A compound has an empirical formula $C_{2} H_{4} O$ . An independent analysis gave a value of $66.08$ for its vapour density. What is the correct molecular formula?

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

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C_{2} H_{2} O

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C_{6} H_{12} O_{3}

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C_{8} H_{16} O_{4}

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Solution

The correct option is C $C_{6} H_{12} O_{3}$
$Empirical formula mass = 44$
$Molecular mass = 2 \times Vapour density$
$= 132.16$
$n = \frac{Molecular mass}{Empirical formula mass} = \frac{132.16}{44} \approx 3$
$∴$ Molecular formula $= (C_{2} H_{4} O)_{3} = C_{6} H_{12} O_{3}$

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Similar questions

C_{2} H_{4} O

X

Y

X Y_{2}

C H_{2} O

45

A_{2} B_{4}

C

H

C = 52 %, H = 13 %

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