If 0.0833 , text{mol} of a carbohydrate conta | vedclass

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(D) Given that $0.0833 \, 	ext{mol}$ of carbohydrate contains $1 \, 	ext{g}$ of hydrogen. Therefore,$1 \, 	ext{mol}$ of carbohydrate contains: $\fr

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$C_6H_{12}O_6$

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(D) Given that $0.0833 \, 	ext{mol}$ of carbohydrate contains $1 \, 	ext{g}$ of hydrogen. Therefore,$1 \, 	ext{mol}$ of carbohydrate contains: $\frac{1}{0.0833} \approx 12 \, 	ext{g}$ of hydrogen. Since the atomic mass of $H$ is $1 \, 	ext{g/mol}$,the number of hydrogen atoms per molecule is $12$. The empirical formula is $CH_2O$,which has $2$ hydrogen atoms. To have $12$ hydrogen atoms,the molecular formula must be $(CH_2O)_n$ where $2n = 12$,so $n = 6$. Thus,the molecular formula is $C_6H_{12}O_6$.

If $0.0833 \, \text{mol}$ of a carbohydrate contains $1 \, \text{g}$ of hydrogen and its empirical formula is $CH_2O$,what is its molecular formula?

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