An organic compound contains 49.3% carbon,6.8 | vedclass

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(B) The percentage of oxygen is $100 - (49.3 + 6.84) = 43.86\%$.ElementMolesSimple Ratio$C$ $(49.3/12 = 4.1)$$4.1/2.7 = 1.5$$1.5 	imes 2 = 3$$H$ $(6.

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$C_6H_{10}O_4$

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(B) The percentage of oxygen is $100 - (49.3 + 6.84) = 43.86\%$.ElementMolesSimple Ratio$C$ $(49.3/12 = 4.1)$$4.1/2.7 = 1.5$$1.5 	imes 2 = 3$$H$ $(6.84/1 = 6.84)$$6.84/2.7 = 2.5$$2.5 	imes 2 = 5$$O$ $(43.86/16 = 2.7)$$2.7/2.7 = 1.0$$1.0 	imes 2 = 2$Empirical formula = $C_3H_5O_2$.Empirical formula weight = $(12 	imes 3) + (1 	imes 5) + (16 	imes 2) = 36 + 5 + 32 = 73$.Molecular weight = $2 	imes 	ext{Vapour Density} = 2 	imes 73 = 146$.$n = \frac{	ext{Molecular weight}}{	ext{Empirical formula weight}} = \frac{146}{73} = 2$.Molecular formula = $(C_3H_5O_2)_2 = C_6H_{10}O_4$.

An organic compound contains $49.3\%$ carbon,$6.84\%$ hydrogen and its vapour density is $73$. The molecular formula of the compound is:

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