Answer to Question #122174 in General Chemistry for Michele Symington

Question

4.7g of ahydrated sample of copper 2sphate cuso4.xh20 was heated to a constant mass the mass reduced by 1.71 g what is the value of x

Answer 1

CuSO₄*xH₂O (s) + heat --> CuSO₄ (s)+ xH₂O (g)

The mass of water that has separated is given and equals 1.71 g.

Therefore, the mass of remaining pure CuSO₄ is 4.7 - 1.71 = 2.99 g.

n(H₂O) = 1.71g / 18g/mol = 0.095 mol

n(CuSO₄) = 2.99g / 160g/mol = 0.019 mol

The H₂O : CuSO₄ mole ratio is 0.095 : 0.019 = 5 : 1.

So the formula is CuSO₄*5H₂O, and x = 5.

Answer: x = 5