In sulphur estimation, 0.471 g of an organic compound gave 1.4439 g of barium sulphate. The percentage of sulphur in the compound is ____________ (Nearest Integer)
(Given : Atomic mass Ba: 137 u, S: 32 u, O: 16 u)

Solution

Chemistry ma’am · 4 steps

All the sulphur ends up in BaSO₄, so find what fraction of that precipitate is sulphur.

1. Where sulphur goes. In the Carius method the compound is heated with fuming HNO₃. Its sulphur becomes H₂SO₄, which is then precipitated as BaSO₄. One mole of BaSO₄ carries one mole of S.
2. Molar mass. BaSO₄ = 137 + 32 + 4 × 16 = 233 g mol⁻¹. So every 233 g of precipitate holds 32 g of sulphur.

   % S = 32/233 × m_(BaSO₄)/m_(compound) × 100

3. Put in numbers. Mass of S = 32/233 × 1.4439 = 0.1983 g. Then % S = 0.1983/0.471 × 100 = 42.1.
4. So. The compound is about 42.1% sulphur, which rounds to 42.

The idea to keep. In gravimetric estimation, multiply the precipitate mass by (atomic mass of the element ÷ molar mass of the precipitate), then divide by the sample mass. The same pattern works for halogens with AgX.

Answer from the NTA final answer key. Solution by PrepWiser.