Chapter 3

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


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.


  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


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.