3.05 g of hydrated copper sulphate produce 1.94 g of anhydrous copper sulphate. Assuming complete removal of water of crystallization. Determine the formula of the hydrated copper sulphate.

Answer 1

First we need to find the mass of water removed:

Mass of water = Mass of hydrated copper sulphate - Mass of anhydrous copper sulphate
Mass of water = 3.05 g - 1.94 g = 1.11 g

Now we need to find the moles of each component:

Moles of anhydrous copper sulphate (CuSO4) = mass / molar mass
Moles of CuSO4 = 1.94 g / (63.6 + 32.1 + 16*4) g/mol
Moles of CuSO4 = 1.94 g / 159.6 g/mol
Moles of CuSO4 ≈ 0.0122 mol

Moles of water (H2O) = mass / molar mass
Moles of H2O = 1.11 g / (1*2 + 16) g/mol
Moles of H2O = 1.11 g / 18 g/mol
Moles of H2O ≈ 0.0617 mol

To find the ratio of moles of water to anhydrous copper sulphate, we divide the moles of water by the moles of CuSO4:

Mole ratio = Moles of H2O / Moles of CuSO4
Mole ratio = 0.0617 / 0.0122
Mole ratio ≈ 5.05

Since the mole ratio is close to 5, we can assume that there are 5 moles of water for each mole of anhydrous copper sulphate. Therefore, the formula of hydrated copper sulphate is:

CuSO4 * 5H2O