Determine the percentage of water of crystallisation, iron , sulphur and oxygen in pure ferrous sulphate `(FeSO_(4).7H_(2)O)`.

Explanation

The formula mass of of ferrous sulphate
=At. Mass of Fe+At. Mass of S+4X At. Mass of oxygen
+7X Mol. Mass of `H_(2)O`
`=56.0+32.0+4xx16.0+7xx18.0=278.0`
So, % of water of crystallisation `=(126)/(278)xx100=45.32`
% of iron `=(56)/(278)xx100=20.14`
% of sulphur=`(32)/(278)xx100=11.51`
% of oxygen`=(64)/(278)xx100=23.02`