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Chemistry Question 72 – JEE-MAIN 2026

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In an estimation of sulphur by Carius method 0.2 g of the substance gave 0.6 g of BaSO4. The percentage of sulphur in the substance is _______ %. (Given molar mass in g mol−1 S:32, BaSO4:231)

The Carius method for sulphur estimation converts all sulphur in the organic compound into barium sulfate (BaSO4).

Step 1: Determine the mass of sulphur✦ Active

In the Carius method, all sulphur in the substance is converted to barium sulfate (BaSO4). From the mass of BaSO4 formed, we can find the mass of sulphur. The molar mass of sulphur (S) is 32 g/mol and the molar mass of BaSO4 is 231 g/mol. Since one mole of BaSO4 contains one mole of S, the mass of S in 0.6 g of BaSO4 is:

Mass of S=Molar mass of SMolar mass of BaSO4×Mass of BaSO4
Mass of S=32231×0.6 g≈0.08311688 g
Step 2: Calculate the percentage of sulphur○ Expand

The percentage of sulphur in the substance is calculated using the formula:

Percentage of S=Mass of SMass of substance×100

Given the mass of the substance is 0.2 g:

Percentage of S=0.08311688 g0.2 g×100≈41.55844%
Step 3: Round the result○ Expand

Rounding the percentage to two decimal places, we get:

Percentage of S≈41.56%
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