Question

A compound on analysis gave the following percentage composition C=54.55%,H=9.09%,O=36.36%. Determine the empirical formula of the
compound.

Answer 1

Parts of C = 55.45 per 100 or 5545 per 10000
Part of H = 9.09 per 100 or 909 per 10000
Part of O = 36.36 per 100 or 3636 per 10000
So by adding,
5545+909+3636=10000
This shows that there is no other element involved
so,
Formula is,
C5455 H909 O3636

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Answer 2

The empirical formula of the given compound is C₂H₄O.

A compound on analysis gave the following percentage composition C = 54.55% , H = 9.09% and O = 36.36%.

We have to find the empirical formula of the compound.

Steps to find the empirical formula of a compound :

1. find percentage by atomic mass ratio of each element.
2. Now, find the simplest ratio of them, it is nothing but empirical formula of the compound.

Required data :

• atomic mass of C = 12 amu
• atomic mass of H = 1 amu
• atomic mass of O = 16 amu

For Carbon,

percentage by atomic mass ratio = 54.55/12 = 4.545

For Hydrogen,

percentage by atomic mass ratio = 9.09/1 = 9.09

For Oxygen,

percentage by atomic mass ratio = 36.36/16 = 2.2725

Now, the simplest ratio of the gotten values,

C : H : O = 4.545 : 9.09 : 2.2725 = 2 : 4 : 1

Therefore the empirical formula of the compound is C₂H₄O.

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