Combustion analysis of 2.796 g of an unknown compound containing carbon, hydrogen, and oxygen produced 5.597 g of CO2 and 2.268 g of H2O. ?

Mass fraction of C in CO₂ = 12/(12 + 16×2) = 12/44
Mass fraction of H in H₂O = 1×2/(1×2 + 16) = 2/18

In 2.796 g of the unknown compound:
Mass of C = Mass of C in CO₂ produced = 5.597 × (12/44) g = 1.526 g
Mass of H = Mass of H in H₂O produced = 2.268 × (2/18) g = 0.252 g
Mass of O = (2.796 - 1.526 - 0.252) g = 1.018 g

Mole ratio C : H : O
= (1.526/12) : (0.252/1) : (1.018/16)
= 0.127 : 0.252 : 0.0636
= (0.127/0.0636) : (0.252/0.0636) : (0.0636/0.0636)
≈ 2 : 4 : 1

Empirical formula = C₂H₄O