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To determine the organic compound with a molecular mass of 60 that contains C = 20%, H = 6.67%, N = 46.67%, and the rest as oxygen, we can follow these steps:
Step 1: Calculate the mass of each element in the compound
Given the percentages, we can calculate the mass of each element in a 100 g sample of the compound:
Mass of Carbon (C) = 20 g
Mass of Hydrogen (H) = 6.67 g
Mass of Nitrogen (N) = 46.67 g
Mass of Oxygen (O) = 100 g - (20 g + 6.67 g + 46.67 g) = 26.66 g
Step 2: Convert mass to moles
Next, we convert the mass of each element to moles using their respective atomic masses:
Moles of C = $\frac{20}{12} = 1.67$
Moles of H = $\frac{6.67}{1} = 6.67$
Moles of N = $\frac{46.67}{14} = 3.33$
Moles of O = $\frac{26.66}{16} = 1.67$
Step 3: Calculate the simplest mole ratio
To find the simplest ratio, we divide each mole value by the smallest number of moles calculated:
Ratio of C = $\frac{1.67}{1.67} = 1$
Ratio of H = $\frac{6.67}{1.67} = 4$
Ratio of N = $\frac{3.33}{1.67} = 2$
Ratio of O = $\frac{1.67}{1.67} = 1$
Step 4: Write the empirical formula
The empirical formula based on the simplest mole ratio is $C{H_4}{N_2}O$.
Step 5: Determine the molecular formula
The empirical weight is calculated as follows:
\[
\text{Empirical weight} = 12 \times 1 + 1 \times 4 + 14 \times 2 + 16 \times 1 = 60
\]
Since the empirical weight equals the molecular weight (60), the molecular formula is also $C{H_4}{N_2}O$.
Step 6: Identify the compound
The compound is urea, which has the formula $(NH_2)_2CO$. On heating, urea decomposes to release ammonia ($NH_3$) and forms a solid residue that gives a violet color with alkaline copper sulfate solution, indicating the presence of the -CONH- group.
Thus, the correct answer is $(NH_2)_2CO$.
