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Assuming air to be composed exclusively of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\), with their partial pressures in the ratio 0.21:0.79, what are their mass fractions?

Short Answer

Expert verified
\( \mathrm{O}_2 \) has a mass fraction of 0.233, and \( \mathrm{N}_2 \) has a mass fraction of 0.767.

Step by step solution

01

Understand the Problem

We are given the ratio of the partial pressures of \( \mathrm{O}_2 \) to \( \mathrm{N}_2 \) as 0.21:0.79. We need to calculate the mass fractions of \( \mathrm{O}_2 \) and \( \mathrm{N}_2 \) in this mixture.
02

Assume Total Pressure

Assume the total pressure \( P \) of the system is 1 atm. Then the partial pressure of \( \mathrm{O}_2 \) is \( P_{O_2} = 0.21 \times P \) and the partial pressure of \( \mathrm{N}_2 \) is \( P_{N_2} = 0.79 \times P \).
03

Use Dalton's Law of Partial Pressures

Dalton's Law states that the total pressure is the sum of the partial pressures. Since we assumed a total pressure of 1 atm, we have: \( P = P_{O_2} + P_{N_2} = 0.21 + 0.79 = 1 \text{ atm}\).
04

Calculate Mole Fractions

The mole fraction \( X \) of a gas in a mixture is the ratio of its partial pressure to the total pressure. Thus, \( X_{O_2} = \frac{0.21}{1} = 0.21 \) and \( X_{N_2} = \frac{0.79}{1} = 0.79 \).
05

Find Molar Mass of Gases

The molar mass of \( \mathrm{O}_2 \) is 32 g/mol, and the molar mass of \( \mathrm{N}_2 \) is 28 g/mol.
06

Calculate Mass Contribution

Calculate the mass contribution of each gas using the formula: \( \text{mass} = \text{molar mass} \times \text{mole fraction} \). For \( \mathrm{O}_2 \), mass contribution is \( 32 \times 0.21 = 6.72 \text{ g/mol} \). For \( \mathrm{N}_2 \), mass contribution is \( 28 \times 0.79 = 22.12 \text{ g/mol} \).
07

Calculate Total Mass

The total mass of the mixture is the sum of the mass contributions: \( 6.72 + 22.12 = 28.84 \text{ g/mol} \).
08

Determine Mass Fractions

The mass fraction is calculated by dividing the mass contribution of each component by the total mass. For \( \mathrm{O}_2 \), the mass fraction is \( \frac{6.72}{28.84} \approx 0.233 \). For \( \mathrm{N}_2 \), the mass fraction is \( \frac{22.12}{28.84} \approx 0.767 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Pressure
In gas mixtures, each individual gas exerts its own pressure, known as partial pressure. It's as if each gas in the mixture is acting independently. Partial pressure is an essential concept in understanding gas behavior in mixtures. When we talk about the partial pressure of a gas, we are referring to the pressure that gas would exert if it occupied the entire volume by itself at the same temperature. This concept helps break down the individual contributions that different gases make to the overall pressure of the mixture. For example, in our exercise, the partial pressures of oxygen (\(\mathrm{O}_2\)) and nitrogen (\(\mathrm{N}_2\)) in the air are given as a ratio, 0.21:0.79. This ratio indicates how the pressure exerted by each gas compares when combined within the same system.
Mole Fraction
The mole fraction provides a simple way to express the composition of a gas mixture based on the number of moles. It is the ratio of the number of moles of a particular component to the total number of moles in the mixture. Mathematically:
  • Mole fraction (\(X\)) = number of moles of gas / total number of moles
In our context, since the partial pressures are given, we can use Dalton’s Law. The mole fraction of oxygen (\(X_{O_2}\)) is calculated as 0.21, and for nitrogen (\(X_{N_2}\)) as 0.79. These values come from the ratio of the partial pressures. The mole fraction plays a crucial role when determining other properties of the gas mixture, such as mass fraction.
Mass Fraction
Mass fraction represents the proportion of a substance's mass compared to the total mass of the mixture. This is useful when you need to know how much of the total mass is made up by each component.
  • Mass fraction = mass of component / total mass of mixture
In the exercise, the mass contributions of both oxygen and nitrogen were calculated based on their mole fractions and molar masses. Using these mass contributions, the mass fraction of oxygen was found to be approximately 0.233, while for nitrogen, it was about 0.767. These fractions help us understand the weight distribution of each component within the mixture.
Oxygen
Oxygen is a diatomic molecule (\(\mathrm{O}_2\)) and a crucial component of the Earth's atmosphere. It is highly reactive, combining with most elements and playing a vital role in processes like combustion and respiration. In our exercise, oxygen was considered with its partial pressure at 0.21 times the total pressure, reflecting its presence in the air mixture. Knowing its molar mass (32 g/mol) is important when determining its mass contribution and exactly understanding its role within the gas mixture. Oxygen's role, both in the exercise and in real life, illustrates its importance in maintaining balance and functionality within different systems, including biological and chemical ones.
Nitrogen
Nitrogen, like oxygen, is diatomic (\(\mathrm{N}_2\)) and comprises a significant portion of the Earth's atmosphere. It is relatively inert compared to other gases, which makes it a stabilizing presence in the air. In our example, its partial pressure, which is 0.79 of the system, demonstrates its dominance in the gaseous mixture of air. With a molar mass of 28 g/mol, nitrogen contributes significantly to the total mass, resulting in a higher mass fraction compared to oxygen. This inertness and abundance make nitrogen essential in processes like the nitrogen cycle and applications requiring non-reactive environments, illustrating its versatility and importance on the global scale.

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Most popular questions from this chapter

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