Problem 49 - Chemistry and Chemical Reactivity

Question

A gas whose molar mass you wish to know effuses through an opening at a rate one third as fast as that of helium gas. What is the molar mass of the unknown gas?

Step-by-Step Solution

Answer

The molar mass of the unknown gas is 36 g/mol.

Step 1: Understanding the Problem

We are asked to find the molar mass of an unknown gas. It effuses at a rate one-third as fast as helium, whose molar mass is known to be 4 g/mol. We'll use Graham's law of effusion to solve this problem.

Step 2: Applying 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 is expressed as:\[\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}\]where \(r_1\) and \(r_2\) are the rates of effusion of the two gases, and \(M_1\) and \(M_2\) are their molar masses.

Step 3: Setting Up the Equation

Let the rate of helium be \(r_{\text{He}} = r_1\) and the rate of the unknown gas be \(r_2\). Given that \(r_2 = \frac{1}{3} r_1\), we substitute into Graham's law:\[\frac{r_{\text{He}}}{\frac{1}{3} r_{\text{He}}} = \sqrt{\frac{M}{4}}\]

Step 4: Simplifying the Equation

Simplify the left side of the equation (rates of effusion):\[\frac{r_{\text{He}}}{\frac{1}{3} r_{\text{He}}} = 3\]This equation becomes:\[3 = \sqrt{\frac{M}{4}}\]

Step 5: Solving for the Molar Mass

Square both sides to eliminate the square root:\[9 = \frac{M}{4}\]Multiply both sides by 4 to solve for \(M\):\[M = 36\]

Step 6: Conclusion

The molar mass of the unknown gas is 36 g/mol.

Key Concepts

Molar mass calculation

Molar mass calculation involves determining the mass of one mole of a substance, often measured in grams per mole (g/mol). In chemistry, the molar mass is crucial for converting between the mass of a substance and the amount of substance in moles.
The molar mass can be calculated simply by adding the atomic masses of each element in a compound. For example, to find the molar mass of water (H₂O), sum the atomic masses of two hydrogen atoms (1 g/mol each) and one oxygen atom (16 g/mol), resulting in 18 g/mol.
In our problem,

• we used Graham's law of effusion to indirectly determine the molar mass of an unknown gas.
• This method leverages the known molar mass of helium (4 g/mol) and uses the relationship of effusion rates to calculate the unknown gas mass, found as 36 g/mol.

By setting the equation according to Graham's law and solving, we effectively translated a physical property (effusion rate) into a molar mass calculation.

Rate of effusion

The rate of effusion refers to the speed at which a gas passes through a tiny opening into a vacuum. It is a concept closely related to diffusion, but effusion involves gas movement through an orifice much smaller than the average distance between molecules.
According to Graham's law of effusion, the rate at which a gas effuses is inversely proportional to the square root of its molar mass. Given by:\[\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}\]

• where \(r_1\) and \(r_2\) are the effusion rates of two gases,
• and \(M_1\) and \(M_2\) are their respective molar masses.

Graham's law provides a useful quantitative relationship that helps in identifying unknown gases through their effusion rates.
In the exercise, we compared the unknown gas's effusion rate to helium's. Since the unknown's effusion rate was one-third that of helium, we applied Graham's law to find its molar mass, ultimately setting up and solving the equation accordingly.

Gas laws

Gas laws are a set of fundamental principles that describe the behavior of gases. They provide relationships between measurable properties of gases such as pressure, volume, temperature, and the number of particles. The most relevant to our discussion is Graham's Law, which is a part of the broader kinetic molecular theory.
Graham's Law of Effusion, for instance, asserts that the rate of gas effusion is inversely proportional to the square root of its molar mass. This is especially useful in comparing different gases under similar conditions.

• In our example, Graham's Law facilitated the comparison between helium and an unknown gas by linking their rates of effusion to their molar masses.
• This law is part of a larger framework, including Boyle's Law, Charles's Law, and Avogadro's Law, all of which describe gas properties.

By using these laws, scientists and students can predict and calculate various gas-related phenomena, which is essential in fields ranging from chemistry to environmental science. Understanding these relationships empowers learners to interpret and solve real-world gas behavior problems effectively.