Question 3.5: Determining the Empirical and Molecular Formulas of a Compou...

1. Determine the mass of each element in a 100.00 g sample.

62.58 g C, 9.63 g H, 27.79 g O

2. Convert each of these masses to an amount in moles.

? mol C = 62.58 g C \times \frac{1  mol  C}{12.011  g  C} = 5.210 mol C

? mol H = 9.63 g H \times \frac{1  mol  H}{1.008  g  H} = 9.55 mol H

? mol O = 27.79 g O \times \frac{1  mol  O}{15.999  g  O} = 1.737 mol O

3. Write a tentative formula based on these numbers of moles.

C_{5.21}  H_{9.55}  O_{1.74}

4. Divide each of the subscripts of the tentative formula by the smallest subscript (1.74), and round off any subscripts that differ only slightly
from whole numbers; that is, round 2.99 to 3.

C_{\frac{5.21}{1.74}}  H_{\frac{9.55}{1.74}}  O_{\frac{1.74}{1.74}} = C_{2.99}  H_{5.49}  O

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 \times 3}  H_{2 \times 5.49}  O_{2 \times 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.