On complete combustion,1.4 , g of a hydrocarb | vedclass

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Question

(B) The combustion reaction of a hydrocarbon $C_xH_y$ is: $C_xH_y + (x + y/4)O_2 \rightarrow xCO_2 + (y/2)H_2O$. Mass of $H_2O$ produced = $1.8 \, g$.

Vedclass · Accepted Answer

$CH_2$

Vedclass · Answer

(B) The combustion reaction of a hydrocarbon $C_xH_y$ is: $C_xH_y + (x + y/4)O_2 ightarrow xCO_2 + (y/2)H_2O$. Mass of $H_2O$ produced = $1.8 \, g$. Moles of $H_2O$ = $\frac{1.8 \, g}{18 \, g/mol} = 0.1 \, mol$. Moles of $H$ atoms = $2 	imes 	ext{moles of } H_2O = 2 	imes 0.1 = 0.2 \, mol$. Mass of $H$ = $0.2 \, mol 	imes 1 \, g/mol = 0.2 \, g$. Mass of $C$ in hydrocarbon = $	ext{Total mass} - 	ext{Mass of } H = 1.4 \, g - 0.2 \, g = 1.2 \, g$. Moles of $C$ = $\frac{1.2 \, g}{12 \, g/mol} = 0.1 \, mol$. Ratio of $C:H$ = $0.1 : 0.2 = 1 : 2$. Therefore,the empirical formula is $CH_2$.

On complete combustion,$1.4 \, g$ of a hydrocarbon gave $1.8 \, g$ of water. The empirical formula of the hydrocarbon is:

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