Complete combustion of X g of an organic compound gave 0.25 g of...

Mass of carbon in the compound is calculated from the mass of $CO_2$ produced:

Mass of C $= \dfrac{12}{44} \times \text{mass of } CO_2 = \dfrac{12}{44} \times 0.25 = \dfrac{3}{44}$ g

Given that the percentage of carbon in the compound is $25\%$:

$\% \text{ of C} = \dfrac{\text{Mass of C}}{\text{Total mass } X} \times 100$

$25 = \dfrac{\dfrac{3}{44}}{X} \times 100$

$X = \dfrac{3}{44} \times \dfrac{100}{25} = \dfrac{3}{44} \times 4 = \dfrac{3}{11}$ g

$X \approx 0.2727$ g

Verifying this using the percentage of hydrogen:

Mass of H $= \dfrac{2}{18} \times \text{mass of } H_2O = \dfrac{2}{18} \times 0.12 = \dfrac{1}{75}$ g

$\% \text{ of H} = \dfrac{\dfrac{1}{75}}{X} \times 100 = 4.89$

$X = \dfrac{100}{75 \times 4.89} \approx 0.2727$ g

Thus, $X = 272.7 \times 10^{-3}$ g.

Rounding to the nearest integer, the required value is $273$.

Answer: $273$