
The study of the quantitative relationships between
reactants and products in a reaction
It is used to answer questions like; If I have this much
reactant, how much product can I make?
 If I want this much product, how much reactant do I need?




These questions have real life application,
particularly in manufacturing.
It allows us to convert the mass of a substance to the
number of particles (atoms, ions or molecules) it
contains.
These numbers can be really large, so they are
counted in groups
 Much like when we count a lot of pennies we stack them in 10’s
and count by 10
Atoms are very tiny, so small
that the grouping we use to count
them must be very large
 MOLE; the group (unit of measure) used to count
atoms, molecules, formula units or ions of a
substance
 1 mole of a substance has a particular number of
particles in it!

Much like 1 dozen always means 12; whether it is
12 eggs
12 oranges
or
12 gold bars

The number of particles in a mole = 6.02 x 10 23
or 602,000,000,000,000,000,000,000 !
This is known as Avogadro’s Number
Using this, We can easily count the number of
particles in all kinds of things !
There are 6.02 x 10 23 Carbon atoms in a mole of carbon
There are 6.02 x 10 23 CO2 molecules in a mole of CO2
There are 6.02 x 10 23 sodium ions in a mole of sodium
There are 6.02 x 10 23 marbles in a mole of marbles
That’s a lot of marbles!
The Size of a mole of a substance changes, the bigger
the substance the more space a mole of the
substance takes up, but the number of particles in a
mole is always the same!

Chemicals do not come bundled in moles, like a dozen
eggs comes in a 1 dozen or 1 ½ dozen package so we
use the mole as a grouping unit.
The mass of 1 mole of a pure substance called it’s molar
mass


If I want to produce 500g of methanol using the
following equation, CO2 +3H2  CH3OH + H20 how
many grams of CO2 and H2 do I need?
These are the questions stoichiometry answers!
If I want to produce 500g of methanol using
the following equation;
CO2 +3H2  CH3OH + H20
How many grams of CO2 and H2 do I need?
This equation relates the molecules of reactants and
products, NOT THEIR MASSES!

1 molecule of CO2 and 3 molecules of H2 will make 1
molecule of CH3OH
We need to relate the masses to the number of molecules.
Remember; The average atomic
masses of the elements are found on the
Periodic Table!

We can use the atomic masses on the PT to
relate the mass of the compound to the
mass of a mole!
Molar mass: The mass (in grams)of one
mole of a molecule or a formula unit
Molecular mass: mass in atomic mass units of just
one molecule
Formula Mass: mass in atomic mass units of one
formula unit of an ionic compound
Steps
1.
Find the average Atomic Mass of the element
on the PT. (state it in grams instead of atomic
units)
a)
b)
2.
Example: molar mass of Fe = 55.847 g
Example: molar mass of Pt = 195.08 g
If the element is a molecule, count the number of
atoms in the molecule then multiply the atomic
mass by the number of atoms.
a)
Example: O2, the mass of O =16.0g There are 2
atoms of O in the O2 molecule , 2 atoms X 16.0g =
32.00g is the molar mass of the molecule.
Calculate the molar mass of each of the
following:
1. N2
2. Cl2
3. Br2
4. I2
5. H2
6. F2
Calculate the molar mass of each of the
following:
1. N2 = 14.007g X 2 =28.014 g/mol
2. Cl2 = 35.453g X 2 =70.906 g/mol
3. Br2 = 79.904g X 2 =159.808 g/mol
4. I2 = 126.904g X 2 =253.808 g/mol
5. H2 = 1.008g X 2 =2.016 g/mol
6. F2 = 18.998g X 2 =37.996 g/mol
Steps
1. Count the number and type of atoms
2. Find the Atomic Mass of each atom type, on
the periodic table. Write it in grams.
3. Multiply the mass times the # of Atoms. Then
add the totals
1.
Count the number and type of atoms
Ethanol (C2H5OH)
2.
3.
Atom type
Amount of each atom
C
2
H
6
O
1
Find the Atomic Mass of each atom type, on the periodic table. Write
it in grams.
Atom type
Amount of atom
Ave. Atomic Mass in g
C
2
12.0
H
6
1.00
O
1
16.0
Multiply The mass X the # of Atoms. Then add the totals.
Atom type
Amount of atom
Ave. Atomic Mass in g
Total
C
2
12.0
=24.0
H
6
1.00
=6.0
O
1
16.0
=16.0
Molar Mass Of Ethanol (C2H5OH)
= 46.0g/mole
Example: Calcium Chloride (CaCl2 )
Atom
Types
Amount of
Atoms
Ave. Atomic
Mass in g
Total
Ca
1
40.1
40.1
Cl
2
35.5
71.0
Mass of 1 mol of CaCl2 (molar mass)
111.1 g/mole
What is the molar mass of each of the following?
1.
Fe2 O3
2.
H2O
3.
CO2
4. NaCl
5. NH3
6. BaI2
Fe2 O3 = 55.85g X 2= 111.7 g
16.0g X 3 = 48.0g
= 159.7 g/mol
_______________________________________________
H2O = 1.01g X 2 = 2.02
16.0g X 1 = 16.0
= 18.02 g/mol
_______________________________________________
CO2 = 12.01g X 1 = 12.01
16.0g X 2 = 32.0
= 44.01 g/mol
________________________________________________
NaCl = 22.99 gX1 = 22.99
35.45g X1 = 35.45
= 58.44 g/mol
________________________________________________
NH3 =14.01g X 1 = 14.01
1.01g X 3 = 3.03
= 17.04 g/mol
________________________________________________
BaI2 = 137.33g X 1 = 137.33
126.90g X 2 = 253.80
= 391.13 g/mol
If I want to produce 500g of methanol using
the following equation;
6CO2 +17H2  3C2H5OH + 9H20
How many grams of CO2 and H2 do I need?
The Molar Mass Of Ethanol (C2H5OH)
= 46.0g/mole

Now we need to find the number of atoms in the
sample.
How many molecules of methanol are in 500g?
Steps to finding the number of atoms in a given
mass of a sample
1. Use PT to find the molar mass of the substance
2. Convert the mass of the substance to number of
moles in the sample (convert using mass of one
mole as conversion factor)
3. Use the number of atoms in a mole to find the
number of atoms in the sample
4. Solve and check answer by canceling out units
The mass of an iron bar is 16.8g. How many iron(Fe)
atoms are in the sample?
Step 1: Use PT to find the molar mass of the
substance : The molar mass of Fe =55.8g/mole
Step 2: Convert the given mass of the substance to
number of moles in the sample: Fe =55.8g/mole
(16.8g Fe) (1 mol Fe) (6.022 X 1023 Fe atoms)
(55.8g Fe)
(1 mol Fe)
=
1.81 X 10
23 Fe atoms
Step 3: Use the number of atoms in a mole to find
the number of atoms in the sample = 1.18 X 1023
1.
25.0 g silicon, Si
2.
1.29 g chromium, Cr
(
25.0 g Si
(
1.29 g Cr
1 mol Si
) ( 28.1g Si ) (
6.02 X 1023 Si atoms
1
= 5.36 X1023 atoms Si
1 mol Cr
) ( 52.0g Cr ) (
1
= 1.49 X1022 atoms Cr
1 mol Si
)
6.02 X 1023 Cr atoms
1 mol Cr
)
1.
2.
3.
4.
98.3g mercury, Hg
45.6g gold, Au
10.7g lithium, Li
144.6g tungsten, W
1 mol Hg
1. 98.3 g Hg
1
200.6g Hg
= 2.95 X1023 atoms Hg
(
)(
)(
6.02 X 1023 Hg atoms
1 mol Hg
23 Au atoms
45.6
g
Au
1
mol
Au
6.02
X
10
2.
1
197.0g Au
1 mol Au
= 1.39 X1023 atoms Au
(
)(
)(
1 mol Li
3. 10.7 g Li
1
6.94g Li
= 9.28 X1023 atoms Li
(
)(
)(
1 mol W
4. 144.6 g W
1
183.8g W
= 4.738 X1023 atoms W
(
)(
6.02 X 1023 Li atoms
1 mol Li
)(
)
6.02 X 1023 W atoms
1 mol W
)
)
)
Steps
1. Use the PT to calculate the molar mass of one
formula unit
2. Convert the given mass of the compound to
the number of molecules in the sample (use
the molar mass as the conversion factor)
3. Multiply the moles of the compound by the
number of the formula units in a mole
(Avagadro’s number) and solve
4. Check by evaluating the units
1.
2.
(
Calculate the molar mass (Fe2O3)
2 Fe atoms 2X 55.8 =
111.6
3 O atoms 3 X 16.0 =
+48.0
molar mass
159.6 g/mol
(change given mass  mole per mass atoms per mole)
16.8 g Fe2O3
1
)(
1 mol Fe2O3
6.02 X 1023 Fe2O3 Formula units
159.6g Fe2O3
1 mol Fe2O3
)(
= 6.34 X1022 Fe2O3 Formula units
)
1.
89.0g sodium oxide (Na2O)
2.
10.8g boron triflouride ( BF3)
89.0g sodium oxide (Na2O)
Calculate the molar mass (Na2O)
2 Na atoms 2X 23.0 =
46.0
1 O atoms 1 X 16.0 = +16.0
molar mass
62.0 g/mol
(change given mass  mole per mass  molecules per
mole)
1.
(
89.0 g Na2O
1
)(
1 mol Na2O
62.0g Na2O
)(
= 8.64 X1023 Na2O molecules
6.02 X 1023 Na2O Molecules
1 mol Na2O
)
10.8g boron triflouride ( BF3)
Calculate the molar mass (Na2O)
1 B atom 1X 10.8 =
10.8
3 F atoms 3 X 19.0 =
+57.0
molar mass
67.8 g/mol
(change given mass  mole per mass  molecules
per mole)
2.
(
10.8 g BF3
1
)(
1 mol BF3
67.8g BF3
)(
6.02 X 1023 BF3 Molecules
1 mol BF3
= 9.59 X1022 BF3 molecules
)
Steps
1. Determine the molar mass
2. Change given mass to moles by using molar
mass as the conversion factor.
Calculate the number of moles in 6.84g sucrose
(C12H22O11)
12 C atoms 12 X 12.0 = 144.0
22 H atoms 22 X 1.0 =
22.0
11 O atoms 11 X 16.0 = +176.0
molar mass
342.0 g/mol
(change molar mass  mole per mass)
(
6.84 g sucrose
1
1 mol sucrose
342.0g sucrose
)(
= 2.0 X10-02 moles of sucrose
)
1.
2.
3.
16.0g sulfur dioxide, SO2
68.0g ammonia, NH3
17.5g copper(II) oxide, CuO
1.
2.
3.
0.250 mol SO2
4.00 mol NH3
0.22 mol CuO