Question

An unknown gas evolved from the fermentation of glucose is found to effuse through a porous barrier in 15.0 min. Under the same conditions of temperature and pressure, it takes an equal volume of \(\mathrm{N}_{2} 12.0 \mathrm{~min}\) to effuse through the same barrier. Calculate the molar mass of the unknown gas, and suggest what the gas might be.

Short answer

The molar mass of the unknown gas is approximately 44 g/mol, likely carbon dioxide (\(\text{CO}_2\)).

Step-by-step solution

Understand Graham's Law

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. It can be expressed with the formula: \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \), where \( r \) is the rate of effusion, and \( M \) is the molar mass of the respective gases.

Calculate the Rates of Effusion

The rate of effusion is inversely proportional to the time it takes for the gas to effuse. For the unknown gas (gas 1): \( r_1 = \frac{V}{15.0} \) and for nitrogen (\(N_2\), gas 2): \( r_2 = \frac{V}{12.0} \).

Set Up Graham's Law Equation

Substitute the rates \( r_1 = \frac{V}{15.0} \) and \( r_2 = \frac{V}{12.0} \) into Graham's law equation: \( \frac{\frac{V}{15.0}}{\frac{V}{12.0}} = \sqrt{\frac{M_{N_2}}{M_{\text{unknown}}}} \).

Simplify and Solve for Molar Mass

The equation simplifies to \( \frac{12.0}{15.0} = \sqrt{\frac{28.0}{M_{\text{unknown}}}} \). Solve for \( M_{\text{unknown}} \) by squaring both sides: \( \left(\frac{12.0}{15.0}\right)^2 = \frac{28.0}{M_{\text{unknown}}} \). This simplifies to \( \frac{144}{225} = \frac{28.0}{M_{\text{unknown}}} \).

Calculate the Molar Mass of the Unknown Gas

Cross-multiplying gives \( 144 \times M_{\text{unknown}} = 28.0 \times 225 \). Solving for \( M_{\text{unknown}} \), we find \( M_{\text{unknown}} = \frac{28.0 \times 225}{144} \approx 43.75 \text{ g/mol} \).

Identify the Gas

A molar mass of approximately 44 g/mol suggests the unknown gas could be carbon dioxide (\( \text{CO}_2 \)), which has a molar mass of approximately 44.01 g/mol.

Key concepts

Molar Mass Calculation

Calculating the molar mass of a gas is a fundamental step in many chemistry problems, such as identifying unknown substances. In the context of effusion, the molar mass can be found by using Graham's Law. This law relates the rate at which gases effuse to their molar masses. The procedure starts by applying the formula \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \). Here, \( r \) represents the rate of effusion, while \( M \) refers to molar mass.

By rearranging this formula, you can solve for the molar mass of an unknown gas if the rates of effusion are known for both the unknown gas and a gas with a known molar mass (like nitrogen). This involves some algebra, including squaring both sides of the equation and cross-multiplying to isolate \( M_{\text{unknown}} \). Ultimately, this process allows you to find the unknown gas's molar mass, paving the way for further analysis like gas identification.

Rate of Effusion

The rate of effusion is a measure of how quickly gas particles pass through a small opening. It is inversely related to the time taken to effuse, so the longer it takes, the slower the rate. According to Graham's Law, this rate is also influenced by the molar mass of the gas. Here's how it works:

• Calculate the rate for each gas based on time: \( r = \frac{V}{t} \), where \( V \) is volume (which cancels out in calculations) and \( t \) is time taken to effuse.
• When comparing two gases, the ratio of their rates can be used to determine unknown molar masses, providing a critical piece of information for the calculations.

Understanding this relationship helps students grasp how molecular weight impacts gas behavior during effusion experiments.

Carbon Dioxide Identification

Identifying carbon dioxide through effusion experiments is an intriguing application of theoretical chemistry in practical scenarios. When the molar mass of an unknown gas is found to be close to 44 g/mol, a common guess is carbon dioxide, given its molar mass of approximately 44.01 g/mol. This identification process involves several steps:

• Calculate the molar mass using the effusion data and Graham’s Law.
• Compare the calculated molar mass to known gases.
• Evaluate other characteristics like smell or reactions with limewater, as supporting evidence.

Thus, linking theoretical predictions with experimental observations gives a comprehensive understanding and confidence in identifying gases like carbon dioxide.

Gas Effusion Experiment

A gas effusion experiment is a powerful way to explore gas properties and verify theoretical principles like Graham's Law. In these experiments, gases flow through tiny openings, allowing us to measure how long the process takes.

This simple setup provides valuable information:

• Measurements of time for effusion lead to calculations of effusion rates.
• These rates are used to determine relationships between gas properties.
• Experiments demonstrate how gas behavior correlates to molecular characteristics, reinforcing theoretical laws with observable data.

Through these experiments, students can visualize abstract concepts, making them engaging and more understandable.