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) \(\mathrm{C}_{3} \mathrm{H}_{10} \mathrm{O}_{5}\) (d) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\)

Step-by-Step Solution

Answer

The molecular formula is \(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\).

Step 1: Determine molar mass of hydrogen

Given that 0.0832 moles of the compound contains 1.0 g of hydrogen, calculate the grams of hydrogen per mole. Since 1 mole of hydrogen atoms is approximately 1 g/mol, the given compound contains about 12 grams of hydrogen per mole (1.0 g / 0.0832 moles).

Step 2: Calculate empirical formula molar mass

The empirical formula of the compound is \(\mathrm{CH}_{2}\mathrm{O}\). Calculate the molar mass: \(12.01\,\text{g/mol (C)} + 2.02\,\text{g/mol (H)} + 16.00\,\text{g/mol (O)} = 30.03\,\text{g/mol}\).

Step 3: Calculate molecular formula molar mass

Since the compound has approximately 12 g of hydrogen per mole, the molecular weight of the compound should be around \( \frac{12}{2} = 6 \) times greater than the empirical formula, meaning it should be \(30.03\,\text{g/mol} \times 6 = 180.18\,\text{g/mol} \).

Step 4: Find molecular formula

Determine the molecular formula by comparing the options with a molar mass close to 180 g/mol. Option (a) \(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\) has a molar mass of \(6 \times 12.01 + 12 \times 1.01 + 6 \times 16.00 = 180.18\,\text{g/mol}\).

Step 5: Conclusion

The molecular formula of the compound is \(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\). This is found by matching the calculated molar mass with the listed options.

Key Concepts

Empirical Formula

The empirical formula provides the simplest whole-number ratio of elements in a compound. It's like a basic blueprint showing the relative quantity of each element. In this problem, the empirical formula given is \( \text{CH}_2\text{O} \). This means that in its simplest form, the compound has one carbon (C), two hydrogens (H), and one oxygen (O).
However, the empirical formula does not show the actual number of atoms in a molecule, just the ratio. More complex compounds can have multiple groups of this basic formula. To determine the exact molecular formula, additional calculations are needed involving molar masses.

Molar Mass Calculation

Calculating molar mass is an essential step when determining a molecular formula from an empirical formula. The molar mass is the mass of one mole of a substance and is usually expressed in grams per mole (g/mol).
The molar mass of the empirical formula \( \text{CH}_2\text{O} \) can be calculated by adding the molar masses of its constituent elements:

• Carbon (C): approximately 12.01 g/mol
• Hydrogen (H): approximately 1.01 g/mol per atom, so 2 H atoms = 2.02 g/mol
• Oxygen (O): approximately 16.00 g/mol

Adding them together, the empirical formula \( \text{CH}_2\text{O} \) has a molar mass of 30.03 g/mol. This mass helps us understand the basic unit before scaling up to the actual molecular formula.

Hydrogen Content Calculation

Calculating the hydrogen content is crucial to figuring out the molecular formula. Given that 0.0832 moles of the compound contains 1.0 g of hydrogen, we need to determine how much hydrogen is present in one mole of the compound.
Calculating hydrogen content allows us to see if the compound consists of multiple sets of the empirical formula. Here, the hydrogen content per mole is calculated by dividing 1.0 g by the 0.0832 moles, resulting in approximately 12 g of hydrogen per mole. This significant value implies that the molecule is likely a larger one, composed of multiple units of \( \text{CH}_2\text{O} \).
This calculation of hydrogen content aids in confidently determining how many times the empirical formula appears in the molecular formula.

Stoichiometry

Stoichiometry helps us observe the quantitative relationships between elements and compounds. In this exercise, stoichiometry is applied to convert between the empirical formula and molecular formula using the calculated molar masses.
Here, the stoichiometric ratio tells us that the calculated molecular mass, linked with the substantial hydrogen content, suggests multiplying the empirical formula molar mass by roughly six. As a result, a larger compound with a molar mass of 180.18 g/mol is deduced, correlating with the option \( \text{C}_6\text{H}_{12}\text{O}_6 \).
Such stoichiometric calculations are key to accurately finding and confirming the actual molecular formulas of chemical compounds.