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?

[1 mark]

C_{2} H_{4} O

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C_{12} H_{24} O_{6}

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

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

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Solution

The correct option is C $C_{6} H_{12} O_{3}$
Given:
Empirical formula = $C_{2} H_{4} O$ ,
Vapour density = 66.08

Empirical formula mass
$= (2 \times M a s s o f C) + (4 \times M a s s o f H) + (1 \times M a s s o f O)$
= (2 × 12) + (4 × 1) + (1 × 16)
= 44

We know that:
Molecular mass = 2 × Vapour density
∴ molecular mass = 2 × 66.08 = 132.16

Now,
molecular mass = n × Empirical mass
Value of n = $\frac{M o l e c u l a r m a s s}{E m p i r i c a l m a s s}$
n = $\frac{133.16}{44}$
∴ n = 3

$M o l e c u l a r f o r m u l a = n \times E m p i r i c a l f o r m u l a$
= 3 × $C_{2} H_{4} O$
= $C_{6} H_{12} O_{3}$

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A compound has an empirical formula C2H4O. An independent analysis gave a value of 66.08 for its vapour density. What is the correct molecular formula? [1 mark]

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?

[1 mark]