A gas has molecular formula \((\mathrm{CH})_{n}\) If vapour density of the gas is 39 , what should be the formula of the compound? (a) \(\mathrm{C}_{3} \mathrm{H}_{3}\) (b) \(\mathrm{C}_{4} \mathrm{H}_{4}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{2}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{6}\)

When scientists speak of a molecule, they often refer to its molecular formula—a notation that provides a direct look into the composition of a compound. The molecular formula is akin to a recipe: it tells us exactly how many atoms of each element are contained in a single molecule of that substance.

Take, for example, the expression \( (CH)_{n} \). This denotes a molecular family characterized by a one-to-one ratio of carbon (C) to hydrogen (H) atoms, where 'n' represents the number of CH units in that molecule. In chemistry, to identify the specific member of this family—like discerning whether a baked treat is a cookie or a cake—we need to define the value of 'n'.

With the exercise at hand, students were tasked to determine this value by utilizing the given vapour density. Once this value is ascertained, it becomes a simple matter of substituting it back into the expression to disclose the molecular formula. Here we discovered that \( n = 6 \) gave us the molecular formula \( C_{6}H_{6} \)—a clear and precise depiction of the molecule's actual structure.

Molar mass is a cornerstone concept in chemistry, offering chemists a bridge between the world of atoms and the scales we use in the lab. Simply put, it's the weight of one mole (Avogadro's Number, or \( 6.022 \times 10^{23} \) entities) of a substance, typically expressed in grams per mole (g/mol).

Arriving at the molar mass of a compound begins with interpreting its molecular formula. By adding up the standard atomic weights of each constituent atom, as sourced from the periodic table, the molar mass of the compound is obtained. In our exercise, vapour density provided a shortcut. Since the vapour density of a gas is half of its molecular mass, doubling the vapour density (39) reveals that the molar mass of our mystery gas is 78 g/mol.

This revelation plays a pivotal role because it sets the stage for matching the molecular formula of the compound to its molar mass, which is the combined mass of all atoms found in the molecule's formula when calculated for a single mole of that substance.

Stoichiometry is the part of chemistry that pertains to the quantitative relationships between reactants and products in chemical reactions. In a broader sense, it involves calculations concerning the masses (or volumes) of reactants and products.

In the context of the given exercise, stoichiometry is employed to align the calculated molar mass of a compound with its empirical formula—or a simplified version of its molecular formula. This balance allows the determination of 'n', the multiplier that unveils how many units of CH are in the molecule. The formula's stoichiometry, \( 12n + n = 13n \), succinctly relates the count of atoms to their collective mass.

Using our molar mass calculation from the previous step and solving the stoichiometric equation for 'n' lead us to understand that the compound consists of six CH units. This direct application of stoichiometry enables the translation of a molar mass to a tangible molecular formula, once again exemplifying how these stoichiometric principles are integral to interpreting chemical phenomena.