AP Chemistry  Empirical Formulas 
When the quantitative analysis of a compound gives us the masses of each element in the samesize 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 formula. An 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

Therefore the formula is C_{0.09176}O_{0.09181}
Pick the smallest of the subscript numbers as the divisor and divide it
into all the other numbers.
C_{1.0000}O_{1.0001} = C_{1}O_{1} 
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

Therefore the formula is C_{0.16678}H_{0.4404}
= C_{1.000}H_{2.667} 
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. 
C_{0.16678 X 2}H_{0.4404
X 2} = C_{2}H_{5.334
}Still not good enough
C_{0.16678 X 3}H_{0.4404
X 3} = C_{3}H_{8.001} = C_{3}H_{8}

Example #3 
A 5.438 gram sample of ludlamite, a greyishblue 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
0.9424 grams of P X 1 mole of P = 0.0304 moles of P

Therefore the formula is Fe_{0.0456}O_{0.1217}P_{0.0304}
= Fe_{1.5}O_{4.0}P_{1} =Fe_{3}O_{8}P_{2}

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

Therefore the formula is Ba_{0.5067}C_{0.5071}O_{1.520} = Ba_{1}C_{1.001}O_{2.999} = BaCO_{3} 
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 
Therefore the formula is Hg_{0.4237}Cl_{0.4237}
= 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 CO_{2} and 0.6369 grams of H_{2}O were obtained.
What is the empirical formulal of the compound? 
First: do a grams to moles calculation to determine
the moles of CO_{2} and H_{2}O produced.
moles of CO_{2 }= g/mm = 1.039 grams/44.01 g/mol = 0.02361 moles of CO_{2} moles of H_{2}O = g/mm = 0.6369
grams/18.02 g/mol = 0.03534 moles of H_{2}O 
Second: do a moles to moles conversion to find
the starting moles of C and H that went into the CO_{2} and H_{2}O
as it burned.
C + O_{2} > CO_{2} The reaction ratio for C to CO_{2} is 1:1 therefore the moles of C in the starting material is 0.02361 mles of C 4 H + O_{2}
> 2 H_{2}O This is the equation
that causes students problems. Up until this moment you have always
used H_{2}. 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 H_{2}O 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 
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 C_{0.02361}H_{0.07073}O_{0.0118}
= C_{2.001}H_{5.994}O_{1} = C_{2}H_{6}O

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, SO_{3}, 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 SO_{3}. Name
the compound as well after determining its formula. 
First: 0.5771 grams of BaO / 153.33 g/mole of BaO = 0.0038 moles of BaO 0.3012 grams of SO_{3} / 80.06
g/mole of SO_{3} = 0.0038 moles of SO_{3} 
Second: 2 Ba + O_{2} > 2 BaO reaction ratio between Ba and BaO is 1:1 therefore moles of Ba = 0.0038 moles 2 S + 3 O_{2} > 2 SO_{3} reaction ratio between S and SO_{3} is 1:1 therefore moles of S = 0.0038 moles 
Third: 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 Ba_{0.038}S_{0.0038}O_{0.0146} = Ba_{1}S_{1}O_{3.84} = BaSO_{4} 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:
C_{1}H_{1} X 8 = C_{8}H_{8} 
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 Hg_{1}Cl_{1} X 2 = Hg_{2}Cl_{2} 
Go to the Empiricals Worksheet 
Go to the Next Stoichiometry Section on Limiting Reagents
and Percentage Yields 