Problem 97 - Chemistry The Molecular Nature of Matter

Question

An unknown gas \(X\) effuses 1.65 times faster than \(\mathrm{C}_{3} \mathrm{H}_{8}\). What is the molecular mass of gas \(X ?\)

Step-by-Step Solution

Answer

The molecular mass of gas \(X\) is approximately 16.18 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. The formula is given by \(\frac{\text{Rate of effusion of gas 1}}{\text{Rate of effusion of gas 2}} = \sqrt{\frac{\text{Molar mass of gas 2}}{\text{Molar mass of gas 1}}}\).

Step 2: Calculate the molar mass of \(\mathrm{C}_{3}\mathrm{H}_{8}\)

First, calculate the molar mass of \(\mathrm{C}_{3}\mathrm{H}_{8}\) by summing the molar masses of its atoms. Carbon has a molar mass of approximately 12.01 g/mol and hydrogen has a molar mass of about 1.008 g/mol. Thus, \(\mathrm{C}_{3}\mathrm{H}_{8}\) has a molar mass of \(3\times12.01\text{g/mol} + 8\times1.008\text{g/mol} = 44.095\text{g/mol}\).

Step 3: Apply Graham's Law to Determine the Molar Mass of Gas \(X\)

Using the provided effusion rate and Graham's Law, set up the equation \(1.65 = \sqrt{\frac{44.095}{M_X}}\) where \(M_X\) is the molar mass of gas \(X\).

Step 4: Solve for the Molar Mass of Gas \(X\)

To find the molar mass of gas \(X\), rearrange the equation to solve for \(M_X\): \[(1.65)^2 = \frac{44.095}{M_X}\], then \[(1.65)^2 \cdot M_X = 44.095\]. Calculate \(M_X\) to get the molar mass of gas \(X\).

Step 5: Calculate the Molar Mass of Gas \(X\)

Calculating yields \(M_X = \frac{44.095}{(1.65)^2}\). Now perform the arithmetic to find \(M_X\): \(M_X = \frac{44.095}{2.7225} = 16.18\text{g/mol}\).

Key Concepts

Rate of Effusion

The concept of rate of effusion is a fundamental aspect of gas behavior described by Graham's Law. It refers to how quickly a gas escapes through a tiny hole or porous barrier into a vacuum or less pressurized space. Imagine you have a balloon with a small puncture; the speed at which the air streams out of this puncture is analogous to the rate of effusion.

To compare two different gases, Graham's Law provides a way to understand their effusion rates relative to each other. According to this law, lighter gases (those with a lower molar mass) effuse more rapidly than heavier gases. This is because lighter gas molecules travel faster on average, making them more likely to collide with and pass through small openings.

Molecular Mass Calculation

When it comes to finding the molecular mass (often referred to as molar mass when calculated per mole of substance), it's about totaling the atomic masses of all the atoms in a molecule. Taking propane, \( \mathrm{C}_{3}\mathrm{H}_{8} \), as an example, we add three times the atomic mass of carbon to eight times the atomic mass of hydrogen to determine its molecular mass.

Here's how you might visualize the process:

• Atomic mass of Carbon (C): 12.01 g/mol
• Atomic mass of Hydrogen (H): 1.008 g/mol
• Molecular mass of \( \mathrm{C}_{3}\mathrm{H}_{8} \): \( 3 \times 12.01 + 8 \times 1.008 = 44.095 \) g/mol

It's this calculated value that plays a crucial part in applying Graham's Law for effusion rates and subsequently identifying unknown gases.

Gas Effusion Rate

The gas effusion rate is an expression of how quickly a gas exits an enclosed space. In Graham's Law, this rate is directly tied to the molar mass of the gas. If you know the effusion rate of one gas and the molar masses of both gases involved, you can find the effusion rate of the second gas.

For instance, if an unknown gas effuses 1.65 times faster than propane, you just need the molar mass of propane to calculate the unknown gas's molar mass using the formula from Graham's Law. Thus, by manipulating the relevant formula, you can derive the gas effusion rate in relation to another known substance.

Molar Mass of Gas

The molar mass of gas directly influences effusion rates according to Graham's Law. This is because the kinetic theory of gases suggests that all gases at a given temperature have the same kinetic energy on average. As a consequence, a lower molar mass means higher velocity, leading to a greater rate of effusion.

To put it in perspective, let's consider the example where an unknown gas effuses 1.65 times faster than propane. Knowing propane's molar mass allows us to deduce the molar mass of this unknown gas. Following the rearranged equation from Graham's Law, you can solve for the unknown gas's molar mass, \( M_X = \frac{44.095}{(1.65)^2} = 16.18 \text{g/mol} \). This calculation shows a direct link between a gas's molar mass and its effusion behavior which is crucial for identifying gases and understanding their physical properties.