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 Empirical Formula and Its Mass

The empirical formula given is \( \mathrm{CH}_{2} \mathrm{O} \). Calculate the molar mass of this empirical formula: \( \mathrm{C} = 12 \text{ g/mol}, \ \mathrm{H} = 1 \times 2 = 2 \text{ g/mol}, \ \mathrm{O} = 16 \text{ g/mol} \). So, the empirical formula mass of \( \mathrm{CH}_{2} \mathrm{O} \) is \( 12 + 2 + 16 = 30 \text{ g/mol} \).

Step 2: Calculate the Molecular Mass Using Vapour Density

Vapour density is given as 30. The molecular mass (molar mass) is twice the vapour density, so molecular mass is \( 2 \times 30 = 60 \text{ g/mol} \).

Step 3: Determine the Multiplication Factor

Divide the molecular mass by the mass of the empirical formula to find the multiplying factor: \( \frac{60}{30} = 2 \). This means the molecular formula is two times the empirical formula.

Step 4: Calculate the Molecular Formula

Multiply the indices in the empirical formula \( \mathrm{CH}_{2} \mathrm{O} \) by 2 to get the molecular formula: \( \mathrm{C}_{1 \times 2} \mathrm{H}_{2 \times 2} \mathrm{O}_{1 \times 2} = \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2} \).

Step 5: Verify Against Given Options

The molecular formula obtained \( \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2} \) matches with option (b).

Key Concepts

Empirical Formula

The empirical formula of a compound represents the simplest whole-number ratio of the elements within it. In this specific case, the empirical formula is \( \mathrm{CH}_{2} \mathrm{O} \). This formula tells us that for every carbon atom, there are two hydrogen atoms and one oxygen atom.
To further understand empirical formulas, it's important to note that they do not represent exact elemental counts in a molecule, but rather show the proportional relationships among the elements. For example, while the empirical formula \( \mathrm{CH}_{2} \mathrm{O} \) provides the ratio, the actual molecule could have multiples of these base units, like \( \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2} \) or \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \).
The key to determining the empirical formula is finding these simplest ratios from either percentages or masses. Once known, it provides the foundation for calculating the molecular formula.

Vapour Density

Vapour density is a practical measure used to determine the molar mass of a gaseous substance. It is defined as the mass of a given volume of substance compared to the same volume of hydrogen gas at identical conditions, making hydrogen a reference point.
To convert vapour density to the molecular mass, simply multiply the vapour density by two. In this exercise, with the vapour density of 30, the molecular mass can be calculated as \( 2 \times 30 = 60 \text{ g/mol} \).
Understanding vapour density helps relate to molecular weights and can be particularly useful in calculating molecular formulas from empirical formulas. The initial factor of two comes from Avogadro's principle, linking volume and number of molecules.

Molecular Mass Calculation

Derived from both empirical formulas and vapour density, the molecular mass is essential for deducing the molecular formula of a compound. After obtaining the molecular mass from the vapour density, which is 60 g/mol in this example, it becomes possible to determine how many times the empirical formula must be multiplied to get the molecular formula.
The calculation steps usually involve:

• Calculating the empirical formula mass. In this example, it is 30 g/mol for \( \mathrm{CH}_{2} \mathrm{O} \).
• Using vapour density to determine molecular mass. Here, molecular mass is 60 g/mol.
• Finding the multiplication factor, which is \( \frac{60}{30} = 2 \) here.

These calculations are foundational for moving from an empirical formula to an accurate molecular formula, ensuring we fully understand the compound's structure.

Chemistry Problem Solving

While solving chemistry problems like these, following a systemic approach ensures accuracy and understanding. This involves integrating knowledge of concepts like empirical formulas, vapour density, and molecular mass calculation. Let's break it down into steps:

• Identifying the empirical formula: This tells you the fundamental ratios of elements.
• Calculating the vapour density: This gives you the pathway to finding molecular mass.
• Solving for the multiplication factor with molecular mass: Provides insight into how many empirical units make up the actual molecule.
• Verifying the molecular formula against given options: Confirm accuracy by comparing it to options provided.

Understanding each one of these steps not only helps solve the problem at hand but also equips you with the tools to tackle similar chemistry problems in the future. Remember, practice and familiarity with concepts result in better problem-solving skills.