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Chapter 13: Problem 136

A pharmaceutical preparation made with ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is contaminated with methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right) .\) A sample of vapor above the liquid mixture contains a \(97 / 1\) mass ratio of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) to \(\mathrm{CH}_{3} \mathrm{OH}\). What is the mass ratio of these alcohols in the liquid mixture? At the temperature of the liquid mixture, the vapor pressures of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) and \(\mathrm{CH}_{3} \mathrm{OH}\) are 60.5 torr and 126.0 torr, respectively.

Short Answer

Expert verified
The mass ratio of ethanol to methanol in the liquid mixture is approximately 1.815:1.

Step by step solution

01

- Understand Raoult's Law

According to Raoult's Law, the partial vapor pressure of a component in a mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction in the liquid mixture. Mathematically, it is given by: \[ P_i = x_i P_i^{\text{pure}} \] where \( P_i \) is the partial vapor pressure of component \( i \), \( x_i \) is the mole fraction of component \( i \) in the liquid mixture, and\( P_i^{\text{pure}} \) is the vapor pressure of the pure component.
02

- Determine the Vapor Pressures

Given in the problem, the vapor pressures for ethanol (\( \text{C}_2\text{H}_5\text{OH} \)) and methanol (\( \text{CH}_3\text{OH} \)) are 60.5 torr and 126.0 torr, respectively: \[ P_{\text{ethanol}}^{\text{pure}} = 60.5 \text{ torr} \] \[ P_{\text{methanol}}^{\text{pure}} = 126.0 \text{ torr} \]
03

- Calculate Mole Fraction in Vapor Phase

The mass ratio in the vapor phase is given as 97:1 for ethanol to methanol. To convert this into mole fractions, we need the molecular weights: \[ M_{\text{ethanol}} = 46.07 \text{ g/mol} \] \[ M_{\text{methanol}} = 32.04 \text{ g/mol} \] Let's assume we have 97 grams of ethanol and 1 gram of methanol. We then convert these masses to moles: \[ n_{\text{ethanol}} = \frac{97 \text{ g}}{46.07 \text{ g/mol}} \approx 2.106 \text{ mol} \] \[ n_{\text{methanol}} = \frac{1 \text{ g}}{32.04 \text{ g/mol}} \approx 0.0312 \text{ mol} \]
04

- Find Mole Fractions in Vapor Phase

The total number of moles in the vapor phase is: \[ n_{\text{total}} = n_{\text{ethanol}} + n_{\text{methanol}} = 2.106 + 0.0312 \approx 2.137 \text{ mol} \] The mole fractions are: \[ y_{\text{ethanol}} = \frac{2.106}{2.137} \approx 0.985 \] \[ y_{\text{methanol}} = \frac{0.0312}{2.137} \approx 0.015 \]
05

- Apply Raoult's Law

Using Raoult's law, the mole fraction in the liquid phase for each component can be calculated using the given total vapor pressure: \[ P_{\text{total}} = y_{\text{ethanol}} P_{\text{ethanol}}^{\text{pure}} + y_{\text{methanol}} P_{\text{methanol}}^{\text{pure}} \] Plug in the values: \[ P_{\text{total}} = 0.985 \times 60.5 + 0.015 \times 126 \] \[ P_{\text{total}} \approx 61.04 \text{ torr} \]
06

- Calculate Liquid Phase Mole Fractions

Use Raoult's Law expressed for the liquid phase to relate to the mole fractions in the liquid: \[ P_{\text{ethanol}} = x_{\text{ethanol}} \times 60.5 \] \[ P_{\text{methanol}} = x_{\text{methanol}} \times 126 \] Using the fact that the vapor pressures must sum to total pressure: \[ 60.5 x_{\text{ethanol}} + 126 x_{\text{methanol}} = 61.04 \] And knowing that \( x_{\text{ethanol}} + x_{\text{methanol}} = 1 \): Solve this system to find: \[ x_{\text{ethanol}} \approx 0.5582 \] \[ x_{\text{methanol}} \approx 0.4418 \]
07

- Determine Mass Ratio in Liquid Phase

Find the mass ratio in the liquid phase considering the mole fractions and the molar masses: \[ \text{mass}_{\text{ethanol}} = x_{\text{ethanol}} \times 46.07 \] \[ \text{mass}_{\text{methanol}} = x_{\text{methanol}} \times 32.04 \] \[ \text{mass ratio} = \frac{\text{mass}_{\text{ethanol}}}{\text{mass}_{\text{methanol}}} \approx \frac{0.5582 \times 46.07}{0.4418 \times 32.04} \approx 1.815 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

partial vapor pressure
Let's start with understanding the concept of partial vapor pressure. When we talk about mixtures of substances, each component exerts a portion of the total pressure, called partial vapor pressure. For example, in our exercise involving ethanol (\text{C}_2\text{H}_5\text{OH}) and methanol (\text{CH}_3\text{OH}), each alcohol exerts its own pressure in the vapor phase. The partial vapor pressure of a component is directly proportional to its presence in the mixture. It's calculated using the equation: \[ P_i = x_i P_i^{\text{pure}} \] where: \* \( P_i \) is the partial vapor pressure, \* \( x_i \) is the mole fraction in the liquid phase, and \* \( P_i^{\text{pure}} \) is the vapor pressure of the pure component. Hence, for two components, the total vapor pressure is the sum of their partial pressures. This principles form the basis for many calculations in chemistry and engineering.
mole fraction
Next, let's delve into the concept of mole fraction. The mole fraction is a way of expressing the ratio of the moles of a component to the total moles in a mixture. It's particularly useful in Raoult's Law calculations to find out how much each component contributes to the total vapor pressure. The mole fraction is calculated using: \[ x_i = \frac{n_i}{n_{\text{total}}} \] Here: \* \( n_i \) represents the moles of the component, and \* \( n_{\text{total}} \) is the total moles in the mixture. For instance, in the given exercise, calculating the moles of ethanol and methanol in the vapor phase gives us their respective mole fractions. These values help us understand the distribution of components in different phases (liquid and vapor) of the mixture.
vapor pressure
Finally, let's explore vapor pressure itself. Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid (or solid) phase at a given temperature. This property is intrinsic to a substance. For instance, at the same temperature, ethanol has a vapor pressure of 60.5 torr, while methanol's vapor pressure is 126.0 torr. These values are critical when using Raoult's Law. They help determine the contribution of each component to the total pressure in a mixture. In real-life applications, understanding vapor pressures enables predicting how mixtures will behave under varying conditions. This is key in fields like pharmaceuticals, where precise control over component ratios can determine the effectiveness and safety of a final product. Summary: To sum up the provided example, knowing the clean definition and behavior of partial pressures, mole fractions, and vapor pressures will empower one to grasp how Raoult's Law is effectively utilized to determine the compositions in both liquid and vapor phases. Our calculation journey exemplifies how to apply these core concepts and see their interplay in practical scenarios.

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Most popular questions from this chapter

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