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

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

${C H}_{2} O_{2}$

${C H}_{2} O$

answer is D.

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Detailed Solution

The mass of the given compound is $60 g$ .

The molar mass of carbon is $12 g$ , the molar mass of hydrogen is $1 g$ and that of oxygen is $16 g$ .

The mass of carbon, hydrogen and oxygen given is $24 g, 4 g$ and $32 g$ .

So, the no. of atoms of carbon $= 24 / 12 = 2$

The no. of atoms of hydrogen $= 4 / 1 = 4$

And the no. of atoms of oxygen $= 32 / 16 = 2$

So, the simple ratio of carbon: hydrogen: oxygen $= 2 : 4 : 2 = 1 : 2 : 1$

Hence, the compound is $C_{2} H_{4} O_{2}$ and the empirical formula is ${CH}_{2} O$ .

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