Problem 69 - A Complete Resource Book in Chemistry for JEE Main

Question

A compound with empirical formula \(\mathrm{CH}_{2} \mathrm{O}\) has a vapour density of \(30 .\) Its molecular formula is (a) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{O}_{2}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)

Step-by-Step Solution

Answer

The molecular formula is \(\mathrm{C}_2\mathrm{H}_4\mathrm{O}_2\) (option b).

Step 1: Understanding Vapour Density

The vapour density of a compound is half of its molar mass. Given that the vapour density is \(30\), we compute the molar mass by multiplying the vapour density by 2, which gives us \(60\).

Step 2: Calculating Molar Mass of Empirical Formula

The empirical formula given is \(\mathrm{CH}_2\mathrm{O}\). Calculate its molar mass: Carbon \(= 12 \), Hydrogen \(= 2 \times 1 = 2 \), and Oxygen \(= 16\). So, the total molar mass is \(12 + 2 + 16 = 30\).

Step 3: Determine the Multiple of the Empirical Formula

Now, we have an empirical formula molar mass of \(30\) and a molecular molar mass of \(60\). The relationship between them is given by the formula: \( n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}} \). Substituting the values, \( n = \frac{60}{30} = 2\).

Step 4: Finding the Molecular Formula

Since \(n = 2\), the molecular formula is \(2\) times the empirical formula. Multiply each subscript in \(\mathrm{CH}_2\mathrm{O}\) by \(2\): \(\mathrm{C}\) becomes \(\mathrm{C}_2\), \(\mathrm{H}_2\) becomes \(\mathrm{H}_4\), and \(\mathrm{O}\) becomes \(\mathrm{O}_2\). Therefore, the molecular formula is \(\mathrm{C}_2\mathrm{H}_4\mathrm{O}_2\).

Key Concepts

Empirical Formula

The empirical formula of a compound is the simplest whole-number ratio of the elements present in it. In simpler terms, it gives the basic skeleton of how many of each type of atom are present in the smallest unit. For example, the given empirical formula for this exercise is \(\mathrm{CH}_2\mathrm{O}\), which indicates that for every carbon atom, there are two hydrogen atoms and one oxygen atom.
It’s crucial to note that the empirical formula does not necessarily indicate the actual number of atoms in a molecule; it just shows the simplest ratio. So, an empirical formula is like a basic recipe that tells you the proportion of ingredients but not the actual quantity needed for the full dish.

Vapour Density

Vapour density is an important concept when determining the molecular formula of a compound. It refers to the mass of a certain volume of a gaseous substance compared to the mass of an equal volume of hydrogen gas under the same conditions. The formula to calculate molar mass from vapour density is:

• Vapour density \( = \frac{\text{Molar Mass}}{2}\)
• Molar Mass \( = \text{Vapour Density} \times 2 \)

In this scenario, the vapour density given is \(30\), which implies that the molar mass of the compound is \(30 \times 2 = 60\). Understanding this relationship is crucial for students aspiring to excel in chemistry, as it allows them to relate easily measurable quantities to molar mass.

Molar Mass

Molar mass can be somewhat daunting at first, but it's simply the mass of one mole of a substance. For any compound, you calculate it by summing the atomic masses of all the atoms in its formula. For example, the empirical formula \(\mathrm{CH}_2\mathrm{O}\) includes:

• Carbon (C) with a mass of \(12\ \text{g/mol}\)
• Hydrogen (H), two with a mass of \(2 \times 1 = 2\ \text{g/mol}\)
• Oxygen (O) with a mass of \(16\ \text{g/mol}\)

So, the empirical formula molar mass adds up to \(30\ \text{g/mol}\). This number is pivotal, as it’s used to find the multiple when determining the molecular formula. Once you have both the empirical formula mass and the molecular mass, calculating the ratio \(n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}}\) determines the factor by which the empirical formula must be scaled to find the molecular formula.

Chemistry Problem Solving

Solving chemistry problems requires a systematic approach, especially in finding molecular formulas. Here’s an easy-to-follow step-by-step method:

• First, identify the empirical formula and calculate its molar mass.
• Next, use the vapour density to find the molecular mass: double the vapour density.
• Figure out the multiple \(n\) by dividing the molecular mass by the empirical formula mass.
• Finally, multiply all subscripts in the empirical formula by \(n\) to get the molecular formula.

This structured approach ensures accuracy while solving problems and identifying the correct molecular formula. It becomes easier to navigate through chemistry challenges when you break them down into simpler steps, ensuring you don’t miss any detail. Remember, practice is key to honing these skills and improving understanding.