Question

A gaseous mixture of \(\mathrm{O}_{2}\) and \(\mathrm{Kr}\) has a density of \(1.104 \mathrm{~g} / \mathrm{L}\) at 355 torr and \(400 \mathrm{~K}\). What is the mole percent \(\mathrm{O}_{2}\) in the mixture?

Short answer

The mole percent of Oxygen (\(\mathrm{O}_2\)) in the gaseous mixture is 75.5%.

Step-by-step solution

1. Write the Ideal Gas Law formula in terms of density and molar mass

The Ideal Gas Law is given by: \(PV = nRT\) where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature. We can write the Ideal Gas Law in terms of molar mass and density as follows: \(P = (\rho \frac{RT}{M})\) where P is the pressure, ρ is the density, R is the gas constant, T is the temperature, and M is the molar mass of the gas.

2. Calculate the molar mass of the mixture given its density, pressure, and temperature

Given the density (ρ) as 1.104 g/L, pressure (P) as 355 torr, and temperature (T) as 400 K, we can calculate the molar mass (M) of the mixture. Make sure to convert the pressure from torr to atm by dividing it by 760. \(P(atm) = \frac{355 \ torr}{760 \ torr/atm} = 0.4671 \ atm\) Now, we can plug in the given values into the Ideal Gas Law and solve for M: \(0.4671 \ atm = (\frac{1.104 \ g/L \cdot 0.0821 \ L \cdot atm/mol \cdot K \cdot 400 \ K}{M})\) Solving for M, we get: \(M = 9.90 \ g/mol\)

3. Write the expression for molar mass in terms of a mole fraction

Let x be the mole fraction of Oxygen in the mixture. Then, the mole fraction of Krypton is (1-x). The expression for the molar mass of the mixture is given by: \(M = x \cdot M_{O_2} + (1 - x) \cdot M_{Kr}\) where \(M_{O_2}\) and \(M_{Kr}\) are the molar masses of Oxygen and Krypton, respectively. The molar masses of Oxygen and Krypton are 32 g/mol and 83.798 g/mol, respectively.

4. Calculate the mole fraction of Oxygen in the mixture

Now we can set up the equation with the value of M that we found in Step 2 and the molar masses of Oxygen and Krypton: \(9.90 \ g/mol = x \cdot 32 \ g/mol + (1 - x) \cdot 83.798 \ g/mol\) Solving this equation for x, we get the mole fraction of Oxygen: \(x = 0.755\)

5. Calculate the mole percent of Oxygen in the mixture

Finally, we can convert the mole fraction of Oxygen to a mole percent by multiplying it by 100: Mole percent O2 = x * 100 = 0.755 * 100 = 75.5% Therefore, the mole percent of Oxygen in the gaseous mixture is 75.5%.

Key concepts

Molar Mass Calculations

Molar mass is an essential property in chemistry that helps us understand the weight of a mole of a substance, usually expressed in grams per mole (g/mol). It's determined by adding up the atomic masses of all the atoms in a molecule. For instance, the molar mass of oxygen (O extsubscript{2}) is calculated by doubling the atomic mass of a single oxygen atom, which results in 32 g/mol. For krypton (Kr), the molar mass is 83.798 g/mol.
Calculating the molar mass of a gas mixture is pivotal in applying the Ideal Gas Law. The mixture's molar mass can be found using the mixture's density, pressure, and temperature. With the aid of the Ideal Gas Law equation reformulated to include density and molar mass, one can solve for the molar mass. By understanding these principles, calculations regarding gas mixtures' properties become more intuitive. This knowledge is particularly useful when exploring more complex gas behaviors and reactions.

Mole Fraction and Mole Percent

The mole fraction is a way to express the composition of a mixture, which measures the proportion of a component in relation to the total mixture. It is a ratio, ranging from 0 to 1, of the number of moles of a specific gas to the total number of moles of all gases in the mixture. The mole fraction (x) can be calculated using the molar masses of the individual gases and the molar mass of the mixture.
In our problem, the mole fraction of oxygen was determined by the equation: \[ M = x \cdot M_{O_2} + (1 - x) \cdot M_{Kr} \]This equation allows us to weigh the contribution of each component based on its fraction and molar mass, leading to the mixed molar mass result. Once the mole fraction is calculated, converting it to a mole percent is straightforward: multiply the mole fraction by 100. This conversion offers a more intuitive percentage format and assists with further analyses. Here, the mole fraction of oxygen is found to be 0.755, translating to a mole percent of 75.5%.

Gas Mixtures and Gas Laws

Gas mixtures are common in various scientific and industrial applications. Understanding how they behave under different conditions is crucial. The Ideal Gas Law, represented by the equation \(PV = nRT\), is a foundational relation for describing how perfect gases behave under varying pressures (P), volumes (V), temperatures (T), and moles of gas (n).
For mixtures, it's important to remember that each component contributes to the total pressure proportionate to its mole fraction, as described by Dalton's Law of Partial Pressures. This law states that in a mixture of non-reacting gases, the total pressure is equal to the sum of the partial pressures of individual gases: \( P_{total} = P_1 + P_2 + \ldots + P_n \). Each partial pressure can be calculated using the mole fraction and the total pressure of the gas mixture.
By incorporating the gas laws, especially the Ideal Gas Law and Dalton's Law, one can effectively evaluate and predict the changes in gas mixtures' behaviors, which helps in understanding and optimizing various practical applications involving gases.