A 1.50-g sample of hydrated copper (ii) sulfatewas heated carefully until it had changed completely to anhydrous copper (ii) sulfate () with a mass of 0.957 g. determine the value of x. [this number is called the number of waters of hydration of copper (ii) sulfate. it specifies the number of water molecules per formula unit of in the hydrated crystal.]

CuSO₄·xH₂O → CuSO₄ + xH₂O

M (CuSO₄) = 159.61 g/mol

M (H₂O) = 18.02 g/mol

m (CuSO₄·xH₂O) = 1.50 g

m (CuSO₄) = 0.957 g

m (CuSO₄·xH₂O) / M (CuSO₄·xH₂O) = m (CuSO₄) / M (CuSO₄)

M (CuSO₄·xH₂O) = M (CuSO₄) + xM (H₂O)

m (CuSO₄·xH₂O) / {M (CuSO₄) + xM (H₂O) }=m (CuSO₄) / M (CuSO₄)

M (CuSO₄) + xM (H₂O) = m (CuSO₄·xH₂O) M (CuSO₄) / m (CuSO₄)

xM (H₂O) = m (CuSO₄·xH₂O) M (CuSO₄) / m (CuSO₄) - M (CuSO₄)

x=M (CuSO₄) / M (H₂O) {m (CuSO₄·xH₂O) / m (CuSO₄) - 1}

x=159.61/18.02*{1.50/0.957-1}=5.0

x=5

CuSO₄·5H₂O