Problem 148 - A Complete Resource Book in Chemistry for JEE Main

Question

An organic compound having molecular mass 60 is found to contain \(\mathrm{C}=20 \%, \mathrm{H}=6.67 \%\) and \(\mathrm{N}=\) \(46.67 \%\) while rest is oxygen. On heating, it gives \(\mathrm{NH}_{3}\) along with a solid residue. The solid residue gives violet colour with alkaline copper sulphate solution. The compound is (a) \(\mathrm{CH}_{3} \mathrm{NCO}\) (b) \(\mathrm{CH}_{3} \mathrm{CONH}_{2}\) (c) \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CONH}_{2}\)

Step-by-Step Solution

Answer

The correct compound is (c) \((\mathrm{NH}_2)_2\mathrm{CO}\).

Step 1: Analyze the given percentages

Given percentages for C, H, N, and the information that the rest is oxygen. Calculate the percentage of oxygen as 100 - (20 + 6.67 + 46.67) = 26.66%.

Step 2: Calculate the moles for each element

Assume 100 g of the compound for simplicity. Calculate the moles: C: \( \frac{20}{12} = 1.67 \), H: \( \frac{6.67}{1} = 6.67 \), N: \( \frac{46.67}{14} = 3.33 \), O: \( \frac{26.66}{16} = 1.67 \).

Step 3: Determine the empirical formula

Divide the moles by the smallest number of moles (1.67 in this case): C: \( \frac{1.67}{1.67} = 1 \), H: \( \frac{6.67}{1.67} = 4 \), N: \( \frac{3.33}{1.67} = 2 \), O: \( \frac{1.67}{1.67} = 1 \). The empirical formula is \( \mathrm{CH}_4\mathrm{N}_2\mathrm{O} \).

Step 4: Calculate molar mass of empirical formula

Calculate molecular mass of \( \mathrm{CH}_4\mathrm{N}_2\mathrm{O} \): \(12 + 4 + 28 + 16 = 60 \). This matches the given molecular mass.

Step 5: Identify properties from reaction products

Given that heating the compound produces \( \mathrm{NH}_3 \) and a solid residue that gives a violet color with alkaline copper sulfate, suggests formation of biuret. Identify which option forms biuret potentially which checks both properties.

Step 6: Choose the correct compound

The compound \( \left( \mathrm{NH}_2 \right)_2 \mathrm{CO} \) (urea) gives biuret upon heating, matching the requirement of evolving \( \mathrm{NH}_3 \). Therefore, the correct answer is (c) \( \left( \mathrm{NH}_2 \right)_2 \mathrm{CO} \).

Key Concepts

Molecular Mass Calculation

Calculating the molecular mass of a compound is a critical step in determining its identity and formula. To find the molecular mass, you sum the atomic masses of all the atoms in a molecule. Atomic masses are typically reflected as integers, rounding from atomic weights found on the periodic table.

In the example problem, we are given the molecular mass as 60. To confirm this with the empirical formula found, which was \( \mathrm{CH}_4\mathrm{N}_2\mathrm{O} \), we compute as follows:

• Carbon \( (\mathrm{C}) \): 1 atom \( \times 12 \) amu = 12 amu
• Hydrogen \( (\mathrm{H}) \): 4 atoms \( \times 1 \) amu = 4 amu
• Nitrogen \( (\mathrm{N}) \): 2 atoms \( \times 14 \) amu = 28 amu
• Oxygen \( (\mathrm{O}) \): 1 atom \( \times 16 \) amu = 16 amu

The total molecular mass is: \( 12 + 4 + 28 + 16 = 60 \) amu.
This confirms the molecular mass matches with the empirical formula, validating the formula derived.

Percent Composition

Percent composition provides the relative amount of each element in a compound expressed in percentages. It's calculated as the mass of each element divided by the total molar mass, multiplied by 100%. This concept helps to confirm the experimental data from chemical analysis.

In the provided exercise, the elements hydrogen, carbon, nitrogens, and oxygen have the following compositions:

• Carbon: 20%
• Hydrogen: 6.67%
• Nitrogen: 46.67%
• Oxygen is calculated as the remainder: \( 100\% - (20 + 6.67 + 46.67)\% \), resulting in 26.66%

These percentages converted to moles helped derive the compound's empirical formula through systematic calculations in the exercise's solution. Understanding percent composition is crucial to coherently discussing the contribution of each element in a compound.

Biuret Test

The biuret test is a chemical assay that detects the presence of peptide bonds, commonly used in protein detection. This test owes its name to biuret, a compound formed when urea is heated and decomposes into a substance with multiple peptide-like bonds.

When biuret is in the presence of alkaline copper sulfate solution, the solution turns violet, indicating peptide bonds. This reaction is why the presence of a solid residue from the given organic compound forming a violet color upon heating suggested the likelihood of a biuret forming compound, such as urea.
This understanding is crucial in experimental chemistry, as it informs the identification of compounds based on reactions like these indicating particular structural features, like peptide links.

Urea Decomposition

Urea is an amide that decomposes upon heating. It forms ammonia \( (\mathrm{NH}_{3}) \) and biuret as a main decomposition product. This process is vital in certain biochemical pathways and industrial applications.

Upon heating, urea breaks down through the following reaction:\[\mathrm{(NH_2)_2CO} \rightarrow \mathrm{NH_3} \ + \ \mathrm{NH_2CONHCONH_2} \text{ (biuret)}\]In the exercise, the mention of \( \mathrm{NH}_3 \) formation and biuret detection by a violet color with copper sulfate strongly supports the decomposition of urea.
This knowledge is significant as it demonstrates how urea's decomposition is not merely a loss of its components, but also a creation of new substances, instrumental in biological systems and laboratory settings.