1. DO NOW Pick up notes. Get out your periodic table and calculator.
Molar Quantities lab due Tuesday.

2. EMPIRICAL FORMULA
The simplest whole number ratio of elements in a compound is called the EMPIRICAL FORMULA.
In contrast, the molecular formula of a compound is the actual number of atoms of each element in the compound.
It can be calculated from the mole ratio or from percent composition data.

3. EXAMPLES
Molecular Empirical Formula Formula C6H12O6 CH2O CO CO P4O10 P2O5

4. MOLE RATIO EXAMPLE
What is the empirical formula for a compound if a g sample contains g of calcium and 1.60 g of chlorine?
Step One: Determine the number of moles of Ca and Cl. (You must first determine the atomic mass of each.)
Ca 0.900g Ca mole Ca = mol
40.08g Ca
Cl 1.60g Cl mole Cl = mol
35.45g Cl

5. MOLE RATIO EXAMPLE
Step Two: Obtain the smallest ratio for the set of moles. Divide both numbers of moles by the smaller number of moles. The smallest is mol.
Ca mol = Cl mol = 2.00
mol mol
These should be close to whole numbers MOST of the time.

6. The empirical formula is CaCl2
MOLE RATIO EXAMPLE
Step Three: Look at the ratio of the two numbers. Round off to whole numbers (there are exceptions to this).
Ca becomes Ca 1
Cl becomes Cl 2
The empirical formula is CaCl2

7. STEPS Convert mass to mole for each element given.
Divide the smallest mole into itself and all others.
Determine the empirical formula.

8. PRACTICE
A. Find the empirical formula of the compound with 63.0g Rb and 5.90g O.

9. ANOTHER WAY TO DO IT
Sometimes you are given percentages, not masses, and asked to determine the empirical formula.
If this is the case, you can change the percentages directly to grams because you assume a 100 g sample (100%)

10. PERCENT COMPOSITION
A compound has a percentage composition of 40.0% C, 6.71% H and 53.3% O. What is the empirical formula?
Assume that there is 100 g (from 100%). In a 100g sample, there would be 40.0g C, 6.71g H, and 53.3g O. Then change the quantities to moles.

11. STEP ONE C 40.0 g C 1 mole C = 3.33 mol C 12.01g C
H 6.71 g H mole H = mol H
1.01g H
O 53.3 g O mole O = mol O
16.00g O

12. STEP TWO
To obtain the smallest ratio, divide both numbers of moles by the smaller number of moles (3.33 mol).
C mol = 1.00 H mol = 1.99
3.33 mol mol
O mol = 1.00
3.33 mol

13. The empirical formula is CH2O
STEP THREE
Look at the ratio of the two numbers. Round off to whole numbers.
C becomes C 1
H becomes H 2
O becomes O 1
The empirical formula is CH2O

14. PRACTICE
A. What is the empirical formula for a compound that has 32.8% Cr and 67.2% Cl?

15. RATIOS ARE NOT WHOLE NUMBERS! What do you do?
What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S?

16. RULES TO FOLLOW If the ratio is: X.1 or X.9, round up or down
If the ratio is X.2 to X.8, determine what to multiply it by to get a whole number.
Examples:
X.3 x 3 = X.9 which can be rounded
X.5 x 2 = X.0 which is a whole number
X.4 x 5 = X.0 which is a whole number

17. TO DO
Percent Composition and Empirical Formula Practice handout is due tomorrow.
Molar Quantities lab is due Tuesday.