Question

(a) Combustion analysis of toluene, a common organic solvent, gives \(5.86 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(1.37 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O}\). If the compound contains only carbon and hydrogen, what is its empirical formula? (b) Menthol, the substance we can smell in mentholated cough drops, is composed of \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{O}\). A \(0.1005-\mathrm{g}\) sample of menthol is combusted, producing \(0.2829 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.1159 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\). What is the empirical formula for menthol? If menthol has a molar mass of \(156 \mathrm{~g} / \mathrm{mol}\), what is its molecular formula?

Short answer

The empirical formula of toluene is CH. The empirical formula of menthol is C₆H₁₃O, and its molecular formula is C₉H₁₉O.

Step-by-step solution

Calculate moles of C and H from the mass of CO₂ and H₂O produced

To determine the empirical formula of toluene, we need to find the ratio of moles of carbon to hydrogen. Given the mass of CO₂ and H₂O produced, we can first determine the moles of carbon and hydrogen. For this step, use the molar mass of carbon (12.01 g/mol) and the molar mass of hydrogen (1.008 g/mol). The molar mass of CO₂ is: \(44.01 \text{ g/mol} \) and for H₂O, it's \(18.02 \text{ g/mol} \). Moles of carbon in CO₂ = \(\frac{5.86\text{ mg}}{44.01\text{ g/mol}} \times \frac{1\text{ g}}{1000\text{ mg}} \times \frac{1\text{ mol C}}{1\text{ mol CO}_2} = 1.33 \times 10^{-4} \text{mol C}\) Moles of hydrogen in H₂O = \(\frac{1.37\text{ mg}}{18.02\text{ g/mol}} \times \frac{1\text{ g}}{1000\text{ mg}} \times \frac{2\text{ mol H}}{1\text{ mol H}_2\text{O}} = 1.52 \times 10^{-4} \text{mol H}\)

Determine the empirical formula

Now that we know the moles of carbon and hydrogen in the given sample of toluene, we can determine the ratio of these elements in the molecule, and thus, the empirical formula. Divide both moles by the smallest value so we can find the ratio between the elements. \( \frac{1.33\times 10^{-4}}{1.33\times 10^{-4}} : \frac{1.52\times 10^{-4}}{1.33\times 10^{-4}} = 1 : 1.14 \) As the ratio is close to 1:1 and compounds have whole numbers for their atomic ratios, we can assume that the empirical formula of toluene is CH. **Part (b):**

Calculate moles of C, H, and O in the sample of menthol

For part (b), we have the mass of CO₂ and H₂O after combusting a sample of menthol. We will determine the moles of C, H, and O in this sample. Moles of carbon in CO₂ = \(\frac{0.2829\text{ g}}{44.01\text{ g/mol}} \times \frac{1\text{ mol C}}{1\text{ mol CO}_2} = 6.427 \times 10^{-3} \text{mol C}\) Moles of hydrogen in H₂O = \(\frac{0.1159\text{ g}}{18.02\text{ g/mol}} \times \frac{2\text{ mol H}}{1\text{ mol H}_2\text{O}} = 12.87 \times 10^{-3} \text{mol H}\) Subtract the mass of C and H from the total mass of menthol to find the mass of O: \(0.1005\text{ g} - (6.427 \times 10^{-3}\text{mol C} \times 12.01\text{g/mol C} + 12.87 \times 10^{-3}\text{mol H} \times 1.008\text{g/mol H}) = 0.0161\text{ g O}\) Moles of oxygen = \(\frac{0.0161\text{ g}}{16.00\text{ g/mol}} = 1.006 \times 10^{-3} \text{mol}\).

Determine the empirical formula for menthol

Now that we know the moles of carbon, hydrogen, and oxygen in the menthol sample, we can determine the ratio of these elements in the molecule. Divide each moles by the smallest value so we can find the ratio. \( \frac{6.427 \times 10^{-3}}{1.006\times 10^{-3}} : \frac{12.87 \times 10^{-3}}{1.006\times 10^{-3}} : \frac{1.006\times 10^{-3}}{1.006\times 10^{-3}} = 6.39 : 12.80 : 1 \) Rounded to the nearest whole number, the ratio is \(6 : 13 : 1\), giving the empirical formula of menthol as C₆H₁₃O.

Determine the molecular formula for menthol

Using the molar mass of menthol (156 g/mol), we can determine the molecular formula: Empirical formula mass of C₆H₁₃O = \(6\times12.01\text{ g/mol C} + 13\times1.008\text{ g/mol H} + 16.00\text{ g/mol O} = 100.18\text{ g/mol}\) Molecular formula ratio = \(\frac{156\text{ g/mol}}{100.18 \text{ g/mol}} = 1.56\), which is approximately 1.5. As the molecular formula ratio is approximately 1.5, we can multiply the empirical formula by 1.5 to get the molecular formula: Molecular formula for menthol = C₉H₁₉O (as we can't have a fraction of an atom in a molecule, we assume the closest whole number) So, the molecular formula for menthol is C₉H₁₉O.