Find the empirical formula of organic compound if it contains \(18.6 \% \mathrm{C}, 1 \cdot 55 \% \mathrm{H}, 55 \cdot 04 \%\) chlorine ? (atomic mass \(\mathrm{C}=12, \mathrm{H}=1, \mathrm{Cl}=35 \cdot 5, \mathrm{O}=16\) )

The sum of the percentage of carbon, hydrogen, and chlorine is not \(100 \%\)
Hence the rest of the part of the compound is oxygen
percentage of oxygen \(=100-(18.6+1.55+55.04)=24.81 \%\)
Simplest ratio \(=1: 1: 1: 1\)
Hence empirical formula \(=\) CHCIO
Empiraical formula weight \(=12 \times 1+1 \times 1+35.5 \times 1+16 \times 1=64.5\)
\(\frac{\text { molecular mass }}{\text { empirical mass }}=\frac{129}{64.5}=2\)
Therefore molecular formula \(=\mathrm{n} \times(\) Empiricalformula \()\)
Molecular formula \(=2 \times(\mathrm{CHCIO})\)
Molecular formula \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{2} \mathrm{O}_{2}\)