Question

A compound containing only C, H, and O, was extracted from the bark of the sassafras tree. The combustion of 58.7 mg produced 159 mg of CO2 and 32.6 mg of H2O. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas.

Answer 1

Answer on Question #78095, Chemistry / General Chemistry

A compound containing only C, H, and O, was extracted from the bark of the sassafras tree. The combustion of 58.7 mg produced 159 mg of CO2 and 32.6 mg of H2O. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas.

Solution

At first find the amounts of the compound, $\mathrm{CO}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$:

$$\begin{array}{l} v_{\text{comp}} = \frac{0.0587}{162} = 3.62 \times 10^{-4} \, (\text{mol}) \\ v_{\text{CO}_2} = \frac{0.159}{44} = 3.61 \times 10^{-3} \, (\text{mol}) \\ v_{\text{H}_2\text{O}} = \frac{0.0326}{18} = 1.81 \times 10^{-3} \, (\text{mol}) \end{array}$$

The equation of the reaction of combustion is

$$\begin{array}{l} 1 \, \mathrm{C}_x\mathrm{H}_y\mathrm{O}_2 + \mathrm{O}_2 \rightarrow x\mathrm{CO}_2 + 0.5Y\mathrm{H}_2\mathrm{O} \\ \text{If } 3.62 \times 10^{-4} - 1 \, \text{mole} \\ 3.61 \times 10^{-3} - X \\ X = 10 \\ \text{If } 3.62 \times 10^{-4} - 1 \, \text{mole} \\ 1.81 \times 10^{-3} - 0.5Y \\ Y = 10 \\ \end{array}$$

So the compound has 10 atoms of Carbon and Hydrogen

$$\begin{array}{l} M \left(\mathrm{C}_{10}\mathrm{H}_{10}\right) = 12 \times 10 + 1 \times 10 = 130 \\ M \left(\mathrm{O}_2\right) = 162 - 130 = 32 \\ Z = 32 / 16 = 2 \\ \end{array}$$

Answer

The molecular formula is $\mathrm{C}_{10}\mathrm{H}_{10}\mathrm{O}_2$.

The empirical formula is $\mathrm{C}_5\mathrm{H}_5\mathrm{O}$.