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 can be

Options

A. `C_(6)H_(8)N_(2)`
B. `C_(7)H_(10)N`
C. `C_(5)H_(6)N_(3)`
D. `C_(4)H_(18)N_(3)`

Answer

To find the molecular formula of the organic compound with a molar mass of 108 g/mol and a mass ratio of C, H, and N as 9:1:3.5, we will follow these steps:

### Step 1: Determine the mass of each element in the ratio
Given the mass ratio of C:H:N = 9:1:3.5, we can assume a total mass of 9 + 1 + 3.5 = 13.5 parts.

- Mass of Carbon (C) = (9/13.5) * 108 g = 72 g
- Mass of Hydrogen (H) = (1/13.5) * 108 g = 8 g
- Mass of Nitrogen (N) = (3.5/13.5) * 108 g = 28 g

### Step 2: Calculate the number of moles of each element
Using the atomic masses:
- Atomic mass of C = 12 g/mol
- Atomic mass of H = 1 g/mol
- Atomic mass of N = 14 g/mol

Now, we calculate the number of moles:
- Moles of C = 72 g / 12 g/mol = 6 moles
- Moles of H = 8 g / 1 g/mol = 8 moles
- Moles of N = 28 g / 14 g/mol = 2 moles

### Step 3: Find the simplest mole ratio
Now we will find the simplest ratio by dividing each by the smallest number of moles:
- Ratio of C = 6 / 2 = 3
- Ratio of H = 8 / 2 = 4
- Ratio of N = 2 / 2 = 1

Thus, the empirical formula is C₃H₄N.

### Step 4: Calculate the molar mass of the empirical formula
Now we calculate the molar mass of the empirical formula C₃H₄N:
- Molar mass = (3 × 12) + (4 × 1) + (1 × 14) = 36 + 4 + 14 = 54 g/mol

### Step 5: Determine the molecular formula
To find the molecular formula, we need to compare the molar mass of the compound (108 g/mol) to the molar mass of the empirical formula (54 g/mol):
- Index = Molar mass of compound / Molar mass of empirical formula = 108 g/mol / 54 g/mol = 2

### Step 6: Write the molecular formula
Now we multiply the subscripts in the empirical formula by the index:
- Molecular formula = C₃H₄N × 2 = C₆H₈N₂

Thus, the molecular formula of the compound is **C₆H₈N₂**.

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