The molecular formula of ethanoic acid is $C_{2} H_{4} O_{2}$ . It's empirical formula is:

C_{4} H_{8} O_{2}

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C H_{2} O

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CHO

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C H O_{2}

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Solution

The correct option is B $C H_{2} O$
$(C H_{2} O)_{x}$ $=$ $C_{2} H_{4} O_{2}$ , where, $x = 2$ .
Therefore the empirical formula $= C H_{2} O$

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Similar questions

The empirical formula and molecular mass of a compound are $C H_{2} O$ and 180 g respectively. What will be the molecular formula of the compound ?

(a) $C_{9} H_{18} O_{9}$ (b) $C H_{2} O$ (c) $C_{6} H_{12} O_{6}$ (d) $C_{2} H_{4} O_{2}$

C H O_{2}

g m o l^{- 1}

C_{2} H_{4} O

C H_{2} O

60

{C H}_{2} O

44

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