Question

Carbon= 40% , hydrogen= 6.66% , oxygen=53.34 . Determine the emperical of compound ,if molecular mass is 60 and determine the molecular formula

Answer 1

Hi!

I'm using the format Element -- Percentage Composition -- Atomic mass -- Atomic ratio.

First,

Carbon -- 40% -- 12 -- 40/12 = 3.33

Hydrogen -- 6.66% -- 1 -- 6.66/1 = 6.66

Oxygen -- 53.34% -- 16 -- 53.34/16 = 3.33

The least atomic ratio from these is 3.33

So, to get the simplest ratio, divide each by 3.33

We get C:H:O = 1:2:1

So, the empirical formula is CH₂O

Empirical formula Weight (EFW) = 12 + 2 + 16 = 30

Now, given that, molecular mass is 60 (MM = 60).

Let the molecular formula be (CH₂O)n where n is a positive integer.

n is given my MM / EFW

So, n = 60/30 = 2

So, Molecular formula = (CH₂O)2 = C₂H₄O₂

which can be written as CH₃COOH (acetic acid)

Answer 2

Answer:

Amprical weight=ch2o

=12*12*1*1*16

=30

Molecular mass=n*amprical weight

60=n*30

N=60/30=2

Molecular formula=n*amprical formula

=2*30

=60

C2h4o2

Explanation: