Determining the Empirical and Molecular Formulas of a Compound from Its Mass Percent Composition
Dibutyl succinate is an insect repellent used against household ants and roaches. Its composition is 62.58% C, 9.63% H, and 27.79% O. Its experimentally determined molecular mass is 230 u. What are the empirical and molecular formulas of dibutyl succinate?
Analyze
Use the five-step approach described above.
Solve
? mol H = 9.63 g H × \frac{1 mol H}{1.008 g H} = 9.55 mol H
? mol O = 27.79 g O × \frac{1 mol O}{15.999 g O} = 1.737 mol O
3. Write a tentative formula based on these numbers of moles. C_{5.210} H_{9.55} O_{1.737}
4. Divide each of the subscripts of the tentative formula by the smallest subscript (1.737), C_{\frac{5.210}{1.737}} H_{\frac{9.55}{1.737}} O_{\frac{1.737}{1.737}} = C_{2.99} H_{5.49} O
and round off any subscripts that differ only slightly from whole numbers; that is, round 2.99 to 3. C_{3} H_{5.49} O
5. Multiply all subscripts by a small whole number to make them integral (here by the factor 2), and write the empirical formula. C_{2×3} H_{2×5.49} O_{2×1} = C_{6} H_{10.98} O_{2}
2 × 5.49 = 10.98 ≈ 11
Empirical formula: C_{6}H_{11}O_{2}
To establish the molecular formula, first determine the empirical formula mass. [(6 × 12.0) + (11 × 1.0) + (2 × 16.0)]u = 115 u
Since the experimentally determined formula mass (230 u) is twice the empirical formula mass, the molecular formula is twice the empirical formula. Molecular formula: C_{12}H_{22}O_{4}
Assess
Check the result by working the problem in reverse and using numbers that are rounded off slightly.
For C_{12}H_{22}O_{4} % C ≈ (12 × 12 u/230 u) × 100% = 63%; % H ≈ (22 × 1 u/230 u) × 100% = 9.6%; and % O ≈ (4 × 16 u/230 u) × 100% = 28%. The calculated mass percents agree well with those given in the problem, so we can be confident that our answer is correct.