During " S " estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is _________ %.
(Given molar mass in g mol⁻¹ of Ba: 137, S: 32, O: 16)

Solution

Chemistry ma’am · 3 steps

Every sulphur atom ends up in one BaSO₄, so find the sulphur share of the precipitate first.

1. Molar mass. BaSO₄ has molar mass 137 + 32 + 4 × 16 = 233 g mol⁻¹. Of this, 32 parts are sulphur.

   Mass of S = 32/233 × 466 mg = 64 mg

2. Percentage. All 64 mg of sulphur came from the 160 mg sample, so % S = 64/160 × 100.
3. So. The compound contains 40 % sulphur.

The common slip: 0.4. Stopping at the fraction 64/160 gives 0.4, not 40. Multiply by 100 to get a percentage.

The idea to keep. In Carius sulphur estimation, % S = 32/233 × (mass of BaSO₄)/(mass of compound) × 100. One mole of BaSO₄ always carries one mole of S.

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