When 7.101 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 21.76 grams of CO2 and 10.39 grams of H2O were produced.

Question

In a separate experiment, the molar mass of the compound was found to be 86.18 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Answer 1

To find the empirical and molecular formulas, we need to first find the number of moles of each element (C and H) present. Let us assume we have 100 g of material, then the % of each element simply becomes the grams of each.

Combustion of a hydrocarbon with oxygen is as follows:

C_xH_y + O₂ ==> _CO₂ + _H₂O ... unbalanced

mols C: 21.76 g CO2 x 1 mol CO2 / 44.01 g CO2 x 1 mol C / mol CO2 = 0.4944 mols C

mols H: 10.39 g H2O x 1 mol H2O / 18.01 g H2O x 2 mol H / mol H2O = 1.154 mols H

Divide both by 0.0.4944 to try to get whole numbers:

mols C = 0.4944 / 0.0.4944 = 1.00 mols C

mols H = 1.154 / 0.4944 = 2.334 mols H

Multiply both by 3 to try to get whole numbers:

mols C = 1.00 x 3 = 3 mols C

mols H = 2.334 x 3 = 7 mols H

Empirical formula = C₃H₇ (this is a propyl group, and not a full hydrocarbon which has the formula C_xH_2x+2)

Molar mass of C3H7 = 3x12 + 7x1 = 36 + 7 = 43

86.18 / 43 = 2

Molecular formula = C₆H₁₄ (this has the correct formula for a hydrocarbon: C_xH_2x+2)