Question

A student performed the Separation by Fractional Crystallization experiment described with the following results: Initial mass of sample mixture: 9.58 g, mass of salicylic acid recovered, 4.35 g; mass of CuSO4•5H2O recovered, 4.68 g. The recovered CuSO4•5H2O was then heated for 20 minutes at 120 °C. The mass of the new hydrate was 3.83 g. (The molar mass of H2O = 18.015 g/mol, and the molar mass of CuSO4= 159.61 g/mol) Calculate the mass of water in the new hydrate (CuSO4•XH2O).

Answer 1

mass of CuSO4•5H2O recovered = 4.68 g

mass of water removed = 4.68-3.83 = 0.85 g

no of mol of water removed = 0.85/18.015 = 0.0472 mol

no of mol of CuSO4•5H2O recovered = 4.68/(159.61+18.015*5) = 0.0187 mol

no of mol of water present in CuSO4•5H2O recovered = 0.0187*5 = 0.0935 mol

no of mol of water present in new hydrate (CuSO4•XH2O) = 0.0935-0.0472 = 0.0463 mol

mass of water present in new hydrate (CuSO4•XH2O) = 0.0463*18.015

                                            = 0.8341 g