Problem 38 - Chemistry and Chemical Reactivity

Question

Mesitylene is a liquid hydrocarbon. Burning 0.115 g of the compound in oxygen gives \(0.379 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(0.1035 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula of mesitylene?

Step-by-Step Solution

Answer

The empirical formula of mesitylene is \(\text{C}_3\text{H}_4\).

Step 1: Determine Moles of Carbon and Hydrogen from Products

From the problem, when mesitylene burns, it produces 0.379 g of \(\text{CO}_2\) and 0.1035 g of \(\text{H}_2\text{O}\). Start by calculating the moles of carbon present using the molar mass of \(\text{CO}_2\), which is about 44.01 g/mol:\[\mathrm{Moles\ of\ C} = \frac{0.379}{44.01} \approx 0.00861 \text{ mol of C}\]Each mole of \(\text{CO}_2\) contains one mole of carbon.Next, calculate the moles of hydrogen in \(\text{H}_2\text{O}\) using its molar mass, which is about 18.02 g/mol:\[\mathrm{Moles\ of\ H} = \frac{0.1035}{18.02}\times 2 \approx 0.0115 \text{ mol of H}\]Each mole of \(\text{H}_2\text{O}\) contains two moles of hydrogen.

Step 2: Calculate the Total Mass of Carbon and Hydrogen in Mesitylene

To find the mass of carbon in mesitylene, multiply the moles of carbon by the molar mass of carbon (12.01 g/mol):\[\mathrm{Mass\ of\ C} = 0.00861 \times 12.01 \approx 0.1034 \text{ g of C}\]For hydrogen, multiply the moles of hydrogen by the molar mass of hydrogen (1.008 g/mol):\[\mathrm{Mass\ of\ H} = 0.0115 \times 1.008 \approx 0.0116 \text{ g of H}\]Add these two masses to verify against the initial mass (0.115 g) of mesitylene burned:\[0.1034 + 0.0116 = 0.115 \text{ g}\]

Step 3: Find the Simplest Whole Number Ratio

Determine the empirical formula by dividing the moles of each element by the smallest mole value among them.\[\mathrm{Ratio\ for\ C} = \frac{0.00861}{0.00861} = 1\]\[\mathrm{Ratio\ for\ H} = \frac{0.0115}{0.00861} \approx 1.34\]Since the exact ratio should be a whole number, multiply both by 3 to get closest whole numbers:For C: 1 x 3 = 3For H: 1.34 x 3 = 4.02 (approximated to 4)This gives an empirical formula of \(\text{C}_3\text{H}_4\).

Key Concepts

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show how reactants transform into products through a rearrangement of atoms. For example, the combustion of hydrocarbons like mesitylene can be expressed through a balanced chemical equation that involves oxygen from the air and releases carbon dioxide and water. The general form is:

• Reactants on the left side
• Products on the right side
• An arrow pointing from reactants to products indicating the reaction direction

In combustion reactions, it's common to see hydrocarbons reacting with oxygen to produce CO2 and H2O. The equation must be balanced, meaning the number of each type of atom in the reactants must equal those in the products.

Combustion Analysis

Combustion analysis is a method used to determine the elemental composition of a compound, usually involving hydrocarbons. In this process, a sample like mesitylene burns in excess oxygen. The products, CO2 and H2O, are then collected and measured. By knowing the amount of these products, we can back-calculate to find the amounts of carbon and hydrogen in the original sample. This is key in identifying an empirical formula. The combined masses of these elements should match the initial mass if done correctly. This method is reliable due to its clear stoichiometric relationships.

Stoichiometry

Stoichiometry is the study of quantitative relationships in chemical reactions. It uses the balanced chemical equation to determine the proportions of reactants and products. For combustion analysis, stoichiometry helps us understand how much of each element is present. For mesitylene's combustion:

• Determine moles of CO2 to find moles of carbon
• Determine moles of H2O to find moles of hydrogen

These calculations are vital for converting mass into moles, a necessary step in finding and comparing these amounts to derive the empirical formula. It's a way to directly apply the chemical equation to a calculation problem.

Molar Mass Calculation

Molar mass calculation is crucial in converting mass into moles, equipping us with a deeper understanding of substance amounts. The molar mass is the mass of one mole of a substance. For CO2 and H2O, known molar masses allow for conversion:

• The molar mass of CO2 is about 44.01 g/mol.
• The molar mass of H2O is about 18.02 g/mol.

Using these values, one can determine the number of moles of each component produced in the combustion. This leads to calculations determining the total mass of carbon and hydrogen in the compound, connecting the moles measured to the masses required for stoichiometry.