Problem 1 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

More challonging questions are indicated by an asterisk '. Note: For some questions, data will be needed from tables within the chapter and from the Periodic Table on the inside front cover. \(3.036 \mathrm{g}\) of a gas occupy a volume of \(426 \mathrm{cm}^{3}\) at \(273 \mathrm{K}\) and 1.00 atm pressure. Calculate the molar mass of the gas. (Section 8.2)

Step-by-Step Solution

Answer

The molar mass of the gas is approximately 159.79 g/mol.

Step 1: Identify the Given Information

We are given the following data: mass of the gas, \( m = 3.036 \text{ g} \); volume of the gas, \( V = 426 \text{ cm}^3 \); temperature, \( T = 273 \text{ K} \); pressure, \( P = 1.00 \text{ atm} \).

Step 2: Convert Units to Standard Units

The volume needs to be converted from cm³ to liters, as standard gas laws use liters. We know that \(1000 \text{ cm}^3 = 1 \text{ L}\), so \( V = \frac{426}{1000} = 0.426 \text{ L}\).

Step 3: Use the Ideal Gas Law Formula

The ideal gas law formula is \( PV = nRT \), where \( n \) is the number of moles and \( R \) is the ideal gas constant \( 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \). We can rearrange to find \( n \), the number of moles: \( n = \frac{PV}{RT} \).

Step 4: Calculate the Number of Moles

Substituting the values into the formula: \[ n = \frac{1.00 \text{ atm} \times 0.426 \text{ L}}{0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \times 273 \text{ K}} \approx 0.019 \text{ mol} \]

Step 5: Determine the Molar Mass

The molar mass (M) is found using the formula \( M = \frac{m}{n} \). Substituting the values we calculated, \[ M = \frac{3.036 \text{ g}}{0.019 \text{ mol}} \approx 159.79 \text{ g/mol} \]

Step 6: Final Result

The molar mass of the gas is approximately \( 159.79 \text{ g/mol} \).

Key Concepts

Molar Mass Calculation

Molar mass is a critical property in chemistry that represents the mass of one mole of a substance. To find the molar mass, you need two pieces of information: the mass of the sample and the number of moles it contains.
To calculate this, use the formula:

• \( M = \frac{m}{n} \)

Here, \( M \) is the molar mass, \( m \) is the mass of the gas, and \( n \) is the number of moles. This formula essentially tells us how much a single mole of the substance weighs.
In our exercise, we found the mass \( m \) and used the ideal gas law to determine \( n \). Once both are known, calculating the molar mass becomes straightforward by dividing the mass by the number of moles. This process is pivotal in identifying substances and understanding their properties.

Gas Laws

Gas laws describe the behavior of gases in response to changes in temperature, volume, and pressure. One of the most essential gas laws is the Ideal Gas Law, represented by the equation \( PV = nRT \). Here:

• \( P \) stands for pressure, usually in atmospheres.
• \( V \) is the volume, typically in liters.
• \( n \) represents the number of moles of the gas.
• \( R \) is the ideal gas constant, \(0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}\).
• \( T \) signifies temperature, measured in Kelvin.

This equation provides a relationship between these properties.
In practice, when you know any three variables of a gas, you can find the fourth using this formula. In our example, knowing \( P \), \( V \), and \( T \), we solved for \( n \), the number of moles. This step is crucial in calculating other properties, such as molar mass.

Unit Conversion

Unit conversion is an essential skill in science, especially when working with gas laws, which require specific units. In our scenario, we needed to convert the volume from cubic centimeters (cm³) to liters because the Ideal Gas Law uses liters.
The conversion process is simple:

• 1 liter = 1000 cubic centimeters

Therefore, to find the volume in liters, divide the volume in cm³ by 1000. So, from the exercise, \( V = \frac{426 \text{ cm}^3}{1000} = 0.426 \text{ L}\). Converting units ensures that calculations align with the necessary formulas, delivering accurate results.
Unit conversion is not only about making numbers match but understanding the relationship between different units, which is crucial for accurate scientific computation.