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In a periodic table, the average atomic mass of magnesium is given as `24.312 u`. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are `._12Mg^(24) (23.98504u)`, `._(12)Mg^(25) (24.98584)` and `._12Mg^(26) (25.98259 u)`. The natural abundance of `._12Mg^(24)` is `78.99%` by mass. Calculate the abundances of the other two isotopes.

Text Solution

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Natural abundance of `""_(12)Mg^(24) = 78.99%`
Let natural abundance of `""_(12)Mg^(26)`
`=(100-(78.99 +x)) = (21.01-x)%`
As average atomic mass is calculated as weighted average of the masses of the isotopes
`therefore 24.312 u =(+(21.01-x)25.98259u)/100`
`rArr 0.99675x = 9.27`
`rArr x=9.27/0.99675 = 9.30%`
So `21.01 -x = 21.01-9.30 = 11.71`%
Hence relative abundance of `""_(12)Mg^(25) = 11.71`%
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