How do you do this mole-to-mole ratio problem?

Question

The ratio of moles of water to one mole of a hydrate Copper(II) Sulfate can be determined experimentally by heating the hydrate to remove water.

a.The mass of Copper(II)Sulfate is 4.13 g. The mass of water is 2.66 g. Determine the number of water molecules associated with one formula unit of the hydrate/

Answer 1

CuSO4.x(H2O) ---> CuSO4 + x H2O

moles CuSO4 = 4.13 g CuSO4/ x g CuSO4/mole

moles H2O = 2.66 g H2O / 18 g H2O / mole

divide both by moles CuSO4 to get 1 mole CuSO4 and # moles of H2O / mole CuSO4

Plug and SOLVE; write as CuSO4.x (H2O)

Answer 2

The molar mass of CuSO4 is 159.7 g/mol.

The molar mass of H2O is 18.0 g/mol.

4.13 g CuSO4 x (1 mol / 159.7 g) = 0.0259 mol CuSO4

2.66 g H2O x (1 mol / 18.0 g) = 0.148 mol H2O

Ratio of water/CuSO4 is 0.148/0.0259 = 5.71 (approximately 6)

CuSO4-6H2O

The actual hydrated form of CuSO4 is closer to CuSO4-5H2O, but it is assumed the above data are experimental.