An organic compound contains `C = 40%, H = 13.33%`, and `N = 46.67%`. Its empirical formula will be

Options

A. `C_(2)H_(7)N_(2)`
B. `CH_(5)N`
C. `CH_(4)N`
D. `C_(2)H_(7)N`

Explanation

To determine the empirical formula of the organic compound with the given percentage composition of elements, we will follow these steps:

### Step 1: List the elements and their percentages
The organic compound contains:
- Carbon (C) = 40%
- Hydrogen (H) = 13.33%
- Nitrogen (N) = 46.67%

### Step 2: Convert percentages to grams
Assuming we have 100 grams of the compound, the mass of each element will be:
- Mass of Carbon = 40 g
- Mass of Hydrogen = 13.33 g
- Mass of Nitrogen = 46.67 g

### Step 3: Convert grams to moles
Now, we will convert the mass of each element to moles using 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

Calculating the moles:
- Moles of Carbon = \( \frac{40 \text{ g}}{12 \text{ g/mol}} = 3.33 \text{ moles} \)
- Moles of Hydrogen = \( \frac{13.33 \text{ g}}{1 \text{ g/mol}} = 13.33 \text{ moles} \)
- Moles of Nitrogen = \( \frac{46.67 \text{ g}}{14 \text{ g/mol}} = 3.33 \text{ moles} \)

### Step 4: Determine the mole ratio
Next, we will find the simplest mole ratio by dividing the number of moles of each element by the smallest number of moles calculated:
- Smallest number of moles = 3.33 (for C and N)

Calculating the ratios:
- Ratio of Carbon = \( \frac{3.33}{3.33} = 1 \)
- Ratio of Hydrogen = \( \frac{13.33}{3.33} \approx 4 \)
- Ratio of Nitrogen = \( \frac{3.33}{3.33} = 1 \)

### Step 5: Write the empirical formula
From the mole ratios, we can write the empirical formula:
- Empirical formula = \( C_1H_4N_1 \) or simply \( CH_4N \)

### Final Answer
The empirical formula of the compound is \( CH_4N \).

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