An organic compound containes `C=74.0%, H=8.65%` and `N=17.3%`. Its empirical formul ais

Options

A. `C_5H_8N`
B. `C_(10)H_(12)N`
C. `C_5H_7N`
D. `C_(10)H_(14)N`

Answer

To determine the empirical formula of the organic compound based on its percentage composition of carbon (C), hydrogen (H), and nitrogen (N), we follow these steps:

### Step 1: Convert percentages to grams
Assume we have 100 g of the compound. This means:
- Carbon (C) = 74.0 g
- Hydrogen (H) = 8.65 g
- Nitrogen (N) = 17.3 g

### Step 2: Convert grams to moles
Next, we convert the mass of each element to moles using their respective molar masses:
- 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 number of moles for each element:
- Moles of Carbon (C) = \( \frac{74.0 \, \text{g}}{12 \, \text{g/mol}} \) = 6.17 moles
- Moles of Hydrogen (H) = \( \frac{8.65 \, \text{g}}{1 \, \text{g/mol}} \) = 8.65 moles
- Moles of Nitrogen (N) = \( \frac{17.3 \, \text{g}}{14 \, \text{g/mol}} \) = 1.23 moles

### Step 3: Find the simplest mole ratio
To find the simplest ratio, we divide the number of moles of each element by the smallest number of moles calculated:
- The smallest number of moles is for Nitrogen (N) = 1.23 moles.

Calculating the ratios:
- Ratio of Carbon (C) = \( \frac{6.17}{1.23} \) ≈ 5.02 ≈ 5
- Ratio of Hydrogen (H) = \( \frac{8.65}{1.23} \) ≈ 7.03 ≈ 7
- Ratio of Nitrogen (N) = \( \frac{1.23}{1.23} \) = 1

### Step 4: Write the empirical formula
Now that we have the simplest whole number ratios, we can write the empirical formula:
- Empirical formula = \( C_5H_7N_1 \) or simply \( C_5H_7N \)

### Conclusion
The empirical formula of the organic compound is \( C_5H_7N \).