Egypt Secondary 3 Chemistry, Week 9: Mass analysis: volatilisation and precipitation

Calculations in analysis: titration and mass analysis

Week 9 · Chapter 2 (D): titration calculations with MaVa/na = MbVb/nb, the percentage of a substance in a mixture from a titration, and mass (gravimetric) analysis by volatilisation and by precipitation (water of crystallisation, mass of an ion from a precipitate) · 3 lessons

Mass analysis: volatilisation and precipitation

Chemistry Sec 3 Term 1, Chapter 2: pp. 43–48

GoalUse mass analysis: find the water of crystallisation by volatilisation, and the amount of an ion from the mass of a precipitate.

New words

water of crystallisation · the water held in the crystals of a hydrated salt, such as 7H₂O in MgSO₄·7H₂Ovolatilisation · driving off a volatile part, here water, and weighing what is leftashless filter paper · paper that burns away without leaving ash, so it adds no mass

Explanation

In mass (gravimetric) analysis we separate the wanted substance and weigh it. There are two ways. In volatilisation, a hydrated salt is strongly heated to constant mass in a laboratory; the mass lost is the water of crystallisation. We find the moles of the dry salt and of the water, and divide to get x in the formula salt·xH₂O. In precipitation, the wanted ion is precipitated in a pure, sparingly soluble form of known formula, such as barium as BaSO₄ or chloride as AgCl. It is filtered on ashless paper, ignited in a crucible, and weighed. From its mass we find the moles of precipitate and then, by the formula or the equation, the mass of the ion or the compound in the sample. All heating, filtering and igniting here are paper only; the atomic masses are given in each task (Mg = 24, S = 32, O = 16, Na = 23, Ba = 137, Ag = 108, Cl = 35.5).

Worked examples

	Mass (g)	Moles
MgSO₄	1.20	0.0100
H₂O	1.26	0.0700
x	—	7

Volatilisation. 2.46 g of MgSO₄·xH₂O loses 1.26 g of water on strong heating (paper only). The dry salt is 2.46 − 1.26 = 1.20 g, and M(MgSO₄) = 24 + 32 + 64 = 120 g/mol, so n = 1.20 ÷ 120 = 0.0100 mol. Water: 1.26 ÷ 18 = 0.0700 mol. x = 0.0700 ÷ 0.0100 = 7, and the formula is MgSO₄·7H₂O. Percentage of water = 1.26 ÷ 2.46 × 100 = 51.2 %.

Precipitation of sulphate. A sample is treated with excess BaCl₂ and gives 1.165 g of BaSO₄ (M = 233 g/mol). n(BaSO₄) = 1.165 ÷ 233 = 0.00500 mol. If the sulphate came from Na₂SO₄ (M = 142 g/mol), the ratio is 1 : 1, so m(Na₂SO₄) = 0.00500 × 142 = 0.710 g.

Precipitation of chloride. A chloride gives 2.87 g of AgCl (M = 143.5 g/mol). n(AgCl) = 2.87 ÷ 143.5 = 0.0200 mol, which is also 0.0200 mol of Cl⁻, so m(Cl) = 0.0200 × 35.5 = 0.710 g. If all of it came from NaCl (M = 58.5 g/mol), m(NaCl) = 0.0200 × 58.5 = 1.17 g. Last, to make 0.466 g of BaSO₄ you need n = 0.466 ÷ 233 = 0.00200 mol of BaCl₂, which is 0.00200 × 208 = 0.416 g.

Class activity

Drying by weighing, with a grown-up. Weigh a clean damp cloth or sponge on a kitchen scale and write the mass. Leave it on a shelf for a day without heating it and weigh it again. The mass lost is the water, the kitchen cousin of volatilisation: mass before − mass after = mass of water. Then write the same steps for a paper problem on hydrated copper(II) sulphate in your notebook, using only numbers given to you. No oven, no hot plate, no crucible.

Practice

1.72 g of hydrated calcium sulphate, CaSO₄·xH₂O, loses 0.36 g of water on strong heating (paper only; Ca = 40, S = 32, O = 16). Find x.

A sample of K₂SO₄ gives 0.932 g of BaSO₄ (M = 233 g/mol) with excess BaCl₂. What mass of K₂SO₄ (M = 174 g/mol) was in the sample?

A solution of KCl gives 1.435 g of AgCl (M = 143.5 g/mol) with excess silver nitrate. What mass of KCl (M = 74.5 g/mol) was present?

What mass of BaCl₂ (M = 208 g/mol) is needed to make 0.699 g of BaSO₄ (M = 233 g/mol) from excess sodium sulphate?