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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.
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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
