Stoichiometry 1
Formulas and the Mole
L. Scheffler
Lincoln High School
1
The Mole
• Chemical reactions involve atoms and molecules.
• The ratios with which elements combine depend
on the number of atoms not on their mass.
• The masses of atoms or molecules depend on
the substance.
• Individual atoms and molecules are extremely
small. Hence a larger unit is appropriate for
measuring quantities of matter.
• A mole is equal to exactly the number of atoms in
exactly 12.0000 grams of carbon 12.
• This number is known as Avogadro’s number.
1 mole is equal to 6.022 x 1023 particles.
2
Definitions of the Mole
• 1 mole of a substance has a mass equal to
the formula mass in grams.
• Examples
• 1 mole H2O is the number of molecules in 18.015 g H2O
• 1 mole H2 is the number of molecules in 2.016 g H2.
• 1 mole of atoms has a mass equal to the atomic weight
in grams.
• 1 mole of particles = 6.02214 x 1023 particles for any
substance!
• The Molar mass is the mass of one mole of a
substance
• Avogadro's number is the number of particles
(molecules) in one mole for any substance
3
The Mole
4
A mole is equivalent to a gram atomic weight or gram
molecular weight
The Formula Mass
• The formula mass is the sum of atomic
masses in a formula.
• If the formula is a molecular formula,
then the formula mass may also be
called a molecular mass.
5
Gram Formula Mass and Molar Mass
• If the formula mass is expressed in
grams it is called a gram formula mass.
• The gram formula mass is also known
as the Molar Mass.
• The molar mass is the number of grams
necessary to make 1 mole of a
substance.
• The units for Molar Mass are g mol-1.
6
Formula Mass and the Mole
• The atomic mass of Carbon 12 is exactly
12.00000.
• 1 atomic mass unit = 1/12 of the atomic mass
of carbon 12.
• The periodic table gives the average atomic
mass for an element relative to Carbon 12.
• 1 mole of a substance is 6.022 x 1023
particles.
• The mole of atomic mass units is equal to
1.000 gram.
7
Gram Formula Mass
• The formula mass is the sum of the
atomic masses in a formula.
• A gram formula mass is the same number
expressed in grams.
• It is also equal to Avogadro’s Number of
particles
• Example: H2O
From the Periodic Table - Atomic Masses: H =1.00797, O = 15.999
The formula mass = 2(1.00797)+15.999 = 18.015
• Adding the unit “grams” to the formula mass
transforms it into a gram formula mass or mole.
8
The Mole
• The mole is connects the macro world
that we can measure with the micro
world of atoms and molecules.
• A Mole is also equal to
– 1 gram formula mass.
– 22.4 dm3 of any gas measured at 0o C and
1.0 atmosphere of pressure.
9
Example 1: Calculating the
Molar Mass of a Compound
• Calculate the gram formula mass or Molar
Mass of Na3PO4.
Atom
Na
P
O
#
3
1
4
Atomic Mass
X 23.0
X 31.0
X 16.0
Total
=
=
=
=
Total
69.0
31.0
64.0
164.0
Therefore the molar mass is 164.0 g mol-1
10
Example 2:
Find the mass of 2.50 moles of Ca(OH)2
Find the molar mass of Calcium hydroxide and
multiply by 2.50 mol
The molar mass of Ca(OH)2 is
1 Ca
1 x 40.08 = 40.08
2 O
2 x 16.00 = 32.00
2 H
2 x 1.01 = 2.02
Molar Mass
= 74.10 g mol-1
2.50 mol x 74.10 g mol-1 = 185.25 g
11
Calculating Moles
• The number of moles in a given mass of a
substance can be determined by dividing the
mass by the molar mass
Moles =
Mass
Molar Mass
12
Example 3:
Find the number of moles in 44.46 grams
of Ca(OH)2
Find the molar mass of and divide it into the
given mass
From the previous example the molar mass
of calcium hydroxide is 74.10 gmol-1.
44.46 g Ca(OH)2 .
= 0. 6000 mol
74.10 g mol-1 Ca(OH)2
13
Example 4: Calculating Moles
• Calculate the number of moles in 20.5
grams of Na3PO4
Moles =
Mass
Molar Mass
Moles =
20.5 g
164.0 g mol-1
=
0.125 mol
Note: Mol is the standard abbreviation for a mole
14
Calculating Mass From Moles
• The mass of a quantity of a substance can
be found by multiplying the number of moles
by the molar mass
Mass =
Moles X Molar Mass
15
Example 5 Calculating Mass
from Moles
• Calculate the mass of 2.50 moles of
Na3PO4
Mass = Moles X Molar Mass
= 2.50 mol x 164.0 g mol-1
= 409 g
16
Percentage Composition
• According to the law of definite proportions,
compounds, contain definite proportions of each
element by mass.
• The sum of all of the atomic masses of elements in
a formula is called the formula mass.
• If it is expressed in grams, then it is called a gram
formula mass or molar mass.
• If it represents the sum of all of the masses of all
of the elements in a molecule then it is called a
molecular mass.
• To find the percentage of each element in a
compound it is necessary to compare the total
mass of each element with the formula mass.
17
Percentage Composition
• The percent by mass of each element in a
compound is equal to the percentage that
its atomic mass is of the formula mass.
• Example: Calculate the percentage of
oxygen in potassium chlorate, KClO3
Atomic masses: K = 39.09, Cl = 35.45 and O = 16.00.
Formula mass = 39.09 + 35.45+ 3(16.00) = 122.54
Percent Oxygen = (3(16.00)/122.54) (100)
= 39.17%
18
Example 2
• Calculate the percentage by mass of each element in
potassium carbonate, K2CO3
First calculate the formula mass for K2CO3 . Find the atomic
mass of each element from the periodic table. Multiply it by
the number of times it appears in the formula and add up
the total
2 Potassium atoms K
1 carbon atom
C
3 Oxygen atoms
O
2 x 39.10 = 78.20
1 x 12.01 = 12.01
3 x 16.00 = 48.00
Total =
138.21
To find the percent of each element divide the part of the
formula mass that pertains to that element with the total formula
mass
Percent of Potassium K = 78.20 X 100
138.21
=56.58 %
Percent of Carbon
C = 12.01 X 100
138.21
= 8.69 %
Percent of Oxygen
O = 48.00
= 34.73 %
X 100
19
Empirical Formula Determination
The empirical formula is the simplest ratio of
the numbers of atoms of each element that
make a compound.
To find the empirical formula of a compound:
1. Divide the amount of each element (either in mass or
percentage) by its atomic mass.
This calculation
gives you moles of atoms for each element that
appears in the formula
2. Convert the results to small whole number ratios.
Often the ratios are obvious. If they are not divide all
of the other quotients by the smallest quotient
20
Example 1
Analysis of a certain compound showed that 32.356
grams of compound contained 0.883 grams of
hydrogen, 10.497 grams of Carbon, and 27.968
grams of Oxygen. Calculate the empirical formula
of the compound.
First divide the amount by the atomic mass to get the number of
moles of each kind of atom in the formula
Hydrogen
H =
0.883 g
1.01 g mol-1
= 0.874 mol
Carbon
C =
10.497 g
12.01 g mol-1
= 0.874 mol
Oxygen
O =
27.968 g
16.00 g mol-1
= 1.748 mol
• Analysis of the ratio s shows that the first two are
identical and that the third is twice the other two.
Therefore the ratio of H to C to O is 1 to 1 to 2. The
empirical formula is HCO2
21
Molecular Formula
• To calculate the molecular formula from the
empirical formula it is necessary to know the
molecular (molar) mass.
• Add up the atomic masses in the empirical formula
to get the factor
• Divide this number into the molecular formula mass.
• If the number does not divide evenly you probably
have a mistake in the empirical formula or its
formula mass
• Multiply each subscript in the empirical formula by
the factor to get the molecular formula
22
Molecular Formula Example
• Example: Suppose the molecular mass of the
compound in the previous example, HCO2 is 90.0.
Calculate the molecular formula.
• The empirical formula mass of is
1H
1.0 x 1 = 1.0
1 C 12.0 x 1 = 12.0
2 O 16.0 x 2 = 32.0
Total
45.0
• Note that 45 is exactly half of the molecular mass of 90.
• So the formula mass of HCO2 is exactly half of the
molecular mass. Hence the molecular formula is double
that of the empirical formula or H2C2O4
23
Part 2: Stoichiometry Problems
• Mass-Mass Problems
• Mass-Volume
24
Stoichiometry Problems
• Stoichiometry problems involve the
calculation of amounts of materials in a
chemical reaction from known quantities in
the same reaction
• The substance whose amount is known is the
given substance
• The substance whose amount is to be
determined is the required substance
25
Mass to Mass Problems
• Goal: To calculate the mass of a substance that
appears in a chemical reaction from the mass of
a given substance in the same reaction.
• The given substance is the substance whose
mass is known.
• The required substance is the substance whose
mass is to be determined.
26
Steps in a Mass to Mass Problem
1. Find the gram formula masses for the given
and the required substances
2. Convert the given mass to moles by
dividing it by the molar mass
3. Multiple the given moles by the mole ratio
to get the moles of the required substance
4. Multiple the number of moles of the
required substance by its molar mass to
get the mass of the required substance
27
Summary of Mass Relationships
28
Example 1 Mass-Mass Problem
• Glucose burns in oxygen to form CO2 and
H2O according to this equation:
C6H12O6 + 6 O2  6 CO2 + 6 H2O
How many grams of CO2 are produced
from burning 45.0 g of glucose?
29
Example 1 Mass-Mass Problem
Glucose burns in oxygen to form CO2 and H2O according to
this equation:
C6H12O6 + 6 O2  6 CO2 + 6 H2O
How many grams of CO2 are produced from burning 45.0 g
of glucose?
1.Make sure that the equations is balanced
2.Divide the mass of the given by its molar mass
45.0 g C6H12O6
x
180.0 g mol-1 C6H12O6
30
Example 1 Mass-Mass Problem
Glucose burns in oxygen to form CO2 and H2O
according to this equation:
C6H12O6 + 6 O2  6 CO2 + 6 H2O
How many grams of CO2 are produced from
burning 45.0 g of glucose?
1.
2.
Make sure that the equations is balanced
Divide the mass of the given by its molar
3. Multiply by the mole ratio
45.0 g C6H12O6
180.0 g mol-1 C6H12O6
x
6 mol CO2
1 mol C6H12O6
= 1.5 moles CO2
31
Example 1 Mass-Mass Problem
Glucose burns in oxygen to form CO2 and H2O according
to this equation:
C6H12O6 + 6 O2  6 CO2 + 6 H2O
How many grams of CO2 are produced from burning 45 g
of glucose?
1.Make sure that the equations is balanced
2.Divide the mass of the given by its molar
3.Multiply by the mole ratio
4.Multiply by the molar mass of the required
45.0 g C6H12O6
180.0 g mol-1
C6H12O6
6 mol CO2
x
1 mol C6H12O6
44.0 g mol-1 CO2
x
= 66.0 g of CO2
32
Example 2 Mass-Mass Problem
What mass of Barium chloride is required to react
with 48.6 grams of sodium phosphate according to
the following reaction:
2 Na3PO4 + 3BaCl2  Ba3(PO4)2 + 6 NaCl
33
Example 2
What mass of Barium chloride is required to
react with 48.6 grams of sodium phosphate
according to the following reaction
2 Na3PO4 + 3BaCl2  Ba3(PO4)2 + 6 NaCl
Molar Masses: Na3PO4 = 3(23.0)+31.0+4(16.0) =164 g mol-1
BaCl2 = 137.3 +2(35.5) = 208.3 g mol-1
48.6g Na3PO4
164.0 g mol-1
Na3PO4
3 mol BaCl2
x
2 mol Na3PO4
= 92.6 g of BaCl2
208.3 g mol-1 BaCl2
x
34
Example 3
What mass of carbon dioxide is produced
from burning 100 grams of ethanol in oxygen
according to the following reaction :
C2H5OH + 3 O2 
2 CO2 + 3 H2O
35
Example 3
What mass of carbon dioxide is produced
from burning 100 grams of ethanol in oxygen
according to the following reaction :
C2H5OH + 3 O2 
2 CO2 + 3 H2O
Molar Masses:
100.0 g C2H5OH
46.0 g mol-1
C2H5OH = 2(12) +6(1)+ 16 = 46
CO2 = 12 + 2(16) = 44.0
x
2 mol CO2
1 mol C2H5OH
X
44.0 g mol-1 CO2
= 191.3 g CO2
36
Mass to Volume Problems
37
Mass to Volume Problems
• Goal: To calculate the volume of a gas that
appears in a chemical reaction from the mass of
a given substance in the same reaction.
• The given substance is the substance whose
mass is known.
• The required substance is the gas whose
volume is to be determined.
• Remember 1 mole of any gas at STP is equal to
22.4 dm3. STP is defined as 0 oC and 1
atmosphere of pressure.
38
Steps in a Mass to Volume Problem
1. Find the gram formula masses for the given
substance.
2. Convert the given mass to moles by dividing it
by the molar mass
3. Multiple the given moles by the mole ratio to get
the moles of the required substance
4. Multiple the number of moles of the required
substance by the molar volume, 22.4 dm3 mol-1,
to get the volume of the required substance.
5. This procedure is only valid if the required
substance is a gas. It does not work for solids,
liquids, or solutions.
39
Example 1 Mass-Volume Problem
• Sucrose burns in oxygen to form CO2 and H2O
according to this equation:
C12H22O11 + 12 O2  12 CO2 + 11 H2O
What volume of CO2 measured at STP is
produced from burning 100 g of sucrose?
40
Example 1 Mass-Volume Problem
Sucrose burns in oxygen to form CO2 and H2O
according to this equation:
C12H22O11 + 12 O2  12 CO2 + 11 H2O
What volume of CO2 measured at STP is
produced from burning 100 g of sucrose?
1. Find the molar mass of the given substance
Molar mass of C12H22O11 = 12 (12.0) +22 (1.0) +
11 (16.0)
= 342.0 g mol-1
41
Example 1: Mass-Volume Problem
Sucrose burns in oxygen to form CO2 and H2O
according to this equation:
C12H22O11 + 12 O2  12 CO2 + 11 H2O
What volume of CO2 measured at STP is
produced from burning 100 g of sucrose?
2. Find moles of the given:
100 g C12H22O11
342 g mol-1 C12H22O11
= 0.292 moles
42
Example 1: Mass-Volume Problem
Sucrose burns in oxygen to form CO2 and H2O
according to this equation:
C12H22O11 + 12 O2  12 CO2 + 11 H2O
What volume of CO2 measured at STP is
produced from burning 100 g of sucrose?
3. Multiply by the mole ratio:
100.0 g C12H22O11
342.0 g mol-1 C12H22O11
x
12 moles CO2
1 mole C12H22O11
= 3.51 moles CO2
43
Example 1: Mass-Volume Problem
Sucrose burns in oxygen to form CO2 and H2O
according to this equation:
C12H22O11 + 12 O2  12 CO2 + 11 H2O
What volume of CO2 measured at STP is
produced from burning 100 g of sucrose?
4. Multiply by the molar volume, 22.4 dm3 mol-1.
100.0 g C12H22O11
342.0 g mol-1
C12H22O11
x
12 moles CO2
1 moles C12H22O11
x
22.4 dm3 mol-1
CO2
=78.6 dm3
44
Example 2 Mass-Volume Problem
What volume of carbon dioxide gas would be
produced by reacting 25.0 g of sodium
carbonate with hydrochloric acid according
to the following reaction:
Na2CO3 + 2 HCl  2 NaCl + CO2 + H2O
45
Example 2 Mass-Volume Problem
• What volume of carbon dioxide gas would be produced
by reacting 25.0 g of Sodium carbonate with
hydrochloric acid according to the following reaction:
Na2CO3 + 2 HCl  2 NaCl + CO2 + H2O
Molar Mass: Na2CO3 =2(23.0)+ 12.0 +3(16.0) =106.0
25.0 g Na2CO3
106.0 g mol-1
Na2CO3
1 mole CO2
x
1 moles
Na2CO3
22.4 dm3 mol-1 CO2
x
= 5.28 dm3 of CO2
46
Summary of Stoichiometric
Relationships
47
Solutions and Stoichiometry
• Many times the reactants and/or products
of chemical reactions are water solutions.
• In these cases the concentration of the
solution must be determined in order to
determine amounts of reactants or
products
• The concentration of a solution is a
measure of the amount of solute that is
dissolved in a given amount of solution
48
Molarity
• The most common concentration unit is
Molarity
Molarity (M) =
Moles of solute
dm3 of solution
49
Molarity Calculations
How many grams of NaOH are required to
prepare 250 cm3 of 0.500 M solution?
– Molar Mass of NaOH = 23+16+1 = 40.0 g/mol
– 250 cm3 = 0.250 dm3
(0.500 mol) x
(
dm3
)x
(40.0 g)
( mol )
x (0.250 dm3 )
= 5.00 g
50
Molarity Calculations
Calculate the concentration of a NaCl
solution that contains 24.5 g of NaCl in
250 cm3 of solution.
– Molar mass of NaCl = 23.0 + 35.5 = 58.5
(24.5 g NaCl)
(58.5 g mol-1 )
X
1
= 1.67 M
(0.250 dm3 )
51
Stoichiometry Calculations
Involving Solutions 1
Copper metal reacts with nitric acid according to the
following reaction:
8 HNO3 (aq) + 3 Cu
 3 Cu(NO3)2 (aq) + 4 H2O (l) + 2 NO (g)
What volume of 8.00 M HNO3 would be required to
consume a copper penny whose mass is 3.08 grams?
(3.08 g Cu )
(8 mol HNO3)
(1 dm3)
( 1000 cm3)
(63.55 g mol-1 Cu )
(3 mol Cu)
(8 mol HNO3)
(1 dm3)
= 16.2 cm3
52
Stoichiometry Calculations
Involving Solutions 2
15.0 cm3 of a 0.500 M AgNO3 solution is required to
precipitate the sodium chloride in 10 cm3 of a salt
solution. What is the concentration of the solution?
AgNO3 (aq) + NaCl (aq) AgCl (s) +KNO3 (aq)
• Molar Mass NaCl = 23.0 + 35.5 = 58.5 g/mol
0.500 mol
0.0150
AgNO3 X
dm3 x
1 mol NaCl X
58.5 g mol-1
dm3
1 mol AgNO3
NaCl
= 0.439 g of NaCl
0.439 g of NaCl
58.5 g mol-1
x
1
0.0100 dm3
= 0.75 mol dm-3 or 0.75 M
53
Cookie Recipe
Recipe Ingredients
• 1 cube butter
• 1 cup canola oil
• 2 cups white sugar
• 1 egg
• 1 teaspoon vanilla extract
• 1/2 teaspoon salt
• 1 teaspoon baking soda
• 4 1/2 cups all-purpose flour
• 1 cup oatmeal
In my cupboard I have:
• 5 cubes butter
• 8 cups canola oil
• 8 cups white sugar
• 12 eggs
• 20 teaspoons vanilla extract
• 1 pound salt
• 40 teaspoons baking soda
• 45 cups all-purpose flour
• 30 cups oatmeal
• 1 (12 ounce) package chocolate
chips
• 5 (12 ounce) packages chocolate
chips
• 5 pounds of dog biscuits
How many cookies I can make
with out going to the store?
Makes 24 cookies
54
Limiting Reagent
• Although we have been basing our calculations thus
far on only one of the reactants in a chemical
reaction, the reaction will only occur if we have all of
the reactants
• The mole ratio determines how much of each
reactant we need for the reaction
• Often we have an excess of one of the reactants.
Then not all of that reactant will be used up. There
will be some left over.
• It is known as the excess reagent.
• The other reactant will be used up and it will
determine the amount of product we can form.
• It is known as the limiting reagent
55
Limiting Reagent
To determine which of the reagents is the limiting reagent
1. Calculate the number of moles of each reactant
2. Multiply first reactant by the appropriate mole ratio to
get the number of moles of the second reactant that
you need.
3. Compare the amount of the second reactant you have
to the amount you need .
4. If you have more than you need it is in excess and the
first reactant is the limiting reagent
5. If you have less of the second reactant than you need
it becomes the limiting reagent
6. Use the number of moles of the limiting reagent to
calculate the required quantity in the problem
56
Limiting Reagent Example 1
Barium chloride reacts with potassium phosphate as follows:
3 BaCl2 (aq) + 2 K3PO4(aq)  6 KCl (aq) + Ba3(PO4)2 (s)
Calculate the mass of barium phosphate that could be formed when a
solution containing 10.00 g of potassium phosphate is added to a
solution containing 12.00 g of barium chloride.
Molar mass potassium phosphate = 3(39.10) + (30.97) + 4(16.00) = 212.27 g mol-1
Molar mass barium chloride
= (137.34) + 2(35.45)
= 208.24 g mol-1
Molar mass barium phosphate
= 3(137.34)+ 2(30.97)+(8)(16.00) = 601.96 g mol-1
Moles barium chloride
= 12.00g / 208.24 g mol-1 = 0.05762 mol
Moles potassium phosphate = 10.00g / 212.27 g mol-1 = 0.04711 mol
The mole ratio is 3 mol BaCl2 to 2 mol K3PO4. While there are more moles of
BaCl2 than K3PO4, It is not 1.5 times greater. Therefore BaCl2 is the limiting
reagent and all other calculations will be based on barium chloride.
(0.05762 mol BaCl2) (1mol Ba3(PO4)2) (601.96 g mol-1 Ba3(PO4)2 )
( 3 mol BaCl2)
= 11.56 g of Ba3(PO4)2
57
Percent Yield
Stoichiometry allows us to calculate the amounts of
reactants required or the amounts of products generated
from a chemical reaction.
Chemical reactions frequently do not proceed to
completion.
Hence the amount of product recovered is often less than
would be predicted from stoichiometric calculations.
In these situations it is helpful to calculate a percent
yield.
58
Percent Yield
The Theoretical Yield is defined as the amount of
product(s) calculated using Stoichiometry
calculations
The Actual Yield is the amount of product that can
actually be recovered when the reaction is done in
a lab.
The Percent Yield is calculated as follows
Actual yield
Theoretical yield
x 100
59
Percent Yield
Iron reacts with copper sulfate in a single replacement
reaction as follows
Fe (s) + CuSO4 (aq)  ZnSO4 (aq) + Cu (s)
30.00 grams of iron metal were added to excess were added
to excess copper sulfate dissolved in a water solution. 22.50
grams of copper were recovered. Calculate the theoretical
yield of copper in this experiment .
1. First calculate the theoretical yield
(30.00 g Fe)
(1 mol Cu )
(55.85 g mol-1 Fe) (1 mol Fe )
(63.55 g mol-1 Cu)
= 34.14
g Cu
2. Divide the actual yield by the theoretical yield and multiply by 100
22.50 g Cu
34.14 g Cu
X 100
= 65.91%
60