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Empirical Formula Tutorial
(aka Elemental Analysis)

 


Below is a tutorial on doing Empirical Formula Problems. Grab a pencil and your notes and work the problem as the tutorial explains it to you. If you follow these simple 5 guidelines shown below, understanding Empirical Formulas should be no problem.

*Guidelines for solving empirical formula problems*
  • 1. Base the formula determination on a 100 gram sample.

  • 2. Divide each mass percentage by the molar mass of the element to obtain the number of moles of the element in the 100 gram sample.

  • 3. Divide each molar amount by whichever amount is the smallest.

  • 4. If some results from Step 3 are far from integers, multiply through by a common factor that converts all molar amounts to integers or near-integers.

  • 5. Round off each molar number to the nearest integer.

Example

(a) Given the following mass percent composition, determine the empirical formula.
49.5%C, 5.2%H, 28.8%N, 16.5%O

        49.5g C, 5.2g H, 28.8g N, 16.5g O

            49.5 g C
        --------------- = 4.121 mol C
        12.011 g/mol C 

             5.2 g H
        --------------- = 5.159 mol H
        1.008 g/mol H

            28.8 g N   
        --------------- = 2.056 mol N
        14.007 g/mol N
 
            16.5 g O
        --------------- = 1.031 mol O
        15.999 g/mol O

        4.121 mol C      4 mol C
        -----------  =  ---------
        1.031 mol O       mol O

        5.159 mol H      5 mol H
        -----------  =  ---------
        1.031 mol O       mol O

        2.056 mol N      2 mol N
        -----------  =  ---------
        1.031 mol O       mol O

        1.031 mol O      1 mol O
        -----------  =  ---------
        1.031 mol O       mol O


Empirical Formula is then C4H5N2O


(b) What is the chemical formula for this compound if the molar mass is 194.2 g/mol?

Remember, empirical formula is C4H5N2O

     Molar Mass  = (4x12.011)+(5x1.008)+(2x14.007)+(1x15.999)
                 =  97.097 g/mol

The empirical formula is NOT the chemical formula in this case. But the chemical formula
must be some integer factor of the empirical formula (thus, the molar mass of the chemical
formula must be some integer factor of the empirical formula).

(Factor) x Molar Mass Empirical Formula = Molar Mass Chemical Formula
(Factor) x 97.097 g/mol = 194.2 g/mol
(Factor) =2

Therefore, the chemical formula is C8H10N4O2

***NOTE: 2C4H5N2O DOES NOT EQUAL C8H10N4O2***


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Last Revised : Tuesday, December 9, 1997

This page was created by Hank Bell as a Chemistry 1201/2 class project.

Copyright © 1997
Louisiana State University, Department of Chemistry.
All rights reserved.

http://www.chem.lsu.edu/lucid/tutorials/empiricaltutorial.html