Question

In \(2.00 \mathrm{~min}, 29.7 \mathrm{~mL}\) of He effuse through a small hole. Under the same conditions of pressure and temperature, \(10.0 \mathrm{~mL}\) of a mixture of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) effuse through the hole in the same amount of time. Calculate the percent composition by volume of the mixture.

Short answer

The percent composition by volume of the mixture is 61.6% CO and 38.4% CO2

Step-by-step solution

Determine the ration of the volume of He and the mixture

We know that in 2.00 minutes, variable volume of helium has effused and the same amount of time also variable volume of CO and CO2 mixture has effused. The ratio of helium’s volume to the mixture’s volume can be computed as follows: \(ratio = \frac{29.7}{10.0} = 2.97\)

Calculate the molar mass of CO and CO2

Calculate the molar masses of CO and CO2. The molar mass of CO is 28.01 g/mol and CO2 is 44.01 g/mol.

Setup of equation to calculate volume percentages

Having the ratio and the molar masses, we'll set up an equation to derive the volume percentages of the two gasses. On the basis of Graham’s law of effusion, \[\frac{1}{\sqrt{M}} = \frac{V_{He}}{V_{mix}} = ratio\] Let \(x\) be the volume percent of CO and \((1-x)\) is volume percent of CO2, substituting these into the equation, we get \( \frac{1}{\sqrt{(x * 28.01) + ((1 - x) * 44.01)}} = 2.97 \).

Solve the equation for x

Solving this equation gives \(x = 0.616\) or 61.6%

Key concepts

Rate of Effusion

The rate of effusion is an essential concept when it comes to understanding how gases pass through small openings. Effusion occurs when gas molecules escape through a tiny hole into an evacuated space. The rate at which this happens depends on several factors, such as the gas's molar mass and the conditions of pressure and temperature.
According to Graham's Law of Effusion, the rate of effusion of gas is inversely proportional to the square root of its molar mass. In formula terms, it can be represented as:

• For gas A: \( Rate_A = \frac{1}{\sqrt{M_A}} \)
• For gas B: \( Rate_B = \frac{1}{\sqrt{M_B}} \)
• Ratio of their effusion rates: \( \frac{Rate_A}{Rate_B} = \frac{\sqrt{M_B}}{\sqrt{M_A}} \)

In our given problem, helium (He) and a mixture of carbon monoxide (CO) and carbon dioxide (CO2) are involved. Helium effuses at a known rate, and the task is to compare this with the effusion rate of the mixture.
The formula aids in calculating how much larger or smaller the volume of one gas is compared to the other. This insight is crucial for determining the proportion of each gas in a mixture.

Molar Mass

Molar mass is a critical component in understanding Graham's Law of Effusion since the rate of effusion is strongly influenced by it. The molar mass of a substance refers to the mass of one mole of its chemical entities (atoms, molecules, etc.), typically expressed in grams per mole (g/mol).
In the context of this exercise, we examine helium, carbon monoxide, and carbon dioxide. Their molar masses are respectively:

• Helium (He): \(4.00\text{ g/mol} \)
• Carbon monoxide (CO): \(28.01\text{ g/mol} \)
• Carbon dioxide (CO2): \(44.01\text{ g/mol} \)

When applying Graham's Law, these values help determine how fast each gas effuses through the small hole. Because helium is much lighter than CO and CO\(_2\), it tends to effuse more quickly.
Understanding molar masses helps set up the equation necessary to solve for the volume percentages of gas mixtures, enabling us to substitute them appropriately into the ratio formula derived from effusion rates.

Volume Percent Composition

Volume percent composition is a useful metric to express concentrations of different components in a gas mixture, typically given in percentages. It indicates how much of the total volume each gas occupies in a mixture.
In problems involving effusion and gas mixtures, Graham's Law allows us to determine these percentages by comparing effusion rates and molar masses. An equation can be set up such that:
\[ \frac{1}{\sqrt{M}} = \text{effusion ratio} \]
Given that \(x\) is the volume percent of CO, the remaining percentage \((1-x)\) represents the volume percent of CO2. The formula derived from the effusion rates enables us to solve for \(x\):

• Equation: \( \frac{1}{\sqrt{(x \cdot 28.01) + ((1-x) \cdot 44.01)}} = 2.97 \)

Evaluating this, we find that CO contributes to about 61.6% of the mixture, demonstrating how Graham's Law provides insights into both effusion and the composition of gas mixtures in terms of volume.