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 \(0.1005\) - \(g\) 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, and the empirical formula of menthol is C₆H₁₃O. The molecular formula of menthol is also C₆H₁₃O.

Step-by-step solution

Calculate the moles of CO₂ and H₂O produced

First, convert the given mass of CO₂ and H₂O to moles using their respective molar masses. The molar mass of CO₂ is \( (12.01 + 2 \times 16.00) \ \mathrm{g/mol} = 44.01 \ \mathrm{g/mol} \) and that of H₂O is \( (2 \times 1.01 + 16.00) \ \mathrm{g/mol} = 18.02 \ \mathrm{g/mol} \). Moles of CO₂ produced = \( \frac{5.86 \ \mathrm{mg}}{44.01 \ \mathrm{g/mol} \times \frac{1 \ \mathrm{g}}{1000 \ \mathrm{mg}}} = 1.33 \times 10^{-4} \ \mathrm{mol} \) Moles of H₂O produced = \( \frac{1.37 \ \mathrm{mg}}{18.02 \ \mathrm{g/mol} \times \frac{1 \ \mathrm{g}}{1000 \ \mathrm{mg}}} = 7.60 \times 10^{-5} \ \mathrm{mol} \)

Calculate the moles of carbon and hydrogen in toluene

Since each CO₂ molecule has 1 carbon atom (C), and each H₂O molecule has 2 hydrogen atoms (H), the moles of carbon and hydrogen in toluene can be calculated as follows: Moles of carbon = Moles of CO₂ = \( 1.33 \times 10^{-4} \ \mathrm{mol} \) Moles of hydrogen = 2 × Moles of H₂O = \( 2 \times 7.60 \times 10^{-5} \ \mathrm{mol} = 1.52 \times 10^{-4} \ \mathrm{mol} \)

Determine the empirical formula

Divide both quantities by the lowest amount to find the whole-number ratio of carbon and hydrogen atoms and then write the empirical formula. Carbon-to-hydrogen ratio = \( \frac{1.33 \times 10^{-4}}{1.52 \times 10^{-4}} = 0.88 \) Since the ratio is approximately 1:1, the empirical formula of toluene is CH. b) For menthol combustion analysis:

Calculate the moles of CO₂ and H₂O produced

Using the same method as in part a, convert the given mass of CO₂ and H₂O to moles. Moles of CO₂ produced = \( \frac{0.2829 \ \mathrm{g}}{44.01 \ \mathrm{g/mol}} = 6.43 \times 10^{-3} \ \mathrm{mol} \) Moles of H₂O produced = \( \frac{0.1159 \ \mathrm{g}}{18.02 \ \mathrm{g/mol}} = 6.43 \times 10^{-3} \ \mathrm{mol} \)

Calculate the moles of carbon and hydrogen in menthol

Again, since each CO₂ molecule contains 1 carbon atom and each H₂O molecule has 2 hydrogen atoms: Moles of carbon = Moles of CO₂ = \( 6.43 \times 10^{-3} \ \mathrm{mol} \) Moles of hydrogen = 2 × Moles of H₂O = \( 2 \times 6.43 \times 10^{-3} \ \mathrm{mol} = 1.29 \times 10^{-2} \ \mathrm{mol} \)

Calculate the moles of oxygen in menthol

Determine the moles of oxygen in menthol by subtracting the moles of carbon and hydrogen from the total moles (by using the given mass of menthol): Moles of menthol = \( \frac{0.1005 \ \mathrm{g}}{156 \ \mathrm{g/mol}} = 6.44 \times 10^{-4} \ \mathrm{mol} \) Moles of oxygen = Moles of menthol - Moles of carbon - Moles of hydrogen = \( 6.44 \times 10^{-4} - 6.43 \times 10^{-3} - 1.29 \times 10^{-2} = 9.97 \times 10^{-4} \ \mathrm{mol} \)

Determine the empirical formula

Divide all moles of elements (carbon, hydrogen, oxygen) by the smallest quantity to find the whole-number ratio: C:H:O ratio = \( \frac{6.43 \times 10^{-3}}{9.97 \times 10^{-4}} \) : \( \frac{1.29 \times 10^{-2}}{9.97 \times 10^{-4}} \) : 1 ≈ 6:13:1 The empirical formula of menthol is C₆H₁₃O.

Determine the molecular formula

Now we use the molar mass of menthol (156 g/mol) to find the molecular formula. Divide the molar mass of menthol by the molar mass of the empirical formula (C₆H₁₃O has a molar mass of 100.2 g/mol). \( \frac{156 \ \mathrm{g/mol}}{100.2 \ \mathrm{g/mol}} ≈ 1.56 \) Since the ratio is nearly 1, the molecular formula of menthol is the same as its empirical formula: C₆H₁₃O.

Key concepts

Combustion Analysis

Combustion analysis is a procedure used in chemistry to determine the elemental composition of a substance. In essence, the unknown substance is burned in the presence of excess oxygen, resulting in the formation of carbon dioxide (CO₂) and water (H₂O) for substances containing carbon and hydrogen, respectively. By measuring the masses of CO₂ and H₂O produced during combustion, we can backtrack to find the amount of carbon and hydrogen in the original sample.

This method is particularly effective due to its simplicity and the accuracy with which we can measure the products of combustion. The masses of CO₂ and H₂O are converted to moles using their molar masses, and since each molecule of CO₂ contains exactly one carbon atom and each molecule of H₂O contains two hydrogen atoms, the moles of carbon and hydrogen in the original sample can be deduced directly.

Molecular Formula

The molecular formula of a compound provides the actual number of atoms of each element present in a molecule of the compound. It is a direct extension of the empirical formula, which shows only the simplest whole-number ratio of atoms within the compound. To determine the molecular formula, we need to know the empirical formula and the molar mass of the compound.

Obtaining the molar mass of the compound is usually done through experimental means such as mass spectrometry or from the given substance's empirical formula and its molar mass. Once we have the molar mass of the compound and the empirical formula's molar mass, the molecular formula can be determined by finding the ratio of the compound's molar mass to the empirical formula's molar mass. If this ratio is an integer or nearly an integer, it can be used to multiply the subscripts in the empirical formula to obtain the molecular formula.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It is a fundamental concept in chemistry that ensures the balance of atoms according to the Law of Conservation of Mass. In stoichiometry, the relationships between the amounts of reactants and products are determined by using the balanced chemical equation.

For a combustion analysis, stoichiometry allows us to relate the moles of carbon dioxide and water produced back to the moles of carbon and hydrogen present in the original compound. By applying stoichiometric principles, we ensure that calculations related to chemical reactions reflect the precise quantitative relationship between reactants and products, enabling us to derive empirical and molecular formulas accurately.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It serves as a bridge between the microscopic world of atoms and molecules and the macroscopic world of grams and liters that we can measure in the lab. The molar mass of a compound is determined by summing the individual molar masses of each element, weighted by the number of atoms of each element present in the compound's formula.

In the context of combustion analysis, molar mass plays a critical role because it allows us to convert the measured masses of carbon dioxide and water into moles, which are needed to calculate the empirical formula. The precise determination of molar mass is crucial for accurate stoichiometric calculations in chemical reactions and is essential for bridging the gap between theoretical concepts and practical laboratory results.