Q. 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, BaSO₄: 233)

Correct Answer: 21.05%

Explanation

In sulphur estimation, the sulphur present in the organic compound is completely converted into barium sulphate. Therefore, the mass of sulphur present can be calculated from the mass of barium sulphate obtained.

Moles of BaSO₄ formed =

0.4813 / 233 = 0.002065 mol

From the formula of BaSO₄, one mole of BaSO₄ contains one mole of sulphur.

Moles of sulphur = 0.002065 mol

Mass of sulphur =

0.002065 × 32 = 0.0661 g

Percentage of sulphur in the compound =

(0.0661 / 0.314) × 100 = 21.05%

Hence, the percentage of sulphur present in the compound is 21.05%.

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