Problem 17 - A Complete Resource Book in Chemistry for JEE Main

Question

The empirical formula of an acid is \(\mathrm{CH}_{2} \mathrm{O}_{2}\), the probable molecular formula of the simplest acid may be (a) \(\mathrm{CH}_{2} \mathrm{O}\) (b) \(\mathrm{CH}_{2} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\) (d) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{4}\)

Step-by-Step Solution

Answer

(b) \( \mathrm{CH}_{2} \mathrm{O}_{2} \)

Step 1: Understand the Empirical Formula

The empirical formula of a compound provides the simplest whole-number ratio of the elements present in the compound. Here, the empirical formula is \( \mathrm{CH}_{2} \mathrm{O}_{2} \), which means for each carbon (C), there are two hydrogens (H) and two oxygens (O).

Step 2: Determine the Molar Mass of the Empirical Formula

Calculate the molar mass of the empirical formula \( \mathrm{CH}_{2} \mathrm{O}_{2} \):- Carbon (C): 12 g/mol- Hydrogen (H): 2 g/mol (1 g/mol \( \times \) 2)- Oxygen (O): 32 g/mol (16 g/mol \( \times \) 2)Adding these gives the molar mass: \( 12 + 2 + 32 = 46 \) g/mol.

Step 3: Compare with the Options Provided

Consider each option and its possible molecular formula:(a) \( \mathrm{CH}_{2} \mathrm{O} \): Has a ratio of 1:2:1 (not matching \( \mathrm{CH}_{2} \mathrm{O}_{2} \))(b) \( \mathrm{CH}_{2} \mathrm{O}_{2} \): Identical to the empirical formula(c) \( \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2} \): Simplified ratio is 1:2:1, matching the empirical formula(d) \( \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{4} \): Simplified ratio is still 1:2:2, matching empirical formula

Step 4: Identify the Possible Molecular Formulas

To be a candidate for the molecular formula, the formula must be either the same as or a multiple of the empirical formula. - \( \mathrm{CH}_{2} \mathrm{O}_{2} \) is equal to the empirical formula- \( \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2} \) is a simple multiple (2x) of the empirical formula

Step 5: Determine the Answer

The simplest molecular formula derived from the empirical formula \( \mathrm{CH}_{2} \mathrm{O}_{2} \) is one of the simplest acids:- \( \mathrm{CH}_{2} \mathrm{O}_{2} \) which is the exact empirical representation and the simplest form.- Therefore, the answer is (b) \( \mathrm{CH}_{2} \mathrm{O}_{2} \).

Key Concepts

Understanding Empirical Formula

The empirical formula of a compound is fundamental in chemistry because it gives the simplest whole-number ratio of atoms of each element in the compound. It's like getting the basic blueprint of a compound without the actual format or size. For the given exercise, the empirical formula of the acid is \( \text{CH}_2\text{O}_2 \), indicating that for every carbon atom, there are two hydrogen atoms and two oxygen atoms.
This helps us understand not the exact number of atoms but the proportions in which these atoms are combined.
Note that while the empirical formula provides the ratio, it does not give information about the actual number of atoms present in a molecule of the compound. Therefore, it serves as a stepping stone for determining the molecular formula, especially when combined with molar mass data.

Molar Mass Calculation

Molar mass is crucial because it connects a substance's mass to the number of particles it contains. Calculating the molar mass involves adding up the atomic masses of each element present in the compound, based on the empirical formula. For \( \text{CH}_2\text{O}_2 \), we need to determine the molar mass as follows:
- Carbon (C) contributes \( 12 \text{ g/mol} \),
- Hydrogen (H) contributes \( 2 \text{ g/mol} \) because there are two hydrogens, and
- Oxygen (O) contributes \( 32 \text{ g/mol} \) since we have two oxygens each weighing \( 16 \text{ g/mol} \).
Therefore, the total molar mass is \( 12 + 2 + 32 = 46 \text{ g/mol} \).
Understanding the molar mass helps evaluate potential molecular formulas by checking if they are simple multiples or equivalents of the empirical formula's mass.

Introduction to Chemical Stoichiometry

Chemical stoichiometry is all about the quantitative relationships between reactants and products in a chemical reaction. Here, it involves understanding how empirical and molecular formulas relate to each other quantitatively.
Given the empirical formula \( \text{CH}_2\text{O}_2 \) and its molar mass, we can deduce potential molecular formulas. Molecular formulas are either the empirical formula itself or integer multiples.
This is where stoichiometry becomes essential as it allows us to compare and contrast options such as \( \text{CH}_2\text{O}_2 \) in option (b), where no further multiplication is needed. \( \text{C}_2\text{H}_4\text{O}_2 \) in option (c) is a simple multiple (2x), making it another candidate. Stoichiometry helps systematically eliminate choices that do not fit the empirical evidence, leading us to the correct possible molecular formula.