A compound contains `38.8%C,16%H`, and `45.2%N` The formula of the compound would be

Options

A. `CH_3NH_2`
B. `C_2H_5CN`
C. `CH_3CH`
D. `CH_2(NH_2)_2`

Explanation

To determine the empirical formula of the compound containing 38.8% carbon (C), 16% hydrogen (H), and 45.2% nitrogen (N), we can follow these steps:

### Step 1: Convert percentages to grams
Assuming we have 100 grams of the compound, we can directly take the percentages as grams:
- Carbon: 38.8 g
- Hydrogen: 16 g
- Nitrogen: 45.2 g

### Step 2: Calculate the number of moles of each element
We will use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
- Molar mass of Carbon (C) = 12 g/mol
- Molar mass of Hydrogen (H) = 1 g/mol
- Molar mass of Nitrogen (N) = 14 g/mol

Calculating the moles:
- Moles of Carbon:
\[ \text{Moles of C} = \frac{38.8 \, \text{g}}{12 \, \text{g/mol}} \approx 3.23 \, \text{mol} \]
- Moles of Hydrogen:
\[ \text{Moles of H} = \frac{16 \, \text{g}}{1 \, \text{g/mol}} = 16 \, \text{mol} \]
- Moles of Nitrogen:
\[ \text{Moles of N} = \frac{45.2 \, \text{g}}{14 \, \text{g/mol}} \approx 3.22 \, \text{mol} \]

### Step 3: Find the simplest mole ratio
Next, we will divide the number of moles of each element by the smallest number of moles calculated:
- Smallest number of moles = 3.22 (for Nitrogen)

Calculating the ratios:
- Ratio of Carbon:
\[ \frac{3.23}{3.22} \approx 1.00 \]
- Ratio of Hydrogen:
\[ \frac{16}{3.22} \approx 4.97 \approx 5 \]
- Ratio of Nitrogen:
\[ \frac{3.22}{3.22} = 1.00 \]

### Step 4: Write the empirical formula
From the ratios, we find:
- C: 1
- H: 5
- N: 1

Thus, the empirical formula is:
\[ \text{Empirical Formula} = \text{C}_1\text{H}_5\text{N}_1 = \text{CH}_5\text{N} \]

### Step 5: Adjust the empirical formula if necessary
Since CH5N does not correspond to a known stable compound, we can rearrange it to a more common structure. The empirical formula can be represented as:
\[ \text{C}_3\text{H}_7\text{N}_2 \]
This is equivalent to the molecular formula of CS3NH2.

### Conclusion
The formula of the compound is:
\[ \text{C}_3\text{H}_7\text{N}_2 \]