1. Unit 2 – Section C
Conserving Matter

2. HW 1
Read & take notes on Section C.1

3. C.1 – Keeping Track of Atoms
The law of conservation of matter – in a chemical reaction matter is neither created nor destroyed.
 C O CO2
1 Carbon oxygen carbon dioxide
atom (C) molecule (O2) molecule (CO2)
Molecules can be converted and decomposed by chemical processes: but atoms are forever.

4. C.1 – Keeping Track of Atoms (continued)
Reactants are placed on the left of the arrow;
Products are placed on the right.
 C O CO2
1 Carbon oxygen carbon dioxide
atom (C) molecule (O2) molecule (CO2)
In a balanced chemical equation, the number of atoms for left side equals the number for the right side.

5. C.1 – Keeping Track of Atoms (continued)
Coefficients indicate the relative number of each unit involved.
 Cu (s) O2 (g) CuO (s)
2 Copper oxygen copper oxide
atoms (Cu) molecule (O2) molecules (CuO)

6. C.1 – Keeping Track of Atoms (continued)
Chemists use the term formula unit when referring to the smallest unit of an ionic compound.
 Cu (s) O2 (g) CuO (s)
2 Copper oxygen copper oxide
atoms (Cu) molecule (O2) molecules (CuO)

7. Classwork
Answer questions 1-5 in Section C.2 pg 155

8. C.2 – Accounting for Atoms
Methane burning with oxygen
CH4 + 2 O2  CO2 + 2 H2O
Reactants Products
C C
H H
O O

9. C.2 – Accounting for Atoms (continued)
Hydrobromic acid reacting with magnesium
HBr + Mg  H2 + MgBr2
Reactants Products
H H
Br Br
Mg Mg

10. C.2 – Accounting for Atoms (continued)
Hydrogen sulfide and metallic silver react
4 Ag + 4 H2S + O2  2 Ag2S + 4 H2O
Reactants Products
Ag Ag
H H
S S
O O

11. C.2 – Accounting for Atoms (continued)
Cellulose burns to form carbon dioxide and water vapor.
C6H10O5 + 6 O2  6 CO2 + 5 H2O
Reactants Products
C C
H H
O O

12. C.2 – Accounting for Atoms (continued)
Nitroglycerin decomposes explosively to form nitrogen, oxygen, carbon dioxide and water vapor.
2 C3H5(NO3) 3  3 N2 + O2 + 6 CO2 + 5 H2O
Reactants Products
C C
H H
N N
O O

13. HW 2 Read & take notes on Section C.3
Address & answer questions 1-6 in section C.4

14. C.3 – Nature’s Conservation: Balancing Chemical Equations
If polyatomic ions (examples NO3-, CO32-) appear as both reactants and product treat them as units.
If water is involved, balance the hydrogen and oxygen atoms last.
Recount all atoms after you think an equation is balanced.

15. C.4 – Writing Chemical Equations
Writing to balance the chemical equations…
Methane Chlorine Chloroform Hydrogen chloride
__ CH4 + __ Cl2  __ CHCl3 + __ HCl
Products
Reactants C H Cl C H Cl

16. C.4 – Writing Chemical Equations (continued)
__ C + __ O2  __ CO
1a.
Reactants
Products C O C O 16

17. C.4 – Writing Chemical Equations (continued)
__ Fe2O3 + __ CO  __ Fe + __ CO2
1b.
Reactants
Products C Fe O C Fe O 17

18. C.4 – Writing Chemical Equations (continued)
__ CuO + __ C  __ Cu + __ CO2 2.
Reactants
Products C Cu O C Cu O 18

19. C.4 – Writing Chemical Equations (continued)
__ O3  __ O2 3.
Reactants
Products O O 19

20. C.4 – Writing Chemical Equations (continued)
__ NH3 + __ O2  __ NO2 + __ H2O 4.
Reactants
Products
NHO
NHO
D-period 2/2/2012 20

21. C.4 – Writing Chemical Equations (continued)
__ Cu + __ AgNO3  __ Cu(NO3)2 + __ Ag 5.
Reactants
Products
CuAgNO
CuAgNO 21

22. C.4 – Writing Chemical Equations (continued)
__ C8H18 + __ O2  __ CO2 + __ H2O 6.
Reactants
Products C H O C H O 22

23. HW 3
Read & take notes on Section C.5

24. C.5 – Introducing the Mole Concept
Chemist have created a counting unit for elements called the mole (symbolized mol).
particles
6.02 x 1023
One mole of ANY element or molecule contains

25. C.5 – Introducing the Mole Concept (continued)
Furthermore, the atomic weight of elements can be used to find the molar mass of a substance.
One mole of boron atoms (6.02 x 1023) would have a molar mass of g 25

26. C.5 – Introducing the Mole Concept (continued)
More examples…
One mole of carbon atoms (6.02 x 1023) would have a molar mass of _______ g 26

27. C.5 – Introducing the Mole Concept (continued)
More examples…
One mole of copper atoms (6.02 x 1023) would have a molar mass of ______ g
One mole of silver atoms (6.02 x 1023) would have a molar mass of ______ g
One mole of gold atoms (6.02 x 1023) would have a molar mass of ______ g 27

28. C.5 – Introducing the Mole Concept (continued)
(Curve ball) How about the molar mass of oxygen gas (O2)?
One mole of oxygen gas (O2) molecules (6.02 x 1023) would have a molar mass of _______ g 28

29. C.5 – Introducing the Mole Concept (continued)
(Curve ball 2) How about the molar mass of water (H2O)?
One mole of water (H2O) molecules (6.02 x 1023) would have a molar mass of _______ g 29

30. C.6 – Molar Masses
HW Questions 1-4,6,8
pg 163

31. C.6 – Molar Masses 1.
One mole of nitrogen (N) atoms (6.02 x 1023) would have a molar mass of _______ g 2.
One mole of nitrogen (N2) molecules (6.02 x 1023) would have a molar mass of _______ g 31

32. C.6 – Molar Masses (continued)3.
One mole of table salt (NaCl) molecules (6.02 x 1023) would have a molar mass of _______ g 32

33. C.6 – Molar Masses (continued)4.
One mole of table sugar (C12H22O11) molecules (6.02 x 1023) would have a molar mass of _______ g 33

34. C.6 – Molar Masses (continued)6.
One mole of magnesium phosphate Mg3(PO4)2 molecules (6.02 x 1023) would have a molar mass of ______ g 34

35. C.6 – Molar Masses (continued)8.
One mole of calcium hydroxyapatite Ca10(PO4)6(OH)2 molecules (6.02 x 1023) would have a molar mass of ______ g 35

36. HW 5
Read & take notes on Section C.7

37. C.7 – Equations and Molar Relationships
Let’s revisit copper-refining…
2 CuO(s) + C(s)  2 Cu(s) + CO2(g)
Alternatively stated…
2 mol CuO + 1 mol C  2 mol Cu + 1 mol CO2
In this example, for every two moles of CuO that react, one mole of CO2 is produced.

38. C.7 – Equations and Molar Relationships (continued)
Sample Problem: A refiner needs to convert g CuO to pure Cu. What mass of C is needed for this reaction?
Using . . .
2 mol CuO + 1 mol C  2 mol Cu + 1 mol CO2
We can reason that . . .
1 mol CuO
79.55 g CuO
X mol CuO
955.0 g CuO = 38

39. C.7 – Equations and Molar Relationships (continued)
Sample Problem: A refiner needs to convert g CuO to pure Cu. What mass of C is needed for this reaction? (continued)
Solving for X . . .
1 mol CuO
79.55 g CuO
X mol CuO
955.0 g CuO =
955.0 g CuO X 1 mol CuO = g CuO X X mol CuO
X mol CuO
955.0 g CuO X
1 mol CuO
79.55 g CuO =
NOTICE 39

40. C.7 – Equations and Molar Relationships (continued)
Sample Problem: A refiner needs to convert g CuO to pure Cu. What mass of C is needed for this reaction? (continued)
Solving for X . . .
X mol CuO
955.0 g CuO X
1 mol CuO
79.55 g CuO =
12.01 mol CuO = X
This all started with . . .
A proportion we created called a conversion factor, both referring to the same number of particles.
1 mol CuO
79.55 g CuO 40

41. C.7 – Equations and Molar Relationships (continued)
Sample Problem: A refiner needs to convert g CuO to pure Cu. What mass of C is needed for this reaction? (continued) So
The refiner knows there are mol CuO in g.
Remembering the equation we started with . . .
2 mol CuO + 1 mol C  2 mol Cu + 1 mol CO2
1 mol C
2 mol CuO
12.01 mol CuO X =
6.005 mol C 41

42. C.7 – Equations and Molar Relationships (continued)
Sample Problem: A refiner needs to convert g CuO to pure Cu. What mass of C is needed for this reaction? (continued)
So Once we know
The refiner knows we need mol C to refine 955 g of CuO we can calculate the actual mass of C needed . . .
Appropriate conversion factors . . .
12.01 g C
6.005 mol C X =
72.12 g C
1 mol C 42

43. C.8 – Molar Relationships
HW 6 Questions 1-4 pg 166

44. C.8 – Molar Relationships1.
CuO(s) + 2 HCl(aq)  CuCl2(aq) + H2O (l)
Molar mass of each… 44

45. C.8 – Molar Relationships (continued)2.
Mass (in grams) of:
a. 1.0 mol HCl
b. 5.0 mol HCl
c mol CuO 45

46. C.8 – Molar Relationships (continued)3.
# of moles represented by
a g CuCl2
b g CuCl2
c g HCl 46

47. C.8 – Molar Relationships4.
CuO(s) + 2 HCl(aq)  CuCl2(aq) + H2O (l)
a. How many moles of CuO are needed to react with 4 mol HCl?
a. How many moles of HCl are needed to react with 4 mol CuO? 47

48. Mole Quiz (explained) Start with ------------------- 39.098 g K
6 K + B2O3  3 K2O + 2 B
Please write out the conversion factor for the reactants and products of the above chemical equation.
1) mol K
g K
2) mol B2O3
g B2O3
3) mol K2O
g K2O
4) mol B
g B

49. Mole Quiz (continued) 3219 g K 6 K + B2O3  3 K2O + 2 B
A processor needs to convert g B2O3 to pure B. What mass of K is needed for this reaction? (show all your work)
955.0 g B2O3 X mol B2O3
= mol B2O3
g B2O3
6 mol K g K
X mol B2O3 = mol K X =
1 mol B2O mol K
3219 g K

50. Mole Quiz (continued) 640.8 g S8 16 Al + 3 S8  8 Al2S3
A processor plans to create g of Al2S3. What mass of pure S is needed for this reaction? (show all your work)
1000. g Al2S3 X mol Al2S3
= mol Al2S3
g Al2S3
3 mol S g S8
X mol Al2S3 = mol S8 X =
8 mol Al2S mol S8
640.8 g S8

51. HW 7
Read & take notes on Section C.9

52. C.9 – Compositions of Materials
The percent mass of each material found in an item is called the percent composition.
Hint: remember solution concentration
Example: post-1982 penny has a mass of g has g zinc & g copper. What is the percent composition?
g zinc
2.500 g total
X 100% =
97.50 % zinc

53. C.9 – Compositions of Materials (continued)
Example: post-1982 penny has a mass of g has g zinc & g copper. What is the percent composition?
g copper
2.500 g total
X 100% =
2.50 % copper

54. C.9 – Compositions of Materials (continued)
Why are the ideas of molar mass & percentage composition so important?
Some Copper-containing Minerals
Common Name Formula
Chalcocite Cu2S
Chalcopyrite CuFeS2
Malachite Cu2CO3(OH)2
It helps us determine which is more profitable to mine.
Mass of copper
Mass of Cu2S
X 100% =
% copper

55. C.10 – Percent Composition
HW 8 – Questions 1-2 on pg 168

56. C.10 – Percent Composition
An atom inventory for Cu3(CO3)2(OH)2
There are :
3 Cu atoms
2 C atoms
8 O atoms
2 H atoms

57. C.10 – Percent Composition (continued)
2. Percent copper in ?
a. Chalcopyrite CuFeS2
Start with . . .
63.55 g g + (2 x 32.07g)
= g
63.55 g g
x 100 %
= % Cu

58. C.10 – Percent Composition (continued)
2. Percent copper in ?
b. malachite Cu2CO3 (OH)2
c. Chalcocite , at 79.85% copper, would be the most profitable to mine.
(2 x g) g + (5 x g) + (2 x 1.008)
= g
(2 x g) g
x 100 %
= % Cu

59. C.11 – Retrieving Copper
HW 9 – please pre-read the lab.

60. HW 10
Read & take notes on Section C.12

61. C.12 – Conservation in the Community
Resources…
Renewable
Fresh water,
Air,
Fertile soil ,
Plants, and
Animals
EVENTUALLY replenished by natural processes.
Nonrenewable
Metals,
Natural gas,
coal and
petroleum
CANNOT be readily replenished.

62. C.13 – Rethinking, Reusing, Replacing & Recycling
HW 11 Questions 1 & 2 on pg. 176

63. C.14 – The Life Cycle of Material
C.15 – Copper Life-Cycle Analysis
Not covered