AP Chemistry - Empirical Formulas
When the quantitative analysis of a compound gives us the masses of each element in the same-size sample, we can convert their masses into proportions by moles by a grams to moles calculation.  With atomic particles, proportions by moles are numerically the same as proportions by atoms, which are just the numbers we need to construct a chemical formula.  The formula obtained in this way is called an empirical formulaAn empirical formula is the one that uses as subscripts the smallest whole numbers that describe the ratios of atoms in a compound.

Calculating an empirical formula from data obtained by quantitative analysis.
Example #1
A sample of an unknown compound with a mass of 2.571 grams was found to contain 1.102 grams of C and 1.469 grams of oxygen.  What is its empirical formula?

1.102 g of C X 1 mole of C  = 0.09176 moles of C
                        12.01 g of C

1.469 g of O X 1 mole of O = 0.09181 moles of O
                         16.00 g of O

Therefore the formula is C0.09176O0.09181   Pick the smallest of the subscript numbers as the divisor and divide it into all the other numbers.

                        C1.0000O1.0001 = C1O1

As a general rule, if the calculated subscript differs from a whole number by only several units in the last place, we can safely round to the whole number.

Example #2
When a sample with a mass of 2.448 grams of compound present in liquified petroleum gas was analyzed, it was found to contain 2.003 grams of carbon and 0.4448 grams of hydrogen.   What is its empirical formula?

2.003 g of C X 1 mole of C  = 0.16678 moles of C
                        12.01 g of C

0.4448 g of H X 1 mole of H = 0.4404 moles of H
                           1.1 g of H

Therefore the formula is C0.16678H0.4404 = C1.000H2.667
The ratios don't come out to the expected nice ratios.  The solution to this is to multiply to see if we can get a better whole number ratio.
      C0.16678 X 2H0.4404 X 2 = C2H5.334             Still not good enough

      C0.16678 X 3H0.4404 X 3 = C3H8.001   =  C3H8

Example #3
A 5.438 gram sample of ludlamite, a greyish-blue mineral sometimes used in ceramics, was found to contain 2.549 grams of iron, 1.947 grams of oxygen, and 0.9424 grams of phosphorus.  What is its empirical formula?

2.549 grams of Fe X 1 mole of Fe = 0.0456 moles of Fe
                                   55.85 g Fe

1.947 grams of O X 1 mole of O = 0.1217 moles of O
                                   16.00 g O

0.9424 grams of P X 1 mole of P = 0.0304 moles of P
                                    30.97 g P

Therefore the formula is Fe0.0456O0.1217P0.0304 = Fe1.5O4.0P1 =Fe3O8P2

Calculating Empirical Formulas from Percentage Compositions
Example #1
Barium carbonate, a white powder used in paints, enamels, and ceramics, has the following composition: Ba, 69.58%; C, 6.090%; O, 24.32%.  What is its empirical formula?
Assume we have a 100 gram sample since it is out of 100 percent.  Therefore all the percentages change to grams and solve it like the above three examples.
69.58  grams of Ba X 1 mole of Ba = 0.5067 moles of Ba
                                    137.33 g Ba

6.090 grams of C X 1 mole of C = 0.5071 moles of C
                                   12.01 g C

24.33 grams of O X 1 mole of O = 1.520 moles of O
                                   16.00 g O

Therefore the formula is Ba0.5067C0.5071O1.520 = Ba1C1.001O2.999 = BaCO3

Example #2
Calomel is the common name of a white powder once used in the treatment of syphilis.  Its composition is 84.98% mercury and 15.02% chlorine.  What is its empirical formula?
84.98 grams of Hg X 1 mole of Hg = 0.4237 moles of Hg
                                    200.59 g Hg

15.02 grams of Cl X 1 mole of Cl =  0.4237 moles of Cl
                                   35.45 g Cl

Therefore the formula is Hg0.4237Cl0.4237 = HgCl

Combustion Analysis
In combustion analysis a sample of the chemical being analysed is burned with oxygen.  The resulting products are used with a balanced combustion equation to stoichiometrically determine the moles of the starting elements.

Example #1
A sample of a liquid consisting of only C, H and O  and having a mass of 0.5438 grams, was burned in pure oxygen and 1.039 grams of CO2 and 0.6369 grams of H2O were obtained.  What is the empirical formulal of the compound?

First:  do a grams to moles calculation to determine the moles of CO2 and H2O produced.

      moles of CO2 = g/mm = 1.039 grams/44.01 g/mol = 0.02361 moles of CO2

     moles of H2O = g/mm = 0.6369 grams/18.02 g/mol = 0.03534 moles of H2O

Second:  do a moles to moles conversion to find the starting moles of C and H that went into the CO2 and H2O as it burned.

      C + O2 -----> CO2    The reaction ratio for C to CO2 is 1:1 therefore the moles of C in the starting material is 0.02361 mles of C

        4 H + O2 ----->  2 H2O    This is the equation that causes students problems.  Up until this moment you have always used H2.    For this example we are using H.  The reason is that the H is in a molecule and is being treated as lone atoms not as a molecule in it's own right.   The reaction ratio for H to H2O is 2:1 therefore the amount of H in the starting material is twice the number of water moles from above so the moles of H in the starting molecule is 0.07073 moles.

Third:  do a moles to grams calculation to determine the starting amounts of C and H in the starting molecule.

       grams of C = 0.02361 moles X 12.01 g/mole = 0.2836 grams of C

       grams of O = 0.07073 moles X 1.01 g/mole = 0.07144 grams of H

Fourth:  Using the original mass of the entire starting material, minus the above two masses calculate the amount of O in the original sample.

                 original sample mass = 0.5438 g
                           - mass of the C = 0.2836 g
                         - mass of the H = 0.07144 g
    mass of O in original sample = 0.18876 g

Fifth: now do the empirical calculation.  You already have the moles of C and H so don't redo them.

moles of O = 0.18876 g / 16.00 g/mole = 0.0118 moles of O

therefore the formula is C0.02361H0.07073O0.0118 = C2.001H5.994O1 = C2H6O

Combustion example:
A compound known to contain only barium, sulphur, and oxygen was analyzed by first heating the sample with a mass of 0.8778 grams at a very high temperature in pure oxygen. All of the barium atoms end up in barium oxide, BaO, which is a solid.  All of the sulphur atoms emerge in molecules of sulphur trioxide, SO3,  a gas that can easily be trapped and weighed.  The weight's of the products were 0.5771 grams of BaO and 0.3012 grams of SO3.  Name the compound as well after determining its formula.

     0.5771 grams of BaO / 153.33 g/mole of BaO = 0.0038 moles of BaO

     0.3012 grams of SO3 / 80.06 g/mole of SO3 = 0.0038 moles of SO3

     2 Ba + O2 -----> 2 BaO  reaction ratio between Ba and BaO is 1:1 therefore moles of Ba = 0.0038 moles
     2 S  + 3 O2 ----> 2 SO3   reaction ratio between S and SO3 is 1:1 therefore moles of S = 0.0038 moles

     grams of Ba = 0.0038 moles X 137.33 g/mole = 0.5219 grams of Ba

     grams of S = 0.0038 moles X 32.07 g/mole = 0.1219 grams of S

Fourth:   grams of O = original mass - mass of Ba - mass of S
                                  =  0.8778 g - 0.5219 g - 0.1219 g
                                  =  0.2341 grams of oxygen

Fifth:   moles of O = 0.2341 g / 16.0g/mole = 0.0146 moles of O

therefore the formula is Ba0.038S0.0038O0.0146 = Ba1S1O3.84 = BaSO4

The molecule is barium sulphate.

Determining a Molecular Formula from an Empricial Formula and a Formula Weight
Example #1
Styrene, the raw material for polystyrene plastics, has the empirical formula CH. Its formula wieght is 104.08 g/mole.  What is its molecular formula?
Find the molecular mass of the empirical formula: the empirical moelcular mass
CH = 13.1 g/mole
Divide the actual molecular mass by the empirical molecular mass:  104.08 g/mole / 13.1 g/mole = 8
Multiply the empirical formula by this number:  C1H1 X 8 = C8H8

Example #2
Calomel has a formula weight of 472.08.  The empirical formula is HgCl.  What is its actual molecular formula?
HgCl = 200.59 + 35.45 = 236.04 g/mole
472.08 g/mole / 236.04 g/mole = 2
Therefore Hg1Cl1 X 2 = Hg2Cl2
Go to the Empiricals Worksheet

Go to the Next Stoichiometry Section on Limiting Reagents and Percentage Yields