A hydrocarbon contains 10.5 ,g of carbon and | vedclass

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(D) Step $1$: Determine the empirical formula. Moles of $C = \frac{10.5 \,g}{12 \,g/mol} = 0.875 \,mol$. Moles of $H = \frac{1 \,g}{1 \,g/mol} = 1 \,m

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None of these

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(D) Step $1$: Determine the empirical formula. Moles of $C = \frac{10.5 \,g}{12 \,g/mol} = 0.875 \,mol$. Moles of $H = \frac{1 \,g}{1 \,g/mol} = 1 \,mol$. Ratio $C:H = 0.875:1 = 7:8$. Thus,the empirical formula is $C_7H_8$. Step $2$: Determine the molar mass using the ideal gas equation $PV = nRT = \frac{w}{M}RT$. $1 \,atm 	imes 1 \,L = \frac{2.4 \,g}{M} 	imes 0.0821 \,L \cdot atm \cdot K^{-1} \cdot mol^{-1} 	imes (127 + 273) \,K$. $M = \frac{2.4 	imes 0.0821 	imes 400}{1} \approx 78.8 \,g/mol$. Step $3$: Determine the molecular formula. Empirical formula mass of $C_7H_8 = (7 	imes 12) + (8 	imes 1) = 92 \,g/mol$. Since the calculated molar mass $(78.8)$ does not match the empirical formula mass $(92)$,and no option matches $C_7H_8$,the correct answer is $D$.

$A$ hydrocarbon contains $10.5 \,g$ of carbon and $1 \,g$ of hydrogen. If $2.4 \,g$ of this hydrocarbon occupies a volume of $1 \,L$ at $1 \,atm$ pressure and $127 \,^{\circ}C$ temperature,what is the molecular formula of the hydrocarbon?

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