Question

A gas mixture contains \(50 \%\) helium and \(50 \%\) methane by volume. What is the per cent by weight of methane in the mixture? a. \(83 \%\) b. \(73.0 \%\) c. \(18.01 \%\) d. \(80.0 \%\) e. \(90.1 \%\)

Short answer

Methane accounts for 80.0% by weight in the gas mixture. (Option d)

Step-by-step solution

Identify the Known Variables

Helium and methane are the components of the gas mixture. We know their volume percentages as follows: helium is 50% by volume, and methane is 50% by volume. To find the percent by weight of methane, we need to calculate the mass of each component and their total mass. We also need the molar masses of helium and methane: the molar mass of helium is 4 g/mol, and methane is 16 g/mol.

Assume a Total Volume

Assume a convenient total volume of the gas mixture for calculation, such as 100 L. This choice simplifies the percentages to 50 L of helium and 50 L of methane, aligning with the given volume percentages.

Calculate the Mass of Helium

Helium's volume is 50 L. Using density, where 1 mole of any ideal gas occupies 22.4 L at Standard Temperature and Pressure (STP), the moles of helium are \( \frac{50}{22.4} \). Hence, the mass is \( \frac{50}{22.4} \times 4 \approx 8.93 \text{ g} \).

Calculate the Mass of Methane

The methane volume is also 50 L. Again using the 22.4 L per mole at STP, the moles of methane are \( \frac{50}{22.4} \). Thus, the mass of methane is \( \frac{50}{22.4} \times 16 \approx 35.71 \text{ g} \).

Calculate the Total Mass of the Mixture

Add the masses of helium and methane to find the total mass: \( 8.93 \text{ g} + 35.71 \text{ g} = 44.64 \text{ g} \).

Calculate the Percent by Weight of Methane

The percent by weight of methane is calculated as \( \frac{35.71}{44.64} \times 100 \approx 80.0\% \).

Verify Against Given Options

Compare the calculated percent by weight of methane, 80.0%, with the given options. It matches option d.

Key concepts

Understanding Molar Mass

Molar mass is an essential concept in chemistry when dealing with gas mixtures. It refers to the mass of one mole of a substance, which helps in calculating the number of moles in a given sample. For gases, knowing the molar mass allows us to determine how much of each gas by mass is present in a mixture. The standard unit for molar mass is grams per mole (g/mol).
In the context of the problem, helium has a molar mass of 4 g/mol and methane's is 16 g/mol. These values are crucial for finding the number of moles of each gas in a given volume and then translating that into a mass, which we need to calculate the percentage by weight of the gases in a mix. By knowing the molar mass, you can bridge the gap between volume measurements typical of gases and mass measurements needed for understanding mixtures better.
Learning to calculate the mass of each gas in a mixture using molar mass and volume can enhance your comprehension of stoichiometry and solution preparation.

Volume Percentage Made Simple

Volume percentage is an intuitive way to express the composition of gas mixtures. It indicates what fraction of the total volume each component of the mixture occupies. For example, if a mixture is composed of 50% helium and 50% methane by volume, it means half of the container's total volume is filled with each gas.
This is particularly useful in contexts involving gases because it aligns with how we usually describe gases by how much space they take up. Indeed, in the problem scenario, using volume percentages simplifies the calculations done later for mass and percentage by weight. It allows for an easy assumption of a total gas volume – like 100 L – ensuring that the percentage numbers directly correlate to volume values.

• Volume percentages offer a straightforward simplification when dealing with gases in mixtures.
• Converting these volumes into masses through molar mass calculations helps find weight percentages, critical in many scientific applications.

The Concept of an Ideal Gas

The concept of an ideal gas forms a fundamental part of understanding gases. An ideal gas is a theoretical notion where the molecules do not interact except during perfectly elastic collisions, and the volume of individual molecules is negligible.
The Ideal Gas Law describes this behavior with the equation: \[ PV = nRT \]
Where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature.
In many problems, including the original exercise, gases are assumed to behave ideally under specific conditions, which simplifies calculations. This simplifies understanding gases in real-world applications and allows direct relationships between pressure, volume, and temperature to ascertain quantities like molar mass and volume percentage.
Understanding the ideal gas concept can help you make better predictions about how real gases behave under various conditions.

Deciphering STP Conditions

STP stands for Standard Temperature and Pressure, which is an important reference point in chemistry. Under STP conditions, the temperature is 0°C or 273.15 K, and the pressure is 1 atm. At these conditions, one mole of any ideal gas occupies 22.4 liters.
This is particularly significant as the calculations performed in the problem assume STP conditions for simplicity. Knowing the volume of one mole allows you to back-calculate the number of moles from a given volume of gas, facilitating the conversion from volume percentage to mass or moles, as seen in the exercise.

• STP provides a consistent foundation for gas calculations.
• This allows chemists to compare results and perform calculations with ease across different experiments.

By using STP conditions, we achieve a standardization in calculations involving gases, helping students and scientists accurately predict outcomes in a controlled environment.