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 \,

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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 $C:H = 0.875:1 \approx 7:8$. Thus,the empirical formula is $C_7H_8$.Step $2$: Determine the molar mass $(M)$ using the ideal gas equation $PV = nRT = \frac{w}{M}RT$.$P = 1 \, atm, V = 1 \, L, w = 2.4 \, g, R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}, T = 127 + 273 = 400 \, K$.$M = \frac{wRT}{PV} = \frac{2.4 	imes 0.0821 	imes 400}{1 	imes 1} = 78.8 \, g/mol \approx 79 \, g/mol$.Step $3$: Compare empirical mass with molar mass.Empirical mass of $C_7H_8 = (7 	imes 12) + (8 	imes 1) = 92 \, g/mol$.Since the calculated molar mass $(79 \, g/mol)$ does not match the empirical formula mass or its multiples,the correct hydrocarbon is not listed among the options.

$A$ hydrocarbon contains $10.5 \, g$ of carbon and $1 \, g$ of hydrogen. If the volume of $2.4 \, g$ of this hydrocarbon is $1 \, L$ at $127 \, ^\circ C$ and $1 \, atm$ pressure,then the hydrocarbon is:

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