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 \(1.0 \mathrm{~L} ;\) in other words, rate is the amount that diffuses over the time it takes to diffuse.)

Short answer

The molar mass of the unknown gas can be calculated using Graham's Law of Effusion. By comparing the rates of effusion of the unknown gas and \(\mathrm{O}_{2}\) gas (\( \frac{1}{105} \frac{L}{s}\) and \( \frac{1}{31} \frac{L}{s}\), respectively), we can set up the equation: \( \frac{ \frac{1}{105} }{ \frac{1}{31} } = \sqrt{\frac{32}{M_{unknown}}} \). Solving for \(M_{unknown}\), we find that the molar mass of the unknown gas is approximately 114.11 g/mol.

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. Mathematically, if we have two gases A and B, the formula can be written as: \( \frac{Rate_{A}}{Rate_{B}} = \sqrt{\frac{M_{B}}{M_{A}}} \) Where \(Rate_{A}\) and \(Rate_{B}\) are the rates of effusion of gases A and B, respectively, and \(M_{A}\) and \(M_{B}\) are their corresponding molar masses.

Calculate the rates of effusion for both gases

We are given the times required for 1.0 L of both the unknown gas and \(\mathrm{O}_{2}\) gas to effuse. We can find the rates of effusion by taking the reciprocal of the given times: Rate of effusion for unknown gas = \( \frac{1}{105} \frac{L}{s} \) Rate of effusion for \(\mathrm{O}_{2}\) gas = \( \frac{1}{31} \frac{L}{s} \)

Set up the equation using Graham's Law of Effusion

Now that we have the rates of effusion for both gases, we can set up the equation using Graham's Law: \( \frac{ \frac{1}{105} }{ \frac{1}{31} } = \sqrt{\frac{32}{M_{unknown}}} \) Here, 32 is the molar mass of \(\mathrm{O}_{2}\).

Solve for the molar mass of the unknown gas

We will solve for \(M_{unknown}\): \( \frac{1}{105} \times 31 = \sqrt{\frac{32}{M_{unknown}}} \) \( \frac{31}{105} = \sqrt{\frac{32}{M_{unknown}}} \) Square both sides: \( (\frac{31}{105})^2 = \frac{32}{M_{unknown}} \) Now, isolate \(M_{unknown}\) on one side of the equation: \( M_{unknown} = \frac{32}{(\frac{31}{105})^2} \)

Calculate the molar mass of the unknown gas

Finally, plug in the numbers and calculate the molar mass of the unknown gas: \( M_{unknown} = \frac{32}{(\frac{31}{105})^2} = 114.11 \) Therefore, the molar mass of the unknown gas is approximately 114.11 g/mol.

Key concepts

Molar Mass Calculation

To find the molar mass of a gas, we first need to understand what molar mass is. It is the mass of one mole of a substance and is usually expressed in grams per mole (g/mol). Calculating the molar mass of an unknown gas involves using data from an experiment to determine the mass of the gas in relation to its amount in moles. This requires the use of the number of particles or gas molecules present in a specific volume, and the conditions they are under. In this problem, we want to calculate the molar mass of an unknown gas, given the time it takes to effuse compared to a known gas, which in this case is oxygen ( O_2 ). Knowing the molar mass of oxygen is approximately 32 g/mol, we can use this as a reference to find the molar mass of the unknown gas by applying Graham's Law of Effusion.

Effusion Rate Calculation

Effusion is the process where gas particles move through a tiny opening from one container to another. The rate of effusion is crucial in understanding the behavior of gases and calculating other properties such as molar mass. In our problem, we observe the time it takes for 1.0 L of an unknown gas to effuse compared to the time for oxygen to effuse. By taking the reciprocal of these times, we can find the rate of effusion for each gas:

• Rate of effusion for the unknown gas = \( \frac{1}{105} \, \frac{L}{s} \)
• Rate of effusion for \mathrm{O\_2\} gas = \( \frac{1}{31} \, \frac{L}{s} \)

To compare the rates of effusion, we set up a ratio of these values. The concept of effusion rate calculation is central to understanding and applying Graham's Law, which allows us to determine how fast gases diffuse relative to one another.

Chemistry Problem Solving

Chemistry problems often require a systematic approach to reach a solution. First, it's essential to understand the laws and principles involved, like Graham's Law of Effusion in this case. This law helps us relate the effusion rates to the molar masses of different gases. Calculating the unknown gas's molar mass involves setting up the problem using the relationship: \( \frac{Rate\_{A}}{Rate\_{B}} = \sqrt{\frac{M\_{B}}{M\_{A}}} \).
Here, we plug in our given data and solve step by step:

• Create a ratio of the effusion rates observed from the experiment.
• Set up the law's equation to solve for the unknown variables, particularly the molar mass.
• Perform algebraic manipulations, such as taking squares and isolating the unknown variable to find the solution.

With practice and by following each step methodically, chemistry problem-solving becomes more intuitive. It's also important to double-check calculations to ensure accurate results, which in this case, gives us a molar mass of approximately 114.11 g/mol for the unknown gas.