Problem 48 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Nitrobenzene is formed as the major product along with a minor product in the reaction of benzene with a hotmixture of nitric acid and sulphuric acid. The minorproduct consists of carbon : \(42.86 \%\), hydrogen : \(2.40 \%\), nitrogen : \(16.67 \%\), and oxygen: \(38.07 \%\) (i) Calculate the empirical formula of the minor product. (ii) When \(5.5 \mathrm{~g}\) of the minor product is dissolved in \(45 \mathrm{~g}\) of benzene, the boiling point of the solution is \(1.84^{\circ} \mathrm{C}\) higher than that of pure benzene. Calculate the molar mass of the minor product and determine its molecular and structural formula. (Molal boiling point elevation constant of benzene is \(\left.2.53 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} .\right)\)

Step-by-Step Solution

Answer

Empirical formula: C₃H₂NO₂. Molecular formula: C₆H₄N₂O₄. Likely structure: Dinitrobenzene.

Step 1: Determine Molar Masses of Elements

First, list the atomic masses: Carbon (C) = 12 g/mol, Hydrogen (H) = 1 g/mol, Nitrogen (N) = 14 g/mol, and Oxygen (O) = 16 g/mol.

Step 2: Convert Percentage to Mass Ratio

Assume 100 g of the compound for ease of calculation: Carbon = 42.86 g, Hydrogen = 2.40 g, Nitrogen = 16.67 g, Oxygen = 38.07 g.

Step 3: Calculate Moles of Each Element

Convert mass to moles: \[\text{Moles of C} = \frac{42.86}{12} = 3.57\]\\[\text{Moles of H} = \frac{2.40}{1} = 2.40\]\\[\text{Moles of N} = \frac{16.67}{14} = 1.19\]\\[\text{Moles of O} = \frac{38.07}{16} = 2.38\]

Step 4: Determine Empirical Formula

Divide each by the smallest mole amount, which is 1.19, resulting in: \[\text{C: } \frac{3.57}{1.19} \approx 3\], \[\text{H: } \frac{2.40}{1.19} \approx 2\], \[\text{N: } \frac{1.19}{1.19} = 1\], \[\text{O: } \frac{2.38}{1.19} \approx 2\].This gives the empirical formula: \(\text{C}_3\text{H}_2\text{N}_1\text{O}_2\).

Step 5: Calculate Molality

Molality is calculated using the change in boiling point and the boiling point elevation constant:\[\Delta T_b = 1.84^{\circ}C, \quad K_b = 2.53 \, \text{K kg mol}^{-1}\] \[\text{Molality} = \frac{\Delta T_b}{K_b} = \frac{1.84}{2.53} = 0.727 \, \text{mol/kg}\]

Step 6: Calculate Moles and Molar Mass of Minor Product

Using the molality and mass of benzene:\[\text{Moles of minor product} = 0.727 \times 0.045 \, \text{kg} = 0.0328 \, \text{mol}\]\[\text{Molar mass} = \frac{5.5}{0.0328} \approx 167.7 \, \text{g/mol}\]

Step 7: Determine Molecular Formula

Calculate n in \(\text{Molar mass} = n \times \text{Empirical formula mass}\). Empirical mass \(\approx \text{C}_3\text{H}_2\text{N}_1\text{O}_2 = (3 \times 12) + (2 \times 1) + (1 \times 14) + (2 \times 16) = 82\;\text{g/mol}\). \(n = \frac{167.7}{82} \approx 2\); thus molecular formula = \(\text{C}_6\text{H}_4\text{N}_2\text{O}_4\).

Step 8: Determine Structural Formula

Based on the molecular formula, the structure could comprise two nitrate (\(NO_2\)) groups attached to a benzene ring, such as dinitrobenzene.

Key Concepts

Boiling Point Elevation

Boiling point elevation occurs when a non-volatile solute, like the minor product in our exercise, is dissolved in a solvent, resulting in an increase in the solvent's boiling temperature. This phenomenon is directly proportional to the amount of solute particles in the solution.

In the exercise, the solute (minor product) is dissolved in benzene, causing the boiling point to rise by 1.84°C. This change can be calculated and related back to the molality of the solution using the formula:

• \(\Delta T_b = i \cdot K_b \cdot m\)

where \(\Delta T_b\) is the increase in boiling point, \(i\) is the van't Hoff factor (equal to 1 for non-electrolytes), \(K_b\) is the ebullioscopic constant of the solvent, and \(m\) is the molality.

This calculation helps us determine the number of moles of the solute in the solution, which guides us in finding the molar mass of the solute, a crucial step in determining the molecular formula.

Nitrobenzene

Nitrobenzene is an aromatic organic compound with the formula \(\text{C}_6\text{H}_5\text{NO}_2\). It is typically formed by the nitration of benzene using a mixture of nitric acid and sulfuric acid, resulting in dinitration products such as the minor product in our problem.

Nitrobenzene is known for its pale-yellow color and almond-like smell. It is primarily used in the industrial synthesis of aniline, which is an important precursor for dyes, drugs, and other chemicals.

In the context of this exercise, understanding the nature and formation of nitrobenzene helps in predicting and calculating the empirical and molecular formulas of nitro-related compounds.

Percentage Composition Analysis

Percentage composition analysis is a method used to determine the elemental composition of a compound by mass. In the exercise, the percentage by mass of each element found in the minor product is given, allowing us to convert these percentages into a usable form for determination of the empirical formula.

The steps typically include:

• Assuming a 100 g sample of the compound, which simplifies conversion as percentages directly translate into grams.
• Converting the mass of each element to moles using their respective atomic masses.
• Using the mole ratio to decide the simplest whole number ratio for each of the atoms in the compound, resulting in the empirical formula.

This analysis serves as the foundation for deeper investigations into a compound's structure and identity, such as empirical and molecular formula calculations.

Empirical Formula Calculation

The empirical formula gives the simplest whole-number ratio of atoms in a compound. It provides crucial foundational information for further analysis and identification of a compound’s molecular formula.

In calculating an empirical formula, as done in the exercise, the percentage composition is first converted into moles for each element. Using the smallest mole value among the elements, each element's moles are divided by this smallest value to find the ratio. For instance, if we calculate:

• \( \text{Moles of C} = \frac{42.86}{12} = 3.57 \)
• \( \text{Moles of H} = \frac{2.40}{1} = 2.40 \)
• \( \text{Moles of N} = \frac{16.67}{14} = 1.19 \)
• \( \text{Moles of O} = \frac{38.07}{16} = 2.38 \)

Each value is divided by the smallest mole amount, yielding the empirical formula \( \text{C}_3\text{H}_2\text{N}_1\text{O}_2 \). Such calculations pave the way toward finding the molecular formula, particularly once a compound's actual molar mass is determined.