An organic compound contains 49.3% carbon,6.8 | vedclass

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(B) $1$. Calculate the percentage of oxygen: $100 - (49.3 + 6.84) = 43.86\%$. $2$. Determine the empirical formula: ElementPercentageAtomic MassMolesS

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$C_6H_{10}O_4$

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(B) $1$. Calculate the percentage of oxygen: $100 - (49.3 + 6.84) = 43.86\%$. $2$. Determine the empirical formula: ElementPercentageAtomic MassMolesSimplest Ratio$C$$49.3$$12$$49.3/12 = 4.11$$4.11/2.74 = 1.5 	imes 2 = 3$$H$$6.84$$1$$6.84/1 = 6.84$$6.84/2.74 = 2.5 	imes 2 = 5$$O$$43.86$$16$$43.86/16 = 2.74$$2.74/2.74 = 1 	imes 2 = 2$Empirical formula = $C_3H_5O_2$. $3$. Calculate empirical formula mass: $(3 	imes 12) + (5 	imes 1) + (2 	imes 16) = 36 + 5 + 32 = 73 \ g/mol$. $4$. Calculate molecular mass: $2 	imes 	ext{Vapor Density} = 2 	imes 73 = 146 \ g/mol$. $5$. Calculate $n$: $n = \frac{	ext{Molecular Mass}}{	ext{Empirical Formula Mass}} = \frac{146}{73} = 2$. $6$. Molecular formula = $n 	imes 	ext{Empirical Formula} = 2 	imes (C_3H_5O_2) = C_6H_{10}O_4$.

An organic compound contains $49.3\%$ carbon,$6.84\%$ hydrogen,and the rest is oxygen. Its vapor density is $73$. What is its molecular formula?

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