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

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Presentation on theme: "For chemists, a mole is NOT a small furry animal."— Presentation transcript:

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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

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

9 196.96655 of these bars would contain a MOLE of gold molecules. One gram bar of gold.

10 A mole is equal to 6.02 X 10 23 of anything. cm 3

11 6.02 X 10 23 is known as Avogadro's number. cm 3

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. cm 3

14 There are 6.02 X 10 23 molecules of water is this cylinder. cm 3

15 There are 6.02 X 10 23 molecules of water is this cylinder. cm 3

16 The formula mass of water is 18 amu. cm 3

17 Water has a density of 1 g/cm 3. cm 3

18 18 cm 3 of water has a mass of 18 grams. cm 3

19 18 grams of water contains a mole of molecules. cm 3

20 The mole concept is important because it allows us to actually WEIGH atoms and molecules in the lab. cm 3

21 What is the mass of a water molecule? cm 3

22 6.02 X 10 23 H 2 O molecules 18 grams = 3 X 10 -23 g / H 2 O molecule

23 2 Important Mole Calculations 1. Convert mass to moles and moles to molecules (particles).

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

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

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

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

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

33 Write these conversion factors on your Paper Periodic Table RIGHT NOW: 1 mole = 6.02 X 10 23 = formula mass particles atoms molecules in grams

34 Practice Problem #1: What is the mass in grams of 2.2 X 10 15 molecules of K 2 S 2 O 8 ? 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 10 15 molecules K 2 S 2 O 8 Write what is given.

37 2.2 X 10 15 molecules K 2 S 2 O 8 Draw these lines.

38 2.2 X 10 15 molecules K 2 S 2 O 8 What does this line mean?

39 2.2 X 10 15 molecules K 2 S 2 O 8 What does this line mean?

40 2.2 X 10 15 molecules K 2 S 2 O 8 What units go here?

41 2.2 X 10 15 molecules K 2 S 2 O 8 molecules Why?

42 2.2 X 10 15 molecules K 2 S 2 O 8 What units go here? molecules

43 2.2 X 10 15 molecules K 2 S 2 O 8 molecules grams Why?

44 2.2 X 10 15 molecules K 2 S 2 O 8 molecules grams Where do we get the numbers?

45 Useful conversion factors: 1 mole = 6.02 X 10 23 = formula mass particles atoms molecules in grams

46 2.2 X 10 15 molecules K 2 S 2 O 8 6.02 X 10 23 molecules K 2 S 2 O 8 formula mass in grams =

47 2.2 X 10 15 molecules K 2 S 2 O 8 6.02 X 10 23 molecules K 2 S 2 O 8 formula mass in grams = These units cancel.

48 2.2 X 10 15 molecules K 2 S 2 O 8 6.02 X 10 23 molecules K 2 S 2 O 8 K = 2 X 39 = 78 S = 2 X 32 = 64 O = 8 X 16 = 128 270 grams = 270 Formula mass calculation.

49 2.2 X 10 15 molecules K 2 S 2 O 8 6.02 X 10 23 molecules K 2 S 2 O 8 K = 2 X 39 = 78 S = 2 X 32 = 64 O = 8 X 16 = 128 270 grams = 270 These are the units are asked for.

50 2.2 X 10 15 molecules K 2 S 2 O 8 6.02 X 10 23 molecules K 2 S 2 O 8 K = 2 X 39 = 78 S = 2 X 32 = 64 O = 8 X 16 = 128 270 grams = 270 The problem is worked - punch buttons to get the number.

51 2.2 X 10 15 molecules K 2 S 2 O 8 6.02 X 10 23 molecules K 2 S 2 O 8 K = 2 X 39 = 78 S = 2 X 32 = 64 O = 8 X 16 = 128 270 grams = 270 this number

52 2.2 X 10 15 molecules K 2 S 2 O 8 6.02 X 10 23 molecules K 2 S 2 O 8 K = 2 X 39 = 78 S = 2 X 32 = 64 O = 8 X 16 = 128 270 grams = 270 times this number

53 2.2 X 10 15 molecules K 2 S 2 O 8 6.02 X 10 23 molecules K 2 S 2 O 8 K = 2 X 39 = 78 S = 2 X 32 = 64 O = 8 X 16 = 128 270 grams = 270 divided by this number

54 9.9 X 10 -7 g K 2 S 2 O 8 2.2 X 10 15 molecules K 2 S 2 O 8 6.02 X 10 23 molecules K 2 S 2 O 8 K = 2 X 39 = 78 S = 2 X 32 = 64 O = 8 X 16 = 128 270 grams = 270 EQUALS

55 9.9 X 10 -7 g K 2 S 2 O 8 2.2 X 10 15 molecules K 2 S 2 O 8 6.02 X 10 23 molecules K 2 S 2 O 8 K = 2 X 39 = 78 S = 2 X 32 = 64 O = 8 X 16 = 128 270 grams = 270 Does the answer have the right number of significant digits?

56 9.9 X 10 -7 g K 2 S 2 O 8 2.2 X 10 15 molecules K 2 S 2 O 8 6.02 X 10 23 molecules K 2 S 2 O 8 K = 2 X 39 = 78 S = 2 X 32 = 64 O = 8 X 16 = 128 270 grams = 270 NOW write this solution under the problem.

57 Practice Problem #2: A sample of CaCO 3 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 CaCO 3 has a mass of 25.5 grams. How many total atoms are in the sample? 7.68 X 10 23 atoms First one with this answer gets 20 points added to their lowest test grade.

59 7.68 X 10 23 atoms 25.5 g CaCO 3

60 7.68 X 10 23 atoms 25.5 g CaCO 3 6.02 X 10 23 molecules 100 g CaCO 3 Ca = 1 X 40 = 40 C = 1 X 12 = 12 O = 3 X 16 = 48 100

61 7.68 X 10 23 atoms 25.5 g CaCO 3 6.02 X 10 23 molecules 5 atoms 100 g CaCO 3 1 molecule Ca = 1 X 40 = 40 C = 1 X 12 = 12 O = 3 X 16 = 48 100

62 Practice Problem #3: Given 100 grams of silver nitrate, how many atoms of silver are in the sample? Set up the factor-label solution for this problem. 4 X 10 23 atoms Ag

63 molecules AgNO 3 4 X 10 23 atoms Ag 100 g AgNO 3 6.02 X 10 23 1 atom Ag 170 g AgNO 3 1 Ag = 1 X 108 = 108 N = 1 X 14 = 14 O = 3 X 16 = 48 170 molecule AgNO 3

64 Practice Problem #4: Calculate the mass, in kilograms, of 0.55 mole of chlorine molecules. Set up the factor-label solution for this problem. 0.039 kg Cl 2

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

66 Practice Problem #5: The density of C 2 H 5 OH is 0.8 g/cm 3. If a sample of this substance contains 3.2 X 10 23 molecules, what is the volume of the sample? Set up the factor-label solution for this problem. 31 cm 3 C 2 H 5 OH

67 molecules C 2 H 5 OH molecules C 2 H 5 OH 31 cm 3 C 2 H 5 OH 3.2 X 10 23 46 g C 2 H 5 OH 1 cm 3 6.02 X 10 23 C - 2 X 12 = 24 H - 6 X 1 = 6 O - 1X 16 = 16 46 0.8 g

68

69 Galvanized Nail

70 End The Mole


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