Question

In the estimation of sulphur by Carius method. x g of an organic compound gave 0.233 g of \(BaSO_{4}\) If the percentage of sulphur in it is 8.89%, the value of x is

• A. 0.12
• B. 0.24
• C. 0.36
• D. 0.48

Hint:

Always ensure you use the molar masses provided within the specific problem question, as they can sometimes vary slightly between exam papers.

Answer 1

The Correct Option is C

Concept: The Carius method for the quantitative estimation of sulfur involves oxidizing the sulfur in an organic compound into sulfuric acid, which is then precipitated as barium sulfate (\(BaSO_4\)). The percentage of sulfur is calculated based on the stoichiometric weight of sulfur in \(BaSO_4\).

Step 1: Define the stoichiometric relationship.
The molar mass of \(BaSO_4\) is \(137 (Ba) + 32 (S) + 4 \times 16 (O) = 233 \text{ g/mol}\). The mass of sulfur in one mole of \(BaSO_4\) is 32 g. The percentage formula is: \[ \% S = \frac{32}{233} \times \frac{\text{mass of } BaSO_4}{\text{mass of compound } (x)} \times 100 \]

Step 2: Substitute the given experimental values.
We are given \(\% S = 8.89\%\) and mass of \(BaSO_4 = 0.233 \text{ g}\). Substituting these: \[ 8.89 = \frac{32}{233} \times \frac{0.233}{x} \times 100 \]

Step 3: Perform the algebraic calculation.
Note that \(\frac{0.233}{233} = 0.001\). The equation becomes: \[ 8.89 = \frac{32 \times 0.001 \times 100}{x} = \frac{3.2}{x} \] Solving for \(x\): \[ x = \frac{3.2}{8.89} \approx 0.36 \text{ g} \]