1. From percentage to formula
Empirical Formula
From percentage to formula

2. Calculating Empirical Formula
Pretend that you have a 100 gram sample of the compound.
That is, change the % to grams.
Convert the grams to mols for each element.
Divide each number by the least number.
Multiply the result to get rid of any fractions.
Write the number of mols as a subscript in a chemical formula.

3. Example
Calculate the empirical formula of a compound composed of % C, % H, and %N.
Assume 100 g so
38.67 g C x 1mol C = mole C gC
16.22 g H x 1mol H = mole H gH
45.11 g N x 1mol N = mole N gN

4. If we divide all of these by the smallest
one it will give us the subscripts for the empirical formula
3.222 mol C = mol N
16.09 mol H = mol N
3.222 mole N = 1
3.222 mol N
Empirical formula: CH5N

5. Learning Check EF-5
Aspirin is 60.0% C, 4.5 % H and 35.5 O. Calculate its simplest formula. In 100 g of aspirin, there are g C, 4.5 g H, and g O.

6. Solution EF-5 60.0 g C x ___________= ______ mol C
4.5 g H x ___________ = _______mol H
35.5 g O x ___________ = _______mol O

7. Solution EF-5 60.0 g C x 1 mol C = 5.00 mol C 12.0 g C
4.5 g H x mol H = mol H
1.01 g H
35.5 g O x 1mol O = mol O
16.0 g O

8. 5.00 mol C = ________________ ______ mol O
Divide by the smallest # of moles.
5.00 mol C = ________________
______ mol O
4.5 mol H = ________________
2.22 mol O = ________________
Are are the results whole numbers?_____

9. Are the results whole numbers? no
Divide by the smallest # of moles.
5.00 mol C = ___2.25_=2 ¼ = 9/4
2.22 mol O
4.5 mol H = ___2.00__
2.22 mol O = ___1.00__
Are the results whole numbers? no

10. Finding Subscripts
A fraction between 0.1 and 0.9 must not be rounded. Multiply all results by an integer to give whole numbers for subscripts.
(1/2) x 2 = 1
(1/3) x 3 = 1
(1/4) x 4 = 1
(3/4) x 4 = 3
(1/5) x 5 =

11. C9H8O4 Multiply everything x 4 to clear the denominator
C: 9/4 mol C x 4 = 9 mol C
H: 2.0 mol H x 4 = 8 mol H
O: mol O x 4 = 4 mol O
Use the whole numbers of mols as the subscripts in the simplest formula
C9H8O4

12. Types of Formulas Empirical Molecular (true) Name CH C2H2 acetylene
The formulas for compounds can be expressed as an empirical formula and as a molecular(true) formula.
Empirical Molecular (true) Name
CH C2H2 acetylene
CH C6H6 benzene
CO2 CO2 carbon dioxide
CH2O C5H10O5 ribose

13. An empirical formula represents the simplest whole number ratio of the atoms in a compound.

14. It is not just the ratio of atoms, it is also the ratio of moles of atoms
In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen
In one molecule of CO2 there is 1 atom of C and 2 atoms of O

15. The molecular formula is the true or actual ratio of the atoms in a compound.

16. Learning Check EF-1 1) CH2O 2) C2H4O2 3) C3H6O3
A. What is the empirical formula for C4H8?
1) C2H ) CH ) CH
B. What is the empirical formula for C8H14?
1) C4H7 2) C6H ) C8H14
C. What is a molecular formula for CH2O?
1) CH2O 2) C2H4O ) C3H6O3

17. Learning Check EF-2
If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain.
1) SN
2) SN4
3) S4N4

18. Solution EF-1 1) CH2O 2) C2H4O2 3) C3H6O3
A. What is the empirical formula for C4H8?
2) CH2
B. What is the empirical formula for C8H14?
1) C4H7
C. What is a molecular formula for CH2O?
1) CH2O 2) C2H4O ) C3H6O3

19. Solution EF-2
If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain.
3) S4N4
If the actual formula has 4 atoms of N, and S is related 1:1, then there must also be 4 atoms of S.

20. Empirical and Molecular Formulas
molar mass = a whole number = n
simplest mass
n = 1 molar mass = empirical mass
molecular formula = empirical formula
n = 2 molar mass = 2 x empirical mass
molecular formula =
2 x empirical formula
molecular formula = or > empirical formula

21. Learning Check EF-3 2) C6H8O6 3) C9H12O9
A compound has a formula mass of and an empirical formula of C3H4O3. What is the molecular formula?
1) C3H4O3
2) C6H8O6
3) C9H12O9

22. Solution EF-3
A compound has a formula mass of and an empirical formula of C3H4O3. What is the molecular formula?
2) C6H8O6
C3H4O3 = g/EF
176.0 g =
88.0

23. Learning Check EF-4 1) C7H6O4 2) C14H12O8 3) C21H18O12
If there are g of O in the molecular formula, what is the true formula if the EF is C7H6O4?
1) C7H6O4
2) C14H12O8
3) C21H18O12

24. Solution EF-4 3) C21H18O12 192 g O = 3 x O4 or 3 x C7H6O4
If there are g of O in the molecular formula, what is the true formula if the EF is C7H6O4?
3) C21H18O12
192 g O = 3 x O4 or 3 x C7H6O4
64.0 g O in EF

25. Homework
Worksheet C: #1-7