Problem 90 - 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 \(111 \mathrm{mg}\) 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: Determine Moles of Carbon in CO2

First, calculate how much carbon comes from the produced \( \text{CO}_2 \). We know the molar mass of \( \text{CO}_2 \) is \( 44.01 \text{ g/mol} \), and each mole contains 12.01 g of carbon (the atomic weight of carbon). To find moles of \( \text{CO}_2 \), use: \[ \text{Moles of } \text{CO}_2 = \frac{269 \, \text{mg of } \text{CO}_2}{44.01 \, \text{mg/mmol}} = 6.11 \, \text{mmol} \] Then, calculate moles of carbon: \[ \text{Moles of C} = 6.11 \, \text{mmol} \]

Step 2: Calculate Mass of Carbon

Using the moles of carbon calculated, convert it to mass. Each mole of carbon has a mass of 12.01 mg. Therefore, mass of carbon is: \[ \text{Mass of C} = 6.11 \, \text{mmol} \times 12.01 \, \text{mg/mmol} = 73.38 \, \text{mg} \]

Step 3: Determine Moles of Hydrogen in H2O

For \( \text{H}_2\text{O} \), with a molar mass of 18.02 g/mol, each mole contains 2 moles of hydrogen. Calculate the moles of \( \text{H}_2\text{O} \): \[ \text{Moles of H}_2\text{O} = \frac{111 \, \text{mg}}{18.02 \, \text{mg/mmol}} = 6.16 \, \text{mmol} \] Then calculate moles of hydrogen (2 moles per molecule of water): \[ \text{Moles of H} = 2 \times 6.16 \, \text{mmol} = 12.32 \, \text{mmol} \]

Step 4: Calculate Mass of Hydrogen

Convert moles of hydrogen to mass. Mass for one mole of hydrogen is 1.008 mg. Therefore: \[ \text{Mass of H} = 12.32 \, \text{mmol} \times 1.008 \, \text{mg/mmol} = 12.42 \, \text{mg} \]

Step 5: Calculate Mass of Oxygen

Initially, we know the total mass of menthol is 95.6 mg. Subtract the masses of carbon and hydrogen: \[ \text{Mass of O} = 95.6 \, \text{mg} - (73.38 \, \text{mg C} + 12.42 \, \text{mg H}) = 9.8 \, \text{mg O} \]

Step 6: Determine Moles of Oxygen

Calculate moles of oxygen using its molar mass (16.00 mg/mmol): \[ \text{Moles of O} = \frac{9.8 \, \text{mg}}{16.00 \, \text{mg/mmol}} = 0.6125 \, \text{mmol} \]

Step 7: Calculate Empirical Formula Ratios

Now, find the empirical formula by dividing moles of each element by the smallest value among them: - C: \( 6.11/0.6125 ≈ 10 \) - H: \( 12.32/0.6125 ≈ 20 \) - O: \( 0.6125/0.6125 = 1 \) Thus, the empirical formula is \( \mathrm{C}_{10}\mathrm{H}_{20}\mathrm{O} \).

Key Concepts

Combustion Analysis

When a compound containing carbon, hydrogen, and oxygen is burned, it reacts completely with oxygen to produce carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)). This type of reaction is known as combustion analysis. By measuring the mass of \(\text{CO}_2\) and \(\text{H}_2\text{O}\) produced, we can deduce the chemical composition of the original substance. In this method, each gram of \(\text{CO}_2\) and \(\text{H}_2\text{O}\) gives direct insight into the amount of carbon and hydrogen present in the substance. By subtracting the summed masses of these elements from the original mass, we can also find the mass of oxygen in the compound.

Chemical Composition

Chemical composition refers to the identity and proportion of elements that comprise a molecule. To determine chemical composition, chemists use data from experiments such as combustion analysis. For a compound containing carbon, hydrogen, and oxygen, like menthol, the chemically essential information is embedded in the masses of the resultants \(\text{CO}_2\) and \(\text{H}_2\text{O}\). These masses inform the calculation of moles of each element, which are then used to establish the empirical formula, showcasing the simplest integer ratio of elements in a compound. Understanding chemical composition provides insights into the molecular structure and reactivity of a compound.

Stoichiometry

Stoichiometry is the balance of elements in a chemical reaction considering the conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction. In combustion analysis, stoichiometry helps us translate the experimental measurements into meaningful mole ratios. With a known amount of \(\text{CO}_2\) and \(\text{H}_2\text{O}\), we determine the moles of carbon and hydrogen present initially. Using these calculations assists chemists in discerning the amount of other elements, like oxygen, via the difference in mass. This method ensures that all calculations align with the stoichiometric coefficients of the chemical equation governing the reaction.

Molecular Weight

Molecular weight is a crucial concept in chemistry that represents the weight of a molecule based on the atomic masses of its constituent atoms. For many chemical analyses, knowing the molecular weights of compounds such as \(\text{CO}_2\) and \(\text{H}_2\text{O}\) is essential. In the context of combustion analysis, it allows for the conversion between mass in milligrams and moles, which is pivotal in determining the empirical formula. By using known molecular weights, we can calculate precisely how much of an element contributed to the formation of the products, thereby piecing together the molecular identity of the compound under investigation.

Chemical Reactions

Chemical reactions, specifically combustion reactions, are central to determining the empirical formula in experiments like the analysis of menthol. In a combustion reaction, organic matter reacts with oxygen to yield \(\text{CO}_2\) and \(\text{H}_2\text{O}\). These reactions release energy and enable the determination of the reactants' composition by analyzing the resulting gaseous products. Understanding these reactions aids in decoding how atoms interact and rearrange during the reaction, allowing researchers to backtrack from the products to the original reactant composition. This reversibility from products to reactants is a fundamental strategy in reactions to determine molecular architecture.