Problem 98
Question
A gas mixture contains \(7.0 \mathrm{g} \mathrm{N}_{2}, 2.0 \mathrm{g} \mathrm{H}_{2},\) and \(16.0 \mathrm{g} \mathrm{CH}_{4}\) What is the mole fraction of \(\mathrm{H}_{2}\) in the mixture?
Step-by-Step Solution
Verified Answer
Answer: The mole fraction of H₂ in the gas mixture is approximately 0.444.
1Step 1: Calculate the moles of each gas
To calculate the moles of each gas, we need to use the formula: moles = mass/molar mass. Let's find the moles of N₂, H₂, and CH₄ in the mixture.
Molar mass of N₂ = 28 g/mol
Moles of N₂ = 7.0 g / 28 g/mol = 0.25 mol
Molar mass of H₂ = 2 g/mol
Moles of H₂ = 2.0 g / 2 g/mol = 1.0 mol
Molar mass of CH₄ = 16 g/mol
Moles of CH₄ = 16.0 g / 16 g/mol = 1.0 mol
2Step 2: Calculate the total moles in the mixture
Add the moles of each gas to find the total moles in the mixture:
Total moles = Moles of N₂ + Moles of H₂ + Moles of CH₄
Total moles = 0.25 mol + 1.0 mol + 1.0 mol
Total moles = 2.25 mol
3Step 3: Calculate the mole fraction of H₂
To find the mole fraction of H₂, divide the moles of H₂ by the total moles in the mixture:
Mole fraction of H₂ = (Moles of H₂) / (Total moles)
Mole fraction of H₂ = (1.0 mol) / (2.25 mol)
Mole fraction of H₂ ≈ 0.444
The mole fraction of H₂ in the gas mixture is approximately 0.444.
Key Concepts
Gas MixtureMole CalculationMolar Mass
Gas Mixture
In the context of chemistry, a gas mixture refers to a combination of different gases that coexist without chemically reacting with each other. Take the air we breathe, for instance; it's a mix of nitrogen, oxygen, carbon dioxide, and several other gases. Similarly, the gas mixture in our original exercise contains nitrogen (\(\text{N}_2\)), hydrogen (\(\text{H}_2\)), and methane (\(\text{CH}_4\)). When we analyze these mixtures, each gas behaves independently, following the same gas laws, such as Boyle's or Charles's Law, as if the other gases aren't there. This independent behavior is key because it allows us to calculate properties like mole fraction for each component gas independently. When tasked with solving gas mixture problems, it involves understanding each component's contribution in terms of moles. The mole is a measurement unit in chemistry used to express amounts of a chemical substance. To truly understand gas mixtures, it is important to explore the concepts of mole calculation and molar mass, which we will delve into next.
Mole Calculation
To determine the mole fraction of a specific gas in a gas mixture, such as \(\text{H}_2\) in the example provided, you first need to calculate the moles of each gas. This involves using the formula:\[\text{moles} = \frac{\text{mass}}{\text{molar mass}}\]Using this equation, we can determine how many moles each component contributes to the overall mixture. Here's how it works:
- Each gas's mass is divided by its molar mass (like Nitrogen's 28 g/mol, Hydrogen's 2 g/mol, etc.).
- This calculation yields the number of moles for each gas, showing which has a greater or lesser presence within the mixture.
- The sum of these individual mole amounts gives us the total moles in the mixture, a crucial step for further calculations.
Molar Mass
Molar mass serves as a bridge between the macroscopic world (what we see and measure) and the molecular world (what's happening at an atomic level). It tells us the mass of one mole of a substance and is expressed in grams per mole (g/mol). Every element in the periodic table has a defined molar mass, which can be compiled to determine the molar mass of compounds (like \(\text{N}_2\) or \(\text{CH}_4\)). The molar mass is crucial because it allows us to convert between the mass of a compound and the number of moles. This conversion is necessary for calculating mole fractions in gas mixtures. To clarify, here are some key points:
- Calculate molar masses from atomic masses found in the periodic table. For example, \(\text{N}_2\) has a molar mass of 28 g/mol (14 g/mol per nitrogen atom multiplied by two).
- Use these molar masses in the mole calculation formula to find the moles present in a mixture.
- This straightforward conversion process reveals the mole ratios within mixtures, imperative for mole fraction calculations.
Other exercises in this chapter
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