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 percentage of sulfur in an organic compound can be estimated using the Carius method. In this method, the sulfur present in the organic compound is converted to barium sulfate (\(BaSO_4\)).

1. Molar Mass Calculations:
The molar mass of \(BaSO_4\) is calculated as:
\(137 + 32 + 4(16) = 137 + 32 + 64 = 233\) g/mol.
The molar mass of S is 32 g/mol.

2. Stoichiometric Relationship:
1 mole of \(BaSO_4\) contains 1 mole of S.
So, 233 g of \(BaSO_4\) contains 32 g of S.

3. Experimental Data:
Given that 160 mg of organic compound gives 466 mg of \(BaSO_4\).
The amount of sulfur in 466 mg of \(BaSO_4\) is:
\[ \frac{32}{233} \times 466 \text{ mg} = 32 \times 2 = 64 \text{ mg} \]
The mass of sulfur in the organic compound is 64 mg.

4. Percentage Calculation:
The percentage of sulfur in the organic compound is:
\[ \frac{\text{mass of sulfur}}{\text{mass of organic compound}} \times 100 = \frac{64 \text{ mg}}{160 \text{ mg}} \times 100 = \frac{64}{160} \times 100 = \frac{2}{5} \times 100 = 40 \]

Final Answer:
The percentage of sulfur in the given compound is 40%.

The final answer is $40$.