The ratio of mass percentage (w/w) of C : H i | vedclass

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(B) $1$. Determine the empirical formula: The ratio of mass percentage of $C : H$ is $12 : 1$. Dividing by atomic masses $(C=12, H=1)$, the mole ratio

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$11.44$

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(B) $1$. Determine the empirical formula: The ratio of mass percentage of $C : H$ is $12 : 1$. Dividing by atomic masses $(C=12, H=1)$, the mole ratio is $(12/12) : (1/1) = 1 : 1$. The empirical formula is $CH$.$2$. Determine the molecular formula: The hydrocarbon has two carbon atoms, so the molecular formula is $(CH)_2 = C_2H_2$ (Ethyne).$3$. Write the balanced combustion reaction: $2C_2H_2 + 5O_2 	o 4CO_2 + 2H_2O$.$4$. Calculate moles of hydrocarbon: Molar mass of $C_2H_2 = (2 	imes 12) + (2 	imes 1) = 26 \ g/mol$. Moles in $3.38 \ g = 3.38 / 26 = 0.13 \ mol$.$5$. Calculate moles of $CO_2$: From the balanced equation, $1 \ mol$ of $C_2H_2$ produces $2 \ mol$ of $CO_2$. Therefore, $0.13 \ mol$ of $C_2H_2$ produces $0.13 	imes 2 = 0.26 \ mol$ of $CO_2$.$6$. Calculate mass of $CO_2$: Mass $= 0.26 \ mol 	imes 44 \ g/mol = 11.44 \ g$.

The ratio of mass percentage (w/w) of $C : H$ in a hydrocarbon is $12 : 1$. It has two carbon atoms. The weight (in $g$) of $CO_2(g)$ formed when $3.38 \ g$ of this hydrocarbon is completely burnt in oxygen is : (Given : Molar mass in $g \ mol^{-1}$, $C : 12, H : 1, O : 16$)

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