The percentage composition of an organic comp | vedclass

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(B) Step $1$: Calculate the empirical formula.$C: \frac{10.06}{12} = 0.838 \approx 1$$H: \frac{0.84}{1} = 0.84 \approx 1$$Cl: \frac{89.10}{35.5} = 2.5

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$CHCl_3$

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(B) Step $1$: Calculate the empirical formula.$C: \frac{10.06}{12} = 0.838 \approx 1$$H: \frac{0.84}{1} = 0.84 \approx 1$$Cl: \frac{89.10}{35.5} = 2.51 \approx 3$Empirical formula is $CHCl_3$.Step $2$: Calculate the molecular weight.$	ext{Molecular weight} = 2 	imes 	ext{Vapour density} = 2 	imes 60.0 = 120$.Step $3$: Calculate the empirical formula weight.$	ext{Empirical formula weight} = 12 + 1 + 3 	imes 35.5 = 13 + 106.5 = 119.5 \approx 120$.Since the molecular weight equals the empirical formula weight,the molecular formula is $CHCl_3$.

The percentage composition of an organic compound is as follows: $C = 10.06\%$,$H = 0.84\%$,$Cl = 89.10\%$. Which of the following corresponds to its molecular formula if the vapour density is $60.0$?

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