Problem 88 - Chemistry and Chemical Reactivity

Question

A 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 \(111 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O},\) what is the empirical formula of menthol?

Step-by-Step Solution

Answer

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

Step 1: Convert Mass of CO2 to Moles of Carbon

First, determine the moles of carbon by using the mass of carbon dioxide produced. The molar mass of \ \ \( \mathrm{CO}_{2} \) is \( 44.01 \, \mathrm{g/mol} \). \ \ Calculate the moles of \( \mathrm{CO}_{2} \): \ \[ \text{Moles of } \mathrm{CO}_{2} = \frac{269 \, \mathrm{mg}}{44.01 \, \mathrm{g/mol}} \times \frac{1 \, \mathrm{g}}{1000 \, \mathrm{mg}} \] \ \[ \text{Moles of } \mathrm{CO}_{2} = 0.00611 \, \mathrm{mol} \] \ Since each \( \mathrm{CO}_{2} \) molecule has one carbon atom, the moles of carbon are equal to the moles of \( \mathrm{CO}_{2} \): \ \[ \text{Moles of C} = 0.00611 \, \mathrm{mol} \] \ nThe mass of carbon is then: \ \[ \text{Mass of C} = 0.00611 \, \mathrm{mol} \times 12.01 \, \mathrm{g/mol} \approx 73.33 \, \mathrm{mg} \]

Step 2: Convert Mass of H2O to Moles of Hydrogen

To find moles of hydrogen, use the mass of water given. The molar mass of \( \mathrm{H}_{2} \mathrm{O} \) is approximately \( 18.02 \, \mathrm{g/mol} \). \ Calculate the moles of \( \mathrm{H}_{2} \mathrm{O} \): \ \[ \text{Moles of } \mathrm{H}_{2} \mathrm{O} = \frac{111 \, \mathrm{mg}}{18.02 \, \mathrm{g/mol}} \times \frac{1 \, \mathrm{g}}{1000 \, \mathrm{mg}} \] \ \[ \text{Moles of } \mathrm{H}_{2} \mathrm{O} = 0.00616 \, \mathrm{mol} \] \ As each \( \mathrm{H}_{2} \mathrm{O} \) contains two hydrogen atoms, \ \[ \text{Moles of H} = 0.00616 \, \mathrm{mol} \times 2 \approx 0.01232 \, \mathrm{mol} \] \ The mass of hydrogen is: \ \[ \text{Mass of H} = 0.01232 \, \mathrm{mol} \times 1.008 \, \mathrm{g/mol} \approx 12.42 \, \mathrm{mg} \]

Step 3: Determine Mass of Oxygen in Menthol

The original mass of menthol was \( 95.6 \, \mathrm{mg} \). We have found the mass of carbon (\( 73.33 \, \mathrm{mg} \)) and hydrogen (\( 12.42 \, \mathrm{mg} \)). Now, find the mass of oxygen by subtracting the mass of carbon and hydrogen from the total mass of menthol: \ \[ \text{Mass of O} = 95.6 \, \mathrm{mg} - (73.33 \, \mathrm{mg} + 12.42 \, \mathrm{mg}) \] \ \[ \text{Mass of O} = 9.85 \, \mathrm{mg} \] \ Convert this mass to moles of oxygen, using the molar mass of oxygen (\( 16.00 \, \mathrm{g/mol} \)): \ \[ \text{Moles of O} = \frac{9.85 \, \mathrm{mg}}{16.00 \, \mathrm{g/mol}} \times \frac{1 \, \mathrm{g}}{1000 \, \mathrm{mg}} \approx 0.000616 \, \mathrm{mol} \]

Step 4: Determine the Empirical Formula

Now that we have the moles of each element, the next step is to find the simplest wholenumber ratio. Divide each by the smallest number of moles obtained (which is for oxygen). \ - C: \[ \frac{0.00611 \, \mathrm{mol}}{0.000616 \, \mathrm{mol}} \approx 9.92 \] - H: \[ \frac{0.01232 \, \mathrm{mol}}{0.000616 \, \mathrm{mol}} \approx 20 \] - O: \[ \frac{0.000616 \, \mathrm{mol}}{0.000616 \, \mathrm{mol}} \approx 1 \] \ Round to the nearest whole number to get the simple ratio \( \mathrm{C}_{10} \mathrm{H}_{20} \mathrm{O} \). Therefore, the empirical formula is \( \mathrm{C}_{10} \mathrm{H}_{20} \mathrm{O} \).

Key Concepts

Combustion Analysis

Combustion analysis is a powerful technique used to determine the empirical formula of compounds that contain carbon, hydrogen, and sometimes oxygen. During combustion, the compound is burned in the presence of excess oxygen, producing carbon dioxide and water as combustion products. These products are then weighed and analyzed to find the amount of each element present in the original compound. This method is particularly useful because it allows chemists to analyze compounds that may not be easily distilled or analyzed by other methods.
In the context of menthol, when it is completely combusted, it forms carbon dioxide and water. By measuring the amounts of these products, we can deduce the amount of carbon and hydrogen in the original sample. Furthermore, stoichiometry and molar mass contribute to determining how many moles of oxygen are derived from the combustion, thereby enabling us to calculate the substance's full empirical formula. It's important to always ensure the measurements are accurate and done under controlled conditions for precise results.

Molecular Composition

Molecular composition refers to the types and numbers of atoms present in a compound. Understanding molecular composition is fundamental in chemistry as it provides the basis for determining how substances behave and react.
For menthol, composed solely of carbon, hydrogen, and oxygen, its combustion analysis allows us to precisely break down its molecular makeup into these elements. By quantifying the carbon and hydrogen from the masses of CO2 and H2O produced, we obtain a clearer picture of its composition. Additionally, determining the leftover mass as oxygen further elucidates the structure. Therefore, molecular composition is not just about deciphering the types of atoms present, but also understanding their arrangement relative to each other.

Stoichiometry

Stoichiometry involves the quantitative relationships within a chemical reaction. In combustion analysis, stoichiometry helps us understand how elements within the original compound are transformed into products.
By analyzing the moles of CO2 and H2O produced, we utilize stoichiometry to calculate the moles of each element in menthol. For example, each mole of CO2 corresponds to one mole of carbon because each CO2 molecule contains one carbon atom. Similarly, each mole of H2O corresponds to two moles of hydrogen. This understanding allows us to translate between the mass of a product and the number of moles, which is crucial to finding out the empirical formula. Practicing stoichiometry skills is essential for anyone studying chemistry, as it forms the basis for calculating yields in reactions, ensuring balanced equations, and even industrial chemical production.

Molar Mass Calculation

Molar mass calculations are a critical component in determining the empirical formula of a compound. Each element has a specific molar mass that is used in converting the measured mass of an element or compound to moles. When dealing with the combustion of menthol, converting the mass of CO2 and H2O into moles involves using the molar masses of CO2 (44.01 g/mol) and H2O (18.02 g/mol) respectively. These calculations are important because they allow us to determine how many moles of each element are present, which can then be used to find the simplest ratio of the elements in the compound, and thus, the empirical formula. Accurate molar mass calculation is crucial in chemistry as it aids in predicting how much product is formed in a chemical reaction or how much reactant is required to produce a certain amount of product. Recognizing the importance of molar mass calculations helps in achieving precise scientific experiments and valid data analysis.