In sulphur estimation. 0.471 g of an organic compound gave 1.4439 g of barium sulphate. The percentage of sulphur in the compound is _____ (Nearest Integer)
(Given: Atomic mass Ba: 137u, S: 32u, O: 16u)

In sulphur estimation, 0.471 g of an organic compound gave 1.4439 g of barium sulphate (BaSO$$_4$$).

Molar mass of BaSO$$_4$$ = 137 + 32 + 64 = 233 g/mol.

$$\text{Mass of S} = \frac{32}{233} \times 1.4439 = \frac{46.2048}{233} = 0.19831 \text{ g}$$

$$\% \text{S} = \frac{0.19831}{0.471} \times 100 = 42.10\%$$

Rounding to the nearest integer: $$\approx 42\%$$.

The percentage of sulphur in the compound is $$\mathbf{42}$$.