Problem 61 - Chemistry and Chemical Reactivity

Question

Menthol, from oil of mint, has a characteristic odor. The compound contains only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O} .\) If \(95.6 \mathrm{mg}\) of menthol burns completely in \(\mathrm{O}_{2},\) and gives \(269 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(110 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O},\) what is the empirical formula of menthol?

Step-by-Step Solution

Answer

The empirical formula is \( \mathrm{C}_{10} \mathrm{H}_{20} \mathrm{O} \).

Step 1: Convert Mass to Moles for CO2

First, we need to calculate the number of moles of carbon present in the produced 269 mg of CO2. The molar mass of CO2 is 44.01 g/mol. Convert the mass to grams (269 mg = 0.269 g) and use the formula: \( \ ext{moles of } \mathrm{CO}_2 = \frac{0.269 \text{ g}}{44.01 \text{ g/mol}} = 0.00611 \text{ mol} \). Since there is one carbon atom in each \( \mathrm{CO}_2 \) molecule, the moles of carbon = 0.00611 mol.

Step 2: Convert Mass to Moles for H2O

Next, calculate the number of moles of hydrogen from the 110 mg of water. The molar mass of H2O is 18.02 g/mol. Convert the mass to grams (110 mg = 0.110 g) and use: \( \ ext{moles of } \mathrm{H}_2 \mathrm{O} = \frac{0.110 \text{ g}}{18.02 \text{ g/mol}} = 0.00611 \text{ mol} \). Each H2O has two hydrogen atoms, so moles of hydrogen = 2 × 0.00611 = 0.01222 mol.

Step 3: Determine Mass of Oxygen in Menthol

Calculate the mass of carbon and hydrogen separately using the moles calculated:\( \text{mass of C} = 0.00611 \text{ mol} \times 12.01 \text{ g/mol} = 0.0734 \text{ g} \). \( \text{mass of H} = 0.01222 \text{ mol} \times 1.008 \text{ g/mol} = 0.0123 \text{ g} \). Convert these to mg (73.4 mg for C and 12.3 mg for H). The total mass of menthol is 95.6 mg, so the mass of oxygen is \( \text{mass of O} = 95.6 \text{ mg} - 73.4 \text{ mg} - 12.3 \text{ mg} = 9.9 \text{ mg} \).

Step 4: Convert Mass of Oxygen to Moles

The molar mass of oxygen is 16.00 g/mol. Convert the oxygen mass to grams (9.9 mg = 0.0099 g) and calculate moles of oxygen: \( \text{moles of O} = \frac{0.0099 \text{ g}}{16.00 \text{ g/mol}} = 0.000619 \text{ mol} \).

Step 5: Determine the Empirical Formula

Find the simplest whole-number mole ratio. Divide all mole quantities by the smallest number of moles calculated: For C: \( \frac{0.00611}{0.000619} \approx 10 \), for H: \( \frac{0.01222}{0.000619} \approx 20 \), and for O: \( \frac{0.000619}{0.000619} = 1 \). Therefore, the empirical formula is \( \mathrm{C}_{10} \mathrm{H}_{20} \mathrm{O} \).

Key Concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the calculation of reactants and products in chemical reactions. It is essential for determining the proportions of elements in compounds and reactions. The stoichiometric calculations allow chemists to predict the amount of each substance involved.
In our problem, stoichiometry helps us calculate the moles of carbon, hydrogen, and oxygen present in menthol after combustion. We begin by converting the masses of the products, like CO2 and H2O, into moles.
This involves using the molar masses of these compounds. For example, the molar mass of CO2 is used to determine the moles of carbon. Stoichiometry allows us to relate these moles back to the original substance, menthol, to find the empirical formula.
In short, these calculations tell us the simplest ratio of atoms in the compound, using the balanced chemical equations and conservation of mass.

Combustion Analysis

Combustion analysis is a method used to determine the elemental composition of a sample, especially organic compounds. By burning the substance in oxygen, we obtain CO2 and H2O, which are then measured to deduce the amounts of carbon and hydrogen in the original compound.
In the case of menthol, by burning it completely, we determine that it produces specific masses of CO2 and H2O. We can then convert these masses into moles to find out how much carbon and hydrogen were originally present in the menthol sample.
This analysis is crucial for identifying the empirical formula, as it allows us to backtrack from the combustion products to the initial composition of the compound.

Molecular Composition

Molecular composition refers to the types and numbers of atoms present in a molecule of a compound. It helps define the proportions of each element within the molecule's structure. Understanding molecular composition is critical in determining the empirical formula, which represents the simplest whole-number ratio of atoms present.
In our exercise, once we've calculated the moles of each element—carbon, hydrogen, and oxygen—from the combustion analysis, we can derive the molecular composition of menthol. This is fundamental in formulating its empirical formula.
By converting the experimental data from combustion into empirical formula ratios, we understand the basic structure and composition of menthol.

Chemical Reaction Calculation

Chemical reaction calculations involve the mathematical predictions on the amounts of different reactants and products that participate in chemical reactions. Through these calculations, chemists can determine how much of each substance is needed or produced.
In the context of menthol analysis, chemical reaction calculations allow us to deduce the resultant moles of CO2 and H2O after complete combustion. This forms the initial step in mapping back to the amounts of each element present before combustion occurred.
The ability to calculate these values precisely is vital for deriving the correct empirical formula. We make use of the known molar masses from the periodic table and the principle of conservation of mass to execute these calculations accurately.