A compound has 50% carbon,50% oxygen and its | vedclass

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(C) The empirical formula is determined by the ratio of moles of each element:						Element			Mole Ratio							$C = 50/12 = 4.16$			$4.16/3.125 \appro

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$C_{12}O_9$

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(C) The empirical formula is determined by the ratio of moles of each element:						Element			Mole Ratio							$C = 50/12 = 4.16$			$4.16/3.125 \approx 1.33$							$O = 50/16 = 3.125$			$3.125/3.125 = 1$			Multiplying by $3$ to get whole numbers: $C = 4, O = 3$. Empirical formula = $C_4O_3$. Empirical formula mass = $(4 	imes 12) + (3 	imes 16) = 48 + 48 = 96$. $n = \frac{	ext{Molecular weight}}{	ext{Empirical formula mass}} = \frac{290}{96} \approx 3$. Molecular formula = $(C_4O_3)_3 = C_{12}O_9$.

$A$ compound has $50\%$ carbon,$50\%$ oxygen and its approximate molecular weight is $290$. Its molecular formula is

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