Problem 106 - A Complete Resource Book in Chemistry for JEE Main

Question

In an organic compound of molar mass \(108 \mathrm{~g} \mathrm{~mol}^{-1} \mathrm{C}\), \(\mathrm{H}\) and \(\mathrm{N}\) atoms are present in \(9: 1: 3.5\) by weight. The molecular formula of the compound can be [2002] (a) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{~N}_{2}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{\mathrm{s}} \mathrm{N}_{2}\) (c) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{~N}_{3}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{12} \mathrm{~N}_{2}\)

Step-by-Step Solution

Answer

The molecular formula is \( \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{N}_{2} \) (choice b).

Step 1: Determine the total mass parts

According to the problem, the atoms are in the ratio 9:1:3.5 by weight. Thus, the total parts by weight is 9 + 1 + 3.5 = 13.5 parts.

Step 2: Calculate the mass contribution of each element

To find the mass contributed by each element, calculate the portion of the compound's molar mass represented by each part. - Mass of Carbon (C): \( \frac{9}{13.5} \times 108 = 72 \) g - Mass of Hydrogen (H): \( \frac{1}{13.5} \times 108 = 8 \) g - Mass of Nitrogen (N): \( \frac{3.5}{13.5} \times 108 = 28 \) g

Step 3: Moles of each element

Using atomic masses: C = 12 g/mol, H = 1 g/mol, and N = 14 g/mol. Calculate moles for each element:- Moles of C: \( \frac{72}{12} = 6 \) mol - Moles of H: \( \frac{8}{1} = 8 \) mol - Moles of N: \( \frac{28}{14} = 2 \) mol

Step 4: Determine the empirical formula

Dividing the number of moles by the smallest number of moles to find the simplest ratio:- C: \( \frac{6}{2} = 3 \)- H: \( \frac{8}{2} = 4 \)- N: \( \frac{2}{2} = 1 \)The empirical formula is \( \mathrm{C}_{3} \mathrm{H}_{4} \mathrm{N} \).

Step 5: Determine the molecular formula

Calculate the molar mass of the empirical formula \( \mathrm{C}_{3} \mathrm{H}_{4} \mathrm{N} \): \( (3 \times 12) + (4 \times 1) + (1 \times 14) = 54 \) g/mol.The multiplier to get from empirical to molecular formula is: \( \frac{108}{54} = 2 \).

Step 6: Find the molecular formula

Multiply the empirical formula by the determined factor (2):Molecular formula: \( \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{N}_{2} \).

Key Concepts

Understanding Empirical Formulas

When you're dealing with chemical compounds, one crucial step in analyzing them is determining the empirical formula. This formula represents the simplest ratio of atoms present in a compound based on their relative masses. To find it, we use the information about the proportion of each element by weight in the compound. For example, in a compound with carbon, hydrogen, and nitrogen present in a 9:1:3.5 ratio by weight, we first calculate the total mass parts of the compound, adding up to 13.5 in this case.

Next, we determine the mass contribution of each element. By dividing each part by the total parts and multiplying by the compound's molar mass, we get how much of each element is present in the compound. Thus, the empirical formula helps us understand the relative number of atoms of each element, providing a foundation for further calculations and understanding of the compound's properties.

Molar Mass Calculation

Once the empirical formula is known, the next step is to calculate its molar mass. The molar mass is the weight of one mole of a compound and is crucial in determining the relationship between empirical and molecular formulas. The process involves summing the product of the atomic masses and the number of atoms of each element in the empirical formula.

For instance, given an empirical formula of \( \text{C}_3\text{H}_4\text{N} \), with atomic masses of \(12 \text{ g/mol} \) for carbon, \(1 \text{ g/mol} \) for hydrogen, and \(14 \text{ g/mol} \) for nitrogen, we calculate its molar mass as: \[ (3 \times 12) + (4 \times 1) + (1 \times 14) = 54 \text{ g/mol} \] This calculation helps us verify the correctness of the empirical formula and aids in determining the compound's molecular formula by comparing it to the given molar mass.

Stoichiometry and the Molecular Formula

Stoichiometry lets us bridge the gap between the empirical formula and the molecular formula, using precise calculations to understand a compound's composition. Once the empirical formula and its molar mass are established, we can use stoichiometry to find the molecular formula.

To do this, we calculate the multiplier, which is the ratio of the compound's given molar mass to the molar mass of the empirical formula. In our example, with a compound molar mass of \(108 \, \text{g/mol}\) and an empirical formula molar mass of \(54 \, \text{g/mol}\):\[ \frac{108}{54} = 2 \] The multiplier indicates how many times we need to scale up the empirical formula to match the compound's molar mass, giving us the molecular formula \( \text{C}_6\text{H}_8\text{N}_2 \). Conclusively, stoichiometry helps us understand the exact number of atoms in a molecule, effectively linking laboratory findings to molecular structure understanding.