An organic compound contains 59.0% C, 7.15% H, 26.20% O, and 7.65% N and has a molar mass of 183 g/mol. What is its molecular formula? (Take C = 12, O = 16, H = 1, N = 14) N.B: Show all working

• For Carbon (C): $$\text{Moles of C} = \frac{59.0\, \text{g}}{12\, \text{g/mol}} = 4.917\, \text{mol}$$
• For Hydrogen (H): $$\text{Moles of H} = \frac{7.15\, \text{g}}{1\, \text{g/mol}} = 7.150\, \text{mol}$$
• For Oxygen (O): $$\text{Moles of O} = \frac{26.20\, \text{g}}{16\, \text{g/mol}} = 1.638\, \text{mol}$$
• For Nitrogen (N): $$\text{Moles of N} = \frac{7.65\, \text{g}}{14\, \text{g/mol}} = 0.546\, \text{mol}$$

The smallest number of moles is $0.546\, \text{mol}$ (for N).

• For C: $$\frac{4.917}{0.546} \approx 9.00$$
• For H: $$\frac{7.150}{0.546} \approx 13.09 \approx 13$$
• For O: $$\frac{1.638}{0.546} \approx 3.00$$
• For N: $$\frac{0.546}{0.546} = 1.00$$

The empirical formula is $C_9H_{13}O_3N$.

Since $n=1$, the molecular formula is the same as the empirical formula.

The molecular formula is $\boxed{C_9H_{13}O_3N}$.

C=8, H=10, N=4, O=2.

The GCD of 8, 10, 4, and 2 is 2.

• For C: $8 \div 2 = 4$
• For H: $10 \div 2 = 5$
• For N: $4 \div 2 = 2$
• For O: $2 \div 2 = 1$

The empirical formula is $\boxed{C_4H_5N_2O}$.