Question

Complete combustion of 4.80 g of a hydrocarbon produced 14.7 g of CO2 and 7.02 g of H2O. What is the empirical formula for the hydrocarbon? Insert subscripts as necessary.

Answer 1

The hydrocarbon empirical formula is C_xH_y. When 1 mol of this hydrocarbon burns, it produces x moles of CO₂ and y/2 moles of H₂O. The number of the moles of CO₂ contained in 14.7 g is:

$n(CO_2) = \frac{m}{M} = \frac{14.7\text{ g}}{44.01 \text{ g/mol}} = 0.334$ mol.

The number of the moles of H2O produced is:

$n(H_2O)= \frac{m}{M} = \frac{7.02 \text{ g}}{18.02\text{ g/mol}} = 0.390$ mol.

As one can see, in this hydrocarbon, for each atom of carbon there are 2·0.390/0.334 = 2.333 atoms of hydrogen. In order to get the simplest empirical formula, we need to have the whole numbers. We can achieve this when calculating that for each 3 atoms of carbon we get 7 atoms of hydrogen. However, C₃H₇ is a radical. Combining two radicals, we get C₆H₁₄. This compound is called hexane.

Answer: The compound's simplest empirical formula is C₆H₁₄.