Question

Calculate the total percentage of oxygen in FeSO4.7H2O.

( Fe=56, S=32, O=16, H=1 )​

Answer 1

The formula mass of ferrous sulphate

At.massofFe+At.,massofS+At.massofoxygen+7×Mol.massofH

2

O

56.0+32.0+4×16.0+7×18.0=278.0

So % of water of crystallisation =

278

126

×100=45.32

% of iron =

278

56

×100=20.14

% of sulphur =

278

32

×100=11.51

% of oxygen =

278

64

×100=23.02

(Oxygen present in water molecules is not taken into account.)

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