A. Volatilization method

The element or compound to be determined is converted to a volatile form; its mass is found either by collecting the volatile substance and weighing it or by the loss in mass of the original sample.

Worked example (water of crystallization). 3.05 g of hydrated barium chloride BaCl₂·xH₂O is heated strongly to constant mass 2.60 g (Ba = 137, Cl = 35.5, H = 1, O = 16).

• Mass of water = 3.05 − 2.60 = 0.45 g; % water = 0.45/3.05 × 100 ≈ 14.75 %.
• Moles BaCl₂ = 2.60/208 = 0.0125 mol; moles H₂O = 0.45/18 = 0.025 mol.
• x = 0.025/0.0125 = 2 → BaCl₂·2H₂O.

Worked example (copper sulphate). 2.495 g of hydrated copper sulphate CuSO₄·xH₂O lose water on heating and leave 1.595 g of anhydrous salt (Cu = 63.5, S = 32, O = 16, H = 1). Mass of water = 0.900 g. Moles CuSO₄ = 1.595/159.5 = 0.0100 mol; moles H₂O = 0.900/18 = 0.0500 mol; x = 5 → CuSO₄·5H₂O. The percentage of water of crystallization = 0.9/2.495 × 100 = 36.1 %.

Worked example (loss of a gas). 5.0 g of an impure limestone sample lose 1.76 g when strongly heated: CaCO₃ → CaO + CO₂. Moles of CO₂ = 1.76/44 = 0.04 mol = moles of CaCO₃ → mass 0.04 × 100 = 4.0 g; purity = 4.0/5.0 × 100 = 80 %.

Percentage of water in a hydrate. For Na₂SO₄·10H₂O (M = 142 + 180 = 322 g/mol): % water = 180/322 × 100 = 55.9 %.