1. 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.

2. EMPIRICAL FORMULA Mass % of Empirical elements Formula Grams of
Assume g sample Calculate mole ratio
Use Atomic Masses
Mass % of
elements
Empirical
Formula
Grams of
each
element
Moles of
each
element

3. 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.

4. Empirical Formula = Na2CO3
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) = moles C
45.3%  45.3 g O (1 mole / 16 g/mol) = moles O
1.887/ =2.00 Na
2.831/ = 3.00 O .
9417/.9417 = 1.00 C
Empirical Formula = Na2CO3

5. Empirical Formula = CaO
2. When 8.00 g of calcium metal is heated in air, g of metal oxide is formed. Calculate the empirical formula.
According to the Law of Conservation of mass,
11.20 g Product g Ca = 3.20 g Oxygen (reactive part of air)
smallest
8.00 g Ca (1 mole / 40 g/mol) = moles Ca
3.20 g O (1 mole / 16 g/mol) = moles O
0.200 / = 1
Empirical Formula = CaO

6. Empirical Formula = SrCl2 . 6 H2O
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) = moles Sr
26.8%  26.8 g Cl (1 mole/35.45 g/mol) = moles Cl
40.2%  40.2 g H2O (1 mole/18.0g/mol) = moles H2O
/ = 2 Cl / = 5.9 = 6 H2O
Empirical Formula = SrCl2 . 6 H2O

7. 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) = moles H
85.7%  85.7 g C (1 mole/12.01 g/mol) = moles C
14.19 / = = 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

8. PRACTICE PROBLEM #12 A K2MnO4 Bi2O3 C6H12O3
______ Which contains the larger number of MOLES of atoms?
a) g KCl b) 25.0 g CaSO c) g of N2
______ What is the empirical formula of the compound whose composition is 39.7% K, 27.8% Mn, and 32.5% O?
______ Determine the empirical formula of a compound that contains 89.7 % bismuth and 10.3 % oxygen.
______ 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?
K2MnO4
Bi2O3
C6H12O3

9. GROUP STUDY PROBLEM #12
______ Which contains the larger number of MOLES of atoms?
a) g HBr b) 25.0 g C6H11O6 c) g of Br2
______ 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?
______ A compound contains 92.3% C and 7.7% H and has a molar mass of 78.0 g/mol. Determine the molecular formula.