Problem 42 - A Complete Resource Book in Chemistry for JEE Main

Question

The molecular formula weight by empirical formula weight of a compound is \(3 .\) The empirical formula is \(\mathrm{CH}_{2} .\) The vapour density of the compound is (a) 21 (b) \(11.5\) (c) 33 (d) 43

Step-by-Step Solution

Answer

The vapour density of the compound is 21.

Step 1: Understanding the Problem

We are given that the ratio of the molecular formula weight to the empirical formula weight is 3. The compound has an empirical formula of \( \mathrm{CH}_{2} \). We need to find the vapour density given this information.

Step 2: Calculating Empirical Formula Mass

The empirical formula \( \mathrm{CH}_{2} \) consists of one carbon atom and two hydrogen atoms. The atomic mass of carbon is approximately 12 atomic mass units (amu), and hydrogen is approximately 1 amu. Therefore, the empirical formula mass is: \[ \text{Empirical formula mass} = 12 + (2 \times 1) = 14 \text{ amu} \]

Step 3: Finding Molecular Formula Mass

We are given that the molecular formula weight is 3 times the empirical formula weight. Thus, the molecular formula mass is: \[ \text{Molecular formula mass} = 3 \times 14 = 42 \text{ amu} \]

Step 4: Understanding Vapour Density

Vapour density is defined as half the molar mass of the substance. Hence, to find the vapour density, we use the formula: \[ \text{Vapour density} = \frac{\text{Molar mass}}{2} \] Since molar mass is equal to the molecular weight, the vapour density of our compound is: \[ \text{Vapour density} = \frac{42}{2} = 21 \]

Step 5: Conclusion and Answer

We have calculated that when the molecular formula mass is 42 amu, the vapour density is 21. Therefore, the correct answer is (a) 21.

Key Concepts

Empirical Formula

The empirical formula of a compound provides the simplest whole-number ratio of the atoms present in it. It doesn't indicate the exact number of each type of atom, but rather the basic proportion. For example, in the empirical formula \( \mathrm{CH}_2 \), it shows that for every carbon atom, there are two hydrogen atoms.
The empirical formula is a foundational concept because it represents the simplest "version" of a compound's composition.
It helps chemists understand the basic structure and is always derived from the relative percentage composition of elements in the compound.

Molecular Formula

The molecular formula of a compound reveals the exact number of each type of atom in a molecule. Unlike the empirical formula, it provides the actual composition of the substance. For example, if the empirical formula is \( \mathrm{CH}_2 \) and the molecular formula mass is 42 amu, we can deduce that the molecular formula must be \( \mathrm{C}_3 \mathrm{H}_6 \).
We determine this because the molecular formula weight is a multiple of the empirical formula weight.
This relationship allows chemists to scale up the empirical formula to get the actual molecular formula by using specific data, like the mass ratios.

Atomic Mass Units

The atomic mass unit (amu) is a standard unit of mass that quantifies the mass of atoms and molecules. It is essential for calculating both empirical and molecular formulas because it allows chemists to measure the weight of individual atoms.
One amu is defined as one-twelfth of the mass of an atom of carbon-12, which is approximately \( 1.66 \times 10^{-24} \) grams.
Understanding amu helps in calculating molecular and empirical weights, as seen in the problem's solution where the atomic masses of carbon and hydrogen (12 amu and 1 amu, respectively) were used.