Final answer:
The formula of the hydrated copper (II) sulphate is calculated by finding the moles of anhydrous copper (II) sulphate and the moles of water. The moles of water (0.050 mol) divided by the moles of copper (II) sulphate (0.010 mol) gives a ratio of 5, indicating five moles of water for every mole of copper (II) sulphate, which is CuSO₄ · 5H₂O.
Step-by-step explanation:
To determine the formula of the hydrated copper (II) sulphate, we first find the mass of the water that was lost during heating. The initial mass of the hydrated copper (II) sulphate was 2.5g, and the mass after heating is 1.6g. Therefore, the mass of the water is 2.5g - 1.6g = 0.9g.
The molar mass of anhydrous copper (II) sulphate (CuSO₄) is approximately 159.6g/mol. Since we have 1.6g of anhydrous copper (II) sulphate, the moles of CuSO₄ are 1.6g / 159.6g/mol ≈ 0.010 mol. The molar mass of water (H₂O) is approximately 18.02g/mol. Therefore, the moles of water are 0.9g / 18.02g/mol ≈ 0.050 mol.
The ratio of moles of water to moles of copper (II) sulphate is 0.050 mol / 0.010 mol = 5. Since the ratio is 5:1, this means that there are five moles of water for every mole of copper (II) sulphate.
Thus, the formula of the hydrated copper (II) sulphate is CuSO₄ · 5H₂O, which means the correct answer is not listed in the multiple choice options given.