•The mole is a counting unit in chemistry.
•Abbreviated mol.
•Defined as the number equal to the number of carbon
atoms in exactly 12 grams of pure 12C .
•Has the value 6.022 x 1023.
•Called Avogradro’s number to honor his contributions to
chemistry.
•One mole of anything consists of 6.022 x 1023 units of that
substance.
•Just like a dozen eggs is 12 eggs; a mole of eggs is
6.022 x 1023 eggs.
•How big is a mole?
•1 mole of seconds represents a span of time 4 million times
as long as the Earth as already existed.
•1 mole of marbles is enough to cover the entire Earth to a
depth of 50 miles.
•Remember that the mole is the number of atoms in exactly
12 grams of 12C therefore 12 grams of 12C contains exactly
6.022 x 1023 atoms.
•Plus it tells us that a 12.01 gram sample of natural carbon
contains 6.022 x 1023 atoms (natural carbon is a mixture of
the isotopes of carbon).
•Notice that the value 12.01 grams of natural carbon is the
same as the atomic mass value (12.01 amu).
•It also tells us that 26.98 grams of aluminum contains
exactly 6.022 x 1023 atoms of aluminum.
•If a mole of natural carbon and aluminum have the same
number of atoms why do they have different masses?
•Consider a dozen doughnuts and a dozen tires.
•Would they have the same mass?
•We can also define the mole such that a sample of a
natural element with a mass equal to the element’s atomic
mass expressed in grams contains 1 mole of atoms.
•Example: Determining Moles of Atoms
•How many moles of copper are in an old copper penny that
has a mass of 3.14 g? How many atoms are in this 3.14 g
penny?(New pennies are made of zinc with only a thin surface
coating of copper.)
•Solution:
•The mass of 1 mole (6.022 x 1023 atoms) of copper is 63.55 g.
•The sample we are considering has a mass of 3.14 g.
•Since the mass is less than 63.55 g, this sample contains less
than 1 mole of copper atoms.
1 mol Cu = 63.55 g Cu (gives two conversion factors)
63.55 g Cu
1 mol Cu
or
1 mol Cu
63.55 g Cu
•Which conversion factor is appropriate? We are converting
from g to mol.
63.55 g Cu
1 mol Cu
or
1 mol Cu
63.55 g Cu
1 mol Cu
# mol Cu  3.14 g Cu x
 0.0494 mol Cu
63.55 g Cu
•How many atoms are in this 3.14 g penny?
•If we have 1 mol Cu then we have 6.022 x 1023 atoms Cu.
1 mol Cu = 6.022 x 1023 atoms Cu
6.022 x 1023 atoms Cu
0.0494 mol Cu x
 2.97 x 1022 atoms Cu
1 mol Cu
•Example: Calculating Moles and Mass
•Uranium is a silvery-white metallic
element in the actinide series of the
periodic table. Calculate both the number
of moles in a sample of uranium
containing 2.530 x 1015 atoms and the
mass of the sample.
•Solution:
•We know 1 mol U = 6.022 x 1023 atoms U.
1 mol U
2.530 x 10 atoms U x

23
6.022 x 10 atoms U
15
4.201 x 10
9
mol U
•How many grams of U are contained in this sample?
•We know if we have 1 mol U we have 238.03 g U.
1 mol U = 238.03 g U
4.201 x 10
9
238.03 g U
6
mol U x
 1.000 x 10 g U
1 mol U
•Molar mass of a substance is the mass in grams of one
mole of the compound.
•Molar mass has units of g/mol.
•Remember a chemical compound is a collection of bonded
atoms.
•If we want to know the mass in grams of one mole of
methane, CH4 (the mass of 6.02 x 1023 atoms of CH4) we
need to take into account that 1 mole of CH4 contains 1
mole of carbon atoms and 4 moles of hydrogen atoms.
•To determine the mass of 1 mole of CH4 we sum the atomic
masses of carbon and hydrogen present:
Mass of 1 mol C
Mass of 4 mol H
Mass of 1 mol CH4
=
12.02 g
= 4 x 1.008 g
16.04 g
•Since 16.04 g represents the mass of 1 mole of methane
molecules it is called the molar mass of methane.
•Example: Calculating Molar Mass of a Compound
•What is the molar mass of calcium phosphate, Ca3(PO4)2,
expressed to the second decimal place?
•Solution: Use atomic masses of 40.08 for Ca, 30.97 for P,
and 16.00 for O.
Ca3(PO4)2 = 3 Ca
+
2P
+
8O
Molar mass of Ca3(PO4)2 =
(3 x 40.08) + (2 x 30.97) + (8 x 16.00) =
120.24
+ 61.94
+ 128.00 =
310.18 g/mol or
1 mol Ca3(PO4)2 = 310.18 g Ca3(PO4)2
•Example: Calculating Grams from Moles for Compounds
•A student has worked out the details for an experiment
that will use 0.115 mol of Ca3(PO4)2, calcium phosphate, as a
starting material. How many grams of Ca3(PO4)2 should be
weighed out?
•Solution: For this problem we need an equality that relates
the number of moles to the number of grams. We know
that 1 mol Ca3(PO4)2 = 310.18 g Ca3(PO )2.
310.18 g Ca3 (PO4 )2
0.115 mol Ca3 (PO4 )2 x
 35.7 g Ca3 (PO4 )2
1 mol Ca3 (PO4 )2
•Notice units of moles cancel and we are left with grams.
•Example: Calculating formula units (molecules if covalent)
from grams.
•How many formula units of Na2CO3 does 45.8 g of Na2CO3
contain?
•Solution: This problem requires 2 equalities.
1 mol Na2CO3 = ? g Na2CO3
1 mol Na2CO3 = 6.022 x 1023 formula units
To determine the number of grams in 1 mol of Na2CO3 sum
the atomic masses of the atoms in the compound.
(2 x 22.99 for Na) + (1 x 12.01 for C) + (3 x 16.00 for O) =
105.99 g/mol
1 mol Na2CO3
6.022 x 1023 formula units
45.8 g Na2CO3 x
x
105.99 g Na2CO3
1 mol Na2CO3
 2.60 x 1023 formula units Na2CO3
•How many sodium atoms were present in the 45.8 g of
Na2CO3?
•Solution:
•In the 45.8 g of Na2CO3 we calculated that we had
2.60 x 1023 formula units. Since each formula unit of Na2CO3
contains 2 Na atoms we can calculate as follows.
23
2.60 x 10
2 atoms Na
formula units Na2CO3 x
1 formula unit Na2CO3
 5.20 x 1023 atoms Na