Explanation:
Copper (II) sulfate is usually present as a hydrous state, which is of the form CuSO4 * nH2O, where n is a whole number.
Mass of sample (CuSO4 * nH2O)
= 152.00g - 128.10g = 23.90g.
Mass of water loss during heating
= 152.00g - 147.60g = 4.40g.
Molar mass of H2O = 18g/mol
Moles of H2O in sample
= 4.40g / (18g/mol) = 0.244mol.
Mass of anhydrous sample (CuSO4)
= 23.90g - 4.40g = 19.50g
Molar mass of CuSO4 = 159.61g/mol
Moles of CuSO4 in sample
= 19.50g / (159.61g/mol) = 0.122mol.
Since mole ratio of CuSO4 to H2O
= 0.122mol : 0.244mol = 1:2, n = 2.
Hence we have CuSO4 * 2H2O.
