Question

When 4.979 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 16.20 grams of CO2 and 4.976 grams of H2O were produced.

In a separate experiment, the molar mass of the compound was found to be 54.09 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Answer 1

combustion analysis is all about recognizing a truth about conservation of mass. When the fuel is burned, all the carbon atoms that used to be in the fuel become bonded to the oxygen and become CO2. Each molecule of fuel contains X number of carbon atoms, but each CO2 molecule only contains 1 carbon atom. So the number of carbon dioxide molecules formed is a direct result of the number of carbon atoms that came from the fuel in the first place.

We can use mass percent to find the mass of carbon that is in the CO2, and deduce that all that carbon came from the sample originally.
\[Mass _C = (Mass _{CO_2})*(Mass_{ \% C}) = 16.20gCO_2*( \frac{Mass_{C}}{Mass_{CO2}}) = 16.20g CO_2 * \frac{12g}{44g} = 4.418g C\]We make the same argument about the hydrogen that's in the water. Or, we can see that the compound contains only C and H atoms, so whatever mass isn't found from the carbon atoms must have come from the hydrogens.
\[Mass_{hydrogen} = Mass_{sample} - Mass_{carbon} = 4.979g - 4.418g = 0.5608g\]
What we have now is the masses of the C and H atoms that came from the fuel sample originally. To make a formula, however, we need to know moles of each. Dividing each mass by the molar mass of the component gets us moles
\[\frac{4.418gC}{12\frac{g}{molC}} = 0.3682mol C\]and\[\frac{0.5608g H}{1\frac{g}{molH}} = 0.5608mol H\]A chemical formula has whole number moles, however, so we can play one more trick to get these fractional numbers to look like whole numbers. If we divide each number of moles by the smaller of the 2 numbers. This guarantees that at least one number turns to a whole number, specifically 1.
\[\frac{0.3681mol C}{0.3681molC} = 1\]and\[\frac{0.5608mol H}{0.3681mol C} = 1.5\]
This means that for every carbon atom, there are 1.5 hydrogen atoms. Since having half an atom is impossible, we double everything and get 2 carbons, and 3 hydrogens per molecule (almost).

The empirical formula of the hydrocarbon is \[C_2H_3\] The empirical formula is the lowest possible whole # ratio for the molecule, but it may not be the actual ratio of atoms in the molecule. Finding the empirical mass and comparing that to the molecular mass lets us see how many times bigger the molecular mass is relative to the empirical mass.

Here, the empirical mass is 27g/mol (2carbons, 3 hydrogens). The accepted molecular mass is 54g/mol. 54 is twice as large as 27, so the actual formula of the molecule is twice as large as the empirical formula found earlier.

The molecular formula is\[C_4H_6\]