1. The
Mole
Concept

2. For chemists, a mole is
NOT a small furry animal.

3. A mole is the SI unit
for amount of substance.

4. This is a dozen eggs -
that's an amount.

5. A mole is like a dozen -
only MORE.

6. One gram bar of gold.
Today, gold
is selling for
$$$ per gram.

7. One gram bar of gold. Actual Size 9 mm X 15 mm X 2 mm
3/8 inch X 3/4 inch X 1/16 in

8. One gram bar of gold. 196.96655 of these bars would contain a MOLE of
gold molecules.

9. One gram bar of gold. 196.96655 of these bars would contain a MOLE of
gold molecules.
How much is a mole
of gold worth?

10. cm3
A mole is
equal to
6.02 X 1023
of anything.

11. cm3
6.02 X 1023
is known as
Avogadro's
number.

12. Avogadro's Hypothesis: equal volumes of gases
at the same temperature and pressure contain
equal numbers of molecules.
He also proposed that oxygen gas and
hydrogen gas were diatomic molecules.

13. A mole of a
substance is
equal to its
formula mass
in grams.
cm3

14. There are
6.02 X 1023
molecules of
water is this
cylinder.
cm3

15. There are 6.02 X 1023 molecules of water is this cylinder.
cm3
How do we know?

16. cm3
The formula
mass of water
is 18 amu.

17. cm3
Water has a
density of
1 g/cm3.

18. cm3
18 cm3 of
water has
a mass of
18 grams.

19. cm3
18 grams of
water contains
a mole of
molecules.

20. The mole concept is important because it allows us to actually WEIGH
atoms and molecules
in the lab.
cm3

21. What is the
mass of a
water molecule?
cm3

22. 18 grams =
6.02 X 1023 H2O molecules
3 X g / H2O molecule

23. 2 Important Mole Calculations
Convert mass to
moles and moles to
molecules (particles).

24. 2 Important Mole Calculations
2. Determine the
concentration of
solutions - Molarity.

25. Most mole calculations use
the Factor-Label method of
problem solving - also called
dimensional analysis.

26. First:
Write what you
are given.

27. Then:
Multiply by fractions
equal to one until all
units cancel except
what you are asked for.

28. Finally:
Punch buttons on the
calculator to get the
number.

29. 1
Setting up the
problem is as
important as
the answer.

30. 2 Form the habit of working neatly, canceling units as you go,
and circling the answer.

31. 3
Remember, units are
just as important
as numbers in the
answer...

32. 3
when the units are
right, the answer
will be right.

33. Write these conversion factors on your Paper Periodic Table RIGHT NOW:
1 mole = 6.02 X 1023 = formula mass
particles
atoms
molecules
in grams

34. What is the mass in grams of 2.2 X 1015 molecules of K2S2O8?
Practice Problem #1:
What is the mass in grams
of 2.2 X 1015 molecules of
K2S2O8?
Write this problem, then put your pen DOWN until told to pick it up.

35. To work this problem,
you would:

36. 2.2 X 1015 molecules
K2S2O8
Write what is given.

37. 2.2 X 1015 molecules
K2S2O8
Draw these lines.

38. What does this line mean?
2.2 X 1015 molecules
K2S2O8
What does this line mean?

39. What does this line mean?
2.2 X 1015 molecules
K2S2O8
What does this line mean?

40. 2.2 X 1015 molecules
K2S2O8
What units go here?

41. 2.2 X 1015 molecules
K2S2O8
molecules
Why?

42. 2.2 X 1015 molecules
K2S2O8
molecules
What units go here?

43. 2.2 X 1015 molecules
K2S2O8
grams
molecules
Why?

44. Where do we get the numbers? 2.2 X 1015 molecules grams K2S2O8

45. Useful conversion factors:
1 mole = 6.02 X 1023 = formula mass
particles
atoms
molecules
in grams

46. formula mass
in grams
2.2 X 1015 molecules
K2S2O8 =
6.02 X 1023
molecules
K2S2O8

47. These units cancel. formula mass 2.2 X 1015 molecules in grams K2S2O8=
6.02 X 1023
molecules
K2S2O8
These units cancel.

48. Formula mass calculation.
2.2 X 1015 molecules
K2S2O8
270 grams =
6.02 X 1023
molecules
K2S2O8
K = 2 X 39 = 78
S = 2 X 32 = 64
O = 8 X 16 = 128
270
Formula mass calculation.

49. These are the units are asked for. 2.2 X 1015 molecules K2S2O8
270 grams =
6.02 X 1023
molecules
K2S2O8
K = 2 X 39 = 78
S = 2 X 32 = 64
O = 8 X 16 = 128
270
These are the units
are asked for.

50. punch buttons to get the number.
2.2 X 1015 molecules
K2S2O8
270 grams =
6.02 X 1023
molecules
K2S2O8
K = 2 X 39 = 78
S = 2 X 32 = 64
O = 8 X 16 = 128
270
The problem is worked -
punch buttons to get the
number.

51. 2.2 X 1015 molecules K2S2O8 270 grams = 6.02 X 1023 molecules K2S2O8
K = 2 X 39 = 78
S = 2 X 32 = 64
O = 8 X 16 = 128
this number
270

52. 2.2 X 1015 molecules K2S2O8 270 grams = 6.02 X 1023 molecules K2S2O8
K = 2 X 39 = 78
S = 2 X 32 = 64
O = 8 X 16 = 128
times this number
270

53. 2.2 X 1015 molecules K2S2O8 270 grams = 6.02 X 1023 molecules K2S2O8
K = 2 X 39 = 78
S = 2 X 32 = 64
O = 8 X 16 = 128
divided by this number
270

54. 9.9 X 10 -7 g K2S2O8 2.2 X 1015 molecules K2S2O8 270 grams =
K = 2 X 39 = 78
S = 2 X 32 = 64
O = 8 X 16 = 128
EQUALS
270
9.9 X g K2S2O8

55. Does the answer have the right number of significant digits?
2.2 X 1015 molecules
K2S2O8
270 grams =
6.02 X 1023
molecules
K2S2O8
K = 2 X 39 = 78
S = 2 X 32 = 64
O = 8 X 16 = 128
Does the answer have the right
number of significant digits?
270
9.9 X g K2S2O8

56. NOW write this solution under 9.9 X 10 -7 g K2S2O8 the problem.
2.2 X 1015 molecules
K2S2O8
270 grams =
6.02 X 1023
molecules
K2S2O8
K = 2 X 39 = 78
S = 2 X 32 = 64
O = 8 X 16 = 128
270
NOW write this
solution under
the problem.
9.9 X g K2S2O8

57. Practice Problem #2:
A sample of CaCO3 has
a mass of 25.5 grams.
How many total atoms
are in the sample?
Write this problem down.

58. Practice Problem #2: A sample of CaCO3 has a mass of 25.5 grams.
How many total atoms
are in the sample?
First one with this answer
gets 20 points added to
their lowest test grade.
7.68 X atoms

59. 25.5 g CaCO3
7.68 X atoms

60. 7.68 X 1023 atoms 25.5 g CaCO3 6.02 X 1023 molecules 100 g CaCO3
Ca = 1 X 40 = 40
C = 1 X 12 = 12
O = 3 X 16 = 48
7.68 X atoms
100

61. 7.68 X 1023 atoms 25.5 g CaCO3 6.02 X 1023 molecules 5 atoms
100 g CaCO molecule
Ca = 1 X 40 = 40
C = 1 X 12 = 12
O = 3 X 16 = 48
7.68 X atoms
100

62. Set up the factor-label solution for this problem.
Practice Problem #3:
Given 100 grams of
silver nitrate, how many
atoms of silver are
in the sample?
4 X 1023
atoms Ag
Set up the factor-label
solution for this problem.

63. 4 X 1023 atoms Ag molecules AgNO3 100 g AgNO3 6.02 X 1023 1 atom Ag
Ag = 1 X 108 = 108
N = 1 X 14 = 14
O = 3 X 16 = 48
4 X atoms Ag
170

64. Set up the factor-label solution for this problem.
Practice Problem #4:
Calculate the mass,
in kilograms, of 0.55
mole of chlorine
molecules.
0.039 kg Cl2
Set up the factor-label
solution for this problem.

65. 0.039 kg Cl2 0.55 mole Cl2 70 g Cl2 1 kg 1 mole Cl2 1000 g
Cl = 2 X 35 = 70
0.039 kg Cl2

66. Set up the factor-label solution for this problem.
Practice Problem #5:
The density of C2H5OH is
0.8 g/cm3. If a sample of this
substance contains 3.2 X 1023
molecules, what is the
volume of the sample?
31 cm3
C2H5OH
Set up the factor-label
solution for this problem.

67. 31 cm3 C2H5OH molecules C2H5OH 3.2 X 1023 46 g C2H5OH 1 cm3
C - 2 X 12 = 24
H - 6 X 1 = 6
O - 1X 16 = 16
molecules
C2H5OH 46
31 cm3 C2H5OH

68. Practice

69. Twelfth Lab
Galvanized Nail

70. End
The Mole