Empirical formula of compound is $C H_{2} O$ . If is vapour density is 90, then molecular formula of compound is:

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Solution

Empiricaz formula is $C H_{2} O$
Mass of Empirical formula $= 12 + 12 + 16 = 30 g$
Vapour density $= \frac{M o l a r m a s s o f c o m p o u n d}{2} ∴ 90 = \frac{M o l a r m a s s}{2}$
Molar mass of compound $= 180 g$
On dividing the $\frac{M o l e c u l a r f o r m u l a m a s s}{E m p i r i c a l f o r m u l a m a s s}$
$n = \frac{180}{30} = 6$
On multiply each atom of empirical formula by $n$ we get actual molecular formula
$C \times + H_{2} \times 6 + O \times 6 = C_{6} H_{12} O_{6}$
The molecular formula of compound is $C_{6} H_{12} O_{6}$ with molar mass $180$ .

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