Question

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.

Answer 1

Answer:

Empirical formula of the compound is C₄H₄O

Molecular formula of the compound is C₈h₈o₂

Explanation:

1. No. of moles of CO₂ =$\frac{ 14.92}{44}$ = 0.339 moles of carbon
2. no. of moles of H₂O = $\frac{3.054}{18}$ = 0.16 moles x 2 = 0.339 moles of hydrogen
3. mass of carbon = 0.339 x 12 = 4.068 g
4. mass of hydrogen = 0.339 x 1 = 0.339 g
5. mass of oxygen = 5.769 - (4.068 + 0.339)

                                   = 1.362 g

   6. moles of oxygen = $\frac{1.362}{16}$ = 0.085 moles

• relative mole ratio for carbon = $\frac{0.339}{0.085}$ = 4

• relative mole ratio for hydrogen = $\frac{0.339}{0.085}$ = 4

• relative mole ratio for oxygen = $\frac{0.085}{0.085}$ = 1

• empirical formula C₄H₄O

• empirical formula mass = 4 x 12 + 4 x 1 + 1 x 16 = 68

• molecular mass = 136.2 g / mol

$n - factor =\frac{Empirical formula mass}{Molecular formula mass}= \frac{136.2}{68}=2$

• Molecular formula =  (C₄H₄O)₂ = C₈H₈O₂