Problem 86 - A Complete Resource Book in Chemistry for JEE Main

Question

An organic compound contains \(49.3 \%\) carbon, \(6.84 \%\) hydrogen and its vapour density is 73. Molecular formula of the compound is (a) \(\mathrm{C}_{3} \mathrm{H}_{10} \mathrm{O}_{2}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{8}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}_{2}\)

Step-by-Step Solution

Answer

The empirical formula is C₃H₅O₂, which doubles to C₆H₁₀O₄. None perfectly matches, but C₆H₁₀O₈ option (c) is closest in mistake.

Step 1: Understanding the Problem

We need to determine the molecular formula of a compound given the percentage composition: 49.3% carbon, 6.84% hydrogen, and the vapor density is 73. We are given multiple-choice options for the molecular formula.

Step 2: Calculate the Empirical Formula

To find the empirical formula, we convert the percentage composition into moles. Assuming 100g of the compound, we have 49.3g of carbon, 6.84g of hydrogen. The rest of the mass is oxygen (100 - 49.3 - 6.84 = 43.86g). The number of moles are calculated as follows:- Moles of Carbon = \( \frac{49.3}{12} = 4.108 \) moles - Moles of Hydrogen = \( \frac{6.84}{1} = 6.84 \) moles - Moles of Oxygen = \( \frac{43.86}{16} = 2.742 \) molesWe simplify the ratios by dividing each by the smallest number of moles (2.742):- Carbon ratio = \( \frac{4.108}{2.742} \approx 1.5 \)- Hydrogen ratio = \( \frac{6.84}{2.742} \approx 2.5 \)- Oxygen ratio = \( \frac{2.742}{2.742} = 1 \)Multiplying each ratio by 2 to remove fractions, we get C₃H₅O₂ as the empirical formula.

Step 3: Determine the Molecular Formula

The molecular formula is a whole-number multiple of the empirical formula. The molecular weight from vapor density is calculated using: \[ \text{Molecular Weight} = 2 \times \text{Vapor Density} = 2 \times 73 = 146 \]The empirical formula weight of C₃H₅O₂ = \(3 \times 12 + 5 \times 1 + 2 \times 16 = 73\).Therefore, the molecular formula is:\[ \text{Molecular Formula} = n \times \text{Empirical Formula} = 2 \times (\text{C}_3\text{H}_5\text{O}_2) = \text{C}_6\text{H}_{10}\text{O}_4 \]Comparing with given options, it is closest to option (c) \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{8} \). However, there may be a mistake in the options as C₃H₅O₂ when doubled equals C₆H₁₀O₄.

Key Concepts

Empirical Formula

An empirical formula represents the simplest whole-number ratio of elements in a compound. It provides insight into the basic structure of the molecule, although it does not always show how the elements are bonded or the exact number of atoms in a molecule.
To determine the empirical formula, you convert the percentage of each element in the compound to grams (assuming a 100 g sample) and then to moles by dividing by atomic masses:

• For carbon, divide by its atomic mass of 12 g/mol.
• For hydrogen, divide by 1 g/mol.
• For oxygen, divide by 16 g/mol.

After converting to moles, you simplify the ratios by dividing each by the smallest number of moles and, if necessary, multiply the ratios to get whole numbers. This process ensures you derive the simplest form of the molecular composition.

Percentage Composition in Chemistry

Percentage composition is the percent by mass of each element in a compound. It allows you to understand what proportion each element contributes to the total mass of a compound.
This is crucial for finding empirical formulas as well. The percentage composition in the problem was given as:

• 49.3% carbon
• 6.84% hydrogen

From this, you can infer the amount of oxygen present by subtracting from 100% what makes up the rest of the compound. A balanced understanding of percentage composition allows you to break down complex chemical formulas to their roots, ultimately assisting in identifying empirical formulas.

Vapor Density

Vapor density helps in determining the molecular weight of gases. It is essential for identifying the molecular formula of a compound. Vapor density is defined as half the molecular weight of the gas.
The formula used is:

• \[\text{Molecular Weight} = 2 \times \text{Vapor Density}\]
• Therefore, for this problem, given a vapor density of 73, the molecular weight is 146.

Once the molecular weight is found, you can compare this with the weight of the empirical formula (73 in this case). The comparison helps you find the multiplier (n), allowing you to determine the molecular formula.This process enables the transition from knowing basic composition to full molecular configuration.