A compound gave 80% carbon and 20% hydrogen o | vedclass

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(C) To determine the empirical formula,we calculate the mole ratio of the elements present in the compound:						$Element$			$Moles \ (\% / 	ext{Atom

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$C_2H_6$

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(C) To determine the empirical formula,we calculate the mole ratio of the elements present in the compound:						$Element$			$Moles \ (\% / 	ext{Atomic Mass})$							$C$			$80 / 12 = 6.67$							$H$			$20 / 1 = 20$			Dividing by the smallest value $(6.67)$:$C = 6.67 / 6.67 = 1$$H = 20 / 6.67 \approx 3$The empirical formula is $CH_3$. The molecular formula must be a multiple of $CH_3$,such as $C_2H_6$ (Ethane). Therefore,the correct option is $C$.

$A$ compound gave $80\%$ carbon and $20\%$ hydrogen on analysis. The compound is possibly

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