CALCULATING AN EMPIRICAL FORMULA AND A MOLECULAR FORMULA FROM A COMBUSTION ANALYSIS Caproic acid Caproic acid, the substance responsible for the aroma of goats, dirty socks, and old shoes, contains carbon, hydrogen, and oxygen. On combustion analysis, a 0.450 g sample of caproic acid gives 0.418 g

Question

CALCULATING AN EMPIRICAL FORMULA AND A MOLECULAR FORMULA FROM A COMBUSTION ANALYSIS

Caproic acid Caproic acid, the substance responsible for the aroma of goats, dirty socks, and old shoes, contains carbon, hydrogen, and oxygen. On combustion analysis, a 0.450 g sample of caproic acid gives 0.418 g of [latex]H_{2}O[/latex] and 1.023 g of [latex]CO_{2}[/latex]. What is the empirical formula of caproic acid? If the molecular mass of caproic acid is 116.2 amu, what is the molecular formula?

STRATEGY
Use the steps outlined in Figure 3.8 to find the empirical formula of caproic acid. Then calculate a formula mass and compare it to the known molecular mass.

Accepted Answer

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

[latex]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}}}[/latex] = 0.023 24 mol C

[latex]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}}[/latex] = 0.0464 mol H

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

[latex]Moles of C = 0.023 24 \cancel{mol C} × \frac{12.01 g C}{1 \cancel{mol C}}[/latex] = 0.2791 g C

[latex]Moles of H = 0.0464 \cancel{mol H} × \frac{1.01 g H}{1 \cancel{mol H}}[/latex] = 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 [latex]\cancel{g O} × \frac{1 mol O}{16.00 \cancel{g O}}[/latex] = 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.

[latex]C_{\left( \frac{0.02324}{0.00775}\right)}H_{\left(\frac{0.0464}{0.00775}\right) }O_{\left(\frac{ 0.00775}{0.00775}\right)} = C_{3}H_{6}O[/latex]

The empirical formula of caproic acid is therefore [latex]C_{3}H_{6}O[/latex], 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

[latex]C_{(2×3)}H_{(2×6)}O_{(2×1)} = C_{6}H_{12}O_{2}[/latex].

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

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