EMPIRICAL FORMULA
A chemical formula relates the number of atoms in a compound. The compound CH2O has one carbon
atom, two hydrogen atoms and one oxygen atom. The formula also can be interpreted on a mole level. It
contains one mole of carbon atoms, two moles of hydrogen atoms and one mole of oxygen atoms. The
formula that has the simplest integer ratios is called the empirical formula. The CH2O formula is an
empirical formula because the subscripts cannot be divided by a common number to yield smaller integer
numbers.
The empirical formula of a compound can be determined from an elemental analysis of the compound
that provides the mass of each atom in the compound. The mass of each atom can be converted to the
mass percent of each element and then to the mole of each atom in the compound. The ratio of the mole
of each atom provides the ratio of formula subscripts.
For CH2O, the percent by mass analysis finds 40.00% carbon, 6.71% hydrogen and 53.29% oxygen. In a
1.00g sample of the compound there are 0.4000 g of carbon, 0.0671 g of hydrogen and 0.5329 g of
oxygen. This corresponds to the 0.0333 mol carbon, 0.0666 mol hydrogen and 0.0333 mol oxygen.
Choosing the smallest number of moles (either carbon or oxygen) as the divisor produces a mole ratio of
1:2:1 that corresponds to the subscript ratio and yields an empirical formula of CH2O.
A molecular formula identifies the number of atoms present in the molecular unit of a compound. In many
cases the molecular formula is the same as the empirical formula but it is not uncommon for the
molecular formula to be a multiple of the empirical formula. The four compounds in the table show that
different compounds may have the same empirical formula. Each of the four compounds is 85.63 % by
mass carbon and 14.37 % by mass hydrogen which in 100 g of compound will yield 7.13 mol of carbon
and 14.26 mol of hydrogen. The ratio of mole is 2 mol hydrogen and 1 mol carbon that yields an empirical
formula of CH2. Note molar mass of each compound in this group is a multiple of the empirical mass and
that the molecular formula is a multiple of the empirical formula. The subscript multiple is determined by
dividing the molar mass by the empirical mass.
Molecular
Formula
Molar Mass (g)
Empirical
Formula
Empirical Mass
(g)
Subscript
Multiple
C2H4
C3H6
C5H10
C6H12
28.054
42.081
70.135
84.162
CH2
CH2
CH2
CH2
14.027
14.027
14.027
14.027
2
3
5
6
The relationship between the empirical formula and molecular formula is useful in determining the
molecular formula for a new compound. Empirical formulas can be determined to great accuracy but
molecular formulas are more difficult to determine. Regardless once the empirical formula of a
compound is known its molecular formula can be ascertained even if only an approximate molar mass is
known. For example consider a compound with an empirical formula of CH2O that is found to have an
experimentally determined molar mass of approximately 62 g. If this approximate molar mass is divided
by the empirical mass (30.03 g), a value of 2.06 is obtained. Since the subscript multiple must be an
integer, the value can be rounded to 2. The molecular formula then is (CH2O)2 or C2H4O2.
Completing an elemental analysis and determining the empirical formula of compound is still the first step
in characterizing a new compound. Analytical analysis data can be presented in a number of ways and
several types of examples are presented to show how the empirical formula can be deduced from the
different presentations.
Binary Compounds
Question: A 0.6412 g sample of sulfur reacts with oxygen to produce a compound containing only sulfur
atoms and oxygen atoms that weighs 1.601 g. Calculate the empirical formula of the compound.
Empirical Formula
50
Solution: Assume all of the sulfur reacts. The mass of the sulfur does not change. The mass of the
oxygen present in the compound can be determined by subtracting the mass of sulfur from the mass of
the compound:
1.601 g Cpd - 0.6412 g S = 0.9598 g O.
Knowing the masses of oxygen and sulfur present, the molar masses of the two elements can be used to
calculate how many moles of each are present:
(0.9598 g O) ⎛⎜⎜
1 mol O ⎞
⎟⎟ = 0.05999 mol O
⎝ 15.999 g O ⎠
(0.6412 g S) ⎛⎜⎜
1 mol S ⎞
⎟⎟ = 0.02000 mol S
⎝ 32.066 g S ⎠
As sulfur has the least number of moles in the compound, divide the number of moles of S into the
number of moles of O to find the number of O represented in the empirical formula.
0.05999
= 2.9995
0.02000
The result is very close to 3, and since atoms must be present in a formula in integer ratios, round to
three. There are three oxygen atoms in the compound for every sulfur atom and the empirical formula is
SO3.
Question: A compound containing only sulfur and fluorine is 45.77% by weight sulfur and 54.23% by
weight fluorine. Determine its empirical formula.
Solution: Recall that mass percent is defined as parts per hundred. These quantities can be used along
with the molar masses of fluorine and sulfur to determine the number of moles present in 100 g of the
sample.
⎛ 45.77 g S ⎞ ⎛ 1 mol S ⎞ ⎛ 1.427 mol S ⎞
⎜⎜
⎟⎟ ⎜⎜
⎟⎟ = ⎜⎜
⎟⎟
⎝ 100 g Cpd ⎠ ⎝ 32.066 g S ⎠ ⎝ 100 g Cpd ⎠
⎛ 54.23 g F ⎞ ⎛ 1 mol F ⎞ ⎛ 2.854 mol F ⎞
⎜⎜
⎟⎟ ⎜⎜
⎟⎟ = ⎜⎜
⎟⎟
⎝ 100 g Cpd ⎠ ⎝ 18.998 g F ⎠ ⎝ 100 g Cpd ⎠
Sulfur has the least number of moles in the compound. It will have the smallest subscript. To find the
number of F represented in the empirical formula, divide the number of moles of S into the number of
moles of F.
⎛
⎜
⎜
⎜
⎜
⎝
2.854 mol F
100 g Cpd
1.427 mol S
100 g Cpd
⎞
⎟
⎟ = 2.000 Note that the 100 g of Cpd cancel.
⎟
⎟
⎠
51
Empirical Formula
The quotient is the number of times F is represented in the empirical formula compared to S. The
empirical formula is SF2.
Polyatomic Compounds
Question: A compound containing only Mo, O, and Cl was found to have percent by mass Mo of 48.25%
and percent by mass O of 16.09%. Calculate the empirical formula of the compound.
Solution: First calculate the percent by weight of the Cl. This is done by adding the percents Mo and O
and subtracting them from 100%.
48.25%
+ 16.09%
64.34%
Mo and O present in the compound.
100%-64.34%=35.66% by weight Cl present in compound.
Recall that percent by mass is parts per hundred. (If there are 100 g of compound which is 48.25% by
weight Mo, there will be 48.25 g of Mo present. Correspondingly, there will also be 16.09 g O and 35.66
g Cl.) Using mol/molar mass ratios calculate the moles of each element.
⎛ 16.09 g O ⎞ ⎛ 1 mol O ⎞ ⎛ 1.006 mol O ⎞
⎜⎜
⎟⎟ ⎜⎜
⎟⎟ = ⎜⎜
⎟⎟
⎝ 100 g Cpd ⎠ ⎝ 16.00 g O ⎠ ⎝ 100 g Cpd ⎠
⎛ 48.25 g Mo ⎞ ⎛ 1 mol Mo ⎞ ⎛ 0.5029 mol Mo ⎞
⎟⎟ = ⎜⎜
⎜⎜
⎟⎟ ⎜⎜
⎟⎟
⎝ 100 g Cpd ⎠ ⎝ 95.94 g Mo ⎠ ⎝ 100 g Cpd ⎠
⎛ 35.66 g Cl ⎞ ⎛ 1 mol Cl ⎞ ⎛ 1.006 mol Cl ⎞
⎜⎜
⎟⎟ ⎜⎜
⎟⎟ = ⎜⎜
⎟⎟
⎝ 100 g Cpd ⎠ ⎝ 35.453 g Cl ⎠ ⎝ 100 g Cpd ⎠
The element with the smallest amount of moles has the smallest subscript. To determine the subscripts of
the other elements divide that number of moles into the moles of the other elements.
⎛ 1.006 mol O
⎜
100 g Cpd
⎜
⎜ 0.5029 mol Mo
⎜
100 g Cpd
⎝
⎞
⎟
⎟ = 2.000
⎟
⎟
⎠
⎛ 1.006 mol Cl
⎜
100 g Cpd
⎜
⎜ 0.5029 mol Mo
⎜
100 g Cpd
⎝
⎞
⎟
⎟ = 2.000
⎟
⎟
⎠
Note the 100 g Cpd cancel. The empirical formula is MoCl2O2.
Question: A 4.150 g sample of a compound containing only P, O, and F atoms was found to contain
1.239 g P and 0.6400 g O. If the molar mass of the compound was determined to be approximately 108
g, what is its molecular formula?
Solution: First determine the mass of F in the sample by adding together the masses of P and O and
subtracting the sum from the total mass of the sample.
Empirical Formula
52
4.150 g Cpd - 1.239 g P - 0.6400 g O = 2.271 g F
Calculate the mol of each element in the compound.
(0.6400
(1.239
⎛ 1 mol O ⎞
⎟⎟ = 0.0400 mol O
g O )⎜⎜
⎝ 16.00 g O ⎠
⎛ 1 mol P ⎞
⎟⎟ = 0.0400 mol Mo
g P ) ⎜⎜
⎝ 30.974 g Mo ⎠
(2.271 g F) ⎛⎜⎜
1 mol F ⎞
⎟⎟ = 0.1195 mol F
⎝ 18.998 g F ⎠
Divide by the smallest number of mol.
0.1195 mol F
= 2.987
0.0400 mol O
0.0400 mol Mo
= 1.000
0.0400 mol O
0.0400 mol O
= 1.000
0.0400 mol O
Round off the values to the nearest integer. The empirical formula is POF3. Compare the molar mass
with the empirical mass: Empirical mass: 103.967. Molar mass: 108 g. The two masses are equal. The
molecular formula is the same as the empirical formula, POF3.
There is another way to solve this problem but it is longer and more cumbersome.
Calculate mass percentages for each element.
⎛
1.239 g P
⎜⎜
⎝ 4.15 g compound
⎞
⎟⎟ (100 ) = 29.9% P
⎠
⎛
0.6400 g O
⎜⎜
⎝ 4.15 g compound
⎞
⎟⎟ (100 ) = 15.4% O
⎠
⎛
2.271 g F
⎜⎜
⎝ 4.15 g compound
⎞
⎟⎟ (100 ) = 54.7% F
⎠
Recall mass percent is part per hundred.
53
Empirical Formula
⎛ 15.4 g O ⎞ ⎛ 1 mol O ⎞ ⎛ 0.962 mol O ⎞
⎟⎟
⎜⎜
⎟⎟ ⎜⎜
⎟⎟ = ⎜⎜
⎝ 100 g Cpd ⎠ ⎝ 16.00 g O ⎠ ⎝ 100 g Cpd ⎠
⎛ 29.9 g P ⎞ ⎛ 1 mol P ⎞ ⎛ 0.965 mol P ⎞
⎟⎟ = ⎜⎜
⎜⎜
⎟⎟ ⎜⎜
⎟⎟
⎝ 100 g Cpd ⎠ ⎝ 30.974 g Mo ⎠ ⎝ 100 g Cpd ⎠
⎛ 54.7 g Cl ⎞ ⎛ 1 mol F ⎞ ⎛ 2.88 mol F ⎞
⎟⎟
⎜⎜
⎟⎟ ⎜⎜
⎟⎟ = ⎜⎜
⎝ 100 g Cpd ⎠ ⎝ 18.998 g F ⎠ ⎝ 100 g Cpd ⎠
Divide by the element with the smallest number of mol per 100 g compound. The P and O have
essentially equivalent values so either would be acceptable. Note the 100 g Cpd cancel
⎛
⎜
⎜
⎜
⎜
⎝
2.88 mol F
100 g Cpd
0.965 mol P
100 g Cpd
⎞
⎟
⎟ = 3.01
⎟
⎟
⎠
⎛ 0.962 mol O
⎜
⎜ 100 g Cpd
⎜ 0.965 mol P
⎜
⎝ 100 g Cpd
⎞
⎟
⎟ = 0.996
⎟
⎟
⎠
⎛ 0.965 mol P
⎜
⎜ 100 g Cpd
⎜ 0.965 mol P
⎜
⎝ 100 g Cpd
⎞
⎟
⎟ = 1.00
⎟
⎟
⎠
Round the results to the nearest integer since atoms are indivisible. The empirical formula is POF3.
Compare the molar mass with the empirical mass: Empirical mass: 103.967. Molar mass: 108 g. The
two masses are equal. The molecular formula is the same as the empirical formula, POF3.
Hydrocarbon Analysis
Question: A 0.4724 g sample of a compound containing C, H, and O was burned to produce a 0.704 g
CO2 and 0.2162 g H2O. Calculate the empirical formula of the compound.
Solution: Determine the grams of C and the grams of H that were in the compound that reacted. Every
gram of carbon in the compound was converted to CO2 and every gram of hydrogen in the compound
was converted to H2O. Consequently we can determine the mass of carbon and hydrogen in the original
sample.
(0.704 g
⎛ 1 mol CO 2 ⎞ ⎛ 1 mol C ⎞ ⎛ 12.011 g C ⎞
⎟⎟ ⎜⎜
⎟⎟ ⎜
CO 2 ) ⎜⎜
⎟ = 0.192 g C
44.001
g
CO
1
mol
CO
2 ⎠⎝
2 ⎠ ⎝ 1 mol C ⎠
⎝
(0.2162 g
⎛ 1 mol H2 O ⎞ ⎛ 2 mol H ⎞ ⎛ 1.0080 g H ⎞
⎟⎟ ⎜⎜
⎟⎟ ⎜
H2 O ) ⎜⎜
⎟ = 0.0242 g H
⎝ 18.015 g H2 O ⎠ ⎝ 1 mol H2 O ⎠ ⎝ 1 mol H ⎠
Determine the grams of O present in the sample by subtracting the grams of C and H from the total mass
of the sample.
0.4724 g Cpd - 0.192 g C - 0.0242 g H = 0.2562 g O
Determine the mol of C, H and O present in the sample.
Empirical Formula
54
(0.192 g C) ⎛⎜⎜
1 mol C ⎞
⎟⎟ = 0.0160 mol C
⎝ 12.011 g C ⎠
(0.0242 g H) ⎛⎜⎜
1 mol H ⎞
⎟⎟ = 0.0240 mol H
⎝ 1.0080 g H ⎠
(0.2562 g O) ⎛⎜⎜
1 mol O ⎞
⎟⎟ = 0.0160 mol O
⎝ 15.999 g O ⎠
Divide by the element with the smallest number of moles to determine the empirical formula. Carbon and
oxygen have the same value so either is appropriate.
0.0240 mol H
= 1.5
0.0160 mol O
0.0160 mol C
= 1.0
0.0160 mol O
0.0160 mol O
= 1.0
0.0160 mol O
The subscript value for H in the formula is not near an integer. But an atom can not be split in half. The
solution is the multiply the value of H by a number that generates and integer. In this case multiple by 2 to
obtain the integer 3. It is important to recognize that all subscripts must be multiplied by the same value.
Both C and O will therefore have subscripts of 2 in the empirical formula. Hydrogen will have a subscript
of 3. The formula is C2H3O2
Inorganic Analysis
Question: A 0.488 g sample of a compound containing only magnesium, sulfate and water of hydration,
after heating to drive off the water of hydration, weighs 0.240 g. A second sample weighing 763 mg is
dissolved in water and excess aqueous barium chloride is added to precipitate 728 mg of barium sulfate.
Calculate the empirical formula of the compound.
Solution: Determine the percentage of water present in the compound.
0.488 g sample - 0.240 g dehydrated sample = 0.248 g H2O
⎛ 0.252 g H2 O
⎜⎜
⎝ 0.488 g sample
⎞
⎟⎟ (100 ) = 50.8% H2 O
⎠
From the precipitation data determine the percentage of SO42- present in the compound.
⎛ 1 mol BaSO 4 ⎞⎛ 1 mol SO 4 2- ⎞⎛ 96.064 g SO 4 2- ⎞
⎟⎜
⎟ = 0.300 g SO 4 2⎟⎟⎜
(0.728 g BaSO 4 ) ⎜⎜
2⎜
⎟
⎜
⎟
⎝ 233.40 g BaSO 4 ⎠⎝ 1 mol BaSO 4 ⎠⎝ 1 mol SO 4
⎠
⎛ 0.300 g SO 4 2- ⎞
⎜
⎟ (100 ) = 39.3% SO 4 2⎜ 0.763 g sample ⎟
⎝
⎠
Determine percentage Mg: 100% - 50.8% H2O - 39.3% SO42- = 9.9% Mg2+.
55
Empirical Formula
Ions are formed from the gain or lose of electrons whose mass is much less that the mass of a proton or
neutron. Hence the mass of Mg2+ is almost exactly the same as that of Mg so the molar mass of the
neutral element can be used to determine the mol of the ion. The same situation applies to SO42-.
The molar mass of sulfate is determined form the molar mass of sulfur ( 32.066 g )and oxygen (16.000 g)
Molar Mass SO42- = 32.066 + 4 (16.000) = 96.066 g. Recall that mass percent is parts per hundred.
⎛ 9.9 g Mg2+ ⎞ ⎛ 1 mol Mg 2+
⎜⎜
⎟⎟ ⎜⎜
2+
⎝ 100 g Cpd ⎠ ⎝ 24.305 g Mg
⎞ ⎛ 0..407 mol Mg2+
⎟⎟ = ⎜⎜
100 g Cpd
⎠ ⎝
⎞
⎟⎟
⎠
⎛ 51.6 g H2 O ⎞ ⎛ 1 mol H2 O ⎞ ⎛ 2.86 mol H2 O ⎞
⎟⎟ = ⎜⎜
⎜⎜
⎟⎟ ⎜⎜
⎟⎟
⎝ 100 g Cpd ⎠ ⎝ 18.016 g H2 O ⎠ ⎝ 100 g Cpd ⎠
⎛ 39.3 g SO 4 2⎜
⎜ 100 g Cpd
⎝
⎞ ⎛ 1 mol SO 4 2⎟⎜
⎟ ⎜ 96.066 g SO 24
⎠⎝
⎞ ⎛ 0.409 mol SO 4 2⎟=⎜
⎟ ⎜ 100 g Cpd
⎠ ⎝
⎞
⎟
⎟
⎠
The mol ratio of the components of the compound water, magnesium ion and sulfate ion can be treated
just like atoms are treated. Divide by the component with the smallest number of moles. The mol value of
the sulfate and the magnesium ion are close enough so that either could be chosen as the smallest
value.
2.86 mol H2 O
= 7.02
0.407 mol Mg 2+
2-
0409 mol SO 4
= 1.004
0.407 mol Mg2+
0407 mol Mg2+
= 1.00
0.407 mol Mg2+
Round off the numbers. The subscripts of SO42- and Mg2+ will be 1 and that for water will be 7. assume
their coefficients will both be equal to 1. Determine the coefficient of H2O. The empirical formula of the
compound is MgSO4 7H2O.
Empirical Formula
56