Question

The rate of effusion of a particular gas was measured and found to be 24.0 mL/min. Under the same conditions, the rate of effusion of pure methane \(\left(\mathrm{CH}_{4}\right)\) gas is 47.8 mL/min. What is the molar mass of the unknown gas?

Short answer

The molar mass of the unknown gas is approximately \(44.97 \: g/mol\), calculated using Graham's law of effusion formula and the given rates of effusion.

Step-by-step solution

Write down given values and Graham's law formula

To solve this problem, we should first identify the given values: Rate of effusion of unknown gas (R1) = 24.0 mL/min, Rate of effusion of methane gas (R2) = 47.8 mL/min, Molar mass of methane gas (M2) = 16 g/mol (since methane is \(\mathrm{CH}_{4}\) with 1 carbon and 4 hydrogen atoms, C = 12 g/mol and H = 1g/mol). Graham's law of effusion formula: \( \frac{R1}{R2} = \sqrt{\frac{M2}{M1}} \) Where R1 is the rate of effusion of the unknown gas, R2 is the rate of effusion of methane gas, M1 is the molar mass of the unknown gas, and M2 is the molar mass of methane gas.

Calculate the ratio of the two rates

According to Graham's law, we have to find the ratio between the rate of effusion of the unknown gas and methane gas: \( \frac{R1}{R2} = \frac{24.0}{47.8} \)

Solve for the molar mass of the unknown gas

We have the formula \( \frac{R1}{R2} = \sqrt{\frac{M2}{M1}} \) and calculated ratio \(\frac{R1}{R2}\) in step 2. Now, we will find the molar mass of the unknown gas (M1): \( \sqrt{\frac{M2}{M1}} = \frac{24.0}{47.8} \) Square both sides of the equation and solve for M1: \( \frac{M2}{M1} = \left(\frac{24.0}{47.8}\right)^2 \) \( M1 = \frac{M2}{(\frac{24.0}{47.8})^2} \) Use the given molar mass of methane gas (M2) = 16 g/mol to calculate M1: \( M1 = \frac{16}{(\frac{24.0}{47.8})^2} \)

Calculate the molar mass of the unknown gas

Now, we can calculate molar mass of the unknown gas (M1): \( M1 = \frac{16}{(\frac{24.0}{47.8})^2} = 44.97 \: g/mol \) The molar mass of the unknown gas is approximately 44.97 g/mol.

Key concepts

Molar Mass Determination

Molar mass determination is an essential concept in chemistry, especially when dealing with gases, as it helps us identify substances. It refers to the process of calculating the molar mass of a substance, which is the mass of one mole of molecules or atoms. Understanding this concept enables chemists to experiment with different substances accurately and make predictions about their behavior.
One common method to determine the molar mass is using Graham's law of effusion, which relates the rates at which different gases effuse. When provided with the rate of effusion of a known gas, like methane, and an unknown gas, you can determine the unknown gas's molar mass by solving Graham's equation. The formula is given by:\[\frac{R1}{R2} = \sqrt{\frac{M2}{M1}}\]where \(R1\) and \(R2\) are the rates of effusion of the unknown and known gases, respectively, and \(M1\) and \(M2\) are their molar masses.
This equation shows that gases with lower molar masses effuse faster than those with higher molar masses. By rearranging and solving the equation, we can find the molar mass of the unknown gas, making molar mass determination straightforward and practical.

Effusion Rates

Effusion rates are all about how fast a gas escapes through a small hole into a vacuum. This concept is crucial in understanding how different gases behave under similar conditions. Graham's law of effusion is used to compare effusion rates of two gases, providing insight into their molar masses.
According to Graham's law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This means that lighter gases effuse more quickly than heavier gases. For instance, methane, which is relatively light, has a faster effusion rate compared to heavier gases. This concept is especially useful in identifying gases and understanding their molecular characteristics based on their observed effusion rates.
In our exercise, we compared the rate of effusion of an unknown gas with methane's rate. By observing that the unknown gas has a slower effusion rate than methane, we used the mathematical relationship provided by Graham's law to find the molar mass, confirming that the unknown gas has a higher molar mass than methane.

Methane (CH₄)

Methane, or \(\mathrm{CH}_{4}\), is a simple molecule consisting of one carbon atom and four hydrogen atoms. It is the main component of natural gas and is known for its light and flammable characteristics.
Because of its simple molecular structure, methane has a low molar mass of 16 g/mol. This low molar mass makes it an excellent reference gas for effusion and diffusion studies in which its effusion rate is often used as a benchmark.
In our exercise, methane's known effusion rate was key to finding the molar mass of an unknown gas. By measuring how fast methane effuses compared to the unknown gas, and using Graham's law, we were able to calculate the molar mass of the unknown gas. Methane's role in this context highlights its significance as an easy-to-use reference due to its uncomplicated molecular composition and predictable behavior.