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Molecular Formula

An organic co...

Question

An organic compound on analysis gave $C$ $=$ $39.9$ %, $H$ $=$ $6.7$ % and $O$ $=$ $53.4$ %. The empirical formula of the compound is:

C H O_{2}

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C H_{2} O

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

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C H O

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Solution

The correct option is C $C H_{2} O$
$C : H : O = \frac{39.9}{12} : \frac{6.7}{1} : \frac{53.4}{16} = 1 : 2 : 1$
Therefore, the empirical formula is $C H_{2} O$ .

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