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(B) The percentage of oxygen $= 100 - (52 + 13) = 35 \%$. Calculating the empirical formula: $C: 52/12 = 4.33 \rightarrow 2$ $H: 13/1 = 13.00 \rightar

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methoxy methane

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(B) The percentage of oxygen $= 100 - (52 + 13) = 35 \%$. Calculating the empirical formula: $C: 52/12 = 4.33 ightarrow 2$ $H: 13/1 = 13.00 ightarrow 6$ $O: 35/16 = 2.18 ightarrow 1$ Empirical formula $= C_2H_6O$. Empirical formula weight $= 2 	imes 12 + 6 	imes 1 + 16 = 46$. Molecular weight $= 2 	imes 	ext{vapour density} = 2 	imes 23 = 46$. Since molecular weight equals empirical formula weight,the molecular formula is $C_2H_6O$. Since the compound reacts with sodium metal to liberate $H_2$ gas,it must be an alcohol,which is $C_2H_5OH$ (ethanol). The functional isomer of ethanol $(C_2H_5OH)$ is methoxy methane $(CH_3OCH_3)$.

An oxygen-containing organic compound was found to contain $52 \%$ carbon and $13 \%$ hydrogen. Its vapour density is $23$. The compound reacts with sodium metal to liberate hydrogen. $A$ functional isomer of this compound is

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