Question 3.18: Limonene is a hydrocarbon that contributes to the odor of ci...

Collect and Organize We know the masses of CO_{2} and H_{2}O produced during the combustion of a hydrocarbon sample, and we are asked to determine its empirical formula and then, from its molecular mass, its molecular formula.

Analyze First we determine the number of moles of C and H in the CO_{2} and H_{2}O produced during combustion. These values are equal to the number of moles of C and H in the combusted sample. Then we calculate the C:H mole ratio and convert it into a ratio of small whole numbers to obtain the empirical formula. Next we calculate the mass of the empirical formula and divide this mass into the molar mass to obtain the multiplier, n, that allows us to convert the empirical formula to a molecular formula .

Solve The moles of C and H in the CO_{2} and H_{2}O collected during combustion are

2.168  \sout{g  CO_{2}} \times \frac{1  \sout{mol  CO_2}}{44.01  \sout{g  CO_2}} \times \frac{1  mol  C}{1  \sout{mol  CO_{2}}} =0.04926  mol  C

0.710  \sout{g  H_{2}O}\times \frac{1  \sout{mol  H_{2}O}}{18.02  \sout{g  H_{2}O}} \times \frac{2  mol  H}{1  \sout{mol  H_{2}O}} =0.0788  mol  H

The mole ratio of the two elements in the sample is

0.04926  mol  C:0.0788  mol  H

Dividing through by the smallest value (0.04926 mol) gives a mole ratio of 1:1.6. We can convert this ratio to whole numbers by multiplying by 5, making the empirical formula of the sample C_{5}H_{8}.
The molar mass of limonene is 136.24 g/mol. The mass of the empirical formula of limonene is 5(12.01  g/mol) + 8(1.008  g/mol) = 68.11  g/mol, so the multiplier n is

n =\frac{molar  mass}{empirical  formula  mass} =\frac{136.24  g/mol}{68.11  g/mol} =2

and the molecular formula is

(C_{5}H_{8})_{n} = (C_{5}H_{8})_{2} = C_{10}H_{16}

Think About It The subscripts in our answer, 5 for C and 8 for H, make sense because the moles C ≈ 5 × 10^{-2} and moles H ≈ 8 × 10^{-2} are essentially at a 5:8 ratio.