<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# Mole Fraction

## Introduction to calculations used to express relative amounts of substances in a mixture

Estimated5 minsto complete
%
Progress
Practice Mole Fraction

MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated5 minsto complete
%
Mole Fraction

Credit: Courtesy of J. D. Griggs, US Geological Survey
Source: http://commons.wikimedia.org/wiki/File:Pahoeoe_fountain_sharpen.jpg

#### The mixed blessing of sulfur dioxide

Sulfur dioxide is a by-product of many processes, both natural and human-made. Massive amounts of this gas are released during volcanic eruptions such as the one seen above on the Big Island (Hawaii). Humans produce sulfur dioxide by burning coal. The gas has a cooling effect when in the atmosphere by reflecting sunlight back away from the earth. However, sulfur dioxide is also a component of smog and acid rain, both of which are harmful to the environment. Many efforts have been made to reduce SO2 levels to lower acid rain production. An unforeseen complication: as we lower the concentration of this gas in the atmosphere, we lower its ability to cool and then we have global warming concerns.

### Mole Fraction

One way to express relative amounts of substances in a mixture is with the mole fraction. Mole fraction (X)\begin{align*}(X)\end{align*} is the ratio of moles of one substance in a mixture to the total number of moles of all substances. For a mixture of two substances, A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}, the mole fractions of each would be written as follows:

XA=mol Amol A+mol BandXB=mol Bmol A+mol B\begin{align*}X_A= \frac{\text{mol} \ A}{\text{mol} \ A+\text{mol} \ B} \quad \text{and} \quad X_B=\frac{\text{mol} \ B}{\text{mol} \ A+\text{mol} \ B}\end{align*}

If a mixture consists of 0.50 mol A\begin{align*}A\end{align*} and 1.00 mol B\begin{align*}B\end{align*}, then the mole fraction of A\begin{align*}A\end{align*} would be XA=0.51.5=0.33\begin{align*}X_A=\frac{0.5}{1.5} = 0.33\end{align*}.  Similarly, the mole fraction of B\begin{align*}B\end{align*} would be XB=1.01.5=0.67\begin{align*}X_B =\frac{1.0}{1.5} = 0.67\end{align*}.

Mole fraction is a useful quantity for analyzing gas mixtures in conjunction with Dalton’s law of partial pressures. Consider the following situation: A 20.0 liter vessel contains 1.0 mol of hydrogen gas at a pressure of 600 mmHg. Another 20.0 liter vessel contains 3.0 mol of helium at a pressure of 1800 mmHg. These two gases are mixed together in an identical 20.0 liter vessel. Because each will exert its own pressure according to Dalton’s law, we can express the partial pressures as follows:

PH2=XH2×PTotalandPHe=XHe×PTotal\begin{align*}P_{H_2}=X_{H_2} \times P_{\text{Total}} \quad \text{and} \quad P_{He}=X_{He} \times P_{\text{Total}}\end{align*}

The partial pressure of a gas in a mixture is equal to its mole fraction multiplied by the total pressure. For our mixture of hydrogen and helium:

XH2=1.0 mol1.0 mol+3.0 mol=0.25andXHe=3.0 mol1.0 mol+3.0 mol=0.75\begin{align*}X_{H_2}=\frac{1.0 \ \text{mol}}{1.0 \ \text{mol}+3.0 \ \text{mol}}=0.25 \quad \text{and} \quad X_{He}=\frac{3.0 \ \text{mol}}{1.0 \ \text{mol} + 3.0 \ \text{mol}}=0.75\end{align*}

The total pressure according to Dalton’s law is 600 mmHg+1800 mmHg=2400 mmHg\begin{align*}600 \text{ mmHg} + 1800 \text{ mmHg} = 2400 \text{ mmHg}\end{align*}. So, each partial pressure will be:

PH2=0.25×2400 mmHg=600 mmHgPHe=0.75×2400 mmHg=1800 mmHg\begin{align*}& P_{H_2}=0.25 \times 2400 \text{ mmHg}=600 \text{ mmHg} \\ & P_{He}=0.75 \times 2400 \text{ mmHg}=1800 \text{ mmHg}\end{align*}

The partial pressures of each gas in the mixture don’t change since they were mixed into the same size vessel and the temperature was not changed.

#### Sample Problem: Dalton’s Law

A flask contains a mixture of 1.24 moles of hydrogen gas and 2.91 moles of oxygen gas. If the total pressure is 104 kPa, what is the partial pressure of each gas?

Step 1: List the known quantities and plan the problem.

Known

• 1.24 mol H2
• 2.91 mol O2
• PTotal=104 kPa\begin{align*}P_{\text{Total}}=104 \ \text{kPa}\end{align*}

Unknown

• PH2=? kPa\begin{align*}P_{H_2}=? \ \text{kPa}\end{align*}
• PO2=? kPa\begin{align*}P_{O_2}=? \ \text{kPa}\end{align*}

First, the mole fraction of each gas can be determined. Then, the partial pressure can be calculated by multiplying the mole fraction by the total pressure.

Step 2: Solve.

XH2=1.24 mol1.24 mol+2.91 mol=0.299PH2=0.299×104 kPa=31.1 kPaXO2=2.91 mol1.24 mol+2.91 mol=0.701PO2=0.701×104 kPa=72.9 kPa\begin{align*}& X_{H_2}=\frac{1.24 \ \text{mol}}{1.24 \ \text{mol} + 2.91 \ \text{mol}}=0.299 && X_{O_2}=\frac{2.91 \ \text{mol}}{1.24 \ \text{mol} + 2.91 \ \text{mol}}=0.701 \\ & P_{H_2}=0.299 \times 104 \text{ kPa}=31.1 \text{ kPa} && P_{O_2}=0.701 \times 104 \text{ kPa}=72.9 \text{ kPa}\end{align*}

The hydrogen is slightly less than one third of the mixture, so it exerts slightly less than one third of the total pressure.

### Summary

• Use of the mole fraction allows calculation to be made for mixtures of gases.

### Explore More

Use the resource below to answer the questions that follow.

1. What is mole percent?
2. Do the mole fractions add up to 1.00?
3. What other way could you calculate the mole fraction of oxygen once you have the mole fraction of nitrogen?

### Review

1. What is mole fraction?
2. How do you determine partial pressure of a gas when given the mole fraction and the total pressure?
3. In a gas mixture containing equal numbers of moles of two gases, what can you say about the partial pressures of each gas?

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes