The elemental composition of a compound is 54 | vedclass

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(C) Step $1$: Calculate the moles of each element in $100 \ g$ of the compound:$n_C = \frac{54.2}{12} = 4.516$$n_H = \frac{9.2}{1} = 9.2$$n_O = \frac{

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$C_6H_{12}O_3$

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(C) Step $1$: Calculate the moles of each element in $100 \ g$ of the compound:$n_C = \frac{54.2}{12} = 4.516$$n_H = \frac{9.2}{1} = 9.2$$n_O = \frac{36.6}{16} = 2.287$Step $2$: Determine the simplest molar ratio by dividing by the smallest value $(2.287)$:$C = \frac{4.516}{2.287} \approx 2$$H = \frac{9.2}{2.287} \approx 4$$O = \frac{2.287}{2.287} = 1$Step $3$: The empirical formula is $C_2H_4O$.Step $4$: Calculate the empirical formula mass:$E.F. 	ext{ mass} = (2 	imes 12) + (4 	imes 1) + (1 	imes 16) = 24 + 4 + 16 = 44 \ g \ mol^{-1}$.Step $5$: Calculate the value of $n$ (where $n = \frac{	ext{Molar mass}}{	ext{E.F. mass}}$):$n = \frac{132}{44} = 3$.Step $6$: Determine the molecular formula:$	ext{Molecular formula} = (C_2H_4O)_3 = C_6H_{12}O_3$.

The elemental composition of a compound is $54.2 \% C$,$9.2 \% H$,and $36.6 \% O$. If the molar mass of the compound is $132 \ g \ mol^{-1}$,the molecular formula of the compound is: [Given: The relative atomic mass of $C:H:O = 12:1:16$]

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