Estimation Of Sulphur (carius Method)
In the Carius method for sulphur, 0.20 g of an organic compound yielded 0.466 g of barium sulphate; what is the percentage of sulphur in the compound (Ba = 137, S = 32, O = 16)?
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About This Question
- Subject
- chemistry
- Chapter
- purification and characterisation of organic compounds
- Topic
- estimation of sulphur (carius method)
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
32%
In the Carius method for sulphur, the organic compound is heated with fuming nitric acid, which oxidises the sulphur completely to sulphuric acid. The sulphate so formed is then precipitated as barium sulphate by adding barium chloride; this precipitate is filtered, washed, dried, and weighed. The percentage of sulphur is computed from %S = (atomic mass of S / molar mass of BaSO4) x (mass of BaSO4 / mass of compound) x 100. The molar mass of BaSO4 is 137 + 32 + 64 = 233 g per mol. Substituting: %S = (32 / 233) x (0.466 / 0.20) x 100 = 0.1373 x 2.33 x 100 = 32.0%. The value 16% halves the correct result, as if only one oxygen-equivalent of sulphate were counted. The value 64% doubles the answer by misusing the oxygen mass instead of sulphur. The value 23.3% mistakenly reports the mass ratio (0.466/0.20) scaled without the sulphur fraction. This calculation follows the NCERT Carius sulphur estimation. Plausibility check: barium sulphate is much heavier than the sulphur it contains, so obtaining 0.466 g of precipitate from 0.20 g of compound consistently yields a sulphur content near 32%.
This hard difficulty chemistry question is from the chapter purification and characterisation of organic compounds, covering the topic of estimation of sulphur (carius method). It appeared in the 2025 exam.
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