Question

Menthol is a flavoring agent extracted from peppermint oil. It contains \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O} .\) In one combustion analysis, \(10.00 \mathrm{mg}\) of the substance yields \(11.53 \mathrm{mg} \mathrm{H}_{2} \mathrm{O}\) and \(28.16 \mathrm{mg} \mathrm{CO}_{2}\). What is the empirical formula of menthol?

Short answer

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

Step-by-step solution

Calculate moles of CO2 produced

First, we need to find out how many moles of \( \mathrm{CO}_2 \) are produced from the combustion of 10.00 mg of menthol. The molar mass of \( \mathrm{CO}_2 \) is 44.01 g/mol. Convert the mass of \( \mathrm{CO}_2 \) to grams: \( 28.16 \: \mathrm{mg} = 0.02816 \: \mathrm{g} \)Calculate moles by dividing by molar mass: \[ \text{moles of } \mathrm{CO}_2 = \frac{0.02816 \: \mathrm{g}}{44.01 \: \mathrm{g/mol}} = 6.40 \times 10^{-4} \: \mathrm{mol} \]

Calculate moles of H2O produced

Now, find out how many moles of \( \mathrm{H}_2\mathrm{O} \) are produced. The molar mass of \( \mathrm{H}_2\mathrm{O} \) is 18.02 g/mol.Convert the mass of \( \mathrm{H}_2\mathrm{O} \) to grams: \( 11.53 \: \mathrm{mg} = 0.01153 \: \mathrm{g} \)Calculate moles by dividing by molar mass:\[ \text{moles of } \mathrm{H}_2\mathrm{O} = \frac{0.01153 \: \mathrm{g}}{18.02 \: \mathrm{g/mol}} = 6.40 \times 10^{-4} \: \mathrm{mol} \]

Determine moles of carbon from CO2

Each mole of \( \mathrm{CO}_2 \) contains one mole of carbon. Therefore, the moles of \( \mathrm{C} \) is equal to the moles of \( \mathrm{CO}_2 \).\[ \text{moles of C} = 6.40 \times 10^{-4} \: \mathrm{mol} \]

Determine moles of hydrogen from H2O

Each mole of \( \mathrm{H}_2\mathrm{O} \) contains two moles of hydrogen. Therefore, the moles of \( \mathrm{H} \) is twice the moles of \( \mathrm{H}_2\mathrm{O} \).\[ \text{moles of H} = 2 \times 6.40 \times 10^{-4} \: \mathrm{mol} = 1.28 \times 10^{-3} \: \mathrm{mol} \]

Find moles of oxygen in menthol

First, determine the mass of carbon and hydrogen:- Mass of \( \mathrm{C} \) = moles of \( \mathrm{C} \) \( \times \) molar mass of \( \mathrm{C} \) = \( 6.40 \times 10^{-4} \: \mathrm{mol} \times 12.01 \: \mathrm{g/mol} = 7.69 \times 10^{-3} \: \mathrm{g} \)- Mass of \( \mathrm{H} \) = moles of \( \mathrm{H} \) \( \times \) molar mass of \( \mathrm{H} \) = \( 1.28 \times 10^{-3} \: \mathrm{mol} \times 1.008 \: \mathrm{g/mol} = 1.29 \times 10^{-3} \: \mathrm{g} \)Mass of oxygen in menthol can be calculated by subtracting mass of C and H from total mass: \[ \text{mass of O} = 0.0100 \: \mathrm{g} - (7.69 \times 10^{-3} \: \mathrm{g} + 1.29 \times 10^{-3} \: \mathrm{g}) = 1.02 \times 10^{-3} \: \mathrm{g} \]

Calculate moles of oxygen

Convert the mass of oxygen to moles using its molar mass (16.00 g/mol):\[ \text{moles of O} = \frac{1.02 \times 10^{-3} \: \mathrm{g}}{16.00 \: \mathrm{g/mol}} = 6.38 \times 10^{-5} \: \mathrm{mol} \]

Determine empirical formula

Now we have the moles of C, H, and O:- C: \( 6.40 \times 10^{-4} \: \mathrm{mol} \)- H: \( 1.28 \times 10^{-3} \: \mathrm{mol} \)- O: \( 6.38 \times 10^{-5} \: \mathrm{mol} \)Divide each by the smallest amount of moles (oxygen) to get the ratio:- C: \( \frac{6.40 \times 10^{-4}}{6.38 \times 10^{-5}} = 10 \)- H: \( \frac{1.28 \times 10^{-3}}{6.38 \times 10^{-5}} = 20 \)- O: \( \frac{6.38 \times 10^{-5}}{6.38 \times 10^{-5}} = 1 \)The empirical formula is \( \mathrm{C}_{10}\mathrm{H}_{20}\mathrm{O} \).

Key concepts

Combustion Analysis

Combustion analysis is a vital experimental procedure used to determine the composition of chemical compounds containing carbon, hydrogen, and often oxygen. The process typically involves burning a known mass of a substance and measuring the amounts of carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)) produced. By capturing these products, we can deduce the content and ratio of elements such as carbon and hydrogen in the original substance. In mentoring, combustion analysis helped ascertain the quantities of both carbon and hydrogen by relating the masses of \(\text{CO}_2\) and \(\text{H}_2\text{O}\) to the moles of these elements. Each molecule of \(\text{CO}_2\) represents one atom of carbon, and each \(\text{H}_2\text{O}\) molecule contains two atoms of hydrogen. This straightforward relationship allows chemists to calculate the necessary ratios needed to determine the empirical formula, providing a glimpse into the chemical structure of a compound.

Molar Mass

Molar mass, also known as molecular weight, is the mass of one mole of a given substance expressed in grams per mole. This crucial concept links the macroscopic property of mass with the amount of substance measured in moles, making it fundamental in stoichiometric calculations. You can find the molar mass of any chemical compound by summing the weighted molar masses of its constituent elements.For example, in the case of menthol combustion, the molar masses of \(\text{CO}_2\) (44.01 g/mol) and \(\text{H}_2\text{O}\) (18.02 g/mol) were used to convert the measured masses in milligrams to moles. This conversion was pivotal in obtaining the relative number of moles of each element involved, which was then used to determine the empirical formula. Understanding molar mass is essential in translating between the mass and amount of substance, a crucial step in accurately determining the chemical composition.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the substances involved in reactions, based on their chemical equations. It allows us to predict the amounts of reactants needed and the products formed in a chemical reaction. By using the coefficients from a balanced chemical equation, stoichiometry provides a map to navigate through the quantitative aspects of chemical transformations.When solving for the empirical formula of menthol, stoichiometry played a key role. After determining the moles of \(\text{CO}_2\) and \(\text{H}_2\text{O}\), stoichiometry was used to calculate the moles of carbon and hydrogen present in menthol. These moles were then normalized to the smallest number of moles (in this case, oxygen) to establish a simple whole-number ratio among carbon, hydrogen, and oxygen atoms, leading to the empirical formula of \(\text{C}_{10}\text{H}_{20}\text{O}\).

Chemical Composition

Chemical composition refers to the identity and relative number of atoms in a substance, providing valuable information about the formula and structure of the compound. It encompasses the type and amount of each element present in a chemical species and is usually expressed as a molecular or empirical formula.The empirical formula is a fundamental concept that represents the simplest whole-number ratio of elements in a compound. In the example of menthol, its composition was deduced by calculating the relative moles of carbon, hydrogen, and oxygen derived from the product of combustion analysis. Through careful analysis and calculation, we determined that menthol’s empirical formula is \(\text{C}_{10}\text{H}_{20}\text{O}\), reflecting its basic chemical makeup. Understanding chemical composition allows scientists to infer the molecular structure and potential reactivity of compounds.