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 34.06 g/mol.

Step-by-step solution

Understand the Given Information

We know that a gas has a density of \(0.355 \mathrm{g/L}\), and it is at temperature of \(17^{\circ} \mathrm{C}\) and pressure of \(189\, \mathrm{mmHg}\). Our goal is to find the molar mass of the compound.

Convert Units

We need to convert the temperature to Kelvin and pressure to atmospheres for use in the ideal gas law. - Temperature: \(17^{\circ} \mathrm{C} = 290.15\, \mathrm{K}\) - Pressure: \(189\, \mathrm{mmHg} = 0.2487\, \mathrm{atm}\) (using the conversion \(1\, \mathrm{atm} = 760\, \mathrm{mmHg}\))

Use the Ideal Gas Law Equation

According to the ideal gas law: \[ PV = nRT \] where:- \(P = 0.2487\, \mathrm{atm}\)- \(V = 1\, \mathrm{L}\) (since density is given in g/L)- \(R = 0.0821\, \mathrm{L}\cdot\mathrm{atm/mol}\cdot\mathrm{K}\)- \(T = 290.15\, \mathrm{K}\)- \(n\) is the number of moles. Rearranging the equation to solve for \(n\): \[ n = \frac{PV}{RT} \]

Calculate the Number of Moles

Plug in the known values into the rearranged ideal gas law:\[ n = \frac{0.2487 \times 1}{0.0821 \times 290.15} \approx 0.01042\, \mathrm{mol} \]

Find the Molar Mass

Since the density is \(0.355\, \mathrm{g/L}\), and we have \(0.01042\, \mathrm{mol}\) of the gas in 1 L, we calculate the molar mass \(M\) using the formula:\[ M = \frac{\text{mass}}{\text{moles}} = \frac{0.355}{0.01042} \approx 34.06\, \mathrm{g/mol} \]

Conclusion

The molar mass of the organofluorine compound is \(34.06\, \mathrm{g/mol}\).

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that describes the behavior of ideal gases. It is expressed as:
\[ PV = nRT \]
This formula highlights the relationship between four key properties of a gas:

• Pressure \(P\)
• Volume \(V\)
• Number of moles \(n\)
• Temperature \(T\)

The constant \(R\) is known as the ideal gas constant, valued at \(0.0821 \, \mathrm{L} \cdot \mathrm{atm/mol} \cdot \mathrm{K}\). To solve gas problems, it often requires converting the given conditions to fit these terms.
For example, pressure is usually needed in atmospheres and temperature in Kelvin.
This set of rules is crucial when we calculate properties such as molar mass, as it allows chemists to predict how the gas will react under various conditions.

Organofluorine Compounds

Organofluorine compounds are organic molecules marked by the presence of carbon-fluorine bonds.
These compounds can be crucial in many areas, including pharmaceuticals, agrochemicals, and materials science.
The carbon-fluorine bond is one of the strongest in organic chemistry. This strength accounts for the exceptional stability and unique properties of organofluorine compounds.
The presence of fluorine can greatly influence the chemical and physical behavior of organic molecules, such as altering solubility and biological activity.
Understanding these compounds' structure and the role of fluorine is vital, especially when looking at their molecular composition and determining properties like molar mass.

Unit Conversion

Unit conversion is an essential skill in chemistry, enabling scientists to ensure that all units in a calculation align properly.
In the problem we examined, the temperature had to be converted from Celsius to Kelvin, requiring:
\[ \text{K} = \text{C}^{\circ} + 273.15 \]
This adjustment ensures compatibility with the ideal gas law. Similarly, pressure was converted from mmHg to atmospheres:
\[ \text{atm} = \text{mmHg} \times \left( \frac{1}{760} \right) \]
These are two examples among many conversions you might encounter. Clear understanding and accurate conversion of units are key to making valid conclusions in scientific analyses, such as in gas calculations.

Gas Density

Gas density is a measure of how much mass a gas has in a given volume.
It's typically expressed in terms of mass per unit volume, like grams per liter \(\mathrm{g/L}\). Understanding this property helps to solve molar mass problems, such as the one presented.
For gases, density ties directly into both the ideal gas law and molar mass.From the ideal gas law, the rearranged equation helps to solve for the number of moles:\[ n = \frac{PV}{RT} \]
Knowing the density, you can calculate molar mass \(M\) by dividing the mass by the number of moles:
\[ M = \frac{\text{mass}}{\text{moles}} \]
In practical terms, this means when you have the density of a gas and use it with other gas law variables, you can uncover the molar mass, a vital step for identifying the unknown gas.