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: 0.887 mol/kg, Mole fraction: 0.0398, Weight percent: 11.89%.

Step 1: Calculate Moles of Solute

The molar mass of camphor (\(\mathrm{C}_{10}\mathrm{H}_{16}\mathrm{O} \)) can be calculated as follows: \(10(12.01) + 16(1.008) + 16.00 = 152.23 \, \mathrm{g/mol}\).Using this, you can find the number of moles of camphor:\[\text{moles of camphor} = \frac{45.0 \, \mathrm{g}}{152.23 \, \mathrm{g/mol}} = 0.296 \, \text{moles}\]

Step 2: Calculate Mass of Solvent in Kg

First, calculate the mass of 425 mL of ethanol using its density:\[\text{mass of ethanol} = 425 \, \mathrm{mL} \times 0.785 \, \mathrm{g/mL} = 333.63 \, \mathrm{g}\]Convert grams to kilograms:\[\text{mass of ethanol in kg} = \frac{333.63 \, \mathrm{g}}{1000} = 0.33363 \, \mathrm{kg}\]

Step 3: Calculate Molality

Molality \(m\) is defined as the number of moles of solute per kilogram of solvent:\[m = \frac{0.296 \, \text{moles of camphor}}{0.33363 \, \mathrm{kg}} = 0.887 \, \mathrm{mol/kg}\]

Step 4: Calculate Mole Fraction

First, calculate the moles of ethanol. Its molar mass is \(2(12.01) + 6(1.008) + 16.00 = 46.08 \, \mathrm{g/mol}\).The number of moles of ethanol:\[\text{moles of ethanol} = \frac{333.63 \, \mathrm{g}}{46.08 \, \mathrm{g/mol}} = 7.24 \, \text{moles}\]The mole fraction of camphor, \(X_\text{camphor}\), can be calculated as:\[X_\text{camphor} = \frac{0.296 \, \text{moles}}{0.296 \, \text{moles} + 7.24 \, \text{moles}} = 0.0398\]

Step 5: Calculate Weight Percent

Weight percent is calculated using the formula:\[\text{Weight Percent} = \left( \frac{\text{mass of solute}}{\text{mass of solute} + \text{mass of solvent}} \right) \times 100\]Substituting the known values:\[\text{Weight Percent} = \left( \frac{45.0 \, \mathrm{g}}{45.0 \, \mathrm{g} + 333.63 \, \mathrm{g}} \right) \times 100 = 11.89\%\]

Key Concepts

Molality Calculation

Molality is a concentration measure that expresses the number of moles of solute per kilogram of solvent. It is particularly useful because it does not change with temperature as volume-based measurements do. To find the molality of camphor in our solution, follow these steps:

• Determine the Moles of Solute: Start by calculating the molar mass of camphor (\( \mathrm{C}_{10}\mathrm{H}_{16}\mathrm{O} \)), which comes out to be 152.23 g/mol. Then, divide the mass of camphor used, 45.0 g, by this molar mass to find the number of moles: \[\frac{45.0 \, \mathrm{g}}{152.23 \, \mathrm{g/mol}} = 0.296 \, \text{moles}.\]
• Calculate Mass of Solvent in Kg: Knowing the density of ethanol is 0.785 g/mL, the mass of 425 mL of ethanol is:\[425 \, \mathrm{mL} \times 0.785 \, \mathrm{g/mL} = 333.63 \, \mathrm{g}.\]Convert this to kg:\[\frac{333.63 \, \mathrm{g}}{1000} = 0.33363 \, \mathrm{kg}.\]
• Calculate Molality: Finally, the molality (\( m \)) is\[m = \frac{0.296 \, \text{moles of camphor}}{0.33363 \, \mathrm{kg}} = 0.887 \, \mathrm{mol/kg}.\]

By considering mass rather than volume, molality remains constant regardless of temperature or pressure changes, making it a reliable measure for various applications.

Mole Fraction Calculation

Mole fraction is a way of expressing the concentration of a component in a mixture. It is defined as the ratio of the moles of one component to the total moles present in the solution. Here's how to calculate the mole fraction for camphor:

• Calculate the Moles of Ethanol: First, determine the molar mass of ethanol (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \)), which is 46.08 g/mol. Then, find the number of moles in 333.63 g of ethanol:\[\frac{333.63 \, \mathrm{g}}{46.08 \, \mathrm{g/mol}} = 7.24 \, \text{moles}.\]
• Formula for Mole Fraction: To find the mole fraction of camphor, divide the moles of camphor by the total moles in the solution (moles of camphor plus moles of ethanol):\[X_\text{camphor} = \frac{0.296 \, \text{moles}}{0.296 \, \text{moles} + 7.24 \, \text{moles}} = 0.0398.\]

Given its dimensionless nature, mole fraction is a versatile unit in thermodynamics since total mole fraction always sums up to 1.

Weight Percent Calculation

Weight percent is the mass of a solute divided by the total mass of the solution, multiplied by 100 to express it as a percentage. Let's explore how to calculate it:

• Formula: The weight percent formula is:\[\text{Weight Percent} = \left( \frac{\text{mass of solute}}{\text{mass of solute} + \text{mass of solvent}} \right) \times 100.\]
• Calculate Weight Percent of Camphor: Substitute the mass of camphor and ethanol into the formula:\[\text{Weight Percent} = \left( \frac{45.0 \, \mathrm{g}}{45.0 \, \mathrm{g} + 333.63 \, \mathrm{g}} \right) \times 100 = 11.89\%.\]

Weight percent is useful for solutions where the actual mass concentrations are needed, such as in industrial formulations.