A sample of phosphorus that weighs \(12.4 \mathrm{~g}\) exerts a pressure 8 atm in a \(0.821\) litre closed vesel at \(527^{\circ} \mathrm{C}\). The molecular formula of the phosphorus vapour is: (a) \(\mathrm{P}_{2}\) (b) \(\mathrm{P}_{4}\) (c) \(\mathrm{P}_{6}\) (d) \(\mathrm{P}_{8}\)

Understanding how to calculate molar mass is crucial in chemistry for converting between grams of a substance and the number of moles. The molar mass is the weight of one mole of a substance, usually expressed in grams per mole (g/mol).

To calculate the molar mass of a compound, you sum the molar masses of all the atoms in its formula. For an element, the molar mass is its atomic weight from the periodic table. However, when dealing with gases, we often have to deduce its molar mass from experimental data, such as in our exercise involving phosphorus vapor.

In this scenario, after finding the number of moles (n) using the ideal gas law, we used the formula:

\[ \text{Molar Mass (M)} = \frac{\text{mass of the sample (m)}}{\text{moles of gas (n)}} \]

By applying this formula, we can understand how much one mole of phosphorus vapor weighs, which is essential in determining its molecular formula.

The molecular formula of a substance provides information about the actual number of atoms of each element in one molecule of the substance. In the case of our phosphorus vapor, we aim to find out if the molecules are made up of \( P_2 \), \( P_4 \), \( P_6 \), or \( P_8 \) units.

Once the molar mass is known, comparing it to the molar mass of a single phosphorus atom allows us to determine how many phosphorus atoms are bound together in a single molecule of the vapor. This comparison is essentially a division problem where we divide the molar mass of the vapor by the atomic weight of a single phosphorus atom.For instance, if the molar mass of the vapor is roughly four times that of the atomic weight of phosphorus, it indicates a \( P_4 \) molecular formula. This step is the culmination of our investigative efforts in the exercise, revealing the molecular complexity of the phosphorus vapor.

Gas law equations are the backbone of many calculations in chemistry, particularly when working with gases' behavior under various conditions of pressure, volume, and temperature. The ideal gas law, stated as \( PV = nRT \), is a widely used equation that summarizes these conditions into a single expression.

In this law, \( P \) stands for pressure in atmospheres, \( V \) for volume in liters, \( n \) for moles of gas, \( R \) for the universal gas constant \( (0.0821 \, L \cdot atm/mol \cdot K) \) and \( T \) for temperature in Kelvin. It's a powerful tool that intuitively connects these variables, allowing us to solve for one if we know the others.

In the given exercise, we were presented with the pressure, volume, and temperature of a phosphorus vapor sample and were asked to apply the ideal gas law. The challenge was to rearrange and solve for the unknown: the number of moles of the vapor. This calculation is pivotal in moving towards the determination of the molar mass and subsequently the molecular formula of the gas.