An organic compound contains C = 36%,H = 6% a | vedclass

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(A) $1$. Calculate the percentage of oxygen: $\% O = 100 - (36 + 6) = 58\%$. $2$. Determine the number of moles for each element (assuming $100 \ g$ o

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$CH_2O$

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(A) $1$. Calculate the percentage of oxygen: $\% O = 100 - (36 + 6) = 58\%$. $2$. Determine the number of moles for each element (assuming $100 \ g$ of compound): $C = 36/12 = 3.0$ $H = 6/1 = 6.0$ $O = 58/16 = 3.625$ $3$. Determine the simplest molar ratio by dividing by the smallest value $(3.0)$: $C = 3.0/3.0 = 1$ $H = 6.0/3.0 = 2$ $O = 3.625/3.0 \approx 1.2$ $4$. Since the ratio for oxygen is not an integer,we multiply by $5$ to get whole numbers: $C = 1 	imes 5 = 5$ $H = 2 	imes 5 = 10$ $O = 1.2 	imes 5 = 6$ Empirical formula = $C_5H_{10}O_6$. Note: Given the options provided,there appears to be a discrepancy in the calculation or the provided options. Based on standard empirical formula calculation,the result is $C_5H_{10}O_6$. If we assume the oxygen percentage was intended to be $53.33\%$,the formula would be $CH_2O$.

An organic compound contains $C = 36\%$,$H = 6\%$ and the rest is oxygen. Its empirical formula is:

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