Question

During "S" estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is               %.
(Given molar mass in g mol\(^{-1}\) of Ba: 137, S: 32, O: 16)

Hint:

In stoichiometry problems, always ensure to use the correct molar mass to calculate moles and then find the percentage of the element based on its molecular composition.

Answer 1

The Carius method estimates the sulfur percentage in organic compounds by converting sulfur to barium sulfate (\(BaSO_4\)).

1. Molar Mass Calculations:
The molar mass of \(BaSO_4\) is calculated as: \(137 + 32 + 4(16) = 233\) g/mol. The molar mass of S is 32 g/mol.

2. Stoichiometric Relationship:
One mole of \(BaSO_4\) contains one mole of S, meaning 233 g of \(BaSO_4\) contains 32 g of S.

3. Experimental Data:
Given 160 mg of organic compound yields 466 mg of \(BaSO_4\), the amount of sulfur in 466 mg of \(BaSO_4\) is: \[ \frac{32}{233} \times 466 \text{ mg} = 64 \text{ mg} \]. Thus, the organic compound contains 64 mg of sulfur.

4. Percentage Calculation:
The percentage of sulfur in the organic compound is calculated as: \[ \frac{64 \text{ mg}}{160 \text{ mg}} \times 100 = 40\% \]

Final Answer:
The sulfur content in the given organic compound is 40%.

The final answer is $40$.