In an organic compound of molar mass `108 gm mol^-1 C, H and N` atoms are presents in `9 : 1 : 3.5` by mass. Molecular formula can be

Options

A. `C_6H_8N_2`
B. `C_7H_10N`
C. `C_5H_6N_3`
D. `C_4H_18N_3`

Answer

To determine the molecular formula of the organic compound with a molar mass of 108 g/mol and the given mass ratio of carbon, hydrogen, and nitrogen, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Mass Ratios**:
The mass ratios of C, H, and N are given as 9:1:3.5.

2. **Convert Mass Ratios to Moles**:
- For Carbon (C):
- Mass = 9 g
- Molar mass of C = 12 g/mol
- Moles of C = \( \frac{9 \text{ g}}{12 \text{ g/mol}} = 0.75 \text{ mol} \)

- For Hydrogen (H):
- Mass = 1 g
- Molar mass of H = 1 g/mol
- Moles of H = \( \frac{1 \text{ g}}{1 \text{ g/mol}} = 1 \text{ mol} \)

- For Nitrogen (N):
- Mass = 3.5 g
- Molar mass of N = 14 g/mol
- Moles of N = \( \frac{3.5 \text{ g}}{14 \text{ g/mol}} = 0.25 \text{ mol} \)

3. **Determine the Simplest Mole Ratio**:
To find the simplest ratio, we divide each mole value by the smallest number of moles calculated:
- Moles of C: \( \frac{0.75}{0.25} = 3 \)
- Moles of H: \( \frac{1}{0.25} = 4 \)
- Moles of N: \( \frac{0.25}{0.25} = 1 \)

Thus, the simplest mole ratio is C: 3, H: 4, N: 1, or C3H4N.

4. **Calculate the Empirical Formula Mass**:
- Empirical formula = C3H4N
- Molar mass = \( (3 \times 12) + (4 \times 1) + (1 \times 14) = 36 + 4 + 14 = 54 \text{ g/mol} \)

5. **Determine the Molecular Formula**:
Given the molar mass of the compound is 108 g/mol, we can find the value of N in the empirical formula:
- \( \frac{108 \text{ g/mol}}{54 \text{ g/mol}} = 2 \)

Therefore, the molecular formula is \( C_{3 \times 2}H_{4 \times 2}N_{1 \times 2} = C6H8N2 \).

### Final Answer:
The molecular formula of the compound is **C6H8N2**.