An organic compound containing C, H, and O ha | vedclass

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(C) To determine the empirical formula, we calculate the molar ratio of the elements:$C: \frac{52.14}{12} = 4.345 \implies \frac{4.345}{2.17} \approx

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$C_2H_6O, C_2H_6O$

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(C) To determine the empirical formula, we calculate the molar ratio of the elements:$C: \frac{52.14}{12} = 4.345 \implies \frac{4.345}{2.17} \approx 2$$H: \frac{13.13}{1} = 13.13 \implies \frac{13.13}{2.17} \approx 6$$O: \frac{34.73}{16} = 2.17 \implies \frac{2.17}{2.17} = 1$Thus, the empirical formula is $C_2H_6O$.Empirical formula mass $= 2(12) + 6(1) + 1(16) = 24 + 6 + 16 = 46 \ g/mol$.Given molar mass $= 46.068 \ g/mol$.$n = \frac{	ext{Molar mass}}{	ext{Empirical formula mass}} = \frac{46.068}{46} \approx 1$.Therefore, the molecular formula is $(C_2H_6O)_1 = C_2H_6O$.

An organic compound containing $C$, $H$, and $O$ has $13.13 \% H$, $52.14 \% C$, and $34.73 \% O$. Its molar mass is $46.068 \ g/mol$. What are its empirical and molecular formulae?

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