Empirical and Molecular Formulas
One of the major uses made of % composition is the determination of the chemical formula
(make-up) of a compound.
Empirical Formula – Formula giving the smallest, whole-number, mole ratio of the atoms in a
compound
Molecular Formula – Formula giving the actual number of atoms, of each element in a
molecule.
Example: diPhosphorous tetroxide
Molecular formula = P2O4
Empirical formula = PO2
The molecular formula is a whole number multiple of the empirical formula.
Molecular formula = X(Empirical formula)
Or
𝑋=
𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝐹𝑜𝑟𝑚𝑢𝑙𝑎
𝐸𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝐹𝑜𝑟𝑚𝑢𝑙𝑎
Or
𝑋=
𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑀𝑎𝑠𝑠
𝐸𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝐹𝑜𝑟𝑚𝑢𝑙𝑎 𝑀𝑎𝑠𝑠
Steps to determine Molecular Formula:
1) Determine the number of moles of each element
2) Determine the smallest, whole number, mole ratio between the components. By setting the
moles as a ratio and then dividing all values by the smallest number of moles.
3) Determine the Empirical Mass
4) Divide Empirical Mass into Molecular Mass to find X
5) Multiply Empirical Formula by X to find Molecular Formula
Examples:
1) A compound contains 49.2 % Phosphorous and 50.8 % oxygen. Find its Empirical Formula.
Since the total % equals 100% assume the sample size = 100 g. The amounts of each element
then are 49.2 g of Phosphorous and 50.8 g Oxygen.
Step 1 - determine the number of moles of each element:
49.2 𝑔 𝑃ℎ𝑜𝑠𝑝ℎ𝑟𝑜𝑢𝑠 𝑥
1 𝑚𝑜𝑙
= 1.59 𝑚𝑜𝑙 𝑃ℎ𝑜𝑠𝑝ℎ𝑜𝑟𝑜𝑢𝑠
30.9738 𝑔
50.8 𝑔 𝑂𝑥𝑦𝑔𝑒𝑛 𝑥
1 𝑚𝑜𝑙
= 3.18 𝑚𝑜𝑙 𝑂𝑥𝑦𝑔𝑒𝑛
15.9994 𝑔
Step 2 – find mole ratio
1.59 mol Phosphorous : 3.18 mol Oxygen
1.59
1.59
Divide both by the smaller number of moles.
1 mol Phosphorous : 2 mol Oxygen
Therefore the empirical formula is PO2
2) A sample of diSilicon hexabromide contains 52.45 g of Silicon and 447.55 g of Bromine. Find
its empirical formula and the value of X.
1 𝑚𝑜𝑙
52.45 𝑔 𝑆𝑖 𝑥
= 1.868 𝑚𝑜𝑙 𝑆𝑖
28.0855 𝑔
447.55 𝑔 𝐵𝑟 𝑥
1 𝑚𝑜𝑙
= 5.6011 𝑚𝑜𝑙 𝐵𝑟
79.904 𝑔
1.868 𝑚𝑜𝑙 𝑆𝑖 5.6011 𝑚𝑜𝑙 𝐵𝑟
∶
1.868
1.868
1 mol Si : 2.998 mol Br (rounds to 1:3)
Empirical formula = SiBr3
Molecular formula of diSilicon hexabromide = Si2Br6 (determined from name)
𝑋=
𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
𝑆𝑖2 𝐵𝑟6
=
=2
𝐸𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
𝑆𝑖𝐵𝑟3
3) Find the Empirical and Molecular formulas for a 10.150 g sample that contains 4.433 g of
Phosphorous. The only other element present in the compound is oxygen. The molecular
mass of this compound is 283.89 g/mol.
Total mass
- mass phosphorous
mass oxygen
= 10.150 g
= 4.433 g
= 5.717 g
4.433 𝑔 𝑃 𝑥
1 𝑚𝑜𝑙
= 0.1431 𝑚𝑜𝑙
30.9738 𝑔
5.717 𝑔 𝑂 =
1 𝑚𝑜𝑙
= 0.3573 𝑚𝑜𝑙
15.9994 𝑔
0.1431 𝑚𝑜𝑙 𝑃 0.3573 𝑚𝑜𝑙
∶
0.1431
0.1431
1 mol P : 2.497 mol O (round to 1:2.5)
To get a whole number ratio multiply by 2 = 2 P : 5 O Empirical formula = P2O5
Formula mass = 2(30.9738) + 5(15.9994) = 141.9446 g/mol
𝑋=
𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑀𝑎𝑠𝑠
283.89 𝑔/𝑚𝑜𝑙
=
=2
𝐸𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝐹𝑜𝑟𝑚𝑢𝑙𝑎 𝑀𝑎𝑠𝑠
141.9446 𝑔/𝑚𝑜𝑙
2(P2O5) = P4O10
4) A hydrate when analyzed contains 29.7% Copper, 15.0% Sulfur, 2.8% Hydrogen and
52.5 % Oxygen. Determine its empirical formula.
Assume sample = 100 g, then:
29.7 𝑔 𝐶𝑢 𝑥
15.0 𝑔 𝑆 𝑥
2.8 𝑔 𝐻 𝑥
52.5 𝑔 𝑂 𝑥
1 𝑚𝑜𝑙
= 0.467 𝑚𝑜𝑙
63.546 𝑔
1 𝑚𝑜𝑙
= 0.468 𝑚𝑜𝑙
32.065 𝑔
1 𝑚𝑜𝑙
= 2.8 𝑚𝑜𝑙
1.0079 𝑔
1 𝑚𝑜𝑙
= 3.28 𝑚𝑜𝑙
15.9994 𝑔
0.467 𝑚𝑜𝑙 𝐶𝑢 0.468 𝑚𝑜𝑙 𝑆 2.8 𝑚𝑜𝑙 𝐻 3.28 𝑚𝑜𝑙 𝑂
∶
∶
∶
0.467
0.467
0.467
0.467
1 Cu : 1 S : 6 H : 7 O giving the formula CuSH6O7
Since this is a hydrate we know it contains H2O, pulling out 3 water molecules leaves CuSO4.
The empirical formula is therefore, CuSO4·3 H2O.
5) A compound contains 53.25 % Yttrium, 21.58 % Carbon and 25.17 % Nitrogen. Find its
Empirical formula. Assume sample size = 100 g, then:
0.5989
0.5989
53.25 𝑔 𝑌 𝑥
1 𝑚𝑜𝑙
= 0.5989 𝑚𝑜𝑙
88.9059 𝑔
21.58 𝑔 𝐶 𝑥
1 𝑚𝑜𝑙
= 1.797 𝑚𝑜𝑙
12.0107 𝑔
25.17 𝑔 𝑁 𝑥
1 𝑚𝑜𝑙
= 1.797 𝑚𝑜𝑙
14.0067 𝑔
1.797
∶ 0.5989 ∶
1.797
0.5989
1 mol Y : 3 mol C : 3 mol N
Empirical Formula = YC3N3 Check the polyatomic ions since CN is a polyatomic ion the
formula would be written Y(CN)3