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Chapter 3: Problem 55

(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 \(0.1005-g\) 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

Expert verified
The empirical formula for toluene is CH. The empirical formula for menthol is C₆H₁₂O, and its molecular formula is C₁₂H₂₄O₂.

Step by step solution

01

Calculate moles of CO₂ and H₂O produced

Using the given mass of CO₂ and H₂O, calculate the moles of each by dividing the mass by its molar mass (44.01 g/mol for CO₂ and 18.02 g/mol for H₂O). \(moles\: of\: CO_{2} = \frac{5.86\:mg}{44.01\:g/mol} = 1.332 \times 10^{-4} mol\) \(moles\: of\: H_{2}O = \frac{1.37\:mg}{18.02\:g/mol} = 7.603 \times 10^{-5} mol\)
02

Calculate moles of C and H in Toluene

Every CO₂ molecule contains 1 carbon atom, and every H₂O molecule contains 2 hydrogen atoms. We can use this information to calculate the moles of carbon and hydrogen in toluene. \(moles\: of\: C = 1.332 \times 10^{-4} mol\) \(moles\: of\: H = 2 \times (7.603 \times 10^{-5} mol) = 1.52 \times 10^{-4} mol\)
03

Find the ratio of C to H in Toluene

Divide the moles of C and H by the smallest value (in this case, moles of C) to find the ratio. \(ratio\: of\: C\: to\: H = \frac{1.332 \times 10^{-4} mol}{1.332 \times 10^{-4} mol} : \frac{1.52 \times 10^{-4} mol}{1.332 \times 10^{-4} mol} = 1:1.14 \approx 1:1\)
04

Determine the Empirical Formula of Toluene

Based on the ratio obtained in Step 3, the empirical formula for toluene is CH. #(b) Finding the Empirical and Molecular Formula of Menthol#
05

Calculate moles of CO₂ and H₂O produced

Using the given mass of CO₂ and H₂O, calculate the moles of each as done in part (a). \(moles\: of\: CO_{2} = \frac{0.2829\:g}{44.01\:g/mol} = 0.00643 mol\) \(moles\: of\: H_{2}O = \frac{0.1159\:g}{18.02\:g/mol} = 0.00643 mol\)
06

Calculate moles of C and H in Menthol

Use the same method applied in part (a) to determine the moles of carbon and hydrogen in menthol. \(moles\: of\: C = 0.00643 mol\) \(moles\: of\: H = 2 \times 0.00643 mol = 0.01286 mol\)
07

Calculate moles of O in Menthol

Subtract the mass of carbon and hydrogen from the total mass of menthol to get the mass of oxygen. Then, calculate the moles of oxygen. \(mass\: of\: O = 0.1005\:g - \{(0.00643\:mol \times 12.01\:g/mol) + (0.01286\:mol \times 1.008\:g/mol)\} = 0.01712\:g\) \(moles\: of\: O = \frac{0.01712\:g}{16.00\:g/mol} = 0.00107\:mol\)
08

Find the ratio of C, H, and O in Menthol

Divide the moles of C, H, and O by the smallest value (in this case, moles of O) to find the ratio. \(ratio\: of\: C:H:O = \frac{0.00643\:mol}{0.00107\:mol} : \frac{0.01286\:mol}{0.00107\:mol} : \frac{0.00107\:mol}{0.00107\:mol} = 6:12:1\)
09

Determine the Empirical Formula of Menthol

Based on the ratio obtained in Step 4, the empirical formula for menthol is C₆H₁₂O.
10

Find the Molecular Formula of Menthol

To find the molecular formula, divide the molar mass of menthol by the molar mass of the empirical formula, and then multiply the empirical formula by this value. \( \text{Molar mass of the empirical formula} = 6 \times 12.01\:g/mol + 12 \times 1.008\:g/mol + 16.00\:g/mol = 100.2\:g/mol \) \( \frac{\text{Molar mass of menthol}}{\text{Molar mass of the empirical formula}} = \frac{156\:g/mol}{100.2\:g/mol} \approx 1.56 \approx 2 \) The molecular formula for menthol is 2 times the empirical formula, so it is C₁₂H₂₄O₂.

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Most popular questions from this chapter

When hydrocarbons are burned in a limited amount of air, both \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) form. When \(0.450 \mathrm{~g}\) of a particular hydrocarbon was burned in air, \(0.467 \mathrm{~g}\) of $\mathrm{CO}, 0.733 \mathrm{~g}\( of \)\mathrm{CO}_{2},\( and \)0.450 \mathrm{~g}\( of \)\mathrm{H}_{2} \mathrm{O}$ were formed. (a) What is the empirical formula of the compound? (b) How many grams of \(\mathrm{O}_{2}\) were used in the reaction? (c) How many grams would have been required for complete combustion?

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\(\mathrm{NO}_{x}\) is a generic term for the nitrogen oxides, \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) \(\mathrm{NO}_{x}\) gases are air pollutants that react to form smog and acid rain. In order to reduce \(\mathrm{NO}_{x}\) emission from vehicle, catalytic converters are installed in car exhausts to decompose NO and \(\mathrm{NO}_{2}\) respectively into \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}(\mathbf{a})\) Write the balanced chemical equations for the decomposition of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) respectively. (b) If the car produces \(100 \mathrm{~g} \mathrm{NO}_{x}\) a day, with equal mole ratio of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\), how many grams of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) are produced respectively?

The allowable concentration level of vinyl chloride, $\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl},\( in the atmosphere in a chemical plant is \)2.0 \times 10^{-6} \mathrm{~g} / \mathrm{L}$. How many moles of vinyl chloride in each liter does this represent? How many molecules per liter?

Balance the following equations: (a) $\mathrm{SiCl}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Si}(\mathrm{OH})_{4}(s)+\mathrm{HCl}(a q)$ (b) $\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+\mathrm{O}_{2}(g)$ (c) $\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)$ (d) $\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$
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