Q18 / 180 · COMEDK 2024 (Evening Shift) · Chemistry
In the estimation of element $\mathrm{X}$ in an organic compound, $0.8 \mathrm{~g}$ of the compound containing $\mathrm{X}$ was heated with fuming $\mathrm{HNO}_3$ and the cooled product was treated with barium chloride. The mass of barium sulphate precipitated was $1.2 \mathrm{~g}$. What is the percentage of element $\mathrm{X}$ and what is the formula of the violet-coloured compound formed when Lassaigne's extract of the organic compound is treated with sodium nitroprusside?
- A$40.27 \quad \& \quad \mathrm{Na}\left[\mathrm{Fe}(\mathrm{CN})_3 \mathrm{NOS}\right]$
- Correct answerB$20.6 \quad \& \quad \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]$
- C$9.156 \quad \& \quad \mathrm{Na}_3\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]$
- D$30.24 \quad \& \quad \mathrm{Na}_3\left[\mathrm{Fe}(\mathrm{CN})_4 \mathrm{NOS}\right]$
Solution
The element X is sulphur because the treatment with fuming $\mathrm{HNO}_3$ followed by $\mathrm{BaCl}_2$ results in the precipitation of $\mathrm{BaSO}_4$.
The molar mass of $\mathrm{BaSO}_4$ is $137 + 32 + 4 \times 16 = 233 \text{ g/mol}$.
The mass of sulphur in $1.2 \text{ g}$ of $\mathrm{BaSO}_4$ is calculated as:
$\text{Mass of S} = \dfrac{32}{233} \times 1.2 \approx 0.1648 \text{ g}$.
The percentage of sulphur in the organic compound is:
$\% \text{S} = \dfrac{\text{Mass of S}}{\text{Mass of compound}} \times 100 = \dfrac{0.1648}{0.8} \times 100 = 20.6 \%$.
When Lassaigne's extract containing $\mathrm{Na}_2\mathrm{S}$ is treated with sodium nitroprusside, $\mathrm{Na}_2[\mathrm{Fe}(\mathrm{CN})_5\mathrm{NO}]$, a violet-coloured complex is formed. The reaction is:
$\mathrm{Na}_2[\mathrm{Fe}(\mathrm{CN})_5\mathrm{NO}] + \mathrm{Na}_2\mathrm{S} \rightarrow \mathrm{Na}_4[\mathrm{Fe}(\mathrm{CN})_5\mathrm{NOS}]$.
Thus, the percentage is $20.6$ and the formula is $\mathrm{Na}_4[\mathrm{Fe}(\mathrm{CN})_5\mathrm{NOS}]$.
Answer: $20.6 \quad \& \quad \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]$
The molar mass of $\mathrm{BaSO}_4$ is $137 + 32 + 4 \times 16 = 233 \text{ g/mol}$.
The mass of sulphur in $1.2 \text{ g}$ of $\mathrm{BaSO}_4$ is calculated as:
$\text{Mass of S} = \dfrac{32}{233} \times 1.2 \approx 0.1648 \text{ g}$.
The percentage of sulphur in the organic compound is:
$\% \text{S} = \dfrac{\text{Mass of S}}{\text{Mass of compound}} \times 100 = \dfrac{0.1648}{0.8} \times 100 = 20.6 \%$.
When Lassaigne's extract containing $\mathrm{Na}_2\mathrm{S}$ is treated with sodium nitroprusside, $\mathrm{Na}_2[\mathrm{Fe}(\mathrm{CN})_5\mathrm{NO}]$, a violet-coloured complex is formed. The reaction is:
$\mathrm{Na}_2[\mathrm{Fe}(\mathrm{CN})_5\mathrm{NO}] + \mathrm{Na}_2\mathrm{S} \rightarrow \mathrm{Na}_4[\mathrm{Fe}(\mathrm{CN})_5\mathrm{NOS}]$.
Thus, the percentage is $20.6$ and the formula is $\mathrm{Na}_4[\mathrm{Fe}(\mathrm{CN})_5\mathrm{NOS}]$.
Answer: $20.6 \quad \& \quad \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]$
Hence, the correct answer is option: 'B'.