Problem 59 - A Complete Resource Book in Chemistry for JEE Main

Question

\(60 \mathrm{~g}\) of a compound on analysis produced \(24 \mathrm{~g}\) carbon, \(4 \mathrm{~g}\) hydrogen and \(32 \mathrm{~g}\) oxygen. The empirical formula of the compound is (a) \(\mathrm{CH}_{4} \mathrm{O}\) (b) \(\mathrm{CH}_{2} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\) (d) \(\mathrm{CH}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

The empirical formula of the compound is (d) \(\text{CH}_2\text{O}\).

Step 1: Find the Number of Moles of Each Element

To determine the number of moles of each element in the compound, use the formula \( ext{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \). - For Carbon (C): Molar mass = 12 g/mol \[ \text{Moles of } C = \frac{24 \, \text{g}}{12 \, \text{g/mol}} = 2 \text{ moles} \]- For Hydrogen (H): Molar mass = 1 g/mol \[ \text{Moles of } H = \frac{4 \, \text{g}}{1 \, \text{g/mol}} = 4 \text{ moles} \]- For Oxygen (O): Molar mass = 16 g/mol \[ \text{Moles of } O = \frac{32 \, \text{g}}{16 \, \text{g/mol}} = 2 \text{ moles} \]

Step 2: Determine the Simplest Whole Number Ratio

Divide the number of moles of each element by the smallest number of moles calculated.- The smallest number of moles is 2 (from both Carbon and Oxygen).- For Carbon: \[ \frac{2}{2} = 1 \]- For Hydrogen: \[ \frac{4}{2} = 2 \]- For Oxygen: \[ \frac{2}{2} = 1 \]

Step 3: Write the Empirical Formula

Use the ratio found in Step 2 to write the empirical formula of the compound. The empirical formula is the simplest whole number ratio of the atoms in the compound.Since the simplest ratio is \( C: 1, H: 2, O: 1 \), the empirical formula is: \[ \text{Empirical formula} = \text{CH}_2\text{O} \]

Step 4: Match with Given Options

Compare the derived empirical formula to the options provided in the problem. The calculated empirical formula \( \text{CH}_2\text{O} \) matches option (d).

Key Concepts

Mole Calculation

To find the empirical formula of a compound, the first step is knowing how to calculate moles for each element involved. This is essential because we're trying to determine the actual number of atoms involved in the compound. Moles are a basic unit in chemistry used to express amounts of a chemical substance.
To calculate moles, use the formula:

• \( ext{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \)

It's like converting the weight of a fruit to how many pieces you have if you know each piece's weight.
For instance, in our exercise, finding moles involves:

• For Carbon (C) with a mass of 24 g and a molar mass of 12 g/mol, you get: \[ \text{Moles of } C = \frac{24}{12} = 2 \text{ moles} \]
• For Hydrogen (H), you have 4 g and a molar mass of 1 g/mol: \[ \text{Moles of } H = \frac{4}{1} = 4 \text{ moles} \]
• For Oxygen (O) with a mass of 32 g and a molar mass of 16 g/mol: \[ \text{Moles of } O = \frac{32}{16} = 2 \text{ moles} \]

Understanding mole calculation is crucial to progress more accurately with the chemical composition analysis.

Simplest Ratio

Once you know the moles for each element, the next step is to find the simplest ratio of these moles. This is because the simplest ratio directly corresponds to the empirical formula. It's like finding the smallest common factor that applies to all quantities involved.
Having calculated the moles, you identify the smallest number of moles among all determined values. Here, both Carbon and Oxygen have 2 moles—which is the smallest value. This number is our baseline for determining the simplest ratio.

• For Carbon, divide its moles by the smallest number of moles: \( \frac{2}{2} = 1 \)
• For Hydrogen, do the same: \( \frac{4}{2} = 2 \)
• For Oxygen: \( \frac{2}{2} = 1 \)

This ratio of 1:2:1 gives the simplest whole number ratio for each element in the compound. These ratios are key components in naturally arriving at the correct empirical formula.

Chemical Formula Analysis

The culmination of finding moles and their simplest ratio is writing the empirical formula—a simplified representation of the compound's composition. This formula tells you exactly which atoms are present and in what basic ratio they combine.
Chemical formula analysis involves taking the simplest ratio determined in the previous step and transforming it into the empirical formula. In our example, the ratio of Carbon (C), Hydrogen (H), and Oxygen (O) as 1:2:1 is reflected in the empirical formula \( \text{CH}_2\text{O} \).
Analyzing a chemical formula provides insight into the identity and potential characteristics of the compound because it establishes the most fundamental molecular structure.

• Compare the derived empirical formula with potential options to confirm accuracy.
• Ensure the empirical formula matches the simplest ratio found.

In this exercise, selecting from given choices, \( \text{CH}_2\text{O} \) aligns perfectly with option (d), confirming the accuracy through cross-verification.