# The Mole Concept.

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The Mole Concept

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

A mole is the SI unit for amount of substance.

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

A mole is like a dozen - only MORE.

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

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

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

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?

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

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

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.

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

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

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

cm3 The formula mass of water is 18 amu.

cm3 Water has a density of 1 g/cm3.

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

cm3 18 grams of water contains a mole of molecules.

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

What is the mass of a water molecule? cm3

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

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

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

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

First: Write what you are given.

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

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

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

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

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

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

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

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.

To work this problem, you would:

2.2 X 1015 molecules K2S2O8 Write what is given.

2.2 X 1015 molecules K2S2O8 Draw these lines.

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

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

2.2 X 1015 molecules K2S2O8 What units go here?

2.2 X 1015 molecules K2S2O8 molecules Why?

2.2 X 1015 molecules K2S2O8 molecules What units go here?

2.2 X 1015 molecules K2S2O8 grams molecules Why?

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

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

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

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

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.

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.

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.

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

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

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

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

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

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

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.

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

25.5 g CaCO3 7.68 X atoms

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

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

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.

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

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.

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

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.

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

Practice

Twelfth Lab Galvanized Nail

End The Mole