A carbon compound contains 12.8 % of carbon,2 | vedclass

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(C) To find the empirical formula,calculate the mole ratio of each element:Carbon: $\frac{12.8}{12.0} = 1.066$Hydrogen: $\frac{2.1}{1.008} = 2.083$Bro

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$C_2H_4Br_2$

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(C) To find the empirical formula,calculate the mole ratio of each element:Carbon: $\frac{12.8}{12.0} = 1.066$Hydrogen: $\frac{2.1}{1.008} = 2.083$Bromine: $\frac{85.1}{79.9} = 1.065$Dividing by the smallest value $(1.065)$:Carbon: $\frac{1.066}{1.065} \approx 1$Hydrogen: $\frac{2.083}{1.065} \approx 2$Bromine: $\frac{1.065}{1.065} = 1$Thus,the empirical formula is $CH_2Br$.Empirical formula mass $= 12.0 + 2(1.008) + 79.9 = 93.996 \approx 94$.Calculate the factor $n = \frac{	ext{Molecular mass}}{	ext{Empirical formula mass}} = \frac{187.9}{94} = 2$.Molecular formula $= n 	imes (	ext{Empirical formula}) = 2 	imes (CH_2Br) = C_2H_4Br_2$.

$A$ carbon compound contains $12.8 \%$ of carbon,$2.1 \%$ of hydrogen and $85.1 \%$ of bromine. The molecular weight of the compound is $187.9$. Calculate the molecular formula of the compound. (Atomic weight: $H=1.008, C=12.0, Br=79.9$)

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