According to its formula FeSO4.7H2O
we can get the percent % by the mass of H2O from this formula
%mass of H2O = (mass of water H2O/ mass of the hydrate)x100
when the mass of water = molar mass x 7 = 18 x 7 = 126
and the mass of hydrate (feSO4) = molar mass = 278
So by substitution:
%mass of H2O = (126/278) x 100 = 45%