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(B) $1$. Calculate the percentage of oxygen: $O = 100 - (57.82 + 3.6) = 38.58 \%$. $2$. Determine the empirical formula: $C: \frac{57.82}{12} = 4.818

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$C_8H_6O_4$

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(B) $1$. Calculate the percentage of oxygen: $O = 100 - (57.82 + 3.6) = 38.58 \%$. $2$. Determine the empirical formula: $C: \frac{57.82}{12} = 4.818 \Rightarrow \frac{4.818}{2.41} = 2$ $H: \frac{3.6}{1} = 3.6 \Rightarrow \frac{3.6}{2.41} = 1.5 = \frac{3}{2}$ $O: \frac{38.58}{16} = 2.41 \Rightarrow \frac{2.41}{2.41} = 1$ Multiplying by $2$ to get whole numbers: $C = 4, H = 3, O = 2$. Empirical Formula ($E$.$F$.) = $C_4H_3O_2$. $3$. Calculate Empirical Formula Mass ($E$.$F$.$M$.): $(4 	imes 12) + (3 	imes 1) + (2 	imes 16) = 48 + 3 + 32 = 83$. $4$. Calculate Molecular Formula Mass ($M$.$F$.$M$.): $2 	imes 	ext{Vapour Density} = 2 	imes 83 = 166$. $5$. Calculate $n$: $n = \frac{M.F.M.}{E.F.M.} = \frac{166}{83} = 2$. $6$. Molecular Formula = $n 	imes E.F. = 2 	imes C_4H_3O_2 = C_8H_6O_4$.

An organic compound on analysis gives the following data: $C = 57.82 \%$,$H = 3.6 \%$ and the rest is oxygen. Its vapour density is $83$. Calculate its molecular formula.

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