A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?

Answer

To determine the empirical and molecular formulas of the compound containing 4.07% hydrogen, 24.27% carbon, and 71.65% chlorine with a molar mass of 98.96 g, we can follow these steps:

### Step 1: Convert percentages to grams
Assuming we have 100 g of the compound, the mass of each element will be:
- Hydrogen: 4.07 g
- Carbon: 24.27 g
- Chlorine: 71.65 g

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

Calculating moles for each element:
- Moles of Hydrogen = 4.07 g / 1 g/mol = 4.07 moles
- Moles of Carbon = 24.27 g / 12 g/mol = 2.02 moles
- Moles of Chlorine = 71.65 g / 35.5 g/mol = 2.02 moles

### Step 3: Determine the simplest mole ratio
Now, we divide the number of moles of each element by the smallest number of moles calculated:
- For Hydrogen: 4.07 moles / 2.02 moles ≈ 2
- For Carbon: 2.02 moles / 2.02 moles = 1
- For Chlorine: 2.02 moles / 2.02 moles = 1

This gives us the ratio:
- Hydrogen: 2
- Carbon: 1
- Chlorine: 1

### Step 4: Write the empirical formula
From the mole ratio, the empirical formula can be written as:
- Empirical formula = C1H2Cl1 = CH2Cl

### Step 5: Calculate the empirical formula mass
Next, we calculate the molar mass of the empirical formula:
- Molar mass of CH2Cl = (1 × 12) + (2 × 1) + (1 × 35.5) = 12 + 2 + 35.5 = 49.5 g/mol

### Step 6: Determine the molecular formula
Now, we compare the empirical formula mass with the given molar mass of the compound (98.96 g/mol):
- Molecular mass / Empirical mass = 98.96 g/mol / 49.5 g/mol ≈ 2

Since the ratio is approximately 2, we multiply the subscripts in the empirical formula by 2:
- Molecular formula = C(1×2)H(2×2)Cl(1×2) = C2H4Cl2

### Final Answer
- Empirical Formula: CH2Cl
- Molecular Formula: C2H4Cl2