By combustion of $0.2475 \ g$ of an organic compound,$0.4950 \ g$ of $CO_2$ and $0.2025 \ g$ of $H_2O$ are obtained. Calculate the percentage of $C, H,$ and $O$. Given atomic masses: $C = 12, H = 1, O = 16$.

(N/A) $1$. Calculate mass of $C$: Mass of $C = (12/44) \times \text{mass of } CO_2 = (12/44) \times 0.4950 = 0.135 \ g$.
$2$. Calculate percentage of $C$: $\% C = (0.135 / 0.2475) \times 100 = 54.54\%$.
$3$. Calculate mass of $H$: Mass of $H = (2/18) \times \text{mass of } H_2O = (2/18) \times 0.2025 = 0.0225 \ g$.
$4$. Calculate percentage of $H$: $\% H = (0.0225 / 0.2475) \times 100 = 9.09\%$.
$5$. Calculate percentage of $O$: $\% O = 100 - (54.54 + 9.09) = 36.37\%$.