1. Percent Composition Can be calculated if given:
masses of elements in compound OR
the chemical formula

2. Percent Composition Can be used to:
 calculate the mass of elements in a compound
 determine the empirical formula of a compound
 determine the molecular formula of a compound

3. Empirical Formula shows the simplest mole ratio of the elements.
CO is a 1:1 ratio of carbon to oxygen
H2O is a 2:1 ratio
CO2 is a 1:2 ratio
Empirical formulas can’t be reduced.

4. Molecular Formula shows the actual number of atoms in a molecule.
The molecular formula for hydrogen peroxide is H2O2. Its empirical formula would be HO.
Often the molecular formula is the same as the empirical formula: H2O, CO2

5. Empirical? CH4O C2H6 C3H10O C6H6O2 yes, cannot be reduced further
no, empirical would be CH3
C3H10O
yes
C6H6O2
no. What would empirical be?
C3H3O

6. Calculating Empirical Formulas
A chemist with an unknown compound can easily figure out its percent composition, but it is much more meaningful to know its formula.
EXAMPLE: What is the empirical formula for a compound that is 25.9% nitrogen and 74.1% oxygen?

7. Method Write the mass (g) of each element in the compound. So….
25.9% N = 25.9g
74.1% O = 74.1g

8. 2. Convert the mass of each element to moles.
N = 25.9g = mol
14.0g/mol
O = 74.1g = mol
16.0g/mol

9. Calculate the simplest whole number ratio by dividing the number of moles by the smallest number of moles.
1.85 : = :
(If the result is not within 0.1 of a whole number, multiply all numbers by a whole number)
2 ( 1 : 2.5) = : 5

10. Write the empirical formula. N2 O5
For inorganic compounds, write the most positive element first.
For organic compounds, write C first, H second and all others alphabetically.

11. A special present just for you……..
Page 135, Problems #20 & 21

12. Molecular Formula
Given the empirical formula and the gram formula mass (gfm) OR
Given the percent composition and the gram formula mass (gfm)

13. Example #1
Calculate the molecular formula for NaO having a gfm of 78g.
 Determine the efm (empirical formula mass).
NaO = 23.0g g = 39.0
Divide the efm into the gfm.
= 2
39.0
This is the conversion factor used to determine the molecular formula.
Na2O2

14. Example #2
Find the molecular formula for a compound having a composition of 58.8% C, 9.8% H and 31.4% O and a gmm of 102g/mol.
Determine the mass of each component.
C = 102g/mol x 58.8% = 60.0g/mol
H = 102g/mol x 9.8% = 10.0g/mol
O = 102g/mol x 31.4% = 32.0g/mol

15.  convert to moles
C = 60.0g/mol = 5
12.0g
H = 10.0g/mol = g
O = 32.0g/mol = 2
16.0g

16.  Use moles as subscripts for components of compound
C5H10O2
Check the gmm of this compound…does it equal 102.0g/mol?
5(12.0) + 10(1.0) + 2(16.0) = 102.0g/mol
YES!

17. Oh Yeah! And there’s more…
And Now…..
Oh Yeah! And there’s more…
Page 136, Problems #22 & 23

18. Now Try page 139, #41 44