A substance contains 52.2% carbon,13% hydroge | vedclass

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(C) ElementPercentageAtomic WeightRatio (\%/At. Wt.)Simple Ratio$C$$52.2$$12$$52.2/12 = 4.35$$4.35/2.175 = 2$$H$$13$$1$$13/1 = 13$$13/2.175 = 6$$O$$34

Vedclass · Accepted Answer

$C_4H_{12}O_2$

Vedclass · Answer

(C) ElementPercentageAtomic WeightRatio (\%/At. Wt.)Simple Ratio$C$$52.2$$12$$52.2/12 = 4.35$$4.35/2.175 = 2$$H$$13$$1$$13/1 = 13$$13/2.175 = 6$$O$$34.8$$16$$34.8/16 = 2.175$$2.175/2.175 = 1$The empirical formula is $C_2H_6O$.Empirical formula mass $= (2 	imes 12) + (6 	imes 1) + 16 = 24 + 6 + 16 = 46$.Molecular mass $= 2 	imes 	ext{Vapor Density} = 2 	imes 46 = 92$.$n = \frac{	ext{Molecular mass}}{	ext{Empirical formula mass}} = \frac{92}{46} = 2$.Molecular formula $= n 	imes (	ext{Empirical formula}) = 2 	imes C_2H_6O = C_4H_{12}O_2$.

$A$ substance contains $52.2\%$ carbon,$13\%$ hydrogen,and $34.8\%$ oxygen. If the vapor density of the substance is $46$,find its molecular formula.

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