Question

In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur (molar mass 32 g mol⁻¹). Molar mass of barium sulphate is 233 g mol⁻¹.

• A. 16.48\%
• B. 10.30\%
• C. 21.97\%
• D. 4.55\%

Hint:

For quantitative analysis problems (like Carius, Dumas, Kjeldahl methods), memorizing the final formula is key to saving time. The core principle is always based on stoichiometry: relating the mass of the final product (like BaSO₄ or AgX) to the mass of the element of interest in the original sample.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks to calculate the percentage of sulphur in an organic compound using the data from the Carius method of estimation. In this method, the sulphur present in the compound is converted into barium sulphate (BaSO₄), which is then weighed.
Step 2: Key Formula or Approach:
The percentage of Sulphur (\%S) in the organic compound can be calculated using the following formula:
\[ \% \text{S} = \frac{\text{Atomic mass of S}}{\text{Molar mass of BaSO}_4} \times \frac{\text{Mass of BaSO}_4 \text{ formed}}{\text{Mass of organic compound taken}} \times 100 \] Step 3: Detailed Explanation:
Given data:
\begin{itemize} \item Mass of the organic compound = 0.75 g
\item Mass of barium sulphate (BaSO₄) formed = 1.2 g
\item Atomic mass of Sulphur (S) = 32 g mol⁻¹
\item Molar mass of barium sulphate (BaSO₄) = 233 g mol⁻¹
\end{itemize} Calculation:
Substitute the given values into the formula:
\[ \% \text{S} = \frac{32}{233} \times \frac{1.2}{0.75} \times 100 \] First, calculate the ratios:
\[ \frac{32}{233} \approx 0.13734 \] \[ \frac{1.2}{0.75} = \frac{120}{75} = \frac{24}{15} = \frac{8}{5} = 1.6 \] Now, multiply the values:
\[ \% \text{S} = 0.13734 \times 1.6 \times 100 \] \[ \% \text{S} = 0.219744 \times 100 \] \[ \% \text{S} = 21.9744\% \] Step 4: Final Answer:
The percentage of sulphur in the given organic compound is approximately 21.97\%. This corresponds to option (C).