In a compound `C, H, N` atoms are present in `9:1:3.5` by weight. Molecular weight of compound is `108`. Its molecular formula is:

Options

A. `C_(2)H_(6)N_(2)`
B. `C_(3)H_(4)N`
C. `C_(6)H_(8)N_(2)`
D. `C_(9)H_(12)N_(3)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the mass of each element
Given the ratio of the masses of Carbon (C), Hydrogen (H), and Nitrogen (N) is 9:1:3.5, we can assign:
- Mass of Carbon (C) = 9 parts
- Mass of Hydrogen (H) = 1 part
- Mass of Nitrogen (N) = 3.5 parts

### Step 2: Calculate the number of moles of each element
Using the atomic weights:
- Atomic weight of Carbon (C) = 12 g/mol
- Atomic weight of Hydrogen (H) = 1 g/mol
- Atomic weight of Nitrogen (N) = 14 g/mol

Now, we can calculate the number of moles for each element:
- Moles of Carbon = \( \frac{9}{12} = 0.75 \) moles
- Moles of Hydrogen = \( \frac{1}{1} = 1 \) mole
- Moles of Nitrogen = \( \frac{3.5}{14} = 0.25 \) moles

### Step 3: Find the simplest mole ratio
To find the simplest ratio, we will divide the number of moles of each element by the smallest number of moles calculated:
- Smallest number of moles = 0.25 (for Nitrogen)

Calculating the ratios:
- Ratio of Carbon = \( \frac{0.75}{0.25} = 3 \)
- Ratio of Hydrogen = \( \frac{1}{0.25} = 4 \)
- Ratio of Nitrogen = \( \frac{0.25}{0.25} = 1 \)

Thus, the simplest ratio of C:H:N is 3:4:1.

### Step 4: Write the empirical formula
From the simplest ratio, we can write the empirical formula as:
- Empirical formula = \( C_3H_4N \)

### Step 5: Calculate the empirical formula mass
Now, we calculate the mass of the empirical formula:
- Mass of \( C_3H_4N = (3 \times 12) + (4 \times 1) + (1 \times 14) \)
- Mass = 36 + 4 + 14 = 54 g/mol

### Step 6: Determine the value of n
To find the molecular formula, we need to calculate n:
- Given molecular weight = 108 g/mol
- \( n = \frac{\text{Molecular weight}}{\text{Empirical formula mass}} = \frac{108}{54} = 2 \)

### Step 7: Write the molecular formula
The molecular formula is obtained by multiplying the empirical formula by n:
- Molecular formula = \( n \times C_3H_4N = 2 \times C_3H_4N = C_6H_8N_2 \)

### Final Answer
Thus, the molecular formula of the compound is \( C_6H_8N_2 \).

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