Purification and Characterisation of Organic Compounds

In sulphur estimation,

0.471\mathrm{~g}

of an organic compound gave

1.4439\mathrm{~g}

of barium sulphate. The percentage of sulphur in the compound is

\underline{\quad\quad\quad}

(Nearest Integer) (Given: Atomic mass

\mathrm{Ba}: 137\mathrm{u}, \mathrm{S}: 32\mathrm{u}, \mathrm{O}: 16\mathrm{u}

)

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Step-by-step solutionView

Correct answer

The percentage of sulphur in the organic compound is 42%.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Mass of organic compound = $0.471\text{ g}$ , Mass of $\mathrm{BaSO_4}$ formed = $1.4439\text{ g}$ (or $\approx 1.44\text{ g}$ ), Atomic masses: $\mathrm{Ba} = 137\text{ u}, \mathrm{S} = 32\text{ u}, \mathrm{O} = 16\text{ u}$ .

02find

Percentage of sulphur in the given organic compound to the nearest integer.

03visualise

All sulphur from the organic compound is converted stoichiometrically into $\mathrm{BaSO_4}$ . Since 1 mole of $\mathrm{BaSO_4}$ contains 1 mole of sulphur atoms, the moles of sulphur equal the moles of $\mathrm{BaSO_4}$ .

04strategise

Molar mass of $\mathrm{BaSO_4} = 137 + 32 + 4(16) = 233\text{ g/mol}$ . Using the Carius formula: $\%\mathrm{S} = \frac{32}{233} \times \frac{m_{\mathrm{BaSO_4}}}{m_{\text{compound}}} \times 100$ .

05execute

Substitute values: $\%\mathrm{S} = \frac{32}{233} \times \frac{1.4439}{0.471} \times 100 = 42.096\% \approx 42\%$ (or using $1.44\text{ g}$ : $\frac{32 \times 1.44}{233 \times 0.471} \times 100 = 41.98\% \approx 42\%$ ).

✓verify

Rounding $42.096\%$ or $41.98\%$ to the nearest integer unambiguously yields 42.

Hints that build this answer step by step

What is the molar mass of barium sulphate ( $\mathrm{BaSO_4}$ )?
$233\mathrm{~g\cdot mol^{-1}}$
Which formula gives the percentage of sulphur from the mass of $\mathrm{BaSO_4}$ obtained?
$\%\mathrm{S} = \dfrac{32}{233} \times \dfrac{m(\mathrm{BaSO_4})}{m(\text{compound})} \times 100$
Substitute $m(\mathrm{BaSO_4}) = 1.4439\mathrm{~g}$ and $m(\text{compound}) = 0.471\mathrm{~g}$ to calculate the percentage of sulphur to the nearest integer.
42%

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Does using 1.44 g instead of 1.4439 g affect the final integer?

No, 1.44 g gives 41.98% and 1.4439 g gives 42.10%, both round directly to 42.

Answer

The percentage of sulphur in the organic compound is 42%.

Why each option works or fails

Step-by-step solution

given: Mass of organic compound = $0.471\text{ g}$ , Mass of $\mathrm{BaSO_4}$ formed = $1.4439\text{ g}$ (or $\approx 1.44\text{ g}$ ), Atomic masses: $\mathrm{Ba} = 137\text{ u}, \mathrm{S} = 32\text{ u}, \mathrm{O} = 16\text{ u}$ .
find: Percentage of sulphur in the given organic compound to the nearest integer.
visualise: All sulphur from the organic compound is converted stoichiometrically into $\mathrm{BaSO_4}$ . Since 1 mole of $\mathrm{BaSO_4}$ contains 1 mole of sulphur atoms, the moles of sulphur equal the moles of $\mathrm{BaSO_4}$ .
strategise: Molar mass of $\mathrm{BaSO_4} = 137 + 32 + 4(16) = 233\text{ g/mol}$ . Using the Carius formula: $\%\mathrm{S} = \frac{32}{233} \times \frac{m_{\mathrm{BaSO_4}}}{m_{\text{compound}}} \times 100$ .
execute: Substitute values: $\%\mathrm{S} = \frac{32}{233} \times \frac{1.4439}{0.471} \times 100 = 42.096\% \approx 42\%$ (or using $1.44\text{ g}$ : $\frac{32 \times 1.44}{233 \times 0.471} \times 100 = 41.98\% \approx 42\%$ ).
verify: Rounding $42.096\%$ or $41.98\%$ to the nearest integer unambiguously yields 42.