Question

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.]

Answer 1

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