Question

Find the empirical formula of an organic compound if it contains
\[ 18.6%\,\mathrm{C},\ 1.55%\,\mathrm{H},\ 55.04%\,\mathrm{Cl} (atomic masses: C = 12, H = 1, Cl = 35.5, O = 16). \]

• A. \(\mathrm{C_2H_2Cl_2O_2}\)
• B. \(\mathrm{CH_2ClO}\)
• C. \(\mathrm{CHClO}\)
• D. \(\mathrm{CHClO_2}\)

Hint:

For empirical formula calculations, always reduce the mole ratio to the simplest whole numbers.

Answer 1

The Correct Option is C

Step 1: Assume 100 g of the compound.
Then, mass of each element present is:
C = \(18.6\,\text{g}\), H = \(1.55\,\text{g}\), Cl = \(55.04\,\text{g}\).
Remaining mass corresponds to oxygen.
Step 2: Calculate mass of oxygen.
\[ 100 - (18.6 + 1.55 + 55.04) = 24.81\,\text{g} \]
Step 3: Convert masses into moles.
\[ \text{Moles of C} = \frac{18.6}{12} = 1.55 \] \[ \text{Moles of H} = \frac{1.55}{1} = 1.55 \] \[ \text{Moles of Cl} = \frac{55.04}{35.5} \approx 1.55 \] \[ \text{Moles of O} = \frac{24.81}{16} \approx 1.55 \]
Step 4: Determine simplest whole-number ratio.
Dividing all mole values by the smallest value (1.55):
C : H : Cl : O = \(1 : 1 : 1 : 1\).
Step 5: Conclusion.
The empirical formula of the compound is \(\mathrm{CHClO}\).