Problem 42 - Chemistry The Central Science In Si Units

Question

A solution is made containing \(20.8 \mathrm{~g}\) of phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\) in \(425 \mathrm{~g}\) of ethanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right) .\) Calculate \((\mathbf{a})\) the mole fraction of phenol, \((\mathbf{b})\) the mass percent of phenol, (c) the molality of phenol.

Step-by-Step Solution

Answer

(a) Mole fraction of phenol is 0.0235; (b) Mass percent is 4.67%; (c) Molality is 0.520 mol/kg.

Step 1: Calculate the Molar Masses

First, find the molar mass of phenol, \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH} \). The molecular weights are: \( \mathrm{C} = 12.01\), \( \mathrm{H} = 1.01\), and \( \mathrm{O} = 16.00 \). The molar mass is \( (6 \times 12.01) + (6 \times 1.01) + 16.00 = 94.11 \ \mathrm{g/mol} \). Next, find the molar mass of ethanol, \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \), which is \( (2 \times 12.01) + (6 \times 1.01) + 16.00 = 46.08 \ \mathrm{g/mol} \).

Step 2: Calculate Moles of Phenol and Ethanol

Calculate the number of moles of phenol: \( \frac{20.8 \ \mathrm{g}}{94.11 \ \mathrm{g/mol}} \approx 0.221 \ \mathrm{mol} \). Calculate the number of moles of ethanol: \( \frac{425 \ \mathrm{g}}{46.08 \ \mathrm{g/mol}} \approx 9.223 \ \mathrm{mol} \).

Step 3: Calculate Mole Fraction of Phenol

The mole fraction of phenol \( X_{\text{phenol}} \) is calculated as: \( X_{\text{phenol}} = \frac{\text{moles of phenol}}{\text{moles of phenol} + \text{moles of ethanol}} = \frac{0.221}{0.221 + 9.223} \approx 0.0235 \).

Step 4: Calculate Mass Percent of Phenol

To find the mass percent of phenol, use: \( \frac{\text{mass of phenol}}{\text{total mass of solution}} \times 100 \ = \frac{20.8}{20.8 + 425} \times 100 \approx 4.67\% \).

Step 5: Calculate Molality of Phenol

Molality \( m \) is calculated using: \( m = \frac{\text{moles of phenol}}{\text{kg of solvent}} \). The mass of ethanol in kg is \( 0.425 \ \mathrm{kg} \). Then, \( m = \frac{0.221 \ \mathrm{mol}}{0.425 \ \mathrm{kg}} \approx 0.520 \ \mathrm{mol/kg} \).

Key Concepts

Mole Fraction

The mole fraction is a way of expressing the concentration of a component within a mixture. It indicates the ratio of moles of one component to the total moles present in the solution. For example, in a solution with phenol and ethanol, the mole fraction helps us understand the proportion of phenol compared to the entire solution.
To calculate the mole fraction of phenol, you first determine the moles of each component. In our case, phenol has 0.221 moles, and ethanol has 9.223 moles. The total moles in the solution is the sum of these two, which equals approximately 9.444 moles.
The formula for the mole fraction of phenol \( X_{\text{phenol}} \) is:\[ X_{\text{phenol}} = \frac{\text{moles of phenol}}{\text{total moles}} \]Inserting the values, you get:\[ X_{\text{phenol}} = \frac{0.221}{9.444} \approx 0.0235 \]

• This means in every 1 mole of the solution, about 0.0235 moles are phenol.
• Mole fractions are dimensionless numbers and are particularly useful in chemical equations and reactions.

You'll notice that mole fractions do not rely on volume or mass, making them a pure count of particles.

Mass Percent

Mass percent, also known as weight percent, provides a way to express the concentration of a component in a mixture based on mass. It shows how many grams of the component are in 100 grams of the solution.
To find the mass percent of phenol in our solution, use the formula:\[ \text{Mass percent} = \frac{\text{mass of phenol}}{\text{mass of phenol} + \text{mass of ethanol}} \times 100 \]Given the mass of phenol is 20.8 grams and the mass of ethanol is 425 grams, the calculation is:\[ \text{Mass percent} = \frac{20.8}{20.8 + 425} \times 100 \approx 4.67\% \]

• This means that phenol makes up approximately 4.67% of the total solution by mass.
• Mass percent is frequently used in laboratory settings to prepare solutions and in industries for quality control.

It is important to remember that mass percent is based on weight, not the number of particles, and it can help analyze the purity of compounds.

Molality

Molality is another way to express the concentration of a solution, but it is distinct from molarity. Molality relates to the moles of solute per kilogram of solvent, providing a useful measure in scenarios where temperature changes are involved, because unlike molarity, it does not change with temperature.
To calculate the molality \( m \) of phenol in ethanol, use the formula:\[ m = \frac{\text{moles of phenol}}{\text{mass of ethanol in kg}} \]We already know phenol is at 0.221 moles, and the mass of ethanol is given as 425 grams. Converting this to kilograms, we have 0.425 kg of ethanol.
Substitute these into the formula:\[ m = \frac{0.221}{0.425} \approx 0.520 \text{ mol/kg} \]

• This result means that for every kilogram of ethanol, there is about 0.520 moles of phenol.
• Molality is particularly useful when studying boiling-point elevation or freezing-point depression.

By using molality, you ensure that concentration measurements remain consistent, even with temperature shifts.