The ratio of the mass percentages of C and H | vedclass

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(D) Given mass ratios: $C : H = 4 : 1$ and $C : O = 3 : 4$.To find the combined mass ratio $C : H : O$,we equate the $C$ part: $C : H = 12 : 3$ and $C

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$5$

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(D) Given mass ratios: $C : H = 4 : 1$ and $C : O = 3 : 4$.To find the combined mass ratio $C : H : O$,we equate the $C$ part: $C : H = 12 : 3$ and $C : O = 12 : 16$.Thus,$C : H : O = 12 : 3 : 16$.Dividing by atomic masses $(C=12, H=1, O=16)$,the mole ratio is $C : H : O = (12/12) : (3/1) : (16/16) = 1 : 3 : 1$.The empirical formula is $CH_3O$. Since the compound is a saturated acyclic organic compound,the molecular formula is $C_2H_6O_2$ (e.g.,ethylene glycol).The combustion reaction is: $C_2H_6O_2 + 2.5 O_2 \longrightarrow 2 CO_2 + 3 H_2O$.For $1$ mole of $X$,$2.5$ moles of $O_2$ are required.For $2$ moles of $X$,the moles of $O_2$ required $= 2 	imes 2.5 = 5$ moles.

The ratio of the mass percentages of $C$ and $H$ and $C$ and $O$ of a saturated acyclic organic compound $X$ are $4 : 1$ and $3 : 4$ respectively. Then,the moles of oxygen gas required for complete combustion of two moles of organic compound $X$ is

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