Quantitative analysis of an organic compound | vedclass

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(B) Total percentage of $C, H, Cl = 14.5 + 1.8 + 64.46 = 80.76\%$.Percentage of Oxygen $(O) = 100 - 80.76 = 19.24\%$.Relative number of moles:$n_C = \

Vedclass · Accepted Answer

$1655$

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(B) Total percentage of $C, H, Cl = 14.5 + 1.8 + 64.46 = 80.76\%$.Percentage of Oxygen $(O) = 100 - 80.76 = 19.24\%$.Relative number of moles:$n_C = \frac{14.5}{12} = 1.208$$n_H = \frac{1.8}{1} = 1.8$$n_{Cl} = \frac{64.46}{35.5} = 1.815$$n_O = \frac{19.24}{16} = 1.2025$Simplest molar ratio (divide by $1.2025$):$C: \frac{1.208}{1.2025} \approx 1$$H: \frac{1.8}{1.2025} \approx 1.5$$Cl: \frac{1.815}{1.2025} \approx 1.5$$O: \frac{1.2025}{1.2025} = 1$Simplest whole number ratio $(	imes 2)$: $C: 2, H: 3, Cl: 3, O: 2$.Empirical formula: $C_2H_3Cl_3O_2$.Empirical formula mass $= (2 	imes 12) + (3 	imes 1) + (3 	imes 35.5) + (2 	imes 16) = 24 + 3 + 106.5 + 32 = 165.5 \ g \ mol^{-1}$.Value $= 1655 	imes 10^{-1}$.

Quantitative analysis of an organic compound $(X)$ shows the following percentage composition: $C: 14.5\%$,$Cl: 64.46\%$,and $H: 1.8\%$. The empirical formula mass of the compound $(X)$ is $.......... \times 10^{-1}$. (Given molar mass in $g \ mol^{-1}$ of $C: 12, H: 1, O: 16, Cl: 35.5$)

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