Question

You add \(0.255 \mathrm{g}\) of an orange, crystalline compound whose empirical formula is \(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{Fe}\) to \(11.12 \mathrm{g}\) of benzene. The boiling point of the benzene rises from \(80.10^{\circ} \mathrm{C}\) to \(80.26^{\circ} \mathrm{C} .\) What are the molar mass and molecular formula of the compound?

Short answer

Molar mass is 362.3 g/mol; molecular formula is \(\mathrm{C}_{20}\mathrm{H}_{16}\mathrm{Fe}_2\).\)

Step-by-step solution

Determine Boiling Point Elevation

Boiling point elevation (\( \Delta T_b \)) can be calculated as the difference in the boiling points after and before the solute is added. Thus, \[ \Delta T_b = 80.26^{\circ} \mathrm{C} - 80.10^{\circ} \mathrm{C} = 0.16^{\circ} \mathrm{C}. \]

Use the Equation for Boiling Point Elevation

The boiling point elevation equation is given by:\[ \Delta T_b = i \cdot K_b \cdot m, \]where \( i \) is the van't Hoff factor (assumed to be 1 since the compound is non-ionizing), \( K_b \) is the ebullioscopic constant of benzene (2.53 °C⋅kg/mol), and \( m \) is the molality of the solution. Rearrange this to find molality: \[ m = \frac{\Delta T_b}{i \cdot K_b} = \frac{0.16^{\circ} \mathrm{C}}{2.53^{\circ} \mathrm{C} \cdot \mathrm{kg/mol}} = 0.0632\, \text{mol/kg}. \]

Calculate Moles of Solute

Since the molality \( m \) is 0.0632 mol/kg, and the mass of the solvent, benzene, is 11.12 g (or 0.01112 kg), use\[ \text{moles of solute} = m \times \text{mass of solvent} \]to find the moles of the compound:\[ \text{moles of compound} = 0.0632 \, \text{mol/kg} \times 0.01112 \, \text{kg} = 0.000703\, \text{mol}. \]

Find Molar Mass of the Compound

The molar mass \( M \) can be calculated from the mass and moles of the compound using the formula:\[ M = \frac{\text{mass of solute}}{\text{moles of solute}}. \]Substituting gives:\[ M = \frac{0.255 \, \text{g}}{0.000703 \, \text{mol}} = 362.3 \, \text{g/mol}. \]

Determine Molecular Formula

The empirical formula mass for \( \mathrm{C}_{10} \mathrm{H}_{8} \mathrm{Fe} \) is calculated by adding atomic weights: \(10 \times 12.01 + 8 \times 1.008 + 1 \times 55.85 = 184.75\, \text{g/mol}.\) The molar mass is almost twice the empirical formula mass: \(362.3 \, \text{g/mol} \approx 2 \times 184.75\, \text{g/mol}.\) Therefore, the molecular formula is \( \mathrm{C}_{20} \mathrm{H}_{16} \mathrm{Fe}_2 \).

Key concepts

Empirical Formula

The empirical formula of a compound is the simplest integer ratio of the atoms present within the compound. It represents the relative number of each type of atom. For the orange, crystalline compound in question, the empirical formula is given as \( \mathrm{C}_{10} \mathrm{H}_{8} \mathrm{Fe} \). This means that for every 10 carbon (C) atoms, there are 8 hydrogen (H) atoms and 1 iron (Fe) atom.

To determine an empirical formula, you typically start by analyzing the percentage composition of each element. Once you have these percentages, you convert them into moles by using the atomic masses of the elements. This avoids any discrepancies resulting from molecular weights.

A trick to remember is that while the empirical formula might not tell you the exact number of atoms in a molecule, it provides a fundamental clue to understanding the molecular structure when combined with molar mass information.

Molar Mass Determination

Molar mass is a very important concept in chemistry as it links the mass of a given substance to the amount in moles. The determination of molar mass can help you realize the nature and potential complexity of the compound you are dealing with.

Using the boiling point elevation method, we determine the molar mass by first finding the number of moles of the solute. This is done through the calculation of boiling point elevation, which depends on properties such as the ebullioscopic constant of the solvent and the mass of the solute added.

For the numerical example, the molar mass was computed using:
\[ M = \frac{\text{mass of solute}}{\text{moles of solute}}. \]
Given the solute mass of 0.255 grams and the calculated 0.000703 moles, the molar mass equated to approximately 362.3 g/mol.

Regularly measuring and comparing empirical versus calculated molar mass aids in determining how simplified or complex the particular formula for a compound might be.

Molecular Formula

The molecular formula of a compound displays the actual number of each type of atom in a molecule. It can be a multiple of the empirical formula. For the compound with an empirical formula of \( \mathrm{C}_{10} \mathrm{H}_{8} \mathrm{Fe} \), the molecular formula turned out to be \( \mathrm{C}_{20} \mathrm{H}_{16} \mathrm{Fe}_2 \).

To determine the molecular formula, compare the molar mass of the compound to the empirical formula mass. This provides insight into the multiplicity of the empirical formula needed to reach the correct molecular formula.

The empirical formula mass here was calculated as 184.75 g/mol. With a measured molar mass of 362.3 g/mol, it was observed as almost twice the empirical formula mass. Therefore, multiplying the empirical formula by 2 delivered the correct molecular formula, illustrating the real structure of the compound.