Problem 57 - A Complete Resource Book in Chemistry for JEE Main

Question

The empirical formula of a compound is \(\mathrm{CH}_{2}\). One mole of the compound has a mass of \(42 \mathrm{~g}\). Its molecular formula is (a) \(\mathrm{C}_{3} \mathrm{H}_{6}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{8}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{2}\) (d) \(\mathrm{CH}_{2}\)

Step-by-Step Solution

Answer

The correct molecular formula is \( \mathrm{C}_{3} \mathrm{H}_{6} \) (option a).

Step 1: Calculate Empirical Formula Mass

The empirical formula of the compound is \( \mathrm{CH}_{2} \). Calculate its empirical formula mass: Carbon (\( \mathrm{C} \)) has an atomic mass of approximately 12 amu, and Hydrogen (\( \mathrm{H} \)) has an atomic mass of approximately 1 amu. Therefore, the empirical formula mass is \( 12 + 2 \times 1 = 14 \) amu.

Step 2: Calculate Number of Empirical Units in Molecular Formula

The molecular weight of the compound is given as \( 42 \mathrm{~g/mol} \). Divide the molecular weight by the empirical formula mass to find the number of empirical units in the molecular formula: \( \frac{42}{14} = 3 \). This means the molecular formula contains 3 empirical units.

Step 3: Determine the Molecular Formula

Using the number of empirical units calculated in Step 2, multiply the subscripts in the empirical formula \( \mathrm{CH}_{2} \) by 3 to get the molecular formula: \( \mathrm{C}_{3} \mathrm{H}_{6} \).

Step 4: Verify the Option

Among the given options, \( \mathrm{C}_{3} \mathrm{H}_{6} \) corresponds to option (a). Therefore, option (a) is correct.

Key Concepts

Understanding the Empirical Formula

The empirical formula is a simplified ratio that represents the simplest whole number ratio of different types of atoms present in a compound. It is not necessarily the actual amount of atoms in the molecule, rather just a ratio. For instance, the empirical formula \( \mathrm{CH}_{2} \) tells us the ratio of carbon to hydrogen is 1:2, but it does not tell us how many carbon and hydrogen atoms are actually in one molecule.

To calculate the empirical formula, the percentage composition of each element is typically used. In simpler exercises, you may start directly with moles of each atom. Once you have the moles, you divide by the smallest value to find the whole number ratio.

• Empirical formula gives the smallest ratio of elements.
• It is not always the same as the molecular formula.
• Useful in determining molecular formulas when combined with molecular weight information.

Calculating Molecular Weight

Molecular weight or molecular mass is the total mass of a single molecule of a compound and is usually expressed in atomic mass units (amu) or grams per mole (g/mol). For molecules, it's calculated by summing the atomic masses of all the atoms present in the molecular formula.

To determine the molecular weight from empirical information, you need both the empirical formula and the actual mass (often in grams per mole) of a mole of the compound. In the example given, the empirical formula mass of \( \mathrm{CH}_{2} \) was found to be 14 amu. Given the molecular weight is 42 g/mol, one can deduce the full structure of the molecule by dividing:

• Divide molecular weight by empirical mass to find how many empirical units fit in the molecule.
• Multiply each element in the empirical formula by this number.

By calculating it as shown: \( \frac{42}{14} = 3 \), the empirical formula must be multiplied by 3.

An Insight into Atomic Mass

Atomic mass is a fundamental property of an element and represents the average mass of atoms of an element. It is expressed in atomic mass units (amu), and accounts for the presence of various isotopes of an element in nature.

For example, carbon (\( \mathrm{C}\)) has an atomic mass of approximately 12 amu due to the natural abundance of its isotopes. Hydrogen (\( \mathrm{H}\)) has an atomic mass of approximately 1 amu.

• Atomic mass is crucial in calculating empirical and molecular formulas.
• Helps in determining molecular weight by summing the atomic masses of constituent atoms.
• Essential in comparing the mass of different atoms and molecules.

Atomic mass provides the foundation for more complex calculations like those needed for molecular formula determination.