Empirical Formula
Thomas M. Moffett Jr., SUNY Plattsburgh, 2008
All compounds are composed of atoms in simple whole numbered ratios. The empirical formula
is the simplest ration of elements in a compound. For example the empirical formula of water is
H2O, indicating that there is a ratio of two hydrogen atom for every oxygen atom. In this case the
empirical formula is the same as the actual formula (also known as the molecular formula or
formula unit). Sometimes the actual formula is a whole number multiple of the empirical
formula, benzene has an empirical formula of CH, and an actual formula of C6H6.
It is practically impossible to work with individual atoms and molecules in lab, instead we use
moles. One mole of anything contains 6.022 X 1023 (Avogadro’s number) objects of that item.
Thus one mole of element X and one mole of element Y contain the same number of atoms. This
fact allows for the determination of empirical formulas from mole ratios. For example in water,
there are two moles of hydrogen atoms for every one mole of oxygen, resulting in the same
empirical formula. Moles can be determined by dividing mass by molar mass (from the periodic
table).
In the following example the empirical formula of an iron-oxygen compound will be
determined. At the start of the experiment the mass of iron was 16.76 g. At the end of the
experiment the mass of the iron-oxygen compound was 23.96 g.
1.) The mass of the oxygen can be found be subtracting the mass of the iron from the
mass of the compound.
23.96 g – 16.76 g = 7.20 g oxygen
(1)
2.) Now calculate the number of moles of oxygen and iron, by dividing the mass of the
element by the molar mass of the element.
Fe: 16.76 g Fe
O: 7.20 g O
1 mol
= 0.300 mol Fe
55.85 g
1 mol
= 0.450 mol O
16.00 g
(2)
(3)
3.) Find the molar ratio by dividing the moles of each element by the smallest number of
moles.
Fe:
0.300 mol
=1
0.300 mol
(4)
O:
0.450 mol
= 1.5
0.300 mol
(5)
4.) The ratio becomes the subscripts in the empirical formula. Since a chemical formula
can’t have a fraction for a subscript (O1.5), the molar ratio needs to be multiplied by a
factor to remove the fraction.
Fe: 1 X 2 = 2
(6)
O: 1.5 X 2 = 3
(7)
5.) Now that the fraction has been eliminated, the empirical formula can be written.
Empirical Formula = Fe2O3
In order to determine the actual formula, the molar mass of the compound must be known. To
determine the actual formula, first determine the mass of one mole of the empirical formula
(empirical mass). If the empirical mass is the same as the molar mass, then the empirical formula
is the actual formula. If they are not the same, then the actual formula is a multiple of the
empirical formula.
In the next example, the molecular formula of acetic acid will be determined. The
empirical formula of acetic acid is CH2O and the molar mass of acetic acid is 60.0 g/mol.
1.) Determine the empirical mass of acetic acid.
C: 1 (12.0g) = 12.0 g
H: 2(1.01g) = 2.02 g
O: 1(16.0g) = 16.0 g
CH2O = 30.0 g/mol
(empirical mass)
(8)
2.) Divide the molar mass by the empirical mass to determine the factor that the
subscripts need to be multiplied by to write the actual formula.
molar mass
60.0 g/mol
=
=2
empirical mass 30.0 g/mol
(9)
The actual formula is therefore C2H4O2
In this experiment you will determine the empirical formula of a compound formed by
magnesium and oxygen. The reaction occurs between magnesium metal and oxygen in the air.
The reaction will be run at high temperature (using a Bunsen burner flame) in order to speed up
the rate of reaction. Unfortunately magnesium will also react with nitrogen at high temperatures
forming a magnesium-nitrogen compound. This compound can be converted to the magnesiumoxygen compound by heating in the presence of water.
Procedure1:
Your instructor will review how to operate a Bunsen burner.
Caution a hot crucible looks the same as a cool crucible.
When weighing your crucible make sure that it has reached room temperature. If you weigh a
warm object, the temperature difference will create air currents around the balance and give you
in incorrect mass.
1.) Wash rinse and dry your crucible and cover. The crucible may have a dark stain that can’t be
removed, this is ok.
2.) Place the crucible in a clay triangle. The bottom of the crucible should be approximately 5
cm above the top of the Bunsen burner. Heat the crucible with the hottest part of the flame
for two minutes. This will serve to further clean out the crucible. After two minutes, turn off
the burner and allow the crucible to cool. When it has reached room temperature determine
the mass of the crucible.
3.) Add approximately 0.5 g of magnesium turnings or magnesium powder to the crucible.
Record the exact mass of the crucible and magnesium.
4.) Cover the crucible and heat it gently for several minutes. Heat with the full flame for 15
minutes.
5.) Let the crucible cool for five minutes, and then adjust the cover so that there is a small
opening.
6.) Heat the crucible with the full flame again. During this time periodically check the contents
of the crucible by lifting the cover. If the contents glow brightly or emit white smoke then the
reaction is not yet complete. Continue with this heating until all of the magnesium has
reacted.
7.) Cool the crucible to room temperature. Add 10 to 15 drops of deionized water. Replace the
cover so that there is a small opening and heat the crucible with a low flame for five minutes.
8.) Cool the crucible to room temperature. Determine the mass of the crucible and magnesiumoxygen compound.
9.) The compound may be scraped into the trash when you are finished with the experiment.
1
Adapted from Dickson, T.R. Laboratory Experiments to Accompany Introduction to Chemistry, 8th ed., New York:
John Wiley & Sons Inc., 2000.
Name: ____________________________________
Empirical Formula – Data Sheet
* Show all calculations for full credit
Section: ____
CHE 101
Mass of crucible and magnesium
_______________________
Mass of crucible
_______________________
Mass of magnesium
_______________________
Mass of crucible and magnesium-oxygen compound
_______________________
Mass of magnesium-oxygen compound
_______________________
Mass of oxygen
_______________________
Use the space below to determine the empirical formula of the magnesium-oxygen compound.
Name: ____________________________________
Empirical Formula – Post Lab Questions
Section: ____
CHE 101
1.) For each of the following procedural errors indicate whether the mass of the compound
would have been too big or too small. Explain your answer.
a.) A student forgot to put the cover on the crucible, when it was heated some of the contents
came flying out of the crucible.
b.) A student closed the air vent all the way on the Bunsen burner, creating a dirty yellow
flame which deposited soot on the crucible.
2.) Determine the empirical formula of calcium phosphide which is comprised of 66.00 %
calcium and 34.00 % phosphorous.
3.) Glucose has a molar mass of 180.1 g/mol and is comprised of 40.0 % carbon, 6.72 %
hydrogen, and 53.3 % oxygen. Determine the molecular formula of glucose.
Name: ____________________________________
Empirical Formula – Pre Lab Questions
Section: ____
CHE 101
1.) Why is water added to the reaction mixture towards the end of the procedure?
2.) What are the molar masses of magnesium and oxygen?
3.) A student determined that the mass of a magnesium ribbon was 0.380 g, how many moles is
this?
4.) Determine the number of magnesium atoms in the 0.380 g ribbon.
5.) Determine the mass of 3.25 X 1022 oxygen atoms.