Ch 10 Notes –
Chemical Quantities:
Working With The Mole
Ch 12: Stoichiometry
.

I. 3 ways that chemists measure the quantity
of matter: mass, volume, amount – number
of, by counting.
A. The “mole” is the unit we need to use
when we want to quantify a chemical
change.
 The mole is the official SI standard base unit
(for the “amount” of a substance). The
symbol for mole is mol.

B. Definition: One mole of any
substance is the amount which
contains the same number of particles
as there are atoms in exactly 12 grams
of carbon-12.
 And, since exactly 12 grams of carbon-12
contains 6.02 x 1023 atoms, 1 mole of
anything contains 6.02 x 10 23 particles.
This number is referred to as “Avogadro’s
number”.
It doesn’t even matter what the “particle” is: For
example, 1 mole of beetles would equal 6.02
x 10 23 beetles. If each beetle has 6 legs,
then 1 mole of beetles has 6 moles of legs
(3.61 x 10 24 legs).
1. How many atoms are present in 1
mole of any element? 6.02 x 10 23
atoms
 How many pretzels are contained in 1
mole of pretzels? 6.02 x 10 23 pretzels



How many molecules of CO2 are in 1 mole of
CO2? 6.02 x 10 23 molecules
How many front wheels are on 1 tricycle?


How many back wheels are on 1 tricycle?





1 front wheel
2 back wheels
How many atoms of carbon (the front wheel) are
in 1 mole of CO2?
In 1 mole of CO2, there is 1 mole of carbon,
which = 6.02 x 10 23 atoms of carbon
How many atoms of oxygen (the back wheel)
are in 1 mole of CO2?
In 1 mole of CO2, there are 2 moles of oxygen,
which = 1.20 x 10 24 atoms of oxygen
Mole road map –see Figure 10.12 p 303.
 Left – Number of “given particles” in a substance
(atoms, ions, molecules, …)
Use Avogadro’s number: 6.02 x 10 23 atoms =
1 mole of those atoms
Center – Volume of a gaseous substance at STP
Use the molar volume (the volume of 1 mole of
any gas) 22.4 L of gas = 1 mol of gas
Right – Mass of a Substance
Use the “molar mass” (the mass of 1 mole of any
substance) # of grams XYZ = 1 mol of XYZ
calculated using the periodic table

 1 mole CO 2

 1 mole C
 1 mole C

 or 

 1 mole CO 2

 means that " in 1 mole of CO 2 , there is 1 mole of C"

 2 mole O

 1 mole O 2

 1 mole O 2 
 or 
 means that " in 1 mole of O 2 , there is 1 mole of O"
 1 mole O 

 5 mole O 
 1 mole N 2 O 5 


 or 

 5 mole O 
 1 mole N 2 O 5 

Formula Factor (or Formula Fraction):

Use the formula of a molecule and/or compound to isolate the
atom or ion in question. NEED TO FIX THESE TYPOS:
 1 mole C 
 1 mole CO 2 
 means that " in 1 mole of CO 2 , there is 1 mole of C"

 or 
 1 mole C 
 1 mole CO 2 
 2 mole O 
 1 mole O 2 

 or 
 means that " in 1 mole of O 2 , there are 2 moles of O"
 2 mole O 
 1 mole O 2 
 5 mole O 
 1 mole N 2 O 5 


 or 

 5 mole O 
 1 mole N 2 O 5 
Means that in 1 mole of N2O5 there are 5 moles of O
1. The mole establishes a connection
between the mass, the amount, and (for
gases) the volume of a substance.
 Mathematically, you will need to follow the
arrows in the figure above as you “go
through” the mole, to get from, say, a #
atoms to the mass and/or volume of a
substance.. Careful – Using the mole road
map for volume is legitimate when working
only with a gas. When working with a
solid or liquid, you will need to use the
substance’s density to arrive at volume.

2. While we were learning how to balance
equations, you may have wondered what
unit belongs to each coefficient in the
balanced equation. The answer is “mole”.
 This equation would be read as
“2 moles of Al combine with 3 moles of Cl2,
in order to produce 2 moles of AlCl3.

Thus it is the mole which establishes the
ratio of one substance to another in a
balanced chemical equation.
The coefficients are whole numbers and are
interpreted as having infinite sig figs.
The mole is the unit we use to relate the
amount of one substance to the amount of
another during a chemical change.
II. Factor label problem solving Using Avogadro‘s
number and chemical formulas:
A. The numerical information in each problem
should be identified:
“What you are given” with its unit
“what you want to know” with its unit
a conversion factor (that relates the two units
– written in the form of a ratio or fraction).
The top part of the factor must equal the
bottom part. Ex: 12 eggs / 1 dozen
This factor is flippable – you can use its reciprocal
whenever you need to: 1 dozen / 12 eggs.



1. Start with what you are given as a numerator
(write down both the number and its unit).
2. Multiply the given by your conversion factor,
which is purposely set up so that the
denominator (bottom) unit is identical with, and
thus will cancel with, the unit of the “given”.
3. Last, use your calculator to multiply all the top
numbers, and divide by all the bottom numbers.


Memory aid: MighTy Duck Bottoms
Multiply All Tops Divide By All Bottoms.
B. Guided Practice #1

Note: 6.02 x 10 23 has been rounded to 3 sig figs. As with any
chemistry calculation, pay attention to sig figs!
How many atoms of silicon are contained in 123 mole Si?
Given: 123 mole Si
Want to know: atoms Si
Conversion factor: 1 mol Si = 6.02 x 10 23 atoms Si
123 mol Si x
25
 6.02 x 10 23 atoms Si 
7.40
x
10
atoms Si

 


1 mol Si


Calculator sequence:
123 x 6.02 “special key” 23 ÷ 1 =
Where “special key” might be “EE”, “exp”, “x10y”, mode, ….

Please do Your Turn 1 and 2 at your seats;
then show your teacher.
Your Turn 1:



Given: 3.70 mol Na
Want to know: atoms Na
Conversion factor:
1 mole Na = 6.02 x 10 23 atoms Na
3.70 mol Na x
24
 6.02 x 10 23 atoms Na 
2.23
x
10
atoms Na

 


1 mol Na




1mole He

1.23 x 10 23 atoms He x 

 6.02 x10 23 atoms He 
Your turn 2:
Given: 1.23 x 10 23 atoms He
 Want to know: moles He
 Conversion factor:
6.02 x 10 23 atoms He = 1 mole He

1.23 x 10 23 atoms He x
1 mole He
=
6.02 x 10 23 atoms He
0.204 mol He
III. Interpreting Formulas of Ionic
Compounds
Follow the examples given. Then have your
teacher check your work.
IV.
Interpreting Formulas of Covalent
Compounds
Follow the examples given. Then have your
teacher correct your work.

a.
b.
c.
d.
e.
f.
g.
h.
Answers to III:
2,1
2 mole K, 1 mole S
2,3
2 mole Al, 3 mole 0
1,2
1 mole Mg, 2 mole C2H3O2
1,3
1 mole Al, 3 mole NO3
3,1
3 mole NH4, 1 mole PO4
2,3
2 mole Mg, 3 mole SO4
3,2
3 mole Ca, 2 mole PO4
1,1
1 mole Na, 1 mole HCO3

a.
b.
c.
d.
e.
f.
g.
h.
Answers to IV:
4,10
4 mol P, 10 mol O
2,1
2 mol H, 1 mol O
1,7
1 mol I, 7 mol F
6,1
6 mol B, 1 mol Si
2,5
2 mol N, 5 mol O
12,22,11 12 mol C, 22 mol H, 11 mol O
1,4
1 mol C, 4 mol H
1,3
1 mol N, 3 mol H
Guided Practice 2

Correct this typo:
Want to know = atoms of Br (NOT Br2)
 2 atoms Br
5.0 molecules Br2 x 
 1 molecule Br
2



  10 atoms Br

Your Turn 5


Given: 7.0 formula units Ca3(PO4)2
Want to know: # of PO4 ions
Conversion factor:
1 formula unit Ca3(PO4)2 = 2 PO4 ions


2 PO 4 ions
7.0 formula units Ca 3 (PO 4 ) 2 x 
 1 formula unit Ca 3 (PO 4 ) 2
14 PO4 ions

 



Your turn 6:
Given: 264 carbon atoms
Want ot know: # molecules C12H22O11
Formula factor:
12 C atoms = 1 molecule C12H22O11
 1 molecule C12 H 22 O11 
264 C atoms x 
 
12 carbon atoms


22 molecules C12H22O11
V.
Putting It All Together:
2 Step Problems
Given a formula, work with # of atoms, #
molecules, and # moles.
 Or, given a formula, work with # ions, #
formula units, and 3 moles.

Guided Practice #3

How many atoms of phosphorus are in 35.0
moles of P2O5?

Given: 35.0 moles P2O5
Want to know: atoms P
Convers Factor: 1 mole P = 6.02 x 10 23 atoms P
Formula Factor: 1 mole P2O5 = 2 moles P



 2 moles P   6.02 x 10 23 atomsP 
 x

35.0 molecules P2 05 x 
 1 mol P2 O 5  

1 mol P

 

Typo: moles
4.21 x 10 25 atoms P

VI. “Average atomic mass” and “molar
mass”:

A. The average mass of 1 atom of an
element is indicated in the element’s square
on the periodic table.
 1.
On our large classroom periodic tables, located
along the BACK WALL, the average atomic mass
is rounded to 2 decimal places to the right of the
point, and is found underneath the symbols and/or
names of the elements.
 2. On your personal periodic table, you can
identify the position of the average atomic mass by
using the key provided on the table itself.
 3. The unit for the mass of 1 atom is called the
“atomic mass unit”, and its symbol is amu.

B. The mass of 1 mole of each element is the
same as the average atomic mass; however, its
unit is grams / mole. The mass of 1 mole of
anything is called its “molar mass”.



1. Ex: The average atomic mass for He is 4.00 amu.
Thus the molar mass of He is 4.00 g/mol. This means:
4.00 grams He = 1 mole He
2. What atom has an average mass of 24.31 amu?
Magnesium.
3. What element has a molar mass of 22.99 g/mol?
sodium


For any compound or molecule, the molar mass is
the addition sum of the molar masses of each
element in the compound or molecule.
Examples below:
molar mass of N2 = 14.01 + 14.01 = 28.01 g / mol
 molar mass of CO2
= 12.01 + 16.00 + 16.00 + 16.00 = 44.01 g/mol
 molar mass P2O3 = 2P + 3O =
30.97 + 30.97 + 16.00 + 16.00 + 16.00 = 109.94 g/mol
 molar mass of Ca(NO3)2
= 1 Ca +2N + 6O =40.08 + 2(14.01) + 6 (16.00)
= 164.10 g/mol


Use the rules for addition while calculating
molar mass:
Use 2 decimal places to the right of the
point in your sum*** (least number of
decimal places to the right, based on the
addends given).
Molar mass Ba(NO3)2
Ba =
N2 =
O6 =
Total =
14.01 x 2 =
16.00 x 6 =
137.34
28.02
96.00
261.36 g/mol
Your Turn 7
Please calculate the molar mass of
aluminum acetate, Al(C2H3O2)3.
*****Using the molar masses given on the
periodic table on the back of your final
exam study guide.
 Show your work.
 Check your answer. Ask for help as
needed.

Al(C2H3O2)3
 C2
24.02
 H3
3.03
 O2
+ 32.00
59.05
26.98 + 3(59.05) =
204.13 grams / mole

VII.
A.
Factor label problems using the molar
mass:
Guided Practice Problem #4 – What is the
mass of 125.0 moles of Ca(OH)2?
 74.10 grams Ca(OH) 2 
  9263 grams Ca(OH) 2
125.0 mol Ca(OH) 2 x 
 1 mol Ca(OH) 2

Please do Your Turn # 8 and #9. Ask for help as needed.
Then, you may begin the HW # 11 – 19.
Your Turn 8
Given: 25.55 g Zn
Want: mol Zn
1 mol Zn = 65.38 g Zn
 1 mol Zn 
  0.3908 mol Zn
25.55 g Zn x 
 65.38 g Zn 
Your Turn 9
Given: 25.55 g Zn(NO3)2
 Want: mol Zn
DO YOU SEE THE NEED FOR A
FORMULA FACTOR?
(yes…whenever there is a different substance involved in
the given/want).
Formula factor: 1 mol Zn = 1 mol Zn(NO3)2

 1 mol Zn(NO 3 ) 2  

1 mol Zn
 x 
  0.1349 mol Zn
25.55 g Zn(NO 3 ) 2 x 



 189.40 g Zn(NO 3 ) 2   1 mol Zn(NO 3 ) 2 
X. Avogadro’s_ hypothesis:

Now recognized as a law of nature:
A. Amedeo Avogadro recognized that unlike
liquids and solids, whose molar volumes differ
dramatically, equal volumes of any gas,
measured at the same temperature and
pressure, will contain an equal number of
particles.
 B. Thus, 1 mole (6.02 x 10 23 particles) of
different gases, measured under the same
physical conditions, will have identical
volumes.

C. The reason why the identity of the gas
is irrelevant is because a gas is mostly
empty space; and since the atoms do NOT
touch each other -- their radius, size,
mass, etc. (all the stuff that determines the
identity of the gas), is not important.
 Instead, what is important with gases, is
their environment (temperature,
atmospheric pressure), and number of
particles or molecules that are in the gas.


D. The mole today is commonly referred
to as “Avogadro’s Number”, as we
posthumously remember and honor
Amedeo Avogadro, not only for his
important work with gases, but also
because he established the difference
between atoms and molecules, and
helped determine the formula for water.
XI. The “molar volume”:
1 mole is equal to the number of particles
of any gas that would fit within a 22.4 L
volume at STP (standard temperature and
pressure).
 So, the molar volume of any gas at STP is
22.4 Liters / mole.

1. Standard temperature = the freezing
point of water = 0 oC = 273 K = 32 oF
 Standard pressure = the pressure of our
atmosphere on the earth at sea level
=
1 atm
= 760 torr = 760 mm Hg
= 101.3 kPa = 14.7 psi

Practice Questions:
A. What is the volume of 1 mole of radon
gas at STP?
22.4 L / mole
How many moles of helium are contained
within a balloon having a volume of 22.4
L?
1 mole He

How many moles of nitrous oxide (N2O)
gas would be present in a cube whose
volume is 22.4 L?
1 mole N2O
What would be the volume of 1 mole of the
metal, sodium?

Gotcha!
Sodium isn’t a gas, it’s a solid – You would need to
use the molar mass and the density of sodium to
calculate this answer.

A container is filled with 22.4 L water. How
many moles of water is this?

Trick question – would need to use density
and molar mass of water to calculate the
answer.
Guided Practice – Molar Volume
Problem #5:
 Problem #6:
What is the volume,
How many moles of He,
given 16.9 moles He
given 0.500 L at STP?
at STP?
 1 mol He 
0.500
L
He
x

  0.0223 mol He
Insert “He” here:
22.4 L He


 22.4 L He 
16.9 mol He x 
  379 L He
1
mol
He




Please do your turn # 10, 11, and 12 now.
Your Turns
10.
Given: 0.247 mol Ar
Want: Liters Ar

 22.4 L Ar 
0.247 mol Ar x 
  5.53 L Ar
 1 mol Ar 
11.
Given: 335.2 L Ar
Want: moles Ar

 1 mol Ar 
335.2 L Ar x 
  15.0 mol Ar
22.4
L
Ar


Putting It Together
Your Turn 12
Given:
Want:
mL to L:
Volume to moles:
Formula factor:

10.0 mL H2
atoms H
1000 mL = 1 L
22.4 L = 1 mole
1 mole H2 = 2 moles H
 1 L H 2   1 mole H 2   2 mole H   6.02 x 10 23 atoms H 
 x 
 x 
 x
10.0 mL H 2 x 



1000
mL
H
22.4
L
H
1
mole
H
1
mole
H
2
2
2





TYPO- Change (fix):
10.0 x 2 x 6.02 x 10 23 ÷ 1000 ÷ 22.4
Answer = 5.38 x 10 20 atoms H
XII. Density Based Problems
Calculating the Molar Mass of a Gas at
STP from its Density;
 And, Calculating the Density of a gas from
its molar mass


You must be more careful than you
normally are, as you will be starting with a
“complex” (two-part) unit.

Start with your “given” in the form of a
fraction, with a unit in its numerator and a
different unit in its denominator.

Then, cross off in a direction opposite from
what you normally do. Given’s bottom top
with top of conversion!
Guided Practice 7, 8


Given: density =
1.964 g / L
Want: molar mass…..
g / mole
 1.964 grams CO 2   22.4 Liter CO 2 

 x 
  44.0 grams / mole
 1 Liter CO 2   1 mole CO 2 


Given: molar mass =
39.9 g Ar / mol Ar
Want: density….
g / Liter
 39.9 grams Ar   1 mole Ar 

 x 
  1.78 grams / Liter
 1 mole Ar   22.4 Liters Ar 

Please do Your Turn 13 and 14; then, you
may begin the HW.
Your Turn 13 and 14


Given = density =
1.24 g N2 / Liter N2
Want = molar mass of
N2… g N2 / mol N2
 1.25 grams N 2   22.4 Liter N 2 

 x 
  28.0 grams / mole
1
Liter
N
1
mole
N
2 
2 




Given = molar mass =
64.1 g / mole
Want = density of SO2
= g / Liter
 64.1 grams SO 2   1 mole SO 2 

 x 
  2.86 grams / Liter
1
mole
SO
22.4
L
SO
2
2

 
