Problem 44 - Chemistry Matter and Change

Question

Calculate the molar mass of a gas that diffuses three times faster than oxygen under similar conditions.

Step-by-Step Solution

Answer

The molar mass of the unknown gas is approximately 3.56 g/mol.

Step 1: Understand Graham's Law of Effusion

Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically: \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \), where \( r_1 \) and \( r_2 \) are the rates of effusion of two gases, and \( M_1 \) and \( M_2 \) are their molar masses.

Step 2: Identify Known Values

We are given that the unknown gas diffuses three times faster than oxygen (\( r_{\text{unknown}} = 3 \times r_{\text{O}_2} \)). The molar mass of oxygen, \( M_{\text{O}_2} \), is 32 g/mol.

Step 3: Set Up the Equation

Using the relationship \( \frac{r_{\text{unknown}}}{r_{\text{O}_2}} = \sqrt{\frac{M_{\text{O}_2}}{M_{\text{unknown}}}} \), substitute \( r_{\text{unknown}} = 3 \times r_{\text{O}_2} \) into the equation, giving \( 3 = \sqrt{\frac{32}{M_{\text{unknown}}}} \).

Step 4: Solve for Molar Mass of Unknown Gas

Square both sides to remove the square root: \( 9 = \frac{32}{M_{\text{unknown}}} \). Solve for \( M_{\text{unknown}} \) by rearranging the equation: \( M_{\text{unknown}} = \frac{32}{9} \).

Step 5: Calculate the Result

Conduct the division: \( M_{\text{unknown}} = \frac{32}{9} \approx 3.56 \text{ g/mol} \).

Key Concepts

Molar Mass Calculation

Molar mass is a critical factor in understanding the properties of gases. It is defined as the mass of one mole of a substance, usually expressed in grams per mole (g/mol). Understanding how to calculate molar mass helps in determining the characteristics of various gases. Here's how you can calculate it:

• First, identify the chemical formula of the compound.
• Next, find the atomic mass of each element present in the compound using the periodic table.
• Multiply the atomic mass of each element by the number of times the element appears in the chemical formula.
• Add all these values together to find the total molar mass of the compound.

In the exercise, we determine the molar mass of an unknown gas by utilizing Graham's Law, alongside known quantities. Performing calculations like these can deepen your understanding of how gases behave under different conditions.

Diffusion Rate

Diffusion rate is a term used to describe the speed at which a gas spreads out and mixes with other gases. This concept is crucial when discussing gas behavior, especially in the context of Graham's Law of Effusion. To understand diffusion rate:

• Recognize diffusion as the process by which gas molecules spread from areas of high concentration to areas of lower concentration.
• The diffusion rate depends on factors like temperature, pressure, and molar mass.

In the context of Graham's Law, the rate of diffusion of a gas is inversely related to the square root of its molar mass. Thus, lighter molecules diffuse more quickly than heavier molecules. In our example, since the unknown gas diffuses three times faster than oxygen, its rate of diffusion is significantly higher. This high rate indicates a lower molar mass for the unknown gas compared to oxygen.

Oxygen

Oxygen is a vital element and a key component in many biological and chemical processes. It is essential for respiration in living organisms and is also involved in combustion and various industrial processes. In chemistry, oxygen usually exists as a diatomic molecule represented by the symbol \( ext{O}_2 \). With a molar mass of 32 g/mol, oxygen serves as a benchmark for comparing the diffusion rates of other gases.In our calculation, oxygen helps demonstrate how differing molar mass affects the rate of effusion for gases. Because the diffusion rate of our unknown gas is three times that of oxygen, we use the known molar mass of oxygen to efficiently solve for the unknown gas's molar mass. Understanding oxygen’s role in these calculations helps in interpreting various physical and chemical phenomena.

Inverse Proportionality

Inverse proportionality is a mathematical principle where one value increases as another value decreases, and vice versa. This concept is central to understanding Graham's Law of Effusion.Graham's Law illustrates inverse proportionality through the equation:\[\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}\]Here, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.For practical applications:

• If the molar mass of a gas doubles, its diffusion rate will be reduced to about 0.71 times its original speed (the square root of 0.5).
• Conversely, if a gas's rate of diffusion triples, its molar mass would be 1/3 of the reference gas’s molar mass, as shown in our exercise calculation.

Understanding this concept allows us to predict how changes in molar mass will impact the rate at which gases diffuse or effuse, providing insights into their behavior under different conditions.