Question

An unknown gas effuses at 0.850 times the effusion rate of nitrogen dioxide, \(NO_{2}.\) Estimate the molar mass of the unknown gas.

Short answer

The molar mass of the unknown gas is approximately 63.67 g/mol.

Step-by-step solution

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. This can be represented by the equation: \ \ \[ \frac{{\text{{Rate}}_1}}{{\text{{Rate}}_2}} = \sqrt{\frac{{M_2}}{{M_1}}} \] \ \ Here, \( \text{{Rate}}_1 \) is the effusion rate of the unknown gas, \( \text{{Rate}}_2 \) is the effusion rate of nitrogen dioxide, \( M_1 \) is the molar mass of the unknown gas, and \( M_2 \) is the molar mass of nitrogen dioxide.

Set Up the Equation with Given Values

We know that the effusion rate of the unknown gas is 0.850 times that of nitrogen dioxide: \ \ \[ \frac{{\text{{Rate}}_1}}{{\text{{Rate}}_2}} = 0.850 \] \ \ The molar mass of nitrogen dioxide \( (NO_2) \) is \( M_2 = 46 \text{{ g/mol}} \). We need to find \( M_1 \).

Solve the Equation for the Unknown Molar Mass

Using Graham's Law, we substitute the given values into the equation: \ \ \[ 0.850 = \sqrt{\frac{{46}}{{M_1}}} \] \ \ Square both sides to get rid of the square root: \ \ \[ (0.850)^2 = \frac{{46}}{{M_1}} \] \ \ Simplify: \ \ \[ 0.7225 = \frac{{46}}{{M_1}} \] \ \ Rearrange to solve for \( M_1 \): \ \ \[ M_1 = \frac{{46}}{{0.7225}} \] \ \ \[ M_1 \approx 63.67 \text{{ g/mol}} \]

Key concepts

Effusion Rate

Effusion is the process by which gas particles pass through a tiny opening. The rate at which this happens is called the effusion rate.
According to Graham's Law of Effusion, the effusion rate of a gas is inversely proportional to the square root of its molar mass. This means that lighter gases effuse faster than heavier gases.
The formula for Graham's Law of Effusion is:
\[ \frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} \]
In this equation:

• \(\text{Rate}_1\) is the effusion rate of one gas (usually the unknown gas)
• \(\text{Rate}_2\) is the effusion rate of the comparison gas
• \(M_1\) is the molar mass of the unknown gas
• \(M_2\) is the molar mass of the comparison gas

This relationship helps us determine the molar mass of an unknown gas by comparing its effusion rate with a known gas.

Molar Mass

The molar mass of a substance is the mass of one mole of that substance. It is usually expressed in grams per mole (g/mol). Knowing the molar mass is crucial for various calculations in chemistry, including the application of Graham's Law of Effusion.
To find the molar mass, we sum the atomic masses of all atoms in the molecule. For example, for nitrogen dioxide \((NO_2)\):

• The atomic mass of nitrogen (N) is approximately 14.01 g/mol
• The atomic mass of oxygen (O) is approximately 16.00 g/mol
• Since \( NO_2 \) has one nitrogen atom and two oxygen atoms, its molar mass is: \[M_{NO_2} = 14.01 + 2 \times 16.00 = 46.01 \text{g/mol}\]

In this exercise, we compared the effusion rate of an unknown gas to nitrogen dioxide to estimate its molar mass. By substituting the known values into Graham's Law, we solved for the molar mass of the unknown gas.

Nitrogen Dioxide

Nitrogen dioxide \((NO_2)\) is a reddish-brown gas with a molar mass of 46 g/mol. It is commonly used in experiments involving effusion or diffusion because its properties and molar mass are well-known.
When comparing the effusion rate of another gas to nitrogen dioxide, we can use the molar mass of \( NO_2 \) to determine the unknown gas's molar mass. In this exercise:

• The effusion rate of the unknown gas was given as 0.850 times that of \( NO_2 \)
• Using Graham's Law, we set up the equation: \[0.850 = \sqrt{\frac{46}{M_1}}\]
• By solving this equation, we found that the molar mass of the unknown gas was approximately 63.67 g/mol

This illustrates how knowing the properties of nitrogen dioxide allows us to estimate the characteristics of other gases.