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

(c) \( \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3} \)

Step 1: Molecular Weight of Empirical Formula

First, determine the molecular weight of the empirical formula \( \mathrm{CH}_{2} \mathrm{O} \). The atomic weights are approximately: \( \mathrm{C} = 12 \), \( \mathrm{H} = 1 \), and \( \mathrm{O} = 16 \). Therefore, the molecular weight of \( \mathrm{CH}_{2} \mathrm{O} \) is calculated as follows:\[12 + (2 \times 1) + 16 = 30\]

Step 2: Determine the Multiple

Next, determine how many empirical formula units are in the molecular formula by dividing the molecular weight of the compound by the weight of the empirical formula:\[\text{Multiple} = \frac{90}{30} = 3\]

Step 3: Calculate the Molecular Formula

Multiply the subscripts in the empirical formula \( \mathrm{CH}_{2} \mathrm{O} \) by the multiple obtained in Step 2. This gives:\[\mathrm{C}_{1 \times 3} \mathrm{H}_{2 \times 3} \mathrm{O}_{1 \times 3} = \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\]

Step 4: Choose the Correct Option

Compare the calculated molecular formula \( \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3} \) from Step 3 to the given options. Option (c) matches the calculated molecular formula.

Key Concepts

Empirical Formula

The empirical formula of a chemical compound represents the simplest whole-number ratio of atoms present in the compound. It does not give the exact number of atoms but provides proportions of the elements within a compound. For instance, the empirical formula \( \mathrm{CH}_{2}\mathrm{O} \) signifies that for every carbon atom, there are two hydrogen atoms and one oxygen atom.

• Does not indicate the actual count of atoms.
• Gives simplified ratios.
• Determined through experimental data, such as elemental composition.

Understanding the empirical formula is crucial as it serves as the foundation in determining the molecular formula, particularly when the molecular weight is known.

Molecular Weight

Molecular weight, also referred to as molecular mass, is the sum of the atomic masses of all the atoms in a molecule. It is measured in atomic mass units (amu). For a given molecular formula, you simply add up the atomic masses of each constituent atom according to its count in the molecule.
For example, to find the molecular weight of the empirical formula \( \mathrm{CH}_{2} \mathrm{O} \):

• Carbon (C) has an atomic weight of approximately 12.
• Hydrogen (H) has an atomic weight of 1, and there are 2 H atoms present, totaling 2.
• Oxygen (O) has an atomic weight of 16.

Therefore, the molecular weight is calculated as:
\[ 12 + 2 + 16 = 30 \]
Knowing the molecular weight is crucial when determining how many empirical formula units are required to match the molecule's true molecular formula.

Chemical Compounds

Chemical compounds are substances composed of two or more different types of atoms bonded together. The composition and structure of these compounds are expressed using chemical formulas, such as the empirical and molecular formulas. Each type of atom contributes to the compound's properties.
When identifying and characterizing a chemical compound:

• Empirical formulas provide insights into the elemental ratios.
• Molecular formulas reveal the exact number of each type of atom.
• The structure of the compound determines its chemical behavior.

By studying chemical formulas, scientists can predict how a compound will interact with other substances and the types of reactions it will undergo.

Stoichiometry

Stoichiometry explores the quantitative relationships and calculations based on chemical formulas and balanced chemical equations. It helps predict the amounts of reactants and products involved in chemical reactions. The fundamental premise of stoichiometry ensures that matter is conserved in these reactions, maintaining precise ratios of atoms.
In the context of molecular formula calculations:

• It's important to use given molecular weights accurately to calculate chemical formulas.
• Understanding atomic proportions aids in determining the necessary multiplicative factor from empirical to molecular formulas.
• It provides the basis for scaling up from simple chemical formulas to balance equations in reactions.

Mastery of stoichiometry is critical for comprehending the full scope of chemical reactions and for performing calculations accurately in practical applications.