Problem 92 - 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 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 \(1.0 \mathrm{~L} ;\) that is, rate and time are inversely proportional.)

Step-by-Step Solution

Answer

The molar mass of the unknown gas is approximately 9.96 g/mol.

Step 1: Write the formula for Graham's Law of Effusion in terms of time and molar mass

Given that rate and time are inversely proportional, we can rewrite Graham's Law of Effusion as the ratio of the square root of the molar masses instead of the effusion rates. \[ \frac{\sqrt{MM_{1}}}{\sqrt{MM_{2}}} = \frac{t_{2}}{t_{1}} \]Where: - \(MM_{1}\) is the molar mass of the unknown gas - \(MM_{2}\) is the molar mass of O₂ gas - \(t_{1}\) is the time for the unknown gas to effuse (105 s) - \(t_{2}\) is the time for the O₂ gas to effuse (31 s)

Step 2: Insert the given values and rearrange the formula to solve for \(MM_{1}\)

We know the molar mass of O₂ gas is 32 g/mol. Plugging in the given values into the formula, we get:\[ \frac{\sqrt{MM_{1}}}{\sqrt{32}} = \frac{31}{105} \]Now, we will rearrange the formula to solve for the molar mass of the unknown gas, \(MM_{1}\):\[ \sqrt{MM_{1}} = \frac{31}{105} \times \sqrt{32} \]Square both sides of the equation to get rid of the square root:\[ MM_{1} = \left(\frac{31}{105} \times \sqrt{32}\right)^2 \]

Step 3: Calculate the molar mass of the unknown gas

Now, we can calculate the value of \(MM_{1}\) using the rearranged formula:\[ MM_{1} = \left(\frac{31}{105} \times \sqrt{32}\right)^2 \Rightarrow MM_{1} \approx 9.96 \] So, the molar mass of the unknown gas is approximately 9.96 g/mol.

Key Concepts

Molar Mass

Molar mass is a critical concept in chemistry. It's defined as the mass of one mole of a substance, typically expressed in grams per mole (g/mol). Understanding molar mass can be immensely helpful when working with reactions and calculations involving gases.
Every element has its own specific molar mass. To determine the molar mass of a compound, you add up the molar masses of all the atoms in that compound.
When it comes to gases, knowing the molar mass is essential for performing various calculations, such as predicting reactions or calculating effusion rates. For example, in our exercise, we needed the molar mass of oxygen to solve for the unknown gas using Graham's Law.

• Molar mass provides a link between a substance's physical mass and the amount of substance represented in moles.
• It helps in converting between grams and moles, supporting stoichiometric calculations.
• The periodic table can be used to find the molar mass of each element, which aids in calculating the molar mass of compounds.

Getting familiar with molar mass enables us to delve deeper into other gas-related calculations and laws, such as the understanding of effusion rates.

Effusion Rate

The effusion rate refers to the speed at which a gas escapes through a small opening, a concept closely examined through Graham's Law. Effusion is specifically the process when gas particles escape through the tiny gaps without collisions among gas molecules.
For gases, the rate of effusion depends significantly on the molar mass of the gas and operates under specific pressure conditions.
The main takeaway is that lighter gases effuse more rapidly than heavier gases. This understanding allows us to predict and compare how various gases will behave under similar conditions.

• Effusion occurs when a gas passes through a small hole into a vacuum, without interacting with other particles.
• The relationship between effusion rates and time is notably inversely proportional: faster effusion rates mean less time required for a given volume of the gas to effuse.
• This concept forms the foundation for using Graham's Law for comparative calculations and solving unknowns in gas equations.

The effusion rate is an integral part of understanding how gases behave, providing experimental and theoretical insights into molecular motion and gas dynamics.

Graham's Law Formula

Graham's Law of Effusion is a fundamental principle in chemistry that relates the effusion rates of two gases to their molar masses. This law is particularly useful for comparing gases under identical conditions.
Graham's Law states that the rate of effusion for a gas is inversely proportional to the square root of its molar mass. Mathematically, this can be expressed as:
\[\frac{r_1}{r_2} = \sqrt{\frac{MM_2}{MM_1}}\]
However, if we know the times taken for the gases to effuse, it can also be expressed in terms of time as:
\[\frac{\sqrt{MM_1}}{\sqrt{MM_2}} = \frac{t_2}{t_1}\]
In our specific exercise, we calculated the molar mass of an unknown gas by manipulating the formula based on the effusion times of the unknown gas and oxygen. This rearrangement helped solve for the unknown gas's molar mass effectively.

• The inverse square root relationship with molar mass means heavier gases effuse slower than lighter ones.
• Using this formula enables predicting unknowns or verifying the behavior of gases in experimental setups.
• The formula is powerful because it links fundamental physical properties (molar mass) with observable experimental data (effusion rates or times).

Understanding and using Graham's Law gives crucial insight into blending theoretical chemistry concepts with practical experimental investigation.