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) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OCl}\) (d) \(\mathrm{ClC}_{2} \mathrm{H}_{5} \mathrm{O}\)

Step-by-Step Solution

Answer

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

Step 1: Determine Moles of Each Element

To find the empirical formula, we need to convert the percentage of each element into moles. We divide the percentage of each element by its respective atomic mass:- Carbon (C): \(\frac{18.5}{12.01} = 1.541 \text{ moles} \)- Hydrogen (H): \(\frac{1.55}{1.008} = 1.537 \text{ moles} \)- Chlorine (Cl): \(\frac{55.04}{35.453} = 1.553 \text{ moles} \)- Oxygen (O): \(\frac{24.81}{15.999} = 1.551 \text{ moles} \)

Step 2: Divide by Smallest Number of Moles

To find the simplest whole number ratio, divide the number of moles of each element by the smallest number of moles calculated in Step 1:- Carbon: \( \frac{1.541}{1.537} \approx 1\)- Hydrogen: \( \frac{1.537}{1.537} = 1\)- Chlorine: \( \frac{1.553}{1.537} \approx 1 \)- Oxygen: \( \frac{1.551}{1.537} \approx 1 \) This simplifies to the ratio: 1:1:1:1.

Step 3: Empirical Formula

Using the whole number ratio determined in Step 2, assign each number to its respective element. The compound's empirical formula is: \( \text{C}_1\text{H}_1\text{Cl}_1\text{O}_1 \) or succinctly \( \text{CHClO} \).

Key Concepts

Percentage Composition

The percentage composition of a compound gives the percentage by mass of each element present in the compound. In the given exercise, the compound's percentage composition is provided as follows: 18.5% Carbon, 1.55% Hydrogen, 55.04% Chlorine, and 24.81% Oxygen. These values represent how much of each element makes up the compound, based on a total mass of 100 grams.

To determine the empirical formula, which is the simplest whole-number ratio of atoms in a compound, it's crucial to start with precise percentage values. This information allows us to perform further calculations that hinge on converting these percentages into moles.

Mole Calculation

The next step involves converting the percentage composition into moles, which are a basic counting unit in chemistry. Moles allow us to compare different elements on an equal basis by considering their atomic masses. Each element's given percentage is divided by its respective atomic mass to find its moles.

For example, Carbon's mole calculation is done by dividing 18.5 by its atomic mass, 12.01, which results in approximately 1.541 moles. Similarly, for Hydrogen, Chlorine, and Oxygen, their moles are: 1.537, 1.553, and 1.551 respectively. Getting these mole values is crucial since it sets the stage for obtaining the simplest whole number ratios.

Atomic Mass

The atomic mass of an element is a weighted average mass of the atoms that compose the element, represented on the periodic table in atomic mass units (amu). Understanding atomic mass is critical for mole calculations because it offers the baseline to convert grams (from percentage composition) into moles.

For instance, Carbon has an atomic mass of 12.01 amu, Hydrogen is 1.008 amu, Chlorine stands at 35.453 amu, and Oxygen is 15.999 amu. These values are derived from the natural abundance of their isotopes. Applying atomic masses in calculations allows for accurate conversion of mass percentages into moles, thus pushing forward the empirical formula determination process.

Simplest Whole Number Ratio

After determining moles, the central task is to find the simplest whole-number ratio of the elements in the compound. We achieve this by dividing each element's mole quantity by the smallest mole value obtained. This ratio gives insight into the proportions of atoms in one molecule of the compound.

In this exercise, dividing each mole calculation by the smallest moles (Hydrogen, 1.537) ensures each step maintains consistency. The resulting values are approximately 1 for each element—Carbon, Hydrogen, Chlorine, and Oxygen. This clear ratio of 1:1:1:1 directly leads us to derive the empirical formula: CHClO.

Understanding this concept is important as it assures that the derived chemical formula accurately represents the compound's elemental make-up in its most reduced form.