1. Empirical Formula
The empirical formula gives the lowest whole-number mole ratio of the atoms or moles of the elements in a compound.

2. Examples:
1) Glucose – Molecular formula C6H12O6.
Empirical formula

3. Examples:
1) Glucose – Molecular formula C6H12O6.
Empirical formula CH2O.

4. Examples:
1) Glucose – Molecular formula C6H12O6.
Empirical formula CH2O.
2)Succinic acid–Molecular formula C4H6O4.
Empirical formula

5. Examples:
1) Glucose – Molecular formula C6H12O6.
Empirical formula CH2O.
2)Succinic acid–Molecular formula C4H6O4.
Empirical formula C2H3O2.

6. Examples:
1) Glucose – Molecular formula C6H12O6.
Empirical formula CH2O.
2)Succinic acid–Molecular formulaC4H6O4.
Empirical formula C2H3O2.
3) Sulfuric acid --Molecular formula H2SO4.
Empirical formula

7. Examples:
1) Glucose – Molecular formula C6H12O6.
Empirical formula CH2O.
2)Succinic acid–Molecular formulaC4H6O4.
Empirical formula C2H3O2.
3) Sulfuric acid --Molecular formula H2SO4.
Empirical formula H2SO4.

8. Practice: Determine the Empirical Formula of Benzopyrene, C20H12.
Find the greatest common factor (GCF) of the subscripts:
factors of 20 = 1, 2, 4, 5,
10, 20
factors of 12 = 1, 2 3, 4, 6, 12
GCF = 4
Divide each subscript by the GCF to find the empirical formula:
C20/4H12/4 =
C5H3
Empirical formula
(C5H3)4 =
C20 H12

9. 1) What is the empirical formula for a compound which contains 0
1) What is the empirical formula for a compound which contains g of iron, g of sulfur and g of oxygen?
1st– Determine the # of moles of each element:
Fe =
g x
= mol (Fe)
S =
g x
= mol (S)
O =
g x
= mol (O)

10. 2nd– Calculate the simplest ratio (dividing
each value by the smallest one):
Fe =
= 1 mol (Fe)
FeSO3
Empirical Formula
S =
= 1 mol (S)
O =
= 3 mol (O)

11. 2) Determine the empirical formula for a
compound that contains %
magnesium, % chlorine, and
57,34 % oxygen?
Assuming 100 g sample of the compound:
10.89 % represent g (Mg).
31.77 % represent g (Cl).
57.34 % represent g (O).

12. 1st– Determine the # of moles of each element:
(Mg = % ; Cl = % & O = %)
Mg =
10.89 g x
= mol (Mg)
Cl =
31.77 g x
= mol (Cl)
O =
57.34 g x
= mol (O)

13. 2nd– Calculate the simplest ratio (dividing each
value by the smallest one):
(Mg = ; Cl = & O = 3.59)
= mol (Mg)
= 1 mol (Mg)
= mol (Cl)
= 2 mol (Cl)
= mol (O)
= 8 mol (O)
Empirical Formula
MgCl2O8

14. 3) A laboratory analysis of aspirin determined the following mass percent composition carbon (60.00 %), hydrogen (4.48 %) and oxygen (35.53 %). Calculate its empirical formula.
C =
= mol (C) = =
2.249 mol (C)
H =
= 4.44 mol (H) = =
2.00 mol (H)
O =
= mol (O) = =
1.000 mol (O)

15. C9H8O4 (Empirical Formula) C = 2.249 mol DO NOT round it !!
(between .3 & .7)
H =
2.00 mol
it should be multiplied by the smallest
number which converts it into a whole number.
O =
1.000 mol
2.249 x 2 =
4.498
(It is not good enough)
2.249 x 3 =
6.747
(It is not good enough)
2.249 x 4 =
8.996
(It is pretty good)
≈ 9
All three numbers should be multiplied by 4
C= x 4 =
H= x 4 =
O= x 4 = 9
C9H8O4 8
(Empirical Formula) 4