Problem 111 - Chemistry An Atoms First Approach

Question

Freon-\(12\) is used as a refrigerant in central home air conditioners. The rate of effusion of Freon-\(12\) to Freon-\(11\) (molar mass \(=137.4 \mathrm{g} / \mathrm{mol})\) is \(1.07: 1 .\) The formula of Freon- \(12\) is one of the following: \(\mathrm{CF}_{4}, \mathrm{CF}_{3} \mathrm{Cl}, \mathrm{CF}_{2} \mathrm{Cl}_{2}, \mathrm{CFCl}_{3},\) or \(\mathrm{CCl}_{4} .\) Which formula is correct for Freon- \(12 ?\)

Step-by-Step Solution

Answer

The correct formula for Freon-12 is \( CF_{2}Cl_{2} \).

Step 1: Recall Graham's Law of Effusion

Graham's Law of Effusion relates the rates of effusion for two gases to the inverse square root of their molar masses: \( \frac{Rate_{1}}{Rate_{2}} = \frac{ \sqrt{Molar\:Mass_{2}}}{\sqrt{Molar\:Mass_{1}}} \) In this case, Rate 1 is the rate of effusion of Freon-12, and Rate 2 is the rate of effusion of Freon-11.

Step 2: Plug in the given rate of effusion

We are given the rate of effusion of Freon-12 to Freon-11 as 1.07 : 1. \( \frac{Rate_{Freon12}}{Rate_{Freon11}} = \frac{1.07}{1} \)

Step 3: Plug in the molar mass of Freon-11

We are given the molar mass of Freon-11 as 137.4 g/mol. Plug this value into the equation: \( \frac{1.07}{1} = \frac{\sqrt{Molar\:Mass_{Freon12}}}{\sqrt{137.4}} \)

Step 4: Solve for the molar mass of Freon-12

To isolate the molar mass of Freon-12, square both sides of the equation and multiply by the molar mass of Freon-11: \( Molar\:Mass_{Freon12} = (1.07)^2 \times 137.4 \) \( Molar\:Mass_{Freon12} = 1.1449 \times 137.4 \) \( Molar\:Mass_{Freon12} ≈ 157.4 \:g/mol \)

Step 5: Determine the correct formula for Freon-12

Now that we have the approximate molar mass of Freon-12, we can compare it to the molar masses of the given formulas to find the correct one: 1. CF4: \(12 + (4 \times 19) = 12 + 76 = 88\:g/mol\) 2. CF3Cl: \(12 + (3 \times 19) + 35.5 = 12 + 57 + 35.5 = 104.5\:g/mol\) 3. CF2Cl2: \(12 + (2 \times 19) + (2 \times 35.5) = 12 + 38 + 71 = 121\:g/mol\) 4. CFCl3: \(12 + 19 + (3 \times 35.5) = 12 + 19 + 106.5 = 137.5\:g/mol\) 5. CCl4: \(12 + (4 \times 35.5) = 12 + 142 = 154\:g/mol\) Comparing the calculated molar mass of each formula with the molar mass we found for Freon-12 (157.4 g/mol), CF2Cl2 appears to be closest to the target value. Thus, the correct formula for Freon-12 is: \( CF_{2}Cl_{2} \).

Key Concepts

Graham's Law of Effusion

When comparing the behavior of gases, one of the fascinating aspects we observe is how different gases escape through tiny openings, which is a process called effusion. The rate of effusion for a gas is inversely proportional to the square root of its molar mass, a relationship beautifully summarized by Graham's Law of Effusion.

To put it into perspective, let's think of two gases, say gas A and B; if gas A has a lower molar mass than gas B, then gas A will effuse faster through a pinhole under the same conditions. The formula given by Graham's Law, \( \frac{Rate_{1}}{Rate_{2}} = \frac{ \sqrt{Molar\:Mass_{2}}}{\sqrt{Molar\:Mass_{1}}} \), lets us precisely compare these rates mathematically.

Molar Mass Calculation

In order to employ Graham's Law effectively, we need to determine the molar mass of gases. Molar mass is the mass of one mole of a substance and it serves as a bridge between the microscopic world of atoms and molecules and the macroscopic world we live in.

It's calculated by summing the atomic masses of all atoms in a molecule as listed on the periodic table. For instance, the molar mass of carbon dioxide (CO₂) is obtained by adding the mass of one carbon atom with the masses of two oxygen atoms. This step is key in solving effusion-related problems, because it allows us to relate a gas's effusion rate to its molar mass.

Rate of Effusion

The rate of effusion, simplistically spelled out, is how fast a gas escapes through an opening into a vacuum. Variables such as temperature, pressure, and the size of the opening can influence this rate, but the molar mass of the gas remains pivotal. In comparative terms, lighter gases effuse more swiftly than heavier ones due to having higher average velocities, which is a concept grounded in kinetic molecular theory.

Graham's Law demonstrates that when you have the rate of effusion of one gas and you seek the other, you can compute it using the mathematical relationship provided, as long as the molar mass is known. This is especially useful in practical contexts like separating isotopes or even in our textbook example, calculating unknown molar masses.

Effusion Process in Gases

The effusion process in gases may seem simple, but it touches on the fundamental characteristics of gases. Effusion occurs when gas molecules move through tiny openings from an area of higher pressure to an area of lower pressure. Unlike diffusion where gas spreads out in another gas or space, effusion is like a directed escape through a hole.

One interesting thing to note is that at a given temperature and pressure, all gas molecules have the same kinetic energy. However, as per the kinetic molecular theory, the lighter the gas, the higher its velocity, which accounts for the variance in effusion rates among gases. This distinction is what makes Graham's Law a valuable tool for scientists and engineers in processes like air purification, manufacturing semiconductors, and even controlling chemical reactions.