An organic compound contains 60% carbon, 13.3% hydrogen and 26.7% oxygen. Calculate the empirical formula. (C=12, H=1, O=16)

Correct answer: C₃H₈O

Explanation: Assume 100 g of the compound.

• C: 60 g → \( \frac{60}{12} = 5 \) mol
• H: 13.3 g → \( \frac{13.3}{1} = 13.3 \) mol
• O: 26.7 g → \( \frac{26.7}{16} \approx 1.66875 \) mol

Divide by the smallest (1.66875):

C: \( \frac{5}{1.66875} \approx 3 \),  H: \( \frac{13.3}{1.66875} \approx 8 \),  O: \( \frac{1.66875}{1.66875} = 1 \).

Empirical ratio = C₃H₈O.