Problem 42 - A Complete Resource Book in Chemistry for JEE Main

Question

The molecular formula weight by empirical formula weight of a compound is \(3 .\) The empirical formula is \(\mathrm{CH}_{2}\). The vapour density of the compound is (a) 21 (b) \(11.5\) (c) 33 (d) 43

Step-by-Step Solution

Answer

The vapor density of the compound is 21.

Step 1: Understanding the Problem

We are given the empirical formula \(\mathrm{CH}_2\) and told that the molecular formula weight by empirical formula weight is \(3\). We want to find the vapor density of the compound.

Step 2: Calculate the Empirical Formula Weight

The atomic weight of carbon (C) is about \(12\) and hydrogen (H) is about \(1\). Therefore, the empirical formula weight of \(\mathrm{CH}_2\) is \(12 + 2 \times 1 = 14\).

Step 3: Find the Molecular Formula Weight

According to the problem, the molecular formula weight is \(3\) times the empirical formula weight. Thus, the molecular formula weight of the compound is \(3 \times 14 = 42\).

Step 4: Relate Vapor Density to Molecular Weight

Vapor density is defined as half of the molecular weight of a gas. Therefore, the vapor density is \(\frac{42}{2} = 21\).

Step 5: Choose the Correct Answer

The answer options were (a) 21, (b) 11.5, (c) 33, and (d) 43. The calculated vapor density of 21 matches option (a).

Key Concepts

Empirical Formula

An empirical formula provides the simplest whole-number ratio of atoms in a compound. For example, if a compound has an empirical formula of \(\mathrm{CH}_2\), it means there are two hydrogen atoms for every carbon atom present. This does not give us the exact number of atoms in the compound but rather the simplest form.

• The empirical formula is derived from mass percent composition data or combustion analysis results.
• It is crucial in analyzing unknown substances since it provides a relative idea of the compound's composition.
• The calculation does not require actual atom quantities but instead, the smallest, simplest ratios of the elements present.

The empirical formula is essential in determining the molecular formula, where the actual number of atoms in a molecule is required.

Molecular Formula

The molecular formula represents the true number of atoms of each element in a molecule of the compound. It is more detailed than the empirical formula, as it reflects the compound's actual molecular structure. For instance, the compound with an empirical formula \(\mathrm{CH}_2\) could have a molecular formula of \(\mathrm{C}_3\mathrm{H}_6\) in our exercise example.

• To calculate the molecular formula, the empirical formula weight is multiplied by a given factor, often derived from the ratio of the compound's molecular formula weight to its empirical formula weight.
• The molecular formula is important in identifying the compound because it provides detailed insight into the molecular architecture.

Once you know the empirical formula and the molecular formula weight, determining the actual molecular formula becomes straightforward.

Vapor Density

Vapor density is a term used in chemistry to describe the density of a vapor in comparison to an assumed standard, usually air or hydrogen. In simple terms, it is defined as half the molecular weight of a gaseous substance.

• To find the vapor density from the molecular weight, you divide the molecular weight by 2.
• It helps in identifying unknown gaseous compounds quickly, especially in laboratory settings.
• In the example problem, the vapor density was calculated as \(\frac{42}{2} = 21\), where 42 is the molecular weight of the compound derived from the empirical formula.

A correct understanding of vapor density is crucial when working with gases, as it is an indirect way of measuring a gas's mass and identifying its composition.

Atomic Weight Calculation

Atomic weight is the average mass of atoms of an element, calculated using the abundance of isotopes in a naturally occurring element. It plays a crucial role in calculating both empirical and molecular formula weights.

• For example, carbon has an atomic weight of approximately 12, and hydrogen has an atomic weight of about 1.
• By knowing the atomic weights, we can calculate the empirical formula weight, as done for \(\mathrm{CH}_2\), giving us \(12 + 2 \times 1 = 14\).
• This weight is essential in determining the molecular formula weight by multiplying it by an integer value derived from empirical-to-molecular formula weight comparisons.

Using atomic weights effectively allows accurate derivation of both empirical and molecular formulas, crucial steps in chemical analysis.