1. Chapter 3
Relative Masses of Atoms and Molecules
LEARNING OUTCOMES
Define relative atomic mass, Ar
Define relative molecular mass, Mr

2. Chapter 3
Relative Masses of Atoms and Molecules
Relative Atomic Mass

The masses of atoms and molecules are
very small and it very hard for us to
compare them or use them in calculations.

For e.g. the mass of a hydrogen atom is
0.000 000 000 000 000 000 000 0014 or
(1.4 x 10-24)g and the mass of a carbon
atom is 1.68 x 10-23 g.

3. Chapter 3
Relative Masses of Atoms and Molecules
Relative Atomic Mass
 Instead of using the true masses of atoms,
scientists compare the masses of atoms with the
mass of a hydrogen atom, which is assigned a
mass of one unit.
 If we take the mass of a hydrogen atom to be 1,
then the mass of a carbon atom will be 12, since a
carbon atom is 12 times as heavy as a hydrogen
atom.

4. Chapter 3
Relative Masses of Atoms and Molecules
12 H atoms
Relative Atomic Mass

The mass of a carbon atom is
12 times as heavy as a hydrogen
atom, so we say that the relative
atomic mass of carbon is 12.
C
 Scientists prefer to use the carbon atom instead of the
hydrogen atom as a standard unit, so if we take 1/12 of the
mass of a carbon atom, we will still get 1 unit.
 Hence we define:
The relative atomic mass of an element is the average mass of
an atom of the element compared to the mass of 1/12 of the
mass of a carbon-12 atom.

5. Chapter 3
Relative Masses of Atoms and Molecules
Relative Atomic Mass
Relative atomic mass
(Ar)
Mass of one atom of the element
Mass of 1 of an atom of carbon-12
=
12

The symbol for Relative Atomic Mass is Ar

Relative atomic mass has no units.

The relative atomic mass of an element can
be obtained from the mass number (nucleon number)
of the element in the Periodic Table.

Examples : 2311Na, 5626Fe
Hence, Ar of Na = 23, Ar of Fe = 56

6. Chapter 3
Relative Masses of Atoms and Molecules
Quick check 1
1.
2.
3.
4.
5.
Define the relative atomic mass of an element.
Using the Periodic Table, find the relative atomic mass of the
following:
Mg, Ca, Cl, O, S, Ne, Br.
What is the mass of a calcium atom compared to the mass
of a helium atom?
How many hydrogen atoms have the same mass as one
potassium atom?
How many oxygen atoms have the same mass as one
bromine atom?
Solution

7. Chapter 3
Relative Masses of Atoms and Molecules
Solution to Quick check 1
1.
2.
3.
4.
5.
The relative atomic mass of an element is the average
mass of an atom of the element compared to the mass of
1/12 of a carbon-12 atom.
Ar: Mg=24, Ca=40, Cl=35.5, O=16, S=32, Ne=20, Br=80.
Mass of a Ca atom =10 x Mass of a He atom
39 hydrogen atoms
5 oxygen atoms
Return

8. Chapter 3
Relative Masses of Atoms and Molecules
Relative Molecular Mass
 The picture shows the mass
of a molecule of water
compared to the mass of
hydrogen atoms.

It shows that one molecule of
water is 18 times as heavy
as one hydrogen atom.
 Therefore, the relative
molecular mass of water is 18.
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9. Chapter 3
Relative Masses of Atoms and Molecules
Relative Molecular Mass

If we use the mass of a carbon-12 atom as the
standard unit of comparison, then the relative
molecular mass of a substance is defined as:
The relative molecular mass of a substance is the average
mass of one molecule of the substance compared to the mass
of 1/12 of a carbon-12 atom.
Relative molecular mass Mass of one molecule of a substance
=
1
(Mr)
Mass of 12 of a carbon-12 atom
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10. Chapter 3
Relative Masses of Atoms and Molecules
Relative Molecular Mass

The symbol for Relative Molecular Mass is Mr

The relative molecular mass of a molecule can be
found by adding up the relative atomic masses of
all the atoms present in the molecule.

E.g. Mr of water, H2O = Mass of 2 H atoms +
mass of 1 O atom
= 2 x 1 + 16
= 18
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11. Chapter 3
Relative Masses of Atoms and Molecules
Relative Formula Mass
 Ionic compounds (e.g. sodium
chloride) are not made up of single
molecules; instead they are made
up of a crystal lattice consisting of
many oppositely charged ions.
 Hence, instead of calculating the relative molecular
mass of an ionic compound, we calculate the mass
based on its formula (formula mass).
 We can take the relative formula mass as equivalent
to the relative molecular mass in our calculations.
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12. Chapter 3
Relative Masses of Atoms and Molecules
Finding Relative Molecular Mass
Worked examples
1.
2.
3.
Find the relative molecular mass of carbon dioxide, CO 2.
Mr of CO2 = 12 + 16 x 2
= 44
Find the relative molecular mass (formula mass) of copper(II)
nitrate, Cu(NO3)2.
Mr of Cu(NO3)2 = 64 + (14 + 16x3 )x2
= 64 + 62x2 = 188
Find the relative molecular mass (formula mass)
of ammonium sulphate, (NH4)2SO4.
Mr of (NH4)2SO4 = (14 + 1x4)x2 + 32 + 16x4
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13. Chapter 3
Relative Masses of Atoms and Molecules
Quick check 2
Find the relative molecular mass (or formula mass) of each
of the following:
(a) Magnesium sulphate, MgSO4
(b) Calcium nitrate, Ca(NO3)2
(c) Ammonium carbonate, (NH4)2CO3
(d) Benzoic acid, C7H6O2
(e) Hydrated sodium carbonate, Na2CO3.10H2O
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Solution

14. Chapter 3
Relative Masses of Atoms and Molecules
Solution to Quick check 2
a)
Mr of MgSO4 = 24 + 32 + 16x4 = 120
b)
Mr of Ca(NO3)2 = 40 + 2(14 + 16x3) = 164
c)
Mr of (NH4)2CO3 = 2(14+4) + 12 + 16x3 = 96
d)
Mr of C7H6O2 = 12x7 + 6 + 16x2 = 122
e)
Mr of Na2CO3.10H2O = 23x2 + 12 + 16x3 + 10(2x1 + 16)
= 286
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Return

15. Chapter 3
Relative Masses of Atoms and Molecules
Percentage composition
Worked example 1
What is the percentage composition of calcium carbonate, CaCO 3?
Solution
% of Ca = [Ca] x 100% =
40
x 100%
[CaCO3]
[40 + 12 + 16x3]
= 40%
% of C = [ C ] x 100% = 12 x 100%
[CaCO3]
100
= 12%
% of O = [ O3 ] x 100% = 16x3 x 100%
[CaCO3]
100
= 48%
Check
15
%Ca + %C + %O = 40 + 12 + 48 = 100%

16. Chapter 3
Relative Masses of Atoms and Molecules
Percentage composition
Worked example 2
What is the percentage of sodium in sodium carbonate, Na2CO3?
Solution
% of Na = [Na2] x 100% =
23x2
x 100%
[Na2CO3]
[23x2 + 12 + 16x3]
= 43.4%
Worked example 3
Find the mass of oxygen in 90 g of water.
Solution
Mr of H2O = 1x2 + 16 = 18
Mass of oxygen = 16 x 90 g = 80 g
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Worked example 4
What mass of magnesium oxide can be made from 6 g of magnesium?
Solution
Mr of MgO = 24 + 16 = 40
16
Mass of MgO = 6 x 40 g = 10 g
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17. Chapter 3
Relative Masses of Atoms and Molecules
Percentage yield
The percentage yield of a product is given by the formula:
Percentage yield = Actual mass of product obtained x 100%
Theoretical mass of product obtained
Worked example 1
In an experiment to prepare magnesium sulphate, 2.4 g of
magnesium was dissolved completely in dilute sulphuric acid.
On crystallisation, 22.0 g of magnesium sulphate crystals,
MgSO4.7H2O, were obtained. What is the percentage yield?
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18. Chapter 3
Relative Masses of Atoms and Molecules
Percentage yield
Solution
Write the chemical equation: Mg + H2SO4  MgSO4 + H2
Number of mole of Mg reacted = 2.4 = 0.1 mol
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From equation,
1 mol of Mg  1 mol of MgSO4
Therefore, 0.1 mol of Mg  0.1 mol of MgSO4.7H2O
Mass of MgSO4.7H2O produced = 0.1 mol x 246 g/mol = 24.6 g
Percentage yield = Actual mass of product obtained x 100%
Theoretical mass of product obtained
= 22.0 x 100 %
24.6
= 89.4 %
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19. Chapter 3
Relative Masses of Atoms and Molecules
Percentage purity
The percentage purity tells us how pure a prepared product
is compared to the pure form of the substance.
 For example, the percentage purity of a gold bar may be
given as 99.99 %.

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20. Chapter 3
Relative Masses of Atoms and Molecules
Percentage purity
Worked example 2
On analysis, 5.00 g of a sample of marble (calcium carbonate)
was found to contain only 4.26 g of pure calcium carbonate.
What is the percentage purity of the marble?
Solution
Percentage purity = 4.26 x 100 %
5.00
= 85.2 %
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21. Chapter 3
Relative Masses of Atoms and Molecules
Quick check 3
1.
Find the percentage composition of each element in sulphuric acid, H2SO4.
2.
Find the percentage of nitrogen in calcium nitrate, Ca(NO3)2.
3.
Find the mass of calcium in 250 g of calcium carbonate, CaCO3.
4.
What mass of iron can be obtained from 320 g of iron(III) oxide, Fe2O3?
5.
24 g of hydrogen combines with 192 g of oxygen to form water.
What mass of hydrogen will combine with 24 g of oxygen?
In the Haber process to manufacture ammonia, it was reported that in a
certain factory, 2.8 tonnes of nitrogen gas produced 0.80 tonne of
ammonia. What is the percentage yield of ammonia?
The equation for the Haber process is:
Solution
N2(g) + 3H2(g)  2NH3(g)
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6.

22. Chapter 3
Relative Masses of Atoms and Molecules
Solution to Quick check 3
1.
2.
3.
4.
5.
6.
% composition of H2SO4 :
H =2.04 %, S =32.7 %, O = 65.3 %
% of nitrogen in Ca(NO3)2 = 17.1 %
Mass of calcium = 40 x 250 = 100 g
100
Mass of iron = 112 x 320= 224 g
160
Mass of hydrogen = 24 x 24 = 3 g
192
Percentage yield = 0.8 x 100 %
3.4
= 23.5 %
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Return

23. Chapter 3
Relative Masses of Atoms and Molecules
To learn more about Relative Atomic and
Molecular Mass, click on the link below!
http://www.ch.cam.ac.uk/magnus/MolWeight.html
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