0.30 g of an organic compound containing C,H | vedclass

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(D) $1$. Calculate the mass of $C$ and $H$ in the compound:Mass of $C = \frac{12}{44} 	imes 0.44 \ g = 0.12 \ g$Mass of $H = \frac{2}{18} 	imes 0.18

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$C_2H_4O_2$

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(D) $1$. Calculate the mass of $C$ and $H$ in the compound:Mass of $C = \frac{12}{44} 	imes 0.44 \ g = 0.12 \ g$Mass of $H = \frac{2}{18} 	imes 0.18 \ g = 0.02 \ g$Mass of $O = 0.30 - (0.12 + 0.02) = 0.16 \ g$$2$. Calculate the molar ratio:Moles of $C = \frac{0.12}{12} = 0.01$Moles of $H = \frac{0.02}{1} = 0.02$Moles of $O = \frac{0.16}{16} = 0.01$$3$. Determine the empirical formula:The ratio $C:H:O = 0.01:0.02:0.01 = 1:2:1$. Thus,the empirical formula is $CH_2O$.$4$. Determine the molecular formula:Empirical formula mass $= 12 + 2(1) + 16 = 30 \ g \ mol^{-1}$.$n = \frac{	ext{Molecular mass}}{	ext{Empirical formula mass}} = \frac{60}{30} = 2$.Molecular formula $= 2 	imes (CH_2O) = C_2H_4O_2$.

$0.30 \ g$ of an organic compound containing $C$,$H$ and $O$ on combustion yields $0.44 \ g \ CO_2$ and $0.18 \ g \ H_2O$. If $1 \ mol$ of the compound weighs $60 \ g$,then the molecular formula of the compound is:

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