Problem 36 - Chemistry and Chemical Reactivity

Question

An unknown compound has the formula \(\mathrm{C}_{x} \mathrm{H}_{y} \mathrm{O}_{z}\) You burn \(0.1523 \mathrm{g}\) of the compound and isolate \(0.3718 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(0.1522 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula of the compound? If the molar mass is \(72.1 \mathrm{g} / \mathrm{mol},\) what is the molecular formula?

Step-by-Step Solution

Answer

The empirical and molecular formula of the compound is \(\mathrm{C}_{4}\mathrm{H}_{8}\mathrm{O}\).

Step 1: Calculate Moles of Carbon

First, determine the moles of carbon in the isolated \(\mathrm{CO}_{2}\). The molar mass of \(\mathrm{CO}_{2}\) is 44.01 g/mol.\[\text{Moles of } \mathrm{CO}_{2} = \frac{0.3718 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.00845 \, \text{mol}\]Each mole of \(\mathrm{CO}_{2}\) contains one mole of carbon, so the moles of carbon are:\[\text{Moles of C} = 0.00845 \, \text{mol}\]

Step 2: Calculate Moles of Hydrogen

Next, find the moles of hydrogen in \(\mathrm{H}_{2}\mathrm{O}\). The molar mass of \(\mathrm{H}_{2}\mathrm{O}\) is 18.02 g/mol.\[\text{Moles of } \mathrm{H}_{2}\mathrm{O} = \frac{0.1522 \, \text{g}}{18.02 \, \text{g/mol}} \approx 0.00845 \, \text{mol}\]Each mole of \(\mathrm{H}_{2}\mathrm{O}\) contains two moles of hydrogen, so:\[\text{Moles of H} = 2 \times 0.00845 \, \text{mol} = 0.0169 \, \text{mol}\]

Step 3: Calculate Mass and Moles of Oxygen in the Compound

The total mass of the compound is 0.1523 g. The mass of carbon and hydrogen combined is:\[\text{Mass of C} = 0.00845 \, \text{mol} \times 12.01 \, \text{g/mol} = 0.1014 \, \text{g}\]\[\text{Mass of H} = 0.0169 \, \text{mol} \times 1.008 \, \text{g/mol} = 0.0170 \, \text{g}\]Calculate oxygen's mass:\[\text{Mass of O} = 0.1523 \, \text{g} - (0.1014 \, \text{g} + 0.0170 \, \text{g}) = 0.0339 \, \text{g}\]Now, convert this mass into moles. The molar mass of oxygen is 16.00 g/mol:\[\text{Moles of O} = \frac{0.0339 \, \text{g}}{16.00 \, \text{g/mol}} = 0.00212 \, \text{mol}\]

Step 4: Determine Empirical Formula

Find the mole ratio by dividing each mole value by the smallest number obtained in Step 3 (0.00212 mol):\[\frac{0.00845}{0.00212} \approx 3.99 \approx 4 \, \text{for C}\]\[\frac{0.0169}{0.00212} \approx 7.98 \approx 8 \, \text{for H}\]\[\frac{0.00212}{0.00212} = 1 \, \text{for O}\]Thus, the empirical formula is \(\mathrm{C}_{4}\mathrm{H}_{8}\mathrm{O}\).

Step 5: Find Molecular Formula Using Molar Mass

Calculate the molar mass of the empirical formula \(\mathrm{C}_{4}\mathrm{H}_{8}\mathrm{O}\):\[4 \times 12.01 + 8 \times 1.008 + 16.00 = 72.14 \, \text{g/mol}\]Since the empirical molar mass (72.14 g/mol) matches the given molar mass (72.1 g/mol), the empirical formula is also the molecular formula.

Key Concepts

Molecular Formula

The molecular formula represents the actual number of atoms of each element in a molecule of a compound. It is related to the empirical formula, which shows the simplest whole-number ratio of the atoms. In this exercise, the empirical formula \[\text{C}_{4}\text{H}_{8}\text{O}\] was derived from combustion analysis data. The molar mass of the empirical formula \(72.14\, \text{g/mol}\) was found to be the same as the provided molar mass \(72.1\, \text{g/mol}\). Thus, the empirical formula \(\text{C}_{4}\text{H}_{8}\text{O}\) is also the compound's molecular formula. A rule of thumb: if the molar mass of the empirical unit matches the given molar mass perfectly or matches after multiplication by an integer, it verifies the molecular formula. This systematic approach connects the empirical formula and molar mass to finalize the molecular formula. Understanding this relationship helps solve problems where the molecular formula is sought from limited data.

Stoichiometry

Stoichiometry plays a vital role in determining the quantities of reactants and products in chemical reactions. It uses balanced chemical equations to predict the amounts of products generated from specific quantities of reactants. In the exercise, stoichiometry helps relate the mass of the unknown compound to the masses of \(\text{CO}_2\) and \(\text{H}_2\text{O}\) produced during combustion analysis. By converting the masses of \(\text{CO}_2\) and \(\text{H}_2\text{O}\) into moles, we determine how much carbon and hydrogen are present in the original compound. Then, the stoichiometric equations are used to trace back to the compound's empirical formula, \(\text{C}_{4}\text{H}_{8}\text{O}\). Stoichiometry is crucial as it ensures that chemical conservation laws, like the conservation of mass, are upheld in analytical calculations, leading to accurate determination of chemical formulas.

Combustion Analysis

Combustion analysis is a synthesis technique employed to determine the empirical formula of an organic compound. It involves burning the compound in excess oxygen and analyzing the resulting products, typically \(\text{CO}_2\) and \(\text{H}_2\text{O}\). The masses of these products are used to back-calculate the quantities of carbon, hydrogen, and often oxygen in the original sample. The exercise illustrates this method by isolating \(0.3718 \, \text{g}\) of \(\text{CO}_2\) and \(0.1522 \, \text{g}\) of \(\text{H}_2\text{O}\), from which the moles of carbon and hydrogen are deduced. Any remaining mass from the original sample likely represents oxygen, allowing us to compute its contribution. This analysis forms the basis to determine the empirical formula, providing a systematic pathway to decomposition analysis where the unknown composition needs to be understood. By employing combustion analysis, students gain insight into the empirical assembly of molecules in an organic compound.