## Q. 3.IE

Molecules of a dicarboxylic acid have two carboxyl groups (—COOH). A 2.250 g sample of a dicarboxylic acid was burned in an excess of oxygen and yielded 4.548 g $CO_{2}$ and 1.629 g $H_{2}O$. In a separate experiment, the molecular mass of the acid was found to be 174 u. From these data, what can we deduce about the structural formula of this acid?

## Verified Solution

Analyze

Our approach will require several steps: (1) Use the combustion data to determine the percent composition of the compound (similar to Example 3-6). (2) Determine the empirical formula from the percent composition (similar to Example 3-5). (3) Obtain the molecular formula from the empirical formula and the molecular mass. (4) Determine how the C, H, and O atoms represented in the molecular formula might be assembled into a dicarboxylic acid. Use molar masses with (at least) one more significant figure than in the measured masses where possible; store intermediate results in your calculator without rounding off.

Solve

1. Determine the percent composition. Calculate the mass of H in 1.629 g $H_{2}O$          ? g H = 1.629 g $H_{2}O × \frac{1 mol H_{2}O}{18.015 g H_{2}O} × \frac{2 mol H}{1 mol H_{2}O} × \frac{1.008 g H}{1 mol H}$ = 0.1823 g H

and then the mass percent H in the 2.250 g sample of the dicarboxylic acid.          % H = $\frac{0.1823 g H}{2.250 g compd.}$ × 100% = 8.102% H

Also, calculate the mass of C in 4.548 g $CO_{2}$,          ? g C = 4.548 g $CO_{2} × \frac{1 mol CO_{2}}{44.009 g CO_{2}} × \frac{1 mol C}{1 mol CO_{2}} × \frac{12.011 g C}{1 mol C}$ = 1.241 g C

followed by the mass percent C in the 2.250 g sample of dicarboxylic acid.          % C = $\frac{1.241 g C}{2.250 g compd.}$ × 100% = 55.16% C

The mass percent O in the compound is obtained as a difference, that is,           % O = 100.00% − 55.16% C − 8 .102% H = 36.74% O

2. Obtain the empirical formula from the percent composition. The masses of the elements in 100.0 g of the compound are        55.16 g C   8.102 g H   36.74 g O

The numbers of moles of the elements in 100.0 g of the compound are            55.16 g C × $\frac{1 mol C}{12.011 g C}$ = 4.592 mol C

8.102 g H × $\frac{1 mol H}{1.008 g H}$ = 8.038 mol H

36.74 g O × $\frac{1 mol O}{15.999 g O}$ = 2.296 mol O

The tentative formula based on these numbers is          $C_{4.592}H_{8.038}O_{2.296}$

Divide all the subscripts by 2.296 to obtain            $C_{2}H_{3.50}O$

Multiply all subscripts by two to obtain the empirical formula          $C_{4}H_{7}O_{2}$

and then determine the empirical formula mass.              (4 × 12.011 u) + (7 × 1.008 u) + (2 × 15.999 u) = 87.098 u

3. Obtain the molecular formula. The experimentally determined molecular mass of 174 u is twice the empirical formula mass. The molecular formula is            $C_{8}H_{14}O_{4}$

4. Assemble the atoms in $C_{8}H_{14}O_{4}$ into a plausible structural formula. The dicarboxylic acid must contain two —COOH groups. This accounts for the two C atoms, two H atoms, and all four O atoms. The remainder of the structure is based on $C_{6}H_{12}$. For example, arrange the six $—CH_{2}$ segments into a six-carbon chain and attach the —COOH groups at the ends of the chain.            $HOOC—CH_{2}(CH_{2})_{4}CH_{2}—COOH$

However, there are other possibilities based on shorter chains with branches, for example            $HOOC—CH_{2}—\overset{\begin{matrix} CH_{3} \\ \mid \end{matrix} }{\underset{\begin{matrix} \mid \\ CH_{3} \end{matrix} }{C}}—CH_{2}—CH_{2}—COOH$

Assess

We have found a plausible structural formula, but there are many other possibilities. For example, the following three isomers have a seven-carbon chain with one methyl group ($—CH_{3}$) substituted for an H atom on the chain:

$HOOCCHCH_{3}(CH_{2})_{4}COOH; HOOCCH_{2}CHCH_{3}(CH_{2})_{3}COOH; HOOC(CH_{2})_{2}CHCH_{3}(CH_{2})_{2}COOH$

In conclusion, we cannot identify a specific isomer with only the data given.