Percentage of Nitrogen in Yellow Product Formed from Reaction Sequence

To determine the percentage of nitrogen in the yellow product $(X)$, we analyze the given reaction sequence step-by-step:

Step 1: Identify the starting material and reaction mechanism

The starting material is diazoaminobenzene with the structure: $\text{Ph}-\text{N}=\text{N}-\text{NH}-\text{Ph}$

When diazoaminobenzene is treated with an acid ($\text{HCl}$) in the presence of aniline at a warm temperature of $40-45^\circ\text{C}$, it undergoes an acid-catalyzed intermolecular rearrangement (diazoamino to aminoazo rearrangement) to form $p$-aminoazobenzene as a yellow dye (Product $X$).

Step 2: Molecular Formula of Product $(X)$

$p$-aminoazobenzene has the structural formula: $\text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_6\text{H}_4-\text{NH}_2$

Counting the total number of constituent atoms:

• Carbon ($\text{C}$): $6 + 6 = 12$
• Hydrogen ($\text{H}$): $5 + 4 + 2 = 11$
• Nitrogen ($\text{N}$): $1 + 1 + 1 = 3$

Thus, the molecular formula of Product $(X)$ is $\text{C}_{12}\text{H}_{11}\text{N}_3$.

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Step 3: Molar Mass Calculation

Using the given atomic masses ($\text{H} = 1\text{ g mol}^{-1}$, $\text{C} = 12\text{ g mol}^{-1}$, $\text{N} = 14\text{ g mol}^{-1}$):

$\text{Molar Mass of } \text{C}_{12}\text{H}_{11}\text{N}_3 = (12 \times 12) + (11 \times 1) + (3 \times 14)$ $\text{Molar Mass} = 144 + 11 + 42 = 197\text{ g mol}^{-1}$

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Step 4: Percentage of Nitrogen

The total mass contributed by nitrogen in 1 mole of $\text{C}_{12}\text{H}_{11}\text{N}_3$ is: $\text{Mass of Nitrogen} = 3 \times 14 = 42\text{ g mol}^{-1}$

The percentage of nitrogen is calculated as: $\% \text{ N} = \left( \frac{\text{Mass of Nitrogen}}{\text{Total Molar Mass}} \right) \times 100$ $\% \text{ N} = \left( \frac{42}{197} \right) \times 100 \approx 21.32\%$

Rounding to the nearest integer, we get 21.