An organic compound on analysis was found to contain 10.06 % Carbon,0.84 % Hydrogen and 89.10 % Chloride .What will be empirical formula of the substance ?

Options

A. `CH_(2)Cl_(2)`
B. `CHCl_(3)`
C. `"CC"l_(4)`
D. `CH_(3)Cl`

Answer

To find the empirical formula of the organic compound based on the given percentages of carbon, hydrogen, and chlorine, we can follow these steps:

### Step 1: Convert percentages to grams
Assuming we have 100 grams of the compound, the percentages can be directly converted to grams:
- Carbon: 10.06 g
- Hydrogen: 0.84 g
- Chlorine: 89.10 g

### Step 2: Convert grams to moles
Next, we convert the mass of each element to moles using their molar masses:
- Molar mass of Carbon (C) = 12 g/mol
- Molar mass of Hydrogen (H) = 1 g/mol
- Molar mass of Chlorine (Cl) = 35.5 g/mol

Calculating moles for each element:
- Moles of Carbon = 10.06 g / 12 g/mol = 0.8383 mol
- Moles of Hydrogen = 0.84 g / 1 g/mol = 0.84 mol
- Moles of Chlorine = 89.10 g / 35.5 g/mol = 2.507 mol

### Step 3: Determine the simplest mole ratio
Now, we need to find the simplest ratio of moles of each element by dividing each by the smallest number of moles calculated:
- Smallest number of moles = 0.8383 mol (for Carbon)

Calculating the ratios:
- Ratio of Carbon = 0.8383 mol / 0.8383 mol = 1
- Ratio of Hydrogen = 0.84 mol / 0.8383 mol ≈ 1.002 (approximately 1)
- Ratio of Chlorine = 2.507 mol / 0.8383 mol ≈ 2.99 (approximately 3)

### Step 4: Write the empirical formula
The simplest whole number ratio of C:H:Cl is 1:1:3. Therefore, the empirical formula of the compound is:
**CHCl3**