0.0833 , mol of a carbohydrate with the empir | vedclass

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(D) Given that $0.0833 \, mol$ of the carbohydrate contains $1 \, g$ of hydrogen. Therefore,$1 \, mol$ of the carbohydrate contains hydrogen $= \frac{

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$C_6H_{12}O_6$

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(D) Given that $0.0833 \, mol$ of the carbohydrate contains $1 \, g$ of hydrogen. Therefore,$1 \, mol$ of the carbohydrate contains hydrogen $= \frac{1}{0.0833} \approx 12 \, g$. The empirical formula is $CH_2O$,which contains $2 \, g$ of hydrogen per mole of the empirical unit. To find the value of $n$ (where $	ext{Molecular Formula} = n 	imes 	ext{Empirical Formula}$),we divide the total mass of hydrogen in the molecular formula by the mass of hydrogen in the empirical formula: $n = \frac{12 \, g}{2 \, g} = 6$. Thus,the molecular formula is $(CH_2O)_6 = C_6H_{12}O_6$.

$0.0833 \, mol$ of a carbohydrate with the empirical formula $CH_2O$ contains $1 \, g$ of hydrogen. What is the molecular formula of the carbohydrate?

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