Problem 25 - Chemistry in 30 days

Question

Equal masses of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\) are mixed in an empty container at \(25^{\circ}\). The fraction of the total pressure exerted by oxygen is: (a) \(1 / 3\) (b) \(1 / 2\) (c) \(2 / 3\) (d) \((1 / 3) \times(273 / 298)\)

Step-by-Step Solution

Answer

The fraction of the total pressure exerted by \(\mathrm{O}_{2}\) is \(\frac{1}{3}\) (answer a).

Step 1: Determine Molar Masses

First, we need the molar masses of methane (\(\mathrm{CH}_{4}\)) and oxygen (\(\mathrm{O}_{2}\)). \(\mathrm{CH}_4\) has a molar mass of about 16 g/mol (\(12 + 4 \times 1 = 16\)), and \(\mathrm{O}_2\) has a molar mass of about 32 g/mol (\(2 \times 16 = 32\)).

Step 2: Calculate Moles of Methane and Oxygen

Since equal masses of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\) are mixed, assume we have \(x\) grams of each. The number of moles of \(\mathrm{CH}_{4}\) is \(\frac{x}{16}\), and the number of moles of \(\mathrm{O}_{2}\) is \(\frac{x}{32}\).

Step 3: Calculate Total Moles in the Mixture

Add the moles of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\). The total moles is \(\frac{x}{16} + \frac{x}{32}\). Simplify to get the total moles: \(\frac{2x}{32} + \frac{x}{32} = \frac{3x}{32}\).

Step 4: Find Fraction of Total Pressure by Oxygen

The pressure contribution by a gas is proportional to its mole fraction. The mole fraction of \(\mathrm{O}_{2}\) is \(\frac{x/32}{3x/32} = \frac{1}{3}\). Therefore, \(\mathrm{O}_{2}\) exerts \(\frac{1}{3}\) of the total pressure.

Key Concepts

Molar Mass

Molar mass is a key concept when dealing with gases and mixtures. It represents the mass of one mole of a substance, usually expressed in grams per mole (g/mol). Understanding molar mass is crucial for converting between mass and moles, which is an essential step in many chemical calculations.
For example, in our exercise, the molar mass of methane (CH$_4$) is 16 g/mol. This is calculated by adding the atomic masses of carbon (12 g/mol) and hydrogen (1 g/mol each). Likewise, the molar mass of oxygen (O$_2$) is 32 g/mol because each molecule contains two oxygen atoms, each with an atomic mass of 16 g/mol.
By understanding and using these values, we can calculate how many moles of each gas are present in a given mass. This forms the foundation for further calculations involving reactions and gas mixtures.

Mole Fraction

The mole fraction is a way to express the concentration of a component in a mixture. It is the ratio of the moles of a component to the total moles in the mixture. This concept is vital in calculating how each component contributes to properties like pressure.
In our problem, the mole fraction of oxygen is calculated as the ratio of its moles to the total moles present.

• The moles of O\(_2\) are calculated using its molar mass (32 g/mol).
• The mole fraction of O\(_2\) becomes \( \frac{x/32}{3x/32} \), which simplifies to \( \frac{1}{3} \).

This calculated mole fraction tells us that oxygen contributes \( \frac{1}{3} \) of the total pressure exerted by the gas mixture in the container.

Pressure

Pressure in the context of gases is the force that gas molecules exert when they collide with the walls of their container. It is measured in units such as atmospheres (atm), pascals (Pa), or torr.
When dealing with mixtures of gases, like in our exercise, the total pressure is the sum of the partial pressures contributed by each component. The partial pressure of each gas in a mixture is directly proportional to its mole fraction.

• In this case, since the mole fraction of oxygen is \( \frac{1}{3} \), oxygen contributes \( \frac{1}{3} \) of the total pressure.

This concept simplifies comparisons and calculations involving gas mixtures, where knowing just the mole fraction and total pressure allows one to determine the individual pressures.

Ideal Gas Law

The Ideal Gas Law is a critical equation in chemistry that relates four properties of gases: pressure (P), volume (V), amount in moles (n), and temperature (T). The mathematical form is \( PV = nRT \), where R is the universal gas constant.
This law assumes that gases behave ideally, meaning they are composed of point-like particles with no volume and experience no intermolecular forces.

• Although real gases don't strictly follow this model, the Ideal Gas Law provides a good approximation under many conditions.

In the context of mixtures, the Ideal Gas Law helps demonstrate how different components affect the overall properties, like total pressure, by relating them to the number of moles and temperature. It plays a crucial role in calculating gas behaviors in containers, making it indispensable for solving exercises dealing with gas mixtures.