Question

In the sulphur estimation, 0.20 g of a pure organic compound gave 0.40 g of barium sulphate. The percentage of sulphur in the compound is

Hint:

To find the percentage of an element in a compound, divide the mass of the element by the mass of the sample and multiply by 100.

Answer 1

Correct Answer: 27.5

Given that the molar mass of sulphur (S) is 32 g/mol and that of barium sulphate (BaSO\(_4\)) is 233 g/mol, the mass of sulphur in the compound can be calculated from the mass of barium sulphate produced: 
The moles of barium sulphate formed: \[ \text{Moles of BaSO}_4 = \frac{0.40 \, \text{g}}{233 \, \text{g/mol}} = 0.00172 \, \text{mol} \] 
The moles of sulphur in the compound are equal to the moles of BaSO\(_4\) because of the 1:1 stoichiometry of BaSO\(_4\) and sulphur. 
The mass of sulphur is: \[ \text{Mass of S} = 0.00172 \, \text{mol} \times 32 \, \text{g/mol} = 0.05504 \, \text{g} \] 
The percentage of sulphur in the compound is: \[ % \text{S} = \frac{0.05504 \, \text{g}}{0.20 \, \text{g}} \times 100 = 27.5% \] Thus, the correct percentage of sulphur is \(27.5\%.\)

Answer 2

Step 1: Write the given data.
Mass of the organic compound = 0.20 g
Mass of barium sulphate (\( \text{BaSO}_4 \)) formed = 0.40 g

Step 2: Recall the relationship between sulphur and barium sulphate.
In the estimation of sulphur, all the sulphur in the organic compound is converted into \( \text{BaSO}_4 \).
1 mole of \( \text{BaSO}_4 \) contains 1 mole of sulphur.

Molar mass of \( \text{BaSO}_4 \):
\[ \text{Ba} = 137, \quad \text{S} = 32, \quad \text{O}_4 = 64 \] \[ \text{Total molar mass of BaSO}_4 = 233 \, \text{g mol}^{-1} \] Thus, 233 g of \( \text{BaSO}_4 \) contain 32 g of sulphur.

Step 3: Calculate the mass of sulphur in 0.40 g of \( \text{BaSO}_4 \).
\[ \text{Mass of S} = \frac{32}{233} \times 0.40 = 0.0549 \, \text{g} \]

Step 4: Calculate the percentage of sulphur in the compound.
\[ \text{Percentage of S} = \frac{0.0549}{0.20} \times 100 = 27.45 \approx 27.5\% \]

Final Answer:
\[ \boxed{27.5\%} \]