JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Practical and Analytical Chemistry

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A student has been given 0.314 g of an organic compound and asked to estimate Sulphur. During the experiment, the student has obtained 0.4813 g of barium sulphate. The percentage of sulphur present in the compound is ______.
(Given Molar mass in

g mol^{- 1}

S: 32,

BaSO_{4}

: 233)

(A) $42.10%$
(B) $63.15%$
(C) $21.05%$ ✓ Correct
(D) $48.24%$

Answer: (C)

Solution

In the Carius estimation the sulphur of the sample is precipitated as

BaS O_{4}

, so

% S = \frac{32}{233} \times \frac{mass of BaS O _{4}}{mass of compound} \times 100 = \frac{32}{233} \times \frac{0.4813}{0.314} \times 100

. With

\frac{32}{233} = 0.1373

and

\frac{0.4813}{0.314} = 1.533

, this is

0.1373 \times 1.533 \times 100 = 21.05

. The sulphur content of the compound is

21.05%

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Practical and Analytical Chemistry

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