Problem 27 - Chemistry and Chemical Reactivity

Question

A 1.007 -g sample of an unknown gas exerts a pressure of \(715 \mathrm{mm}\) Hg in a 452 -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 57.87 g/mol.

Step 1: Identify the Given Values

We have a sample of gas with a mass of 1.007 g, a pressure of \( 715 \, \text{mm Hg} \), a volume of \( 452 \, \text{mL} \) (or 0.452 L), and a temperature of \( 23^{\circ} \text{C} \) (which is \( 296 \text{K} \) when converted to Kelvin).

Step 2: Convert Pressure to Atmospheres

To use the ideal gas law in standard units, convert the pressure from mm Hg to atmospheres using the conversion factor 1 atm = 760 mm Hg: \[ P = \frac{715 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} \approx 0.941 \text{ atm} \]

Step 3: Apply Ideal Gas Law

Use the ideal gas law \( PV = nRT \) to find the number of moles \( n \). Substitute in \( P = 0.941 \text{ atm} \), \( V = 0.452 \text{ L} \), \( R = 0.0821 \text{ L atm / mol K} \), and \( T = 296 \text{ K} \): \[ n = \frac{PV}{RT} = \frac{0.941 \times 0.452}{0.0821 \times 296} \approx 0.0174 \text{ moles} \]

Step 4: Calculate Molar Mass

Now, calculate the molar mass \( M \) using the formula \( M = \frac{m}{n} \), where \( m \) is the mass of the gas sample (1.007 g) and \( n \) is the number of moles (0.0174 moles): \[ M = \frac{1.007 \text{ g}}{0.0174 \text{ moles}} \approx 57.87 \text{ g/mol} \]

Key Concepts

Molar Mass Calculation

Calculating the molar mass of a gas involves determining how much one mole of that gas weighs. This is crucial for identifying unknown gases by comparing their molar mass to known values.
Molar mass (\( M \)) is calculated using the formula: \( M = \frac{m}{n} \), where \( m \) is the mass of the sample and \( n \) is the number of moles.

Steps to calculate:

• Determine the mass (\( m \)) of your sample.
• Use the Ideal Gas Law to find the number of moles (\( n \)).
• Substitute \( m \) and \( n \) into the formula to find the molar mass.

This procedure is beneficial in chemistry to identify gases in a given sample by translating measurable properties into molar mass.

Pressure Conversion

Pressure conversion is vital when working with the ideal gas law, as calculations are typically done using standard units such as atmospheres (atm).
Converting measurements into standard units allows for consistent results that can be broadly applied. Here, we convert pressure from millimeters of mercury (\( ext{mm Hg} \)) to atmospheres using the conversion factor: 1 atm = 760 mm Hg.
To convert pressure, use the formula: \[ P = \frac{\text{Pressure in mm Hg}}{760} \]
Why It's Important:

• Ensures compatibility with the ideal gas constant \( R \), which uses atm.
• Avoids errors in calculations due to unit mismatches.

Converting to atmospheres allows you to accurately apply the ideal gas equation across different scenarios.

Volume Conversion

Volume conversion might be needed when working with gases, especially when dealing with the ideal gas law. The typical unit for volume in these calculations is liters (L).
The conversion from milliliters (mL) to liters is simple: \[ \text{Volume in L} = \frac{\text{Volume in mL}}{1000} \]Converting volume ensures that each variable in the ideal gas law is expressed in compatible units.
Key Points:

• Volumetric data in mL needs conversion to L for compatibility with \( R \).
• Prevents miscalculations by standardizing the volume measurement.

Accurate conversion helps confirm that volume measurements harmonize with other variables in the calculation process.

Ideal Gas Constant R

The ideal gas constant \( R \) is a fundamental component in the ideal gas law, pivotal for calculations involving gases. The value of \( R \) depends on the units used for pressure, volume, and temperature. In atmospheric calculations, \( R = 0.0821 \) L atm / mol K.
Why is \( R \) Important?

• It provides the necessary link between volume, pressure, and temperature when calculating the number of moles.
• Ensures that the calculation aligns with physical constants and standard conditions.

When using \( R \), it is critical to ensure that all other measurements are in compatible units, secure the accuracy of calculations, and achieve reliable results using the ideal gas law formula: \[ PV = nRT \] Understanding and properly using \( R \) is essential for solving problems involving gas calculations efficiently.