Question

When 4.088 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 13.82 grams of CO2 and 2.829 grams of H2O were produced.

In a separate experiment, the molecular weight of the compound was found to be 78.11 amu. Determine the empirical formula and the molecular formula of the hydrocarbon.

Answer 1

Empirical formula: CH

Molecular formula: C₆H₆

Step-by-step explanation:

Based on the combustion of a hydrocarbon, the moles of CO₂ = Moles of Carbon in the hydrocarbon and the moles of H₂O = 1/2 moles of hydrogen in the hydrocarbon.

The empirical formula is the simplest whole number of atoms present in a molecule. With the moles of C and H we can find empirical formula:

Moles C -Molar mass CO₂ = 44.01g/mol-:

13.82g * (1mol / 44.01g) = 0.314 moles C

Moles H -Molar mass H₂O = 18.01g/mol-:

2.829g H₂O * (1mol / 18.01g) = 0.157 moles H₂O * (2mol H / 1mol H₂O) = 0.314 moles of H

The ratio of moles H: moles C:

0.314 moles / 0.314 moles = 1

That means empirical formula is:

CH

With the molecular weight and empirical formula we can find the molecular formula:

Molar mass CH = 12.01g/mol+1.01g/mol = 13.02g/mol

As the molecular weight of the molecule is 78.11amu = 78.11g/mol, there are:

78.11g/mol / 13.02g/mol = 6 times the empirical formula in the molecular formula

That means molecular formula is:

C₆H₆