Question

Azulene is a beautiful blue hydrocarbon. If 0.106 g of the compound is burned in oxygen, \(0.364 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(0.0596 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are isolated. (a) What is the empirical formula of azulene? (b) If a separate experiment gave \(128.2 \mathrm{g} / \mathrm{mol}\) as the molar mass of the compound, what is its molecular formula?

Short answer

(a) Empirical formula: \(\text{C}_5\text{H}_4\) (b) Molecular formula: \(\text{C}_{10}\text{H}_8\)

Step-by-step solution

Calculate moles of carbon in CO2

First, determine the moles of carbon in the carbon dioxide ((\mathrm{CO}_2) formed using its mass.The molar mass of \(\mathrm{CO}_2\) is 44.01 g/mol. Subsequently, the moles of carbon, which is directly from CO2, corresponds to: \[\text{Moles of CO}_2 = \frac{0.364 \text{ g}}{44.01 \text{ g/mol}} \approx 0.00827 \text{ mol}\]Since each mole of CO2 contains one mole of carbon, the moles of carbon is also 0.00827 mol.

Calculate moles of hydrogen in H2O

Determine the moles of hydrogen in the water ((\mathrm{H}_2\mathrm{O}) produced. The molar mass of \(\mathrm{H}_2\mathrm{O}\) is 18.02 g/mol. The moles of water can be calculated as follows:\[\text{Moles of H}_2\mathrm{O} = \frac{0.0596 \text{ g}}{18.02 \text{ g/mol}} \approx 0.00331 \text{ mol}\]Each mole of water has two moles of hydrogen atoms, hence:\[\text{Moles of hydrogen} = 2 \times 0.00331 \approx 0.00662 \text{ mol}\]

Find moles of oxygen by difference

Since azulene was burned, it initially contains only carbon, hydrogen, and oxygen. The initial mass of azulene is 0.106 g, unclouded by the mass of carbon and hydrogen to account for oxygen:\[\text{Mass of carbon} = 0.00827 \text{ mol} \times 12.01 \text{ g/mol} = 0.0993 \text{ g}\]\[\text{Mass of hydrogen} = 0.00662 \text{ mol} \times 1.01 \text{ g/mol} \approx 0.00668 \text{ g}\]\[\text{Mass of oxygen in azulene} = 0.106 \text{ g} - (0.0993 \text{ g} + 0.00668 \text{ g}) = 0.00002 \text{ g}\]Hence this mass is too tiny compared against measurement accuracy, typically signifying no oxygen atoms are present.

Determine empirical formula

Normalize the moles of each element: Carbon:C = 0.00827 mol Hydrogen: H = 0.00662 molFind the simplest ratio dividing through by the smallest: \[\text{Carbon to Hydrogen ratio}: \frac{0.00827}{0.00662} \approx 1.25 \text{ ; If this simplifies further to: } \times 4 \rightarrow 5\text{C}:4\text{H}\]This suggests the empirical formula could be \(\text{C}_5\text{H}_4\).

Calculate molecular formula

Verify if the empirical formula mass fits into the given molar mass of azulene: Empirical formula mass = \(5(12.01) + 4(1.01) = 64.08\) g/mol.\[ \frac{128.2 \text{ g/mol}}{64.08 \text{ g/mol}} \approx 2 \]Giving a molecular formula that doubles the empirical: \(\text{C}_{10}\text{H}_8\).

Key concepts

Molecular Formula

Understanding the concept of a molecular formula is essential in chemistry. The molecular formula provides the exact number of each type of atom in a molecule. While the empirical formula gives the simplest whole-number ratios of atoms in a compound, the molecular formula is a more detailed representation. Knowing the molecular formula helps us comprehend the actual composition of a chemical substance.

For example, if we determine through experimental data that the empirical formula of a substance is \( ext{C}_5 ext{H}_4 \), like in our azulene case, we then explore the molar mass to outline the molecular formula. Given in this problem, the molar mass is 128.2 g/mol, and the empirical formula mass equals 64.08 g/mol.

To find the molecular formula, divide the molar mass by the empirical formula mass. Thus, \( \frac{128.2 \text{ g/mol}}{64.08 \text{ g/mol}} \approx 2 \), signaling that each empirical formula unit is doubled. Consequently, the molecular formula is \( \text{C}_{10} ext{H}_8 \). This kind of insight into molecular structures is vastly beneficial when dealing with chemical synthesis or reactivity studies.

Combustion Analysis

Combustion analysis is a widely used technique to determine the composition of an unknown hydrocarbon. It involves burning the organic compound in oxygen and measuring the amounts of \( ext{CO}_2 \) and \( ext{H}_2 ext{O} \) produced. This method effectively reveals the carbon and hydrogen content of a substance.

In the given problem, burning 0.106 g of azulene results in the formation of 0.364 g of \( ext{CO}_2 \) and 0.0596 g of \( ext{H}_2 ext{O} \). By calculating the moles from these masses using the molar masses of \( ext{CO}_2 \) (44.01 g/mol) and \( ext{H}_2 ext{O} \) (18.02 g/mol), the moles of carbon and hydrogen in azulene can be determined. These calculations inform us about the carbon and hydrogen proportions.

Through this analysis, we establish that azulene comprises these elements in definite amounts, thereby guiding us to the empirical formula. In this scenario, combustion analysis is a vital step to quantify empirical ratios before deducing the molecular formula. It's strikingly useful for determining unknown compositions in organic chemistry settings.

Stoichiometry

Stoichiometry in chemistry involves the quantitative relationship between reactants and products in a chemical process. It is pivotal when assessing reactions, including combustion, to predict product formation and reactant usage. In our case, stoichiometry helps uncover elemental proportions in azulene post-combustion.

In the problem, moles of carbon from 0.364 g \( ext{CO}_2 \) and moles of hydrogen from 0.0596 g \( ext{H}_2 ext{O} \) were computed. These steps underscore stoichiometry as it quantifies elemental content and underpins determining the empirical formula, \( ext{C}_5 ext{H}_4 \), from the ratio of moles calculated.

By leveraging stoichiometry, once the moles per element are known, the simplest whole-number ratio can be found among them, enlightening the empirical formula. Stoichiometry thus acts as a bridge connecting experimental data with theoretical chemistry predictions, essential for constructing formulas and understanding chemical behavior.