Problem 81 - A Complete Resource Book in Chemistry for JEE Main

Question

What is the molecular formula of a compound, whose empirical formula is \(\mathrm{CH}_{2} \mathrm{O}\) and molecular weight is \(90 ?\) (a) \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{3}\) (c) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{4}\)

Step-by-Step Solution

Answer

The molecular formula is \( \mathrm{C}_3\mathrm{H}_6\mathrm{O}_3 \), option (c).

Step 1: Determine the Empirical Formula Weight

First, calculate the molar mass of the empirical formula \( \mathrm{CH}_2\mathrm{O} \) by adding up the atomic masses of each element: \( \mathrm{C} = 12 \text{ g/mol} \), \( \mathrm{H} = 1 \text{ g/mol} \) (2 hydrogen atoms), and \( \mathrm{O} = 16 \text{ g/mol} \). The empirical formula weight is thus calculated as follows: \( 12 + 2 \times 1 + 16 = 30 \text{ g/mol} \).

Step 2: Calculate the Number of Empirical Units

To find out how many empirical units make up the molecular formula, divide the given molecular weight by the empirical formula weight. The molecular weight provided in the problem is \( 90 \text{ g/mol} \). Divide \( 90 \text{ g/mol} \) by \( 30 \text{ g/mol} \) to get the number of empirical formula units (\( n \)): \( n = \frac{90}{30} = 3 \).

Step 3: Determine the Molecular Formula

Multiply the empirical formula \( \mathrm{CH}_2\mathrm{O} \) by the number of empirical units (\( n = 3 \)) obtained in the previous step: \[ \mathrm{C}_1 \times 3 = \mathrm{C}_3, \quad \mathrm{H}_2 \times 3 = \mathrm{H}_6, \quad \mathrm{O}_1 \times 3 = \mathrm{O}_3 \]. Thus, the molecular formula is \( \mathrm{C}_3\mathrm{H}_6\mathrm{O}_3 \).

Step 4: Match the Molecular Formula to the Given Options

Compare the obtained molecular formula \( \mathrm{C}_3\mathrm{H}_6\mathrm{O}_3 \) with the provided options:(a) \( \mathrm{C}_2\mathrm{H}_6\mathrm{O}_2 \)(b) \( \mathrm{C}_6\mathrm{H}_6\mathrm{O}_3 \)(c) \( \mathrm{C}_3\mathrm{H}_6\mathrm{O}_3 \)(d) \( \mathrm{C}_4\mathrm{H}_8\mathrm{O}_4 \)Option (c) is \( \mathrm{C}_3\mathrm{H}_6\mathrm{O}_3 \), which matches our calculated molecular formula.

Key Concepts

Empirical Formula

An empirical formula represents the simplest whole number ratio of elements in a compound. It tells you the relative number of atoms of each element within a molecule, but doesn't define the exact number of atoms as a molecular formula does. For example, the empirical formula for glucose is the same as that of formaldehyde:

• Both have the empirical formula \(\mathrm{CH}_2\mathrm{O}\).

The empirical formula is useful because it provides the foundational ratio from which you can derive the molecular formula. Calculating the empirical formula involves reducing the number of atoms in the molecular formula to the simplest set of ratios. This often involves dividing by the greatest common divisor that the quantities of elements share.

Molecular Formula Determination

The molecular formula derived from the empirical formula represents the actual number of atoms of each element in a molecule. It is obtained by multiplying the subscripts in the empirical formula by a whole number. This scale-up factor is determined by comparing the molecular weight of the compound to the empirical formula weight. To determine the molecular formula:

• First, calculate the empirical formula weight, which is the sum of the atomic masses of all atoms in the empirical formula.
• Then, divide the given molecular weight of the substance by this empirical formula weight to find the whole number scale factor.
• This factor tells you how many times the empirical formula must be multiplied to reach the molecular formula.

For instance, if the empirical formula is \(\mathrm{CH}_2\mathrm{O}\) and the molecular weight is \(90\ \text{g/mol}\), with the empirical formula weight being \(30\ \text{g/mol}\), divide \(90\ \text{g/mol}\) by \(30\ \text{g/mol}\) to find that the molecular formula is three times the empirical formula, giving \(\mathrm{C}_3\mathrm{H}_6\mathrm{O}_3\).

Stoichiometry

Stoichiometry is the study of the quantitative relationships between the amounts of reactants used and products formed in a chemical reaction. It's like the recipe math in chemistry, letting us predict the outcomes of chemical reactions and how much of each substance is involved. Key concepts include:

• Mole ratios derived from the balanced chemical equation.
• Conversions between mass, moles, and molecule counts.

The approach to solving stoichiometric problems boils down to understanding these concepts and making sure all units are consistent. In the context of determining molecular formulas, stoichiometry helps to identify the ratios in which atoms combine to form compounds.

Molar Mass Calculation

Molar mass is a key concept for understanding chemical formulas because it links the microscopic world of atoms with the macroscopic world of grams and liters. The molar mass of a compound is the mass of one mole of that compound's molecules. It is expressed in grams per mole (\(\text{g/mol}\)). To calculate it, you sum the atomic masses of all the atoms in a compound's empirical or molecular formula, using the periodic table for atomic masses. When determining a molecular formula, like converting an empirical formula \(\mathrm{CH}_2\mathrm{O}\) to \(\mathrm{C}_3\mathrm{H}_6\mathrm{O}_3\), accurate molar mass calculations allow you to understand how many empirical formula units fit into the molecular weight of a compound. Proper calculation of these values is crucial for both research and practical applications in chemistry.