1. Love, Life and Stoichiometry

2. Stoichiometry Greek for “measuring elements”
The calculations of quantities in chemical reactions based on a balanced equation.
We can interpret balanced chemical equations several ways.

3. In terms of Particles Element- atoms
Molecular compound (non- metals)- molecule
Ionic Compounds (Metal and non-metal) - formula unit

4. 2H2 + O2 ® 2H2O
Two molecules of hydrogen and one molecule of oxygen form two molecules of water.
2 Al2O3 ® 4Al + 3O2 2
formula units
Al2O3
form 4
atoms Al
and 3
molecules O2
2Na + 2H2O ® 2NaOH + H2

5. Look at it differently 2H2 + O2 ® 2H2O
2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen molecules of water.
2 x (6.02 x 1023) molecules of hydrogen and 1 x (6.02 x 1023) molecules of oxygen form 2 x (6.02 x 1023) molecules of water.
2 moles of hydrogen and 1 mole of oxygen form 2 moles of water.

6. In terms of Moles 2 Al2O3 ® 4Al + 3O2 2Na + 2H2O ® 2NaOH + H2
The coefficients tell us how many moles of each kind

7. In terms of mass
The law of conservation of mass applies
We can check using moles
2H2 + O2 ® 2H2O
2.02 g H2
2 moles H2
4.04 g H2 =
1 moles H2
32.00 g O2 =
1 moles O2
32.00 g O2
1 moles O2
36.04 g H2
36.04 g Reactants

8. No mass has been created or destroyed
Therefore…
2H2 + O2 ® 2H2O
18.02 g H2O
36.04 g H2O
2 moles H2O =
1 mole H2O
2H2 + O2 ® 2H2O
36.04 g H2 + O2 =
36.04 g H2O
No mass has been created or destroyed

9. Your turn
Show that the following equation follows the Law of conservation of mass.
2 Al2O3 ® 4Al + 3O2

10. Mole to mole conversions
2 Al2O3 ® 4Al + 3O2
every time we use 2 moles of Al2O3 we make 3 moles of O2
2 moles Al2O3
3 mole O2 or
3 mole O2
2 moles Al2O3

11. Mole to Mole conversions
How many moles of O2 are produced when 3.34 moles of Al2O3 decompose?
2 Al2O3 ® 4Al + 3O2
3.34 moles Al2O3
3 mole O2 =
5.01 moles O2
2 moles Al2O3

12. Your Turn
2C2H2 + 5 O2 ® 4CO2 + 2 H2O
If 3.84 moles of C2H2 are burned, how many moles of O2 are needed?
How many moles of C2H2 are needed to produce 8.95 mole of H2O?
If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed?

13. How do you get good at this?

15. Stoichiometry Practice

16. Given the following equation:
2 KClO3  2KCl + 3 O2
How many moles of O2 can be produced by letting moles of KClO3 react?

17. 2 KClO3  2KCl + 3 O2
The strategy will be to convert to Remember, conversion factors are what you over what you
moles of KClO3
moles of O2
want
have

18. KClO3  2KCl + O2 2 3
12.00 moles of KClO3
moles of O2
moles of KClO3
= 12 x (3/2)
= 18

19. If ammonia, NH3, is burned in air, the following reaction takes place:
4NH3 + 3O2  2 N2 + 6H2O
Given that you started with 51.0 g of NH3, how many grams of water will be produced?

20. 4NH3 + 3O2  2 N2 + 6H2O
The overall strategy will be to convert the of ammonia of ammonia, then convert the of ammonia of water, and then finally convert the of water of water.
grams
to moles
moles
to moles
moles
to grams

21. 4NH3 + 3O2  2 N2 + 6H2O = 81 g of H2O 51.0 g NH3   
grams NH moles NH moles H2O grams of H2O
51.0 g NH3   
1 mole NH3
18 g H2O
6 mole H2O
4 mole NH3
1 mole H2O
17 g NH3
= 81 g of H2O

22. 20.0 g of silver(I)nitrate is reacted with an excess of sodium choride to produce silver(I) chloride
AgNO3 + NaCl => AgCl + NaNO3
What mass of silver(I) chloride is produced?

23. AgNO3 + NaCl => AgCl + NaNO3
The overall strategy will be to convert the of silver nitrate of silver nitrate , then convert the of silver nitrate of silver chloride, and then finally convert the of silver chloride of silver chloride.
grams
to moles
moles
to moles
moles
to grams

24. AgNO3 + NaCl => AgCl + NaNO3
grams AgNO3
20.0 g AgNO3 
moles AgNO3
moles AgNO3 
moles AgCl
moles AgCl 
grams AgCl
1 mole AgNO3
1 mole AgCl
143.4 g AgCl
169.9 g AgNO3
1 mole AgNO3
1 mole AgCl
= 16.9
g AgCl

25. Now go to mass to mass problems

26. Mass in Chemical Reactions
How much do you make?
How much do you need?

27. We can’t measure moles!! What can we do?
We can convert grams to moles.
Periodic Table
Then do the math with the moles.
Balanced equation
Then turn the moles back to grams.
Periodic table

28. For example...
If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form?
Fe + CuSO4 ® Fe2(SO4)3 + Cu
2Fe + 3CuSO4 ® Fe2(SO4)3 + 3Cu
1 mol Fe
10.1 g Fe
0.181 mol Fe =
55.85 g Fe

29. 2Fe + 3CuSO4 ® Fe2(SO4)3 + 3Cu
3 mol Cu
0.272 mol Cu
0.181 mol Fe =
2 mol Fe
63.55 g Cu
0.272 mol Cu =
17.3 g Cu
1 mol Cu

30. Could have done it 1 mol Fe 63.55 g Cu 10.1 g Fe 3 mol Cu 55.85 g Fe=
17.3 g Cu

31. More Examples
To make silicon for computer chips they use this reaction
SiCl4 + 2Mg ® 2MgCl2 + Si
How many grams of Mg are needed to make 9.3 g of Si?
How many grams of SiCl4 are needed to make 9.3 g of Si?
How many grams of MgCl2 are produced along with 9.3 g of silicon?

32. For Example The U. S. Space Shuttle boosters use this reaction
3 Al(s) + 3 NH4ClO4 ® Al2O3 + AlCl3 + 3 NO + 6H2O
How much Al must be used to react with 652 g of NH4ClO4 ?
How much water is produced?
How much AlCl3?

33. We can also change Liters of a gas to moles At STP
25ºC and 1 atmosphere pressure
At STP 22.4 L of a gas = 1 mole
If 6.45 moles of water are decomposed, how many liters of oxygen will be produced at STP?

34. For Example
If 6.45 grams of water are decomposed, how many liters of oxygen will be produced at STP?
H2O ® H2 + O2
1 mol H2O
1 mol O2
22.4 L O2
6.45 g H2O
18.02 g H2O
2 mol H2O
1 mol O2

35. Your Turn
How many liters of CO2 at STP will be produced from the complete combustion of 23.2 g C4H10 ?
What volume of oxygen will be required?

36. Gases and Reactions
A few more details

37. Example
How many liters of CH4 at STP are required to completely react with 17.5 L of O2 ?
CH4 + 2O2 ® CO2 + 2H2O
22.4 L O2
1 mol O2
1 mol CH4
22.4 L CH4
1 mol O2
1 mol CH4
22.4 L CH4
17.5 L O2
22.4 L O2
2 mol O2
1 mol CH4
= 8.75 L CH4

38. Avagadro told us
Equal volumes of gas, at the same temperature and pressure contain the same number of particles.
Moles are numbers of particles
You can treat reactions as if they happen liters at a time, as long as you keep the temperature and pressure the same.

39. Example
How many liters of CO2 at STP are produced by completely burning L of CH4 ?
CH4 + 2O2 ® CO2 + 2H2O
2 L CO2
17.5 L CH4
= 35.0 L CH4
1 L CH4

40. Limiting Reagent
If you are given one dozen loaves of bread, a gallon of mustard and three pieces of salami, how many salami sandwiches can you make.
The limiting reagent is the reactant you run out of first.
The excess reagent is the one you have left over.
The limiting reagent determines how much product you can make

41. How do you find out? Do two stoichiometry problems.
The one that makes the least product is the limiting reagent.
For example
Copper reacts with sulfur to form copper ( I ) sulfide. If 10.6 g of copper reacts with 3.83 g S how much product will be formed?

42. If 10. 6 g of copper reacts with 3. 83 g S
If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed?
2Cu + S ® Cu2S
Cu is Limiting Reagent
1 mol Cu
1 mol Cu2S
g Cu2S
10.6 g Cu
63.55g Cu
2 mol Cu
1 mol Cu2S
= 13.3 g Cu2S
= 13.3 g Cu2S
1 mol S
1 mol Cu2S
g Cu2S
3.83 g S
32.06g S
1 mol S
1 mol Cu2S
= 19.0 g Cu2S

43. Your turn
If 10.1 g of magnesium and 2.87 L of HCl gas are reacted, how many liters of gas will be produced?
How many grams of solid?
How much excess reagent remains?

44. Your Turn II
If 10.3 g of aluminum are reacted with 51.7 g of CuSO4 how much copper will be produced?
How much excess reagent will remain?

46. Yield The amount of product made in a chemical reaction.
There are three types
Actual yield- what you get in the lab when the chemicals are mixed
Theoretical yield- what the balanced equation tells you you should make.
Percent yield = Actual x 100 % Theoretical

47. Example
6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate.
2Al + 3 CuSO4 ® Al2(SO4)3 + 3Cu
What is the actual yield?
What is the theoretical yield?
What is the percent yield?

48. Details Percent yield tells us how “efficient” a reaction is.
Percent yield can not be bigger than 100 %.

49. Homework!!