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 combine these,we equate the parts of $C$: $C:H = 12:3$ and $C:O = 12:16$.Thus,the mass ratio is

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$5$

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(D) Given mass ratios: $C:H = 4:1$ and $C:O = 3:4$.To combine these,we equate the parts of $C$: $C:H = 12:3$ and $C:O = 12:16$.Thus,the mass ratio is $C:H:O = 12:3:16$.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 $C_2H_6O_2$ (Ethylene glycol).The combustion reaction is: $C_2H_6O_2 + \frac{5}{2}O_2 \longrightarrow 2CO_2 + 3H_2O$.For $1$ mole of $C_2H_6O_2$,we need $2.5$ moles of $O_2$.For $2$ moles of $C_2H_6O_2$,we need $2 	imes 2.5 = 5$ moles of $O_2$.

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 the complete combustion of two moles of organic compound $X$ is:

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