A hydrocarbon contains 85.7 % C. If 42 , mg o | vedclass

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(B) $1$. Calculate the molar mass of the hydrocarbon:Given $42 \, mg = 0.042 \, g$ contains $3.01 	imes 10^{20}$ molecules.Number of moles $n = \frac

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$C_6H_{12}$

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(B) $1$. Calculate the molar mass of the hydrocarbon:Given $42 \, mg = 0.042 \, g$ contains $3.01 	imes 10^{20}$ molecules.Number of moles $n = \frac{3.01 	imes 10^{20}}{6.022 	imes 10^{23}} \approx 0.5 	imes 10^{-3} \, mol = 5 	imes 10^{-4} \, mol$.Molar mass $M = \frac{	ext{mass}}{	ext{moles}} = \frac{0.042 \, g}{5 	imes 10^{-4} \, mol} = 84 \, g/mol$.$2$. Determine the number of atoms of $C$ and $H$:Mass of $C = 85.7 \% 	ext{ of } 84 = 0.857 	imes 84 \approx 72 \, g$.Number of $C$ atoms $= \frac{72 \, g}{12 \, g/mol} = 6$.Mass of $H = 84 - 72 = 12 \, g$.Number of $H$ atoms $= \frac{12 \, g}{1 \, g/mol} = 12$.$3$. Conclusion:The molecular formula is $C_6H_{12}$.

$A$ hydrocarbon contains $85.7 \% C$. If $42 \, mg$ of the compound contains $3.01 \times 10^{20}$ molecules,the molecular formula of the compound will be

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