1. Find the molecular formula from percentage composition

Given:

• C = $20\%$
• H = $6.67\%$
• N = $46.67\%$
• O = $100 - (20 + 6.67 + 46.67) = 26.66\%$

Take $100$ g of compound:

• C = $20$ g
• H = $6.67$ g
• N = $46.67$ g
• O = $26.66$ g

Convert into moles:

$\text{Moles of C} = \frac{20}{12} \approx 1.67$ $\text{Moles of H} = \frac{6.67}{1} = 6.67$ $\text{Moles of N} = \frac{46.67}{14} \approx 3.33$ $\text{Moles of O} = \frac{26.66}{16} \approx 1.67$

Now divide by the smallest value $1.67$:

$\text{C} : \text{H} : \text{N} : \text{O} = 1 : 4 : 2 : 1$

So the empirical formula is:

$CH_4N_2O$

Its empirical formula mass is:

$12 + 4 + 28 + 16 = 60$

Since molecular mass is also $60$, the molecular formula is:

$CH_4N_2O$

2. Match with the options

Check option formulas:

• A: $CH_3NCO = C_2H_3NO$ → not $CH_4N_2O$
• B: $CH_3CONH_2 = C_2H_5NO$ → not $CH_4N_2O$
• C: $(NH_2)_2CO = CH_4N_2O$ → matches
• D: $CH_3CH_2CONH_2 = C_3H_7NO$ → not $CH_4N_2O$

So from composition alone, the compound is urea:

$(NH_2)_2CO$

3. Verify with the heating behavior

On heating, urea decomposes to give ammonia and a solid residue:

$2(NH_2)_2CO \rightarrow NH_2CONHCONH_2 + NH_3$

The solid residue is biuret.

4. Confirm the biuret test clue

Biuret gives a violet color with alkaline copper sulphate solution. Since the residue gives violet color, this confirms that the original compound is urea, because heating urea forms biuret.

5. Correct option

Therefore, the compound is:

$\boxed{(NH_2)_2CO}$

So, Option C is correct.