Page 115 - Chemistry ICSE Class IX

Atomic Structure and Chemical Bonding 103
Fractional Atomic Masses

Mass numbers are in whole numbers, whereas atomic masses, in general,
are found to have fractional values. Why is it so?
This is because a sample of any element generally contain a mixture
of all the isotopes of that element in certain proportion. The atomic mass
represents the weighted average of all the naturally-occurring isotopes of
that element.
This is illustrated below.

How to calculate the atomic mass of an element from the
mass numbers of its isotopes
The atomic mass of an element is the weighted arithmetic mean of the
atomic masses of its isotopes present in the sample of the element.
Let us consider a sample of an element X containing its two isotopes
X and X . Then
2
1
(Percentage of X × Mass number of X ) + (Percentage of X × Mass number of X )
Atomic mass of the element X = 1 1 2 2
100
37
35
This method is illustrated by taking the case of chlorine. The two isotopes of chlorine, Cl and Cl
17
17
occur in the ratio 3 : 1. Then,
(35 u × 3) + (37 u × 1) 105 u + 37 u 142 u
Atomic mass of chlorine = = = = 35.5 u
3 + 1 4 4
The calculated atomic mass of chlorine (or any other element) is thus
a weighted average of the mass numbers of all the isotopes present in any
sample.
This explains why the atomic masses are fractional, although the mass
numbers are whole numbers.

79
EXAMPLE 4.5. If bromine occurs in the form of two isotopes, say Br (49.7%)
35
81
and Br (50.3%), then calculate the atomic mass of bromine atom.
35
79 u × 49.7 + 81 u × 50.3
Solution: Atomic mass of bromine = = 80.0 u
49.7 + 50.3
So, the atomic mass of bromine is 80.0 u.
EXAMPLE 4.6. The atomic mass of a sample of an element X is 16.2 u. What are
16
18
the percentages of isotopes X and X in the sample?
8
8
Solution: Atomic mass of X = 16.2 u
16
Let, the percentage of the isotope X be P . Then,
8
1
18
Percentage of the isotope X = (100 – P )
1
8
We can then write,
16
16
(Percentage of X × 16 u) + [(100 – Percentage of X) × 18 u]
Atomic mass of X = 8 8
100
(P × 16 u) + (100 – P ) × 18 u
16.2 u = 1 1
100
1620 u = 16 P u + 1800 u – 18 P u
1
1
This gives,
(18 P u – 16 P u) = 1800 u – 1620 u
1
1
2 P u = 180 u
1
180
P = = 90
1
2