Download presentation

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

Similar presentations

Presentation is loading. Please wait....

OK

Stoichiometry.

Stoichiometry.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google