Find the empirical formula of organic compound if it contains $18.6 \% \mathrm{C}, 1 \cdot 55 \% \mathrm{H},…

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 $\mathrm{C}=12, \mathrm{H}=1, \mathrm{Cl}=35 \cdot 5, \mathrm{O}=16$ )
  1. $\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{2} \mathrm{O}_{2}$
  2. $\mathrm{CH}_{2} \mathrm{ClO}$
  3. CHClO
  4. $\mathrm{CHClO}_{2}$

Solution

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 $\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{2} \mathrm{O}_{2}$

Asked in: MHT CET 2020 (19 Oct Shift 1)

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