Question

The elemental analysis of an organic solid extracted from gum arabic (a gummy substance used in adhesives, inks, and pharmaceuticals) showed that it contained 40.0 percent \(\mathrm{C}, 6.7\) percent \(\mathrm{H}\), and 53.3 percent \(\mathrm{O}\). A solution of \(0.650 \mathrm{~g}\) of the solid in \(27.8 \mathrm{~g}\) of the solvent diphenyl gave a freezing-point depression of \(1.56^{\circ} \mathrm{C}\). Calculate the molar mass and molecular formula of the solid. \(\left(K_{\mathrm{f}}\right.\) for diphenyl is \(8.00^{\circ} \mathrm{C} / \mathrm{m} .\) )

Short answer

The molar mass is approximately 119.90 g/mol, and the molecular formula is \( \text{C}_4\text{H}_8\text{O}_4 \).

Step-by-step solution

Calculate Moles of Solute from Freezing Point Depression

Using the formula for freezing point depression, \[\Delta T_f = K_f \cdot m\]where \( \Delta T_f = 1.56^{\circ} \mathrm{C} \), \( K_f = 8.00^{\circ} \mathrm{C} / \mathrm{m} \), solve for the molality \( m \). \[m = \frac{\Delta T_f}{K_f} = \frac{1.56}{8.00} = 0.195 \frac{\text{mol}}{\text{kg}}\]Since the mass of the solvent is 27.8 g or 0.0278 kg, we can calculate the moles of solute using:\[\text{Moles of solute} = 0.195 \times 0.0278 = 0.005421 \, \text{mol}\]

Determine Molar Mass

Given the moles of solute and the mass of the sample (0.650 g), the molar mass \( M \) can be calculated:\[M = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{0.650}{0.005421} \approx 119.90 \, \text{g/mol}\]

Determine Empirical Formula

Start with a 100 g sample, which means we have:- 40.0 g C- 6.7 g H- 53.3 g OFind the moles of each element:\[\text{Moles of C} = \frac{40.0}{12.01} \approx 3.33 \, \text{mol}\]\[\text{Moles of H} = \frac{6.7}{1.008} \approx 6.65 \, \text{mol}\]\[\text{Moles of O} = \frac{53.3}{16.00} \approx 3.33 \, \text{mol}\]The ratios are approximately 1:2:1 (C:H:O) which gives an empirical formula of \( \text{CH}_2\text{O} \).

Determine Molecular Formula from Empirical Formula

Determine the molar mass of the empirical formula \( \text{CH}_2\text{O} \):\[M_{\text{empirical}} = 12.01 + (2 \times 1.008) + 16.00 = 30.03 \, \text{g/mol}\]Divide the molar mass by the empirical formula mass:\[n = \frac{119.90}{30.03} \approx 4\]Thus, the molecular formula is \((\text{CH}_2\text{O})_4 = \text{C}_4\text{H}_8\text{O}_4\).

Key concepts

Freezing Point Depression

Freezing point depression occurs when a solute is added to a solvent, lowering the freezing point of the solution compared to the pure solvent. This happens because the solute particles disrupt the solvent particles, making it harder for the solvent to form a solid lattice. The relationship can be expressed with the formula:

• \[\Delta T_f = K_f \cdot m\]

where \( \Delta T_f \) is the decrease in freezing temperature, \( K_f \) is the freezing point depression constant, and \( m \) is the molality of the solution.
In this problem, by knowing the constant \( K_f = 8.00^{\circ} \text{C} / \text{m} \) and measuring \( \Delta T_f = 1.56^{\circ} \text{C} \), we can calculate the molality. Understanding this concept is vital for determining the moles of solute, leading us to the calculation of molar mass.

Empirical Formula

The empirical formula represents the simplest whole-number ratio of atoms in a compound. To find it, we start with the mass percentages of each element, which in this problem are given as: 40.0% C, 6.7% H, and 53.3% O.
By converting these percentages to moles:

• C: \( \frac{40.0}{12.01} \approx 3.33 \text{ mol} \)
• H: \( \frac{6.7}{1.008} \approx 6.65 \text{ mol} \)
• O: \( \frac{53.3}{16.00} \approx 3.33 \text{ mol} \)

we find the simplest ratio is 1:2:1, yielding the empirical formula \( \text{CH}_2\text{O} \). This formula reflects the proportion between different elements without indicating the actual number of atoms in the molecule.

Molecular Formula

The molecular formula provides the actual number of atoms of each element in a molecule. It is often a multiple of the empirical formula.
To determine it, we need the molar mass of the empirical formula, calculated as \( 30.03 \text{ g/mol} \) for \( \text{CH}_2\text{O} \).
Then, we divide the given molar mass of the compound \( 119.90 \text{ g/mol} \) by the empirical formula mass:

• \( n = \frac{119.90}{30.03} \approx 4 \)

This tells us that the molecular formula is \( \text{C}_4\text{H}_8\text{O}_4 \), showing that the actual molecule contains four times the number of atoms suggested by the empirical formula.

Elemental Analysis

Elemental analysis is a process used to determine the percentage composition of elements in a compound. In this problem, it provided the percentages for carbon, hydrogen, and oxygen.
Using this information, we found the moles of each element, essential for deriving the empirical formula. This step bridges our understanding of the compound's basic make-up from a mass perspective.
It is a foundational technique in chemistry, enabling the conversion from mass percentages to quantitative chemical formulas, thereby revealing the compound's empirical and molecular structures.