1. A.P. Ch. 3 Review Work
Stoichiometry

2. Atomic Mass Average of isotope masses based on their abundance
Ex. Carbon has atomic mass of amu
12C has mass of amu and 98.89%
13C has mass of amu and 1.11%
( x 12.00) + ( x ) = amu
Diamond (pure Carbon) in Kimberlite

3. Calculate the Atomic Mass
Copper has two isotopes: 65Cu is 30.91%, 63Cu is 69.09%

4. Mole
Like a dozen represents a certain number of an object, a mole represents a certain amount of particles
A mole is equal to 6.02x1023 representative particles
Ex. 1 mole of carbon contains 6.02x1023 atoms
1 mole of H2O contains 6.02x1023 molecules
All of the atomic masses on the periodic table are equivalent to 1 mole of those elements in grams

5. Molar Mass The mass of one mole of any substance
For compounds, add all of the atomic masses of every element together
Example: CH4
12.01 g/mol + 4(1.008 g/mol) = g/mol

6. Percent Composition The % of each element in a compound
Take total masses of each element, divide by molar mass of compound, multiply by 100

7. Empirical/Molecular Formulas
Molecular formula: shows how many of each element are present in a compound
Glucose is C6H12O6
Empirical formula: shows smallest whole number ratio
Glucose is CH2O

8. Determining Chemical Formulas
Empirical Formula:
Starting with % composition of each element
Change % to mass (10% of C is 10 grams C)
Convert each mass to moles using molar mass
Divide each elements # of moles by smallest #
If you get whole numbers you are done, if not multiply each by a factor to get whole numbers
74.8% C, 25.2% H = 74.8g C, 25.2g H =
6.23 mol C, 25.0 mol H
6.23/6.23 = 1 C, 25.0/6.23 = 4 H  CH4

9. Determine Empirical Formula
87.4% Nitrogen, 12.6% Hydrogen

10. Molecular Formula: Using emp. Form., molar mass
Take molar mass of emp. form
Divide molar mass by emp. form mass
Multiply Empirical formula coefficients by that factor
Mol. Form Mol. Form Mass
Emp. Form Emp. Form Mass =
C3H6O3
____X____ ____90 g____
CH2O ? =
30 g

11. Calculate Molecular Formula
Empirical formula: NH2
Approximate molar mass: 32 g/mole

12. Chemical Equations CH4 (g) + O2 (g)  CO2 (g) + H2O (g) 2 2
Reactants
Products
***Note: Remember back to Dalton’s 4th theory, that reactions are simply the rearranging of atoms ***
And, according to the conservation of mass, the masses and elements must be the same before and after
Balancing Equations: to ensure equal atomic quantities
Start with atoms appearing least # of times per side
Put coefficients in front of entire molecule to balance
If you have to use a half # to balance an atom, double the entire equation at the end

13. Half-Number Example
C2H6 + O2  CO H2O

14. Stoichiometric Conversions
Typically involve using a balanced chemical equation to change from a reactant amount to a product amount
Steps (Usually):
Convert any given amounts to moles
Determining limiting reactant (if any)
Use limiting reactant moles to convert to moles of product using mole/mole ratio
Convert moles product to desired units

15. Limiting/Excess Reagents
Limiting reagent: reactant that runs out first, must use to determine how much product can theoretically be made
Excess reagent: left-over reactant
Can determine which is which by comparing given moles to stoichiometric ratios in equation
Percent Yield: experimental mass x 100
theoretical mass

16. Example
Using 2 H2 + O2  2 H2O how many grams of H2O can be made with 5.00 grams of H2 and 32.0 grams of O2? If the experimental mass of H2O is 32.5 grams, what is the percent yield?