Question

An unknown compound has the formula \(\mathrm{C}_{x} \mathrm{H}_{2} \mathrm{O}_{2} .\) You burn \(0.0956 \mathrm{g}\) of the compound and isolate \(0.1356 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(0.0833 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula of the compound? If the molar mass is 62. I g/mol, what is the molecular formula?

Short answer

Empirical formula: CH₃O, Molecular formula: C₂H₆O₂.

Step-by-step solution

Convert CO2 and H2O to moles of C and H

First, we calculate the number of moles of carbon in CO2. From the molecular weight of CO2 (44.01 g/mol), we find:\[\text{Moles of } \text{C} = \frac{0.1356 \text{ g}}{44.01 \text{ g/mol}} = 0.00308 \text{ mol}\ \]Since 1 mole of CO2 contains 1 mole of C, there are 0.00308 mol of C. For H2O, the molecular weight is 18.02 g/mol, so:\[\text{Moles of } \text{H}_2 = 2 \times \frac{0.0833 \text{ g}}{18.02 \text{ g/mol}} = 0.00924 \text{ mol} \]This gives us the moles of hydrogen, as there are 2 moles of H per mole of H2O.

Calculate moles of O in the compound

To find the moles of oxygen, we calculate the mass of oxygen from the given compound:Mass of the original compound is 0.0956 g. Thus:\[\text{Mass of O} = 0.0956 \text{ g} - (\text{mass of C} + \text{mass of H})\]Since we have 0.00308 mol of C (12.01 g/mol) and 0.00924 mol of H (1.008 g/mol):\[\text{Mass of C} = 0.00308 \text{ mol} \times 12.01 \text{ g/mol} = 0.0370 \text{ g}\ \text{Mass of H} = 0.00924 \text{ mol} \times 1.008 \text{ g/mol} = 0.0093 \text{ g}\ \]\[\text{Mass of O} = 0.0956 \text{ g} - (0.0370 \text{ g} + 0.0093 \text{ g}) = 0.0493 \text{ g} \]\[\text{Moles of O} = \frac{0.0493 \text{ g}}{16.00 \text{ g/mol}} = 0.00308 \text{ mol}\]

Determine the empirical formula

Now, we have the moles of C, H, and O:\[\text{C: } 0.00308 \text{ mol} \ \text{H: } 0.00924 \text{ mol} \ \text{O: } 0.00308 \text{ mol} \ \]Divide by the smallest number of moles, which is 0.00308:\[\text{C: } \frac{0.00308}{0.00308} = 1 \ \text{H: } \frac{0.00924}{0.00308} = 3 \ \text{O: } \frac{0.00308}{0.00308} = 1 \]This gives the empirical formula \(\text{CH}_3\text{O}\).

Calculate the molecular formula

Given that the molar mass of the compound is 62 g/mol, we determine the molecular formula using the empirical formula mass. The empirical formula mass of \(\text{CH}_3\text{O}\) is:\[\text{Mass of } \text{CH}_3\text{O} = 12.01 \times 1 + 1.008 \times 3 + 16.00 \times 1 = 31.028 \text{ g/mol} \]Divide the molar mass by the empirical formula mass to find the multiplication factor:\[\frac{62}{31.028} = 2\ \]Thus, we multiply the subscripts in the empirical formula by 2 to get the molecular formula \(\text{C}_2\text{H}_6\text{O}_2\).

Key concepts

Molecular Formula

The molecular formula represents the exact number of each type of atom in a molecule of the compound. It provides a way of understanding the composition of a compound at the molecular level. The molecular formula allows chemists to comprehend how elements combine and the specific number of atoms involved.
In our given exercise, we started with the empirical formula, which was determined to be \(\text{CH}_3\text{O}\). With a known molar mass of 62 g/mol, we then calculated the molecular formula. This process involved calculating the empirical formula mass and using a multiplication factor to determine the molecular formula \(\text{C}_2\text{H}_6\text{O}_2\).
This calculation shows that doubling the subscripts in the empirical formula allowed us to match the molar mass, highlighting the correct composition of the compound.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions, crucial for understanding how substances interact chemically. It's like a recipe in chemistry, where knowing the right proportions ensures successful reactions.
By employing stoichiometry, we converted mass to moles, which is essential as chemical reactions occur in mole ratios rather than mass ratios. For instance, when we determined the moles of carbon in \(\text{CO}_2\) and the moles of hydrogen in \(\text{H}_2\text{O}\), we applied stoichiometric principles to find out the elemental composition.
Stoichiometry ensures we reach precise reactions, which helps accurately determine the empirical formula, an essential step before finding the molecular formula.

Combustion Analysis

Combustion analysis is a useful method for determining the molecular makeup of hydrocarbons and related compounds. In this process, a sample of the compound is burned, and the resulting carbon dioxide \((\text{CO}_2)\) and water \((\text{H}_2\text{O})\) are measured.
This analysis helps chemists figure out the proportion of carbon and hydrogen in the sample. By calculating these amounts, we determine empirical formulas, which is the simplest whole-number ratio of the atoms in a compound. In our scenario, combustion analysis allowed us to find the moles of carbon and hydrogen based on the mass of \(\text{CO}_2\) and \(\text{H}_2\text{O}\) produced.
Significantly, this laid the foundation for determining the molar quantity of each element, a critical step in ascertaining the empirical formula.

Molar Mass Calculation

Molar mass calculation is crucial for bridging the empirical formula to the molecular formula. It represents the mass of one mole of a substance, typically expressed in grams per mole. This conversion is foundational in comparing the mass of substances involved in chemical reactions.
In the exercise, we calculated the empirical formula's mass by summing the atomic masses of the elements involved. With this value, we cross-referenced the provided molar mass of the compound to derive the multiplication factor, leading to the molecular formula. This step was essential as the empirical formula mass was found to be 31.028 g/mol.
Molar mass calculations directly influence the accuracy of molecular formula determination, confirming that the calculated molecular formula aligns with the compound's given molar mass.