1. Mrs. Crowley’s AP Chemistry Class Chapter 3
Reaction Types, Balancing Reactions, Empirical Formulas, Stoichiometry: LR, TY, left overs, %Yield and % error
Mrs. Crowley’s
AP Chemistry Class
Chapter 3

2. Reaction Types (the basics) We will learn MUCH more in ch. 4
Synthesis: A + B  AB
Decomposition: AB  A + B
Single replacement: AB + C  AC + B
Look at metal or nonmetal reactivity to determine if single element will react.
Double replacement: AB + CD  AD + CB
Look at products for a precipitate (a solid product), a gas, or a liquid (including water) to be formed to determine if the reaction will work.
Combustion: CxHx + O2  CO2 + H2O

3. Balancing Chemical Reactions
The law of conservation of mass, which states that matter can neither be created nor destroyed in an ordinary chemical reaction, must be upheld when writing a chemical equation.

4. Use coefficients to balance reaction:

5. Coefficients: add more compounds

6. Sandwich Factory
You just got hired to a sandwich factory. In order for you to keep your job, there are a few things you need to know about making sandwiches.

7. The Sandwich
Each sandwich needs to be put together with the following ingredients:
2 Slices of bread
4 slices of meat
1 slice cheese
1 piece of lettuce
2 slices of tomato

8. How many “complete” sandwiches can you make? What is left over?
Your Supplies
You need to make the maximum number of “complete” sandwiches that you can with the following supplies:
16 slices of bread
5 slices of cheese
19 slices of meat
10 pieces of lettuce
15 slices of tomato
How many “complete” sandwiches can you make? What is left over?
Best strategy is to figure out how many sandwiches each ingredient CAN make

9. Limiting Reagents in Chemistry
In chemistry, a limiting reagent/reactant is the substance that will limit the amount of product that can be formed in a reaction.
Example:
2 H O --> 2 H + O 2 2 2

10. 2 H O --> 2 H + O 2 2 2
2 moles of water will decompose to make 2 moles of hydrogen gas and 1 mole of oxygen gas
Coefficients = # moles or molecules/atoms/formula units
In the above BALANCED reaction, if I give you 2 moles of water, how many moles of H2 gas will you produce? How many moles of O2 will you produce?
What if I only give you 1 mole of water. How many moles of H2 gas will be produced? How many moles of O2 gas will be produced?

11. 2 H O --> 2 H + O 2 2 2
In the above BALANCED reaction, if I give you 2 moles of water,
how grams of H2 gas (as STP) can you make?
How many liters of O2 gas (at STP) can you make?
How many molecules of oxygen gas (at STP) can you make?

12. 2 Na + Cl --> 2 NaCl (s) 2 (g) (s) Sodium Chloride
Sodium chloride can be prepared by the reaction of sodium metal with chlorine gas.
If 6.70 moles of Na reacts with 3.20 moles of Cl2, what is the limiting reagent?
How many moles of NaCl can we make?
2 Na + Cl > 2 NaCl
(s) 2
(g)
(s)

13. This ratio MUST come from a balanced reaction!
The math: Part one
6.70 moles Na
What ratio goes here?
This ratio MUST come from a balanced reaction!

14. The ratio
6.70 moles Na
2 Moles NaCl
2 Moles Na

15. The math: part 2
3.20 moles Cl2
What ratio goes here?

16. The Ratio
3.20 moles Cl2
2 moles of NaCl
1 mole Cl2

17. Limiting reagent
Whichever REACTANT makes the least amount of NaCl is the “limiting reagent”
Which is the limiting reagent? Na or Cl2?
3.20 moles Cl2 produces 6.40 moles of NaCl
6.70 moles Na produces 6.70 moles of NaCl.

18. Maximum Produced
The maximum we can produce is whatever amount the “limiting” reagent can make. The maximum we can make is also sometimes called the "theoretical yield."
The most NaCl we can make is 6.40 moles NaCl.

19. Another example
2Na (s) + S (s)  Na2S (s)
If you have 14.2 g of sodium and 12.9 g of sulfur, which one is the limiting reagent and what is the maximum amount (grams) of sodium sulfide you can make?
You have 5 minutes…

20. The solution 14.2 g Na 1 mole Na 1 mole Na2S 78.1 g Na2S 23.0 g Na
The reactant that produced lesser amount is called the limiting reagent.
14.2 g Na
1 mole Na
1 mole Na2S
78.1 g Na2S
23.0 g Na
2 moles Na
1 mole Na2S
= 24.1 grams
Lesser amount is the theoretical yield, AKA: max we can make
12.9 g S
1 mole S
1 mole Na2S
78.1 g Na2S
32.1 g S
1 moles S
1 mole Na2S
31.4 grams

21. % yield vs. % error
In the lab, you will not always get what you were "supposed" to get…AKA: the theoretical or "accepted" value.
% yield is a # that reflects how close you got to the actual value. % error reflects how far away you were from the true value.
Remember: you can sometimes get MORE than what you were supposed to get…giving you a % Yield > 100%!
Note: there is no such thing as negative %error or negative % yield.

22. Example:
What would be the % Yield and % Error in the previous stoichiometry problem if you performed this reaction and actually obtained 25.4 grams of sodium sulfide?
Hint: you were supposed to get 24.1 grams.

23. One last question: How much excess do you have?
You used up sulfur to produce 24.1 grams of Na2S. Did you use all of it? NO.
You did use ALL of the 14.2 grams of Na, though. That’s all gone.
The question is: how much of the grams of Na2S did you actually use? How much is still left over?

24. Start a new problem…starting with the theoretical yield:
You didn’t use all of the 24.1 grams of S. Only 9.91 grams of it.
24.1 g Na2S
1 mole Na2S
1 moles S
32.1 g S
78.1 g Na2S
1 mole Na2S
1 moles S
=9.91 grams S

25. How much is left over? 24.1 grams of S (amount you started with)
grams of S (amount you actually used)

26. % composition of compounds
Mass of element
Mass of compound
Example: what is the % by mass of hydrogen in water?
X 100

27. Chemical formulas are really just Mole/Mole ratios!
In sodium chloride there is a 1:1 ratio between the Na atom and the Cl atom. In water, there is a 2:1 ratio between hydrogen and water.
These ratios are based not off of mass, but off of MOLE ratios.
If given the % composition of a compound, you CAN determine the formula of the compound. Here's how…

28. Example: Determine the % composition of sodium and chlorine in sodium chloride.
Look up their masses on the PTE:
% Na = g / ( g) x 100
= _______ % Na
% Cl = g / ( g) x 100
= _______ % Cl

29. Step 1: assume you have 100g of mysterious compound
Assuming you have 100 grams of the mystery compound makes your math EASY.
That means you have GRAMS of Chlorine and GRAMS of sodium in 100g of sodium chloride.

30. Step #2: change those masses to MOles
Chemical formulas are based off of mole to mole ratios. Not mass to mass ratios!
So you’ll need to convert those two masses into moles.

31. Step 2: mass to moles 39.34 g Na 1.711178773 moles Na 22.99 g Na
60.66 g Cl
moles Na
35.45 g Cl
You can reduce your answers to the correct # sig figs, but
I think you can see that you can now reduce that ratio to one to one (which is step #3: reduce ratios).

32. One more harder example: let’s see if you’ve got it!
What is the empirical formula for a mysterious compound that has the following mass spec information:
104.9gFe and 45.15g oxygen
You have a MASS ratio…you need MOLE ratios in order to determine the formula.

33. Skip step 1 and go straight to step 2: determine mole ratio
104.9 g Fe
1 mole Fe
= moles Fe
55.85 g Fe
45.15 g O
1 mole O
= moles O
16.00 g O

34. Step 3: Reduce ratios 1.878 moles Fe to 2.822 moles O
Not pretty…here's the strategy: divide BOTH values by the lowest # moles you've got:
1.878 moles Fe / = 1 mole Fe
2.822 moles O / = moles O

35. What now?
1 mole Fe : 1.5 moles O (notice I rounded a bit…this is fine as long as you're careful not to round past the 10th place)
Sometimes, you need STEP 4: multiply both ratios as necessary to get least WHOLE NUMBER ratio:
[1 mole Fe : 1.5 moles O] x 2 = Fe2O3. "Iron (III) oxide"

36. Combustion Analysis
When a hydrocarbon is “completely” combusted in the apparatus below, the carbon is converted to CO2, the hydrogen is converted to H2O, and the amount of CO2 and H2O are measured. From this information alone, the empirical formula of the hydrocarbon can be determined.

37. Determining formula from Combustion analysis
Look at example 3.15 on page 97:
Isopropyl alcohol, a substance sold as subbing alcohol, is composed of C, H, and O. Combustion of g of isopropyl alcohol produces g of CO2 and g of H2O. Determine the empirical formula of isopropyl alcohol.

38. Strategies:
Both CO2 and H2O have oxygen in them, so we will determine how much oxygen was in the original sample LAST. Let's figure out how much C and H was in the original sample FIRST.
Remember: formulas come from MOLE RATIOS…not mass ratios. Find the masses first, then convert to moles.

39. Mass CO2 to Mass Carbon in unknown
0.561 g CO2
1 mole C
12.01 g C
1 mole CO2
= g C
44.01 g CO2
1 mole CO2
1 mole C
In every ONE CO2 molecule, there is ONE C atom

40. Mass H2O to Mass Hydrogen in unknown
0.306 g H2O
2 moles H
1.01 g H
1 mole H2O
= g H
18.02 g H2O
1 mole H2O
1 mole H
In every ONE H2O molecule, there is ONE O atom

41. Best for last: mass Oxygen in sample
The original unknown sample had a total mass of g, and it had C, H, and O in it. Let's calculate mole of oxygen in that original sample:
0.255 g sample – (0.153 g C g H) = g O

42. Finally, convert all masses to mole and find that empirical formula

43. Get the mole ratio Now fine the ratio between all these values
We can round this to 8. We usually round as long as the 10th place is not larger than .3 or higher.