Question

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

Short answer

The mixture is 54.86% CO and 45.14% CO₂ by volume.

Step-by-step solution

Understand Graham's Law of Effusion

Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The formula can be expressed as \( \frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \), where \( r \) is the rate of effusion and \( M \) is the molar mass of the gases.

Calculate the Rate of Effusion for Each Gas

First, determine the rate of effusion for helium and the unknown gas mixture. Use the fact that \( r = \frac{V}{t} \), where \( V \) is volume and \( t \) is time. For Helium: \( r_{He} = \frac{29.7 \text{ mL}}{2.00 \text{ min}} = 14.85 \text{ mL/min} \). For the mixture: \( r_{mixture} = \frac{10.0 \text{ mL}}{2.00 \text{ min}} = 5.00 \text{ mL/min} \).

Re-arrange Graham's Law

Based on Graham's Law, solve for the molar mass of the mixture using Helium's known molar mass (4 g/mol). Rearrange the equation for the molar mass of the mixture (\(M_{mixture}\)): \( \sqrt{\frac{M_{mixture}}{4}} = \frac{14.85}{5} \). Solve for \(M_{mixture}\):

Solve for Molar Mass of Mixture

Square both sides of the equation from Step 3 to isolate \(M_{mixture}\): \(\frac{M_{mixture}}{4} = \left(\frac{14.85}{5}\right)^2\). Calculate to find \(M_{mixture} = 35.2225 \text{ g/mol} \).

Set Up Molar Mass Equation for Mixture Composition

Use the weighted average molar mass formula for the mixture of \(CO\) (28 g/mol) and \(CO_2\) (44 g/mol). Let \( x \) be the volume fraction of \(CO\), then the volume fraction of \(CO_2\) is \( 1-x \). So, \(M_{mixture} = 28x + 44(1-x)\).

Plug the Known Molar Mass into the Mixture Equation

Set up the equation with the calculated molar mass of the mixture, 35.2225 g/mol: \( 35.2225 = 28x + 44(1-x) \).

Solve for x

Solve the equation \( 35.2225 = 28x + 44 - 44x \), which simplifies to \( 35.2225 = 44 - 16x \). Rearrange for \(x\): \(16x = 44 - 35.2225 \), and finally, \( x = \frac{8.7775}{16} = 0.5486\).

Calculate Percent Composition

The percentage of \(CO\) in the mixture is \( 0.5486 \times 100\% = 54.86\%\). Thus, \( %CO_2 = 100\% - 54.86\% = 45.14\% \).

Key concepts

Rate of Effusion

The rate of effusion refers to how quickly a gas can escape from a container through a small opening. Elementary to understanding effusion is Graham's Law, which provides a mathematical relationship explaining this phenomenon. Graham's Law of Effusion states that the rate at which a gas effuses is inversely proportional to the square root of its molar mass. Mathematically, this is expressed as: \[\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}\]where \( r_1 \) and \( r_2 \) are the rates of effusion for gases 1 and 2, respectively, and \( M_1 \) and \( M_2 \) are their respective molar masses. In simple terms, lighter gases effuse faster than heavier gases. To calculate the rate of effusion for any gas, you can use the formula \( r = \frac{V}{t} \), where \( V \) is the volume of gas and \( t \) is the time it takes to effuse. This basic understanding enables us to compare rates across different gases, as was done in the exercise to derive meaningful relationships between helium and the gas mixture.

Molar Mass Calculation

Calculating the molar mass of a gas mixture involves using the relationship established by Graham's Law. Once you've determined the rates of effusion for the gases in question, you can rearrange this relationship to solve for the unknown molar mass. Given that the rate of effusion for helium was found to be \( 14.85 \text{ mL/min} \) and for the mixture \( 5.00 \text{ mL/min} \), Graham's Law rearranges to allow the calculation of the molar mass of the gas mixture \( M_{mixture} \):\[\sqrt{\frac{M_{mixture}}{4}} = \frac{14.85}{5}\]This step involves isolating \( M_{mixture} \) by squaring both sides and solving the resulting equation, leading us to a calculated molar mass of \( 35.2225 \text{ g/mol} \). Understanding how to manipulate the equation to find unknown variables is crucial in the application of Graham's Law to real-world problems, especially in gaseous samples.

Gas Mixture Composition

The composition of a gas mixture can be understood in terms of its percentage by volume. For a mixture containing different gases, knowing the molar mass of each component allows us to calculate the exact percentage of each gas in the mixture.In the exercise, the mixture is comprised of carbon monoxide (CO) and carbon dioxide (CO\(_2\)). The weighted average molar mass formula allows us to correlate the known and unknown masses:\[ M_{mixture} = 28x + 44(1-x) \]where \( x \) represents the volume fraction of CO. The molar mass values of CO (28 g/mol) and CO\(_2\) (44 g/mol) are substituted into this equation. By solving the equation with \( 35.2225 \text{ g/mol} \) as the effective molar mass of the mixture, you can determine \( x \), leading to the conclusion that CO accounts for \( 54.86\% \) of the mixture by volume, and CO\(_2\) makes up the remaining \( 45.14\% \). This comprehensive approach simplifies the understanding of gas mixtures and their properties.