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* For an element E with the naturally occurring isotopes aE, bE,
cE, and with the respective abundances of A%, B%, C% etc,
the relative atomic mass (r.a.m.) = (A/100 x a) + (B/100 x b) +
(C/100 x c)
* Given the relative atomic mass (r.a.m.) of an element and the
mass of each of its isotopes, the relative abundance of each
isotope can be calculated:
let y/100 = abundance of isotope 1
and (100 - y)/100 = abundance of isotope 2
then, r.a.m = [y/100 x mass isotope 1] + [(100 - y)/100 x mass
isotope 2]
and solve for y
The Abundance of Some Naturally Occurring Isotopes
Element Isotopes Mass Number Relative Abundance (%)
hydrogen 1H 1 99.99
2H 2 0.01
lithium 6Li 6 7.59
7Li 7 92.41
boron 10B 10 19.6
11B 11 80.4
carbon 12C 12 98.89
13C 13 1.11
nitrogen 14N 14 99.63
15N 15 0.37
oxygen 16O 16 99.76
17O 17 0.04
18O 18 0.20
magnesium 24Mg 24 78.99
25Mg 25 10.00
26Mg 26 11.01
silicon 28Si 28 92.23
29Si 29 4.68
30Si 30 3.09
chlorine 35Cl 35 75.78
37Cl 37 24.22
argon 36Ar 36 0.34
38Ar 38 0.06
40Ar 40 99.60
copper 63Cu 63 69.17
65Cu 65 30.83
silver 107Ag 107 51.84
109Ag 109 48.16
barium 130Ba 130 0.11
132Ba 132 0.10
134Ba 134 2.42
135Ba 135 6.59
136Ba 136 7.85
137Ba 137 11.23
138Ba 138 71.70
lead 204Pb 204 1.5
206Pb 206 23.6
207Pb 207 22.6
208Pb 208 52.3
uranium 234U 234 0.01
235U 235 0.72
238U 238 99.27
Calculating Relative Atomic Mass
Naturally occurring silver is 51.84% silver-107 and 48.16%
silver-109.
Calculate the relative atomic mass of silver.
r.a.m. (Ag) = (51.84/100 x 107) + (48.16/100 x 109)
= 55.469 + 52.494
= 107.96
Calculating the Abundance of an Isotope
Copper consists of two isotopes, copper-63 and copper-65.
Its relative atomic mass is 63.62.
Find the abundance of each isotope.
Let y/100 = the abundance of copper-63
and (100 - y)/100 = the abundance of copper-65
63.62 = (y/100 x 63) + [(100 - y)/100 x 65]
63.62 = 63y/100 + 6500/100 - 65y/100
6362 = 63y + 6500 - 65y
-135 = -2y
y = 69
Abundance of copper-63 = 69/100 = 69%
Abundance of copper-65 = 100 - 69 = 31%
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