1. Chapter 3
Stoichiometry

2. Mass Spectrometer
Most accurate method currently available for comparing the masses of atoms
See process in chapter 3 for a more accurate description

4. Average Atomic Mass (Isotope) Calculation
Move decimal of percent abundance two to the left
Multiply resulting number by its given, corresponding mass
Add the products
This gives us the molar mass of one mole of a particular element

5. Isotope Example
If carbon-13 ( amu) represents 35% natural abundance and carbon-12 ( amu) represents the other 65%, what is the average atomic mass?
Answer: ( X 0.35) + ( X 0.65) = amu

6. Keep in mind…
You can be given different parts of an isotope problem and be asked to solve for a new variable. Example:
The average atomic mass of a mixture of 75% carbon-12 (11.999amu) and carbon-13 is What is the mass of the carbon-13?
Carbon
Carbon X = X
AVERAGE ATOMIC MASS-->12.01
X = (solve for X)
X=12.04amu

7. Mole Equal to 6.022 X 1022 units aka Avogadro’s number
Molar Mass is found by adding the average atomic masses (found on the periodic table) of each element in a compound
Ex: molar mass of calcium chloride (CaCl2) is Ca: X 1 = 40.08
Cl: X 2 = 70.90
TOTAL = g/mol

8. Mole Roadmap!
You will not be given a mole roadmap anymore…

9. Calculations…
You can not use a calculator on a portion of the AP exam. Watch out for easy ratios!
Wait to round off your answer UNTIL THE END!

10. Problem Solving
*There is not always a cut and dry way to do a problem…here are significant things to consider:
Where are we going?
How do we get there?
Does it make sense?

11. Percent Composition Part (individual element masses)_ X 100
Whole (molar mass of compound)
Percents should add up to 100
Again, watch out for different givens/variables!
Example: Calculate the mass percent of each element in glucose, C6H12O6
Answer: 40.00%C, 6.71%H, 53.29%O

12. Using Percent Composition
If you know that an element makes up a certain percent of a compound, and how much there is of the compound, you can calculate the amount of a specific element.
Example: If carbon makes up 42.2% of a 435g sample of a compound (you can calculate this percent this using percent composition too), how much carbon does it contain?
435g X = 184g Carbon
SAME AS: X = (solve for X)

13. Empirical/Molecular Formula
Empirical: Smallest whole-number ratio of a chemical formula (ex. CaCl2)
Molecular: Some multiple of the empirical formula (can be the same or __ times as big)

14. Empirical Formula Steps
Assume grams (switch from % -> g)
Convert each to moles (use the molar mass and roadmap calculation)
Divide all by the smallest
Need whole numbers - ONLY very close numbers can be rounded (*NEW: multiply all by whatever number it takes to get whole numbers - may need 3/4/5/etc)
See examples in your book

15. Molecular Formula Steps
EFM (empirical formula mass - use P.T.)
Given molar mass/EFM
Multiply the subscripts of the given empirical formula by the previous answer (from #2)
See examples in your book

16. Chemical Equations
Reactants are on the left of the arrow, products are on the right of the arrow
Law of Conservation of Mass: all atoms must be accounted for among both the reactants and products
Skeleton equation gets balanced using coefficients (NOT subscripts)
(s) = solid, (l) = liquid, (g) = gas, (aq) = aqueous/dissolved in water

17. Balancing Hints H2O can split into H and OH Do oxygen last
If elements show up in more than one compound on either side, break them up further

18. Stoichiometry Chemical equation MUST be balanced!
Mole to mole ratio connects moles of two different elements/compounds in a chemical reaction and is found from the coefficients
? mole trying to find
? mole trying to cancel
? Determined by the coefficients in the balanced equation
To use the roadmap, units must be LITERS, GRAMS, OR ATOMS/MOLECULES/R.P./FORM UNITS
First step is to convert to MOLES

19. Limiting Reagent
In a chemical reaction, the reagent that runs out first. The other(s) is the excess reagent.
Determines how much (mole or g) of the first product can be formed with each reactant
Usually you will be given at least TWO numbers for limiting reactant problems
Often, it is assumed that you have to find the LR to do a problem (ex. how much excess remains)
Start by using the roadmap to go to MOLES

20. L.R. Example
17.80 g of sodium hydroxide reacts with g of phosphoric acid to give sodium phosphate and water. How many grams of the excess reactant remain unreacted?
Answer: 0.86 g H3PO4 excess

21. Percent Yield Actual Yield X 100% Theoretical Yield
The closer to 100%, the more efficient the reaction
Units for actual and theoretical must match!
Actual is generally given to you, theoretical is usually calculated (use LR if possible)