Problem 76 - A Complete Resource Book in Chemistry for JEE Main

Question

The empirical formula of a compound is \(\mathrm{CH}_{2} \mathrm{O}\). If \(0.0832\) mole of the compound contains \(1.0 \mathrm{~g}\) of hydrogen, then the molecular formula of the compound is (a) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (b) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) (c) \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5}\) (d) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\)

Step-by-Step Solution

Answer

The molecular formula is \( \text{C}_6\text{H}_{12}\text{O}_6 \) (option a).

Step 1: Determine the Molar Mass of Hydrogen

The atomic mass of hydrogen is approximately 1 g/mol. Given that the compound contains 1.0 g of hydrogen, the number of moles of hydrogen is directly equal to 1.0 moles.

Step 2: Relate Moles of Compound to Moles of Empirical Formula

The empirical formula \( \text{CH}_2\text{O} \) indicates that there are 2 moles of hydrogen per mole of the compound (since \( \text{H}_2 \) is part of it). Therefore, \( 0.0832 \) mole of the compound would contain \(2 \times 0.0832 = 0.1664\) moles of hydrogen given the empirical formula.

Step 3: Calculate the Ratio to Find the Molecular Formula

Since we observe 1.0 mole of hydrogen instead of 0.1664 moles, we need to find the multiplication factor. This factor is \( \frac{1.0}{0.1664} \approx 6 \). This means the molecular formula is 6 times the empirical formula.

Step 4: Derive the Molecular Formula

Multiply each subscript in \(\text{CH}_2\text{O}\) by 6: \( \text{C}(1 \times 6) \text{H}(2 \times 6) \text{O}(1 \times 6) = \text{C}_6\text{H}_{12}\text{O}_6 \).

Step 5: Match Molecular Formula to Answers

The molecular formula \( \text{C}_6\text{H}_{12}\text{O}_6 \) is the same as option (a).

Key Concepts

Mole Concept

The mole concept is an essential foundation in chemistry. It helps chemists express amounts of a chemical substance using a unit defined as 'mole'. Think of the mole as a bridge between the atomic world and the macroscopic world we can see. One mole of any substance contains exactly Avogadro's number of entities, which is approximately \(6.022 \times 10^{23}\) particles, be it atoms, molecules, ions, or others.

The mole is used because chemical reactions occur in terms of moles. It allows conversion between mass, the amount of substance, and the number of particles. For instance, in the given exercise, we determined how many moles of hydrogen were involved by considering their mass. The atomic mass of hydrogen is approximately 1 g/mol, so when we have 1 g of hydrogen, it directly correlates to 1 mole of hydrogen.

To put it simply:

• Mass gives us the number of moles (using the molar mass).
• Moles can then tell us how many particles are present (using Avogadro's number).

This concept is critical in determining how chemical compounds form during reactions.

Chemical Composition

Understanding chemical composition is key to analyzing compounds. A chemical formula such as \(\text{CH}_2\text{O}\) tells us the types and ratios of atoms in a molecule. The empirical formula presents the simplest whole-number ratio of elements within a compound.

In the formula \(\text{CH}_2\text{O}\):

• There is 1 carbon (C) atom,
• 2 hydrogen (H) atoms, and
• 1 oxygen (O) atom.

This implies that for every carbon and oxygen, there are two hydrogens.

Chemical composition is critical when converting empirical formulas to molecular formulas, especially if you're trying to determine the compound's true identity. By knowing the composition and total mass involved, one can establish how many empirical units fit into the actual molecular structure. As demonstrated in the problem, the empirical formula was used to derive the molecular formula \(\text{C}_6\text{H}_{12}\text{O}_6\) by recognizing that the composition repeats six times to match the observed quantity of hydrogen.

Stoichiometry

Stoichiometry is the quantitative aspect of chemistry concerned with the relationships in chemical reactions. It involves using the molar relationships defined by balanced equations and chemical formulas.

Consider stoichiometry as the roadmap for converting and relating different quantities in chemical processes, such as moles to grams or moles of one substance to moles of another. In our exercise, stoichiometry was used to solve for the actual number of hydrogen moles in the molecular formula.

This was done by determining how many more moles of hydrogen were present than predicted by the empirical formula. The calculation \( \frac{1.0}{0.1664} \approx 6 \) shows that the molecular formula has six times the quantity of atoms as the empirical formula did.

Key components of stoichiometry include:

• Balancing chemical equations.
• Using ratios from equations to find relationships between reactants and products.
• Converting mass to moles and vice versa.

In essence, stoichiometry connects measurements with chemical reactions, enabling precise calculations necessary for predicting yields and determining unknowns in a chemical reaction.