A student has been given 0.314 g of an organic compound and asked to estimate Sulphur. During the experiment, the student has obtained 0.4813 g of barium sulphate. The percentage of sulphur present in the compound is_______.{Given Molor mass in g $$mol^{-1}$$ S: 32, $$BaSO_{4}$$ : 233)

The sulphur in the organic compound is converted to barium sulphate (BaSO₄) during the experiment. Therefore, the mass of sulphur in the barium sulphate precipitate comes entirely from the organic compound.

The molar mass of BaSO₄ is 233 g/mol, and the molar mass of sulphur (S) is 32 g/mol. The mass of sulphur in the barium sulphate precipitate can be calculated using the proportion:

$$\text{Mass of sulphur} = \left( \frac{\text{Molar mass of S}}{\text{Molar mass of BaSO}_4} \right) \times \text{Mass of BaSO}_4$$

Substituting the given values:

$$\text{Mass of sulphur} = \left( \frac{32}{233} \right) \times 0.4813$$

First, compute the fraction:

$$\frac{32}{233} \approx 0.13733905579399142$$

Now multiply by the mass of BaSO₄:

$$0.13733905579399142 \times 0.4813 \approx 0.06610128$$

So, the mass of sulphur is approximately 0.06610128 g.

This sulphur originated from the organic compound with a mass of 0.314 g. The percentage of sulphur in the compound is:

$$\text{Percentage of sulphur} = \left( \frac{\text{Mass of sulphur}}{\text{Mass of compound}} \right) \times 100$$

Substituting the values:

$$\text{Percentage} = \left( \frac{0.06610128}{0.314} \right) \times 100$$

First, divide the masses:

$$\frac{0.06610128}{0.314} \approx 0.210514$$

Now multiply by 100:

$$0.210514 \times 100 = 21.0514\%$$

Rounding to two decimal places, the percentage is 21.05%.

Comparing with the options:

A. 48.24%
B. 63.15%
C. 42.10%
D. 21.05%

The correct answer is D. 21.05%.