Problem 88 - Chemistry The Central Science In Si Units

Question

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required \(105 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of the gas to effuse. Under identical experimental conditions it required \(31 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of \(1.0 \mathrm{~L} ; \mathrm{in}\) other words, rate is the amount that diffuses over the time it takes to diffuse.)

Step-by-Step Solution

Answer

The molar mass of the unknown gas is approximately 348.58 g/mol.

Step 1: 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. Mathematically, it can be written as: \[ \frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} \]where \( \text{Rate}_1 \) and \( \text{Rate}_2 \) are the effusion rates of two gases, and \( M_1 \) and \( M_2 \) are their molar masses, respectively.

Step 2: Calculate Effusion Rates

The rate of effusion for a gas can be calculated using the formula:\[ \text{Rate} = \frac{\text{Volume}}{\text{Time}} \]For the unknown gas, the rate is:\[ \text{Rate}_{\text{unknown}} = \frac{1.0\,\text{L}}{105\,\text{s}} \approx 0.00952\,\text{L/s} \]For \( \mathrm{O}_2 \) gas, the rate is:\[ \text{Rate}_{\mathrm{O}_2} = \frac{1.0\,\text{L}}{31\,\text{s}} \approx 0.03226\,\text{L/s} \]

Step 3: Apply Graham’s Law to Find Molar Mass

Using Graham's Law:\[ \frac{0.00952}{0.03226} = \sqrt{\frac{32.00}{M_1}} \]Solve this equation for \( M_1 \), where \( 32.00 \) g/mol is the molar mass of \( \mathrm{O}_2 \).

Step 4: Solve the Molar Mass Equation

First, calculate the ratio:\[ \left(\frac{0.00952}{0.03226}\right)^2 = \frac{32.00}{M_1} \]\[ 0.0918 = \frac{32.00}{M_1} \]Rearrange to find \( M_1 \):\[ M_1 = \frac{32.00}{0.0918} \approx 348.58\,\text{g/mol} \]

Step 5: Conclude Molar Mass

The molar mass of the unknown gas, calculated using Graham’s law, is approximately \( 348.58 \) g/mol.

Key Concepts

Effusion Rates

Effusion is the process by which gas molecules escape through a tiny hole into a vacuum, and it plays a key role in understanding various properties of gases. 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. This means that lighter gases effuse faster than heavier gases.

To find the effusion rate, we use the formula:

• Effusion Rate = Volume / Time

In this exercise, for the unknown gas, the effusion rate is calculated as approximately 0.00952 L/s. For oxygen, it is approximately 0.03226 L/s. These values indicate that the unknown gas diffuses slower than oxygen, suggesting the unknown gas is heavier. This thought process is crucial for determining the molar mass of gases based on their effusion behaviour.

Molar Mass Calculation

Calculating the molar mass of an unknown gas is an important application of Graham's Law. Using the rates derived, we apply Graham's Law to find the molar mass of the unknown gas in relation to a known gas such as oxygen. We use the equation:

• \( \frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} \)

Here, \( \text{Rate}_1 \) and \( \text{Rate}_2 \) are the rates of the unknown gas and oxygen respectively.

By rearranging and solving the equation \( \frac{0.00952}{0.03226} = \sqrt{\frac{32.00}{M_1}} \), we find the molar mass of the unknown gas to be approximately 348.58 g/mol. This calculation highlights the inverse relationship between gas effusion rates and molar masses, helping us deduce the molecular weight of an unknown gas sample.

Gas Laws

Gas laws are fundamental in understanding how gases behave under different conditions. Graham's Law of Effusion is one aspect of these laws, providing insight into how effusion rates depend on molar masses. The broader set of gas laws, such as Boyle's Law, Charles's Law, and Avogadro's Law, also help predict gas behavior in various environments.

Understanding gas laws involves:

• Predicting how gas volume changes with pressure and temperature
• Calculating the amount of gas needed to exert a specific pressure
• Applying gas constants to solve related chemical problems

In this particular exercise, applying Graham's Law allows us to deduce molecular properties from observed effusion rates. Mastering these laws enhances our ability to approach complex chemical reactions and predict the behavior of gas under experimental conditions.