Question 29 of 34intermediate🔧 ApplyNumerical3 marks

In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.

Correct Answer

The percentage of sulphur in the given compound is 19.60%.

Exercise: EXERCISES | Q: 8.35
For More Understanding

Explanation

In the Carius method, sulphur is oxidised to sulphuric acid and precipitated as barium sulphate. The percentage of sulphur is calculated using the mass of the organic compound and the mass of the precipitated barium sulphate, based on the fact that 1 mol of BaSO₄ (233 g) contains 32 g of sulphur.

Solution Steps

  1. Given data: Mass of organic compound (mm) = 0.468 g, Mass of barium sulphate (m1m_1) = 0.668 g.

  2. Molar mass of BaSO₄: The molecular mass of BaSO₄ is 233 g. Therefore, 233 g of BaSO₄ contains 32 g of sulphur.

  3. Mass of sulphur: m1m_1 g of BaSO₄ contains 32×m1233\frac{32 \times m_1}{233} g of sulphur. Thus, 0.668 g of BaSO₄ contains 32×0.668233\frac{32 \times 0.668}{233} g of sulphur.

  4. Percentage of sulphur: Percentage of sulphur = 32×m1×100233×m=32×0.668×100233×0.468=19.60%\frac{32 \times m_1 \times 100}{233 \times m} = \frac{32 \times 0.668 \times 100}{233 \times 0.468} = 19.60\%.

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