An organic compound on analysis gave `C=42.8%`, `H=7.2%, and`N=50%` volume of 1 gm of the compoun was found to be 200 ml at STP. Molecular formula of the compound is:

Options

A. `C_4H_8N_4`
B. `C_(16)H_(32)N_(16)`
C. `C_(12)H_(24)N_(12)`
D. `C_2H_4N_2`

Explanation

To find the molecular formula of the organic compound based on the provided percentage composition and volume at STP, we can follow these steps:

### Step 1: Convert the percentage composition to grams
Assuming we have 100 grams of the compound, we can convert the percentages to grams:
- Carbon (C): 42.8 g
- Hydrogen (H): 7.2 g
- Nitrogen (N): 50 g

### Step 2: Convert grams to moles
Next, we convert the grams of each element to moles using their atomic masses:
- Moles of Carbon (C) = \( \frac{42.8 \text{ g}}{12 \text{ g/mol}} = 3.57 \text{ moles} \)
- Moles of Hydrogen (H) = \( \frac{7.2 \text{ g}}{1 \text{ g/mol}} = 7.2 \text{ moles} \)
- Moles of Nitrogen (N) = \( \frac{50 \text{ g}}{14 \text{ g/mol}} = 3.57 \text{ moles} \)

### Step 3: Determine the simplest mole ratio
Now, we need to find the simplest whole number ratio of the moles:
- For Carbon: \( \frac{3.57}{3.57} = 1 \)
- For Hydrogen: \( \frac{7.2}{3.57} \approx 2 \)
- For Nitrogen: \( \frac{3.57}{3.57} = 1 \)

Thus, the empirical formula is \( C_1H_2N_1 \) or simply \( CH_2N \).

### Step 4: Calculate the empirical formula mass
Now, we calculate the mass of the empirical formula:
- Mass of \( CH_2N = 12 \text{ g/mol (C)} + 2 \text{ g/mol (H)} + 14 \text{ g/mol (N)} = 28 \text{ g/mol} \)

### Step 5: Calculate the molecular weight using the gas volume
Given that 1 g of the compound occupies 200 mL at STP, we can find the molecular weight:
- At STP, 1 mole of gas occupies 22400 mL.
- Therefore, the weight of 1 mole of the gas can be calculated as:
\[ \text{Molecular weight} = \frac{22400 \text{ mL}}{200 \text{ mL}} \times 1 \text{ g} = 112 \text{ g/mol} \]

### Step 6: Determine the number of empirical units in the molecular formula
Now, we find \( n \) (the number of empirical units in the molecular formula):
\[ n = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} = \frac{112 \text{ g/mol}}{28 \text{ g/mol}} = 4 \]

### Step 7: Write the molecular formula
Finally, we can write the molecular formula:
\[ \text{Molecular formula} = n \times \text{Empirical formula} = 4 \times (CH_2N) = C_4H_8N_4 \]

Thus, the molecular formula of the compound is **C₄H₈N₄**.

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