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 NH3 along with a solid residue. The solid residue gives violet colour with alkaline copper sulphate solution. The compound is?

Options

A. `CH_3 CH_2 CONH_2`
B. `(NH_2)_2 CO`
C. `CH_3 CONH_2`
D. `CH_3 NCO`

Explanation

To solve the problem step by step, we will analyze the given data and calculate the empirical formula of the compound, followed by determining its molecular formula.

### Step 1: Analyze the given percentages
The organic compound has the following composition:
- Carbon (C) = 20%
- Hydrogen (H) = 6.67%
- Nitrogen (N) = 46.67%
- Oxygen (O) = 100% - (20% + 6.67% + 46.67%) = 26.66%

### Step 2: Calculate the number of moles of each element
To find the number of moles, we will divide the percentage of each element by its atomic mass:
- Moles of C = 20 g / 12 g/mol = 1.67 moles
- Moles of H = 6.67 g / 1 g/mol = 6.67 moles
- Moles of N = 46.67 g / 14 g/mol = 3.33 moles
- Moles of O = 26.66 g / 16 g/mol = 1.66 moles

### Step 3: Find the simplest mole ratio
Next, we will divide the number of moles of each element by the smallest number of moles calculated (1.67 moles):
- C: 1.67 / 1.67 = 1
- H: 6.67 / 1.67 = 4
- N: 3.33 / 1.67 = 2
- O: 1.66 / 1.67 = 1

This gives us the ratio:
C : H : N : O = 1 : 4 : 2 : 1

### Step 4: Write the empirical formula
From the mole ratio, the empirical formula can be written as:
\[ \text{Empirical formula} = \text{C}_1\text{H}_4\text{N}_2\text{O}_1 = \text{CH}_4\text{N}_2\text{O} \]

### Step 5: Calculate the empirical formula mass
Now we calculate the empirical formula mass:
- Mass of C = 12 g/mol
- Mass of H = 4 × 1 g/mol = 4 g/mol
- Mass of N = 2 × 14 g/mol = 28 g/mol
- Mass of O = 16 g/mol

Total empirical mass = 12 + 4 + 28 + 16 = 60 g/mol

### Step 6: Determine the molecular formula
Since the molecular mass given is also 60 g/mol, we can find the value of n:
\[ n = \frac{\text{Molecular weight}}{\text{Empirical weight}} = \frac{60}{60} = 1 \]

Thus, the molecular formula is the same as the empirical formula:
\[ \text{Molecular formula} = \text{CH}_4\text{N}_2\text{O} \]

### Step 7: Confirm the properties of the compound
The compound produces NH3 upon heating, indicating the presence of nitrogen. The solid residue giving a violet color with alkaline copper sulfate solution suggests the presence of a protein or amino compound, which aligns with the determined molecular formula.

### Conclusion
The compound is:
\[ \text{CH}_4\text{N}_2\text{O} \]