Problem 2 - Chemistry and Chemical Reactivity

Question

Assume you dissolve \(45.0 \mathrm{g}\) of camphor, \(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O},\) in \(425 \mathrm{mL}\) of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\). Calculate the molality, mole fraction, and weight percent of camphor in this solution. (The density of ethanol is \(0.785 \mathrm{g} / \mathrm{mL}\).)

Step-by-Step Solution

Answer

Molality is 0.885 mol/kg, mole fraction is 0.0392, weight percent is 11.88%.

Step 1: Calculate the Mass of Ethanol

First, find the mass of ethanol using its density and volume. The formula is: \( \text{mass} = \text{density} \times \text{volume} \). Here, \( \text{density} = 0.785 \, \text{g/mL} \) and \( \text{volume} = 425 \, \text{mL} \). Thus, \( \text{mass of ethanol} = 0.785 \, \text{g/mL} \times 425 \, \text{mL} = 333.625 \, \text{g} \).

Step 2: Calculate Moles of Camphor

Determine the moles of camphor by using its molar mass. Camphor's formula is \( \mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O} \). Its molar mass is \( (10 \times 12.01) + (16 \times 1.01) + (16.00) = 152.23 \, \text{g/mol} \). The moles of camphor are given by \( \frac{45.0 \, \text{g}}{152.23 \, \text{g/mol}} = 0.2955 \, \text{mol} \).

Step 3: Calculate Moles of Ethanol

Calculate the moles of ethanol using its molar mass. Ethanol’s formula is \( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \). Its molar mass is \( (2 \times 12.01) + (6 \times 1.01) + (16.00) = 46.08 \, \text{g/mol} \). The moles of ethanol are given by \( \frac{333.625 \, \text{g}}{46.08 \, \text{g/mol}} = 7.239 \, \text{mol} \).

Step 4: Calculate Molality

Calculate the molality of the solution, which is defined as \( \frac{\text{moles of solute}}{\text{kilograms of solvent}} \). Here, the moles of camphor is \( 0.2955 \, \text{mol} \) and the mass of ethanol is \( 333.625 \, \text{g} = 0.333625 \, \text{kg} \). Thus, the molality is \( \frac{0.2955}{0.333625} = 0.885 \, \text{mol/kg} \).

Step 5: Calculate Mole Fraction of Camphor

The mole fraction of camphor is given by \( \frac{\text{moles of camphor}}{\text{total moles}} \). Total moles = moles of camphor + moles of ethanol = \( 0.2955 + 7.239 = 7.5345 \). Thus, the mole fraction of camphor is \( \frac{0.2955}{7.5345} = 0.0392 \).

Step 6: Calculate Weight Percent of Camphor

The weight percent of camphor is \( \frac{\text{mass of camphor}}{\text{total mass of solution}} \times 100 \). The total mass of the solution = mass of camphor + mass of ethanol = \( 45.0 \, \text{g} + 333.625 \, \text{g} = 378.625 \, \text{g} \). Therefore, the weight percent is \( \frac{45.0}{378.625} \times 100 = 11.88\% \).

Key Concepts

Molar Mass

Molar mass is a fundamental concept in chemistry that represents the mass of one mole of a substance. It is expressed in grams per mole (g/mol) and can be calculated using the atomic masses of the elements in a compound. For example, to find the molar mass of camphor (C\(_{10}\)H\(_{16}\)O), you sum the atomic masses of all the atoms present in its formula.

• Carbon (C) has an atomic mass of 12.01 g/mol.
• Hydrogen (H) has an atomic mass of 1.01 g/mol.
• Oxygen (O) has an atomic mass of 16.00 g/mol.

The calculation would be \((10 \times 12.01) + (16 \times 1.01) + 16.00 = 152.23 \, \text{g/mol}\). Knowing the molar mass is crucial for converting between mass and moles in any chemical calculation.

Molality

Molality is a way of expressing the concentration of a solution by considering the moles of solute per kilogram of solvent. Unlike molarity, which depends on the volume of the solution, molality depends solely on the mass of the solvent, making it particularly useful when temperature changes occur.
To calculate molality, you use the formula: \( \frac{\text{moles of solute}}{\text{kilograms of solvent}} \). In the given exercise, the moles of camphor are 0.2955 mol, and the mass of ethanol (the solvent) is 333.625 g, which is equivalent to 0.333625 kg. Thus, the molality is \( \frac{0.2955}{0.333625} = 0.885 \, \text{mol/kg} \).
This unit of concentration remains unchanged under varying physical conditions, providing more consistency in experiments.

Mole Fraction

The mole fraction is a measure of the concentration of a component in a mixture and is expressed as a ratio of the moles of that component to the total moles in the mixture. It provides an important way to quantify the composition of a solution without needing mass or volume.
To find the mole fraction of camphor in our example, calculate the total moles:

• Moles of camphor = 0.2955 mol.
• Moles of ethanol = 7.239 mol.

Total moles equals 7.5345 mol. The mole fraction of camphor is then \( \frac{0.2955}{7.5345} = 0.0392 \).
This dimensionless quantity is useful for thermodynamic calculations and predicting how mixtures will behave under different conditions.

Weight Percent

Weight percent, also known as mass percent, is another way to express the concentration of a solution, indicating how much of the total mass is made up by a particular component. It is calculated using the formula: \( \frac{\text{mass of solute}}{\text{total mass of solution}} \times 100 \).
In the provided exercise:

• Mass of camphor = 45.0 g.
• Mass of ethanol = 333.625 g.

Total mass of the solution is 378.625 g. Thus, the weight percent of camphor is \( \frac{45.0}{378.625} \times 100 = 11.88\% \).
The weight percent gives a straightforward percentage comparison, valuable for quick analyses and discussions about the relative content of the mixtures.