Determine the percentage of water of crystallisation, iron, sulphur and oxygen in pure ferrous sulphate `(FeSO_4.7H_2O)`.

Explanation

The formula mass of ferrous sulphate
= At mass of Fe + At mass of S +4 `xx `At mass of O +7 `xx` Mol. mass of `H_2 O = 56.0 + 32.0 + 4 xx 16.0 + 7 xx 18.0 = 278.0`
So, % of water of crystallisation =`(126 )/(278 ) xx 100 =45.32`
% of iron `= (56)/(278 )xx 100 = 20.14`
% of sulphur `= (32)/(278 ) xx 100 = 11.51`
% of oxygen =` (64 )/(278 )xx 100 = 23.02`
(Oxygen present in water molecules is not taken into account.)