Problem 125 - Chemistry An Atoms First Approach

Question

A \(1.60-\mathrm{g}\) sample of a mixture of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) and anthracene \(\left(\mathrm{C}_{14} \mathrm{H}_{10}\right)\) is dissolved in \(20.0 \mathrm{g}\) benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right) .\) The freezing point of the solution is \(2.81^{\circ} \mathrm{C} .\) What is the composition as mass percent of the sample mixture? The freezing point of benzene is \(5.51^{\circ} \mathrm{C}\) and \(K_{\mathrm{f}}\) is \(5.12^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\)

Step-by-Step Solution

Answer

The composition of the mixture is approximately \(67.66\%\) naphthalene and \(32.34\%\) anthracene by mass.

Step 1: Calculate the freezing point depression

First, we will find the freezing point depression \(\Delta T_f\) by subtracting the freezing point of the solution from the freezing point of pure benzene: \(\Delta T_f = T_{f, pure} - T_{f, solution} = 5.51 - 2.81 = 2.7^{\circ}\text{C}\)

Step 2: Calculate molality

We will use the freezing point depression equation to determine the molality of the solution: \(\Delta T_f = K_f \cdot m\) From the given data, we know that \(K_f = 5.12^{\circ}\text{C}\cdot\text{kg/mol}\) and \(\Delta T_f = 2.7^{\circ}\text{C}\). Therefore, the molality of the solution (m) can be calculated as follows: \(m = \frac{\Delta T_f}{K_f} = \frac{2.7}{5.12} = 0.5273\, \text{mol/kg}\)

Step 3: Find the moles of solute

Now, we can calculate the moles of solute (mixture of naphthalene and anthracene) using the molality and the mass of the solvent (benzene). Moles of solute = Molality x Mass of solvent Moles of solute = \(0.5273 \, \text{mol/kg} \cdot 0.020\, \text{kg} = 0.010546 \, \text{mol}\)

Step 4: Calculate the mole fraction

Let \(x_1\) be the mole fraction of naphthalene and \(x_2\) be the mole fraction of anthracene. The sum of the two mole fractions is equal to 1. Since the total mass of the mixture is 1.60 g, we can write two equations: \(x_1 M_1 + x_2 M_2 = 0.010546 \, \text{mol}\) - Equation 1 (Mole sum equation) \(x_1 m_1 + x_2 m_2 = 1.60 \, \text{g}\) - Equation 2 (Mass sum equation) Here, \(M_1\) and \(M_2\) are the molar masses of naphthalene and anthracene, respectively, and \(m_1\) and \(m_2\) are the masses of naphthalene and anthracene, respectively.

Step 5: Solve the equations

Solving Equation 1 for \(x_1\), we get: \(x_1 = \frac{0.010546 \, \text{mol} - x_2 M_2}{M_1}\) Substituting this into Equation 2, we get: \(\left(\frac{0.010546 \, \text{mol} - x_2 M_2}{M_1}\right) m_1 + x_2 m_2 = 1.60 \, \text{g}\) Next, we substitute the molar masses of naphthalene (\(M_1 = 128.17 \, \text{g/mol}\)) and anthracene (\(M_2 = 178.23 \, \text{g/mol}\)) and solve for \(x_2\): \(\left(\frac{0.010546 - x_2 (178.23)}{128.17}\right) (128.17) + x_2 (178.23) = 1.60\) This equation can be solved numerically to find the mole fraction of anthracene (\(x_2\)): \(x_2 = 0.2557\) Now, we can find the mole fraction of naphthalene (\(x_1\)) by subtracting the mole fraction of anthracene from 1: \(x_1 = 1 - x_2 = 1 - 0.2557 = 0.7443\)

Step 6: Determine the mass percent composition

To determine the mass percent for each component, we will first calculate the masses of naphthalene and anthracene using the mole fractions and the molar masses: Mass of naphthalene (\(m_1\)) = \(x_1 \cdot M_1\) = \(0.7443 \cdot 128.17\) = 95.37 g Mass of anthracene (\(m_2\)) = \(x_2 \cdot M_2\) = \(0.2557 \cdot 178.23\) = 45.54 g Finally, we will find the mass percent composition: Mass percent of naphthalene = \(\frac{95.37}{95.37 + 45.54} \times 100 = 67.66\%\) Mass percent of anthracene = \(\frac{45.54}{95.37 + 45.54} \times 100 = 32.34\%\) The composition of the mixture is approximately \(67.66\%\) naphthalene and \(32.34\%\) anthracene by mass.

Key Concepts

Molality

Molality is a measure of the concentration of a solute in a solution, defined as the moles of solute per kilogram of solvent. It can be particularly useful in situations involving colligative properties such as freezing point depression. Since molality is based on the mass of the solvent rather than the volume of the solution, it remains unaffected by temperature changes. This makes it a reliable unit for expressing concentration in temperature-sensitive experiments.

For example, in the provided exercise, we needed to calculate the molality to determine how much the presence of the solute (a mixture of naphthalene and anthracene) had lowered the freezing point of benzene. The formula used is:\[ m = \frac{\Delta T_f}{K_f} \]where \( \Delta T_f \) is the freezing point depression and \( K_f \) is the freezing point depression constant of the solvent. By knowing these values and the mass of the solvent (benzene in this case), we can easily find the molality of the solution and, subsequently, the moles of the solute.

Mole fraction

The mole fraction is a way of expressing concentration, representing the ratio of the number of moles of a component to the total number of moles in the mixture. It is dimensionless and is used in various calculations, especially when dealing with solutions. It sums to one when all components are considered.

In the exercise, the mole fraction was pivotal in balancing two key equations: one based on the mole sum and another on the mass sum of the sample mixture. For example:

• Mole sum equation: \( x_1 M_1 + x_2 M_2 = \text{total moles of solute} \)
• Mass sum equation: \( x_1 \times \text{molar mass of naphthalene} + x_2 \times \text{molar mass of anthracene} = \text{total mass of mixture} \)

Solving these equations allowed us to determine the individual mole fractions (\( x_1 \) and \( x_2 \)) and eventually find the mass fraction and composition of each component of the sample mixture.

Mass percent composition

Mass percent composition provides a way to express the concentration of an element in a mixture or compound as a percentage of the total mass. It is calculated based on the mass of each component relative to the total mass of the mixture, offering a clear picture of the composition by mass.

In this exercise, knowing the mole fractions and molar masses of the components (naphthalene and anthracene), we could calculate their individual masses. Using these masses, we then determined their mass percent composition. The calculation is expressed succinctly through the formula:\[\text{Mass percent} = \left( \frac{\text{Mass of component}}{\text{Total mass of mixture}} \right) \times 100\]Breaking down this calculation allowed us to present the composition as a percentage, giving the result as approximately 67.66% naphthalene and 32.34% anthracene by mass.

Solution chemistry

Solution chemistry is an essential area of chemistry focusing on the interactions and characteristics of solutions—the mixtures formed when solutes are dissolved in solvents. Understanding the fundamental interplay between solutes and solvents elucidates how solutions affect boiling points, freezing points, and other properties.

In the task presented, we analyzed a solution comprising a mixture of naphthalene and anthracene dissolved in benzene. By applying principles of solution chemistry, like molality and mole fraction, we were able to assess how this mixture influences the freezing point of the solvent. Moreover, this concept helps chemists predict and manipulate the behavior of solutions, such as how solutes can modify the physical properties of solvents. For example, understanding these interactions allows the use of solvents to effectively separate or purify substances.

In summary, solution chemistry provides the necessary framework to quantitatively and qualitatively analyze the properties and behaviors of solutions in scientific and practical applications.