1. 4S6
Welcome To the
Chemistry Class

3. WHAT IS AN
EMPIRICAL
FORMULA?
The simplest ratio of the number of
atoms for each element in a compound.

4. EXAMPLE:
 The empirical formula for a glucose molecule
(C6H12O6) is CH2O. All the subscripts are
divisible by six.
C6 H12 O6
6 6 6
C H2 O

5. WHAT IS
MOLECULAR
FORMULA?
Formula which shows the actual number
of each type of atom in a molecule of a particular
compound

6. Molecular Formula Empirical Formula
C2H6 CH3
C6H12O6 CH2O

7. 5 STEPS TO GO….
 Step 1: Write the element symbol
 Step 2: Show the mass,or percentage, of
each element
 Step 3. Divide each by the R,A.M. of the
element
 Step 4. Divide each answer by the smallest
 Step 5: Get the simplest ratio

8. CALCULATION OF EMPIRICAL
FORMULA
Example
Analysis of an oxide of copper shows that it contains
3.175 g of copper and 0.4 g of oxygen. What is the
empirical formula of this compound?
Step 1. Write the symbols for the elements Cu O
Step 2. Show the mass,or percentage, of each element. 3.175 0.4
3.175 0.4
Step 3. Divide each by the R,A.M. of the element
63.5 16
= 0.05 0.025
0 .05 0 .05
Step 4. Divide each answer by the smallest
0.025 0.025
= 2 1
Step 5. Write the formula Cu2O

9. DRAW A TABLE:

10. Calculations for you to try.
1. Calculate the empirical formula for the compound which
contains 2.40g of carbon and 0.60g of hydrogen.
Element C H
Mass(g) 2.40 0.60
Number of moles 2.40/12 0.60/1
= 0.2 = 0.60
Ratio of moles of 0.2/0.2 0.60/0.2
atoms =1 =3
Simplest ratio 1 3
Empirical formula = CH3.

11. Analysis of a compound showed that it contained 40% calcium,
12% carbon and 48% oxygen by mass.
Calculate the empirical formula of the compound.
Element Ca C O
Percentage(%) 40 12 48
Number of 40/40 12/12 48/16
moles =1 =1 =3
Ratio of moles 1/1 1/1 3/1
of atoms =1 =1 =3
Simplest ratio 1 1 3
Empirical formula = CaCO3.

12. TRY THE EXERCISE ON
WORKSHEET
Na2O

13. NOTE: IF FINAL RATIO
1.7 : 1 => 2:1
1.4 : 1 => 1:1
0.5 : 1 => ½ : 1 => 1:2
1.5 : 1 => 3/2:1 => 3:2

14. 1.22 G OF PHOSPHORUS(P=31) COMBINE
WITH 0.95 G OF OXYGEN. WHAT IS THE
EMPIRICAL FORMULA FOR
PHOSPHORUS OXIDE?
Element P O
Mass(g) 1.22 0.95
Number of 1.22/31 0.95/16
moles = 0.04 = 0.0594
Ratio of moles 0.04/0.04 0.0594/0.04
of atoms =1 =1.5
= 3/2
Simplest ratio 2 3
Empirical formula = P2O3.

15. A COMPOUND CONSISTS OF 29.08% OF
SODIUM, 40.65% OF SULPHUR AND
30.27% OF OXYGEN. FIND THE
EMPIRICAL FORMULA OF THE
COMPOUND. [RAM: NA,23 ; S,32; O,16]
Element Na S O
Percentage 29.08 40.65 30.27
(%)
Number of 29.08/23 40.65/32 30.27/16
moles = 1.264 = 1.270 = 1.892
Ratio of 1.264/1.264 1.270/1.264 1.892/1.264
moles of =1 =1 =1.5
atoms =3/2
Simplest 2 2 3
ratio
Empirical formula = Na2S2 O3.

16. WATCH THE VIDEO..

17. REMINDER!!
 Do ALL the exercises (mole conversion).
 Revise the mole and chemical formula.

Will have a short quiz any time after
holiday.

18. AND….
Have a look on the chemistry blog
http://interestingchemistryworld.blogspot.com/

19. LASTLY…