91影视

To determine the empirical formula of toluene, we need to find the ratio of moles of carbon to hydrogen. Given the mass of CO鈧� and H鈧侽 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 CO鈧� is: \(44.01 \text{ g/mol} \) and for H鈧侽, it's \(18.02 \text{ g/mol} \). Moles of carbon in CO鈧� = \(\frac{5.86\text{ mg}}{44.01\text{ g/mol}} \times \frac{1\text{ g}}{1000\text{ mg}} \times \frac{1\text{ mol C}}{1\text{ mol CO}_2} = 1.33 \times 10^{-4} \text{mol C}\) Moles of hydrogen in H鈧侽 = \(\frac{1.37\text{ mg}}{18.02\text{ g/mol}} \times \frac{1\text{ g}}{1000\text{ mg}} \times \frac{2\text{ mol H}}{1\text{ mol H}_2\text{O}} = 1.52 \times 10^{-4} \text{mol H}\)

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):**

For part (b), we have the mass of CO鈧� and H鈧侽 after combusting a sample of menthol. We will determine the moles of C, H, and O in this sample. Moles of carbon in CO鈧� = \(\frac{0.2829\text{ g}}{44.01\text{ g/mol}} \times \frac{1\text{ mol C}}{1\text{ mol CO}_2} = 6.427 \times 10^{-3} \text{mol C}\) Moles of hydrogen in H鈧侽 = \(\frac{0.1159\text{ g}}{18.02\text{ g/mol}} \times \frac{2\text{ mol H}}{1\text{ mol H}_2\text{O}} = 12.87 \times 10^{-3} \text{mol H}\) 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 = \(\frac{0.0161\text{ g}}{16.00\text{ g/mol}} = 1.006 \times 10^{-3} \text{mol}\).

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 C鈧咹鈧佲們O.

Using the molar mass of menthol (156 g/mol), we can determine the molecular formula: Empirical formula mass of C鈧咹鈧佲們O = \(6\times12.01\text{ g/mol C} + 13\times1.008\text{ g/mol H} + 16.00\text{ g/mol O} = 100.18\text{ g/mol}\) 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 = C鈧塇鈧佲倝O (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 C鈧塇鈧佲倝O.