EMPIRICAL FORMULA
• The empirical formula represents the
smallest ratio of atoms present in a
compound.
• The molecular formula gives the total
number of atoms of each element present
in one molecule of a compound.
The empirical formula is the simplest
formula and the molecular formula is the
“true” formula.
EMPIRICAL FORMULA
Mass % of
elements
Assume
Empirical
Formula
100g sample
Calculate mole ratio
Grams of
each
element
Moles of
each
element
Use Atomic Masses
EMPIRICAL FORMULA
Step 1: If given the % composition, assume a 100g
sample then convert % to grams.
Step 2: Use the atomic masses to convert grams to
moles.
Step 3: Divide the moles of each element by the
SMALLEST mole fraction.
Step 4: The results from step 3 should be a whole
number, if not, make it so by multiplying by a
common factor.
EMPIRICAL FORMULA
1. Calculate the empirical formula from a sample
containing 43.4% Na, 11.3% C, and 45.3% O.
smallest
43.4%  43.4 g Na (1 mole / 23 g/mol) =1.887 moles Na
11.3%  11.3 g C (1 mole / 12 g/mol) = 0.9417 moles C
45.3%  45.3 g O (1 mole / 16 g/mol) = 2.831 moles O
1.887/0.9417 =2.00 Na
2.831/0.9417 = 3.00 O .
9417/.9417 = 1.00 C
Empirical Formula = Na2CO3
EMPIRICAL FORMULA
2. When 8.00 g of calcium metal is heated in air, 11.20 g of
metal oxide is formed. Calculate the empirical formula.
According to the Law of Conservation of mass,
11.20 g Product - 8.00 g Ca = 3.20 g Oxygen (reactive part of air)
smallest
8.00 g Ca (1 mole / 40 g/mol) = 0.200 moles Ca
3.20 g O (1 mole / 16 g/mol) = 0.200 moles O
0.200 / 0.200 = 1
Empirical Formula = CaO
EMPIRICAL FORMULA
3. A compound was found to have a composition of 33.0 % Sr, 26.8
% Cl, and 40.2 % water. Calculate the empirical formula of this
hydrate.
smallest
33.0%  33.0 g Sr (1 mole/87.6 g/mol) = 0.3767 moles Sr
26.8%  26.8 g Cl (1 mole/35.45 g/mol) = 0.7560 moles Cl
40.2%  40.2 g H2O (1 mole/18.0g/mol) = 2.233 moles H2O
0.7560 / 0.3767 = 2 Cl
2.233 / 0.3767 = 5.9 = 6 H2O
Empirical Formula = SrCl2 . 6 H2O
EMPIRICAL FORMULA
& Molecular Formula
4. Propylene contains 14.3 % H, 85.7% C, and has a molar mass of
42.0 g/mol. What is its molecular formula?
smallest
14.3%  14.3 g H (1 mole/1.01 g/mol) = 14.19 moles H
85.7%  85.7 g C (1 mole/12.01 g/mol) = 7.142 moles C
14.19 / 7.142 = 1.987 = 2 H
Empirical Formula = CH2
Molar mass / empirical mass = multipier
(42.0 g/mol / 14.0 g/mol) = 3
3 x CH2 becomes the molecular formula 
C3H6
PRACTICE PROBLEM #12
A
______
1. Which contains the larger number of MOLES of
atoms?
a) 125.0 g KCl
b) 25.0 g CaSO4
c) 17.0 g of N2
K
2MnO4 2. What is the empirical formula of the compound
______
whose composition is 39.7% K, 27.8% Mn, and 32.5% O?
Bi2O3 3. Determine the empirical formula of a compound
______
that contains 89.7 % bismuth and 10.3 % oxygen.
C6______
H12O3 4. Write the molecular formula for a compound
that contains 54.5 % C, 9.1% H, and 36.4 % O and has a
molar mass of 132 amu?
GROUP STUDY PROBLEM #12
______ 1. Which contains the larger number of MOLES of
atoms?
a) 125.0 g HBr
b) 25.0 g C6H11O6
c) 17.0 g of Br2
______ 2. A sample of a compound weighing 4.18 g
contains 1.67 g of sulfur and the rest is oxygen. What is
the empirical formula?
______ 3. What is the empirical formula of the compound
whose composition is 28.7% K, 1.4% H, 22.8 % P, and
47.1% O?
______ 4. A compound contains 92.3% C and 7.7% H and
has a molar mass of 78.0 g/mol. Determine the
molecular formula.