1. Empirical and Molecular Formulas
Section 11.4
Chemistry

2. Objectives
Explain what is meant by the percent composition of a compound.
Determine the empirical and molecular formulas for a compound from mass percent and actual mass data.

3. Key Terms
Percent composition
Empirical formula
Molecular formula

4. Percent Composition Percent by mass of each element in a compound.
Mass of Element x 100
Mass of Compound

5. Example Calculate percent composition of H in H2O
Molar mass of water: g/mol
Determine mass of H in 1 mol of H2O

6. Mass of H 1.01 x 2 = 2.02 g H in 1 mol water Atomic mass of H
from periodic table
Number of H in H2O

7. Percent Composition of H
2.02 g of H x 100 = 11.2% H
18.02 g of H2O

8. Empirical Formula
Formula for a compound with the smallest whole-number ratio of elements.
Percent composition can be used to find the chemical formula.

9. Empirical Formula When given percent composition, assume:
The total mass of the compound is 100 g
The percent composition of the element is equal to the mass in grams of the element.

10. Example A compound has a percent composition of 40.05% S and 59.95% O.
So in 100 g of the compound, g are S and g are O.
Find the amount of mol for each element.

11. Empirical Formula 40.05g S x 1 mol S = 1.249 mol S 32.07g S
59.95g O x 1 mol O = mol O
16.00g O

12. Empirical Formula
The element with the smallest number of mol gets the subscript 1.

13. Empirical Formula S has a subscript 1
Then divide the mol O by the mol S
3.747 mol O/ mol S = 3 mol O
Then write your empirical formula using your subscripts: SO3

14. Practice Problems
Pg. 333

15. Molecular Formula
Formula that specifies the actual number of atoms of each element in one molecule or formula unit of a substance.
n = molecular formula mass
empirical formula mass

16. Example
Compound is composed of 40.68% carbon, 5.08% hydrogen, and 54.24% oxygen and has a mass of g/mol. Determine the empirical and molecular formulas for succinic acid.

17. Example 40.68 g C x 1 mol C = 3.387 mol C 12.01 g C
5.08 g H x 1 mol H = 5.04 mol H
1.008 g H
54.24 g O x 1 mol O = mol O
16.00 g O

18. Example
3.387 mol C/ = 1 mol C
5.040 mol H/ = 1.49 = 1.5 mol H
3.390 mol O/ = = 1 mol O
Ratio of C : H : O = 1 : 1.5 : 1

19. Example
Empirical Formula: C2H3O2

20. Example n = molecular formula mass empirical formula mass
To find molecular formula, calculate n.
n = molecular formula mass
empirical formula mass
Molecular mass is in the problem!
Calculate molar mass of empirical

21. Example
n = = 2
59.04
Multiply the subscripts of the empirical by n to find the molecular formula.
Molecular Formula: C4H6O4

22. Practice Problems
Pg 335

23. Homework
Section 11.4 Problems on page 877