Problem 27 - Chemistry and Chemical Reactivity

Question

A 1.007 -g sample of an unknown gas exerts a pressure of \(715 \mathrm{mm} \mathrm{Hg}\) in a \(452-\mathrm{mL}\) container at \(23^{\circ} \mathrm{C} .\) What is the molar mass of the gas?

Step-by-Step Solution

Answer

The molar mass of the gas is approximately 56.9 g/mol.

Step 1: Convert Pressure to Atmospheres

The problem gives pressure in mmHg. We need to convert this to atmospheres because we will use the ideal gas law, which requires pressure in atmospheres. Use the conversion: \(1 \, \text{atm} = 760 \, \text{mmHg}\).\[P = \frac{715 \, \text{mmHg}}{760 \, \text{mmHg/atm}} = 0.9408 \, \text{atm}\]

Step 2: Convert Volume to Liters

The volume of the container is in milliliters and needs to be converted to liters for the ideal gas law.\[V = \frac{452 \, \text{mL}}{1000 \, \text{mL/L}} = 0.452 \, \text{L}\]

Step 3: Convert Temperature to Kelvin

The temperature is given in degrees Celsius. Convert this to Kelvin by adding 273.15.\[T = 23 + 273.15 = 296.15 \, \text{K}\]

Step 4: Use the Ideal Gas Law to Find Moles

Use the ideal gas law equation: \(PV = nRT\), where \(R = 0.0821 \, \text{L atm/mol K}\). Solve for \(n\) (moles of gas).\[n = \frac{PV}{RT} = \frac{(0.9408 \, \text{atm})(0.452 \, \text{L})}{(0.0821 \, \text{L atm/mol K})(296.15 \, \text{K})} \approx 0.0177 \, \text{mol}\]

Step 5: Calculate Molar Mass

The molar mass \(M\) is the mass of the gas divided by the number of moles. Given the mass is \(1.007 \, \text{g}\).\[M = \frac{1.007 \, \text{g}}{0.0177 \, \text{mol}} \approx 56.9 \, \text{g/mol}\]

Key Concepts

Molar Mass Calculation

The molar mass of a gas plays a critical role in understanding its identity and behavior. In this exercise, we discovered the molar mass by first determining the number of moles present in the sample using the ideal gas law. Once we had the number of moles, we divided the given mass of the gas by these moles to find the molar mass. The formula for molar mass is:

• Molar Mass (M) = \( \frac{\text{mass}}{\text{moles}} \)

For our specific gas sample, we used a mass of 1.007 grams and divided it by 0.0177 moles, resulting in a molar mass of approximately 56.9 grams per mole.
This method is invaluable for practical applications in chemistry, such as identifying unknown substances.

Unit Conversion

Unit conversion is a fundamental skill in any scientific discipline, especially chemistry. Proper conversion is essential for accurate calculations. In this exercise, we converted pressure, volume, and temperature into the units required by the ideal gas law.

• Pressure was converted from mmHg to atmospheres using the conversion factor: \( 1 \text{ atm} = 760 \text{ mmHg} \).
• Volume was converted from milliliters to liters, recognizing that \( 1000 \text{ mL} = 1 \text{ L} \).
• Temperature was converted from degrees Celsius to Kelvin by adding 273.15, as Kelvin is the SI base unit for thermodynamic temperature.

Each converted unit allowed us to accurately apply the ideal gas law formula and ensure the consistency of units in our calculations.

Gas Laws

Gas laws describe the behavior of gases under various conditions. The ideal gas law, represented by the equation \(PV = nRT\), is a cornerstone of these laws. Each variable in this equation has a specific meaning:

• \(P\): Pressure of the gas in atmospheres (atm)
• \(V\): Volume of the gas in liters (L)
• \(n\): Number of moles of the gas
• \(R\): Universal gas constant, approximately 0.0821 L·atm/mol·K
• \(T\): Temperature in Kelvin (K)

In the provided scenario, we rearranged the formula to solve for "n," the number of moles, allowing for the determination of molar mass. Understanding how these variables interact is key in predicting how a gas will respond to changes in temperature, pressure, or volume.

Pressure Conversion

Pressure conversion is often necessary in scientific calculations because different systems of measurement are used across fields. In our problem, the pressure given was in millimeters of mercury (mmHg), an older but still common unit of pressure, especially in medical and meteorological settings.
To use the ideal gas law, pressure in mmHg needed to be converted to atmospheres, the standard unit used in chemistry. The conversion involves dividing the mmHg value by 760, as:

• \(1 \text{ atm} = 760 \text{ mmHg} \)

This conversion ensures that the units are consistent with the universal gas constant, making calculations straightforward and reliable. Proper pressure conversion lays the groundwork for accurately using gas laws in chemistry.