Empirical Formula Tutorial |
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.
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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
Remember, empirical formula is C4H5N2O
Molar Mass = (4x12.011)+(5x1.008)+(2x14.007)+(1x15.999) = 97.097 g/mol
(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.
http://www.chem.lsu.edu/lucid/tutorials/empiricaltutorial.html