The empirical formula of a compound is C_2H_5 | vedclass

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(C) Given,empirical formula of compound $= C_2H_5O$Vapour density $= 45$Molecular weight $= 2 	imes 	ext{vapour density} = 2 	imes 45 = 90$Empirica

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$C_4H_{10}O_2$

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(C) Given,empirical formula of compound $= C_2H_5O$Vapour density $= 45$Molecular weight $= 2 	imes 	ext{vapour density} = 2 	imes 45 = 90$Empirical formula mass $= (2 	imes 12) + (5 	imes 1) + (1 	imes 16) = 24 + 5 + 16 = 45$Calculate the value of $n$ using the formula: $n = \frac{	ext{Molecular weight}}{	ext{Empirical formula mass}} = \frac{90}{45} = 2$Molecular formula $= n 	imes (	ext{empirical formula}) = 2 	imes (C_2H_5O) = C_4H_{10}O_2$

The empirical formula of a compound is $C_2H_5O$ and its vapour density is $45$. What is the molecular formula of the compound?

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