Problem 63 - A Complete Resource Book in Chemistry for JEE Main

Question

The empirical formula of a compound is \(\mathrm{CH}_{2}\). One mole of the compound has a mass of \(42 \mathrm{~g}\), its molecular formula is (a) \(\mathrm{C}_{2} \mathrm{H}_{2}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (c) \(\mathrm{C}_{3} \mathrm{H}_{\mathrm{g}}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{\mathrm{s}}\)

Step-by-Step Solution

Answer

The molecular formula is \(\mathrm{C}_{3}\mathrm{H}_{6}\), option (c).

Step 1: Calculate the Molar Mass of Empirical Formula

First, determine the molar mass of the compound's empirical formula, \(\mathrm{CH}_{2}\). Carbon (C) has an atomic mass of approximately 12 g/mol, and hydrogen (H) has an atomic mass of 1 g/mol. Thus, the molar mass of \(\mathrm{CH}_{2}\) is calculated as follows: \[ 12 \text{ g/mol (for C)} + 2 \times 1 \text{ g/mol (for H)} = 14 \text{ g/mol}\]

Step 2: Determine the Multiplication Factor

The compound's empirical formula molar mass is 14 g/mol, while one mole of the compound actually has a mass of 42 g. To find the multiplication factor that determines the molecular formula, divide the molar mass of the compound by the molar mass of the empirical formula:\[ \text{Multiplication Factor} = \frac{42 \text{ g/mol}}{14 \text{ g/mol}} = 3\]

Step 3: Calculate the Molecular Formula

Multiply the empirical formula \(\mathrm{CH}_{2}\) by the multiplication factor of 3. This gives:\[ \mathrm{C}_{1 \times 3}\mathrm{H}_{2 \times 3} = \mathrm{C}_{3}\mathrm{H}_{6} \] This is the molecular formula of the compound.

Step 4: Find the Correct Option

Compare the calculated molecular formula \(\mathrm{C}_{3}\mathrm{H}_{6}\) with the given options: (a) \(\mathrm{C}_{2}\mathrm{H}_{2}\)(b) \(\mathrm{C}_{2}\mathrm{H}_{4}\)(c) \(\mathrm{C}_{3}\mathrm{H}_{6}\)(d) \(\mathrm{C}_{4}\mathrm{H}_{8}\)The correct answer is option (c) \(\mathrm{C}_{3}\mathrm{H}_{6}\).

Key Concepts

Empirical Formula

The empirical formula is the simplest expression of a chemical compound. It represents the ratio of atoms of each element in the compound in the simplest whole numbers. For instance, the empirical formula of ethene is \(\mathrm{CH}_{2}\), indicating a ratio of one carbon atom to two hydrogen atoms.

It doesn't convey the actual number of atoms in the molecules, just the simplest ratio. This is important in chemistry as it lays the groundwork to determine a compound's molecular formula when combined with additional data like molecular mass. To find the empirical formula from given data:

• Determine the amount of each constituent element in the compound.
• Convert these amounts from grams to moles using atomic masses.
• Simplify the moles ratio to the smallest whole numbers to get the empirical formula.

Molar Mass Calculation

Calculating the molar mass is a cornerstone of determining molecular formulas from empirical ones. To find the molar mass of a compound, sum upthe atomic masses of all atoms in the empirical formula. For example, given \(\mathrm{CH}_{2}\), the molar mass calculation involves:

Adding the mass of one carbon atom (12 g/mol) and two hydrogen atoms (each 1 g/mol), giving a total molar mass of: \[ 12\,\text{g/mol (C)} + 2 \times 1\,\text{g/mol (H)} = 14\,\text{g/mol} \]

This calculated molar mass is crucial in the next steps, helping to link the empirical formula with the specific molecular formula by comparing it against the molecular mass provided.

Multiplication Factor

After figuring out the molar mass of the empirical formula, the multiplication factor helps us determine the molecular formula. This factor decides how many times the empirical formula needs to be repeated to match the actual molar mass of the compound.

Derived by dividing the given molecular mass (42 g/mol in this problem) by the calculated empirical formula mass (14 g/mol), it is expressed as:\[ \text{Multiplication Factor} = \frac{42\,\text{g/mol}}{14\,\text{g/mol}} = 3 \]

Thus, the multiplication factor of 3 indicates that the empirical formula needs to be multiplied three times to equate with the compound’s molecular structure, leading us to the final molecular formula.

Chemical Composition Analysis

Analyzing a compound's chemical composition helps reveal its true molecular structure. Once you have the right multiplication factor, transforming the empirical formula into the molecular formula becomes straightforward. In our example, multiplying the empirical formula by 3 gives:\[ \text{For}\,\mathrm{CH}_{2}: \text{C}_{1\times 3}\,\text{H}_{2\times 3} = \mathrm{C}_{3}\,\mathrm{H}_{6} \]
This endpoint molecular formula represents the actual number of atoms in each element of the compound, giving a detailed insight into the compound's nature.

• Confirms the compound’s mass aligns with the calculated molecular formula.
• Shows how simple ratios expand to actual complex molecular counts.

This step effectively bridges the gap between theoretical stoichiometry and real-world chemical formulations.