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"Empirical formula" is a REAL IMPORTANT concept. Here's the definition:
the formula of a compound expressed as the smallest possible whole-number ratio of subscripts of the elements in the formula
For example, CH3COOH has two carbons, four hydrogens and two oxygens. So we could write the formula like this: C2H4O2 and so it reduces to CH2O.
Contrast the above definition to this one for "molecular formula:"
the formula of a compound in which the subscripts give the actual number of each element in the formula
Here are the four formulas being used as examples:
|Molecular Formula||Empirical Formula|
Notice two things:
1. The molecular formula and the empirical formula can be identical.
2. You scale up from the empirical formula to the molecular formula by a whole number factor.
The tutorial below will focus on empirical formulas, but molecular formulas will return very, very soon. You will need the "scale up" idea when molecular formula questions get joined up with empirical formula questions.
The article below describes the steps involved in calculating an empirical formula. Read it carefully. The ChemTeam makes some comments below.
When teaching the method for converting percentage composition to an empirical formula, I have devised the following rhyme:
Percent to mass
Mass to mole
Divide by small
Multiply 'til whole
Here's an example of how it works. A compound consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula?
(1) Percent to mass:
Assume 100 g of the substance, then 72.2 g magnesium and 27.8 g nitrogen.
(2) Mass to moles:
for Mg: 72.2 g Mg x (1 mol Mg/24.3 g Mg) = 2.97 mol Mg
for N: 27.8 g N x (1 mol N/14.0 g N) = 1.99 mol N
(3) Divide by small:
for Mg: 2.97 mol / l.99 mol = 1.49
for N: 1.99 mol / l.99 mol = 1.00
(4) Multiply 'til whole:
for Mg: 2 x 1.49 = 2.98 (i.e., 3)
for N: 2 x 1.00 = 2.00
and the formula of the compound is Mg3N2.
Students enjoy this device and have discovered that they have both the rhyme and reason for working chemistry problems of this type.
The above article is copyright © 1988 by the Division of Chemical Education of the American Chemical Society, Inc.
ChemTeam comments on the article:
(1) Percent to mass: the assumption of 100 grams is purely for convenience sake. This means that the percentages transfer directly into grams. If you assumed that 36.7 grams were present, you would have to multiply 36.7 by each percentage. Assuming 100 grams makes it much easier.
(2) Mass to moles: this is a technique you should ALREADY know. If you do not, then my advice is to blame the teacher, whine alot about how hard the class is and complaim about how your self-esteem isn't being cared for. (That was sarcasm, folks! My real advice is to go back and learn the material you should have learned before.)
(3) Divide by small: make sure you divide ALL answers from #2 by the smallest value. This may seem obvious, but the ChemTeam has had students who neglect to divide the smallest value by itself to get one.
(4) Multiply 'til whole: multiply ALL values from #3 by the same factor. This factor is selected so as to produce ALL whole numbers as answers. Often this factor is chosen by trial-and-error.
Example Problem: A compound is analyzed and found to contain 68.54% carbon, 8.63% hydrogen, and 22.83% oxygen. The molecular weight of this compound is known to be approximately 140 g/mol. What is the empirical formula? What is the molecular formula? 1) Percent to mass. Assume 100 grams of the substance is present, therefore its composition is:
carbon: 68.54 grams
hydrogen: 8.63 grams
oxygen: 22.83 grams
(2) Mass to moles. Divide each mass by the proper atomic weight.
carbon: 68.54 / 12.011 = 5.71 mol
hydrogen: 8.63 / 1.008 = 8.56 mol
oxygen: 22.83 / 16.00 = 1.43 mol
(3) Divide by small:
carbon: 5.71 ÷ 1.43 = 3.99
hydrogen: 8.56 ÷ 1.43 = 5.99
oxygen: 1.43 ÷ 1.43 = 1.00
(4) Multiply 'til whole. Not needed since all values came out whole.
The empirical formula of the compound is C4H6O.
Next we need to determine the molecular formula, knowing the empirical formula and the molecular weight.
1) Calculate the "empirical formula weight." This is not a standard chemical term, but the ChemTeam believes it is understandable.
C4H6O gives an "EFW" of 70.092.
2) Divide the molecular weight by the "EFW."
140 ÷ 70 = 23) Multiply the subscripts of the empirical formula by the factor just computed.
C4H6O times 2 gives a formula of C8H12O2.
This is the molecular formula.
Empirical Formula Practice Problems
1) A compound is found to have (by mass) 48.38% carbon, 8.12% hydrogen and the rest oxygen. What is its empirical formula?
2) A compound is found to have 46.67% nitrogen, 6.70% hydrogen, 19.98% carbon and 26.65% oxygen. What is its empirical formula?
3) A compound is known to have an empirical formula of CH and a molar mass of 78.11 g/mol. What is its molecular formula?
4) Another compound, also with an empirical formula if CH is found to have a molar mass of 26.04 g/mol. What is its molecular formula?
5) A compound is found to have 1.121 g nitrogen, 0.161 g hydrogen, 0.480 g carbon and 0.640 g oxygen. What is its empirical formula? (Note that masses are given, NOT percentages.)
Go to the answers for 1 - 5
6) A compound containing only carbon, hydrogen and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. What is its empirical formula?
Go to answer for problem number 6
7) When 0.55 grams of Magnesium is heated in a nitrogen atmosphere, a chemical reaction occurs. The product of the reaction weights 0.76 grams. Calculate the empirical formula of the compound containing Mg and N.
Go to a video of the answer to 7
8) Determine the empirical formula for a compound that is 70.79% carbon, 8.91% hydrogen, 4.59% nitrogen, and 15.72% oxygen.
There is an empirical formula calculator on-line. You might try to do this one by hand before using the on-line calculator. By the way, it's capsaicin.
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