1. The Mole Concept Avagadro’s Number = 6.022 x 1023 Objectives:
To state the value of Avagadro’s number: x1023.
To state the mass of Avagadro’s number of atoms for any element by referring to the periodic table.
To relate the moles of a substance to the number of particles.

2. Counting Atoms
Chemistry is a quantitative science - we need a "counting unit."
The MOLE
1 mole is the amount of substance that contains as many particles (atoms or molecules) as there are in 12.0 g of C-12.

3. The Mole is Developed Carbon Atoms Hydrogen Atoms Mass Ratio
Number Mass (amu)
Number Mass (amu)
Mass carbon / Mass hydrogen 12 1
12 amu = 12
1 amu 24
[2 x 12] 2
[2 x 1]
24 amu = 12
2 amu
120
[10 x 12] 10
[10 x 1]
120 amu = 12
10 amu
600
[50 x 12] 50
[50 x 1]
600 amu = 12
50 amu
(6.02 x 1023) x (12)
Avogadro’s
number (6.02 x 1023) x (1)
(6.02 x 1023) x (12) = 12
(6.02 x 1023) x (1)
Avogadro’s
number

4. Amedeo Avogadro (1776 – 1856) Particles in a Mole
A more precise value of Avagadro’s number = E23.
How could you determine the number of sugar crystals in a bowl of sugar?
Is it possible to get a reasonably close estimate? ?
quadrillions
thousands
trillions
billions
millions
1 mole =
or x 1023
There is Avogadro's number of particles in a mole of any substance.

5. Careers in Chemistry - Philosopher
Q: How much is a mole?
A: A mole is a quantity used by chemists to count atoms and
molecules. A mole of something is equal to 6.02 x 1023 “somethings.”
1 mole =
Q: Can you give me an example to put that number in perspective?
A: A computer that can count 10,000,000 atoms per second would
take 2,000,000,000 years to count 1 mole of a substance.

6. Counting to 1 Mole
Is that right? A computer counting 10 million atoms every second
would need to count for 2 billion years to count just a single mole.
Lets look at the mathematics.
Therefore 1 year has 31,536,000 seconds or x 107 sec.
A computer counting 10,000,000 atoms every second could count
3.153 x 1014 atoms every year.
Finally, 6.02 x 1023 atoms divided by x 1014 atoms every year
equals 1,908,929,477 years or approximately 2 billion years!

7. How Big is a Mole? One mole of marbles would cover the entire Earth
(oceans included) for a depth of three miles.
One mole of $100 bills stacked one on top of another
would reach from the Sun to Pluto and back 7.5 million
times.
6.02 x 1023 softballs = volume of Earth
6.02 x 1023 Olympic shot puts = mass of Earth
6.02 x 1023 atoms H laid side by side would encircle Earth ~1,000,000 times
It would take light 9500 years to travel from the
bottom to the top of a stack of 1 mole of $1 bills.

8. Avogadro’s Number
A MOLE of any substance contains as many elementary units (atoms and molecules) as the number of atoms in 12 g of the isotope of carbon-12.
This number is called AVOGADRO’s number NA = 6.02 x 1023 particles/mol
The mass of one mole of a substance is called MOLAR MASS symbolized by MM
Units of MM are g/mol
Examples
H2 hydrogen g/mol
He helium g/mol
N2 nitrogen g/mol
O2 oxygen g/mol
CO2 carbon dioxide g/mol

9. 1 Mole of Particles

10. Molar Mass Objectives:
To calculate the molar mass of a substance given its chemical formula.

11. Molecular Weight and Molar Mass
Molecular weight is the sum of atomic weights
of all atoms in the molecule.
example: NaCl has a molecular weight of 58.5 a.m.u.
this is composed of a single molecule of NaCl
Molar mass = molecular weight in grams.
example: NaCl has a molar mass of 58.5 grams
this is composed of a 6.02 x1023 molecules of NaCl

12. The Molar Mass and Number of Particles in One-Mole Quantities
Substance Molar Mass Number of Particles in One Mole
Carbon (C) g x 1023 C atoms
Sodium (Na) g x 1023 Na atoms
Iron (Fe) g x 1023 Fe atoms
NaF (preventative g x 1023 NaF formula units
for dental cavities)
CaCO3 (antacid) g x 1023 CaCO3 formula units
C6H12O6 (glucose) g x 1023 glucose molecules
C8H10N4O2 (caffeine) g x 1023 caffeine molecules

13. Chemical Equations

14. Molar Mass and Mole Calculations
n M

15. Molar Mass
Molar mass is the mass in grams of one mole of particles (atoms, ions, molecules, formula units).
Equal to the numerical value of the average atomic mass (get from periodic table)
1 mole of C atoms = g
1 mole of Mg atoms = 24.3 g
1 mole of Cu atoms = 63.5 g
So, Molar Mass is the atomic mass expressed in units grams/mole (g/mol)

16. The number is the same, just change the unit.
Molar Mass
Chlorine has an atomic mass of amu.
So what is its molar mass?
grams per mole
The number is the same, just change the unit.

17. Molar Mass How do we find the molar mass of a compound?
Find the molar mass of each element in the compound
Determine how many of each element are in the compound.
Multiply the number of each element by its molar mass.
Add these mumbers.

18. Molar Mass
Aluminum Bromide
AlBr3
In one mole of Aluminum Bromide, there is 1 mole of aluminum atoms and 3 moles of bromine atoms.

19. Molar Mass To find the molar mass of AlBr3, we do the following:
1 mol Al x g/mol Al = g Al
3 mol Br x g/mol Br = g Br
g Al g Br = g AlBr3 per mole or g/mol AlBr3

20. Mole Calculations Where, n is the # of moles (mol) m is the mass (g)
n = m / M
M = m/n
m = n x M m
n M
Where, n is the # of moles (mol)
m is the mass (g)
M is the molar mass (g/mol)

21. Relationship Between Moles, Mass and Particles

22. Mole Calculations
What is the mass of 3.0 moles of chromium (III) iodide? CrI3

23. Mole and # of Particles
6.02x1023 molecules in a mole

24. A mole contains 6.02 x 1023 particles.
How many particles are in 2.0 moles of particles?
2 mol x 6.02 x 1023 particles/mol
= x 1023 particles
= 1.2 x 1024 particles
When changing moles to number of particles simply
Multiply the number of moles of particles by
6.02 x 1023 particles/mol

25. A mole contains 6.02 x 1023 particles.
How many particles are in 4.0 moles of particles?
4 mol x 6.02 x 1023 particles/mol
= x 1023 particles
= 2.4 x 1024 particles
How many particles are in 0.35 moles of particles?
0.35 mol x 6.02 x 1023 particles/mol
= 2.1 x 1023 particles

26. A mole contains 6.02 x 1023 particles.
How many moles of particles are in
1.02 x 1025 particles
= 1.02 x 1025 particles / 6.02 x 1023 particles/mole
= 16.9 moles
To change # of particles to moles simply
Divide the # of particles by x 1023 particles/mole

27. A mole contains 6.02 x 1023 particles.
How many moles of particles are in
4.74 x 1024 particles
= 4.74 x 1024 particles / 6.02 x 1023 particles/mole
= moles = 7.87mol
3.24 x 1022 particles
= 3.24 x 1022 particles / 6.02 x 1023 particles/mole
= moles = 5.38x10-2 moles

28. A mole contains 6.02 x 1023 particles.
How many particles are in 45.6 moles of particles?
44.6 moles x 6.02 x 1023 particles/mol
= x 1023 particles
= 2.68 x 1025 particles
How many moles are in 4.5 x 1021 particles?
4.5 x 1021 particles / 6.02 x 1023 particles/mol
= mol

29. How many molecules are there in 4.78 g of chlorine dioxide? ClO2
a) Find the number of moles of ClO2 first
n = m / M = 4.78 g / 67 g/mol
= mol
b) Now convert # of mol into particles
= mol x 6.02 x 1023 particles/mol
=4.3 x 1022 molecules of ClO2

30. How many molecules are there in 2.3 g of aluminum bromide? AlBr3
a) Find the number of moles of AlBr3
n = m / M = 2.3 g / 267 g/mol
= mol of AlBr3
b) Now convert # of mol into particles
= mol x 6.02 x 1023 particles/mol
= 5.2 x 1021 molecules of AlBr3

31. What is the mass of 8.2 x 1021 molecules of C2H5?
a) First find the # of moles.
8.2 x 1021 molecules / 6.02 x 1023 molecules/mol
= mol
b) Convert from moles to mass by multiplying the number of moles by the molar mass of C2H5.
MM = (12 x x 1 ) = 29 g/mol
= ( mol)x 29 g/mol
= g

32. What is the mass of 8.2 x 1021 molecules of C2H5?
a) First find the # of moles.
8.2 x 1021 molecules / 6.02 x 1023 molecules/mol
b) Convert from moles to mass by multiplying the number of moles by the molar mass of C2H5.
MM = (12 x x 1 ) = 29 g/mol
= (8.2 x 1021 molecules / 6.02 x 1023) x 29 g/mol
= g

33. What is the mass of 3.7 x 1024 molecules of C4H10?
First find the # of moles, then find the mass.
3.7 x molecules / 6.02 x 1023 molecules/mol
(= 6.15mol)
b) Convert from moles to mass by multiplying the number of moles by the molar mass of C4H10.
MM = (12 x x 1 ) = 58 g/mol
1 step:
(3.7 x 1024 molecules / 6.02 x 1023) x 58 g/mol = 356 g
= 3.6x102g

34. Converting Particles, Moles and Mass
1. Find the mass of each of the following:
a) 7.8 x 1026 molecules of N2O4
b) 5.6 moles of C2H5OH
c) 1.56 x 1021 molecules of C3H6
d) 7.4 moles of ClS3
e) 3.6 x 1025 molecules of CO2
2. Find the number of particles in each of the following:
a) 75 g of NF5
b) moles of C3H7OH
c) 2.8 g of C5H10
d) 2.5 moles of FBr3
e) 435 g of CO2