Problem 69 - A Complete Resource Book in Chemistry for JEE Main

Question

A compound with empirical formula \(\mathrm{CH}_{2} \mathrm{O}\) has a vapour density of 30 . Its molecular formula is (a) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{O}_{2}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)

Step-by-Step Solution

Answer

The molecular formula is \( \mathrm{C}_{2}\mathrm{H}_{4}\mathrm{O}_{2} \) (option b).

Step 1: Calculate Molar Mass of Empirical Formula

First, determine the molar mass of the empirical formula \( \mathrm{CH}_{2}\mathrm{O} \). Carbon (C) has an atomic mass of 12, hydrogen (H) has an atomic mass of 1, and oxygen (O) has an atomic mass of 16. So, \( 12 + 2 \times 1 + 16 = 30 \). The molar mass of \( \mathrm{CH}_{2}\mathrm{O} \) is 30 g/mol.

Step 2: Determine Molecular Mass from Vapour Density

The vapour density of a compound is half of its molar mass. Therefore, the molar mass of the compound is \( 2 \times 30 = 60 \) g/mol.

Step 3: Find the Relationship Between Empirical and Molecular Formulas

Divide the molar mass of the molecular formula by the molar mass of the empirical formula to find the ratio: \( \frac{60}{30} = 2 \). This means the molecular formula contains 2 times the number of atoms in the empirical formula.

Step 4: Calculate the Molecular Formula

Multiply the subscripts in the empirical formula \( \mathrm{CH}_{2}\mathrm{O} \) by 2. This gives \( \mathrm{C}_{2}\mathrm{H}_{4}\mathrm{O}_{2} \). This is the molecular formula of the compound, matching option (b).

Key Concepts

Empirical Formula

An empirical formula represents the simplest whole-number ratio of atoms within a compound. It's a way to express the most reduced ratios of elements. For instance, in our exercise, the empirical formula provided is \(\mathrm{CH}_{2}\mathrm{O}\). This formula tells us that for every carbon atom, there are two hydrogen atoms and one oxygen atom.

The empirical formula does not inform us of the actual number of atoms in a molecule. Instead, it gives a simplified version of these numbers. To determine the empirical formula, you typically need the percent composition by mass of a compound.

In this case, knowing the empirical formula is a crucial first step in finding the molecular formula, as it forms the basis for understanding the simplest structure of a compound.

Vapour Density

Vapour density is an essential concept in chemistry. It refers to the density of a vapor in comparison to the density of hydrogen. Mathematically, it's half the molar mass of a compound.

In the exercise, the given vapour density was 30. This value allows us to calculate the molar mass of the compound by using the formula:

• Molar Mass = 2 \( \times \) Vapour Density

Substituting the values, we find a molar mass of \( 2 \times 30 = 60 \ \text{g/mol}\). This calculation indicates how much mass one mole of the compound's vapor would weigh compared to an equivalent volume of hydrogen. It is a stepping stone for determining the molecular formula.

Molar Mass Calculation

Calculating the molar mass is a vital step in finding the molecular formula of a compound from its empirical formula. Molar mass is essentially the mass of one mole of a given substance and is usually expressed in grams per mole. It’s calculated by summing the atomic masses of all the atoms in a molecule.

Given the empirical formula \( \mathrm{CH}_{2}\mathrm{O} \), the molar mass calculation is simple:

• Carbon (C) contributes 12 g/mol.
• Hydrogen (H), with 2 atoms, contributes \( 2 \times 1 = 2 \ \mathrm{g/mol} \).
• Oxygen (O) contributes 16 g/mol.

Adding these together gives a total molar mass for the empirical formula of 30 g/mol.

Once the vapour density has given us the molar mass of the compound as 60 g/mol, we can determine that the molecular formula is twice the empirical formula. Following these calculations, each subscript in the empirical formula is multiplied by 2 to yield the molecular formula \( \mathrm{C}_{2}\mathrm{H}_{4}\mathrm{O}_{2} \). This process exemplifies how understanding the molar mass helps translate the empirical formula into the more detailed molecular formula.