate of diffusion can be set up as shown below.
The particles in hydrogen chloride gas are twice as heavy as those in ammonia gas.
 Cotton wool soaked in ammonia solution is put into one end of a long tube (at A). It gives off
ammonia gas.
 At the same time, cotton wool soaked in hydrochloric acid is put into the other end of the tube (at B).
It gives off hydrogen chloride gas.
 The gases diffuse along the tube. White smoke forms where they meet:
The white smoke forms closer to B. So the ammonia particles have travelled further than the hydrogen
chloride particles – which mean they have travelled faster.
MEASUREMENTS AND EXPERIMENTAL TECHNIQUES
Objectives:
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 Name apparatus for measuring time, temperature, mass and volume. (For volume you need to know burette,
pipette, measuring cylinder, gas syringe)
 Suggest apparatus for simple experiments, including collection of gases and measuring rate of reaction.
Experiments are an essential aspect of the scientific enquiry and learning. In chemistry, various instruments and
apparatus are used to measure physical quantities.
Four basic physical quantities measured in experiments:
 Volume
 Temperature
 Mass
 Time
S.I UNITS
The International System of Units (S.I units) is a system of different units used to measure quantities of
different things. In order to communicate scientific information easily, scientists all over the world use S.I
units as common standard for their measurements.
a. Measuring mass
Mass is measured using a beam balance or a digital balance.The S.I units for mass are the kilograms. Kg
b. measuring volume of liquids and gases
SI unit: m3
Other units: cm3, dm3 (1 dm3 = 1000 cm3)
Apparatus
Beaker
Measuring cylinder
Burette
Pipette
Accuracy /usage
Used to estimate the volume of a liquid, e.g. approximately 100cm3
More accurate than a beaker
Measures up to the nearest cm3,e.g 99cm3
Accurately measures out the volume of liquid to nearest 0.1cm3
Accurately measures out fixed volumes of liquids, e.g. 20.0cm3or 25.0cm3
How to read the volume of a liquid
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When water or solution is placed in a glass container, it forms a curved surface called a meniscus. A meniscus may be
concave or convex.
Collecting and measuring volume of gases
The three methods for collecting gases are shown below. The methods for collecting gases are based upon:
1. Solubility of a gas
2. Density of a gas
Displacement of water is suitable for
collecting gases that are insoluble or
slightly soluble in water.
Downward delivery is used to
collect gases that are soluble in
water and denser than air.
Upward delivery is used to
collect gases that are soluble in
water and less dense than air.
Measuring the volume of gas
A gas syringe is used to measure the volume of a gas. A gas syringe measures volume of gas up to 100cm3.
c. Measuring time
In the laboratory, a stopwatch or stop clock is used to measure time. The S.I unit for time is the second (s).
Other units, such as the minute (min) and the hour (h), are used to measure longer intervals of time.
d. Measuring temperature
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

The mercury thermometer can be used to measure temperature in the laboratory. The S.I unit for
temperature is the Kelvin (K). Another unit of temperature, the degree Celsius, is also commonly used.
The mercury thermometer measures temperature from -10℃ to 110℃. Each division on the scale
represents 1℃.
Activity:
1. The table below shows three ways of collecting gases in the laboratory.
(a) For each method:
(i) Give the essential property the gas must have to allow collection in this way
(ii) Name a gas that could be collected in this way.
2. Find the volume used in experiment 1,2 and 3 below
Experiment
Final volume
Initial volume
Volume used
1
2
3
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CRITERIA FOR PURITY
Criteria of purity refer to the standards of determining the purity of different substances using the physical
properties which are a characteristic of pure substances. A pure substance is made up of only one substance and is
not mixed with any other substance. In nature, many substances are not pure as they exist as mixtures. A mixture is a
substance that contains two or more substances that are not chemically combined. The physical properties of pure
substances do not show variations.
Determining purity of substances
The physical properties such as boiling point, melting point, freezing point, and density can be used to determine the
purity of substances.
Boiling point
The boiling point of a substance is the temperature at which a liquid boils. Different liquids have different boiling
points. Example pure water boils at 100℃ at its boiling point, a substance exists as a liquid and gas.
Pure substances have sharp and fixed boiling points; they do not boil between a certain ranges of temperature. An
impure substance however, boils between a certain ranges of temperature.


Impurities raise the boiling point of a substance.
Coolant in car engines increases the boiling point of water
Melting point
This is the temperature at which a solid changes to liquid. During melting a substance exists as a solid and as a liquid.
Pure substances have sharp melting points

Impurities lower the melting point of a substance
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Density
Density is mass per unit volume of a substance. Pure substances have fixed densities, for example, pure water has a
density of 1g/cm3.
Importance of purity of substances
Purity of substances is important in the processes of



Food
Water and drinks
Manufacture of medicines
Activity:
A substance was removed from a freezer and heated. The temperature was recorded at regular intervals. The results
are shown on the graph below.
(a) The letters V–Z represent the states of matter listed below. Write the five letters at the correct places on the
graph above. V = solid, W = solid + liquid, X = liquid, Y = liquid + gas, Z = gas [5]
(b) Explain what is happening to the particles at those five places on the graph:
(c) The substance that was heated was pure. How can you tell this from the graph?
(d) Name the substance that was heated.
TECHNIQUES FOR SEPARATION:
Objectives:
 describe methods of purification using a suitable solvent, filtration and crystallisation.
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 describe methods of purification using distillation and fractional distillation (to separate crude oil, liquid air
and fermented liquor)
 suggest methods of purification when given suitable information
 describe paper chromatography and interpret chromatograms (including use of Rf values)
 explain the use of locating agents in chromatography
 use melting and boiling point data to identify substances and assess purity
 recognise the importance of measuring purity in everyday life, e.g. drugs, foods
The methods used for separating mixtures depend on the physical properties of components of the mixture. Some
and not all of the methods are listed below.
1.
2.
3.
4.
5.
6.
7.
8.
Filtration
Crystallization
Distillation
Fractional Distillation
Chromatography
Separating funnel
Magnetism
Sublimation
Filtration
It is used for separation of insoluble solid from a liquid. Filtration is used to remove solid impurities from
water/liquid. An example of mixture that can be separated by filtration is sand and water.



The liquid will pass through as filtrate
Sand as insoluble will remain on the filter paper as residue
Filtration is used in the purification of water in water treatment plants
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Crystallisation:
It is a method used to recover a solute from its solution. The solution is evaporated to the crystallization
point i.e. to a point at which crystals of the solute will form on cooling, which can then be filtered out and
dried. E.g. copper (II) sulphate crystals from a solution.
Experiment:
1. Solution is heated, the water evaporates and the solution becomes concentrated.
2. The saturated solution is then allowed to cool, crystals will start to form.
3. The saturated solution is cooled to form, crystals that can be dried on a filter paper.
Uses of Crystallisation:


Crystallization is used in obtaining pure sugar.
Purification of antibiotics is by crystallization.
Evaporation:
It is a process of obtaining a solute from a solution by heating the solution. In the case which we do not need to
collect the solvent. The solvent is boiled off and escape into the air while the solute is left behind in the holding
container.
Note that this method is not suitable for use on solutes which can decompose by heating (e.g. Copper II sulfate).
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Sublimation
Sublimation is used to separate a mixture of solids containing one which sublimes and one (or more than one) which
does not, by heating the mixture. Examples of substances that sublime include, ammonium chloride, iodine crystals,
dry ice.
Distillation:
Distillation is the method of obtaining a pure liquid (solvent) from a solution of a solid (solute).
Distillation Flask:
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In this the solution boils, liquid vaporizes and enters the condenser.
Condenser:
In the condenser vapour cools and it turns into pure liquid.
Important steps to be taken during distillation:




The thermometer should not be dipped into the solution; this ensures that the thermometer measures
the boiling point of the substance that is being distilled.
The condenser consists of two tubes. The cold running water is allowed to enter the condenser from
bottom and leave from the top.
The condenser slopes downwards, so that the pure solvent formed can run into the beaker.
If the distillate is volatile, the receiver can be put in a large container filled with ice.
Uses of Distillation:

It is used to obtain pure water from sea water (desalination).
Fractional Distillation:
It is used to separate mixture of miscible liquids with different boiling points. Miscible liquids are liquids
which form uniform mixture .e.g. water and ethanol with boiling points of 100℃ and 78℃ respectively.
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Fractionating Column:




Fractionating Column is a column containing glass beads wool and it increases the surface area,
ensuring 100% separation and purification.
The water vapour condenses in the fractionating column and drops back into the flask, therefore
insuring more purity.
The liquid with the lowest boiling will distill off first, as the fractionating column will condense the
liquids with higher boiling point and therefore it will drop back into the flask
The thermometer will be used to identify which liquid is getting distilled through its boiling point.
Uses of Fractional Distillation:




It is used in industries to obtain Nitrogen, Argon and Oxygen from Air.
It is used to separate crude oil into petrol, kerosene and other useful components.
Fractional distillation is used to separate fermented liquor into ethanol and water.
It is used to separate a mixture of miscible liquids.
Separating immiscible liquids
Immiscible liquids are those liquids which do not form uniform mixtures when they are mixed. Examples of
immiscible liquids are water and oil. When these two liquids are mixed they form two layers, the denser
liquid at the bottom and less dense on top.
The best method to separate such liquids is by using a separating funnel as shown below.
Separating using magnets
Magnets can be used to separate magnetic substances from non-magnetic substances. For example a mixture
of iron fillings and sulphur can be separated by using a magnet. Iron fillings are attracted to magnets while
sulphur does not get attracted to magnets.
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Chromatography
Chromatography is a separation technique where a solvent is used to separate a mixture into its components.
Chromatography separates the substances according to their solubility in the solvent used.

If paper is used as a separation medium then it is called paper chromatography
Types of chromatography
1. Ascending chromatography
2. Descending chromatography
Terminologies in chromatography



Chromatogram: a chromatography paper showing the results of a chromatography experiment
Solvent front: the furthest distance moved by the solvent
Rf value: the ratio between the distance travelled by the substance (solute) and the distance travelled
by the solvent. The Rf values for substances does not change it is a constant. The Rf values are used
to identify substances on the chromatograms.
Rf =
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑏𝑦 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑏𝑦 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
𝑋
=𝑌
Calculating the Rf value
The figure below shows the chromatograms of the same substance. The chromatograms were
obtained at different times of the experiment and the Rf values calculated.
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Ascending paper chromatography
In ascending paper chromatography, the components of the mixture move up the chromatography paper
as they dissolve. The common solvent used is ethanol or water.
The most soluble substance in the mixture move the further up while the least soluble substance moves
just a shorter distance from the base line.
Descending chromatography
In descending paper chromatography, the components of the sample move down the paper as thy
dissolve. The most soluble component moves further down. Descending paper chromatography is faster
than ascending paper chromatography. This is because descending paper chromatography is helped by
gravity while in ascending chromatography, the components move against gravity.
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Interpreting chromatograms
Different types of chromatograms are obtained from experiments. Below are some of the chromatograms
that can be obtained.
1. Chromatogram of a mixture
 In this chromatogram, more than one spot is formed
2. Chromatogram of a pure substance
 In this chromatogram, only one spot is produced from the original spot
3. Chromatogram of an insoluble substance
In this case, the chromatogram only shows the original spot.
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4. Chromatogram of a colourless substance
Chromatography can also be used for colourless substances such as amino acids. To separate and
analyse colourless substances, we apply a locating agent on a chromatogram. E.g. in order to
separate and analyse amino acids, the steps below can be followed.
i.
Separate the mixture of amino acids by chromatography using a suitable solvent.
ii.
Before the solvent reaches the top of the paper, stop the chromatography and then dry the
paper.
iii.
Spray a locating agent on to the paper (Ninhydrin is usually a chemical substance that is used
as a locating agent when identifying amino acids it produces a blue colour)
iv.
The locating agent reacts with each of the amino acids on the paper to form coloured spots.
By checking the Rf value of each spot the amino acids can be identified.
Uses of chromatography



Identifying substances in a sample
Separating substances in a sample
To determine whether a sample is pure
LANGUAGE OF CHEMISTRY
Chemical and physical change
a. Chemical change
It is a change where a new substance is formed or produced. Chemical changes are also called chemical
reactions.
E.g. burning paper or reaction of magnesium with oxygen produces new substances ash and magnesium
oxide respectively.
b. physical change
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It is a process or change in which there are no new substances which are formed or produced
E.g. boiling water or melting ice or wax
Differences between chemical change and physical change
Chemical change
Mass differs from original substance
Heat is absorbed or released
The change is irreversible
New substance is formed
Physical change
Mass remains the same
No heat is usually evolved
The change is reversible
No new substances are formed
Elements, Compounds and Mixtures
a. Element
An element is a pure substance that cannot be broken down into simpler substances through any chemical or
physical means. E.g. copper, sodium, magnesium, sulphur, iron
b. Compound
A compound is a pure substance that contains two or more elements which are chemically combined in a
fixed ratio. It can consist of either molecules or ions. The properties of a compound differ from its
constituent elements. E.g. water, sugar, salt, iron (II) sulphide
c. Mixture
A mixture consists of two or more substances that are physically combined. These substances can be
elements or compounds. The ratio of these substances in a mixture is not fixed. The components in a mixture
can easily be separated through physical methods. E.g. sugar solution, alloys, air
Differences between a compound and a mixture

Compound
Substances are chemically combined

Mixture
Substances are physically combined

Substances are combined in fixed ratio

Substances in a mixture are in varying
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proportions

Properties of a compound are different
from the substances that make it

Properties of a mixture are a sum of
properties of its components

Cannot be separated into its
components by any means

Can be separated by physical means
into its components
Atoms and Molecules
Atoms
An atom is the smallest indivisible particle of an element that can take part in a chemical change.
Molecules
It is a smallest particle of an element or compound that can exist independently in a free state.
Diatomic molecules
These are molecules that are made up of two atoms of the same element chemically combined. Examples of
diatomic molecules include: oxygen, hydrogen, nitrogen
These molecules are written with the number 2 subscript to show that, two atoms are present in a single
molecule. Example N2, H2, N2
Monoatomic molecules
These molecules are made up of only one atom of that particular element. These do not combine with other
atoms to form compounds. Examples of these molecules are the elements in group VIII of the periodic table
of elements. E.g. Neon, xenon, argon
Triatomic molecules
These are molecules made up or composed of three atoms chemically combined. E.g. water, carbon dioxide,
nitrogen dioxide, ammonia
THE ATOMIC STRUCTURE
Though atoms are very small, they are made up of other very small particles called the sub atomic particles
of an atom. These sub atomic particles are: Protons, Neutrons and Electrons.
The table below summarises the sub atomic particles
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


1 atomic mass unit (amu) is approximately 1.67 × 10–27 kg.
Protons and neutrons are found in the nucleus of an atom. They are collectively known as nucleon
particles.
Electrons are found outside the nucleus. They are arranged in shells, also referred to as energy
levels, which surround the nucleus.
Arrangement of electrons in the various shells
The way electrons are arranged in the various shells or energy levels of an atom is called electron
configuration. The diagrams drawn to show how electrons are arranged in the shells of atoms are called
electron configuration diagrams.


The various shells are assigned letters as K, L, M etc. the formula for the maximum number of
electrons in the shells is 2n2 where, n represents the number of the shell.
The electrons fill the energy levels starting from the energy level nearest to the nucleus, which has
the lowest energy. When this is full (with two electrons) the next electron goes into the second
energy level. When this energy level is full with eight electrons, then the electrons begin to fill the
third and fourth energy levels as stated above
E.g. an atom of sodium has 11 electrons and its electron configuration is as below. That is 2, 8, 1 in the
shells.
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Activity: complete and draw diagrams for the elements in the table below.
Protons, electrons and neutrons
In a neutral atom, the numbers of protons are equal to the number of electrons. The number of protons in the
nucleus of an atom is called the atomic number (Z). The sum of the protons and neutrons is called the mass
number (A). The mass number is sometimes referred to as the nucleon number of the atom.
The mass number then can be calculated by:
A = Z + N; where, N is the number of neutrons and Z the atomic number.
Isotopes
Isotopes are atoms of the same element which have the same number of protons but different number of
neutrons. They can also be defined as atoms of the same element having the same atomic number but
different mass number. E.g. uranium has two isotopes uranium 238 and uranium 235. These are written as
U- 238 and U-235, isotopes are detected by an instrument called the mass spectrometer.
Isotope notation
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In order to distinguish between different isotopes of the same element in writing symbols and formulae, a
simple system is used.
Example: write the notation of the atoms of sodium and magnesium.
Relative atomic mass
The relative atomic mass is given by the symbol Ar. The relative atomic mass of an element is the average
mass of atoms of an element compared to 1/12th the mass of a carbon- 12 atom.

The relative atomic mass has no units
Calculating relative atomic mass
The relative atomic mass of an atom of element can be calculated if the percentage isotopic composition is
known.
Example: calculate the relative atomic mass of chlorine from the isotopic composition of Cl-35 and Cl-37
with isotopic percentages of 75.78% and 24.22% respectively.
Solution
𝟕𝟓. 𝟕𝟖
𝟐𝟒. 𝟐𝟐
𝑿 𝟑𝟓 +
𝑿 𝟑𝟕
𝟏𝟎𝟎
𝟏𝟎𝟎
= (0.7578 X 35) + (0.2422 X 37)
= 35.48
Or
more simple
(75.78 X 35) + (24.22 X 37)
=
𝟑𝟓𝟒𝟖.𝟒𝟒
𝟏𝟎𝟎
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= 35.48
Uses of isotopes
Isotopes have many applications in the real world; some of the uses of isotopes are;





For sterilizing medical instruments
Used as fluid tracers; added to liquids to show if they flowing correctly. E.g. monitoring the flow of
blood ( Technetium- 99)
Treatment of cancer ( Cobalt- 60)
For dating organic fossils. E.g. Carbon- 14
In nuclear reactors to produce energy E.g. Uranium-235
Activity:
1. Given that the percentage abundance of Ne- 20 is 90% and that of Ne-22 with 10%, calculate the
relative atomic mass (Ar) of neon.
2. Naturally occurring silicon consists of 92.2% which has a mass of 27.9769 amu, 4.67% which has a mass of
28.9765 amu, and 3.18% which has a mass of 29.9738 amu. Calculate the atomic mass of silicon.
3. Complete the table of common isotopes below.
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CHEMICAL BONDING
Atoms of elements are not chemically stable until they have 8 valence electrons (octet rule). Atoms gain,
lose or share electrons with other atoms to become chemically stable ( have 8 valence electrons)

Atoms form bonds in order to be stable. A stable atom has a filled up outer most shell or in short has
the electronic configuration of noble gases.
Formation of ions
Non-metals usually gain electrons to form negative ions (anions) while metals usually lose electrons to form
positive ions (cations).
The electrons which metals donate are the ones in the outermost shell (valence electrons)


An ion is a charged particle.
The charge of an ion can be found by finding the difference between the number of electrons and the
number of protons
E.g. formation of chloride ion and sodium ion
Activity:
Write the formulae of the ions of; magnesium, oxygen and aluminium
Ionic bonding
This type of bonding takes place between oppositely-charged ions. This usually occurs for compounds made
from a metal and a non-metal.

Ionic bonds are formed by electron transfer, where metal atoms donate electrons to non-metal atoms.
The ions are arranged in an ionic lattice and are held together by electrostatic forces of attraction.
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
Two examples of dot-and-cross diagrams that illustrate the formation of ionic bonds are as shown.
Magnesium chloride and Sodium chloride
Properties of ionic compounds




They are usually solids at room temperature, with high melting points. This is due to the
strong electrostatic forces holding the crystal lattice together. A lot of energy is therefore
needed to separate the ions and melt the substance.
They are usually hard substances.
They mainly dissolve in water. This is because water molecules are able to bond with both the
positive and the negative ions, which breaks up the lattice and keeps the ions apart.
They usually conduct electricity when in the molten state or in aqueous solution because the
ions become mobile.
Covalent Bonding
Covalent bonds are formed between non-metal atoms. The bond is formed by sharing of electrons between
atoms. A single covalent bond is formed by the sharing of two electrons between two atoms, with the atoms
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contributing one electron each. Covalent substances can be found as simple molecules or as large molecules.
Some of the common covalent compounds are shown below with their electron sharing arrangements.
 Note that only the outermost electrons are used for electron sharing
Covalent bonds are also formed between atoms of the same elements. Hydrogen, oxygen, nitrogen and
halogen (Group VII) elements exist as diatomic molecules by forming covalent molecules of two atoms
bonded together. The covalent bonds in hydrogen and oxygen molecules are shown below.
Properties of covalent compounds




Simple covalent compounds have low melting points due to week intermolecular forces between
the molecules.
They have low boiling points
They do not conduct electricity in any state
They are not soluble in water mostly but soluble in organic solvents
Giant Molecular Structures
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

Covalent substances with giant molecular structures consist of broad network of atoms held together
by covalent bonds.
Substances with giant molecular structures have high melting and boiling points as a lot of energy is
required to overcome the strong covalent bonds holding the atoms together.
Diamond and Graphite
Diamond and graphite are allotropes of carbon which have giant molecular structures.
Allotropy: The existence of an element in two or more different forms in the same physical state.

The carbon atoms in these substances are arranged in different ways, hence giving them different
properties.
Structures of diamond and graphite are shown below
Uses



Due to its rigid structure, diamond is a very hard substance and is used for drill tips or cutting tools.
Graphite consists of many layers of carbon atoms which are held together by weak van der Waals’
forces of attraction. These layers of carbon atoms can slide past each other, making graphite a soft
and slippery substance. This makes graphite suitable for use as a lubricant.
Graphite conducts electricity due to the delocalised electron
Metallic bonding
Atoms in a metal are held by metallic bonding in a giant lattice structure. These atoms lose their valence
electrons, which are then delocalised across the metal lattice. The metal lattice structure consists of
lattice of positive ions surrounded by a ‘sea of electrons’. The electrostatic forces of attraction between
the positive ions and the mobile electrons hold the structure together.
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Properties of metals due to metallic bonding




Metals are good conductors of electricity and heat due to the presence of mobile electrons.
Metals have high melting and boiling points as a lot of energy is required to overcome the strong
electrostatic forces of attraction between the ‘sea of electrons’ and the lattice of positive ions
As atoms in metals are packed tightly in layers, they usually have high densities.
The well-ordered arrangement of atoms also makes metals malleable and ductile, which means
that metals can be shaped by applying pressure and stretched without breaking.
Activity:
Draw dot and cross diagrams for the following covalent compounds.
1. Carbon dioxide
2. Hydrogen chloride
SYMBOLS OF ELEMENTS, FORMULAE AND CHEMICAL EQUATIONS
 state the symbols of the elements named in the syllabus
 state the formulae of the compounds named in the syllabus
 work out the formula of a simple compound from the ratio of the number of atoms present and vice
versa
Symbols of elements
A symbol of an element is a letter (capital letter) or letters (capital and small) written to represent that
particular element. The symbols for elements are derived from their names in English or other languages
names.
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FORMULAE OF COMPOUNDS
A chemical formula of a compound is shorthand way of representing a compound. The steps below are
useful for writing a chemical formula of a compound.
Writing the chemical formula
Every compound is represented by a chemical formula. The following steps should be followed in
writing the chemical formula.
Step 1: the symbol of the positive atom or basic radical is placed on the left-hand side and the symbol
of the negative atom or acid radical is placed on the right-hand side.
Step 2: The valences of the symbols are written below them.
Step 3: Interchange the valences of the symbol and shift it to the lower right corner of the symbol.
The valency numbers become subscripts of the corresponding symbols.
Step 4: Reduce the valency numbers to a simple ratio by dividing with a common factor, if required
Determining the valence of an element
The valence of an element or radical is its combining power.

The valence of an element is the charge an atom takes when it loses or gains electrons and becomes
an ion.
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Examples: what are the valencies for the following elements?
a. Sodium
b. Oxygen
Solution
a. Na has 11 electrons its electron configuration is 2, 8, 1. It has one electron in the outer shell that will
be lost when it forms an ion. Its valency is 1.
b. O has 8 electrons its electronic configuration is 2, 6 it has 6 electrons in the outer most shell and it
will receive 2 electrons to make it stable. Its valency is 2.
Activity:
What are the valences of the following elements?
Calcium, Chlorine, Aluminium
Examples
Write the chemical formulae for the following
a. Aluminum oxide
b. Calcium sulphide
c. Sodium fluoride
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Some common radicals and their valencies
A radical is a group of atoms which are found in different compound but cannot exist independently.
Some radicals are shown below
Example: write the formulae of the following compounds
Calcium phosphate and sodium sulphate
Solution
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Activity: complete the table below.
Cation
Ca2+
Mg2+
Ba2+
Anion
NO3OHSO42-
compound
Mg3(PO4)2
Ba(OH)2
Some elements have more than one valency examples are:
Roman numerals are used in the compound to show the valency on the cation in the compound.
E.g. Iron (II), Iron (III), Copper (I), Copper (II), Manganese (II), Manganese (IV)



Iron (III) oxide means the valency of iron is 3
Copper (II) phosphide means the valency of copper is 2
Manganese(IV) oxide the valency of manganese is 4
Chemical equations
Objectives
 state the symbols of the elements and formulae of the compounds mentioned in the syllabus ( deduce
the formulae of simple compounds from the relative numbers of atoms present and vice versa
 deduce the formulae of ionic compounds from the charges on the ions present and vice versa
 interpret chemical equations with state symbols
 construct chemical equations, with state symbols, including ionic equations
1. Chemical Equations
 A chemical equation shows the reactants and products in a reaction and may include state
symbols, which show the physical states of each substance. Solid, liquid and gaseous states
are represented by the state symbols (s), (l) and (g) respectively. Substances that are
dissolved in water to form an aqueous solution are represented with the state symbol (aq).
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The reactants are on the left hand side and the products are on the right hand side as shown
below for a general reaction.
Writing chemical equations from word equations
In order to be able to write a balanced chemical equation, there are a number of important things that
need to be done:
1. Know the chemical symbols for the elements involved in the reaction
2. be able to write the chemical formulae for different reactants and products
3. Balance chemical equations by understanding the laws that govern chemical change
4. Know the state symbols for the equation
Examples
1. Nitrogen gas reacts with hydrogen gas to form ammonia. Write a balanced chemical equation for this
reaction.
Answer
Step 1: Identify the reactants and the products, and write their chemical formulae the reactants are
nitrogen (N2) and hydrogen (H2), and the product is ammonia gas (NH3).
Step 2: Write the equation so that the reactants are on the left and products on the right of the arrow .The
equation is as follows: N2 + H2 → NH3
Step 3: show the state of the reactants and the products by writing the state symbols.
N2 (g) + H2 (g) → NH3 (g)
2. Hydrogen fuel cells are extremely important in the development of alternative energy sources. Many
of these cells work by reacting hydrogen and oxygen gases together to form water, a reaction which
also produces electricity. Write a chemical equation for the reaction.
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3. Solid Zinc metal reacts with aqueous hydrochloric acid to form an aqueous solution of Zinc chloride
and hydrogen gas. Write a chemical equation for this reaction.
4. When calcium chloride solution is mixed with silver nitrate solution, a white precipitate (solid) of
silver chloride appears. Calcium nitrate (Ca (NO3)2) is also produced in the solution. Write the
equation for the reaction.
Answers
Follow the steps as in example number 1 above
H2 (g) + O2 (g) → H2O (l)
Zn(s) + HCl (aq) → ZnCl2 (aq) + H2 (g)
CaCl2 (aq) + AgNO3 (aq) → AgCl(s) + Ca (NO3)2(aq)
Balancing chemical equations
In a balanced chemical equation, the number of atoms of each element is the same in the reactants and in the
products.
Steps to balance a chemical equation
When balancing a chemical equation, there are a number of steps that need to be followed.

STEP 1: Identify the reactants and the products in the reaction, and write their chemical formulae.

STEP 2: Write the equation by putting the reactants on the left of the arrow, and the products on the
right.

STEP 3: Count the number of atoms of each element in the reactants and the number of atoms of
each element in the products.
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
STEP 4: If the equation is not balanced, change the coefficients of the molecules until the number of
atoms of each element on either side of the equation balance.

STEP 5: Check that the atoms are in fact balanced.
Examples
1. Balance the following equation:
Mg + HCl → MgCl2 + H 2
Answer
Step 1: Because the equation has been written for you, you can move straight on to counting the number of
atoms of each element in the reactants and products
Reactants: Mg = 1; H = 1 and Cl = 1
Products: Mg = 1 atom; H = 2 atoms and Cl = 2 atoms
Step 2: Balance the equation
The equation is not balanced since there are 2 chlorine atoms in the product and only 1 in the reactants. If we
add a coefficient of 2 to the HCl to increase the number of H and Cl atoms in the reactants, the equation will
look like this: Mg + 2HCl → MgCl2 + H2
Step 3: Check that the atoms are balanced If we count the atoms on each side of the equation, we find the
following:
Reactants: Mg = 1; H = 2; Cl = 2
Products: Mg = 1; H = 2; Cl = 2 the equation is balanced.
The final equation is: Mg + 2HCl → MgCl2 + H2
2. Balance the following equation:
CH4 + O2 → CO2 + H2O
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Answer
Step 1: Count the number of atoms of each element in the reactants and products Reactants: C = 1;
H = 4; O = 2 Products: C = 1; H = 2; O = 3
Step 2: Balance the equation
If we add a coefficient of 2 to H2O, then the number of hydrogen atoms in the reactants will be 4,
which is the same as for the reactants.
The equation will be: CH4 + O2 → CO2 + 2H2O
Step 3: Check that the atoms balance
Reactants: C = 1; H = 4; O = 2 Products: C = 1; H = 4; O = 4
Although the number of hydrogen atoms now balances, there are more oxygen atoms in the products.
You now need to repeat the previous step. If we put a coefficient of 2 in front of O2, then we will
increase the number of oxygen atoms in the reactants by 2. The new equation is: CH4 + 2O2 → CO2
+ 2H2O
Activity:
1. In our bodies, sugar (C6H12O6) reacts with the oxygen we breathe in to produce carbon dioxide,
water and energy. Write the balanced equation for this reaction.
2. Ethane a gas (C2H6) reacts with oxygen to form carbon dioxide and steam. Write a balanced
chemical equation for the reaction.
3. Mg + P4 → Mg3P2
4.
Ca + H2O → Ca(OH)2 + H2
5. CuCO3 + H2SO4 → CuSO4 + H2O + CO2
6. CaCl2 + Na2CO3 → CaCO3 + NaCl
7. C12H22O11 + O2 → CO2 + H2O
Ionic equations
Ionic equations are chemical equations that show the reaction involving ions.
Steps to writing ionic equations
 A balanced chemical equation with state symbols is first written.
 Take note of the physical states of all substances in the reaction.
 Break this equation down further by writing the dissolved (aqueous) substances in terms of ions. It is
useful to note the solubility rules of salts in water before rewriting the equation.
 The ionic equation is then obtained by cancelling out spectator ions.
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Spectator ions remain unchanged at the end of the reaction, showing that they do not take part in
the reaction.
Examples:
1. Solid zinc metal reacts with aqueous hydrochloric acid to form an aqueous solution of zinc chloride
(ZnCl2) and hydrogen gas. Write the ionic equation for the reaction.
2. Write the ionic equation for the reaction below
3. Aluminium oxide can react with dilute sulphuric acid to form aluminium sulphate and water.
a. Write a well-balanced equation for the reaction including state symbols.
b. Write the ionic equation for the reaction
4. When zinc metal is added to a solution of copper (II) sulphate, copper metal and a solution of zinc
sulphate are the products. From this information, write the ionic equation for the reaction.
Activity:
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1. Sodium hydroxide reacts with a solution of iron (II) sulphate to give a green precipitate (solid) of
iron (II) hydroxide and a solution of sodium sulphate.
a. Write a well-balanced chemical equation for the reaction. Include state symbols
b. Write the ionic equation for the reaction
2. Write ionic equations from chemical equations below
a. CaCO3(s) + H2SO4(aq) → CaSO4(aq) + H2O(l) + CO2(g)
b. Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
c.
2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)
d. Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)
THE PERIODIC TABLE
Objectives

describe the Periodic Table as an arrangement of the elements in the order of increasing proton
(atomic) number

describe how the position of an element in the Periodic Table is related to proton number and
electronic structure

describe the relationship between group number and the ionic charge of an element

explain the similarities between the elements in the same group of the Periodic Table in terms of their
electronic structure

describe the change from metallic to non-metallic character from left to right across a period of the
Periodic Table

describe the relationship between group number, number of valence electrons and
metallic/nonmetallic character

predict the properties of elements in Group I and Group VII using the Periodic Table

describe chlorine, bromine and iodine in Group VII (the halogens) as a collection of diatomic
nonmetals showing a trend in colour, state and their displacement reactions with solutions of other
halide ions

describe the elements in Group 0 (the noble gases) as a collection of monatomic elements that are
chemically unreactive and hence important in providing an inert atmosphere

describe the transition elements as metals having high melting points, high density, variable
oxidation state and forming coloured compounds
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1. Features of the Periodic Table
Elements are arranged in order of increasing atomic numbers in the Periodic Table. They are
organised into horizontal rows known as periods, and vertical columns known as groups.
Periods
The periodic table has seven (7) periods. The periods run horizontally from left to right.
Each element in a period has a proton number which is one less than the element after it.

In a period, metals are found on the left side while the non-metals are found on the right side. The
metallic character of elements decreases as we move from left to right of a period. Elements in the
same period have the same number of electron shells. The number of electron shells corresponds
with the period number of the element. For example, aluminium belongs to Period 3 and has three
electron shells.
Groups
There are eight (8) groups on the periodic table. Each group is indicated by Roman numerals (I to
VIII).
Elements in the same group have the same number of valence electrons. The number of valence
electrons for each of the elements corresponds with the group number. For instance, Group I
elements (e.g. lithium, sodium, potassium) each have one valence electron, while Group II elements
(e.g. Magnesium, calcium) have two valence electrons each.
Group I elements (alkali metals)
Elements in Group I are also known as alkali metals. The atoms of these elements have one valence
electron each.
Physical properties
 These metals are soft and can be cut easily with a knife.
 They have relatively low melting and boiling points.
 Their densities are relatively low. Lithium, sodium and potassium have densities lower than
water, enabling them to float.
 Moving down the group, the melting and boiling points decrease while the densities increase.
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Chemical properties


Alkali metals are highly reactive metals. They react easily with oxygen and water.
The reactivity of these metals increases as we move down the group. This is due to an
increase in the atom size, which means that the valence electrons are further away
from the nucleus and are more easily lost.
Alkali metals react with water to form an alkali and hydrogen gas. The trend in
reactivity can be observed from their reactions with water.
Alkali metal
Lithium
Sodium
Potassium
Reaction with water
Reacts quickly with water, no flame is seen and floats on water
Reacts very quickly , melts and yellow flame is seen
Reacts violently and always catches fire

Lithium reacts quickly with cold water, but potassium reacts very violently with cold
water. As alkali metals easily give away their valence electrons, they are strong
reducing agents. These metals react with non-metals to form ionic salts which are
soluble in water.
 Because metals of group I are very reactive, they are usually kept in oil to prevent
their reaction with water or air.
Group II (alkaline earth metals)
 Group II consists of the five metals beryllium, magnesium, calcium, strontium and barium, and the radioactive
element radium.


They are harder than those in Group I.
They are silvery-grey in colour when pure and clean. They tarnish quickly, however, when left in air due to
the formation of a metal oxide on their surfaces
Group VII (halogens)
Elements in Group VII are also known as halogens. Atoms of these elements have seven valence electrons
each. These are non-metals that are found as diatomic molecules (e.g. Cl2, Br2, I2).
Physical properties
 Since they are found as simple covalent molecules, they have low melting and boiling points.
 Moving down the group, the melting and boiling points increase.
Element
Chlorine
Bromine
Iodine
State at room temperature
Gas
Liquid
Solid
Colour
Greenish-yellow
Reddish-brown
Black
Chemical properties
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

Halogens are highly reactive non-metals as they only need to gain one electron for a noble gas
electronic configuration.
The reactivity in group VII decreases down the group. Of the three halogens chlorine is the most
reactive.
Displacement reactions of the halogens
A displacement reaction is a reaction in which one element takes the place of another element in a
compound. A more reactive halogen will displace a less reactive halogen from its halide solution.
 For instance, when chlorine gas is bubbled into sodium bromide solution, bromide ions get displaced.
The solution changes from colourless to reddish-brown as bromine molecules are produced in the
reaction.
Uses of halogens




Fluorine is used in the form of fluorides in drinking water and toothpaste because it reduces tooth decay by
hardening the enamel on teeth.
Chlorine is used to make PVC plastic as well as household bleaches. It is also used to kill bacteria and viruses
in drinking water
Bromine is used to make disinfectants, medicines and fire retardants.
Iodine is used in medicines and disinfectants and also as a photographic chemical.
Group VIII elements
The elements in Group VIII or 0 are called the noble gases. They are the least reactive elements in the periodic table.
Apart from helium, the other noble gases have eight valance electrons. Their full electronic structures make the noble
gases unreactive.
Physical properties
 They are monoatomic elements
 Colourless gases at room temperature
 Have low melting and boiling points that increases down the group
 Are insoluble in water
 Are unreactive due to filled up outer shells
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Uses of noble gases
Element
Helium
Neon
Argon
Krypton
Xenon
Application
Weather balloons
Advertising signs or lights
Light bulbs and welding
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Transition elements
Transition Elements Transition elements are a block of metals found between Groups II and III in the
Periodic Table.
Properties of transition elements




Uses

These metals have high melting and boiling points
They have high densities.
Compounds of transition elements are usually coloured.
Transition elements have variable oxidation states. For example, iron commonly forms Fe2+ and Fe3+
ions.
Transition elements and their compounds are good catalysts and are commonly used in industrial
processes. For example, nickel is used in the manufacture of margarine (hydrogenation of vegetable
oil) and iron is used in the Haber process (manufacture of ammonia).
Activity:
1. Lithium is in Group I of the periodic table. Lithium reacts with water to form lithium hydroxide and
hydrogen.
a. Describe what you would observe when a small piece of lithium is dropped into the surface of
cold water.
b. Write the equation for the reaction between lithium and water
c. Rubidium Rb is another element in Group I.
Predict what would happen when a small piece of rubidium is dropped into cold water.
ACIDS, BASES AND SALTS
Acids and Bases
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a. Acids
Definition: An acid is a substance that dissolves in water to produce hydrogen ions (H+).
Examples of acids
Mineral acids

Hydrochloric acid

Sulphuric acid

Nitric acid

Phosphoric acid
Organic acids
Ehanoic acid
Physical properties of acids

They have a sour taste

Turn blue litmus paper red

They a PH value less than 7
Acid Strength and Concentration
Acid strength is determined by the degree of ionisation of an acid in water. A strong acid fully ionises in
water to form hydrogen (H+) ions. Such acids include hydrochloric acid, sulfuric acid and phosphoric acid.
E.g. HCl (aq) → H+ (aq) + Cl-(aq)
A weak acid partially ionises in water. The partial dissociation is represented in an equation with a ⇌
symbol. Examples of weak acids include carboxylic acids, such as ethanoic acid (CH3COOH).
CH3COOH (aq) ⇌ CH3COO-(aq) + H+ (aq)
Chemical properties of acids
1. Reaction with metals
2. Reaction with bases
3. Reaction with carbonates
1. Reaction with metals
Dilute acids react with metals that lie above hydrogen in the reactivity series. The reaction produces salt and hydrogen
gas.
Zinc + dilute hydrochloric acid → zinc chloride + hydrogen
Zn(s) + 2HCl (aq) → ZnCl2 (aq) + H2 (g)
Metals such as copper and gold do not react with dilute acids as they are unreactive
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2. Reaction with bases
Acids react with bases to form salts and water as the only products. A base could be a metal oxide or an alkali. The
reaction of an acid and a base is called a neutralisation reaction. Examples are shown below
Al2O3(s) + 6HCl (aq) → 2AlCl3 (aq) + 3H2O (l)
i.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
ii.
3. Reaction with carbonates and hydrogen carbonates
Acids react with carbonates (and hydrogen carbonates) to produce salt, water and carbon dioxide.
i.
ii.
CaCO3(s) + H2SO4 (aq) → CaSO4 (aq) + H2O (l) + CO2 (g)
NaHCO3(s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)
Uses of acids
Acids are important substances in the chemical industries.
 manufacture of detergents
 Manufacture of fertilisers.
 used in car batteries as an electrolyte
b. Bases
A base is a substance that produces hydroxides ions as the only negatively charged ions when it is dissolved in water.
 A base can be an oxide or hydroxide of a metal
 Some bases are soluble in water and are called alkalis
 Ammonia is also a weak base
Examples of bases
1. Sodium hydroxide
2. Calcium hydroxide
3. Potassium hydroxide
 insoluble bases are formed by transition elements
Physical properties of bases
 They have a bitter taste
 Have a pH value greater than 7
 They turn red litmus paper blue
Chemical properties of bases
1. Reaction with acids
2. Reaction with ammonium salts
3. Precipitation of insoluble metal hydroxides
1. Reaction with acids
Alkalis undergo neutralisation with acids to produce salt and water only. Neutralisation involves the reaction between
a base and acid to produce water. This can be described in the following ionic equation.
H+ (aq) + OH-(aq) → H2O (l)
2. Reaction with ammonium salts
Heating alkalis with ammonium salts produces salt, water and ammonia gas.
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NaOH (aq) + NH4Cl (aq) → NaCl (aq) + H2O (l) + NH3 (g)
3. Reaction with aqueous salt solutions
Alkalis react with aqueous salts to precipitate insoluble hydroxides. For example, copper(II) hydroxide can be formed
by reacting aqueous copper(II) sulphate with an alkali .
CuSO4 (aq) + 2NaOH (aq)
Cu (OH) 2(s) + Na2SO4 (aq)
Oxides
Oxides are compounds formed from oxygen and another element. These can be categorised into four types of oxides,
namely acidic oxides, basic oxides, amphoteric oxides and neutral oxides.
Acidic oxides
 These are oxides of non-metals
 These oxides can dissolve in water to give acids. Acidic oxides react with bases to form salt and water. For
example, carbon dioxide reacts with calcium hydroxide to form calcium carbonate and water.
Examples: carbon dioxide, sulphur dioxide, nitrogen dioxide
Basic oxides
 Metals usually form basic oxides. Some of these oxides dissolve in water to give alkalis. Basic oxides react
with acid to form salt and water.
 For example, magnesium oxide reacts with sulfuric acid to form magnesium sulphate and water.
Examples: calcium oxide, sodium oxide
Amphoteric oxides
 These oxides display both acidic and basic properties and as such, can react with both acids and bases. Such
oxides include aluminium oxide (Al2O3), zinc oxide (ZnO) and lead (II) oxide (PbO).
Neutral oxides
 These exhibit neither basic nor acidic properties. Instances of such oxides are water (H2O), carbon monoxide
(CO) and nitrogen oxide (NO).
pH scale
 pH of a substance is the measure of alkalinity or acidity of a solution. The pH scale has a set of numbers from
0 to 14.
 Acids have a pH value less than 7; alkalis have a pH value greater than 7; and a neutral substance has a pH
value of exactly 7.



Strong acids have very small pH values less than 7 while strong alkalis have greater pH values above 7.
Importance of pH
The knowledge of pH is important in agriculture. When the soil pH is acidic, agriculture lime (calcium
hydroxide) is added to neutralise it.
In the manufacture of soaps, detergents and house hold bleaches.
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Salts
A salt is a compound formed when the hydrogen of an acid is fully or partially replaced by a metallic or
ammonium ion.
Salts can be classified into the following types:
a) Normal salt
A salt that does not contain any replaceable hydrogen ions is a normal salt. It is obtained by replacing all the
hydrogen ions of an acid by metal ions or ammonium ions. Examples: NaCl, Na2SO4, Na3PO4, NH4Cl,
K2CO3, CuSO4 etc.
b) Bisalts or Acid salts
A bisalt, or an acid salt contains replaceable hydrogen ions in association with the acidic radicals. It is
obtained by the incomplete replacement of hydrogen ions by metal ions or ammonium ions.
Examples: NaHCO3, NaH2PO4 and Na2HPO4.
Since such salts also contain hydrogen ion, the term 'bi' or acid can be substituted by hydrogen, e.g. sodium
bicarbonate or sodium acid carbonate is also called sodium hydrogen carbonate.
When bisalts or acid salts dissociate in water, they yield hydrogen or hydronium ions.
Since an acid salt liberates hydrogen or hydronium ions, it exhibits some acidic properties. Hence it behaves
like an acid.
i) An acid salt reacts with bases to form salt and water only.
iii) They react with carbonates to yield carbon dioxide.
Remember :
Some Common Acid Salts
NaHCO3, NaHSO3, NaHSO4, Na2HPO4, Ca(HCO3)2, Ca(HSO4)2, KHCO3, KHSO3, KHSO4, NaH2PO4and
Mg(HCO3)2
c) Basic Salts
A basic salt is formed by the action of an acid with higher proportion of the base, than is necessary for the
formation of a normal salt.
Example: Basic copper carbonate CuCO3.Cu(OH)2 ,Basic lead carbonate (white lead) PbCO3.Pb(OH)2
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A basic salt may also be formed by the partial replacement of the hydroxyl groups of a diacidic or triacidic
base, by an acid radical. Basic salts are usually insoluble in water.
Solubility of salts
The table below summarises the solubility of salts in water.
Methods of preparing salts
The methods used for preparing a particular salt depends on whether the salt is soluble or not and also on the nature of
the reactants.
Preparation of soluble salts
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1. by titration
E.g. sodium chloride
Reagents: dilute hydrochloric acid and aqueous sodium hydroxide
Procedure:
The preparation of any salt by titration involves
a. Titration to determine the volume of reactants required.
b. Actual preparation of the salt
Procedure:
1. Fill up the burette with dilute hydrochloric acid. Note the initial burette reading (V1cm3)
2. Pipette 25.0cm3 of the aqueous sodium hydroxide into a conical flask
3. Add one or two drops of indicator to the sodium hydroxide in the conical flask and swirl.
4. Add dilute hydrochloric acid from the burette slowly until the solution just changes colour. Add dilute
hydrochloric acid drop wise until a permanent colour change.
5. Record the final burette reading (V2). Hence the volume of dilute hydrochloric acid needed is given by V2-V1.
Making sodium chloride

Repeat the steps 1 and 2 ( do not add any indicator)

Add the volume of hydrochloric acid determined in step 5 using burette

Heat the solution to evaporate the water until it is saturated.

Allow the solution to cool in order for crystals to form.

Filter the crystals and dry them on filter paper
2. Dilute acid and relatively reactive metal (Direct synthesis )
E.g. Zinc Sulphate
Reagents: Zinc metal (powder) and dilute sulphuric acid
Procedure:
1. Fill half a beaker with dilute sulphuric acid
2. Add Zinc powder a little at a time while stirring until no more Zinc dissolves or until no more effervescence
3. Filter the mixture to remove the excess Zinc to obtain a solution of Zinc sulphate
4. Heat the filtrate to evaporate the water and saturate it
5. Allow the solution to cool and crystals to form
6. Filter off the crystals, wash them with a little cold distilled water and dry them on a filter paper.

Excess metal must be used to make sure that all the acid is used up
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3. Reaction of a dilute acid with an insoluble base ( digestion method)
Example: copper (II) sulphate
Reagents: copper (II) oxide and dilute sulphuric acid
Procedure:
1. Fill half a beaker with dilute sulphuric acid and warm the acid
2. Add the black copper (II) oxide powder a little at a time while stirring until it is in excess
3. Filter the mixture to remove the excess copper (II) oxide and obtain the filtrate ( copper(II) sulphate solution)
4. Heat the filtrate in order to evaporate the water and concentrate it
5. Allow the solution to cool or cool in an ice bath to form crystals
6. Filter off the crystals and dry them on a filter paper
Note: the procedure for preparation of a salt from an insoluble carbonate is similar to the procedure above, except for
the heating of acid part.
Preparation of insoluble salts
Insoluble salts are prepared by the precipitation reaction of two soluble salt solutions.
Method: precipitation
Example: Barium sulphate
Reagents: Barium nitrate and sodium sulphate
Procedure:
1. Pour about 50cm3 of barium nitrate solution into a beaker
2. Add sodium sulphate solution(in excess) and stir until no more precipitate forms
3. Filter to collect the precipitate( barium sulphate)
4. Wash the precipitate with a small amount of cold distilled water.
5. Allow the precipitate to dry on a piece of filter paper
Note: the reagents are always soluble salt solutions
Qualitative analysis (Identification of ions)
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STOICHIOMETRY AND THE MOLE CONCEPT
Relative molecular mass
The relative molecular mass (Mr) of an element or compound is the average mass of the molecule compared
to
1
12
the mass of one atom of carbon- 12.

The relative molecular mass is calculated by adding atomic masses of elements present in a
compound.
Ionic compounds are not made up of molecules; therefore, it is better to say formula mass when dealing with
ionic compounds.
 Relative molecular mass and relative formula mass have no units
Example:
Calculate the relative molecular mass of the following
a. Sulphuric acid
b. Water
c. Carbon dioxide
Solution
Relative atomic mass
The relative atomic mass of an atom is the number of times the mass of one atom of an element is greater
1
than 12 the mass of an atom of carbon -12.



The symbol for relative atomic mass is Ar
It has no units because it is a ratio
Relative atomic numbers seldom occur as whole numbers, this is because most elements occur as
mixtures of isotopes.
Example:
Chlorine exists in two isotopic forms: chlorine-35 and chlorine- 37. A sample of chlorine is made up of 75%
of chlorine- 35 atoms and 25% of chlorine-37 atoms.
Hence, relative atomic mass of chlorine
=(
75
100
x 35) + (
25
100
x 37)
= 26.25 + 9.25
= 35.5
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The mole
 A mole is a measure of the amount of substance.
It is the amount of substance containing as many elementary particles as in 12g of carbon -12.
 Elementary particles could be; atoms, molecules, ions or electrons
 The units for a mole are the mol
One mole of a substance contains approximately 6.02 x 1023 particles. These particles could be atoms,
molecules, ions or electrons. 6.02 x 10 is known as the Avogadro’s number or Avogadro’s constant.
Relationship between number of moles and number of particles
Number of moles =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠
𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜′ 𝑠𝑛𝑢𝑚𝑏𝑒𝑟
Example:
1. Convert 1x 1023 of neon atoms to moles of neon atoms.
Solution
Number of moles of neon =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑜𝑓 𝑛𝑒𝑜𝑛
𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜′ 𝑠𝑛𝑢𝑚𝑏𝑒𝑟
2. How many iron atoms are in 0.5 mol of iron?
Solution
Molar mass
Molar mass (M) is the mass of 1 mole of a chemical substance. The unit for molar mass is grams per mole or
g mol−1.
mass
𝑚
Number of moles =
; n = where n is the number of moles, m is the
Molar mass
M
mass and M is molar mass.
Converting moles to mass and mass to moles
Examples
1. What is?
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a. the mass of 1 mole ethanol
b. the relative formula mass (RFM) of ethanol, C2H5OH?
(Ar: H = 1; C = 12; O = 16)
a One mole of C2H5OH contains 2 moles of carbon atoms, 6 moles of hydrogen atoms and 1 mole of
oxygen atoms. Therefore: mass of 1 mole of ethanol
= (2 × 12) + (6 × 1) + (1 × 16)
= 46g
2. Calculate the number of moles of magnesium oxide, MgO, in a 80 g and b 10 g of the compound.
(Ar: O = 16; Mg = 24)
a. One mole of MgO contains 1 mole of magnesium atoms and 1 mole of oxygen atoms.
3. Calculate the mass of 3.01 x1023 atoms of copper.
Number of moles of copper =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑜𝑓 𝑐𝑜𝑝𝑝𝑒𝑟
𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜′ 𝑠𝑛𝑢𝑚𝑏𝑒𝑟
Molar volume of gases
Avogadro’s law:
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Equal volumes of gases, at the same temperature and pressure, contain the same number of particles.
One mole of any gas occupies a volume of approximately 24 dm3 (24 litres) at room temperature and
pressure (rtp) and 22.4 dm3 at standard temperature and pressure (stp). This quantity is also known as the
molar gas volume, Vm.
1dm3 = 1litre = 1000cm3
Examples:
1. Calculate the number of moles of ammonia gas, NH3, in a volume of 72 dm3 of the gas measured at
rtp.
2. Calculate the volume of carbon dioxide gas, CO2, measured at rtp occupied by;
i. 5 moles
ii. 0.5 mole
3
3. A jar contains 120 cm of oxygen gas. Find the number of moles of oxygen at rtp.
4. What is the volume of 7.1 g of chlorine gas at rtp?
Solution
Concentration of solutions
 Concentration refers to the amount of solute dissolved in a solvent.
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

It tells us the amount of solute per unit volume of a solution.
We can measure concentration in g/dm3 or mol/dm3 which is also called the solution’s molarity (M).
Converting to g/dm3
A solution of glucose contains 0.45g of glucose in 75cm3 of solution. What is the concentration of glucose
solution in g/dm3?
Molar concentration
The concentration of a solution when expressed in mol/dm3 is called molar concentration. A solution of 1
mol/dm3 has one mole of solute dissolved in 1dm3 of solution.
Examples:
1. Calculate the concentration in mol/dm3 of a solution of sodium hydroxide made by dissolving 10g of
sodium hydroxide pallets in water making a volume of 250cm3.
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2. Calculate the mass of potassium hydroxide, KOH, which needs to be used to prepare 500 cm3 of a 2
mol dm−3 solution in water. (Ar: H = 1; O = 16; K = 39)
3. Calculate the volume of a 2moldm-3 solution that contains 0.5 mol of solute.
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Dilution
Dilution is the decrease in concentration of a solution.
 The dilution formula can be used in dilution calculations
M1V1 = M2V2; where, M1 and M2 are the initial and final concentrations while V1 and V2 are the
initial and final volumes.
Examples:
1. To what volume must 300cm3 of 0.60M sodium hydroxide solution be diluted to give a 0.40 M
solution?
2. You are given a stock solution of concentrated ammonia with a concentration of 17.9M. What
volume of concentrated ammonia is needed to make up 1dm3 of 0.1M ammonia solution?
3. Given a stock solution of 2.0 mol/dm3sodium chloride, how would you prepare 250cm3 of a 0.5
mol/dm3 solution?
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Chemical calculations
Chemical calculations are important when chemicals are produced on a large scale. In chemistry
chemical calculations are used to predict the amount of reactants and products expected from a chemical
reaction.
 The relationship between the amounts (measured in moles) of the reactants and products involved in
a chemical reaction is called the stoichiometry of the reaction.
 Whenever calculations are performed from chemical equations, the equations have to be balanced.
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Calculating masses
Examples:
1. What mass of magnesium oxide would be produced from 16 g of magnesium in the reaction
between magnesium and oxygen?
i Write the full balanced equation
2Mg(s) + O2(g) → 2MgO(s)
ii Read the equation in terms of moles
2 moles of magnesium reacts to give 2 moles of magnesium oxide
iii Convert the moles to masses using the Mr values
∴ (2 x 24g) of magnesium gives 2 x (24+16) = 80 g of Magnesium oxide
𝑋 16
∴ 16 g of magnesium gives 80
= 𝟐𝟔. 𝟕 𝐠 𝐨𝐟 𝐦𝐚𝐠𝐧𝐞𝐬𝐢𝐮𝐦 𝐨𝐱𝐢𝐝𝐞
2 𝑋 24
2. What mass of lead (II) sulphate would be produced by the action of excess dilute sulphuric acid
on 10 g of lead nitrate dissolved in water?
Pb(NO3)2(aq) + H2SO4(aq) → PbSO4(s) + 2HNO3(aq)
∴ 1 mole of lead(II) nitrate gives 1 mole of lead sulphate
∴ 331 g of lead(II) nitrate gives 303 g of lead sulphate
∴ 10 g of lead nitrate gives
303𝑔 𝑋 10𝑔 𝑜𝑓 𝑙𝑒𝑎𝑑(𝐼𝐼) 𝑠𝑢𝑙𝑝ℎ𝑎𝑡𝑒
331 𝑔
= 9.15 g of lead (II) sulphate
3. When excess carbon dioxide is passed into sodium hydroxide solution, sodium carbonate
solution is formed. This can be crystallised out as Na2CO3.10H2O. What mass of crystals would
be produced from 5 g of sodium hydroxide in excess water?
Note. You need the water as moles in the equation.
2NaOH(aq) + CO2(g) + 9H2O → Na2CO3(aq) + 10H2O(l) → Na2CO3.10H2O(s)
∴ 2 moles of sodium hydroxide give 1 mole of the crystals of sodium carbonate
∴ 2 x 40 g of sodium hydroxide give 286 g of crystals
∴ 5 g of sodium hydroxide give
286 𝑋 5
2 𝑋 40
= 17.88 g of crystals
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Percentage Yield
When you write an examination, the highest grade that you can earn is usually 100%. Most people, however,
do not regularly earn a grade of 100%. There are several factors that come into play to prevent from scoring
a 100%. A farmer may plant seed and expect a certain yield of crop of up to 100% his input! But many
things come into play which make it impossible to realize 100% yield. For example, rainfall, type of soil,
general care etc. similarly, in a chemical process or reaction, a 100% yield is desired but not attainable due
to several factors.
Theoretical Yield and Actual Yield
Chemists use stoichiometry to predict the amount of product that can be expected from a chemical reaction.
The amount of product that is predicted by stoichiometry is called the theoretical yield. This predicted
yield, however, is not always the same as the amount of product that is actually obtained from a chemical
reaction. The amount of product that is obtained in an experiment is called the actual yield.
Calculating Percentage Yield
The percentage yield of a chemical reaction compares the mass of product obtained by experiment (the
actual yield) with the mass of product determined by stoichiometric calculations (the theoretical yield). It is
calculated as follows:
Actual yield
Percentage yield = Theoretical yield × 100%
Example
Ammonia can be prepared by reacting nitrogen gas, taken from the atmosphere, with hydrogen gas.
N2(g) + 3H2(g) → 2NH3(g)
When 7.5 g of nitrogen reacts with sufficient hydrogen, the theoretical yield of ammonia is 9.10 g. (You can
verify this by doing the stoichiometric calculations.) If 1.72 g of ammonia is obtained by experiment, what is
the percentage yield of the reaction?
What Is Required?
You need to find the percentage yield of the reaction.
What Is Given?
actual yield = 1.72 g
theoretical yield = 9.10 g
Solution
Actual yield
Percentage yield = Theoretical yield × 100%
1.72𝑔
9.10𝑔
𝑋 100%
= 18. 9 %
TASK
1.
20.0 g of bromic acid, HBrO3, is reacted with excess HBr.
HBrO3(aq) + 5HBr(aq) → 3H2O() + 3Br2(aq)
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(a) What is the theoretical yield of Br2 for this reaction?
(b) If 47.3 g of Br2 is produced, what is the percentage yield of Br2?
2.
Barium sulfate forms as a precipitate in the following reaction:
Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaNO3(aq)
When 35.0 g of Ba(NO3)2 is reacted with excess Na2SO4, 29.8 g of BaSO4 is recovered by the
chemist.
(a) Calculate the theoretical yield of BaSO4.
(b) Calculate the percentage yield of BaSO4.
3.
Yeasts can act on a sugar, such as glucose, C6H12O6, to produce ethyl alcohol, C2H5OH, and carbon
dioxide. C6H12O6 → 2C2H5OH + 2CO2
If 223 g of ethyl alcohol are recovered after 1.63 kg of glucose react, what is the percentage yield of
the reaction?
Percentage purity
The percentage purity of a sample describes what proportion, by mass, of the sample is composed of a
specific compound or element. For example, suppose that a sample of gold obtained from Kasengele mine in
North Western province has a percentage purity of 98%. This means that every 100 g of the sample contains
98 g of gold and 2 g of impurities. You can apply your knowledge of stoichiometry and percentage yield to
solve problems related to percentage purity.
Percentage purity =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑝𝑢𝑟𝑒 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 𝑎 𝑠𝑎𝑚𝑝𝑙𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒
𝑋 100%
Examples
Iron pyrite, FeS2, is known as “fool’s gold” because it looks similar to gold. Suppose that you have a 13.9 g
sample of impure iron pyrite. (The sample contains a non-reactive impurity.) You heat the sample in air to
produce iron (III) oxide, Fe2O3, and sulfur dioxide, SO2.
If you obtain 8.02 g of iron (III) oxide, what was the percentage of iron pyrite in the original sample?
Assume that the reaction proceeds to completion. That is, all the available iron pyrite reacts completely
Steps:
1. Write a well-balanced equation for the reaction
4FeS2(s) + 11O2(g) → 2Fe2O3(s) + 8SO2(g)
2. Use your stoichiometry problem-solving skills to find the mass of Fe2S expected to have produced
8.02 g Fe2O3.
3. Determine percentage purity of the Fe2S using the formula:
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Task
1. An impure sample of sodium sulphate (Na2SO4) has a mass of 1.56 grams. This sample is dissolved
and allowed to react with barium chloride (BaCl2) solution. The precipitate has a mass of 2.15
grams. Calculate the percentage of sodium sulphate (Na2SO4) in the original sample.
2. An impure, 0.500 grams sample of sodium chloride (NaCl) was dissolved in 20.0 mL of water. The
chloride ions were precipitated completely by addition of a silver nitrate (AgNO3) solution. The
dried silver Chloride (AgCl) precipitate has a mass of 1.15 grams.
a. How many moles of silver Chloride (AgCl) formed?
b. How many moles of sodium Chloride (NaCl) were in the sample?
c. How many grams of sodium Chloride were in the sample?
d. What was the percentage of sodium Chloride in the impure sample?
3. An impure sample of sodium sulphate (Na2SO4) has a mass of 1.65 grams and is dissolved in water.
Addition of barium chloride (BaCl2) solution produced a precipitate of barium sulphate with mass
2.32 grams. What is the percentage of sodium sulphate (Na2SO4) in the impure sample?
The Limiting Reactant
The reactant that is completely used up in a chemical reaction is called the limiting reactant. In
other words, the limiting reactant determines how much product is produced. When the limiting
reactant is used up, the reaction stops. In real-life situations, there is almost always a limiting
reactant.
A reactant that remains after a reaction is over is called the excess reactant. Once the limiting
reactant is used, no more product can be made, regardless of how much of the excess reactants may
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be present.
When you are given amounts of two or more reactants to solve a stoichiometric problem, you first
need to identify the limiting reactant. One way to do this is to find out how much product would be
produced by each reactant if the other reactant were present in excess. The reactant that produces the
least amount of product is the limiting reactant.
Example
Lithium nitride reacts with water to form ammonia and lithium hydroxide, according to the
following balanced chemical equation:
Li3N(s) + 3H2O(l) → NH3(g) + 3LiOH(aq)
If 4.87 g of lithium nitride reacts with 5.80 g of water, find the limiting reactant.
Strategy
Convert the given masses into moles. Use the mole ratios of reactants and products to determine
how much ammonia is produced by each amount of reactant. The limiting reactant is the reactant
that produces the smaller amount of product.
Steps:
4.87 g
Number of moles for Li3N = 34.8 g/mol
= 0. 140 mol of Li3N
5.80 g
Number of moles for H2O = 18.0 g/mol
= 0.322 mol of H2O
Now calculate the amount of ammonia produced based on Li3N
1 𝑚𝑜𝑙 NH3
Number of moles of ammonia = 1 𝑚𝑜𝑙 Li3N ( 0.140 𝑚𝑜𝑙 𝑜𝑓 Li3N)
= 0.140 mol of NH3
Now calculate the amount of ammonia produced based on H2O
1 𝑚𝑜𝑙 NH3
Number of moles of ammonia = 3 𝑚𝑜𝑙
H2O
𝑋 ( 0.322 𝑚𝑜𝑙 H2O)
= 0.107 mol of NH3
The water would produce less ammonia than the lithium nitride. Therefore, the limiting reactant is
water.
Task
1. The following balanced chemical equation shows the reaction of aluminum with copper (II)
chloride. If 0.25 g of aluminum reacts with 0.51 g of copper (II) chloride, determine the limiting
reactant.
2Al(s) + 3CuCl2(aq) → 3Cu(s) + 2AlCl3(aq)
2. Hydrogen fluoride, HF, is a highly toxic gas. It is produced by the double displacement reaction
of calcium fluoride, CaF2, with concentrated sulfuric acid, H2SO4.
CaF2(s) + H2SO4 (aq) → 2HF (g) + CaSO4(s)
Determine the limiting reactant when 10.0 g of CaF2 reacts with 15.5 g of H2SO4.
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3. Acrylic, a common synthetic fibre, is formed from acrylonitrile, C3H3N. Acrylonitrile can be
prepared by the reaction of propylene, C3H6, with nitric oxide, NO.
4C3H6 (g) + 6NO (g) → 4C3H3N (g) + 6H2O (g) + N2 (g)
What is the limiting reactant when 126 g of C3H6 reacts with 175 g of NO?
4. 3.76 g of zinc reacts with 8.93 × 1023 molecules of hydrogen chloride. Which reactant is present
in excess?
5. Chloride dioxide, ClO2, is a reactive oxidizing agent. It is used to purify water.
6ClO2 (g) + 3H2O (l) → 5HClO3 (aq) + HCl (aq)
(a) If 71.00 g of ClO2 is mixed with 19.00 g of water, what is the limiting reactant?
(b) What mass of HClO3 is expected in part (a)?
(c) How many molecules of HCl are expected in part (a)?
6. Hydrazine, N2H4, reacts exothermically with hydrogen peroxide, H2O2
N2H4 (g) + 7H2O2 (aq) → 2HNO3 (g) + 8H2O (g)
(a) 120 g of N2H4 reacts with an equal mass of H2O2. Which is the
limiting reactant?
(b) What mass of HNO3 is expected?
(c) What mass, in grams, of the excess reactant remains at the end of the reaction?
7. In the textile industry, chlorine is used to bleach fabrics. Any of the toxic chlorine that remains
after the bleaching process is destroyed by reacting it with a sodium thiosulfate solution,
Na2S2O3 (aq).
Na2S2O3 (aq) + 4Cl2(g) + 5H2O (l) → 2NaHSO4 (aq) + 8HCl (aq)
135 kg of Na2S2O3 reacts with 50.0 kg of Cl2 and 238 kg of water.
How many grams of NaHSO4 are expected?
8. Manganese (III) fluoride can be formed by the reaction of manganese (II) iodide with fluorine.
2MnI2(s) + 13F2 (g) → 2MnF3(s) + 4IF5 (g)
(a) 1.23 g of MnI2 reacts with 25.0 g of F2. What mass of MnF3 is expected?
(b) How many molecules of IF5 are produced in part (a)?
(c) What reactant is in excess? How much of it remains at the end of the reaction?
Calculations involving acid – base titration
Suppose that a solution of an acid reacts with a solution of a base. You can determine the
concentration of one solution if you know the concentration of the other. (This assumes that the
volumes of both are accurately measured.) Use the concentration and volume of one solution to
determine the amount (in moles) of reactant that it contains. The balanced chemical equation for the
reaction describes the mole ratio in which the compounds combine. In the following Sample
Problems and Practice Problems, you will see how to do these calculations.
Examples :
1. Finding the concentration
13.84 mL of hydrochloric acid, HCl(aq), just neutralizes 25.00 mL of a 0.1000 mol/L solution of
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sodium hydroxide, NaOH(aq). What is the concentration of the hydrochloric acid?
What Is Required?
You need to find the concentration of the hydrochloric acid.
What Is Given?
Volume of hydrochloric acid, HCl = 13.84 mL
Volume of sodium hydroxide, NaOH = 25.00 mL
Concentration of sodium hydroxide, NaOH = 0.1000 mol/L
Plan Your Strategy
Step 1: Write the balanced chemical equation for the reaction.
Step 2: Find [HCl (aq)], based on the amount and volume of hydrochloric acid solution needed.
The balanced chemical equation is
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Using the formula: MaVa = MbVb where MaVa molarity and volume of the acid, MbVb is the
molarity and volume of the base. We calculate the concentration of the acid. It is important to take
note of the mole ratios in the balanced chemical equation when using the formula above.
Ma =
MbVb
Ma =
Va
25 𝑋 0.1
13.84
Ma = 0.1086 mol/ L
An alternative method is shown below.
Step 1 The balanced chemical equation is
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Step 2 Amount (in mol) = Concentration (in mol/L) × Volume (in L)
Amount NaOH (in mol) added = 0.1000 mol/L × 0.02500 L
= 2.500 × 10−3 mol
Step 3 HCl reacts with NaOH in a 1:1 ratio, so there must be 2.500 × 10-3 mol HCl.
𝐴𝑚𝑜𝑢𝑛𝑡 ( 𝑖𝑛 𝑚𝑜𝑙)
𝑉𝑜𝑙𝑢𝑚𝑒 ( 𝑖𝑛 𝐿)
𝟎.𝟎𝟎𝟐𝟓 𝒎𝒐𝒍
=
𝟎.𝟎𝟏𝟑𝟖𝟒 𝑳
Step 4 concentration ( in mol/L) =
= 0.1086 mol/L
Task
17.85 mL of nitric acid neutralizes 25.00 mL of 0.150 mol/L NaOH (aq). What is the concentration of the
nitric acid?
11. What volume of 1.015 mol/L magnesium hydroxide is needed to neutralize 40.0 mL of 1.60 mol/L
hydrochloric acid?
12. What volume of 0.150 mol/L hydrochloric acid is needed to neutralize each solution below?
(a) 25.0 mL of 0.135 mol/L sodium hydroxide
(b) 20.0 mL of 0.185 mol/L ammonia solution
(c) 80 mL of 0.0045 mol/L calcium hydroxide
13. What concentration of sodium hydroxide solution is needed for each neutralization reaction?
(a) 37.82 mL of sodium hydroxide neutralizes 15.00 mL of 0.250 mol/L hydrofluoric acid.
(b) 21.56 mL of sodium hydroxide neutralizes 20.00 mL of 0.145 mol/L sulfuric acid.
(c) 14.27 mL of sodium hydroxide neutralizes 25.00 mL of 0.105 mol/L phosphoric acid
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Additional questions and solutions
1. 3.88 g of a monoprotic acid was dissolved in water and the solution made up to 250 cm3. 25.00 cm3
of this solution was titrated with 0.0950 mol dm-3 NaOH solution, requiring 46.50 cm3.
a) Write the chemical equation using HA as the formula of the monoprotic acid.
HA + NaOH → NaA + H2O
b) Calculate the moles of NaOH used in the titration.
Moles of NaOH used in the titration = C x V, = 0.0950 x 0.0465 = 0.00442 moles
c) Calculate the number of moles of the monoprotic acid used in the titration.
1:1 ratio, so 0.00442 moles of acid used in the 25cm3 titration
d) Use your answer in (c) to calculate the number of moles of the monoprotic acid in the 250cm3
solution.
0.00442 moles in 25cm3, so x 10 ( /25, x250) = 0.0442 moles in 250cm3 (of which the 3.88g was
dissolved)
e) Calculate the relative molecular mass of the acid.
Mr = mass / moles, 3.88g / 0.0442 mol = 87.8 g mol-1
2. A 1.575 g sample of ethanedioic acid crystals, H2C2O4.nH2O, was dissolved in water and made up
to 250 cm3. One mole of the acid reacts with two moles of NaOH. In a titration, 25.00 cm3 of this
solution of acid reacted with exactly 15.60 cm3 of 0.160 mol dm-3 NaOH.
a) Calculate the moles of NaOH used in the titration.
Moles of NaOH used in the titration = C x V, = 0.160 x 0.0156 = 0.00250 moles
b) Calculate the number of moles of ethanedioic acid used in the titration.
2:1 ratio of NaOH : acid, so 0.00125 moles of acid used in the 25cm3 titration
c) Use your answer in (b) to calculate the number of moles of ethanedioic acid in the 250cm3 solution.
0.00125 moles in 25cm3, so x 10 (/25, x250) = 0.0125 moles in 250cm3 (of which the 1.575g was
dissolved)
d) Calculate the relative molecular mass of ethanedioic acid.
Mr = mass / moles, 1.575 / 0.0125 = 126 g mol-1
e) Use your answer to (d) and the formula H2C2O4.nH2O to calculate the value of n.
H2C2O4 has a Mr = 90, 126 - 90 = 36. 36 is the Mr of all the H2O's. Each H2O has a Mr of 18.
36 / 18 = 2. So n = 2: H2C2O4.2H2O
3. A solution of a metal carbonate, M2CO3, was prepared by dissolving 7.46 g of the anhydrous solid
in water to give 1000 cm3 of solution. 25.00 cm3 of this solution reacted with 27.00 cm3 of 0.100
mol dm-3 hydrochloric acid.
a) Write the chemical equation for the reaction
2HCl + M2CO3 → 2MCl + H2O + CO2 (M must be a 1+ ion as CO32- has a 2- charge. 1+ ions
are found in Group1 so must be a Group1 metal
b) Calculate the moles of HCl used in the titration.
Moles of HCl used in the titration = C x V, = 0.100 x 0.0270 = 0.00270 moles
c) Calculate the number of moles of M2CO3 used in the titration. Hence calculate the number of
moles in 7.46g
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2:1 ratio of HCl: M2CO3, so 0.00135 moles of M2CO3 used in the 25cm3 titration.
so x 40 ( /25, x1000) M2CO3 = 0.0540 moles in 1000cm3 (of which the 7.46g was dissolved)
d) Calculate the relative molecular mass of M2CO3.
Mr = mass / moles, 7.46 / 0.0540 = 138.1 g mol-1
e) Use your answer to (d) and the Mr of M2CO3 and identify the metal M. CO3 has a Mr = 60,
138.1 - 60 = 78.1. 78.1 is the Ar of M x 2. Each M has an Ar of 39.1 (78.1 / 2). M must be
potassium, K
Percentage composition
The percentage composition of a compound refers to the relative mass of each element in the compound.
In other words, percentage composition is a statement of the values for mass percent of every element in
the compound.
calcium carbonate
calcium
12%
40%
48%
Oxygen
For example, the compound calcium carbonate, CaCO3, has a
percentage composition of 40% calcium, 12% carbon, and 48%
oxygen, as shown in the figure. A compound’s percentage
composition is an important piece of information. For example,
percentage composition can be determined experimentally, and
then used to help identify the compound.
Carbon
Examine the following example to learn how to calculate the percentage composition of a compound from the mass
of the compound and the mass of the elements that make up the compound. Then do the Practice Problems to try
expressing the composition of substances as mass percents.
Example
A compound with a mass of 48.72 g is found to contain 32.69 g of zinc and 16.03 g of sulfur. What is the percentage
composition of the compound?
What Is Required?
You need to find the mass percents of zinc and sulfur in the compound.
What Is Given?
You know the mass of the compound. You also know the mass of each element in the compound.
Mass of compound = 48.72 g
Mass of Zn = 32.69 g
Mass of S = 16.03 g
Plan Your Strategy
To find the percentage composition of the compound, find the mass percent of each element. To do this, divide the
mass of each element by the mass of the compound and multiply by 100%.
Act on your strategy
Mass percent of Zn =
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑍𝑛
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
𝑥 100%
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=
32.69 𝑔
48.72 𝑔
𝑋 100%
= 67.10 %
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑆
Mass percent of S = 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 𝑥 100%
16.03 𝑔
= 48.72 𝑔 𝑋 100%
= 32.90%
The percentage composition of the compound is 67.10% zinc and 32.90 % sulphur
Task
1. A sample of a compound is analyzed and found to contain 0.90 g of calcium and 1.60 g of chlorine. The
sample has a mass of 2.50 g. Find the percentage composition of the compound.
2. Find the percentage composition of a pure substance that contains 7.22 g nickel, 2.53 g phosphorus, and 5.25
g oxygen only.
3. A sample of a compound is analyzed and found to contain carbon, hydrogen, and oxygen. The mass of the
sample is 650 mg, and the sample contains 257 mg of carbon and 50.4 mg of hydrogen. What is the
percentage composition of the compound?
4. A scientist analyzes a 50.0 g sample and finds that it contains 13.3 g of potassium, 17.7 g of chromium, and
another element.
Later the scientist learns that the sample is potassium dichromate, K2Cr2O7. Potassium dichromate is a bright
orange compound that is used in the production of safety matches. What is the percentage composition of
potassium dichromate?
5. Calculate the mass percent of nitrogen in each compound.
(a) N2O (c) NH4NO3
(b) Sr(NO3)2 (d) HNO3
6. Sulfuric acid, H2SO4, is an important acid in laboratories and industries. Determine the percentage
composition of sulfuric acid.
7. Potassium nitrate, KNO3, is used to make fireworks. What is the mass percent of oxygen in potassium nitrate?
8. A mining company wishes to extract manganese metal from pyrolusite ore, MnO2.
(a) What is the percentage composition of pyrolusite ore?
(b) Use your answer from part (a) to calculate the mass of pure manganese that can be extracted from 250 kg
of pyrolusite ore.
Empirical and Molecular formulae
(a) Empirical formula
The empirical formula (also known as the simplest formula) of a compound shows the lowest whole number
ratio of the elements in the compound.
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