A 5.769 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 14.92 grams of CO2 and 3.054 grams of H2O are produced. In a separate experiment, the molar mass is found to be 136.2 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

Empirical formula of the compound is C₄H₄O

Molecular formula of the compound is C₈h₈o₂

Explanation:

No. of moles of CO₂ = = 0.339 moles of carbonno. of moles of H₂O = = 0.16 moles x 2 = 0.339 moles of hydrogenmass of carbon = 0.339 x 12 = 4.068 gmass of hydrogen = 0.339 x 1 = 0.339 gmass of oxygen = 5.769 - (4.068 + 0.339)

                                   = 1.362 g

   6. moles of oxygen = = 0.085 moles

relative mole ratio for carbon = = 4relative mole ratio for hydrogen = = 4relative mole ratio for oxygen = = 1 empirical formula C₄H₄Oempirical formula mass = 4 x 12 + 4 x 1 + 1 x 16 = 68molecular mass = 136.2 g / mol

Molecular formula =  (C₄H₄O)₂ = C₈H₈O₂