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%
