Question 3.17: CALCULATING AN EMPIRICAL FORMULA AND A MOLECULAR FORMULA FRO...

First, find the molar amounts of C and H in the sample:

$Moles  of  C  =  1.023  \cancel{g  CO_{2}}  ×  \frac{1  \cancel{mol  CO_{2}}}{44.01  \cancel{g  CO_{2}}}  ×  \frac{1  mol  C}{1  \cancel{mol  CO_{2}}}$ = 0.023 24 mol C

$Moles  of  H  =  0.418  \cancel{g  H_{2}O}  ×  \frac{1  \cancel{mol  H_{2}O}}{18.02  \cancel{g  H_{2}O}}  ×  \frac{2  mol  H}{1  \cancel{mol  H_{2}O}}$ = 0.0464 mol H

Next, find the number of grams of C and H in the sample:

$Moles  of  C  =  0.023 24  \cancel{mol  C}  ×  \frac{12.01  g  C}{1  \cancel{mol  C}}$ = 0.2791 g C

$Moles  of  H  =  0.0464  \cancel{mol  H}  ×  \frac{1.01  g  H}{1  \cancel{mol  H}}$ = 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 find the number of moles of oxygen in the sample:

Moles of O = 0.124 $\cancel{g  O}  ×  \frac{1  mol  O}{16.00  \cancel{g  O}}$ = 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.

The empirical formula of caproic acid is therefore $C_{3}H_{6}O$, and the empirical formula mass is 58.1 amu. Because the molecular mass of caproic acid is 116.2, or twice the empirical formula mass, the molecular formula of caproic acid must be

$C_{(2×3)}H_{(2×6)}O_{(2×1)}  =  C_{6}H_{12}O_{2}$.