Download 2.2 X 10 15 molecules K 2 S 2 O 8 6.02 X 10 23 molecules K 2 S 2 O 8

For chemists, a mole is
NOT a small furry animal.
A mole is the SI unit
for amount of substance.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
This is a dozen eggs that's an amount.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
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/
3/ inch X 1/ in
inch
X
8
4
16
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.
cm3
A mole is
equal to
23
6.02 X 10
of anything.
cm3
23
10
6.02 X
is known as
Avogadro's
number.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
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
23
6.02 X 10
molecules of
water is this
cylinder.
cm3
There are
23
6.02 X 10
molecules of
water is this
cylinder.
cm3
cm3
The formula
mass of water
is 18 amu.
cm3
Water has a
density of
3
1 g/cm .
cm3
3
cm
18
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 10-23 g / H2O molecule
2 Important
Mole Calculations
1. 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.
Qu ic kTi me™ a nd a
TIFF (Unc om pres se d) de co mp re ss or
are n ee de d to s ee th is pi ctu re .
1
Setting up the
problem is as
important as
the answer.
Qu ic kTi me™ a nd a
TIFF (Unc om pres se d) de co mp re ss or
are n ee de d to s ee th is pi ctu re .
2
Form the habit of
working neatly,
canceling units as you go,
and circling the answer.
Qu ic kTi me™ a nd a
TIFF (Unc om pres se d) de co mp re ss or
are n ee de d to s ee th is pi ctu re .
3
Remember, units are
just as important
as numbers in the
answer...
Qu ic kTi me™ a nd a
TIFF (Unc om pres se d) de co mp re ss or
are n ee de d to s ee th is pi ctu re .
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
in grams
atoms
molecules
Practice Problem #1:
What is the mass in grams
15
of 2.2 X 10 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.
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
molecules
What units go here?
2.2 X 1015 molecules
K2S2O8
grams
molecules
Why?
2.2 X 1015 molecules
K2S2O8
grams
molecules
Where do we get
the numbers?
Useful conversion factors:
1 mole = 6.02 X 1023 = formula mass
particles
in grams
atoms
molecules
2.2 X 1015 molecules
K2S2O8
formula mass
in grams
=
6.02 X 1023
molecules
K2S2O8
2.2 X 1015 molecules
K2S2O8
formula mass
in grams
=
6.02 X 1023
molecules
K2S2O8
These units cancel.
2.2 X 1015 molecules
K2S2O8
K = 2 X 39 = 78
S = 2 X 32 = 64
O = 8 X 16 = 128
270
270 grams
=
6.02 X 1023
molecules
K2S2O8
Formula mass calculation.
2.2 X 1015 molecules
K2S2O8
K = 2 X 39 = 78
S = 2 X 32 = 64
O = 8 X 16 = 128
270
270 grams
6.02 X 1023
molecules
K2S2O8
These are the units
are asked for.
=
2.2 X 1015 molecules
K2S2O8
K = 2 X 39 = 78
S = 2 X 32 = 64
O = 8 X 16 = 128
270
270 grams
6.02 X 1023
molecules
K2S2O8
The problem is worked punch buttons to get the
number.
=
2.2 X 1015 molecules
K2S2O8
K = 2 X 39 = 78
S = 2this
X 32
= 64
number
O = 8 X 16 = 128
270
270 grams
6.02 X 1023
molecules
K2S2O8
=
2.2 X 1015 molecules
K2S2O8
K = 2 X 39 = 78
number
S =times
2 X 32 this
= 64
O = 8 X 16 = 128
270
270 grams
6.02 X 1023
molecules
K2S2O8
=
2.2 X 1015 molecules
K2S2O8
270 grams
6.02 X 1023
K = 2 X 39 = 78
molecules
S = 2 X 32 = 64
K2S2O8
O = 8 X 16 = 128
divided by270
this number
=
2.2 X 1015 molecules
K2S2O8
K = 2 X 39 = 78
S = 2 X 32 = 64
O = 8 X 16 = 128
EQUALS
270
270 grams
6.02 X 1023
molecules
K2S2O8
-7
9.9 X 10
g K2S2O8
=
2.2 X 1015 molecules
K2S2O8
270 grams
=
6.02 X 1023
molecules
have the right
K2S2O8
K = 2 X 39 = 78
S = 2 X the
32 = answer
64
Does
O = 8 X 16 = 128
number of270
significant
digits?
-7
9.9 X 10
g K2S2O8
2.2 X 1015 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.
270 grams
6.02 X 1023
molecules
K2S2O8
9.9 X 10 -7
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 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 = 40
C = 1 X 12 = 12
O = 3 X 16 = 48
100
7.68 X 1023
atoms
25.5 g CaCO3
6.02 X 1023 molecules 5 atoms
100 g CaCO3
Ca = 1 X 40 = 40
C = 1 X 12 = 12
O = 3 X 16 = 48
100
1 molecule
7.68 X 1023
atoms
Practice Problem #3:
Given 100 grams of
silver nitrate, how many
atoms of silver are
4 X 1023
in the sample?
atoms Ag
Set up the factor-label
solution for this problem.
100 g AgNO3
6.02 X 1023
molecules
AgNO3
170 g AgNO3
Ag = 1 X 108 = 108
N = 1 X 14 = 14
O = 3 X 16 = 48
170
1 atom Ag
1 molecule
AgNO3
4 X 1023
atoms Ag
Practice Problem #4:
Calculate the mass,
in kilograms, of 0.55
mole of chlorine
0.039 kg Cl2
molecules.
Set up the factor-label
solution for this problem.
0.55 mole Cl2
70 g Cl2
1 kg
1 mole Cl2
1000 g
Cl = 2 X 35 = 70
0.039 kg Cl2
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
31 cm3
volume of the sample? C2H5OH
Set up the factor-label
solution for this problem.
3.2 X 1023
molecules
46 g C2H5OH
C2H5OH
C - 2 X 12 = 24
H-6X 1= 6
O - 1X 16 = 16
1 cm3
6.02 X 1023
molecules
C2H5OH
0.8 g
46
31
3
cm C
2H5OH
Galvanized Nail
End
The Mole