MOLE 10
A frustration,
a concept, a
unit
WHY?
Look at the two
recipes.
What is the
same?
What is
d i f fe r e n t ?
EARLY CHEMISTRY
 If you were given a pile of this chemical and a pile of that
chemical how do you know how much to put into were?
 This is the issue of the early chemist.
 Compounds react, but why and how?
 On an atomic view
 C + O 2  CO 2
 2H 2 + O 2  2H 2 O
HOW DO WE SCALE THAT UP?
 How do we go from atoms to something we can touch and
measure?
 Avogadro expanded on the theories around him
 It theorized that the number of particles in an atom will be
similar regardless of which element.
 Avogadro’s number is based on the number of atoms in 12 grams
of carbon-12.
CONVERTING BETWEEN MOLES AND
PARTICLES





6.02 x 10 23 is the number of particles in one mole
One mole is the same as saying
One dozen
One ream
One pair




These are all numbers that are associated with the label.
One dozen= 12
One ream= 500
One pair= 2
If you have one mole of
lithium how many atoms do
you have?
6.02x10 23
If you have 3 moles of
Fluorine how many atoms
are there?
1.81x10 24
THE MATH
 3.92x10 24 molecules of carbon dioxide gas
will be how many moles?
 How many moles of carbon are there in
that problem?
 6.5 moles
 How many moles of of oxygen are there?
 13 moles
SETTING UP THE
CONVERSION
 6.5 moles
MOLES TO GRAMS
Something
we can
measure
THE IDEA OF A MOLE
 Avogadro based the number of atoms on the number of atoms
found in 12 grams of carbon -12.
 One mole is equal to the atomic weight of the substance in
grams.
 How many grams are in one mole of Phosphorus?
 31 grams
 What is the mass of one mole of silver?
 108 grams
How many grams in 3.75
moles of Carbon?
45 grams C
4.05 moles of Zirconium
has how much mass?
369 grams Zr
How many moles are in 200
grams of Vanadium?
3.92 moles V
How many moles are in 8
grams of Helium?
2 moles He
MASS TO
MOLE AND
MOLE TO
MASS
 When we start talking about compound and
molecules then we change it up.
 Carbon dioxide
 CO 2
 C (12) + O (16)*2
MOLAR MASS
 Molar mass is the weight of a substance
with one mole of the substance.
 When dealing with elements this is easy.
 Carbon is 12g
 Oxygen is 16g
 Hydrogen is 1g
 And so on.
 WHAT IS THE MOLAR MASS FOR…
• Calcium Oxide
• 56 grams
• Magnesium Iodide
• 278 grams
• Sodium Nitrate
• 85 grams
• Sodium Nitrite
• 69 grams
Practicing
molar mass
VOLUME
 Avogadro’s principle
 All gases at the same temperature and pressure contain equal
volumes must contain equal numbers of particle
 One mole contains 6.02x10 23 particles
 Avogadro’s principle explains that it should take up the same
amount of space.
 1 mole of a gas = 22.4 L of that gas
3.72L of He is how many
moles?
0.166 moles Helium
How many L would you have
if you had 1.6 moles of
Chlorine?
35.8 L of Cl
How much space would 2
moles of Xe take up?
44.8 L of Xe
HOW MUCH
SPACE?
MULTI-STEP CONVERSIONS
 If I asked you how many Cups of water are in a gallon of water
how would you do that?
 What if you needed to give 10000 seconds in the number of
hours how would you do that?
 These are both multi step conversions.
 Now if I gave you that you had 3.25X10 23 atoms of Ne gas and
asked you how many much space it takes up in L. We have the
same issue as above.
If 22.4 L of He is stored at
STP what is the mass in
grams?
If you have 275 grams of
Silver. How many particles
of silver do you have?
How many grams are there
in 2.78x10 24 particles of
CO 2?
How much volume does it
have?
EXAMPLES
If 22.4 L of He is stored at
STP what is the mass in
grams?
4 g He
If you have 275 grams of
Silver. How many particles of
silver do you have?
1.53x10 24 part. Ag
How many grams are there in
2.78x10 24 particles of CO 2 ?
203 g CO 2
How much volume does it
have?
103 L CO 2
EMPIRICAL AND
MOLECULAR FORMULAS
PERCENT COMPOSITION
The percent by mass
of each element in a
compound is the
percent composition
of a compound.
100 grams of
xy
55g x
45g y
55x/100xy=
55% element
x
45y/100xy=
45% element
y
 If you have one Mole of NaCHO 3
 First what is the molar mass of
NaCHO 3 ?
 84g
 Now we take each element
mass divided by the molar
mass.
 Na/NaCHO 3
 (23/84)x 100= 27.38%
 C/NaCHO 3
 (12/84) x100= 14.29%
 H/NaCHO 3
 1.19%
 3O/NaCHO 3
 57.14%
CALCULATING
What is the
percent
composition
for each
element in
NaHCO 3 ?
MINI LAB
 Analyze chewing gum
 Are the sweetening and flavoring added as a coating or mixed
throughout?
 Unwrap two pieces of chewing gum. Place each piece on a
weighing paper. Measure and record the mass of each using the
balance.
 Add 150 ml of cold tap water into a 250 ml beaker. Place one
piece of chewing gum in the water, and stir with stirring rod for 2
minutes.
 Pat the gum dry using paper towels. Measure and record the
mass of the dried gum.
 Use the scissors to cut the second piece of gum into small
pieces. Repeat step 3 with new tap water. Try to keep the pieces
from clumping together.
 Strain the water and pat the gum dry on a paper towel. Measure
and record the mass of the dried gum.
ANALYSIS
 Calculate: Calculate the % mass of the flavorings and
sweeteners for both pieces. That would be the amount of
mass lost (sweeteners and flavorings) / by the total mass of
the gum.
 Infer: What can you infer from the two percentages? Is the
gum sugar coated or are the sweeteners mixed in?
 As a Pair you and your lab partner(s) will need to do the
calculation and analysis questions from the book.
 What is the percent by mass of Carbon in
CH 4 ?
 75%
 H= 3.7%
 C= 44.44%
 N= 51.85%
 What is the % by mass of Zn in zinc
iodide?
 20.38%
 Which element has the smallest %
composition in tin nitrate?
 N 15.25% or 11.52%
EXAMPLES
 What is the percent composition of
Hydrocyanic acid (HCN)?
 If you are given % composition
but not the actual formula you
can use the % to find how many
moles of each element are in
the compound.
 Example
 If you are given that a
compound has
 48.64% C
 8.16% H
 43.20% O
 You can find out how much
many carbon, to hydrogen, to
oxygen are in your compound.
EMPIRICAL
FORMULA
The smallest
w h o l e - n um b er
mole ratio of
the elements.
 When finding Empirical formulas
from percent composition
 Look at the % given to you.
 Assume that they are out of 100
grams total.
 100%= 100g
 For Carbon we would then state
that 48.64%= 48.64g
 H=8.16 g
 O= 43.20g
 Next you find how many moles
there are of each element
 Carbon
48.64g(1mole/12g)=4.05moles
 H 8.16g (1mole/ 1g) = 8.1 moles
 O 43.20g (1 mole/ 16g) =2.7 moles
HOW DO
WE DO IT
48.64% C
8.16% H
43.20% O
Now we simplify
C=4.05, H= 8.1, O= 2.7
We divide them all by the
lowest number of moles.
C 4.05/2.7= 1.5 moles C
H 8.1/2.7= 3 moles H
O 2.7/2.7= 1 mole O
If they had all come out in
round able numbers we
would be done, but 1.5 is
half way so there is another
step.
SIMPLIFYIN
G DOWN TO
THE
LOWEST #
OF MOLES
To make every thing a
whole number we need to
multiply to get rid of the .5.
So we multiply everyone by
2.
C 1.5(2)= 3 moles Carbon
H 3(2)= 6 moles Hydrogen
O 1(2)= 2 moles O
So we would write the
empirical formula: C 3H 6O 2
HAS TO BE
A WHOLE
NUMBER
3.08% H, 31.61% P, 65.31%
O
H 3PO 4
36.11% Ca, 63.89% Cl
CaCl 2
42% C, 6% H, 52% O
CH 2O
PROBLEMS
Find the
e m p i r i ca l
f o r m ul a s f o r
t h e f o l l owi n g
 Sometimes substances with very different
properties and structures have the same
empirical formula.
C2H2
C4H4
are all very different, how do you draw each
 The Molecular formula, specifies the actual number of
atoms of each element in one molecule or formula unit of
the substance.
Experimentally determined molar mass = n
Mass of empirical formula
Molecular
Formula
•
MOLECULAR FORMULA = (EMPIRICAL
FORMULA)n
• If you have 40.68% C, 5.08% H, and 54.24% O in
a chemical analysis of Succinic acid, and an
experimental mass of 118.1 g/mole.
• Find the empirical formula
• (40.68g/12) …
• You should find 1 C, 1.5 H, and 1 O (multiplied by
2 to get out of the 1.5
• Gives you 2 C, 3 H, and 2 O: C2H3O2
• No we find the molar mass of the empirical
formula. (59g/mole)
example
• We take the experimental molar mass divided by the
empirical molar mass
• 118.1/59=n~ 2
• We then multiply our empirical formula by our n
2(C2H3O2)=
• C4H6O4 molecular formula
Cont.
The mineral ilmenites is usually mined and
processed for titanium. A sample of ilmenite
contains 5.41 G of iron, 4.64 G of titanium, and
4.65 G of oxygen.
What is the empirical formula?
A chemical analysis shows that a compound commonly
used as a bleaching agent is composed of 13.79 g Na,
21.27g Cl, and 9.60 g O. Determine the empirical
formulas for this compound.
If you are given an experimental value of 222.08
grams/mole, what is the molecular formula?
Practice
problems