Question

A compound containing only carbon, hydrogen and oxygen was analyzed by combustion analysis. Combustion of a 10.68 g sample of the compound in excess oxygen gives 16.01 g CO2 and 4.37 g H¬2O. The molar mass of the compound is between 170 and 180 g/mol. What is the molecular formula of this compound?

Answer 1

Question #63812, Chemistry / Other

A compound containing only carbon, hydrogen and oxygen was analyzed by combustion analysis. Combustion of a 10.68 g sample of the compound in excess oxygen gives 16.01 g CO₂ and 4.37 g H₂O. The molar mass of the compound is between 170 and 180 g/mol. What is the molecular formula of this compound?

Solution:

$$\mathrm{C_4H_2O_2} + \mathrm{O_2} = \mathrm{xCO_2} + \mathrm{y/2H_2O}$$ $$n(C) = n(CO_2) = \frac{16.01\,g}{44.04\,g/mol} = 0.364\,mol$$ $$n(H) = 2n(H_2O) = \frac{2 \times 4.37\,g}{18.00\,g/mol} = 0.486\,mol$$ $$m(C) = 0.364\,mol \times 12.001\,\frac{g}{mol} = 4.368\,g$$ $$m(C) = 0.486\,mol \times 1.002\,\frac{g}{mol} = 0.486\,g$$ $$m(O) = 10.68\,g - 4.368\,g - 0.486\,g = 5.826\,g$$ $$n(O) = \frac{5.826\,g}{15.999\,g/mol} = 0.364\,mol$$ $$C: H: O = 0.364: 0.486: 0.364 = 1: 1.33: 1 = 3: 4: 3$$ $$\left(\mathrm{C_3H_4O_3}\right)_n$$ $$M\left(\mathrm{C_3H_4O_3}\right) = 12 \times 3 + 4 \times 1 + 3 \times 16 = 88\,g/mol$$ $$n = \frac{180\,g/mol}{88\,g/mol} = 2.045 \approx 2$$

Formula is $\left(\mathrm{C_3H_4O_3}\right)_n = \mathrm{C_6H_8O_6}$

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