Estimation Of Sulphur (carius Method)
A 0.30 g sample of a sulphur-containing organic compound was oxidised by the Carius method and the sulphate produced was precipitated completely as 0.60 g of barium sulphate. What is the percentage of sulphur?
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About This Question
- Subject
- chemistry
- Chapter
- principles related to practical chemistry
- Topic
- estimation of sulphur (carius method)
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
27.5%
In the Carius estimation of sulphur, the compound is heated with fuming nitric acid so that all the sulphur is oxidised to sulphate, which is then precipitated as barium sulphate by adding barium chloride. Barium sulphate has a molar mass of 233 g mol^{-1} and contains 32 g of sulphur per mole. The mass of sulphur in the precipitate = (32 / 233) × 0.60 = 0.0824 g. Percentage of sulphur = (0.0824 / 0.30) × 100 = 27.5%. Option 13.7% is the fraction of sulphur within barium sulphate and neglects the sample mass entirely. Option 55.0% results from using half the correct sample mass, doubling the apparent percentage. Option 32.0% mistakes the atomic mass of sulphur for a percentage. The calculation applies the NCERT relation %S = (32 × mass BaSO_4)/(233 × mass of compound) × 100. A key requirement is that the oxidation by fuming nitric acid be complete, converting every sulphur atom to sulphate, since any sulphur escaping as a volatile gas would be lost from the precipitate and lower the apparent percentage. Plausibility check: a sulphur content of 27.5% is reasonable for a thiol or thioether of modest molar mass, and since the percentage is well below 100% the result is internally consistent.
This hard difficulty chemistry question is from the chapter principles related to practical chemistry, covering the topic of estimation of sulphur (carius method). It appeared in the 2025 exam.
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