Problem 82 - Chemistry: The Central Science

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 \(1.0 \mathrm{Lof} \mathrm{O}_{2}\) 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 \(1.0 \mathrm{~L} ;\) that is, rate and time are inversely proportional.)

Step-by-Step Solution

Answer

The molar mass of the unknown gas is approximately \(290.92 \text{ g/mol}\).

Step 1: 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, it can be represented as: \( \frac{\text{Rate 1}}{\text{Rate 2}} = \sqrt{\frac{\text{Molar Mass 2}}{\text{Molar Mass 1}}} \) In our problem, we will use the given effusion times to determine the rates of effusion for both gases and their relationship to find the molar mass of the unknown gas.

Step 2: Calculate the rates of effusion for both gases

We are given the effusion times for both the unknown gas and oxygen gas, which are inversely proportional to their rates of effusion. We can calculate the rates of effusion as follows: \( \text{Rate of Unknown gas} = \frac{1}{105} \) \( \text{Rate of O}_{2}\text{ gas} = \frac{1}{31} \)

Step 3: Apply Graham's Law of Effusion to relate the rates and molar masses

Now, we will use Graham's Law of Effusion to relate the calculated rates of effusion to the molar masses of the unknown gas and oxygen gas: \( \frac{\frac{1}{105}}{\frac{1}{31}} = \sqrt{\frac{\text{Molar Mass of O}_{2}}{\text{Molar Mass of Unknown gas}}} \) Simplify the equation: \( \frac{31}{105} = \sqrt{\frac{\text{Molar Mass of O}_{2}}{\text{Molar Mass of Unknown gas}}} \)

Step 4: Solve for the molar mass of the unknown gas

We will now solve the equation above to find the molar mass of the unknown gas. We know the molar mass of oxygen gas is approximately 32 g/mol. Substitute this value into the equation and solve for the molar mass of the unknown gas: \( \frac{31}{105} = \sqrt{\frac{32}{\text{Molar Mass of Unknown gas}}} \) Square both sides of the equation: \(\left(\frac{31}{105}\right)^2 = \frac{32}{\text{Molar Mass of Unknown gas}} \) Now, solve for the molar mass of the unknown gas: \( \text{Molar Mass of Unknown gas} = \frac{32}{(\frac{31}{105})^2} \) \( \text{Molar Mass of Unknown gas} = 290.92 \)

Step 5: State the answer

The molar mass of the unknown gas is approximately 290.92 g/mol.

Key Concepts

Effusion

Effusion is a fascinating concept in chemistry that explains how gas molecules escape through a tiny opening. Imagine you have a balloon with a pinhole in it. Effusion is the escape of gas molecules from the balloon through that small hole into a vacuum or another container with very low pressure. The rate of effusion depends on the speed at which molecules travel and is influenced by the size of the molecule—the smaller and lighter the molecule, the faster it escapes.
In simpler terms, think of effusion as a race of gas molecules trying to squeeze through a door into freedom. Their success in doing so quickly is what we call the rate of effusion. This concept becomes especially important when understanding how gases behave under different conditions and comparing the movement of different gases.

Molar Mass

Molar mass is a crucial term used in chemistry to describe how much a mole of a substance weighs. Specifically, it's the mass of one mole of a particular element or compound, expressed in grams per mole (g/mol). A mole is a basic unit in chemistry that represents a specific number of particles, usually atoms or molecules, which is around Avogadro's number—approximately

• 6.022 x 10²³ particles.

To find the molar mass of a compound, you simply add together the molar masses of each element, which you can find on the periodic table.
In our exercise, calculating the molar mass of the unknown gas was pivotal in applying Graham's Law. By comparing it to the known molar mass of oxygen, which is about 32 g/mol, we were able to accurately determine the heavier unknown gas's molar mass.

Oxygen Gas

Oxygen gas, often denoted as O extsubscript{2}, is essential for life on Earth. It is one of the most abundant elements in the Earth's atmosphere and crucial for processes such as respiration and combustion. The molecular weight of oxygen gas is approximately 32 g/mol, with each molecule comprising two oxygen atoms.
In the context of our problem, oxygen gas serves as a reference point. By knowing its molar mass and effusion rate, we can compare an unknown gas with oxygen to find its molar mass. It is important because oxygen is a common standard when using Graham's Law to compare the behavior and properties of different gases.

Gas Laws

Gas laws are a set of rules that explain how gases behave in response to changes in pressure, temperature, and volume. They are foundational in understanding gas behavior in various scientific and real-world applications. Here are a few key concepts:

• Boyle's Law: States that the pressure of a gas is inversely proportional to its volume when temperature is constant.
• Charles's Law: Describes how gas volume directly increases with temperature if pressure is constant.
• Avogadro's Law: Relates volume of gas to the number of molecules, keeping temperature and pressure consistent.

Graham's Law of Effusion, specifically used in our problem, tells us the rate at which a gas will effuse is inversely proportional to the square root of its molar mass. This law helps in comparing effusion rates of different gases, leading us to the discovery of unknown gases’ molar mass through comparison with known gases like oxygen.