$0.45 \ g$ of an organic compound on combustion gives $0.792 \ g$ of $CO_2$ and $0.324 \ g$ of $H_2O$. In the Kjeldahl method,$0.24 \ g$ of the same compound yields ammonia,which is absorbed in $50 \ mL$ of $0.25 \ N \ H_2SO_4$. The excess acid requires $77.0 \ mL$ of $0.25 \ N \ NaOH$ for neutralization. Determine the empirical formula of the compound. (Atomic masses: $C = 12, H = 1, O = 16, N = 14, S = 32$)

(D) $1$. Calculate percentage of $C$: $\text{Mass of } C = (12/44) \times 0.792 = 0.216 \ g$. $\% C = (0.216 / 0.45) \times 100 = 48\%$.
$2$. Calculate percentage of $H$: $\text{Mass of } H = (2/18) \times 0.324 = 0.036 \ g$. $\% H = (0.036 / 0.45) \times 100 = 8\%$.
$3$. Calculate percentage of $N$: $\text{Total } H_2SO_4 = 50 \times 0.25 = 12.5 \ \text{meq}$. $\text{Acid used for } NH_3 = 12.5 - (77.0 \times 0.25 / 2) = 12.5 - 9.625 = 2.875 \ \text{meq}$. $\text{Mass of } N = (2.875 \times 14) / 1000 = 0.04025 \ g$. $\% N = (0.04025 / 0.24) \times 100 \approx 16.77\%$.
$4$. Calculate percentage of $O$: $\% O = 100 - (48 + 8 + 16.77) = 27.23\%$.
$5$. Empirical formula calculation: $C: 48/12 = 4, H: 8/1 = 8, N: 16.77/14 \approx 1.2, O: 27.23/16 \approx 1.7$. Dividing by smallest $(1.2)$: $C: 3.33, H: 6.66, N: 1, O: 1.4$. Multiplying by $3$: $C: 10, H: 20, N: 3, O: 4$. The empirical formula is $C_{10}H_{20}N_3O_4$.