Chapter 3

CHAPTER 3 : CHEMICAL FORMULAE AND EQUATIONS

1. All living things on earth, directly or indirectly, depend on photosynthesis for food, energy and oxygen, making it one
of the most important biochemical processes known. This complex process can be represented by the following
equation.

6C02(g) + 6H2O(l)  C6H12O6(aq) + 6O2(g)

A Relative Atomic Mass and Relative Molecular Mass

The mass of an atom is very small. Therefore, how can we determine its mass? We cannot weigh a single atom, but we
can determine the mass of one atom relative to another.

Relative atomic mass, Ar

1. Every atom has its own unique atomic mass based on a standard comparison or relative scale eg. it has been
based on hydrogen H = 1, oxygen O = 16. Since hydrogen atom is the lightest atom, chemists first started comparing
masses of other atoms with the mass of one hydrogen atom.

Relative Atomic Mass

Every atom has a relative atomic mass (R.A.M.) which tells you how
heavy the atom is when compared to hydrogen. For instance, every
helium atom is 4 times as heavy as a hydrogen atom. Thus, helium is
said to have a relative atomic mass of 4.

Well, if a mole of hydrogen atoms has a mass of 1g, and helium atoms
are 4 times as heavy as hydrogen atoms, then a mole of helium has a 4 hydrogen atoms weigh the same as 1
helium atom
mass of 4 g.

Figure 3.1 A helium atom is 4 times as heavy as a hydrogen atom.

2. This method, however, was not always convenient. Oxygen was then used as a standard to compare the masses of
atoms. However, this also posed some problems.

3. Finally, in year 1961, the international unions of chemists and physicists agreed on a single new standard, the
isotope carbon-12. It is also used as a reference standard in the mass spectrometer.
Carbon- 12 is chosen because it is a solid and can be easily handled. Furthermore, 98.9 % of carbon is in the
form of 12C.

4. Carbon-12 isotope is assigned a mass of exactly 12.000 units.
5. The relative atomic mass is that Ar of an element is the average mass of one atom present in the element

compared to 1/12th the mass of a carbon-12 isotope.

The average mass of one atom of an element

Relative atomic mass of an element =

1  the mass of an atom of carbon -12
12

6. For example, the relative atomic mass of helium is 4. This means that the average mass of a helium atom is 4 times

1

larger than of the mass of a carbon-12 atom. Since relative atomic mass compares the masses of atoms, it does

12

not have any unit.

Relative molecular mass, Mr

1. The idea of relative atomic mass can be extended to molecules.

1

The relative molecular mass of a molecule is the average mass of the molecule when compared with

12

of the mass of one atom of carbon-12.

Relative molecular mass of a molecule = The average mass of one molecule
1  the mass of an atom of carbon -12
12

1

2. For example, water has a relative molecular mass of 18. This means that one molecule of water is 18 times heavier

1

than one atom of carbon-12.

12

3. Since a molecule is made up of atoms, the relative molecular mass of a molecule can be calculated by adding up the
relative atomic masses of all the atoms that are present in the molecule. Let us check out some examples….

 Example 1: Water H2O
o relative atomic masses are H=1 and O=16
Mr = (1x2) + 16 = 18 (molecular mass of water)

 Example 2: Sulphuric acid H2SO4
o relative atomic masses are H=1, S=32 and O=16
o Mr = (1x2) + 32 + (4x16) = 98 (molecular mass of sulphuric acid)

Relative formula mass

1. Some substances consist of ions and not molecules. For these ionic substances, the relative formula masses
are used in place it relative molecular masses. It has no unit.

2. The relative formula mass of an ionic compound is the sum of the relative atomic masses of all the atoms in the
chemical formula of that ionic compound.

3. Example 1 : Calcium hydroxide Ca(OH)2 (formula mass of calcium hydroxide, ionic)
o relative atomic masses are Ca=40, H=1 and O=16
o Relative formula mass, Mr = 40 + 2(16+1) = 74

Example 2 : Hydrated copper(II) sulphate, CuSO4.5H2O
o relative atomic masses are Cu=64, S=32,O=16 and H = 1
o Relative formula mass, Mr = (64x1) + (32x1) +(16x4) + 5[(1x2) + (16x1)] = 250

Example 3 : Aluminium sulphate, Al2(SO4)3
o relative atomic masses are Al = 27, S=32, and O=16
o Relative formula mass, Mr = (27x2) + 3[(32x1) +(16x4)] = 342

B The Mole and the Number of Particles

Pairs and dozens are examples of units that we use in measuring the amount of objects in our daily lives. The word
‘pair’ and ‘dozen’ represent a fixed number of objects. In a similar way, chemists use the unit mole to measure the
amount of substances. The symbol of mole is mol.

What is a mole?

1. Since carbon-12 is the standard chosen for relative atomic mass, we use the number of atoms in exactly 12 g of
carbon-12 as a standard in measuring the amount of substances.

2. Thus, we define a mole as the following.

A mole is an amount of substance that contains as many particles as the number of atoms in exactly 12 g of
carbon-12.

3. Scientists had determined the number of atoms in 12 g of carbon-12 experimentally as 6.02 x 1023 which is a very
large number!

4. The value of 6.02 x 1023 is called the Avogadro constant or the Avogadro number in honour of a famous Italian
scientist, Amedeo Avogadro (1776-1856).

5. The Avogadro constant, NA is defined as the number of particles in one mole of a substance.
6. One mole of any substance is the amount of substance that contains NA number of particles.

In other words, one mole of substance contains 6.02 x 1023 particles.
• 1 mol of atomic substance contains 6.02 x 1023 atoms.
• 1 mol of molecular substance contains 6.02 x 1023 molecules.
• 1 mol of ionic substance contains 6.02 x 1023 formula units.

2

7. We can convert the number of moles of any substance to the number of particles in it and vice versa by using the
following relationship.

Figure 3.2 Relationship between the number of moles and the number of particles

Example 3.1

A closed glass bottle contains 1.5 mol of carbon dioxide gas, C02. [Avogadro constant: 6.02 x 1023 mol-1]
(a) How many carbon dioxide molecules, C02 are there in the bottle?
(b) Calculate the number of atoms found in 1.5 mol of carbon dioxide gas, C02

Solution:

(a) The number of carbon dioxide molecules, C02 in 1.5 mol of gas
= 1.5 mol x 6.02 x 1023 mol-1.
= 9.03 x 1023 Number of particles = Number of moles x NA

(b) Each carbon dioxide molecule, C02 consists of three atoms (1 carbon atom + 2 oxygen atoms).

Therefore, the number of atoms in 1.5 mol of gas

= The number of carbon dioxide molecules x 3
= 9.03 x 1023 x 3
= 2.709 x 1024

Example 3.2
Find the number of moles of ammonia molecules that contain 3.01 x 1022 of ammonia, NH3 molecules.
[Avogadro constant: 6.02 x 1023 mol-1]

Solution:

The number of moles of 1.505 x 1022 ammonia, NH3 molecules

3.011022

=

6.02 1023

= 0.05 mol

C The Mole and the Mass of Substances

How do we measure a substance in moles? Instead of counting the number of particles, we can just weigh its mass.
This is easily done if we know its molar mass.
The mass of one mole of any substance is called its molar mass.

The molar mass of a substance = The mass of 1 mol of the substance
= The mass of NA number of particles
= The mass of 6.02 x 1023 particles

Molar mass has the unit of grams per mole or g mol-1.

(1) mole of Z = g of Z / atomic or formula mass of Z,

(2) or g of Z = mole of Z x atomic or formula mass of Z

(3) or atomic or formula mass of Z = g of Z / mole of Z

where Z represents atoms, molecules or formula of the particular
element or compound defined in the question.

1 mol of carbon-12 has a mass exactly 12 g. Therefore, the molar mass of carbon-12 is 12 g mol-1.
Notice that the molar mass of carbon-12 is numerically equal to the relative atomic mass of carbon-12, which is 12.
Similarly, the molar masses of all other substances are numerically equal to their relative masses.

3

Table 3.3 Molar masses of substances

Element Relative mass Mass of 1 mol Molar mass
4 g mol-1
Helium 4 4g 24 g mol-1
16 g mol-1
Aluminium 27 27 g 44 g mol-1
28 g mol-1
Methane, CH4 12 + 4(1) 16 g
Carbon dioxide, CO2 12 + 2(16) 44 g
Nitrogen, N2 28 g
2(14)

Molar mass and the mole

By knowing the molar mass of a substance, we cars weigh any fraction of a mole of the substance. For example, when
12 g of carbon is 1 mol of carbon, 6 g of carbon is 0.5 mol of carbon and 24 g of carbon is 2 mol of carbon.
Molar mass is used as the conversion factor between the number of moles of any substance and its mass. The
relationship in Figure 3.3 shows how this can be done.

Figure 3.3 Relationship between the number of moles and the mass of a substance

Example 3.3

What is the mass of
(a) 0.3 mol of magnesium?
(b) 2.408 x 1023 atoms of magnesium?
[Relative atomic mass: Mg, 24. Avogadro constant : 6.02 x 1023 mol-1]

Solution: (1. Determine the molar mass of Mg.)
( 2. Calculate its mass. Mass = Number of moles x molar mass)
(a) The molar mass of magnesium = 24 g mol-1
Therefore, the mass of 0.1 mol of magnesium
= 0.3 mol x 24 g mo1-1 = 7.2 g

(b) The number of moles of magnesium atoms

= (1. Calculate the number of moles of Mg atoms.

Number of moles = Number of particles ÷ NA)

= 0.4 mol
The mass of 2.408 x 1023 atoms of magnesium
= 0.4 mol x 24 g mol-1 (2. Calculate its mass.

= 9.6 g Mass = Number of moles x molar mass )

Example 3.4

How many moles of molecules are there in 11 g of carbon dioxide gas, CO2 [ Relative atomic mass: C, 12 ; 0, 16]
Solution:

The relative molecular mass of carbon dioxide, CO2 (1. Calculate the relative molecular based on its chemical formula.)

= 12 + 2(16)

= 44
Therefore, the molar mass of carbon dioxide, CO2 = 44 g mol-1 (2. Determine its molar mass)
The number of moles of molecules in 16 g of carbon dioxide, CO2

= 11 g (3. Calculate its number of moles.
44 g mol-1
Number of moles = Mass in grams  molar mass

= 0.25 mol

4

Example 3.5

How many chloride ions are there in 26.7 g of aluminium chloride, AlCl3 {Relative atomic mass: Cl, 35.5; Al, 27.
Avogadro constant: 6.02 x 1023 mol-1}

Solution:

The relative formula mass of aluminium chloride, AlCl3 = 27 + 3(35.5) = 133.5
Therefore, the molar mass of aluminium chloride = 136 g mol-1
(1. Determine the mass of AlCl3)

The number of moles of 27.2 g of zinc chloride, ZnCl2

= 26.7 g [2. Calculate its number of moles
133.5 g mol-1
Number of moles = Mass in grams  molar mass ]

= 0.2 mol

The number of formula units of AlCl3 [ 3. Calculate the formula units of AlCl3 ]
= 0.2 mol x 6.02 x 1023 mol-1 Number of particles = Number of moles x NA
= 1.204 x 1023

Each formula unit of AlCl3 has 3 chloride ions.,

Therefore, the number of chloride ions

= The number of formula units of AlCl3 x 3 [ 4. Calculate the number of chloride ions based on the proportion
= 1.204 x 1023 x 3 = 3.612 x 1023

D The Mole and the Volume of Gas

In the previous lesson, you have learnt how to find the amount of substances in moles just by weighing them out.
However, this is not practical for gases as most gases are very light and difficult to be weighed. Therefore, it is easier to
measure their volumes.

Molar volume of a gas

The molar volume of a gas is defined as the volume occupied by one mole of the gas. One mole of any gas (or the
formula mass in g), at the same temperature and pressure occupies the same volume . This volume is known as the
molar volume of gases.
The molar volume of any gas is 22.4 dm3 at STP or 24dm3 (24 litres) or 24000 cm3, at room temperature and
pressure.
Avogadro's Law states that equal volumes of gases under the same conditions of temperature and pressure
contain the same number of molecules. So the volumes have equal moles of separate particles in them.

Two handy relationships for substance Z below:

 moles Z = [mass of Z (g)] / [atomic or formula mass of Z (g/mol)]
o mass of Z in g = moles of Z x atomic or formula mass of Z
o atomic or formula mass of Z = mass of Z / moles of Z

 gas volume of Z = moles of Z x molar volume
o moles of Z = gas volume of Z / molar volume

 Note (i): In the following examples, assume you are dealing with room temperature and pressure ie 25oC and 1

atmosphere pressure.

We can use the molar volume to convert the volume of a gas to the number of moles and vice versa.

Figure 3.4 Relationship between the number of moles and the volume of a gas
Example 3.6
(a) What is the volume of 2.3 mol of ammonia gas, NH3 at STP? [Molar volume: 22.4 dm3 mol-1 at STP]
(b) Calculate the number of moles of ammonia gas, NH3 which occupies a volume of 720 cm3 at room conditions.

[molar volume : 24 dm3 mol-1 at room conditions]

5

Solution :

(a) The volume of 2.3 mol of ammonia gas, NH3 at STP
= 2.3 mol x 22.4 dm3 mol-1 [Volume of gas = Number of moles x molar volume ]
= 51.52 dm3

(b) The number of moles of 720 cm3 ammonia gas, NH3

= 720 [ Number of moles = Volume of gas ÷ molar volume ]

24000

= 0.03 mol

Relationships between the number of particles, number of moles, mass of substances and volume of gases

1. The relationships between the number of particles, number of moles, mass and volume of gases

2. The table below shows the relationship between the mass and the number of moles with the volume of the gases.

Gas Chemical Relative Mass of Number of Volume at STP/ Volume at room
formula molecular sample/g dm3 temperature and
Nitrogen moles
N2 mass 5.6 pressure/ dm3
Carbon 5.6 0.2 x 22.4 = 4.48 0.2 x 24 = 4.8
monoxide CO 2(24) = 28 7 0.25 x 22.4 = 5.6
Chlorine = 0.2 0.3 x 22.4 = 6.72 0.25 x 24 = 6
Cl2 12 + 16 21.3
= 28 28 0.3 x 24 = 7.2
7
2(35.5)
= 71 = 0.25

28
21.3

= 0.3

71

Example 3.7

What is the volume
(a) of 12.8 g of oxygen gas, 02 in cm3
(b) containing 1.5 X 1023 molecules of oxygen gas at STP?
[ relative atomic mass: 0, 16. Molar volume: 22.4 dm3 mol-1 at STP]

Solution:

(a)

Mass of O2 Volume of O2 at STP

12.8 g ?

The number of moles of 12.8 g of oxygen gas, O2

12.8 g
= 2(16) g mol-1

= 0.4 mol (1. Convert the number of moles of O2 to volume.
The volume of oxygen gas, 02 at STP Volume of gas = Number of moles x molar volume )
= 0.4 mol x 22.4 dm3 mol-1 .
= 8.96 dm3
= 8960 cm3

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(b) The number of moles of 1.5 X 1023 molecules of oxygen gas, O2

1.5 1023
= 6 1023

= 0.25 mol

The volume of oxygen gas, 02 at STP (1. Convert the number of moles of O2 to volume.
= 0.25 mol x 22.4 dm3 mol-1 . Volume of gas = Number of moles x molar volume )
= 5.6 dm3
= 5600 cm3

Example 3.8
Calculate the number of carbon dioxide, CO2 molecules at room conditions for 240 cm3 of the gas.
[molar volume : 24 dm3 mol-1 at room conditions. Avogadro constant: 6.02 x 1023 mol-1

Solution:

Volume of CO2 at room conditions Number of molecules
120 cm3 ?

Volume of carbon dioxide, CO2 = 240 cm3 = 0.24 dm3
The number of moles of 0.24 dm3 carbon dioxide, CO2

0.24 dm3

= = 0.01 mol

24 dm3 mol-1

The number of molecules of carbon dioxide, CO2
= 0.01 mol x 6.02 x 1023 mol-1

[4. Convert the number of moles to the number of molecules.

Number of particles = Number of moles x NA]
= 6.02 x 1021

E Chemical Formulae

1. A chemist does not have to write down a long description of a substance. An easy way to represent a substance is to
use its chemical formula.

2. A chemical formula is a the symbol or symbol of elements and ratio of the combination of atoms which are
combined and written together to represent a chemical compound.

3. The chemical formula of an element represents its atoms. Some elements, however, naturally exist as molecules.
Therefore, their chemical formulae represent their molecules.

Figure 3.6 Chemical formula of hydrogen gas, H2
4. The chemical formula of a compound shows all the elements that are present in the compound and the number of

atoms of each element.

Figure 3.7 Chemical formula of water molecule, H2O
5. Compounds can be represented by two types of chemical formulae, namely the empirical formula and the molecular

formula.

7

Empirical formula

1. The empirical formula of a compound is the chemical formula that gives the simplest whole number ratio of
atoms of each element in the compound.
For example, the formula of a glucose molecule is C6H12O6. The ratio of carbon to hydrogen to oxygen atoms in the
molecule is 6 : 12 : 6.
This can be simplified as 1: 2 : 1. Therefore, the empirical formula of glucose is CH2O.
The following example shows how we can calculate the empirical formulae of substances based on experimental
values.

Example 3.9 :

1.15g of sodium reacted with 0.8 g of sulphur. Calculate the empirical formula of sodium sulphide.

RATIOS ... Sodium Na (Ar = 23) Sulphur S (Ar = 32) Comments and tips

Mass 1.15g 0.80g not the real atom ratio

moles (mass in g / Ar) 1.15 / 23 = 0.05 mol 0.8 / 32 = 0.025 mol can now divide by
0.05 / 0.025 = 2 0.025 / 0.025 = 1 smallest ratio number or
simplest whole number scale up by x factor to get
ratio by trial and error or 0.05 x 40 = 2 or 0.025 x 40 = 1 simplest whole number

ratio

therefore the simplest ratio = empirical formula for sodium sulphide = Na2S

Example 3.10 :

1.35g of aluminium was heated in oxygen until there was no further gain in weight. The white oxide ash formed weighed
2.55g. Deduce the empirical formula of aluminium oxide. Note: to get the mass of oxygen reacting, all you have to do is
to subtract the mass of metal from the mass of the oxide formed.

RATIOS ... Aluminium Al (Ar = 27) Oxygen O (Ar = 16) Comments and tips
Mass 1.35g 2.55 - 1.35 = 1.2g not the real atom ratio

moles (mass in g / Ar) 1.35 / 27 = 0.05 mol 1.2 / 16 = 0.075 mol can now divide by
0.05 / 0.05 = 1 0.075 / 0.05 = 1.5 smallest ratio number or
simplest whole number scale up by x factor to get
ratio (then x 2 = 2) (then x 2 = 3) simplest whole number

or 0.05 x 40 = 2 or 0.075 x 40 = 3 ratio

therefore the simplest ratio = empirical therefore the formula for aluminium oxide = Al2O3

Example 3.11 :

A chlorinated hydrocarbon compound when analysed, consisted of 24.24% carbon, 4.04% hydrogen, 71.72% chlorine.
The molecular mass was found to be 99 from another experiment. Deduce the empirical and molecular formula. (you
can 'treat' the %'s as if they were grams, and it all works out like examples 1 and 2)

RATIOS ... Carbon (Ar = 12) Hydrogen (Ar = 1) Chlorine (Ar = 35.5) Comments and tips

Mass or % mass 24.24 4.04 71.72 just think of it as
based on 100g

No. of moles (mass 24.24 / 12 = 2.02 4.04 / 1 = 4.04 mol 71.72 / 35.5 = 2.02
in g / Ar) mol 4.04 / 2.02 = 2 mol can now divide by
smallest ratio number
simplest whole 2.02 / 2.02 = 1
number ratio 2.02 / 2.02 = 1

therefore the simplest ratio = empirical empirical formula for the chlorinated hydrocarbon = CH2Cl

8

Assume that the molecular formula is (CH2Cl)n
Based on the molecular formula, the relative molecular mass = n[12 + 2(1) + 35.5]

= 49.5n

However, its relative molecular mass is 99.

Therefore, 49.5n = 99 [ 2. Equate the calculated relative molecular mass and the given

relative molecular mass.]

n = 99 =2 [ 3. Replace the value of n into the assumed molecular formula.]
49.5

Hence, the molecular formula of the compound is (CH2Cl)2 or C2H4Cl2

Molecular formula

1. The molecular formula of a compound is the chemical formula which shows the actual number of atoms of

each element that are present in a molecule of the compound.

2. In a molecular formula, the subscript numbers are simple multiples of those in the empirical formula.

Molecular formula = (Empirical formula)n
whereby n is a positive integer.

Table 3.4 Relating empirical formula to molecular formula

Compound Empirical formula Molecular formula n

Water H2O (H2O)1 = H2O 1

Ethene CH2 (CH2)2 = C2H4 2

Glucose CH2O (CH2O)6 = C6H12O6 6

3. To find the molecular formula of a compound, we need to know the relative molecular

mass and the empirical formula of the compound. Let us check out the following example.

Example 3.12

The empirical formula of a compound is CH2. Its relative molecular mass is 42. Find its molecular formula. [Relative
atomic mass: H, 1; C, 12]

Solution:

Calculate the relative molecular mass

Assume that the molecular formula is (CH2)n
Based on the molecular formula, the relative molecular mass = n[12 + 2(1)]

= 14n

However, its relative molecular mass is 42.

Therefore, 14n = 42 [ 2. Equate the calculated relative molecular mass and the given

relative molecular mass.]

n = 42 =3 [ 3. Replace the value of n into the assumed molecular
14

formula.]

Hence, the molecular formula of the compound is (CH2)3 or C3H6
We can also solve various numerical problems involving empirical and molecular formulae.

Example 3.13:

Calculate the % of copper in copper sulphate, CuSO4

Relative atomic masses: Cu = 64, S = 32 and O = 16
a. relative formula mass = 64 + 32 + 4x16 = 160
b. only one copper atom of relative atomic mass 64
c. % Cu = 64 x 100 / 160 = 40% copper by mass in the compound

9

Example 3.14:

Calculate the % of oxygen in aluminium sulphate, Al2(SO4)3
Relative atomic masses: Al = 27, S = 32 and O = 16

a. relative formula mass = 2x27 + 3x(32 + 4x16) = 342
b. there are 4 x 3 = 12 oxygen atoms, each of relative atomic mass 16, giving a total mass of oxygen in the

formula of 12 x 16 = 192
c. % O = 192 x 100 / 342 = 56.1% oxygen by mass in aluminium sulphate

Example 3.15

6.24 g of element X combines with 1.28 g of oxygen to produce a compound with an empirical formula of X2O.

What is the relative atomic mass of X? [relative atomic mass: 0, 16]

Solution:

Assume that the relative atomic mass of X is x.

Element X O

Mass (g) 6.24 1.28

Number of moles 6.24 1.28
x
= 0.08

16

Based on the given empirical formula X2O the mole ratio of atoms X to O is 2 : 1.

Therefore, 6.24 (The calculated mole ratio should be the same as the mole ratio given in

: 0.08 = 2 : 1

x

the empirical formula.)

6.24  0.08 = 2 : 1
x

6.24 2
=
0.08x 1

0.08x 1

=

6.24 2

x= 1x6.24 = 39

2x0.08

Therefore, the relative atomic mass of X is 39.

Exercise

A student carried out an experiment to determine the empirical formula of magnesium oxide.
The steps and set-up of apparatus of the experiment are shown in the diagram below.

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(a) Complete the following table by stating the observations and related inferences in the experiment.

Observations Inferences

(i) .............................................................................. (i) ................................................................
............................................................................. ................................................................

(ii) ............................................................................ (ii) ...............................................................
............................................................................ ................................................................

(iii) ........................................................................... (iii) .................................................................

.......................................................................... ..................................................................
.......................................................................... ..................................................................

(b) Round off the reading to two decimal places and record it in the table below.

Description Mass / g
The crucible and lid.
The crucible, lid and magnesium powder.
The crucible, lid and magenesium oxide.

(c) (i) Calculate the mass of magnesium that has been used.

(ii) Calculate the mass of oxygen which reacted with magnesium.

(iii) Determine the empirical formula of magnesium oxide.
Use the information that the relative atomic mass, O = 16, Mg = 24

(d) The student wants to determine the empirical formula of lead(II) oxide. He used the steps and set-up of apparatus
as the experiment before. Predict whether the empirical formula of lead(II) oxide can be determined.
Explain your answer.
...................................................................................................................................................................................................
...................................................................................................................................................................................................

11

(e) Explain why the students need to lift the lid of the crucible at intervals.
…………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………..

(f) Explain how would you ensure that the reaction between magnesium and oxygen is complete.
…………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………..

4. An experiment on the reduction of copper oxide by hydrogen is shown in the diagram above. A combustion tube

and an asbestos paper which are empty are weighed using a chemical balance. A little copper oxide powder is

placed on the folded asbestos paper and is weighed again. Before heating, the apparatus is flowed with hydrogen

gas for several minutes. At the end of the experiment, a clear liquid is collected at A and copper is formed in the

combustion tube.

(a) Write the formula for the preparation of hydrogen in the laboratory.
……………………………………………………………………………………………………………………………….

(b) What is the function of the dehydrated calcium chloride?
……………………………………………………………………………………………………………………………….

(c) (i) Why does the setup need to be completely filled with hydrogen before it is heated?
……………………………………………………………………………………………………………………………

(ii) How would you know if the setup was completely filled with hydrogen?
…………………………………………………………………………………………………………………………….

(d) (i) Name the clear liquid that is collected at A.
…………………………………………………………………………………………………………………………….

(ii) How would you ensure that the liquid collected is pure?
…………………………………………………………………………………………………………………………….
…………………………………………………………………………………………………………………………….

(e) (i) If magnesium oxide is used to replace copper(II) oxide, will the reaction take place?
……………………………………………………………………………………………………………………………..

(ii) Give a reason for your answer in (e) (i).
……………………………………………………………………………………………………………………………..

Ionic formulae

1. Ionic compounds are made up of positively-charged ions, cations and negatively-charged ions, anions.
2. Before you can determine the formulae of these compounds, you need to know the formulae of the ions.
3. The table below shows the formulae of some common cations and anions.

(a) Table of cations

+1 H+ +2 Mg2+ +3 Al3+ +4 Pb4+
Hydrogen ion Na+ Magnesium ion Ca2+ Aluminium ion Lead(IV) Sn4+
Sodium ion K+ Calcium ion Fe2+ Chromium(III) ion Cr3+ Tin(IV)
Potassium ion Li+ Iron(II) ion Iron(III) ion Fe3+
Lithium ion Cu+ Barium ion Ba2+
Copper(I) ion Copper(II) ion Cu2+
Silver ion Ag+ Lead(II) ion
Ammonium ion NH4+ Tin(II) ion Pb2+
Zinc Sn2+
Manganese(II) Zn2+
Mn2+

12

(b) Table of anions

-1 OH- -2 S2- -3 N3-
Hydroxide ion NO2- Sulphide ion SO32- Nitride ion Cr3-
Nitrite ion NO3- Sulphite ion SO42- Chromium(III) ion Fe3-
Nitrate ion Sulphate ion CO32- Iron(III) ion PO43-
Fluoride ion F- Carbonate ion Phosphate ion
Chloride ion Cl- Oxide ion O2-
Iodide ion I- Chromate(VI) ion CrO42-
Manganate(VII) ion MnO4- Dichromate(VI) Cr2O72-
Chlorate(I) ion ClO- Thiosulphate ion S2O32-
Hydrogen carbonate // HCO3-
Bicarbonate ion

WORKING OUT FORMULAE OF IONIC COMPOUNDS (THE CROSS-OVER METHOD)

 Step 1 - In the ionic compounds to be learnt in junior science, there are two parts to the ionic compound - the
first is a positive ion (usually a metal e.g. Na1+) and the second is a negative ion (e.g. Cl1-).

 Step 2 - Using the valency table, write the two ions and their valencies.
 Step 3 - Now ignore the positive and negative signs. Cross-over the top valency number to the bottom of the

other ion symbol. Do this for both.
 Step 4 - Write the completed formulae with those same numbers at the bottom.
 Step 5 - If the numbers on each part are the same (e.g. Na1 Cl1 or Mg2 O2), ignore them and rewrite the

formulae without them (e.g. Na Cl or Mg O).
 Step 6 - Brackets may be used around radicals (groups of atoms that are charged e.g CO3).
The table below shows an example on how the ionic formula of an ionic compound is constructed.

'A' (charge=valency) 'B' (charge=valency) deduced formula

2 of Na+ (1) balances 1 of O2- (2) 2 x 1 = 1 x 2 = Na2O
1 of Mg2+ (2) balances 2 of Cl- (1) 1 x 2 = 2 x 1 = MgCl2
1 of Fe3+ (3) balances 3 of F- (1) 1 x 3 = 3 x 1 = FeF3
2 of Fe3+ (3) balances 3 of SO42- (2) 2 x 3 = 3 x 2 = Fe2(SO4)3
1 of Zn2+ (2) balances 2 of NO3- (1) 1 x 2 = 2 x 1 = Zn(NO3)2

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Naming of chemical compounds

1. Chemists have devised clear and systematic ways of naming chemical compounds, based on the
recommendation of the International Union of Pure and Applied Chemistry (IUPAC).

2. For ionic compounds, the name of the cation comes first, followed by the name of the anion.
A few examples are given in table below

Cation Anion Name of ionic compound

Potassium ion Bromide ion Potassium bromide

Magnesium ion Sulphate ion Magnesium sulphate

Zinc ion Nitrate ion Zinc nitrate

3. Certain metals can form more than one type of ions. Thus, Roman numerals are used in their naming to distinguish
the different types of ions. For example, iron can form two cations, namely iron(II) ions, Fe2+ and iron(III) ions, Fe3+.

Thus, the names of the compounds formed by these ions with chlorine would be iron(II) chloride and iron(III) chloride

respectively.

4. In the naming of simple molecular compounds, the name of the first element is maintained

as it is. The name of the second element is added with an ‘ide’. Take a look at a few examples below.

HCl — Hydrogen chloride {ide’ is added to the name

HBr — Hydrogen bromide of the second element.}

EXAMPLES OF CHEMICAL NAMES OF COMPOUNDS CHEMICAL NAME
CHEMICAL FORMULA

CO2 carbon dioxide
CO carbon monoxide

NaCl sodium chloride

CuO Copper(II) oxide

AgBr silver bromide

KI potassium iodide

HCl hydrogen chloride (hydrochloric acid)

NH4Cl ammonium chloride
KOH potassium hydroxide

14

Ca(OH)2 calcium hydroxide
CaS calcium sulphide
NaNO3 sodium nitrate
HNO3 hydrogen nitrate (nitric acid)
NaHCO3 sodium bicarbonate/sodium hydrogen carbonate
(NH4)2CO3 ammonium carbonate
CaSO4 calcium sulphate
CuCO3 Copper(II) carbonate
Al PO4 aluminium phosphate
FeSO4 Iron(II) sulphate
FeCO3 Iron(II) carbonate
NH4HCO3 ammonium bicarbonate
H2SO4 hydrogen sulphate (sulphuric acid)
Na2SO4 sodium sulphate

F Chemical Equations

Chemical equation is a shorthand description of a chemical reaction.
Chemical equations describe a chemical reaction which occurs qualitatively and quantitatively.

Qualitative aspect of chemical equations

1. From the qualitative aspect, chemical equations give the following information :

(a) Types of reactants (b) Types of products formed (c) The physical states of reactants and products

2. Take the equation that represents the combustion of carbon in charcoal as an example.

C(s) + 02(g)  CO2(g)

Reactants Product

3. The arrow in the equation tells us which way the reaction is occurring. The starting substances are called reactants.

They are shown on the left-hand side of the equation.

The new substances produced are called products. They are shown on the right-hand side of the equation.

4. A chemical equation also shows the state of each substance. It shows whether a

substance is a solid (s), liquid (l), gas (g) or is dissolved in water as an aqueous solution (aq).

C(s) + O2(g)  CO2(g)

Meaning: Solid carbon reacts with oxygen gas to give carbon dioxide gas

Writing chemical equations

1. Based on the law of conservation of mass, matter can neither be created nor destroyed in a chemical reaction.
This means that atoms are neither created nor destroyed in any reaction.

2. Therefore, a chemical equation must be balanced. There must always be the same
number of atoms of each element on each side of the equation.

15

THE CONSTRUCTION OF CHEMICAL EQUATIONS

1. Chemical Symbols and Formula
Chemical word equations

 ==> means the direction of change from reactants =to=> products
 No symbols or numbers are used in word equations.
 Always try to fit all the words neatly lined up from left to right, especially if its a long word equation.

2. Below shows several symbols that are usually used along with the chemical equation.

Symbol Representing

(s) The physical state of chemical is solid

(l) The physical state of chemical is liquid

(g) The physical state of chemical is gas

(aq) The physical state of chemical is an aqueous solution

 Irreversible reaction

 Gas is released

 Precipitation or sedimentation

∆ Heat is supplied

Reversible reaction

3. Write the balanced chemical equations for the following reactions.

(a) iron + oxygen  iron(III) oxide

(b) zinc + dilute sulphuric acid → zinc sulphate + hydrogen

(c) potassium oxide + dilute sulphuric acid  potassium nitrate + water

(d) iron + sulphur  iron sulphide

(e) Sodium hydroxide + sulphuric acid  sodium sulphate + water

(f) Lead(II) nitrate  lead(II) oxide + nitrogen dioxide + oxygen

Quantitative aspect of chemical equations

1. Stoichiometry is a study of quantitative composition of substances involved in chemical reactions. A chemical

equation acts as an important tool for this quantitative work. The coefficients in a balanced equation tell us the exact

proportions of reactants and products in a chemical reaction. Let us study the following equation as an example.

2H2(g) + O2(g) → 2H2O(l)

2 molecules or 2 mol 1 molecule or 1 mol 2 molecules or 2 mol

2. The coefficients tell us that every two molecules of hydrogen gas, H2 react with one molecule of oxygen gas,
O2 to produce two molecules of water, H2O.
Therefore, we can also say that for every two moles of hydrogen gas, H2 that react with one mole of oxygen
gas, O2 , two moles of water, H2O are formed.

3. Now that you know how to interpret a chemical equation quantitatively, try interpreting the

following equation in the same manner.

2KI(aq) + Br2(aq) → I2(aq) + 2KBr(aq)
1 molecule 2 formula units
2 formula units 1 molecule

or or or or

2 mol 1 mol 1 mol 2 mol

Numerical problems involving chemical equations

16

Whether manufacturing industrial products or studying a reaction in laboratory, chemists need to know the relative
amounts of substances involved. The chemical equations provide useful imformation for this purpose. We can always
make use of the stoichiometric coefficients in a chemical equation to solve various numerical problems. The following
are some examples.

Example 3.16

Copper(II) oxide, CuO reacts with aluminium according to the following equation.
3CuO(s) + 2Al(s) → Al2O3(s) + 3Cu(s)

Calculate the mass of aluminium required to react completely with 12 g of copper(II) oxide, CuO. [relative atomic mass:

0, 16; Al, 27; Cu, 64 ]

Solution:

3CuO(s) + 2Al(s) → Al2O3(s) + 3Cu(s)
3 mol 2 mol ( Compare the mole ratio of the related substances )
The number of moles of 12 g of copper(II) oxide, CuO

12 g

=

( 64  16 ) g mol-1

12 g

= = 0.15 mol

80 g mol-1

Based on the chemical equation, 3 mol of copper(II) oxide, CuO requires 2 mol of aluminium. Therefore, the number of

moles of aluminium required by 0.15 mol of copper(II)oxide, CuO

= 0.15 mol  2 mol ( 3. Calculate the number of moles of Al based on the mole ratio
3 mol

= 0.1 mol in the equation )

Thus, the mass of aluminium required
= 0.1 mol x 27 g mol-1
(4.Convert the number of moles of Al to mass.

= 2.7 g Mass = Number of moles x molar mass )

Example 3.17

What is the volume of carbon dioxide released when 20 g of calcium carbonate, CaCO3 is reacts with excess dilute
hydrochloric acid at room conditions?

Solution:

CaCO3(s) + 2HCl  CaCl2(aq) + H2O(l) + CO2(g)

1 mol 1 mol

The number of moles of 10 g of calcium carbonate, CaCO3

10g

= = 0.1 mol

[40 12  3(16)]g mol1

Based on the chemical equation, 1 mol of calcium carbonate, CaCO3 produces 1 mol of CO2
Therefore, 0.1 mol of calcium carbonate, CaCO3 produces 0.1 mol of CO2
Thus the volume of carbon dioxide, CO2 collected at room conditions

= 0.1 mol  24 dm3 mol-1

= 2.4 dm3

Example 3.18
Calculate the mass of zinc, Zn reacting with dilute nitric acid, HNO3 if 360 cm3 of hydrogen gas, H2 gas is released.
[ Relative atomic mass: Zn, 65; 0, 16; Ca, 40. Molar volume; 24 dm3 mol-1 at room conditions]

Solution

17

Zn(s) + 2HNO3(aq)  Zn(NO3)2(aq) + H2(g)

? g 360 cm3

The number of moles of carbon dioxide, H2

360

= = 0.015 mol

24000

Based on the chemical equation, 1 mol of zinc, Zn produces 1 mol of H2
Therefore, 0.015 mol of zinc, Zn produces 0.015 mol of H2
Thus the mass of zinc, Zn used

= 0.015 mol  65 g mol-1 = 0.975 g

CHEMICAL FORMULA AND EQUATION EXERCISES

1. Write the chemical formula of the ionic compound in the table below.

Ion Chloride Nitrate Hydroxide Sulphate Carbonate oxide

Sodium NaCl

Magnesium

Ammonium

Copper (II)

Iron (II)

Iron (III)

Aluminium Al(OH)3

2. Write balanced chemical equation for the following chemical reactions.

(a) Heat solid copper (II) carbonate to form copper(II) oxide and carbon dioxide
………………………………………………………………………………………….

(b) Nitric acid reacts with sodium hydroxide to form sodium nitrate and water
………………………………………………………………………………………….

(c) Hydrochloric acid reacts with zinc metal to form zinc chloride and hydrogen gas.
………………………………………………………………………………………….

(d) Copper(II)nitrate reacts with magnesium to produce magnesium nitrate and copper
………………………………………………………………………………………….

(e) Chlorine gas reacts with lithium hydroxide to form lithium chloride, lithium chlorate(I) and water
………………………………………………………………………………………….

(f) Hydrogen gas reacts with lead(II) oxide to produce lead and water
………………………………………………………………………………………….

3. Avogadro constant, NA is defined as the number of particles in one mole of a substance
[1 Mol any substance consist of 6.02 X 1023 particles]. Calculate the number of particles in:

(a) 0.1 mol of calcium : ………………………………………………………..…………………………..

(b) 1.5 mol of iron : ……………………………………………………………………………………..

(c) 2.0 mol of oxygen gas : …………………………………………………………………………………….

(d) 1.5 mol of helium : ……………………………………………………………………………………..

(e) 2.0 mol hydrogen chloride : ………………………………………………………………………………….

4. Calculate the number of moles of the following substances.
(a) 6.0 x 1023 aluminium

18

(b) 1.8 x 1021 argon
(c) 1.2 x 1023 bromine gas
(d) 2.4 x 1020 carbon dioxide
(e) 3.0 x 1023 ammonia (NH3)

5. Calculate the mass of following substances.
(a) 1.5 mol of chlorine gas
(b) 2.0 moles of sulphuric acid
(c) 0.5 moles of ammonia
(d) 2.5 moles of lead (II) carbonate
(e) 0.5 moles copper(II) nitrate, Cu(NO3)2

Molar volume: The volume of one mole of the gas
[22.4 dm3 mol-1 at stp, standard temperature and pressure] or [24 dm3 mol-1 at room condition]

6. Calculate the volume of gases below.
(a) 0.5 mol of chlorine at STP.
(b) 0.2 mol of carbon dioxide at room condition.

(c) 1.5 mol of methane at room condition.

(d) 0.5 mol of helium at room condition.

(e) 2.5 mol of ammonia at standard temperature and pressure (stp)
7. Calculate the number of moles of the gases below.

(a) 250 cm3 of carbon dioxide in room temperature.
(b) 500 cm3 of hydrogen sulfide at stp.
(c) 200 cm3 of hydrogen chloride in room temperature.
(d) 750 cm3 of neon in room temperature.
(e) 300 cm3 of ammonia at stp.

19

#### Refer to your periodic table for the relative atomic masses. [ Avogadro constant: 6.02 x 1023 mol-1
Molar volume : 22.4 dm3 mol-1 at STP; 24 dm3 mol-1 at room conditions ] ####

8. What is the mass of 0.6 dm3 of chlorine gas, Cl2 at room temperature and pressure?

9. Calculate the volume of the following gases measured at STP.
(a) 1.806 x 1024 atoms of neon
(b) 18.25 g of hydrogen chloride, HCl

10. A sample of nitrogen gas, N2 has a volume of 1800 cm3 at room conditions.
What is the mass of the sample and how many molecules of nitrogen gas, N2 are in it?

11. Find the volume of these gases at room conditions.
(b) 9.03 x 1023 atoms of helium
(a) 0.8 mol of methane, CH4

12. Calculate the mass of 1.505 x 1024 molecules of hydrogen gas, H2 at STP.

13. 3 dm3 of an unknown gas has a mass of 6.0 g at room conditions. Find the molar mass of the gas.

14. 1.12 dm3 of hydrogen gas, H2 and 1.12 dm3 of oxygen gas, O2 are mixed together in a closed container at STP.
What is the total number of molecules in the container? What is the total mass of the gases in the container?

15. Magnesium powder reacts with hydrochloric acid to produce salt and hydrogen gas.
(a) Write the balance chemical equation for this reaction.
………………………………………………………………………………………….
(b) If 2.4 g of magnesium powder is added into excess hydrochloric acid, calculate,
(i) The mass of salt formed.

(ii) The volume of hydrogen gas liberated at room temperature.

20

16. Lead is extracted according to the following equation.
__C + __PbO → __CO2 + __Pb

(a) Write the balanced chemical equation for the reaction.

(b) Determine the number of moles of lead extracted from 0.5 mole of lead (II) oxide.

(c) Calculate the number of moles of carbon required to extracts 0.5 mole of lead(II) oxide.

(d) What is the mass of lead are produced if 44.6 g of lead (II) oxide is heated with excess carbon.
[ RAM : Pb = 207, O = 16, C = 12 ]

17. Excess sodium carbonates is added into 50 cm3 of 0.5 mol dm-3 copper(II) chloride.
The chemical equation for the reaction is as follows:
_CuCl2 + _Na2CO3 → _NaCl + _CuCO3
(a) Balance the chemical equation for the reaction.
(b) Calculate the mass of copper(II)carbonate.

18. 5g of calcium carbonate is added to excess hydrochloric acid.
(a) Write balanced chemical equation for the reaction.
……………………………………………………………………………………….…….
(b) Determine the volume of carbon dioxide gas evolved at room temperature

19. When sodium bicarbonate is heated, it decomposed to sodium carbonate, carbon dioxide and water.
(a) Write a balanced chemical equation for the decomposition of sodium bicarbonate on heating.
……………………………………………………………………………………………..….
(b) If 8.4 g of sodium bicarbonate decomposes, calculate
(i) The volume of carbon dioxide gas evolved at room temperature.

(ii) The mass of sodium carbonate formed.

Concentration of acids and alkalis (STOICHEIOMETRY)

The concentration of a solution is the mass (in gram) or the number of mole of solute dissolved in a solvent to form 1.00
dm3 of solution.

Concentration, g dm-3 = Massof solute, g
Volume of solution , dm3

Concentration, mol dm-3 / molarity = Number of mole of solute, mol
Volume of solution , dm3

Titration formula  MaVa = a
M bVb b

Ma = Concentration of the acid, Mb = Concentration of the alkali

Va = Volume of used acid, Vb = Volume of used alkali

a = the number of mole of acid, b = the number of mole of alkali

Dilution formula  M1V1 = M2V2

21

13. 100cm3 of HCl 2.0 mol dm-3 reacts completely with excessive zinc powder.
Calculate the volume of gas released at room temperature? (Molar volume = 24 dm3 at room temperature).
14. 5 g of calcium carbonate dissolves in 100 cm3 of nitric acid 0.5 mol dm-3 to produce salt, carbon dioxide gas and
water.
(i) Write out the equation.
………………………………………………………………………………………..……
(ii) Calculate the moles of calcium carbonates used. (RAM Ca =40, C = 12, O = 16).

……………………………………………………………………………………………………………………………..
(iii) Calculate the moles of acid.

………………………………………………………………………………………………………………………………
(iv) Calculate the volume of gas produced.

15. X g sodium carbonate reacts completely with 100 cm3 hydrochloric acid, 1.0 mol dm-3 to produce sodium chloride,
carbon dioxide and water. Find X.
(relative atomic mass: Na =23, C = 12, O = 16)

16. 10 g of copper(II) oxide reacts completely with 100 cm3 hydrochloric acid to form copper(II) chloride and water.
Calculate the molarity of acid used.
(Ar Cu = 64, O = 16 )

17. The following equation is not balanced.
___Na2SO4(aq) + ___BaCl2(aq) → ___BaSO4(s) + ___NaCl(aq)

(a) Identify the reactants and products of the reaction.
Reactants : _____________ and __________ Products : ____________ and ______________

(b) Balance the equation. Then, calculate
(i) the number of moles of sodium sulphate, Na2SO4 that react completely with a solution containing 2.08 g of
barium chloride, BaCl2.

(ii) the mass of barium sulphate, BaSO4 formed when 1 mol of sodium sulphate, Na2SO4 is reacted with 1 mol of
barium chloride, BaCl2.

PAPER 2 : STRUCTURE

1. Diagram 1 shows the set-up of the apparatus for an experiment to determine the empirical formula of an oxide of
copper.
Table 1 shows the results of an experiment after heating, cooling and weighing are repeated until a constant weight
is obtained.

Substance Mass(g)
Mass of combustion tube + porcelain dish 18.75 g
Mass of combustion tube + porcelain dish + oxide of copper 20.75 g
Mass of combustion tube + porcelain dish + copper 20.35 g

22

(a) What is meant by empirical formula?

………..………………………………………………………………………………………………………………….

………………………………………………………………………………………………………………..[1 mark]

(b) Based on Table 1 results,

(i) Calculate the mass of copper and the mass of oxygen used in the experiment. [2 marks]

(ii) Calculate the mole ratio of copper atoms to oxygen atoms. [2 marks]
Given that the relative atomic mass of Cu, 64; O ,16.

……………………………………………………………………………………………….

(iii) State the empirical formula of this oxide of copper.
……….………………………………………………………………………………………[1 mark]

(iv) Write the chemical equation for the reaction in this experiment.
……………...………………………………………………………………………………[1 mark]

Diagram 1

(c) Describe the steps that should be taken to ensure that all the air in the combustion tube has been expelled , before

any heating is carried out.
…………………………..…………………………………………..…………………………
. ……………………………….………………………………………………………………
……………………………………...…………………………………………………………[3 marks]

(d) Why does the setup need to be completely filled with hydrogen before it is heated?
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………

(e) (i) Name the clear liquid that is collected at A.
……………………………………………………………………………………………………………………………..

(ii) How would you ensure that the liquid collected is pure?
……………………………………………………………………………………………………………………………..

(e) (i) Predict whether the empirical formula of copper(II) oxide can be determined using this reduction method.
……………………………………………………………………………………………………………………………..

(ii) Give a reason for your answer in (e) (i).
……………………………………………………………………………………………………………………………..

2. (a) Calculate the relative molecular or formula masses of the following substances. [1 mark]
(i) Ethanol, C2H5OH. Given that relative atomic mass of H = 1, C = 12, O = 16.

(ii) Zinc nitrate, Zn(NO3)2 [1 mark]
Given that relative atomic mass of O = 16 , N = 14, Zn = 65 .

(b) A closed glass bottle contains 4 mol molecules of oxygen, O2. [1 mark]
(i) What is the number of oxygen molecules in the bottle?

……………………………………………………………………………………………………………………………

(ii) How many oxygen atoms are there in the bottle? [1 mark]

…………………………………………………………………………………………………………………………….
(c) Find the number of moles of atoms in a sample containing 9.03 × 1020 atoms of copper.
[1 mark]

23

…………………………………………………………………………………………………………………………………

(d) Calculate the mass, in gram, of 3.5 moles of copper(II) carbonate, CuCO3. [1 mark]
Given that relative formula mass of CuCO3 = 124

…………………………………………………………………………………………………………………………………

(e) When silver carbonate, Ag2CO3 is heated, it will decompose to produce silver metal, carbon dioxide gas and

oxygen gas as shown in the equation below.

2Ag2CO3(s) → 4Ag(s) + 2CO2(g) + O2(g)

A student heats 8.28 g silver carbonate. Calculate the volume of carbon dioxide gas, CO2 collected at room

temperature. Given that relative atomic mass of C = 12, O =16, Ag = 108, [3 marks]

……………………………………………………………………………………………………………………………………
3. Diagram 3 shows 3.1 g of copper (II) carbonate being heated strongly in a test tube.

Diagram 3
The gas given out is passed into limewater in a test tube through a delivery tube.

(a) State the observation made when copper(II)carbonate powder is heated until the reaction is complete.
……………………………………………………………………………………………………………………[1 mark]

(b) Write the chemical equation to represent the reaction that takes place.
……………………………………………….………………………………………………………………..…[1 mark]

(c) Calculate the number of moles of copper(II)oxide produced.

…………………………………………………………………………………………………………………… [2 marks]
(d) Calculate the volume of gas produced at STP.

…………………………………………………………………………………………………………………….
……………………………………………………………………………………………………………………[2 marks]

(e) (i) What can be observed if the product is heated in a stream of hydrogen gas ?
……………………....………………………………………………………………………………………[1 mark]

(ii) Calculate the mass of the substance produced.

……………………………………………………………………………………………………………….[2 marks]

4. The element chlorine forms a compound with oxygen with the molecular formula Cl2On
The relative molecular mass of the compound is 183. What is the value of n?

5. Butane has the empirical formula of C2H5 and relative molecular mass of 58. Find its molecular formula.

24

6. Ethanoic acid is an important ingredient of vinegar. The empirical formula of this acid is CH2O.
Given that its molar mass is 60 g mol-1 , find its molecular formula.

7. Determine the percentage composition by mass of water in hydrated copper(II) sulphate, CuSO4.5H2O

8. Rust contains iron(III) oxide, Fe2O3. How many moles of iron(III) ions are present in 4.0 g of rust?

9. Element Y reacts with oxygen to produce a compound with molecular formula YO3.
Given that the mass of 1 mol of the compound is 80 g, determine the relative atomic mass of Y.

10. What is the mass of lead, in grams, needed to combine with 0.5 mol of chlorine atoms to produce a compound with
an empirical formula of PbCl2.

Exercise

Construct a chemical formula for each of the following ionic compounds.

(a) Magnesium chloride (e) Silver iodide (i) Lead(II) sulphite

(b) Potassium carbonate (f) Zinc nitrate (j) Potassium chlorate(V)

(c) Calcium sulphate (g) Aluminium oxide (k) Ammonium phosphate

(d) Copper(I) oxide (h) Iron(II) hydroxide (l) Sodium bromate(I)

Quick Review
Refer to the Periodic Table for the relative atomic masses of related elements. [ Avogadro constant: 6.02 x 1023 mol-1]

1. How many atoms of nitrogen have the same mass as one atom of iron?

2. The mass of an atom of element Y is ten times greater than the mass of an atom of

beryllium. What is the relative atomic mass of Y?

3. Calculate the relative molecular or formula masses of the following substances.

(a) I2 (b) SF6 (c) CaCO3 (d) H3PO4 (e) Cu(N03)2

4. Gypsum, CaSO4.2H2O is widely used in plastering of walls and ceilings. Show that the relative formula mass of

gypsum is 172.

5. The element chlorine forms a compound with oxygen with the molecular formula Cl2On

25

The relative molecular mass of the compound is 183. What is the value of n?

Calculation

1. Calculate the number of atoms in

(a) 1 mol of aluminium. (b) 3.2 mol of helium atoms. (c) 0.5 mol of iodine molecules, I2
(d) 0.4 mol of ozone gas, 03.
2. A sample contains 6.02 x 1025 molecules of water. How many moles of water are there in the sample?

3. Find the number of mole of hydrogen gas, H2 containing
(a) 3.01 x 1024 hydrogen molecules, H2 (b) 6.02 x 1023 hydrogen atoms.

4. A beaker contains 0.1 mol of zinc chloride, ZnCl2
(a) Calculate the number of moles of chloride ions in the beaker.

(b) Find the total number of ions in the beaker.
5. A container contains 1.806 x 1023 oxygen molecules, 02. A sample of 0.5 mol of oxygen gas, 02 is added to the

container. How many molecules are there altogether in the container?

Quick Review

1. (a) Define the terms ‘mole’ and ‘Avogadro constant’.
(b) Relate the number of moles of a substance to the number of particles in it.

2. How many moles of bromine gas, Br contain 1.505 x 1024 molecules of bromine gas,Br2?
3. Calcium is needed for the formation of bones and teeth. How many calcium ions are there in a serving of cereal that

contains 0.007 mol of calcium ions?
4. Given 1 mol of hydrogen gas, H2 and 0.7 mol of sulphur dioxide, SO2 which one has more atoms? Give your reasons.
5. What is the difference between 1 mol of oxygen atoms and 1 mol of oxygen molecules, 02?

Calculation

The relative atomic masses are given on page 177. [Avogadro constant: 6.02 x 1023 mol-1]

1. Calculate the mass, in grams, of each of the following.

(a) 0.3 mol aluminium (b) 2 mol sodium nitrate, NaNO3

2. How many moles of atoms or molecules are there in

(a) 5.6 g of iron? (b) 3.2 g of methane, CH4
3. What is the mass of carbon that contains 6.02 x 1021 carbon atoms?
4. What is the meaning of ‘molar mass’?

2. What is the mass of
(a) 0.01 mol of ammonia gas, NH3 (b) 6.02 x 1024 nitrogen molecules, N2?

3. How many moles of molecules are there in

(a) 2.8 g of carbon monoxide, CO? (b) 4 g of bromine gas, Br2?

Calculation

[ Molar volume: 22.4 dm3 mol-1 at STP; 24 dm3 mol-1 at room conditions]
1. Calculate the volume of the following gases.

(a) 0.3 mol of oxygen gas, 02 measured at room conditions
(b) 4 mol of helium gas measured at STP
(c) 1.5 mol of hydrogen gas, H measured at room conditions
2. Find the number of moles of the following gases.
(a) 48 dm3 of chlorine gas, Cl2 at room conditions
(b) 560 cm3 of carbon dioxide, CO2 at STP
(c) 960 cm3 of hydrogen chloride, HCl at room conditions.

Calculation

Refer to your Periodic Table for the relative atomic masses. [ Avogadro constant: 6.02 x 1023 mol-1
Molar volume : 22.4 dm3 mol-1 at STP; 24 dm3 mol-1 at room conditions ]
1. What is the mass of 0.6 dm3 of chlorine gas, Cl2 at room temperature and pressure?

2. Calculate the volume of the following gases measured at STP.
(a) 1.806 x 1024 atoms of neon
(b) 18.25 g of hydrogen chloride, HCl
3. A sample of nitrogen gas, N2 has a volume of 1800 cm3 at room conditions.

26

What is the mass of the sample and how many molecules of nitrogen gas, N2 are in it?

4. Find the volume of these gases at room conditions.
(b) 9.03 x 1023 atoms of helium
(a) 0.8 mol of methane, CH4
5. Calculate the mass of 1.505 x 1024 molecules of hydrogen gas, H2 at STP.
6. 3 dm3 of an unknown gas has a mass of 6.0 g at room conditions. Find the molar mass of the gas.
7. 1.12 dm3 of hydrogen gas, H2 and 1.12 dm3 of oxygen gas, O2 are mixed together in a closed container at STP.

What is the total number of molecules in the container? What is the total mass of the gases in the container?

Calculation

Refer to Periodic Table for the relative atomic masses.
1. 0.20 g of calcium reacts with fluorine to give 0.39 g of calcium fluoride.

Find the empirical formula of the fluoride compound produced.
2. Find the empirical formula of a compound that consists of 32.4% of sodium, 22.6% of sulphur and 45.0% of oxygen.
3. 60 g of aluminium sulphide contains 38.4 g of sulphur. Find the empirical formula of the compound.

Calculation

1. Butane has the empirical formula of C2H5 and relative molecular mass of 58. Find its molecular formula.
2. Ethanoic acid is an important ingredient of vinegar. The empirical formula of this acid

is CH2O Given that its molar mass is 60 g mol-1 , find its molecular formula.
3. Determine the percentage composition by mass of water in hydrated copper(II) sulphate, CuSO4.5H2O
4. Rust contains iron(III) oxide, Fe2O3. How many moles of iron(III) ions are present in 4.0 g of rust?
5. Element Y reacts with oxygen to produce a compound with molecular formula YO3.

Given that the mass of 1 mol of the compound is 80 g, determine the relative atomic mass of Y.

6. What is the mass of lead, in grams, needed to combine with 0.5 mol of chlorine atoms

to produce a compound with an empirical formula of PbCl2.

Quick Review

1. Name the ionic compound with each of the following formulae.

(a) Ba(N03)2 (d) NaOH (g) KBr

(b) MgO (e) ZnSO4 (h) Ca(OH)2

(c) LiCl (f) NaHCO3 (i) AlPO4

2. Name the following molecular compounds.

(a) CCl4 (c) BF3 (e) NO

(b) CS2 (d) SO2(f) N2O4

3. The molecule of a compound consists of one nitrogen atom and three chlorine atoms.

What is the name of this compound?

Quick Review

1. Name the ionic compound with each of the following formulae.

(a) Ba(N03)2 (d) NaOH (g) KBr

(b) MgO (e) ZnSO4 (h) Ca(OH)2

(c) LiCl (f) NaHCO3 (i) AlPO4

2. Name the following molecular compounds.

(a) CCl4 (c) BF3 (e) NO
(b) CS2 (d) SO2(f) N2O4

3. The molecule of a compound consists of one nitrogen atom and three chlorine atoms.

What is the name of this compound?

Calculation

Refer to page 177 of your textbook for the relative atomic masses. [molar volume: 22.4 dm3 mol-1 at STP]
1. Hydrogen peroxide, H2O2 decomposes according to the following equation.

2H2O2(l) → 2H2O(l) + O2(g)
Calculate the volume of oxygen gas, O2 measured at STP that can be obtained from the decomposition of 34 g of
hydrogen peroxide, H2O2
2. Silver carbonate, Ag2CO3 breaks down easily when heated to produce silver metal.

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Ag2CO3(s) 4Ag(s) + 2CO2(g) + O2(g)

Find the mass of silver carbonate, Ag2CO3 that is required to produce 10 g of silver.

3. 16 g of copper(II) oxide, CuO is reacted with excess methane, CH4. Using the equation below, find the mass of

copper that is produced.

4CuO(s) + CH4(g) → Cu(s) + CO2(g) + 2H2O(l)

4. Write balanced chemical equations for the following reactions.

(a) Iron + sulphuric acid → iron(III) sulphate solution + hydrogen gas

(b) Copper wire burns in oxygen gas to produce black powder of copper(II) oxide

Summary

Relative Atomic Mass and Relative Molecular Mass
/ The relative atomic mass of an element is the average mass of one atom of an element when compared with of the

mass of an atom of carbon-12.
/ The relative molecular mass of a molecule is the average mass of one molecule when compared with - of the mass of

an atom of carbon-12.
/ The relative molecular mass of a molecule is calculated by adding up the relative atomic masses of all the atoms that

are present in the molecule.
/ The relative formula mass of an ionic substance is the average mass of one formula unit

of the ionic substance when compared with - of the mass of an atom of carbon-12.

The Mole and the Number of Particles
/ One mole of substance contains 6.02 x 1023 particles.
/ 6.02 x 1023 mol-1 is known as the Avogadro constant.

The Mole and the Mass of Substances
/ The molar mass of a substance is the mass of 1 mol of the substance with the unit grams per mol or g mol-1

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/ The mass of one mole of atoms is numerically equal to its relative atomic mass in grams.
/ The mass of one mole of molecules is numerically equal to its relative molecular mass in grams.
The Mole and the Volume of Gas
/ The molar volume of a gas is the volume occupied by one mole of the gas.
/ One mole of a gas occupies a volume of 24 dm3 at room conditions and 22.4 dm3 at STP.

Chemical Formulae
/ A chemical formula is a representation of a chemical substance using letters for atoms and subscript numbers to show

the numbers of each type of atoms that are present in the substance.
/ The empirical formula of a compound gives the simplest whole number ratio of atoms of each element in the

compound.
/ The molecular formula of a compound gives the actual number of atoms of each element present in the molecule of

the compound.
Chemical Equations
/ A chemical equation is a representation of a chemical reaction in words or using chemical formulae.
/ Qualitatively, a chemical equation shows the reactants and products of a reaction as well as the physical state of each

of them.
/ Quantitatively, a chemical equation shows the exact proportions of reactants and products in a chemical reaction.

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