Complete combustion of 0.44 g of an organic compound produced 0.88 g of CO₂ and 0.36 g of H₂O. What is the empirical formula of the compound?
- CH
- CH₂
- CH₂O
- C₂H₄O
📝 Detailed explanation
Moles C = 0.88/44 = 0.02. Moles H = 2 × (0.36/18) = 0.04.Mass C+H = (0.02×12) + (0.04×1) = 0.24 + 0.04 = 0.28 g.
Mass O = 0.44 - 0.28 = 0.16 g. Moles O = 0.16/16 = 0.01.
Ratio C:H:O = 0.02:0.04:0.01 = 2:4:1. Thus, the formula is C₂H₄O.
Wait, let me recheck the options.
If 0.02 C, 0.04 H, 0.01 O -> C₂H₄O.
Actually, if the mass was different... let's check CH₂O: 12+2+16=30.
C:H:O ratio is 2:4:1. My calculation is correct. C₂H₄O.