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The Mole Concept

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

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A mole is the SI unit for amount of substance.

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

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A mole is like a dozen - only MORE.

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

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**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

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

gold molecules.

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**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?

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

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

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**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.

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

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

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

cm3 How do we know?

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cm3 The formula mass of water is 18 amu.

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cm3 Water has a density of 1 g/cm3.

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

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cm3 18 grams of water contains a mole of molecules.

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**The mole concept is important because it allows us to actually WEIGH**

atoms and molecules in the lab. cm3

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What is the mass of a water molecule? cm3

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

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**2 Important Mole Calculations**

Convert mass to moles and moles to molecules (particles).

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**2 Important Mole Calculations**

2. Determine the concentration of solutions - Molarity.

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**Most mole calculations use**

the Factor-Label method of problem solving - also called dimensional analysis.

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First: Write what you are given.

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

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

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

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**2 Form the habit of working neatly, canceling units as you go,**

and circling the answer.

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

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

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**Write these conversion factors on your Paper Periodic Table RIGHT NOW:**

1 mole = 6.02 X 1023 = formula mass particles atoms molecules in grams

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**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.

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To work this problem, you would:

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2.2 X 1015 molecules K2S2O8 Write what is given.

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2.2 X 1015 molecules K2S2O8 Draw these lines.

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**What does this line mean?**

2.2 X 1015 molecules K2S2O8 What does this line mean?

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**What does this line mean?**

2.2 X 1015 molecules K2S2O8 What does this line mean?

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2.2 X 1015 molecules K2S2O8 What units go here?

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2.2 X 1015 molecules K2S2O8 molecules Why?

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2.2 X 1015 molecules K2S2O8 molecules What units go here?

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2.2 X 1015 molecules K2S2O8 grams molecules Why?

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

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**Useful conversion factors:**

1 mole = 6.02 X 1023 = formula mass particles atoms molecules in grams

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

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**These units cancel. formula mass 2.2 X 1015 molecules in grams K2S2O8**

= 6.02 X 1023 molecules K2S2O8 These units cancel.

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**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.

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**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.

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**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.

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**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

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**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

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**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

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**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

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**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

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**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

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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.

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**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

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25.5 g CaCO3 7.68 X atoms

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**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

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**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

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**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.

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**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

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**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.

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**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

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**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.

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**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

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Practice

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Twelfth Lab Galvanized Nail

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End The Mole

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Chapter 10 The Mole.

Chapter 10 The Mole.

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