1. Stoichiometry
Molar mass, Percent composition, Moles, Conversions, Empirical formulas, Molecular formulas

2. 1 mole = 6.022 x 10 23 atoms = molar mass
The Mole
SI unit for amount
A counting unit for measuring large quantities of small items
Atoms
molecules
Particles
1 mole = x atoms = molar mass

3. Molar Mass For compounds
The amount of mass (grams) in one mole of a substance
For elements
Average atomic mass  periodic table
For compounds
Must add up each element’s mass
Subscripts are factored in by multiplication
Units are g/mol
May be called formula mass or molecular mass

4. Steps to solve Find elements on periodic table Write down the mass
Multiply the mass by the subscript
Add together

5. Sample: Determine Molar Mass
Calculations are FUN!
40.08 g/mol
g/mol Ca Fe
NaCl 
Na: g/mol
Cl: g/mol
g/mol

6. Molar Mass (cont.) 5. Na2O Fe: 55.847 x (2) = 111.694 g/mol
4. Fe2O3
5. Na2O
Fe: x (2) = g/mol
O: x (3)= g/mol
g/mol
Na: x(2) =  g/mol
O: = g/mol
g/mol

7. Percent Composition Steps to solve Find the total mass of the COMPOUND
Take the mass of each element and divide it by the total mass
Multiply by 100

8. Sample: Calculate the percent Composition
C: g mol
g mol
O: g mol
.2730
100
27.3 %
.727
100
72.7%
CO2 C: g/mol O: (2) g/mol = g/mol g/mol

9. Sample 2 .327 .653 H2SO4 H: (2)1.008 g/mol = 2.016 g/mol
S: g/mol
O: (4) g/mol = g/mol
g/mol
H: g mol
g mol
S: g mol
O: g mol
.0206
100
2.06 %
.327
100
32.7 %
.653
100
65.3%

10. Composition of Hydrates
Hydrate: crystal contains water within
Water can fit into the salt crystal
Happens in fixed ratios
Each salt that forms a hydrate has a DIFFERENT water ratio
Each salt crystal is unique
Anhydrous: water has been removed
Achieved through drying
Steps to solve
Calculate the total mass of the compound INCLUDING THE water
Divide the water mass by total mass
Multiply by 100

11. Sample Problem: Hydrate
That was nothing. Time for Coco!
Sample Problem: Hydrate
What percentage of water is found in CuSO4 •5H2O
Cu: g mol
S: g mol
O: (4) g mol = g mol
H2O: (5)18.01 g mol = g mol
g mol
g mol
g mol
.361
100
36.1% H2O

12. The Mole conversion ÷ by 6.022 x 10 23 Mass Particles (grams) Volume
Ions
Atoms
Molecules
Formula units
÷ by molar mass
X by x 10 23
Mole
X by Molar mass
÷ by x 10 23
÷ by 22.4 L
x by 22.4 L
Volume
(Liters)
Gases at STP

13. Stoichiometric Conversions: Mass to Mass
You can relate any unit to another by this method
Relate one unit to another
conversion factors
Steps to convert
Identify what you start with
Determine what units you end in
Might have to do side calculations
Set up conversion factor
Evaluate
Do the Math!

14. Example 28 grams CO2 _______ moles CO2 .636
How many moles are in 28 grams of CO2?
1 mole CO g CO2
28 grams CO2
_______ moles CO2
.636
C: g/mol
O: (2) g/mol = g/mol
g/mol

15. Empirical Formula Steps to solve Steps to solve When given percentages
Change percent sign to grams (NO MATH)
Convert masses to moles
Using conversions
Re-divide ALL moles amounts by the smallest mole amount
Multiply if not whole numbers
Write compound with subscripts
When the elements are in the smallest whole number ratio within compound
Steps to solve
When given a compound
Divide out by the common multiple

16. Example: What’s my Empirical Formula
Piece of Cake!
C6H6
C8H18
       
C2H6O2
    
X39Y13    CH
C4H9
CH3O
X2Y

17. Example: with percent Na2SO3 1.587 moles 0.792 moles = 2.01 2
A compound is found to contain 36.48% Sodium, 25.41% Sulfur, and 38.11% Oxygen. Find its empirical formula.
Na2SO3
Element
Percent
Mass
Convert to moles
Re-divide
Subscripts Na
36.48%
36.48 g
36.48 g x 1 mole g = moles
1.587 moles moles = 2.01 2 S
25.41%
25.41 g
25.41 g x 1 mole g = moles
0.792 moles moles = 1.00 1 O
38.11 %
38.11 g
38.11 g x 1 mole g = moles
2.382 moles moles = 3.01 3

18. So many compounds with the same Empirical formula
Molecular Formula
Multiple of an empirical formula
So many compounds with the same Empirical formula

19. Steps to solve Complete an empirical formula process if needed
Find the mass of the EMPIRICAL FORMULA
Divide the Empirical formula mass by the molecular mass
Molecular mass is normally given in the problem
Answer is the common multiple
Multiple the SUBSCRIPTS by the common multiple

20. All done! Time to go sledding! Yippee!!
Example
A compound with an empirical formula of C4H4O and a molecular mass of grams. What is the molecular formula of this compound?
C: 4 ( g mol ) = g mol
H:4 ( g mol ) = g mol
O: g mol
g mol
136 g g =1.997
C4H4O x 2 = C8H8O2