A compound of carbon, hydrogen and nitrogen contains these elements in the ratio 9:1: 3.5. If its molecular mass is 108, what is the molecular formula?

Options

A. `C_2H_5 N `
B. `C_3 H_4 N`
C. `C_2 HN_2`
D. `C_6H_8N_2`

Explanation

To find the molecular formula of the compound containing carbon, hydrogen, and nitrogen in the ratio of 9:1:3.5 with a molecular mass of 108, we can follow these steps:

### Step 1: Convert the Ratio to Whole Numbers
The given ratio of carbon, hydrogen, and nitrogen is 9:1:3.5. To convert this to whole numbers, we can multiply each part of the ratio by 2 (since 3.5 is the only non-integer).

- Carbon: \( 9 \times 2 = 18 \)
- Hydrogen: \( 1 \times 2 = 2 \)
- Nitrogen: \( 3.5 \times 2 = 7 \)

So, the new ratio is 18:2:7.

### Step 2: Determine the Number of Atoms
Next, we will find the number of atoms of each element based on their atomic masses:
- Atomic mass of Carbon (C) = 12 g/mol
- Atomic mass of Hydrogen (H) = 1 g/mol
- Atomic mass of Nitrogen (N) = 14 g/mol

Now, we can express the empirical formula based on the ratio:
- C: 18/6 = 3
- H: 2/1 = 2
- N: 7/7 = 1

Thus, the empirical formula is \( C_3H_2N_1 \) or simply \( C_3H_2N \).

### Step 3: Calculate the Empirical Formula Mass
Now, we will calculate the mass of the empirical formula:
- Mass of \( C_3 \): \( 3 \times 12 = 36 \)
- Mass of \( H_2 \): \( 2 \times 1 = 2 \)
- Mass of \( N_1 \): \( 1 \times 14 = 14 \)

Adding these together gives:
\[ 36 + 2 + 14 = 52 \text{ g/mol} \]

### Step 4: Determine the Molecular Formula
Now that we have the empirical formula mass (52 g/mol) and the molecular mass (108 g/mol), we can find the ratio of the molecular mass to the empirical formula mass:
\[ \text{Ratio} = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{108}{52} = 2.08 \approx 2 \]

This means that the molecular formula is \( 2 \times \) the empirical formula.

### Step 5: Write the Molecular Formula
Thus, the molecular formula is:
\[ C_{3 \times 2}H_{2 \times 2}N_{1 \times 2} = C_6H_4N_2 \]

### Final Answer
The molecular formula of the compound is \( C_6H_4N_2 \).

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