Question

A gaseous organofluorine compound has a density of \(0.355 \mathrm{g} / \mathrm{L}\) at \(17^{\circ} \mathrm{C}\) and \(189 \mathrm{mm}\) Hg. What is the molar mass of the compound?

Short answer

The molar mass of the compound is approximately 34.5 g/mol.

Step-by-step solution

Identify the Ideal Gas Law

The Ideal Gas Law equation is given by \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. The molar mass \( M \) can be calculated using the formula \( M = \frac{dRT}{P} \), where \( d \) is the density.

Convert Temperature to Kelvin

Convert the given temperature from Celsius to Kelvin using the formula: \( T(K) = T(^{\circ}C) + 273.15 \). So, \( 17^{\circ}C \) translates to \( 17 + 273.15 = 290.15 \text{ K} \).

Convert Pressure to Atmospheres

Convert the pressure from millimeters of mercury to atmospheres using: \( 1 \text{ atm} = 760 \text{ mm Hg} \). Therefore, \( 189 \text{ mm Hg} \times \frac{1 \text{ atm}}{760 \text{ mm Hg}} \approx 0.2487 \text{ atm} \).

Use the Ideal Gas Law to Find Molar Mass

Substitute the known values into the molar mass equation: \( M = \frac{dRT}{P} \). Here, \( d = 0.355 \text{ g/L} \), \( R = 0.0821 \text{ L atm/mol K} \), \( T = 290.15 \text{ K} \), and \( P = 0.2487 \text{ atm} \). Calculate: \[ M = \frac{0.355 \times 0.0821 \times 290.15}{0.2487} \approx 34.5 \text{ g/mol} \].

Calculate and Verify the Result

Carry out the calculation to ensure accuracy: \( M \approx 34.5 \text{ g/mol} \). Double-check all conversions and calculations to confirm that no errors were made.

Key concepts

Molar Mass Calculation

When dealing with gases, it's often important to know the molar mass of a compound since it helps to identify what the compound is. Molar mass is the mass of one mole of a substance and is usually expressed in grams per mole (g/mol). To calculate the molar mass of a gas using the Ideal Gas Law, we utilize the formula:
\[ M = \frac{dRT}{P} \]
Where:

• \( M \) is the molar mass.
• \( d \) is the density of the gas.
• \( R \) is the ideal gas constant, approximately \(0.0821\) L atm/mol K for most calculations related to gases.
• \( T \) is the temperature in Kelvin.
• \( P \) is the pressure.

To identify the molar mass in the example, measure the density and obtain the temperature and pressure of the gas. Then, convert temperature to Kelvin and pressure to atmospheres to properly use them in the formula. Substitute the given values to solve for the molar mass.

Density of Gases

Understanding the density of gases is essential in diverse fields like chemistry and environmental science. Density is defined as the mass per unit volume and for gases, it is typically measured in grams per liter (g/L). The density can give clues about the molecular size and composition of a gas.
To calculate the molar mass from density, use the ideal gas law rearranged into the molar mass equation:
\[ M = \frac{dRT}{P} \]
Here, the measured density of the gaseous organofluorine compound was given as \(0.355\) g/L. This value helped determine the molar mass when substituted into the equation along with the other transformed values for pressure and temperature.

Pressure Conversion

Pressure conversion is a crucial step when dealing with the Ideal Gas Law, as the pressure needs to be in atmospheres (atm) for the calculations to be accurate. In our example, the pressure was initially provided in millimeters of mercury (mm Hg).
The conversion is made using the relationship:

• \[ 1 \text{ atm} = 760 \text{ mm Hg} \]

Given the pressure of \(189 \text{ mm Hg}\) in the problem, we convert it to atmospheres by dividing it by \(760\):
\[ 189 \text{ mm Hg} \times \frac{1 \text{ atm}}{760 \text{ mm Hg}} \approx 0.2487 \text{ atm} \]
Performing this conversion correctly is essential for calculating accurate results in subsequent steps.

Temperature Conversion

Converting temperature to Kelvin is another critical step when applying the Ideal Gas Law. Since this law operates with absolute temperatures, you must convert Celsius to Kelvin by adding \(273.15\) to the Celsius temperature.
For example, the temperature given in the problem was \(17^{\circ}C\). We convert it to Kelvin as follows:
\[ T(K) = T(^{\circ}C) + 273.15 = 17 + 273.15 = 290.15 \text{ K} \]
This conversion ensures that you are using the proper temperature units for the Ideal Gas Law, as Kelvin is the standard unit for temperature in the context of gas calculations.