A gas effuses \(1.55\) times faster than propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) at the same temperature and pressure. (a) Is the gas heavier or lighter than propane? (b) What is the molar mass of the gas?

Understanding the molar mass of a substance is crucial when studying chemistry, especially when working with gases. The molar mass is the mass of one mole of a particular substance and is usually expressed in grams per mole (g/mol).

To calculate the molar mass, you need to know the formula of the substance and the molar mass of each element in the formula. Atoms of each element have a unique molar mass, typically found on the periodic table. For a molecule like propane (\texttt{C}\(_3\)\texttt{H}\(_8\)), the calculation involves the sum of the molar masses of the constituent atoms: three carbon atoms and eight hydrogen atoms. By multiplying the molar mass of carbon (about 12.01 g/mol) by three and the molar mass of hydrogen (about 1.01 g/mol) by eight, and then adding these together, you get the molar mass of propane.

The rate of effusion in gases is described by Graham's law, which states that the rate at which a gas effuses is inversely proportional to the square root of its molar mass. Therefore, lighter gases effuse faster than heavier gases under the same conditions of temperature and pressure.

This relationship can be mathematically expressed as follows: the rate of effusion of gas 1 divided by the rate of effusion of gas 2 is equal to the square root of the inverse ratio of their molar masses. When you're given the rate of effusion of one gas compared to another, as in the exercise example, you can use this relationship to calculate the unknown molar mass.

Gas effusion problems often involve the use of Graham's law to compare the rates at which different gases escape through a tiny opening, or effuse. When solving these problems, you typically need the rate of effusion of one gas and either the molar mass of that gas or the other gas in question.

In the provided exercise, we worked with an unknown gas that effuses 1.55 times faster than propane. By employing Graham's law, we set up a ratio and used algebra to solve for the molar mass of the unknown gas. As the law indicates an inverse relationship between rate of effusion and molar mass, you can deduce that a gas that effuses faster than another has a lower molar mass.

Once the molar mass of an unknown gas is determined through calculations using Graham's law, you can compare it to the molar mass of a known gas to identify whether it's lighter or heavier. This comparison is straightforward: if the molar mass of the unknown gas is lesser than that of the known gas, the unknown gas is lighter, and vice versa.

In our original exercise, after calculating, we found that the molar mass of the unknown gas (approximately 18.34 g/mol) was less than that of propane (44.11 g/mol), enabling us to conclude that the unknown gas is lighter than propane. Not only does this indicate how the gases will behave in terms of effusion, but it also provides insights into how they might behave in a mixture or when subjected to other gas laws.