An organic compound contains 60 % C, 4.48 % H | vedclass

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(A) To find the empirical formula, we calculate the molar ratio of the elements:$1$. Moles of $C = \frac{60}{12} = 5.0$$2$. Moles of $H = \frac{4.48}{

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$C_9H_8O_4$

Vedclass · Answer

(A) To find the empirical formula, we calculate the molar ratio of the elements:$1$. Moles of $C = \frac{60}{12} = 5.0$$2$. Moles of $H = \frac{4.48}{1} = 4.48$$3$. Moles of $O = \frac{35.5}{16} \approx 2.22$Divide each by the smallest value $(2.22)$:$C: \frac{5.0}{2.22} \approx 2.25$$H: \frac{4.48}{2.22} \approx 2.0$$O: \frac{2.22}{2.22} = 1.0$Multiplying by $4$ to get whole numbers: $C = 9, H = 8, O = 4$.Thus, the empirical formula is $C_9H_8O_4$.

An organic compound contains $60\ \%\ C$, $4.48\ \%\ H$, and $35.5\ \%\ O$. Its empirical formula is:

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