1. Stoichiometry
Spring 2016

2. What is stoichiometry? Greek for ”measuring elements”
Defined as calculations of the quantities in chemical reactions, based on a balanced equation.
In this chapter, we will be interpreting balanced equations.

3. Balance the Chemical Equations
Al(NO3)3 + NaOH  Al(OH)3 + NaNO3
C3H8 + O2  CO2 + H2O
KClO3  KCl + O2
BaF2 + K3PO4  Ba3(PO4)2 + KF

4. Answers Al(NO3)3 + 3NaOH  Al(OH)3 + 3NaNO3 C3H8 + 5O2  3CO2 + 4H2O
2KClO3  2KCl + 3O2
3BaF2 + 2K3PO4  Ba3(PO4)2 + 6KF

5. Law of Conservation of Matter
Why do we have to balance chemical equations?
The mass of the system must remain constant over time.
Matter is neither created nor destroyed
So what do we think about when we balance equations?
We have to make sure that there are the same number of each atom on the left hand side as the right hand side of the equation.
For example:
2KClO3  2KCl + 3O2
2 potassium chlorate formula units decompose to form 2 potassium chloride formula units and 3 oxygen molecules.

6. Law of Conservation of Matter
However, chemists don’t work with reactions on such a small scale.
We use a molar scale because it is measurable.
The coefficients in balanced equations also tell us the molar ratio of each substance to each other.
For example:
2KClO3  2KCl + 3O2
2 moles of potassium chlorate decompose to form 2 moles of potassium chloride and 3 moles of oxygen gas.

7. Law of Conservation of Matter
We can check to make sure that mass is conserved using the molar ratios.
For example:
2KClO3  2KCl + 3O2
2 mol KClO3 * ( g KClO3 / 1 mol KClO3 ) = g KClO3
2 mol KCl * ( g KCl / 1 mol KCl ) = g KCl
3 mol O2 * (32 g O2 / 1 mol O2 ) = 96 g O2
149.1 g KCl + 96 g O2 = g KClO3
The mass of the reactant equals the mass of the products.

8. Another Example CaCl2 + Na3PO4  Ca3(PO4)2 + NaCl
3 mol CaCl2 * ( g CaCl2 / 1 mol CaCl2 ) = g CaCl2
2 mol Na3PO4 * ( g Na3PO4 / 1 mol Na3PO4 ) = g Na3PO4
1 mol Ca3(PO4)2 = g Ca3(PO4)2
6 mol NaCl * (58.44 g NaCl / 1 mol NaCl ) = g NaCl
g CaCl g Na3PO4 = g Ca3(PO4) g NaCl
g reactants = g products

9. Practice
Balance the equation and show that the mass of the reactant(s) is equal to the mass of the product(s):

10. Mole to Mole Conversions
Al2O3  Al + O2
2Al2O3  4Al + 3O2
When 2 moles of aluminum oxide decompose, it makes 4 moles of aluminum metal and 3 moles of oxygen gas.
We can use this information to make conversion factors. For example:

11. Mole to Mole Conversions
How many moles of oxygen gas are produced when 3.34 moles of aluminum oxide is decomposed?
3.34 mol Al2O3 * (3 mol O2 / 2 mol Al2O3 ) = 5.01 mol O2

12. Mole to Mole Conversions
If you know the amount (moles, grams, or formula units) of any chemical in the reaction, you can find the amount of all the others.
6.022 x 1023
6.022 x 1023
Formula Units A
Formula Units B

13. Practice C2H2 + O2  CO2 + H2O 2C2H2 + 5O2  4CO2 + 2H2O
If 3.84 mol of C2H2 are burned, how many mol O2 are needed? How many mol C2H2 are needed to produce 8.95 mol of H2O? If 2.47 mol C2H2 are burned, how many mol CO2 will be produced?
9.6 mol O2, 8.95 mol H2O, 4.94 mol O2

14. Do Now
Li2O  O2 + Li
If 1.8 x 1024 Li2O formula units decompose, how many molecules of O2 are created?
How many formula units Li2O are needed to produce 2.4 x 1024 of Li?

15. Answers 2Li2O  O2 + 4Li 9.03 x 1023 molecules O2
1.2 x 1024 formula units Li2O

16. Class March 18th, 2016
Based on exit tickets yesterday, here’s what we need to do today:
Review how to set up these problems
Review the mol ratio from the balanced chemical equation
Practice

17. Put this on a notecard:

18. How to set up a stoichiometry problem
Propane (C3H8) is burned in many fireplaces and gas grills. If 8.16 g of water is formed, how many grams of propane were burned?
What is this question even asking?
Given: grams of water
Asked to find: grams of propane
Basic roadmap:
Mass A  Mol A  Mol B  Mass B
Mass water  Mol water  Mol propane  Mass propane

19. Make the basic roadmap for the following problems:
6 g of Al2O3 decomposes to form Al and O2. How many grams of Al are formed?
LiOH reacts with HCl to form LiCl and 2.4 x 1024 molecules of water. How many formula units of LiOH reacted?
1.8 x 1024 formula units of NaCl react with Pb(NO3)2 to form NaNO3 and PbCl2. How many grams of PbCl2 is formed?

20. Answers
6 g of Al2O3 decomposes to form Al and O2. How many grams of Al are formed?
grams Al2O3  mol Al2O3  mol Al  grams Al
LiOH reacts with HCl to form LiCl and 2.4 x 1024 molecules of water. How many formula units of LiOH reacted?
molecules water  mol water  mol LiOH  formula units LiOH
1.8 x 1024 formula units of CaCl2 react with Pb(NO3)2 to form Ca(NO3)2 and PbCl2. How many grams of PbCl2 is formed?
formula units CaCl2  mol CaCl2  mol PbCl2  grams PbCl2

21. What are the molar ratios needed?
Molar ratio = Mol B / Mol A
6 g of Al2O3 decomposes to form Al and O2. How many grams of Al are formed?
LiOH reacts with HCl to form LiCl and 2.4 x 1024 molecules of water. How many formula units of LiOH reacted?
1.8 x 1024 formula units of NaCl react with Pb(NO3)2 to form NaNO3 and PbCl2. How many grams of PbCl2 is formed?

22. Answers
6 g of Al2O3 decomposes to form Al and O2. How many grams of Al are formed?
4 Mol Al / 2 Mol Al2O3
LiOH reacts with HCl to form LiCl and 2.4 x 1024 molecules of water. How many formula units of LiOH reacted?
1 Mol LiOH / 1 Mol water
1.8 x 1024 formula units of NaCl react with Pb(NO3)2 to form NaNO3 and PbCl2. How many grams of PbCl2 is formed?
1 Mol PbCl2 / 2 Mol NaCl

23. Solve. Show work.
6 g of Al2O3 decomposes to form Al and O2. How many grams of Al are formed?
LiOH reacts with HCl to form LiCl and 2.4 x 1024 molecules of water. How many formula units of LiOH reacted?
1.8 x 1024 formula units of CaCl2 react with Pb(NO3)2 to form Ca(NO3)2 and PbCl2. How many grams of PbCl2 is formed?

24. Answers
6 g of Al2O3 decomposes to form Al and O2. How many grams of Al are formed?
3.16 g Al
LiOH reacts with HCl to form LiCl and 2.4 x 1024 molecules of water. How many formula units of LiOH reacted?
2.4 x 1024 FUN LiOH
1.8 x 1024 formula units of CaCl2 react with Pb(NO3)2 to form Ca(NO3)2 and PbCl2. How many grams of PbCl2 is formed?
415 g PbCl2

25. Exit Ticket: Solve this problem. Show work.
3.6 x 1024 formula units of NaCl react with AgNO3 to form NaNO3 and AgCl. How many grams of AgCl is formed?
Potentially helpful info:
MM NaCl = g/mol
MM AgNO3 = g/mol
MM NaNO3 = 85 g/mol
MM AgCl = g/mol

26. Limiting Reagents
If I have 18 hamburger buns and 12 hamburger patties, how many hamburgers can I make?
If I have 26 tires and 16 bike frames, how many bikes can I make?

27. Limiting Reagents
If I have 18 hamburger buns and 12 hamburger patties, how many hamburgers can I make?
12 hamburgers
The hamburger patties are the limiting reagent
If I have 26 tires and 16 bike frames, how many bikes can I make?
13 bikes
The tires are the limiting reagent
The limiting reagent is the reactant that I will run out of first. It limits the amount of product I can make.

28. How to know if I’m being asked to find the LR
How to find the Limiting Reagent (LR)
To find the limiting reagent, calculate the amount of product produced for each of the reagents given. The one that produces the smallest amount of product is the LR.
How to know if I’m being asked to find the LR
If you are given the amount (moles, grams, formula units) of both reactants and asked for the amount of the products, you must first identify the limiting reagent.

29. Example Problem: Step 1
A 50.6 g sample of Mg(OH)2 is reacted with 45.0 g of HCl to yield MgCl2 and H2O. How many grams of MgCl2 are produced? What is the limiting reagent?
I am given the amount (grams) of both reactants. This is a limiting reagent problem.
Step 1: Write out and balance the equation
Mg(OH) HCl  MgCl H2O
50.6g g ?g N/A

30. Example Problem: Step 2
A 50.6 g sample of Mg(OH)2 is reacted with 45.0 g of HCl to yield MgCl2 and H2O. How many grams of MgCl2 are produced? What is the limiting reagent?
Mg(OH) HCl  MgCl H2O
50.6g g ?g N/A
Step 2: Calculate the grams of MgCl2 produced when 50.6g of Mg(OH)2 are fully reacted. Calculate the grams of MgCl2 produced when 45.0g of HCl are fully reacted.

31. Example Problem: Step 2 Mg(OH)2 + 2HCl  MgCl2 + 2H2O
50.6g g ?g N/A
50.6 g Mg(OH)2* * * = g MgCl2
45.0 g HCl * * * = g MgCl2
1 mol Mg(OH)2
1 mol MgCl2
95.21 g MgCl2
58.32 g Mg(OH)2
1 mol Mg(OH)2
1 mol MgCl2
1 mol HCl
1 mol MgCl2
95.21 g MgCl2
36.46 g HCl
2 mol HCl
1 mol MgCl2

32. Example Problem: Step 3
A 50.6 g sample of Mg(OH)2 is reacted with 45.0 g of HCl to yield MgCl2 and H2O. How many grams of MgCl2 are produced? What is the limiting reagent?
Mg(OH) HCl  MgCl H2O
50.6g g ?g N/A
Step Three: Compare. The reactant that yields the smaller amount of product when fully reacted is the limiting reagent.

33. Example Problem: Answer
A 50.6 g sample of Mg(OH)2 is reacted with 45.0 g of HCl to yield MgCl2 and H2O. How many grams of MgCl2 are produced? What is the limiting reagent?
58.76 g MgCl2 are produced, HCl is the limiting reagent
These are basically two stoichiometry problems in one.
The smallest answer is the correct answer.

34. Practice
3.4 g NaOH react with 5.0 g H2SO4 to form Na2SO4 and H2O. How many grams of H2O are formed? What is the limiting reagent?
2.3 g C6H6 react with 2.3 g Br2 to form C6H5Br and HBr. How many grams of HBr are formed? What is the limiting reagent?
4.0 g CaCl2 react with 2.1 g NaOH to form NaCl and Ca(OH)2. How many grams of NaCl are formed? What is the limiting reagent?

35. Answers
3.4 g NaOH react with 5.0 g H2SO4 to form Na2SO4 and H2O. How many grams of H2O are formed? What is the limiting reagent?
1.53 g H2O, NaOH is the LR
2.3 g C6H6 react with 2.3 g Br2 to form C6H5Br and HBr. How many grams of HBr are formed? What is the limiting reagent?
1.16 g HBr, Br2 is the LR
4.0 g CaCl2 react with 2.1 g NaOH to form NaCl and Ca(OH)2. How many grams of NaCl are formed? What is the limiting reagent?
3.07 g NaCl, NaOH is the LR

36. Exit Ticket
If 4.95 g of ethylene (C2H4) are combusted with 3.25 g of oxygen. How many grams of CO2 are formed? What is the limiting reagent?

37. Announcements Last day of the quarter is Friday, April 1st
Last day to finish a test is Wednesday, March 30th
All the times I am available after school are posted on the class website. Make sure we finish those tests ASAP!!
We will do a lab tomorrow (3/24) and Wednesday (3/30). Lab notebooks are due Monday (4/11).
We will have a quiz on Tuesday (3/29). Review day will be Monday (3/28).

38. Exit Ticket
If 4.95 g of ethylene (C2H4) are combusted with 3.25 g of oxygen. How many grams of CO2 are formed? What is the limiting reagent?
2.98 g CO2
O2 is the LR

39. 3. 0 g propane (C3H8) is combusted in 30 g oxygen gas
3.0 g propane (C3H8) is combusted in 30 g oxygen gas. How many grams of water are formed? What is the LR?
12 g of hydrogen sulfide comes in contact with 10 g silver metal to produce silver sulfide and hydrogen gas. How many grams of silver sulfide are formed? What is the LR?
5.0 g zinc metal reacts with 10 g copper (II) sulfate to yield copper metal and zinc sulfate. How many grams of copper metal are formed? What is the LR?
16 g NaOH reacts with 12 g H3PO4 to yield Na3PO4 and H2O. How many grams of water are formed? What is the LR?

40. 3. 0 g propane (C3H8) is combusted in 30 g oxygen gas
3.0 g propane (C3H8) is combusted in 30 g oxygen gas. How many grams of water are formed? What is the LR?
Propane = LR, 4.9 g water
12 g of hydrogen sulfide comes in contact with 10 g silver metal to produce silver sulfide and hydrogen gas. How many grams of silver sulfide are formed? What is the LR?
Silver Metal = LR, g Ag2S
5.0 g zinc metal reacts with 10 g copper (II) sulfate to yield copper metal and zinc sulfate. How many grams of copper metal are formed? What is the LR?
Copper (II) Sulfate = LR, 3.98 g Cu
16 g NaOH reacts with 12 g H3PO4 to yield Na3PO4 and H2O. How many grams of water are formed? What is the LR?
H3PO4 = LR, 6.61 g water

41. End of Material for Quiz 1
Quiz is March 29, 2016

42. Percent Yield Simple Stoichiometric Conversions: given one number
Limiting Reagent Problems: given two numbers
Percent Yield Problems: given three numbers
You will be given the amounts both reagents and a product.
Percent Yield = Experimental Yield * 100
Experimental yield will be given to you. You must calculate the theoretical yield using the limiting reagent and compare the two.
Theoretical Yield

43. Example
3.0 g calcium chloride solution and 5.0 g sodium carbonate solution are mixed together to form calcium carbonate and sodium chloride. When the dry calcium carbonate product had a mass of 0.5 g. What is the percent yield?
Molar Masses:
CaCl2 = g/mol
Na2CO3 = g/mol
CaCO3 = g/mol

44. Answer
16.8 % yield

45. Practice Problems 3/31 DUE 4/11
3.0 g LiOH reacts with 10.0 g HCl to form LiCl and H2O. The dry lithium chloride product is found to have a mass of 1.5 g. What is the percent yield?
5.0 g MgCl2 reacts with 6.0 g Na2CO3 to form MgCO3 and NaCl. The dry magnesium carbonate product is found to have a mass of 4.0 g. What is the percent yield?
7.0 g Mg reacts with 35 g AuCl to form MgCl2 and Au. The dry gold product is found to have a mass of 20 g. What is the percent yield?

46. Answers
28% yield
90% yield
68% yield

47. Error Analysis What if I get a percent yield bigger than 100%?
Something went wrong.
Most likely: excess reagent or a side product is mixing with your desired product.
For example, in our percent yield lab, some unreacted iron may be mixed with your copper product.
Side product example: carbon monoxide is an undesired side product of a combustion reaction that occurs when not enough oxygen gas is present. If you were only measuring the mass of gas produced by your combustion reaction, you could overestimate your percent yield.

48. Solution Stoichiometry
So far, all of our stoichiometry problems have given the amount of reactant(s) in moles, grams, or formula units.
What if we were working with a solution (aq)?
We use molarity to determine the number of moles in a solution.

49. Molarity
Molarity (M) is defined as the amount of moles of a compound dissolved in an amount of solvent (usually water).
Molarity = moles of solute
M = mol / L
liters of solution

50. Examples
If you have 500 mL of a 6M solution of HCl, how many moles of HCl do you have?
If you pour 8M HCl into a beaker until you have 2 moles, how many milliliters would be in the beaker?
You place 3 moles of HCl in a graduated cylinder and then add 400 mL of water. What is the molarity of the solution you made?

51. Answers
If you have 500 mL of a 6M solution of HCl, how many moles of HCl do you have?
3 mol HCl
If you pour 8M HCl into a beaker until you have 2 moles, how many milliliters would be in the beaker?
250 mL of 8M HCl
You place 3 moles of HCl in a graduated cylinder and then add 400 mL of water. What is the molarity of the solution you made?
7.5M HCl

52. How does this fit into our roadmap?
Molarity
Molarity
M = mol / L
M = mol / L

53. Example
500 mL of 6M NaOH is added to an excess of HCl to form NaCl and H2O. How many grams of NaCl are formed?

54. Answer
500 mL of 7M NaOH is added to an excess of HCl to form NaCl and H2O. How many grams of NaCl are formed?
g NaCl

55. Practice -- Just in your notebook
5 g of Pb(OH)2 is added to 250 mL of 6M HCl to yield PbCl2 and H2O. The dry PbCl2 is found to have a mass of 4 g. What is the percent yield?
300 mL of 2M AgNO3 is added to 250 mL of 3M HCl to yield AgCl and HNO3. The dry AgCl is found to have a mass of 80g. What is the percent yield?

56. Answers
5 g of Pb(OH)2 is added to 250 mL of 6M HCl to yield PbCl2 and H2O. The dry PbCl2 is found to have a mass of 4 g. What is the percent yield?
69% yield
300 mL of 2M AgNO3 is added to 250 mL of 3M HCl to yield AgCl and HNO3. The dry AgCl is found to have a mass of 80 g. What is the percent yield?
93% yield

57. Announcements 4/12
If you haven’t turned in your percent yield problems do so today
If you haven’t turned in your lab notebook do so today
Quiz is Monday
Review materials are posted and we will review in class on Friday
Dilutions lab on Thursday, lab notebooks are due Friday
New seats when we start gas laws next Wednesday

58. Dilutions
To dilute a solution means to add more solvent without the addition of more solute. Of course, the resulting solution is thoroughly mixed so as to ensure that all parts of the solution are identical.
For example: If I add 300 mL of water to 500 mL of 6 M HCl, what is the molarity of my solution now? My moles of HCl remain unchanged, but more water has been added. The solution has been diluted.
The fact that the solute amount stays constant allows us to develop calculation techniques.

59. Dilution Equation First, we write:
moles solute before dilution = moles solute after dilution
From rearranging the equation that defines molarity, we know that the moles of solute equals the molarity times the volume.
mol = (mol / L) * L
So we can substitute MV (molarity times volume) into the above equation, like this:
M1V1 = M2V2
The "sub one" refers to the situation before dilution and the "sub two" refers to after dilution.

60. Example
53.4 mL of a 1.50 M solution of NaCl is on hand, but you need some M solution. How many mL of M can you make?

61. Answer Using the dilution equation, we write:
(1.50 mol/L) (53.4 mL) = (0.800 mol/L) (x)
x = 100 mL
Notice that the volumes need not be converted to liters. Any volume measurement is fine, just so long as the same one is used on each side.
However, if you are calculating how many moles of solute are present, you need to have the volume in liters  mol = (mol/L) * L

62. Practice
How many milliliters of 5.0 M copper(II) sulfate solution must be added to 160 mL of water to achieve a 0.30 M copper(II) sulfate solution?
10 mL
20 mL of water is added to 80 mL of 6 M HCl. What is the molarity of the final solution?
4.8 M
Extension: To what volume should you dilute 133 mL of an 7.90 M CuCl2 solution so that 51.5 mL of the diluted solution contains 4.49 g CuCl2?
1620 mL

63. Extension Solved
1) Find moles: (4.49g CuCl2) (1 mole CuCl2 / grams) = moles CuCl2 2) Find the molarity of the 51.5 mL of the diluted solution that contains 4.49g CuCl2: ( moles CuCl2) / ( liters) = M 3) Use the dilution formula: M1V1 = M2V2 (7.90 M) (133 mL) = (0.648 M) (V2) V2 = 1620 mL You should dilute the 133 mL of an 7.90 M CuCl2 solution to 1620 mL.

64. More Dilutions What if I mix two solutions together?
Calculate total moles
Calculate total volume
Divide moles by volume to get molarity
You can also think of it like this: M1V1 + M2V2 = M3V3

65. Example
Calculate the molarity of the resulting solution if 2.00 L of 3.00 M NaCl and 4.00 L of 1.50 M NaCl are mixed.

66. Answer Using the first method, we have this:
x = (3.00 mol/L) (2.00 L) = 6 moles
x = (1.50 mol/L) (4.00 L) = 6 moles
Total moles = 12 moles
The total volume calculation is = 6.00 L.
Divide total moles by total volume to get the final answer.
12 moles / 6.00 L = 2.00 M
Using M1V1 + M2V2 = M3V3, we have this:
(3.00 mol/L) (2.00 L) + (1.50 mol/L) (4.00 L) = (x) (6.00 L)
6.00 mol mol = (x) (6.00 L)
x = 12.0 mol / 6.00 L = 2.00 M

67. Practice
300 mL of 12 M NaOH and 500 mL of 6 M NaOH are mixed. What is the molarity of the resulting solution?
200 mL of 3 M H2SO4 and 350 mL of 9 M H2SO4 are mixed. What is the molarity of the resulting solution?
600 mL of 2 M LiOH and 200 mL of 10 M LiOH are mixed. What is the molarity of the resulting solution?

68. Answers
300 mL of 12 M NaOH and 500 mL of 6 M NaOH are mixed. What is the molarity of the resulting solution?
8.25 M
200 mL of 3 M H2SO4 and 350 mL of 9 M H2SO4 are mixed. What is the molarity of the resulting solution?
6.82 M
600 mL of 2 M LiOH and 200 mL of 10 M LiOH are mixed. What is the molarity of the resulting solution?
4 M

69. Mixed Practice
250 mL of 3 M NaOH is diluted to a final volume of 600 mL. What is the molarity of the resulting solution?
300 mL of 0.5 M HCl and 200 mL of 1 M HCl are mixed. What is the molarity of the resulting solution?
250 mL of 10 M Na2SO4 is added to 500 mL of 6 M CaCl2 to form NaCl and CaSO4. How much CaSO4 is formed?

70. Answers
250 mL of 3 M NaOH is diluted to a final volume of 600 mL. What is the molarity of the resulting solution?
1.25 M
300 mL of 0.5 M HCl and 200 mL of 1 M HCl are mixed. What is the molarity of the resulting solution?
0.7 M
250 mL of 10 M Na2SO4 is added to 500 mL of 6 M CaCl2 to form NaCl and CaSO4. How much CaSO4 is formed?
g CaSO4

71. End of Material for Quiz 2
Quiz is April 19th, 2016

72. Provided on Quiz:
Molarity
Molarity

73. Review Materials Attached
Stoic. Quiz 2 Study Guide
Stoic. Quiz 2 Study Guide KEY
We will work in these in class on Friday, April 15th