An organic compound having molecular formula | vedclass

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(B) The organic compound with molecular formula $C_2H_6O$ is ethanol $(CH_3CH_2OH)$,as it undergoes oxidation to form a carboxylic acid.$CH_3CH_2OH \x

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$C_2H_4O_2$

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(B) The organic compound with molecular formula $C_2H_6O$ is ethanol $(CH_3CH_2OH)$,as it undergoes oxidation to form a carboxylic acid.$CH_3CH_2OH \xrightarrow{K_2Cr_2O_7 / H_2SO_4} CH_3COOH$ $(X)$To determine the molecular formula of $X$,we calculate its empirical formula based on the given percentage composition:ElementPercentageMoles $(Percentage / Atomic Mass)$Simplest Ratio$C$$40\%$$40 / 12 = 3.33$$3.33 / 3.33 = 1$$H$$6.7\%$$6.7 / 1 = 6.7$$6.7 / 3.33 = 2$$O$$53.3\%$$53.3 / 16 = 3.33$$3.33 / 3.33 = 1$The empirical formula is $CH_2O$. The empirical formula mass is $12 + (2 	imes 1) + 16 = 30$.The molar mass of acetic acid $(CH_3COOH)$ is $60$.$n = \frac{	ext{Molecular Mass}}{	ext{Empirical Formula Mass}} = \frac{60}{30} = 2$.Therefore,the molecular formula is $(CH_2O)_2 = C_2H_4O_2$.

An organic compound having molecular formula $C_2H_6O$ undergoes oxidation with $K_2Cr_2O_7 / H_2SO_4$ to produce $X$ which contains $40\,\%$ carbon,$6.7\,\%$ hydrogen and $53.3\,\%$ oxygen. The molecular formula of the compound $X$ is

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