Question

Menthol \((\mathscr{A} \ell=156.3 \mathrm{~g} / \mathrm{mol}),\) the strong- smelling substance in many cough drops, is a compound of carbon, hydrogen, and oxygen. When \(0.1595 \mathrm{~g}\) of menthol was burned in a combustion apparatus, \(0.449 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.184 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) formed. What is menthol's molecular formula?

Short answer

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

Step-by-step solution

- Determine the moles of CO\textsubscript{2} and H\textsubscript{2}O

Calculate the moles of \ \( \text{CO}_2 \ \) and \ \( \text{H}_2O \ \) produced. Use the molar masses: \ \( \text{Molar mass of CO}_2 = 44.01 \text{ g/mol} \ \) and \ \( \text{Molar mass of H}_2O = 18.015 \text{ g/mol} \ \). \ \( n(\text{CO}_2) = \frac{0.449 \text{ g}}{44.01 \text{ g/mol}} \ \approx 0.0102 \text{ mol} \ \). \ \( n(\text{H}_2O) = \frac{0.184 \text{ g}}{18.015 \text{ g/mol}} \ \approx 0.0102 \text{ mol} \ \).

- Determine moles of C and H from CO\textsubscript{2} and H\textsubscript{2}O

Each mole of \ \( \text{CO}_2 \ \) contains 1 mole of carbon atoms. Each mole of \ \( \text{H}_2O \ \) contains 2 moles of hydrogen atoms. \ \( n(\text{C}) = n(\text{CO}_2) = 0.0102 \text{ mol} \ \). \ \( n(\text{H}) = 2 \times n(\text{H}_2O) = 2 \times 0.0102 = 0.0204 \text{ mol} \ \).

- Determine the mass of carbon and hydrogen

Calculate the mass of C and H using their atomic masses: \ \( \text{Atomic mass of C} = 12.01 \text{ g/mol} \ \) and \ \( \text{Atomic mass of H} = 1.008 \text{ g/mol} \ \). \ \( m(\text{C}) = 0.0102 \text{ mol} \times 12.01 \text{ g/mol} = 0.1224 \text{ g} \ \). \ \( m(\text{H}) = 0.0204 \text{ mol} \times 1.008 \text{ g/mol} = 0.0206 \text{ g} \ \).

- Determine the mass of oxygen

Use the total mass of menthol and subtract the mass of carbon and hydrogen to find the mass of oxygen. \ \( m(\text{O}) = 0.1595 \text{ g} - m(\text{C}) - m(\text{H}) = 0.1595 \text{ g} - 0.1224 \text{ g} - 0.0206 \text{ g} = 0.0165 \text{ g} \ \).

- Determine the moles of oxygen

Calculate the moles of oxygen using its atomic mass: \ \( \text{Atomic mass of O} = 16.00 \text{ g/mol} \ \). \ \( n(\text{O}) = \frac{0.0165 \text{ g}}{16.00 \text{ g/mol}} = 0.00103 \text{ mol} \ \).

- Determine the empirical formula

Find the ratio of moles of each element to the smallest number of moles. \ \( \text{Ratio of } \text{C} = \frac{0.0102}{0.00103} \ \approx 10 \ \). \ \( \text{Ratio of } \text{H} = \frac{0.0204}{0.00103} \ \approx 20 \ \). \ \( \text{Ratio of } \text{O} = \frac{0.00103}{0.00103} = 1 \ \). Thus, the empirical formula is \ \( \text{C}_{10}\text{H}_{20}\text{O} \ \).

- Determine the molecular formula

Compare the empirical formula mass with the molar mass of menthol. Empirical formula mass: \ \( 10 \times 12.01 \text{ g/mol} + 20 \times 1.008 \text{ g/mol} + 16.00 \text{ g/mol} = 156.3 \text{ g/mol} \ \). Since the empirical formula mass is the same as the molar mass, the molecular formula is \ \( \text{C}_{10}\text{H}_{20}\text{O} \ \).

Key concepts

Combustion Analysis

Combustion Analysis is a common method used to determine the elemental composition of a carbon-containing compound. During combustion, the compound burns in excess oxygen, producing carbon dioxide \(CO_2\) and water \(H_2O\). By measuring the mass of \(CO_2\) and \(H_2O\) produced, we can determine the amounts of carbon and hydrogen in the original compound.
This process helps in calculating the empirical formula of the compound.
For example, in the provided menthol problem, the masses of \(CO_2\) and \(H_2O\) produced were used to find the moles of carbon and hydrogen.
Start by calculating the moles of \(CO_2\) and \(H_2O\) produced using their respective molar masses. Here is the step wise working:
- \(n(\text{CO}_2) = \frac{0.449\text{ g}}{44.01\text{ g/mol}} \approx 0.0102\text{ mol}\)
- \(n(\text{H}_2O) = \frac{0.184\text{ g}}{18.015\text{ g/mol}} \approx 0.0102\text{ mol}\)
Each mole of \(CO_2\) contains one mole of carbon, and each mole of \(H_2O\) contains two moles of hydrogen.
It's crucial to correctly interpret these results to further calculate the empirical and molecular formulas.

Empirical Formula

The empirical formula of a compound represents the simplest whole-number ratio of elements present in it.
To find the empirical formula, first, calculate the moles of each element present. This involves using data from combustion analysis or other means.
In the menthol problem, moles of carbon (\(C\)) and hydrogen (\(H\)) were found through combustion analysis:
- Moles of C: 0.0102 mol
- Moles of H: 0.0204 mol
Next, the mass of oxygen (\(O\)) is calculated by subtracting the masses of carbon and hydrogen from the total mass of the compound.
Finally, the number of moles of each element is converted to the simplest ratio:
- Ratio of C: \( \frac{0.0102}{0.00103} \approx 10\)
- Ratio of H: \( \frac{0.0204}{0.00103} \approx 20\)
- Ratio of O: \( \frac{0.00103}{0.00103} = 1\)
These ratios give the empirical formula \(C_{10}H_{20}O\), which is the simplest representation of the compound's composition.

Mole Calculations

In chemistry, mole calculations are crucial for converting between the mass of a substance and the number of particles it contains.
The most fundamental mole calculation formula is:
\( n = \frac{m}{M} \)
where \(n\) is the number of moles, \(m\) is the mass of the substance, and \(M\) is the molar mass.
In our menthol problem, moles of \(CO_2\) and \(H_2O\) were calculated to find the amount of carbon and hydrogen:
- Moles of \(CO_2\): \( \frac{0.449 \text{ g}}{44.01 \text{ g/mol}} = 0.0102 \text{ mol} \)
- Moles of \(H_2O\): \( \frac{0.184 \text{ g}}{18.015 \text{ g/mol}} = 0.0102 \text{ mol} \)
These calculations allow further determination of the empirical formula and, subsequently, the molecular formula.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions based on the balanced chemical equations.
In the analysis of menthol, stoichiometry helps in determining the relationship between quantities of substances consumed and produced.
During combustion, menthol reacts to produce \(CO_2\) and \(H_2O\), and stoichiometric calculations are used to ensure the proportions align with the chemical equations.
This accuracy is essential for deriving the correct empirical and molecular formulas.
By understanding stoichiometry, students can ensure that their calculations align with chemical laws and provide an accurate picture of the compound's composition. For example, to find how much \(O\) present in menthol, we carefully calculate the moles as below:
\( n(\text{O}) = \frac{0.0165 \text{ g}}{16.00 \text{ g/mol}} = 0.00103 \text{ mol} \)