An organic compound was found to contain 40.0 | vedclass

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(D) First,calculate the percentage of oxygen: $\% O = 100 - (40.0 + 6.66) = 53.34 \%$. Construct the table for empirical formula calculation: Element%

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$C_{6} H_{12} O_{6}$

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(D) First,calculate the percentage of oxygen: $\% O = 100 - (40.0 + 6.66) = 53.34 \%$. Construct the table for empirical formula calculation: Element% CompositionAtomic MassMole RatioSimple Ratio$C$$40.0$$12$$40/12 = 3.33$$1$$H$$6.66$$1$$6.66/1 = 6.66$$2$$O$$53.34$$16$$53.34/16 = 3.33$$1$ Empirical formula $= CH_{2} O$. Empirical formula mass $= 12 + (2 	imes 1) + 16 = 30$. Calculate $n = 	ext{Molar mass} / 	ext{Empirical formula mass} = 180 / 30 = 6$. Molecular formula $= n 	imes (	ext{Empirical formula}) = 6 	imes (CH_{2} O) = C_{6} H_{12} O_{6}$.

An organic compound was found to contain $40.0 \% C$ and $6.66 \% H$. Find its molecular formula (molar mass $= 180$).

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