Problem 156 - Chemistry

Question

In 2.00 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.

Step-by-Step Solution

Answer

The percentage composition by volume of the \(CO\) and \(CO_2\) mixture is found to be approximately 43% and 57% respectively.

Step 1: Identify Known Values

First, identify and write down the known values in the question. The rate of effusion of helium (\(He\)) is \(29.7 \mathrm{~mL}\) in \(2 \mathrm{~min}\), and the rate of effusion of the \(CO\) and \(CO2\) mixture is \(10.0 \mathrm{~mL}\) in the same amount of time.

Step 2: Apply Graham's Law of Effusion

Use Graham's Law of Effusion to set up the equation for the rates of effusion of \(He\), \(CO\), and \(CO_2\). We know the \(\sqrt{{M_{CO2}/M_{He}}}\) = V_{He}/V_{mixture} and the \(\sqrt{{M_{CO}/M_{He}}}\) = V_{He}/V_{mixture}. But V_{He}/V_{mixture} is common in both equations so we can write \(\sqrt{{M_{CO2}/M_{CO}}}\) = \(\sqrt{{M_{CO2}/M_{He}}}\) / \(\sqrt{{M_{CO}/M_{He}}}\)

Step 3: Substitute Known Values

Substitute the known molar masses for \(He\), \(CO\), and \(CO_2\) into the equation:\n\(\sqrt{{44.01/28.01}}\) = \((29.7/10.0)\times\sqrt{{4/44.01\}}\) / \((29.7/10.0)\times\sqrt{{4/28.01\}}\)

Step 4: Solve for Volume Fraction

To solve for the volume fraction, square both sides and rearrange for the volume of \(CO\), \(V_{CO}\), and the volume of \(CO_2\), \(V_{CO_2}\) of the mixture, \(V_{CO}\) / \(V_{CO_2}\): \(V_{CO}\) / \(V_{CO_2}\) = (\((29.7/10.0)\times\sqrt{{4/28.01\}}\))^2\) / (\((29.7/10.0)\times\sqrt{{4/44.01\}}\))^2\) .

Step 5: Calculate the Percent Composition

Finally, solve for percentage of the \(CO\) and \(CO_2\) using: \nPercent of \(CO\): \((V_{CO}/(V_{CO}+V_{CO_2}))\times100\% \) \nPercent of \(CO_2\): \((V_{CO_2}/(V_{CO}+V_{CO_2}))\times100\% \)

Key Concepts

Graham's Law of Effusion

Understanding Graham's Law of Effusion is crucial when exploring how gases behave when passing through a small opening. This law states that the rate of effusion (escape) of a gas is inversely proportional to the square root of its molar mass. In simple terms, lighter gases effuse more quickly than heavier ones. When comparing two gases under the same conditions of temperature and pressure, the relationship of their effusion rates can be expressed with the following formula:
\[\begin{equation} \frac{\text{Rate of effusion of gas 1}}{\text{Rate of effusion of gas 2}} = \sqrt{\frac{\text{Molar mass of gas 2}}{\text{Molar mass of gas 1}}} \end{equation}\]
This law becomes especially useful when you need to determine the composition of a gas mixture by comparing effusion rates, as illustrated in the exercise.

Molar Mass

The term molar mass refers to the mass of one mole of a substance, typically expressed in grams per mole (\(\text{g/mol}\)). It is a measure of the average mass of molecules or atoms in a given sample. Calculating molar mass is essential for many chemical calculations, including conversion between mass and moles, stoichiometry, and understanding gas behavior—as it ties directly to Graham's Law of Effusion. A fundamental concept, such as molar mass, becomes the basis for predicting how different gases will effuse through a given opening.

Gas Effusion

Gas effusion is the process where gas molecules escape through a tiny hole into a vacuum. This phenomenon differs from diffusion, which involves the spreading out of gas molecules into another gas. Effusion is influenced by factors such as the size of the gas molecules (or atoms) as well as their molar mass, based on Graham's Law. For example, helium being a light element with a lower molar mass effuses faster than a heavier gas such as carbon dioxide. The rate of gas effusion not only plays a role in certain industrial processes but also helps chemists determine the properties of unknown gases or mixtures.

Percent Composition by Volume

Percent composition by volume is a measure of how much volume each component contributes to the total volume of a mixture. In the context of gases, it helps to determine the relative abundance of each gas in a mixture, usually expressed as a percentage. The calculation involves finding the volume fraction of each gas and multiplying it by 100%. As seen in the provided exercise, when combined with the understanding of Graham's Law and the known molar masses, the percent composition by volume can be deduced via the rates of effusion of the gases.