Question

In sulphur estimation, \(2.0 \times 10^{-3}\) mol of an organic compound (X) (molar mass 76 \(g \cdot mol^{-1}\)) gave 0.4813 g of barium sulphate (molar mass 233 \(g \cdot mol^{-1}\)). The percentage of sulphur in the compound (X) is _______ \(\times 10^{-1}\) % (Nearest integer)

Answer 1

Correct Answer: 435

Step 1: Understanding the Concept:
This problem uses the Carius method for the quantitative estimation of sulfur. All the sulfur present in the known mass of the organic compound is converted quantitatively into a precipitable sulfate ($BaSO_4$). By weighing the precipitate, we determine the exact mass of sulfur, which is then expressed as a percentage of the original compound's mass.
Step 2: Key Formula or Approach:
Moles of S = Moles of $BaSO_4 = \frac{\text{Mass of } BaSO_4}{\text{Molar mass of } BaSO_4}$
Mass of S = Moles of S $\times$ Atomic mass of S (32 g/mol)
Percentage of S = $\frac{\text{Mass of S}}{\text{Mass of compound X}} \times 100$
Step 3: Detailed Explanation:
First, find the actual mass of the organic compound (X) used in the experiment:
Mass of X = Moles of X $\times$ Molar mass of X
Mass of X = $(2.0 \times 10^{-3} \text{ mol}) \times 76 \text{ g/mol} = 0.152 \text{ g}$.
Next, find the mass of sulfur recovered from the $BaSO_4$ precipitate:
Moles of $BaSO_4$ produced = $\frac{0.4813 \text{ g}}{233 \text{ g/mol}} = 0.0020657 \text{ mol}$.
Since each mole of $BaSO_4$ contains exactly one mole of Sulfur atoms:
Moles of S = $0.0020657 \text{ mol}$.
Mass of S = $0.0020657 \text{ mol} \times 32 \text{ g/mol} = 0.0661 \text{ g}$.
Now, calculate the percentage of Sulfur in the compound:
\[ % \text{S} = \frac{0.0661 \text{ g}}{0.152 \text{ g}} \times 100 \] \[ % \text{S} = 0.434868 \dots \times 100 = 43.4868\dots % \] The question asks for the answer in the format $\alpha \times 10^{-1} %$.
$43.4868% = 434.868 \times 10^{-1} %$.
Rounding to the nearest integer, we get 435.
(Self-check alternative reasoning): Since $0.002$ moles of compound yielded roughly $0.002$ moles of S, there is exactly 1 Sulfur atom per molecule of X.
Theoretical percentage = $\frac{\text{Mass of 1 S atom}}{\text{Molar mass of X}} \times 100 = \frac{32}{76} \times 100 = 42.1%$.
The experimental yield data gives $43.5%$, which means the question specifically tests the calculation from the empirical $BaSO_4$ data rather than the theoretical molecular formula deduction. Always trust the empirical data given.
Step 4: Final Answer:
The value is 435.