1) The document discusses how Avogadro's number allows chemists to determine the number of atoms or molecules in a given number of moles of a substance. 2) It provides examples of using molar mass to convert between mass and moles, and using Avogadro's number to convert between moles and number of particles. 3) Key concepts covered include the mole as a unit for counting particles, molar mass, and conversion factors involving moles, mass, and number of particles.

1. Moles, Molar Mass, and
Avogadro’s Number, Part 1
By Shawn P. Shields, Ph.D.
This work is licensed by Shawn P. Shields-Maxwell under a Creative Commons
Attribution-NonCommercial-ShareAlike 4.0 International License.

2. A Practical Problem Concerning
Atom Size…
Recall: Atomic Mass Units (amu)
One amu is equal to 1.6605  1027 kg
This is a VERY small amount of mass.
We know that the average mass of a water (H2O)
molecule is 18.02 amu…
…which would be 2.992  1026 kg, and very hard to
measure!

3. A Practical Problem Concerning
Atom Size…
We can’t count them either…(too small)
Plus, we’d need a whole lot of them to be able to
weigh them on a balance.

4. A Practical Problem…
So, how could we solve this problem?
Observe:
The ratio of atomic masses for one atom each of O
and H is
15.999 amu
1.008 amu
= 15.872
The ratio of atomic masses for 100 atoms each of O
and H is
1599.9 amu
100.8 amu
= 15.872 !

5. A Practical Problem…
We could keep on using larger numbers, and still get
the same result, so…
Let’s go ahead and add enough atoms to equal 15.999 g
O and 1.008 g H; matching the average atomic masses
shown on the periodic table for O and H.)
15.999 g
1.008 g
= 15.872
The number of atoms required to have the mass
(in g) the same as that shown on the periodic table
is a very special number…
!
8
O
15.999
1
H
1.008

6. Avogadro’s Number!
It takes 6.022  1023 atoms of Oxygen to have a mass
of 15.999 g.
It takes 6.022  1023 atoms of Hydrogen to have a
mass of 1.008 g.
How many P atoms would it take to have 30.973 g?
15
P
30.973
1
H
1.008
8
O
15.999

7. Avogadro’s Number!
It takes 6.022  1023 atoms of Oxygen to have a mass
of 15.999 g.
It takes 6.022  1023 atoms of Hydrogen to have a
mass of 1.008 g.
How many P atoms would it take to have 30.973 g?
Avogadro’s Number! 
15
P
30.973
1
H
1.008
8
O
15.999

8. The Mole Concept
(A Chemist Counting Unit)
A “mole” of atoms is 6.022  1023 atoms.
The “mole” is a counting unit directly
analogous to a “dozen.”
12 pencils = 1 dozen pencils
6.022  1023 pencils = 1 mole pencils

9. The Mole Concept
(A Chemist Counting Unit)
Here are a few other ways you’ll see this number expressed:
6.022  1023 mol-1
This just means that there are 6.022  1023 “particles” per mol.
(“particles” is implied)
You’ll also see NA = 6.022  1023 mol-1 and N0 = 6.022  1023 mol-1
Remember, “particles” can be atoms, electrons, cations,
anions…anything!
Avogadro’s number is just the number of
particles in a mole!

10. Molar Mass
Molar Mass is the amount of mass (in grams) in one
mole of particles (atoms, ions, etc).
A “mole” of atoms is 6.022  1023 atoms.
One mole of any element has a molar mass in grams
numerically equal to the atomic mass of that
element. (Look on the Periodic Table)
74
W
183.84
6
C
12.011
8
O
15.999

11. Molar Mass
Look at your periodic table:
Now, the average atomic mass (atomic weight) can be
interpreted in grams.
So, one mole of tungsten (W) has a mass of 183.84 g
6.022  1023 “tungsten atoms” has a mass of 183.84 g
74
W
183.84
6
C
12.011
8
O
15.999

12. Calculating the Molar Mass of a
Compound
To calculate the molar mass of a compound:
• Add the molar masses for all elements in the compound
using the atomic mass given (in grams).
• Be sure to multiply the mass by the number of moles of
that element in the compound before adding them up.
1
H
1.008
6
C
12.011
53
I
126.90
15
P
30.973

13. Calculating the Molar Mass of a
Compound
Ex: Calculate the molar mass of propane (C3H8)
Molar Mass = 3(Cavg mass) + 8(Havg mass) = ? g
1
H
1.008
6
C
12.011
53
I
126.90
15
P
30.973

14. Calculating the Molar Mass of a
Compound
Ex: Calculate the molar mass of propane (C3H8)
Molar Mass = 3(Cavg mass) + 8(Havg mass) = ? g
Molar Mass = 3(12.011) + 8(1.008) = 44.097 g
1
H
1.008
6
C
12.011
53
I
126.90
15
P
30.973

15. Mini Quiz: Calculating the Molar Mass
of a Compound
Calculate the molar mass of phosgene (COCl2)
18
Ar
39.948
6
C
12.011
8
O
15.999
7
N
14.007
17
Cl
35.453
74
W
183.84

16. Mini Quiz Solution
Calculate the molar mass of phosgene (COCl2)
Add up the molar mass of each element multiplied by the
number of moles of that element in the compound:
Molar Mass = 1(12.011) + 1(15.999) + 2(35.453) = 98.916 g
18
Ar
39.948
6
C
12.011
8
O
15.999
7
N
14.007
17
Cl
35.453
74
W
183.84

17. Converting from Mass to Moles
We use the molar mass to convert from grams
(which we can measure) to moles of that substance.
Example: Suppose we are going to use 35.5 g of
potassium chloride (KCl) in a reaction. How moles
of KCl are there is 35.5 g?
Step 1: Calculate the molar mass of KCl.
Molar Mass = 1(39.098) + 1(35.453) = 74.551 g
17
Cl
35.453
6
C
12.011
8
O
15.999
19
K
39.098

18. Converting from Mass to Moles
Example: Suppose we are going to use 35.5 g of
potassium chloride (KCl) in a reaction. How moles
of KCl are there is 35.5 g?
Step 1: Calculate the molar mass of KCl.
Molar Mass = 1(39.098) + 1(35.453) = 74.551 g
Step 2: Use the molar mass of KCl as a conversion
factor to moles…
1 mol KCl
74.551 g KCl
𝐨𝐫
74.551 g KCl
1 mol
17
Cl
35.453
6
C
12.011
8
O
15.999
19
K
39.098

19. Converting from Mass to Moles
Example: Suppose we are going to use 35.5 g of
potassium chloride (KCl) in a reaction. How moles of
KCl are there is 35.5 g?
Step 1: Calculate the molar mass of KCl.
Molar Mass = 1(39.098) + 1(35.453) = 74.551 g
Step 2: Use the molar mass of KCl as a conversion
factor to moles…
Step 3: Write the quantity that you need to convert,
then use the appropriate conversion factor (with the
unit you need to cancel out on the bottom), then do the
math. 
35.5g KCl
1 mol KCl
74.551 g KCl
= 𝟎. 𝟒𝟕𝟔 𝐦𝐨𝐥𝐞𝐬 𝐊𝐂𝐥
17
Cl
35.453
6
C
12.011
8
O
15.999
19
K
39.098

20. Converting Moles to Number of
Molecules (Atoms, Particles, etc.)
We use Avogadro’s number to convert from moles to
number of molecules of that substance.
Example: Convert 0.267 moles of sulfur trioxide
(SO3) to number to molecules.
Remember, there are 6.022  1023 “particles”
(molecules) in one mole.
Written another way…
6.022 × 1023 SO3 molecules
1 mole SO3
16
S
32.065
7
N
14.007
8
O
15.999

21. Converting Moles to Number of
Molecules (Atoms, Particles, etc.)
Example: Convert 0.267 moles of sulfur trioxide (SO3)
to number to molecules.
So,
0.267 mol SO3
6.022 × 1023 SO3 molecules
1 mole SO3
= 1.61 × 1023 SO3 molc
You try one…
How many atoms are in 0.23 moles N atoms?
16
S
32.065
7
N
14.007
8
O
15.999

22. Converting Moles to Number of
Molecules (Atoms, Particles, etc.)
You try one…
How many atoms are in 0.23 moles N atoms?
0.23 mol N
6.022 × 1023 N molecules
1 mole N
= 1.4 × 1023 N atoms
Did you notice that we did not need the molar
mass for N ?
16
S
32.065
7
N
14.007
8
O
15.999

23. Converting Moles to Number of
Molecules (Atoms, Particles, etc.)
Example 2: How many O atoms are in 0.82 moles of SO3?
So,
0.82 mol SO3
6.022 × 1023 SO3 molecules
1 mole SO3
= 4.94 × 1023 SO3 molc
Are we done? NO!
16
S
32.065
7
N
14.007
8
O
15.999

24. Converting Moles to Number of
Molecules (Atoms, Particles, etc.)
Example 2: How many O atoms are in 0.82 moles of SO3?
So,
0.82 mol SO3
6.022 × 1023 SO3 molecules
1 mole SO3
= 4.94 × 1023 SO3 molc
Multiply the number of O atoms per molecule by
the number of SO3 molecules.
4.94 × 1023 SO3 molc
3 O atoms
1 molecule SO3
= 1.5 × 1024 O atoms
remember sig figs!
16
S
32.065
7
N
14.007
8
O
15.999

25. Converting Moles to Number of
Molecules (Atoms, Particles, etc.)
Finally, how many moles of O atoms is this?
Convert from number of atoms to moles using Avogadro’s
number:
1.5 × 1024 O atoms
1 mol O atoms
6.022 × 1023 O atoms
= 2.5 mol O atoms
remember sig figs!
16
S
32.065
7
N
14.007
8
O
15.999