The empirical formula of the compound is $C H O_{2}$ . The molecular weight of the compound is 90. Calculate the molecular formula of the compound.

C_{2} H_{2} O

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C H O

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C H_{2} O_{4}

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C_{2} H_{2} O_{4}

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Solution

The correct option is D $C_{2} H_{2} O_{4}$
The weight of empirical formula is $(12 + 1 + 32) = 45$ .
$n = \frac{M o l e c u l a r w e i g h t}{E m p i r i c a l w e i g h t} = 2$
Now, multiplying empirical formula by $n$ gives $C_{2} H_{2} O_{4}$ .

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