In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur.
Q. 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.
Correct Answer: D (21.97%)

Step-by-Step Explanation

Step 1: Note the given data
Mass of organic compound (\(w\)) = 0.75 g
Mass of barium sulphate (\(w_1\)) = 1.2 g
Molar mass of S = 32 g/mol
Molar mass of BaSO₄ = 233 g/mol
Step 2: Understand the relationship
In BaSO₄, there is 1 atom of Sulphur.
So, 233 g of BaSO₄ contains 32 g of Sulphur.
Step 3: Calculate the mass of Sulphur in 1.2 g of BaSO₄
Mass of Sulphur = \( \left( \frac{32}{233} \right) \times 1.2 \) g
Step 4: Calculate the percentage of Sulphur
Formula: \( \text{\% of S} = \frac{32 \times \text{Mass of BaSO}_4 \times 100}{233 \times \text{Mass of organic compound}} \)

\( \text{\% of S} = \frac{32 \times 1.2 \times 100}{233 \times 0.75} \)

\( \text{\% of S} = \frac{3840}{174.75} \)

\( \text{\% of S} \approx 21.974\% \)
Step 5: Final Result
The percentage of sulphur is 21.97%.

Detailed Related Theory

1. What is the Carius Method?
The Carius method is a classical analytical technique used for the quantitative estimation of halogens (Chlorine, Bromine, Iodine) and sulphur in organic compounds. It involves heating a known mass of the organic compound with fuming nitric acid (\(HNO_3\)) in a sealed hard glass tube (Carius tube). If sulphur is present, it is oxidized to sulphuric acid (\(H_2SO_4\)).

2. Chemistry of Sulphur Estimation
When the organic compound containing sulphur is heated with fuming nitric acid, the carbon and hydrogen are oxidized to \(CO_2\) and \(H_2O\) respectively, while sulphur is converted to sulphuric acid. \[ \text{S} + \text{H}_2\text{O} + 3\text{O} \xrightarrow{\text{fuming } HNO_3} \text{H}_2\text{SO}_4 \] To estimate the amount, an excess of barium chloride (\(BaCl_2\)) solution is added. This results in the precipitation of barium sulphate (\(BaSO_4\)): \[ \text{H}_2\text{SO}_4 + \text{BaCl}_2 \rightarrow \text{BaSO}_4 \downarrow + 2\text{HCl} \] The precipitate is filtered, washed, dried, and weighed carefully. Since we know the molar mass of \(BaSO_4\) and the atomic mass of Sulphur, we can use stoichiometry to find the original amount of sulphur.

3. Why is BaSO₄ chosen?
Barium sulphate is used because it is highly insoluble in water and dilute acids, making it easy to recover quantitatively without significant loss during the washing process. Its high molar mass (233) compared to Sulphur (32) also reduces the percentage error in weighing.

4. Estimation of Halogens (Brief)
For halogens, the process is similar, but silver nitrate (\(AgNO_3\)) is added instead of \(BaCl_2\). This forms silver halides (\(AgX\)): - AgCl (White ppt) - AgBr (Pale yellow ppt) - AgI (Yellow ppt) The formula for halogen percentage follows the same logic: \( \text{\% X} = \frac{\text{Atomic Mass of X} \times \text{Mass of AgX} \times 100}{\text{Molar Mass of AgX} \times \text{Mass of Compound}} \).

5. Limitations of Carius Method
- The method is time-consuming as it requires several hours of heating. - It is not suitable for compounds containing both sulphur and phosphorus simultaneously without further separation, as phosphorus also forms precipitates. - Handling a sealed glass tube under high pressure (Carius tube) requires extreme safety precautions to prevent explosions.

6. Calculation Tips for Competitive Exams
- Stoichiometric Factor: For Sulphur, always remember the ratio \( \frac{32}{233} \approx 0.137 \). Multiplying the mass of \(BaSO_4\) by 0.137 gives the mass of sulphur quickly. - Unit Consistency: Ensure all masses are in the same units (usually grams). - Significant Figures: In many entrance exams, the options are very close. Always carry out calculations up to at least 2 or 3 decimal places before rounding off at the final step.

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Related Covered Topics

Carius method quantitative analysis organic chemistry basics percentage of sulphur barium sulphate precipitation analytical chemistry stoichiometry molar mass calculations fuming nitric acid oxidation chemical estimation JEE chemistry prep NEET chemistry prep precipitate analysis halogens estimation stoichiometric factors

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