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CALCULATIONS OF AVERAGE ATOMIC MASS (Ar)
With given isotopic masses say m1, m2 and m3 with respective intensities I1, I2 and I3, the average atomic
mass is given by the following formula:
Average atomic mass (Ar) =
=
∑( Isotopic mass × intensity)
Total intensity
M1 I1 +M2 I2 +M3 I3
I1 +I2 +I3
While for given percentage abundances, say P1 %, P2 % and P3 % for isotopic masses m1, m2 and m3
respectively, the above formula may be rewritten as follows;
Ar =
∑(Isotopic mass ×percentage abundance)
=
100
m1 P1 +m2 P2 +m3 P3
100
Where ∑ means ‘summation of’
Worked examples
Example 1
Magnesium has three stable isotopes of masses 23
...
985, 25
...
m
...
The relative abundances of
the three isotopes are 39
...
065, and 5
...
Calculate the average atomic mass
...
985 ×39
...
985×5
...
982×5
...
35+5
...
585
= 24
...
m
...
3 a
...
u
Example 2
A sample of pure unknown element X was analysed and the data is given in the table below
...
Symbol
Mass (a
...
u)
Natural abundance (%)
20
10X
19
...
22
21
10X
20
...
257
22
10X
21
...
82
Solution
Ar =
=
∑(Isotopic mass × percentage adunbance)
Total percentage adunbances
19
...
22+20
...
257+21
...
82
90
...
257+8
...
17
So the relative atomic mass of X is 20
...
Solution
Ar =
∑(isotopic mass × proportion)
Total proportions
=
69×60+71×40
60+40
= 69
...
8
Example 4
The mass spectrum of an element enables the relative abundance of each isotope o f the element to be
determined
...
5%
81
49
...
(The readers should recognise a little difference between
definition of isotopes and isotope)
(ii)
79
35X
and 81
35X
∑(isotopic mass × percentage abundance)
79×50
...
5
(iii) Ar =
=
= 79
...
0 (To three significant figures)
Example 5
109
Silver consists of two isotopes 107
47Ag and 47Ag of atomic masses 106
...
91g/mol
109
respectively
...
88% for 107
47Ag and 48
...
Calculate the average atomic mass of Ag
...
88×106
...
12×108
...
87g/mol
= 107
...
Cu − 63 Have an atomic mass of 62
...
15%
...
546 amu
...
15)% or 30
...
546 =
69
...
9296+30
...
9277 amu
Hence the atomic mass of the second isotope is 64
...
The heights of the three peaks and the mass
number of isotopes are shown
...
63
Intensity
(Isotopic percentage)
9
...
1
24
25
26
Mass number
Solution
Ar =
∑(Isotopic mass × intensity)
Total intensity
=
24×63+25×8
...
1
63+8
...
1
= 24
...
3
Example 8
In a sample of 400 lithium atoms, it is found that 30 atoms are lithium−6 (6
...
016g/mol)
...
Solution
Using Ar =
∑(Relative abundance ×isotopic mass)
∑ Abundance
But abundance of isotopic is directly proportional to the number of atoms of the isotopes
It follows that: Ar =
∑(Number pf atoms of isotope x isotopic mass)
Total number of atoms in the sample
370×7
...
015
) g/mol
400
=(
= 6
...
941g/mol
Alternative solution:
Percentage abundance of Li − 6 =
Number of atoms of Li−6 in sample
×
Total number of atoms in the sample
100%
30
= 400 × 100% = 7
...
5)% = 92
...
Example 9
Natural rubidium has the average mass 85
...
9117amu) and Rb − 87
...
591
...
Solution
Using Ar =
Given that:
∑(Number of atoms of isotope × isotopic mass)
Total number of atoms of isotopes
Number of atoms of Rb−85
Number of atoms of Rb−87
=
2
...
591 atoms of Rb − 85 there is 1 atom of Rb − 87
...
4678 =
2
...
9117+1×m
2
...
9087 amu
Hence the mass of Rb − 87 is 86
...
m
...
The heights of the peaks are in the
ratio of 9:6:1 respectively
...
Solution
Cl – 35Cl show peak at mass of 70
35
Cl – 37Cl shows peak at mass of 72
35
Cl – 37Cl shows peak at mass of 74
37
Thus % o f 35Cl = (
% of 37Cl = (
1
2
9+( ×6)
1
2
9+6+1
) × 100% =
1+( ×6)
9+6+1
) × 100% =
4
×
16
12
16
× 100% = 75%
100% = 25%
Thus the relative abundance of 35Cl is 75%
The relative abundance of 37Cl is 25%
Using;
Ar =
∑(isotopic mass × percentage abundance)
100
=
75×35+25×37
100
=
2625+925
100
= 35
...
m
...
5 a
...
u
In this question it was instructed to find firstly relative abundances before finding average atomic mass,
otherwise the following alternative solution would be applicable
...
m
...
m
...
5 a
...
u
Thus average atomic mass is 35
...
m
...
Then x+y = 100 (Total % abundances must be 100)
From which, x = 100 - y
So using Ar =
Then 35
...
, limit us to find relative abundances
before calculating the average atomic mass and not vice-versa