For chemists, a mole isNOT a small furry animal.

A mole is the SI unitfor 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.

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

196.96655of these barswould containa MOLE ofgold molecules.

One gram bar of gold.

196.96655of these barswould containa MOLE ofgold molecules.

One gram bar of gold.

A mole is equal to6.02 X 1023 of anything.

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6.02 X 1023 is known asAvogadro'snumber.

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Avogadro's Hypothesis: equal volumes of gasesat 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 asubstance isequal to itsformula massin grams.

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There are6.02 X 1023 molecules ofwater is thiscylinder.

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There are6.02 X 1023 molecules ofwater is thiscylinder.

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The formula mass of water is 18 amu.

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Water has a density of 1 g/cm3.

cm3

18 cm3 of water has a mass of18 grams.

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18 grams ofwater containsa mole ofmolecules.

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The mole concept isimportant becauseit allows us toactually WEIGHatoms and moleculesin the lab.

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What is the mass of a water molecule?

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6.02 X 1023 H2O molecules

18 grams=

3 X 10-23 g / H2O molecule

2 ImportantMole Calculations

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

2. Determine the concentration of solutions - Molarity.

2 ImportantMole Calculations

Most mole calculations usethe Factor-Label method ofproblem solving - also calleddimensional analysis.

First:

Write what you are given.

Then:

Multiply by fractionsequal to one until allunits cancel except what you are asked for.

Finally:

Punch buttons on thecalculator to get thenumber.

Setting up the problem is as important as the answer.

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

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Remember, units arejust as importantas numbers in theanswer...

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when the units areright, the answerwill be right.

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Write these conversion factors on your PaperPeriodic Table RIGHT NOW:

1 mole = 6.02 X 1023 = formula massparticles

atomsmolecules

in grams

Practice Problem #1:

What is the mass in gramsof 2.2 X 1015 molecules ofK2S2O8?

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.

2.2 X 1015 molecules K2S2O8

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

What units go here?

molecules

2.2 X 1015 molecules K2S2O8

molecules

grams

Why?

2.2 X 1015 molecules K2S2O8

molecules

grams

Where do we get the numbers?

Useful conversion factors:

1 mole = 6.02 X 1023 = formula massparticles

atomsmolecules

in grams

2.2 X 1015 molecules K2S2O8

6.02 X 1023 moleculesK2S2O8

formula massin grams

=

2.2 X 1015 molecules K2S2O8

6.02 X 1023 moleculesK2S2O8

formula massin grams

=

These units cancel.

2.2 X 1015 molecules K2S2O8

6.02 X 1023 moleculesK2S2O8

K = 2 X 39 = 78S = 2 X 32 = 64O = 8 X 16 = 128

270 grams=

270

Formula mass calculation.

2.2 X 1015 molecules K2S2O8

6.02 X 1023 moleculesK2S2O8

K = 2 X 39 = 78S = 2 X 32 = 64O = 8 X 16 = 128

270 grams=

270

These are the unitsare asked for.

2.2 X 1015 molecules K2S2O8

6.02 X 1023 moleculesK2S2O8

K = 2 X 39 = 78S = 2 X 32 = 64O = 8 X 16 = 128

270 grams=

270

The problem is worked -punch buttons to get thenumber.

2.2 X 1015 molecules K2S2O8

6.02 X 1023 moleculesK2S2O8

K = 2 X 39 = 78S = 2 X 32 = 64O = 8 X 16 = 128

270 grams=

270

this number

2.2 X 1015 molecules K2S2O8

6.02 X 1023 moleculesK2S2O8

K = 2 X 39 = 78S = 2 X 32 = 64O = 8 X 16 = 128

270 grams=

270

times this number

2.2 X 1015 molecules K2S2O8

6.02 X 1023 moleculesK2S2O8

K = 2 X 39 = 78S = 2 X 32 = 64O = 8 X 16 = 128

270 grams=

270divided by this number

9.9 X 10 -7 g K2S2O8

2.2 X 1015 molecules K2S2O8

6.02 X 1023 moleculesK2S2O8

K = 2 X 39 = 78S = 2 X 32 = 64O = 8 X 16 = 128

270 grams=

270EQUALS

9.9 X 10 -7 g K2S2O8

2.2 X 1015 molecules K2S2O8

6.02 X 1023 moleculesK2S2O8

K = 2 X 39 = 78S = 2 X 32 = 64O = 8 X 16 = 128

270 grams=

270

Does the answer have the rightnumber of significant digits?

9.9 X 10 -7 g K2S2O8

2.2 X 1015 molecules K2S2O8

6.02 X 1023 moleculesK2S2O8

K = 2 X 39 = 78S = 2 X 32 = 64O = 8 X 16 = 128

270 grams=

270

NOW write thissolution underthe problem.

Practice Problem #2:

A sample of CaCO3 hasa mass of 25.5 grams.How many total atomsare in the sample?

Write this problem down.

Practice Problem #2:

A sample of CaCO3 hasa mass of 25.5 grams.How many total atomsare in the sample?

7.68 X 1023 atoms

First one with this answer

gets 20 points added to their lowest test grade.

7.68 X 1023 atoms

25.5 g CaCO3

7.68 X 1023 atoms

25.5 g CaCO3 6.02 X 1023 molecules

100 g CaCO3

Ca = 1 X 40 = 40C = 1 X 12 = 12O = 3 X 16 = 48

100

7.68 X 1023 atoms

25.5 g CaCO3 6.02 X 1023 molecules 5 atoms

100 g CaCO3 1 molecule

Ca = 1 X 40 = 40C = 1 X 12 = 12O = 3 X 16 = 48

100

Practice Problem #3:

Given 100 grams of silver nitrate, how manyatoms of silver arein the sample?

Set up the factor-labelsolution for this problem.

4 X 1023

atoms Ag

moleculesAgNO3

4 X 1023 atoms

Ag

100 g AgNO3 6.02 X 1023 1 atom Ag

170 g AgNO3 1

Ag = 1 X 108 = 108N = 1 X 14 = 14O = 3 X 16 = 48

170

moleculeAgNO3

Practice Problem #4:

Calculate the mass,in kilograms, of 0.55 mole of chlorine molecules.

Set up the factor-labelsolution for this problem.

0.039 kg Cl2

0.039 kg Cl2

0.55 mole Cl2 70 g Cl2 1 kg

1 mole Cl2 1000 g

Cl = 2 X 35 = 70

Practice Problem #5:

The density of C2H5OH is 0.8 g/cm3. If a sample of thissubstance contains 3.2 X 1023 molecules, what is thevolume of the sample?

Set up the factor-labelsolution for this problem.

31 cm3

C2H5OH

moleculesC2H5OH

moleculesC2H5OH

31 cm3 C2H5OH

3.2 X 1023 46 g C2H5OH 1 cm3

6.02 X 1023

C - 2 X 12 = 24H - 6 X 1 = 6O - 1X 16 = 16

46

0.8 g

Galvanized Nail

End

The Mole