1. Chapter 9 Chemical Quantities

2. In order to avoid waste and to be economic, a chemist has to know how much to make, a doctor must know how much to prescribe, an engineer has to know how much fuel to mix with others, etc., etc., etc...
Questions we answer here in Ch 9? How much can I get? How much do I need?
The engineers who develop airbags have to make sure exactly the right amount of gas is made - no more, no less.

3. Ch 9 is a compilation of Chs 1-8!
must name, balance, g to mol, mol to g…
Notice that for the whole chapter, I’ll give you info on one thing, you’ll give me info on another
but the only place they can meet is the balanced equation, the only language they speak is “molese”

4. 9.1 Information Given by Chemical Equations
The cook is a chemist!
Cooking is: mixing ingredients (reactants) in a recipe (equation) to get a certain amount of food (product)
Importance here? that info in a recipe can be used proportionally
double amts? double product!

5. The chemist is a cook!
A 2:1 mole ratio from a balanced equation is just that - as long as the two substances are in 2:1 number ratio (not mass ratio!) it doesn’t matter how much you use

6. 9.2 Mole-Mole Relationships
Remember!
The most common number ratio in chemistry is the
mole ratio!

7. Example
How many moles of KClO3 must decompose to produce 9 moles of oxygen gas? (The other product is KCl.)
1) As always! First, get a balanced equation!
2KClO3  2KCl + 3O2
2) What is the mol ratio between KClO3 & O2?

8. from balanced equation!
Example (cont’d)
Remember this?
9 mol O2
2 mol KClO3
= 6 mol KClO3
3 mol O2
from balanced equation!
Given

9. Example
When 5 mols sodium react with oxygen gas, how many mols of Na2O are formed?
4Na + O2  2Na2O
5 mol Na
2 mol Na2O
= 2.50 mol Na2O
4 mol Na

10. Example
For this unbalanced equation, how many mols of NO do you need to make 7.5 mol H2O?
NO + H2  N2 + H2O
2NO + 2H2  N2 + 2H2O
it’s a 1:1 ratio!
you need 7.5 mol!

11. Example
For this unbalanced equation, how many mols of SiH4 do you need to use with 4.2 mol NH3?
SiH4 + NH3  Si3N4 + H2
3SiH4 + 4NH3  Si3N4 + 12H2
it’s a 3:4 ratio!
you need 3.15 mol!

12. 9.3 Mass Calculations
The study of the amount of substances consumed/produced = stoichiometry
Remember that chemical equations tell you number ratios, not mass amounts!
Your 3 friends now: 1) balanced chemical equations 2) mole ratios 3) math relationships

13. We’ll start off simple with a problem in which I give you moles of one but you are asked about mass of another…
But remember: the only place they can meet is the balanced equation, the only language they speak is molese

14. CaC2 + 2H2O  C2H2 + Ca(OH)2 Balanced equation!!!
CaC2 and water get together to make acetylene, C2H2, and calcium hydroxide. How many grams of water do you need to make 1.55 mol C2H2?
Balanced equation!!!
CaC2 + 2H2O  C2H2 + Ca(OH)2 g
mol H2O
1.55 mol C2H2

15. Example cont’d mol bridge
They can only talk mole language from the balanced equation!
1.55 mol C2H2
2 mol H2O
18.0 g H2O
1 mol C2H2
1 mol H2O
= 55.8 g H2O
mol bridge

16. Example
How many grams of potassium chlorate (KClO3) must decompose to produce KCl and 1.45 mol oxygen gas?
First! balance the equation!!!
2KClO3  2KCl + 3O2

17. 2KClO3  2KCl + 3O2 g
mol KClO3
1.45 mol

18. Example cont’d mol bridge
They can only talk mol language from the balanced equation!
1.45 mol O2
2 mol KClO3
122.6 g KClO3
3 mol O2
1 mol KClO3
= 119 g KClO3
mol bridge

19. Example
How many moles of copper must react with AgNO3 to produce 5.5 g silver and copper(II) nitrate?
First! balance the equation!!!
Cu + 2AgNO3  2Ag + Cu(NO3)2

20. example cont’d
Cu + 2AgNO3  2Ag + Cu(NO3)2
5.5g
mol Cu
mol Ag

21. Example cont’d mol bridge
They can only talk mol language from the balanced equation!
5.5 g Ag
1 mol Ag
1 mol Cu
107.9 g Ag
2 mol Ag
= mol Cu
mol bridge

22. (gramsA  molsA )  (molsB  gramsB )
This is for big people only; here I give you mass of one you tell me mass of the other
key? start and end w/ grams; but must cross thru mol bridge!!!
Basic method?
(gramsA  molsA )  (molsB  gramsB ) PT
BEq PT

24. example
How many grams of oxygen gas are required to completely react with 14.6 g of solid sodium to form sodium oxide, Na2O?

25. 4Na O2  2Na2O g
14.6 g
mol Na
mol O2
notice that there are always non-players. but they are needed for balancing the equation!

26. example cont’d mol bridge
remember: they can only talk mol language from the balanced equation!
14.6 g Na
1 mol Na
1 mol O2
32.0 g O2
23.0 g Na
4 mol Na
1 mol O2
mol bridge
= 5.08 g O2

27. example
When 20.4 grams of sodium metal are mixed with chlorine gas, are 52.0 g of sodium chloride produced?

28. 2Na Cl2  2NaCl g
20.4 g
mol NaCl
mol Na

29. example cont’d mol bridge
They can only talk mol language from the balanced equation!
20.4 g Na
2 mol Na
2mol NaCl
= 51.9 g NaCl
1 mol NaCl
58.5 g NaCl
mol bridge
23.0 g Na
1 mol Na

30. example
Limestone, CaCO3, is heated to produce lime, calcium oxide, CaO, and carbon dioxide. How much limestone is required to produce 10.0 g of lime?

31. CaCO3  CaO + CO2 g
10.0 g
mol CaO
mol CaCO3

32. example cont’d mol bridge
They can only talk mol language from the balanced equation!
10. g CaO
1 mol CaO
1mol CaCO3
= 17.8 g CaCO3
1 mol CaCO3
100.1g CaCO3
mol bridge
56.1 g CaO

33. summary gA gB PT PT
molA
molB BE

34. 9.5 Mass Calculations: Comparing Two Reactions
we can use all this to compare; for example, which store-bought antacid is more effective by weight…
sodium bicarbonate and magnesium hydroxide are both used as antacids
if you have 1.00 g of both which will eat the most acid?

35. NaHCO3 + HCl  NaCl + H2O + CO2
1) sodium bicarb?
NaHCO3 + HCl  NaCl + H2O + CO2
1.00 g NaHCO3
1 mol NaHCO3
1 mol HCl
= 1.19 x 10-2 mol
mol bridge
84.01 g NaHCO3
1mol NaHCO3

36. 2) magnesium hydroxide? mol bridge Mg(OH)2 + 2HCl  MgCl2 + 2H2O
1.00 g Mg(OH)2
1 mol Mg(OH)2
2 mol HCl
58.33 g Mg(OH)2
1 mol Mg(OH)2
mol bridge
= 3.42 x 10-2 mol

37. Sorry, but equations Do Not describe:
1) Exact conditions
* Temperature
* Pressure
* Volume
2) What the atoms are doing (e.g. not all of them react)

38. 9.6 The Concept of Limiting Reactants
Just as if you were to make sandwiches there will be something left over,
Impossible for every atom/molecule to react, so…
must add one reactant in there in excess (xs) to react as much of the other as possible

39. what is the limiting ingredient here. i. e
what is the limiting ingredient here? i.e. what will stop you from making more sandwiches?

40. the “limiting reactant” here were hammers
how many sets of 1 pliers, 2 screwdrivers, and 1 hammer can you make, and what is left over?

41. Mg reacts w/ O2 in 2:1 ratio
So the 6 Mg on left only need 3 O2’s; but there are 7 O2’s!
what happens to the remaining oxygens???

42. 4 O2’s will not play, but they assure that (essentially) all Mg’s will be found and destroyed!
Mg’s will run out first; they limit the amount of product that can be made; Mg is the limiting reactant

43. Here N2 is in excess, H2 was the limiting reactant

44. 9.6 EOCs 29

45. 9.7 Calculations Involving a Limiting Reactant
If reactants in a reaction are not present in their mol ratios, one will be used up before the other = limiting reactant; the other one = excess
but you can still predict how much prodruct will be made; yippee!
[hint: limiting reactant probs betray themselves in asking how much product you can get when they give amts of 2 reactants]

46. You must determine which of the 2 is limiting and finish the prob w/ that one only, like this:
1) change amts of both reactants (masses) to mols (old stuff)
2) check mol ratios for which one will be used up first (limiting reactant), (new thing!)
3) use that one only to finish prob (forget about the xs reactant) (old stuff)

47. example We’ll start simple:
1.21 mol zinc are added to 2.65 mol HCl. Zinc chloride and hydrogen gas are formed. Which reactant is in XS? Calculate the amt of mols of zinc chloride produced.
START W/ BALANCED EQUATION!

48. example cont’d Zn + 2HCl  ZnCl2 + H2 the Zn:HCl is 1:2
Only one reactant can take us to product (= LR)
Will it be Zn or HCl???

49. example cont’d Zn + 2HCl  ZnCl2 + H2
One way is to pretend you don’t know about one of the reactants, e.g...
If there are 1.21 mol Zn, how many mols of HCl would you need?
the Zn:HCl is 1:2, so 1.21 mol Zn need twice as much HCl = 2.42 mol HCl

50. example cont’d Zn + 2HCl  ZnCl2 + H2 Do we have 2.42 mol HCl???
YESSIREEBOB!, and more! (2.65 mol)
We have plenty of HCl (= XS) :)
but we will run out of Zn (= LR) :(
Zn will take us to products!...

51. example cont’d Zn + 2HCl  ZnCl2 + H2 then, back to the old ways...
1.21 mol Zn
1 mol Zn
1 mol ZnCl2
= 1.21 mol ZnCl2

52. example Zn + 2HCl  ZnCl2 + H2 What if reactants are given in grams?
No problem! Just change to mols first...
79.1 g Zn react with 76.5 g HCl. 1) Which is Limiting Reactant? 2) How much H2 will be formed?

53. example cont’d Zn + 2HCl  ZnCl2 + H2 Balanced equation!
Change the reactants to mols and compare: 79.1 g Zn  1.21 mol Zn 76.5 g HCl  2.10 mol HCl

54. example cont’d Zn + 2HCl  ZnCl2 + H2
pretend not to know about one of the reactants, this time Zn
if there is a 2 HCl : 1 Zn ratio, and we have 2.10 mol of HCl, we should have at least half as much Zn (1.05 mol) for a complete reaction
we have 1.21 mol Zn = PLENTY! = XS

55. example cont’d HCl will take us to product! = 2.12 g H2 2 mol HCl

56. example
1.00 g Zn reacts with 6.2 x 10-3 mol Pb(NO3)2 to form Zn(NO3)2 and Pb… 1) Which is LR? 2) How much Pb will be formed?
first, the balanced equation!...

57. example cont’d Zn + Pb(NO3)2  Pb + Zn(NO3)2
find the mols of both players:
1.00 g Zn  mol Zn
(done already!)  6.2 x 10-3 mol Pb(NO3)2

58. example cont’d Zn + Pb(NO3)2  Pb + Zn(NO3)2
again, pretend we have only one (Zn)
if we have only mol Zn, we need mol Pb(NO3)2 (1:1 ratio)
we don’t have enough!!! So Pb(NO3)2 is the Limiting Reactant and will take us to the products...

59. example cont’d Zn + Pb(NO3)2  Pb + Zn(NO3)2
same old way to our destination...
.0062 mol Pb(NO3)2
1 mol Pb(NO3)2
1 mol Pb
= 1.3 g Pb
207.2 g Pb

60. what happens when there is not enough of one reactant (l)
not enough O2 means sooty flame

61. 9.7 EOCs
35 all, 38

62. 9.8 Percent Yield
Up to this point, all the calculations we’ve done only predicts what we’d get in a perfect world, but…
Reactants aren’t always 100% pure; there are spills, etc., therefore, what we really get will be less

63. The predicted yield is called a theoretical yield (the one we do on paper)
What we actually get is called actual yield (the one we get from the lab)
percent yield is...
percent yield =
actual yield
theoretical yield
• 100

64. example
In the reaction of Zn with HCl, g ZnCl2 are formed, although the theoretical yield was 143 g. What was the % yield?
percent yield = g
143 g
• 100
percent yield = 98.0 %

65. Big Person example For the reaction: K2CO3 + 2HCl  2KCl + H2O + CO2
if 45.8 g K2CO3 are added to xs HCl, 46.3 g of KCl are recovered.
1) What is the theoretical yield?
2) What is the percent yield?

66. example cont’d theoretical yield: = 49.4 g KCl 45.8 g K2CO3
1 mol K2CO3
2 mol KCl
= 49.4 g KCl
1 mol KCl
74.6 g KCl
138.2 g K2CO3

67. example cont’d Now we can find % yield... percent yield = 46.3 g
• 100
percent yield = 93.7 %

68. chemists shoot for 100% but are often satisfied with < 50%
chemists involved in manufacturing, medicine, plastics, etc., constantly strive for the highest profit margin; i.e. they seek the elusive 100% yield

69. 9.8 EOCs
42, 45, 52ab, 58, 59XC