Question

A compound has CBr3 as its empirical formula. Find the molecular formula, if the vapour density is 252 (C =12, Br = 80).​

Answer 1

Answer:

The empirical formula is CH

2

O. Therefore, the empirical weight = 30 g.

Given, the molecular weight = 2×V.D = 2×45=90 g.

x=

Empirical weight

Molecular weight

=

30

90

=3

Molecular Formula=(Empirical Formula)

x

∴Molecular Formula=(CH

2

O)

3

=C

3

H

6

O

3

Answer 2

Given: A compound has CBr3 as its empirical formula. Its vapour density is 252.

To find: Molecular formula of the compound

Explanation: Empirical mass of CBr3

=molar mass of C+ 3* molar mass of Br

= 12 + 3* 80

= 252 g/mol

Molecular mass= 2* Vapour density

= 2*252

= 504 g/mol

Now, Molecular mass = n * Empirical mass

=> 504 = n* 252

=> n = 504/252

=> n= 2

Now, molecular formula= (empirical formula)2

= (CBr3)2

= C2Br6

Checking, molecular mass of C2Br6

= 2* molar mass of C+6* molar mass of Br

= 2* 12 + 6* 80

= 24 +480

= 504 g/mol which justifies the answer

Therefore, the molecular formula of the compound is C2Br6.