Question

Calculate the percentage loss of mass of hydrated copper [II] sulphate [CuSO₄.5H₂O] when it is completely dehydrated.

\[\ce{[CuSO4.5H2O -> CuSO4 + 5H2O]}\]

[At. wts. are Cu = 64, S = 32, O = 16, H = 1]

Answer 1

Mol. wt. of CuSO₄.5H₂O = At. wt. of Cu + At. wt. of S + 4 × At. wt. of O + 5 × [2 × At. wt.of H + At. wt. of O]

= 64 + 32 + 4 × 16 + 5 × [2 × 1 + 16]

= 64 + 32 + 64 + 5 × 18

= 160 + 90

= 250 amu

Mass of water of crystallization lost (5H₂O) = 5 × 18 = 90 amu

Percentage loss of mass = `90/250 xx 100` = 36.00%