Calculating Empirical Formula of Organic Compounds December 3rd, 2010

The empirical formula of an organic compound can be found by:

-burning the compound(reacts with O2)
- collecting and weighing products
-from mass of products, moles of each element in the original organic compound can be calculated.
What is the difference between empirical and molecular?

What is the empirical formula of a compound whose composition is 45.28% Carbon, 6.289% Hydrogen, 35.22% O and 13.21% Nitrogen?

mol C = 45.28 g x 1 mol/12.0 g C = 3.777 mol

mol H = 6.289 g H x 1 mol/ 1 g = 6.289 mol H

mole O = 35.22 g O x 1 mole/16.0 g = 2.2 mol

mole N = 13.21 g N x 1 mole/ 14.0 g = 0.944 mol



Divide by smallest molar amount

3.777/0.944= 4 x 3 = 12

6.289/ 0.944 = 6.6 x 3 = 20

2.2/ 0.944= 2.33 x3 = 7

0.944/0.944 = 1 x3 =3

= C12H20O7N3






A compound contains 92.26% Carbon and 7.74 % H

What is the empirical formula?

*Assume there are 100 g

92.26 g C x 1 mole/ 12.0 g = 7.688 mol

7.74 g H x 1 mole/1.0 g = 7.74 mol


Divide by smallest amount

7.74/7.688 = 1

7.688/7.688 = 1

Ratio of C to H = 1:1

= C1H1

= CH

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Cheers!