Question

find the molecular formula for the following compound with given percentage composition C-40% H-6.6% O-53.4% [molar mass is 90]​

Answer 1

Answer:

By using element percentage composition - Atomic mass - Atomic ratio

Carbon ⟶40%÷12=

12

40

=3.33

Hydrogen ⟶6.66%÷1=

1

6.66

=6.66

Oxygen ⟶53.34%÷16=

16

40

=3.33

The least atomic ratio is 3.33.

For simplest ratio, we get C:H:O = 1:2:1

Empirical formula = CH

2

O of mass 30 g.

As molar mass = 60, Molecular formula = (CH

2

O)

2

=CH

3

COOH

I hope it will help you