Problem 48 - Chemistry: Principles and Reactions

Question

The Rast method uses camphor \(\left(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}\right)\) as a solvent for determining the molar mass of a compound. When \(2.50 \mathrm{~g}\) of cortisone acetate is dissolved in \(50.00 \mathrm{~g}\) of camphor \(\left(k_{\mathrm{f}}=40.0^{\circ} \mathrm{C} / \mathrm{m}\right)\), the freezing point of the mixture is determined to be \(173.44^{\circ} \mathrm{C}\); that of pure camphor is \(178.40^{\circ} \mathrm{C}\). What is the molar mass of cortisone acetate?

Step-by-Step Solution

Answer

Question: Using the Rast method, calculate the molar mass of cortisone acetate given the following information: - Mass of cortisone acetate: 2.50 g - Mass of camphor: 50.00 g - Freezing point depression constant: 40.0 °C/molal - Freezing point of the mixture: 173.44 °C - Freezing point of pure camphor: 178.40 °C Answer: The molar mass of cortisone acetate is approximately 403.23 g/mol.

Step 1: Calculate the freezing-point depression

Delta T = Freezing point of pure camphor - Freezing point of the mixture = \(178.40^{\circ} \mathrm{C}\) - \(173.44^{\circ} \mathrm{C}\) = \(4.96^{\circ} \mathrm{C}\)

Step 2: Use the freezing-point depression equation

The equation for freezing-point depression is: \(\Delta T = k_f \cdot m\), where \(\Delta T\) is the freezing-point depression, \(k_f\) is the freezing-point depression constant, and \(m\) is the molality. We need to find the molality, so we will rearrange the equation: \(m = \frac{\Delta T}{k_f}\). Substitute the given values and we get: \(m = \frac{4.96^{\circ} \mathrm{C}}{40.0 \frac{^{\circ} \mathrm{C}}{\mathrm{molal}}}\) = \(0.124 \mathrm{~molal}\).

Step 3: Convert the mass of camphor to kg

We have to convert the given mass of camphor (50.00 g) to kilograms to find the number of moles of solute in the solution. 50.00 g \(\times\) \(\frac{1 \mathrm{~kg}}{1000 \mathrm{~g}}\) = 0.05000 kg.

Step 4: Find the number of moles of cortisone acetate

To find the number of moles, we will use the molality formula which is: \(m= \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\). So, number of moles = molality \(\times\) mass of solvent in kg = \(0.124 \mathrm{~molal} \times 0.05000 \mathrm{~kg}\) = 0.00620 moles of cortisone acetate.

Step 5: Calculate the molar mass of cortisone acetate

Molar mass is equal to the mass of the substance divided by the number of moles of the substance: Molar mass = \(\frac{\text{Mass of cortisone acetate}}{\text{No. of moles of cortisone acetate}}\) = \(\frac{2.50 \mathrm{~g}}{0.00620 \mathrm{~moles}}\) = 403.23 g/mol. The molar mass of cortisone acetate is 403.23 g/mol.

Key Concepts

Rast Method

The Rast method is a classic technique utilized for the determination of the molar mass or molecular weight of a non-volatile compound. This technique involves dissolving the compound in a suitable solvent and measuring the depression in freezing point of the solution. The choice of solvent is crucial—camphor is a popular choice due to its large freezing-point depression constant and non-reactive nature.

Through the Rast method, the direct effect of a solute on the freezing point of a solvent is observed. It is particularly useful because it only requires a small amount of the compound to determine its molar mass and it's not necessary for the solute to be in its pure form, as long as it dissolves completely in the solvent without reacting.

Freezing-Point Depression

Freezing-point depression is one of several colligative properties that are dependent on the solute concentration, not its identity. When a solute is dissolved in a solvent, it generally decreases the temperature at which the solvent will freeze. This phenomenon is steming from the solute particles disrupting the formation of the crystalline structure of the solvent, which is essential for the freezing process.

The degree of freezing-point depression can be predicted using the formula \( \Delta T = k_f \cdot m \), where \( \Delta T \) is the change in freezing point, \( k_f \) is the freezing-point depression constant characteristic of the solvent, and \( m \) is the molality of the solution. This formula allows for the calculation of molar mass of dissolved substances by measuring the freezing-point depression they cause.

Molality

Unlike molarity, which is concentration expressed as moles of solute per liter of solution, molality is defined as the moles of solute per kilogram of solvent. Represented with \( m \), it is expressed as \( \text{molal} \), and is crucial for colligative properties calculations because it doesn't change with temperature and doesn't require knowledge of the volume of the solution.

Molality is derived using the formula \( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \). This measure is important when dealing with high precision colligative properties because it ensures consistent calculations regardless of temperature changes or the expansion of the solvent, making molality an ideal concentration unit for these types of problems.

Colligative Properties

Colligative properties, such as boiling point elevation, freezing point depression, and vapor pressure lowering, depend solely on the number of solute particles in a solvent, not the type of particles. This is a key concept in chemistry because it means that these properties can be used to determine molar masses and degrees of dissociation for ionic compounds.

Each colligative property has a corresponding constant and formula that relates to the molarity or molality of the solution. For instance, freezing-point depression and boiling point elevation constants are unique to each solvent. Hence, in the educational context, understanding colligative properties paves the way for students to grasp how the presence of a solute can alter the physical properties of a solution in predictable ways, which is fundamental in both theoretical and practical chemistry.