A 0.866-g sample of hydrated copper(ii) sulfate was heated carefully until it had changed completely to anhydrous copper(ii) sulfate () with a mass of 0.554 g. determine the value of . [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.] (express your answer as an integer.)

Respuesta :

CuSO4 * nH2O ------>   CuSO4 + nH2O
0.866 g                         0.554 g     0.312 g

1) m(H2O) = m(
CuSO4 * nH2O) - m(CuSO4) = 0.866 g - 0.554 g = 0.312 g
2) M(H2O) = 2*1.0 + 16.0 = 18.0 g/mol

 0.312 g * 1 mol/18.0 g = 0.0173  mol H2O

3) M(CuSO4) = 63.5 + 32.1 + 4*16.0 = 159.6 g/mol

   0.554 g * 1 mol/159.6 g≈0.00347 mol CuSO4

4) 0.00347 mol CuSO4 : 0.0173  mol H2O ≈ 1 mol CuSO4 : 5 mol H2O

CuSO4 *5H2O

5 molecules of water per 1 CuSO4*5H2O