Question 6.13: Calculating an Empirical Formula from a Combustion Analysis ......

First, find the molar amounts of C and H in the sample:
Moles of C = 1.023 \ \cancel{g \ CO_2} \times \frac{1 \ \cancel{mol \ CO_2}}{44.01 \ \cancel{g \ CO_2}} \times \frac{1 \ mol \ C}{1 \ \cancel{mol \ CO_2}} = 0.023 \ 24 \ mol \ C \\ \text{Moles of H} = 0.418 \ \cancel{g \ H_2O} \times \frac{1 \ \cancel{mol \ H_2O}}{18.02 \ \cancel{g \ H_2O}} \times \frac{2 \ mol \ H}{1 \ \cancel{mol \ H_2O}} = 0.0464 \ mol \ H
Next, find the number of grams of C and H in the sample:
\text{Mass of C} = 0.023 \ 24 \ \cancel{mol \ C} \times \frac{12.01 \ g \ C}{1 \ \cancel{mol \ C}} = 0.2791 \ g \ C \\ \text{Mass of H} = 0.0464 \ \cancel{mol \ H} \times \frac{1.01 \ g \ H}{1 \ \cancel{mol \ H}} = 0.0469 \ g \ H
Subtracting the masses of C and H from the mass of the starting sample indicates that 0.124 g is unaccounted for:
0.450 g – (0.2791 g + 0.0469 g) = 0.124 g
Because we are told that oxygen is also present in the sample, the “missing” mass must be due to oxygen, which can’t be detected by combustion. We therefore need to fnd the number of moles of oxygen in the sample:
\text{Moles of O} = 0.124 \ \cancel{g \ O} \times \frac{1 \ mol \ O}{16.00 \ \cancel{g \ O}} = 0.007 \ 75 \ mol \ O
Knowing the relative numbers of moles of all three elements, C, H, and O, we divide the three numbers of moles by the smallest number (0.007 75 mol of oxygen) to arrive at a C:H:O ratio of 3:6:1.
\mathrm{C}_{\left(\frac{0.02324}{0.00775}\right)} \mathrm{H}_{\left(\frac{0.0464}{0.00755}\right)} \mathrm{O}_{\left(\frac{0.00775}{0.00775)}\right)}=\mathrm{C}_3 \mathrm{H}_6 \mathrm{O}
The empirical formula of caproic acid is therefore C_3H_6O, and the empirical formula weight is 58.1. Because the molecular weight of caproic acid is 116.2, or twice the empirical formula weight, the molecular formula of caproic acid must be C_{(2 \times 3)}H_{(2 \times 6)}O_{(2 \times 1)} = C_6H_{12}O_2.