1. NaCl
H2O
C6H12O6
Chemical Formulas
NaHCO3

2. What information is in a chemical formula?
A ratio of ATOMS. In 1 molecule of H2O there are 2 atoms of H and 1 atom of O.
A ratio of MOLES. In 1 mole of H2O there are 2 moles of H and 1 mole of O.
Formulas are NOT ratios of grams. In 1 mole H2O there are NOT 2 g of H for every g of O.

3. How do chemists determine the formula of a compound?
Step 1: Elemental Analysis
Identify ELEMENTS present
Determine % COMPOSITION by MASS for each element.

4. How do chemists determine the formula of a compound?
Step 2: Determination of empirical formula
Empirical Formula = SIMPLEST Ratio of atoms in a formula

5. IONIC COMPOUND (+ metal ion/ nonmetal – ion or polyatomic ion)
Empirical Formula = “Final” Formula
Ions bond together in a lattice – giant “molecule”
Actual formula would be to difficult write, e.g. Na 1023 Cl 1023
So formula is written as simplest ratio, NaCl

6. Covalent Compound (2 or more nonmetals sharing pairs of electrons exist as molecules.
Empirical Formulas (simplest ratio) may or may not be true ratio
True or Actual Formula is called MOLECULAR Formula
Example:
H2O: empirical = molecular
Benzene : empirical ≠ molecular
Empirical = CH Molecular C6H6

7. MOLECULAR FORMULA DETERMINATION
Requires empirical formula + MOLAR MASS
Molar Mass is Determined by Experiment (in problems, molar mass must be provided).

8. Empirical Formula vs Molecular Formula
Empirical Formula = SIMPLEST ratio of atoms in a compound.
Molecular Formula = ACTUAL ratio of atoms in a compound.
Ionic compound → Empirical Formula is “actual” formula
Molecular compound (covalent bonds) → Molecular formula is usually different than empirical formula.
Determining empirical formula is still a key step in the process of the determining the formula of a Molecular Compound.

9. Examples of Empirical and Molecular Formulas for covalently bonded compounds:
Carbon dioxide: Empirical = CO2 Molecular = CO2
Benzene: Empirical CH Molecular = C6H6
Glucose:
Molecular = C6H12O6
Empirical =
CH2O

10. % Composition by Mass Mass of Element x 100% Total Mass of Compound
Definition of %:
Part out of ;
To calculate % : x 100 %
Definition of Mass % =
100
Part
Total
Mass of Element x 100%
Total Mass of Compound
Mass % of each element must add up to 100

11. Example #1: % from mass data
What is the % by mass of a compound containing 28.0 g Fe and 8.00 g O?
% Fe =
% O =
Mass of Fe x 100 %
Total Mass of Compound (Fe + O)
28.0 g Fe
x 100% = 77.8%
= 36.0
8.00 g O
x 100% = 22.2%
36.0 g Fe + O

12. Example #2: % from a chemical formula
What is the % by mass of each element in water, H2O?
Step 1 – Calculate grams of each element and molar mass:
H: 2 mol x 1.01 g/mol = g
O: 1 mol x g/mol = g
18.02 g
Step 2 – Calculate %
H: / x 100 % = 11.2 %
O: 16.00/ x 100 % = 88.8 %
2nd method to calculate % O = 100 – 11.2 = 88.8%
HW Self check 8.7

13. Determining Empirical Formula from Elemental Analysis Data
Example #1: Elemental Analysis (% by mass) for a compound is:
Al: % F: 67.87%
Recall = Simplest Ratio of Atoms: AlxFy, where x,y = integers
Step 1: Find masses of elements. Since data in mass %, assume 100 g sample; then can replace % with g.
Assume 100 g sample:
32.13% of 100 = g Al
67.87% of 100 = g F

14. Step #2: Convert to moles(
1 mole Al )
Al: g
F: g
= moles Al
26.98 g Al
1 mole F ) (
= moles F
19.00 g F
Can we leave final Answer as: Al F ?
No! Final Answer must be simplest ratio!

15. Step 3: Divide by smallest # of moles:
Al: moles F: moles F
3.572 > moles so divide both by 1.191
Al: ÷ = 1
F: ÷ = 3
Final Answer: AlF3
HW: Self check 8-8

16. Determining Empirical Formula from Elemental Analysis Data
Example #2: An oxide of aluminum is formed by the reaction of g of Al with g of oxygen. Calculate the empirical formula of the compound.
Al: g O: g
Recall = Simplest Ratio of Atoms:
AlxOy, where x,y = integers
Step 1: Find masses of elements: Given here

17. Step #2: Convert to moles(
1 mole Al )
Al: g
O: g
= moles Al
26.98 g Al
1 mole O ) (
= moles O
16.00 g O
Can we leave final Answer as: Al O ?
No! Final Answer must be simplest ratio!

18. Step 3: Divide by smallest # of moles:
Al: moles O: moles O
> moles so divide both by
Al: ÷ = 1
O: ÷ = 1.5
Can’t leave answer as: AlO1.5

19. 1.500 moles O x 2 = 3 moles O 1.000 moles Al x 2 = 2 moles Al
Step 4: Multiply the numbers from step 3 by the smallest integer that convert them both to whole numbers.
1.5 = 3/2 so x 2 will get rid of 2 in denominator
1.500 moles O x 2 = 3 moles O
1.000 moles Al x 2 = 2 moles Al
Final Answer: Al2O3
HW: Self check 8-9 and 8.10

20. Common Decimals/Fractions in our problems
1/4
1/3
1/2
2/3
3/4
0.25
0.33
0.50
0.67
0.75

21. Moles X
Moles Y
Multiply by
Final Answer
1.00
2.00 NA
XY2
1.25
1.00 4
X5Y4
1.33
1.00 3
X4Y3
1.50
1.00 2
X3Y2
1.67
1.00 3
X5Y3
X7Y4
1.75
1.00 4
1.99
1.00
Round to 2
X2Y
1.89
1.00
Error check your math