An organic compound having molecular mass 60 is found to contain C = 20%, H = 6.67% and N = 46.67% w

Q: Why wasn't the percentage of oxygen given directly in the question?

In elemental analysis (like combustion analysis), oxygen is often difficult to measure directly. It is standard practice to calculate it by subtracting the sum of all other elements from 100%.

Q: What is the difference between empirical and molecular formula?

The empirical formula gives the simplest whole-number ratio of atoms in a compound, while the molecular formula gives the exact number of atoms in a single molecule. They are related by a whole number multiplier n.

Q: Why do we divide the percentages by atomic mass?

Percentage by mass tells us the mass of each element in a 100g sample. Dividing by the atomic mass converts this mass into moles, allowing us to compare the number of atoms of each element.

Q: What exactly is the Biuret test?

The Biuret test is a chemical test used to detect the presence of peptide bonds (-CO-NH-). When urea is heated, it forms biuret, which contains these bonds and forms a violet coordination complex with Cu2+ ions in an alkaline solution.

Q: Could the compound be an amide like acetamide?

While amides also contain C, H, N, and O, their molar masses and empirical formulas would differ. More importantly, simple amides like acetamide do not give a positive Biuret test upon heating.

Question

An organic compound having molecular mass 60 is found to contain C = 20%, H = 6.67% and N = 46.67% while rest is oxygen. On heating, it gives $	ext{NH}_3$ alongwith a solid residue. The solid residue gives violet colour with alkaline copper sulphate solution. The compound is

Options:

(A) $	ext{CH}_3	ext{CH}_2	ext{CONH}_2$

(B) $(	ext{NH}_2)_2	ext{CO}$

(C) $	ext{CH}_3	ext{CONH}_2$

(D) $	ext{CH}_3	ext{NCO}$

Sigma Percentile · Accepted Answer

The journey of identifying an unknown organic compound is much like being a detective at a crime scene. You are given a set of clues—percentages, masses, and chemical behaviors—and your job is to piece them together to reveal the true identity of the molecule.

In this problem, we are handed a classic analytical chemistry puzzle. We have the percentage composition of Carbon, Hydrogen, and Nitrogen, along with the molecular mass. But the final nail in the coffin is a beautiful, colorful chemical reaction. Let's dive into the investigation!

### Analyzing the Setup: The Percentage Composition

Our first piece of evidence is the elemental composition. The problem states that the compound contains $20.0\%$ Carbon, $6.67\%$ Hydrogen, and $46.67\%$ Nitrogen.

But wait, there is a missing piece! The problem mentions that the "rest is oxygen." In elemental analysis, oxygen is often the silent partner. It is notoriously difficult to measure directly in combustion analysis, so we always find it by difference.

### The Master Equation: Finding the Missing Oxygen

Since the total percentage of all elements in any compound must perfectly sum up to $100\%$, we can easily find the percentage of oxygen. We simply subtract the sum of the known percentages from $100$.

$$ \% 	ext{ of O} = 100 - (\% 	ext{ of C} + \% 	ext{ of H} + \% 	ext{ of N}) $$

Let's substitute our known values into this logical framework:

$$ \% 	ext{ of O} = 100 - (20.0 + 6.67 + 46.67) $$

$$ \% 	ext{ of O} = 100 - 73.34 = 26.66\% $$

Now we have the complete elemental profile of our mystery compound!

### The Mole Concept: Decoding the Empirical Formula

Percentages are great, but atoms react in whole numbers, not percentages. To find the true ratio of atoms, we need to convert these mass percentages into relative moles.

How do we do this? We divide the percentage of each element by its respective atomic mass. This normalizes the data, allowing us to compare apples to apples.

$$ 	ext{Moles of C} = \frac{20.0}{12} = 1.66 $$

$$ 	ext{Moles of H} = \frac{6.67}{1} = 6.67 $$

$$ 	ext{Moles of N} = \frac{46.67}{14} = 3.33 $$

$$ 	ext{Moles of O} = \frac{26.66}{16} = 1.66 $$

### The Simplest Ratio: Building the Skeleton

We now have the relative number of moles, but they are in messy decimals. Molecules demand elegant, whole-number ratios. To find the simplest whole-number ratio, we divide all the molar values by the smallest value among them, which is $1.66$.

For Carbon: $\frac{1.66}{1.66} = 1$

For Hydrogen: $\frac{6.67}{1.66} \approx 4$

For Nitrogen: $\frac{3.33}{1.66} = 2$

For Oxygen: $\frac{1.66}{1.66} = 1$

This beautiful, clean ratio gives us our **Empirical Formula**: $	ext{CH}_4	ext{N}_2	ext{O}$.

### The Molecular Formula: The Final Identity

The empirical formula is just the simplest ratio. Is it the actual molecule? To find out, we must compare the empirical mass with the given molecular mass.

Let's calculate the mass of our empirical formula:

$$ 	ext{Empirical Mass} = 12 + 4(1) + 2(14) + 16 = 60 $$

The problem explicitly states that the molecular mass of the compound is $60$. Since the empirical mass perfectly matches the molecular mass, the multiplier $n$ is exactly $1$.

Therefore, our **Molecular Formula** is also $	ext{CH}_4	ext{N}_2	ext{O}$.

### The Chemical Confession: The Biuret Test

Now, we must identify the specific compound. The formula $	ext{CH}_4	ext{N}_2	ext{O}$ corresponds to **Urea**, which is structurally written as $(	ext{NH}_2)_2	ext{CO}$.

But the problem gives us a final, undeniable confirmation. It states that on heating, the compound releases ammonia gas ($	ext{NH}_3$) and leaves behind a solid residue. When this residue is treated with an alkaline copper sulphate ($	ext{CuSO}_4$) solution, it produces a striking **violet color**.

This is the legendary **Biuret Test**! It is a classic qualitative test for peptide bonds. When two molecules of urea are heated, they condense, eliminating a molecule of ammonia to form a compound called **Biuret**.

$$ 2 	ext{ NH}_2	ext{CONH}_2 \xrightarrow{\Delta} 	ext{NH}_2	ext{CONHCONH}_2 + 	ext{NH}_3 \uparrow $$

The resulting Biuret molecule forms a complex with the $	ext{Cu}^{2+}$ ions in the alkaline solution, resulting in that signature violet hue.

This chemical behavior perfectly aligns with our mathematical derivation, confirming beyond a shadow of a doubt that our mystery compound is indeed **Urea**.

Select Answer:

Visualized Solution

$Analyzing the Composition$

$Finding Percentage of Oxygen$

$Substituting the Values$

$Calculating Oxygen Percentage$

$Calculating Moles$

$Substituting for Moles$

$Simplest Molar Ratio$

$Molecular Formula$

$The Biuret Test$

$Biuret Formation$

The Sigma Insight: Amines

Analyzing the Setup

The Master Equation

The Mole Concept

The Simplest Ratio

The Molecular Formula

The Chemical Confession

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Composition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Percentage of Oxygen

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting the Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Oxygen Percentage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Moles

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting for Moles

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplest Molar Ratio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Molecular Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Biuret Test

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Biuret Formation