Empirical FormulaHardQuestion 600062
Question
Combustion of 0.3 g of an organic compound gives 0.44 g CO₂ and 0.18 g H₂O. If no other element is present, its empirical formula is:
- ACH₂Correct
- BCH₃
- CCH₄
- DCH
Correct answer
CH₂
Explanation
n(C) = 0.44/44 = 0.01 mol → C mass = 0.12 g. n(H) = 2×0.18/18 = 0.02 mol → H mass = 0.02 g. O = 0.3−0.12−0.02 = 0.16 g? But problem says no other element → Check: 0.12+0.02=0.14 ≠ 0.3. Hmm, there must be oxygen: O = 0.3−0.14=0.16 g → O moles=0.01. Ratio C:H:O = 0.01:0.02:0.01 = 1:2:1 → CH₂O. Revise: empirical formula is CH₂O (not CH₂ — error in options; CH₂O is correct for these numbers).