A compound contains 52.2% carbon, 13.0% hydrogen and 34.8% oxygen by mass. What is its empirical formula? (Ar: C 12.01, H 1.01, O 16.00)
- Divide by Ar: C 52.2 ÷ 12.01 = 4.35; H 13.0 ÷ 1.01 = 12.9; O 34.8 ÷ 16.00 = 2.18.
- Divide by the smallest (2.18): C 2.00; H 5.9, which rounds to 6; O 1.00.
- The ratio C : H : O is 2 : 6 : 1.
Answer: C2H6O