Problem 68 - A Complete Resource Book in Chemistry for JEE Main

Question

If a compound on analysis was found to contain \(\mathrm{C}=\) \(18.5 \%, \mathrm{H}=1.55 \%, \mathrm{Cl}=55.04 \%\) and \(\mathrm{O}=24.81 \%\) then its empirical formula is (a) \(\mathrm{CHClO}\) (b) \(\mathrm{CH}_{2} \mathrm{ClO}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OCl}\) (d) \(\mathrm{ClC}_{2} \mathrm{H}_{5} \mathrm{O}\)

Step-by-Step Solution

Answer

The empirical formula is \( \text{CHClO} \).

Step 1: Convert Percentages to Masses

Assume we have 100 grams of the compound. Then, the mass of each element in the compound is equal to its percentage: - Carbon (C): 18.5 grams - Hydrogen (H): 1.55 grams - Chlorine (Cl): 55.04 grams - Oxygen (O): 24.81 grams.

Step 2: Calculate Moles of Each Element

To find the moles of each element, divide the mass by the atomic mass:- Moles of C = \( \frac{18.5 \text{ g}}{12.01 \text{ g/mol}} \approx 1.54 \text{ mol} \) - Moles of H = \( \frac{1.55 \text{ g}}{1.01 \text{ g/mol}} \approx 1.53 \text{ mol} \)- Moles of Cl = \( \frac{55.04 \text{ g}}{35.45 \text{ g/mol}} \approx 1.55 \text{ mol} \)- Moles of O = \( \frac{24.81 \text{ g}}{16.00 \text{ g/mol}} \approx 1.55 \text{ mol} \)

Step 3: Find the Simplest Ratio of Moles

Divide each of the mole values by the smallest number of moles (1.53 in this case) to get the simplest ratio:- C: \( \frac{1.54}{1.53} \approx 1\)- H: \( \frac{1.53}{1.53} = 1\)- Cl: \( \frac{1.55}{1.53} \approx 1\)- O: \( \frac{1.55}{1.53} \approx 1\)

Step 4: Write the Empirical Formula

The ratio of carbon, hydrogen, chlorine, and oxygen is approximately 1:1:1:1. Therefore, the empirical formula of the compound is \( \text{CHClO} \).

Key Concepts

Percent Composition

The percent composition of a compound gives us the percentage by mass of each element present in it. This information is crucial for determining the empirical formula.
In the given problem, we analyze a compound that contains carbon (C), hydrogen (H), chlorine (Cl), and oxygen (O). Their respective percent compositions are 18.5%, 1.55%, 55.04%, and 24.81%.
When considering percent composition, we assume 100 grams of the compound to easily convert percentages to gram amounts. For instance:

• 18.5 grams of carbon
• 1.55 grams of hydrogen
• 55.04 grams of chlorine
• 24.81 grams of oxygen

This approach simplifies the calculation of empirical formulas by directly translating percentages into mass units.

Mole Calculation

Mole calculation is a fundamental concept in chemistry used to convert mass into a measurable number of particles in moles.
To calculate the number of moles for each element in the compound, divide their masses by their respective atomic masses. The atomic mass acts as a conversion factor.
For this problem, the moles of each element are:

• Carbon: \( \frac{18.5}{12.01} \approx 1.54 \text{ mol} \)
• Hydrogen: \( \frac{1.55}{1.01} \approx 1.53 \text{ mol} \)
• Chlorine: \( \frac{55.04}{35.45} \approx 1.55 \text{ mol} \)
• Oxygen: \( \frac{24.81}{16.00} \approx 1.55 \text{ mol} \)

These calculations are essential as they help in determining the simplest ratio of atoms in a compound.

Atomic Mass

Atomic mass is a key concept in chemistry representing the mass of an atom, typically expressed in atomic mass units (amu).
It serves as a crucial conversion factor between mass in grams and moles.
For example, the atomic mass of carbon (C) is approximately 12.01 g/mol, which means one mole of carbon atoms weighs 12.01 grams.
In mole calculations, atomic masses are used as follows:

• Carbon: 12.01 g/mol
• Hydrogen: 1.01 g/mol
• Chlorine: 35.45 g/mol
• Oxygen: 16.00 g/mol

Understanding atomic mass is essential for converting between mass and moles, which in turn helps determine the empirical formula.

Elemental Analysis

Elemental analysis involves breaking down a compound into its constituent elements and determining the mass percentage of each element.
This process is key in identifying the empirical formula, which gives the simplest ratio of elements in a compound.
In this exercise, elemental analysis provided percentages of carbon, hydrogen, chlorine, and oxygen. By converting these percentages to masses and then to moles, we can derive the simplest molecular ratio.
This ratio unveils the empirical formula, a fundamental aspect of understanding the composition and proportions of elements within any given compound.