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 CH2. The empirical formula of menthol is C₁₀H₂₀O, and its molecular formula is also C₁₀H₂₀O.

Step-by-step solution

Find the moles of carbon and hydrogen in toluene

To find the moles of carbon (C) in toluene, we will use the mass of CO2 produced during combustion. The mass of CO2 produced is 5.86 mg. Now, we need to convert that mass into moles by dividing it by the molar mass of CO2 (44.01 g/mol). Keep in mind that 1 gram equals 1000 milligrams. So, we get: moles of C = \(\frac{5.86 \mathrm{mg} \times \frac{1 \mathrm{g}}{1000 \mathrm{mg}}}{44.01 \mathrm{g/mol}}\) = 0.0001332 mol Similarly, we find the moles of hydrogen (H) using the mass of H2O produced (1.37 mg) and the molar mass of H2O (18.02 g/mol): moles of H = \(\frac{1.37 \mathrm{mg} \times \frac{1 \mathrm{g}}{1000 \mathrm{mg}}}{18.02 \mathrm{g/mol}}\) = 0.0000760 mol

Determine the empirical formula

To find the empirical formula, we take the ratio of moles of H to moles of C: \(\frac{0.0000760 \ \mathrm{mol} \ \mathrm{H}}{0.0001332 \ \mathrm{mol} \ \mathrm{C}}\) ≈ 0.57 From this, we can determine that the empirical formula contains approximately two hydrogen atoms for every carbon atom. Therefore, the empirical formula of toluene is CH2. (b)

Find the moles of carbon, hydrogen, and oxygen in menthol

First, let's find the moles of carbon (C) and hydrogen (H) using the same process as before: moles of C = \(\frac{0.2829 \mathrm{g}}{44.01 \mathrm{g/mol}}\) = 0.00643 mol moles of H = \(\frac{0.1159 \mathrm{g}}{18.02 \mathrm{g/mol}}\) = 0.00643 mol To find the moles of oxygen (O), we will use the molar mass of menthol and the given sample mass. Based on the results for C and H, we know that the remaining mass is due to O. So, we subtract the masses of C and H from the mass of the sample and then divide by the molar mass of oxygen (16.00 g/mol) to find the moles: Mass of O = 0.1005 g - (0.00643 mol C × 12.01 g/mol C) - (0.00643 mol H × 1.008 g/mol H) = 0.0148 g moles of O = \(\frac{0.0148 \mathrm{g}}{16.00 \mathrm{g/mol}}\) = 0.000926 mol

Determine the empirical formula

Now let's find the ratio of moles of H, C, and O: \(\frac{\mathrm{moles \ of} \ \mathrm{H}}{\mathrm{moles \ of} \ \mathrm{C}} = \frac{0.00643 \ \mathrm{mol}}{0.00643 \ \mathrm{mol}}\) = 1 \(\frac{\mathrm{moles \ of} \ \mathrm{O}}{\mathrm{moles \ of} \ \mathrm{C}} = \frac{0.000926 \ \mathrm{mol}}{0.00643 \ \mathrm{mol}}\) ≈ 0.144 The empirical formula of menthol is then C₁₀H₂₀O (since 0.144 is close to 1/7, and 10 is the smallest integer multiple that allows this ratio).

Find the molecular formula

To determine the molecular formula, we need to compare the empirical formula molar mass to the given molar mass of menthol. The molar mass of the empirical formula is: C₁₀H₂₀O = (10 × 12.01 g/mol C) + (20 × 1.008 g/mol H) + (1 × 16.00 g/mol O) = 152.3 g/mol Now, divide the given molar mass of menthol (156 g/mol) by the empirical formula molar mass (152.3 g/mol) to get a scaling factor: Scaling factor = \(\frac{156 \mathrm{\ g/mol}}{152.3 \mathrm{\ g/mol}}\) ≈ 1.02 Since the scaling factor is approximately 1, the molecular formula for menthol is therefore the same as the empirical formula: C₁₀H₂₀O.

Key concepts

Combustion Analysis

Combustion analysis is a fundamental technique used in chemistry to determine the empirical formula of an organic compound. This method involves burning the compound in the presence of excess oxygen to convert all of its carbon into carbon dioxide (\(\mathrm{CO}_2\)) and all hydrogen into water (\(\mathrm{H}_2\mathrm{O}\)).

By measuring the amounts of \(\mathrm{CO}_2\) and \(\mathrm{H}_2\mathrm{O}\) produced, we can deduce the amounts of carbon and hydrogen that were originally present in the compound. This process lets us establish the ratio of carbon and hydrogen atoms in the compound, leading to its empirical formula.

For instance, consider a combustion analysis where 5.86 mg of \(\mathrm{CO}_2\) and 1.37 mg of \(\mathrm{H}_2\mathrm{O}\) are produced. You first convert these masses into moles using their respective molar masses. For \(\mathrm{CO}_2\), the molar mass is 44.01 g/mol, and for \(\mathrm{H}_2\mathrm{O}\), it is 18.02 g/mol. This conversion reveals the moles of carbon and hydrogen present and ultimately gives us the necessary data to determine the empirical formula of the compound.

Moles Calculation

In chemistry, calculating moles is a crucial step in finding the empirical or molecular formulas of compounds. A mole is a basic unit in chemistry that represents an amount of a substance containing 6.022 x 10²³ constituent particles, usually atoms or molecules.

To calculate the number of moles from a given mass, divide the mass of the substance by its molar mass (grams per mole). This formula is expressed as:

• \( \text{Moles} = \frac{\text{Mass in grams}}{\text{Molar mass in g/mol}} \)

Applying this to a situation like combustion analysis, you would use the given mass of \(\mathrm{CO}_2\) or \(\mathrm{H}_2\mathrm{O}\), along with their molar masses, to determine the moles of carbon or hydrogen.

Consider menthol, for example. When 0.2829 g of \(\mathrm{CO}_2\) is produced, you convert it to moles to find the number of carbon atoms. Similarly, converting the 0.1159 g of \(\mathrm{H}_2\mathrm{O}\) to moles gives you the number of hydrogen atoms. Such calculations help to establish the mole ratio needed to construct empirical formulas.

Organic Compounds Analysis

Organic compounds are primarily composed of three elements: carbon, hydrogen, and often oxygen. Analyzing these compounds involves determining which elements are present and in what proportions.

In the context of combustion analysis, you burn the compound to convert all carbon to \(\mathrm{CO}_2\) and hydrogen to \(\mathrm{H}_2\mathrm{O}\). If oxygen is also present, its amount is often determined indirectly by subtracting the mass of carbon and hydrogen from the total mass of the compound.

This approach was used in the analysis of menthol's composition. Combining the mole calculations for carbon and hydrogen with the remaining mass converted to moles for oxygen provides the full picture of the compound's composition. The empirical formula reflects these mole ratios in the smallest whole-number terms. In the case of menthol, after determining the molecule's composition, the empirical formula of \(\mathrm{C}_{10}\mathrm{H}_{20}\mathrm{O}\) was deduced, and the molecular formula turned out to be the same due to the similar mass potential.