A mixture containing \(4.33 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(3.11 \mathrm{~g}\) of \(\mathrm{CH}_{4}\) has a total pressure of \(1.09\) atm. What is the partial pressure of \(\mathrm{CO}_{2}\) in the mixture?

Short Answer

Expert verified
The partial pressure of \(\mathrm{CO}_{2}\) is the mole fraction of \(\mathrm{CO}_{2}\) times the total pressure, which is 0.589 atm.

Step by step solution

01

Calculate the number of moles of each gas

Use the ideal gas law and the molar mass of each gas to calculate the number of moles. The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. However, for this calculation, we only need the molar mass of each gas to find n, since n = mass / molar mass. The molar mass of \(\mathrm{CO}_{2}\) is 44.01 g/mol and that of \(\mathrm{CH}_{4}\) is 16.04 g/mol. So, the moles of \(\mathrm{CO}_{2}\) is 4.33 g / 44.01 g/mol and for \(\mathrm{CH}_{4}\) it is 3.11 g / 16.04 g/mol.
02

Calculate the total number of moles of gas

Add the number of moles of \(\mathrm{CO}_{2}\) and \(\mathrm{CH}_{4}\) together to get the total moles of gas in the mixture.
03

Determine the mole fraction of \(\mathrm{CO}_{2}\)

The mole fraction of \(\mathrm{CO}_{2}\) is the ratio of the number of moles of \(\mathrm{CO}_{2}\) to the total number of moles of the gas mixture.
04

Calculate the partial pressure of \(\mathrm{CO}_{2}\)

Apply Dalton's Law of Partial Pressures which states that the total pressure of a gas mixture is the sum of the partial pressures of each component gas. The partial pressure of \(\mathrm{CO}_{2}\) is the mole fraction of \(\mathrm{CO}_{2}\) times the total pressure of the gas mixture.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Ideal Gas Law
The ideal gas law is a crucial equation in chemistry and physics that relates the pressure, volume, temperature, and amount of a gas. It can be written as: \[\begin{equation}PV = nRT\text{,}\text{where:}\begin{itemize} \text{\begin{description}} \text{\item[P = pressure]} of the gas (in atmospheres, atm) \text{\item[V = volume]} of the gas (in liters, L) \text{\item[n = number of moles]} of the gas \text{\item[R = universal gas constant]} which is approximately 0.0821 L atm/K mol \text{\item[T = temperature]} of the gas (in Kelvin, K) \text{\end{description}}\text{\end{itemize}}\text{\end{equation}\]}

It commonly comes into play when we are trying to calculate the amount of gas (as moles) given its volume, pressure, and temperature. For this calculation, however, knowing the molar mass of the gas enables us to find the number of moles directly from the mass using the simple rearrangement of the ideal gas law:

\[\begin{equation}n = \frac{mass}{molar~mass}\text{.}\text{\end{equation}\]}

This can be done because we are looking at a fixed volume and temperature, leaving the number of moles as the only unknown.

Mole Fraction
The mole fraction is a dimensionless number expressing the ratio of the number of moles of a component to the total number of moles of a mixture. When you have more than one gas mixed together, understanding the composition of the mixture becomes important.

To calculate the mole fraction, you take:

\[\begin{equation}mole~fraction = \frac{number~of~moles~of~a~component}{total~number~of~moles~of~all~components}\text{.}\text{\end{equation}\]}

For our problem, we use the mole fraction to determine how much of the mixture's total pressure is due to \(\text{CO}_2\). Knowing the mole fraction of a component allows us to figure out its partial pressure, which is the pressure the component would exert if it were alone in the container at the same temperature.

Dalton's Law of Partial Pressures
Dalton's Law of Partial Pressures is vital for understanding how gas mixtures behave. It states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of individual gases.

Dalton's Law Equation

The partial pressure of each gas is proportional to its mole fraction:

\[\begin{equation}P_{total} = P_1 + P_2 + P_3 + \dots\text{,}\text{\end{equation}\]}\[\begin{equation}P_{individual} = mole~fraction \times P_{total}\text{,}\text{\end{equation}\]}

where \(P_{total}\) is the total pressure of the gas mixture, and \(P_{individual}\) is the partial pressure of each gas. This concept is extremely useful when we want to determine the pressure contributed by a specific gas within a mixture, as in the exercise at hand.

Molar Mass
Molar mass, an essential concept in chemistry, is the mass of one mole of a substance (usually measured in grams per mole). It is a fundamental property that links the mass of a substance to the amount of substance (moles).

To find the molar mass, we sum up the atomic masses of all atoms in a molecule. For instance:

\[\begin{equation}Molar~mass~of~\text{CO}_2 = 12.01~g/mol~(C) + 2 \times 16.00~g/mol~(O) = 44.01~g/mol\text{.}\text{\end{equation}\]}

Similarly, we can calculate the molar mass of any compound by using the atomic masses from the periodic table. Knowing the molar mass helps us convert between the mass of a substance and the number of moles, which is a necessary step in solving many chemistry problems, including the calculation of partial pressures in a gas mixture.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free