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.
8
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
9
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
10
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.
11
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
12
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
13
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
14
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
18
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
24
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?
17
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
24
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 %
18
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 %.
19
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 %
20
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)
21
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 %
22
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
23
