A compound ( $60$ g) on analysis gave $C = 24$ g, $H = 4$ g, and $O = 32$ g. Its empirical formula is :

C_{2} H_{2} O

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

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C H_{2} O

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C H_{2} O_{2}

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Solution

The correct option is A $C H_{2} O$
The ratio of number of gram atoms gives,
$\begin{matrix} C & : & H & : & O \frac{24}{12} & : & \frac{4}{2} & : & \frac{32}{16} 2 & : & 4 & : & 2 \end{matrix}$
Therefore, the empirical formula is $= C_{2} H_{4} O_{2} = C H_{2} O$ .

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Similar questions

60 g m

C = 24 g, H = 4 g

O = 32 g

60 g of a compound on analysis produced 24 g carbon, 4 g hydrogen, and 32 g oxygen. The empirical formula of the compound is

A g = 108, S = 32, O = 16

A g = 108, S = 32, O = 16

C

=

39.9

H

=

6.7

O

=

53.4

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