The Mole Concept
Mr. Matthew Totaro
Legacy High School
Honors Chemistry
Why Is Knowledge of
Amounts so Important?
• To know the amount of a material
in a sample, you need to know what
fraction of the sample it is.
• Some Applications:
 The amount of sodium in sodium
chloride for a diet.
 The amount of iron in iron ore for
steel production.
 The amount of hydrogen in water for
a hydrogen fuel cell.
 The amount of chlorine in freon to
estimate ozone depletion.
2
Stoichiometry
• Stoichiometry = comes from the Greek words
‘stoicheon’ meaning ‘element’ and ‘metron’
meaning ‘to measure’.
• System used to measure quantities in chemistry
Two Branches of Stoichiometry
Composition Stoichiometry = describes the quantitative
relationships among elements in compounds.
Reaction Stoichiometry = describes the quantitative
relationships among substances as they participate in
chemical reactions
3
Calculation
of
Formula
(Molecular)
Mass
4
Formula (Molecular) Mass
• The mass of an individual molecule or formula unit.
• Also known as molecular mass or molecular weight.
• Sum of the masses of the atoms in a single molecule
or formula unit.
 Whole = Sum of the parts.
Ex: Mass of 1 molecule of H2O
= 2(1.01 amu H) + 16.00 amu O = 18.02 amu.
5
Practice—Calculate the Formula Mass of
Al2(SO4)3.
6
Practice—Calculate the Formula Mass of
Al2(SO4)3, Continued.
Al
S
O
Al 2 (SO 4 )3
 2
 3
 12 

26.98 amu
32.07 amu
16.00 amu
342.17 amu
7
Groupings into Units
The Dozen = a Group of 12
8
Counting Atoms by Moles
• The number of atoms we
use in chemistry is 6.02 x
1023 and we call this a mole.
1 mole = 6.02 x 1023
things.
Like 1 dozen = 12
things.
Avogadro’s number.
9
The Mole
Two large helium
balloons contain
approximately 1 mol of
helium (He) atoms.
10
Amadeo Avogadro
The mole is named in honor of him. ‘Mole’ in Latin means ‘mass’.
11
The Mole - Avogadro’s Number
How Big is a Mole?
13
Chemical Packages—Moles
• Mole = Number of things equal to the number
of atoms in 12 g of C-12.
1 atom of C-12 weighs exactly 12 amu.
1 mole of C-12 weighs exactly 12 g.
• In 12 g of C-12 there are 6.02 x1023 C-12
atoms.
6.02 10 atoms
1 mole
1 mole
carbon
12.01 g
23
14
1 mole
23
6.02  10 atoms
Counting by Weighing: Nails
by the Pound
Some hardware stores sell nails by
the pound, which is easier than
selling them by the nail.
This problem is similar to asking
how many atoms are in a given mass
of an element.
A customer buys 2.60 lb of medium-sized nails,
and a dozen of these nails weigh 0.150 lb. How
many nails did the customer buy?
• The solution map for the problem is as follows:
• We convert from pounds to number of nails:
One Step
Stoichiometry
Conversions
17
Example—A Silver Ring Contains 1.1 x 1022 Silver
Atoms. How Many Moles of Silver Are in the Ring?
1.1 x 1022 atoms Ag
moles Ag
Given:
Find:
Solution Map:
atoms Ag
mol Ag
1 mol
6.022  1023 atoms
1 mol = 6.022 x 1023 atoms
Relationships:
Solution:
1.1 10
22
1 mol
atoms Ag 
23
6.022  10 atoms
 1.8266  10- 2 mol Ag  1.8  10- 2 mol Ag
Check:
Since the number of atoms given is less than
Avogadro’s number, the answer makes sense.
18
Practice—Calculate the Number of Atoms
in 2.45 mol of Copper.
19
Relationship Between
Moles and Mass
•
•
•
•
The mass of one mole of
atoms is called the molar mass
The molar mass of an element,
in grams, is numerically equal to 1 mole
sulfur
the element’s atomic mass, in
32.06 g
amu
The lighter the atom, the less a
mole weighs
The lighter the atom, the more
atoms there are in 1 g
20
Mole and Mass Relationships
1 mole
sulfur
32.06 g
1 mole
carbon
12.01 g
21
Example—Calculate the Moles of Sulfur
in 57.8 g of Sulfur.
Given:
Find:
Solution Map:
57.8 g S
mol S
gS
1 mol S
32.07 g
mol S
Relationships: 1 mol S = 32.07 g
Solution:
Check:
1 mol
57.8 g S 
32.07 g
 1.80 mol S
Since the given amount is much less than 1 mol S,
the number makes sense.
22
Practice—Calculate the Grams of Carbon
in 0.0265 mols of Pencil Lead, Continued.
Given:
Find:
Solution Map:
0.0265 g C
mol C
mol C
gC
12.01 g
1 mol
Relationships: 1 mol C = 12.01 g
Solution:
Check:
12.01 g
0.0265 g C 
1 mol
 3.18 x 10-1 mol C
Since the given amount is much less than 1 mol C,
the number makes sense.
23
Molar Volume of a Gas
There is so much
empty space
between molecules
in the gas state that
the volume of the
gas is not effected
by the size of the
molecules (under
ideal conditions).
24
Molar Volume
25
Practice—Calculate the Volume of
2.79 mols of He gas.
26
Two Step
Stoichiometry
Conversions
27
Example — How Many Aluminum Atoms
Are in a Can Weighing 16.2 g?
Given:
Find:
Solution Map:
16.2 g Al
atoms Al
g Al
mol Al
1 mol
26.98 g
Relationships:
Solution:
atoms Al
6.02 10 23 atoms
1 mol
1 mol Al = 26.98 g, 1 mol = 6.022 x 1023
1 mol Al 6.02 10 23 atoms
16.2 g Al 

26.98 g Al
1 mol
 3.62 10 23 atoms Al
Check: Since the given amount is much less than 1 mol Cu,
the number makes sense.
28
Molar Mass of Compounds
• The relative weights of molecules can be
calculated from atomic weights.
Formula mass = 1 molecule of H2O
= 2(1.01 amu H) + 16.00 amu O = 18.02 amu.
• Since 1 mole of H2O contains 2 moles of H
and 1 mole of O.
Molar mass = 1 mole H2O
= 2(1.01 g H) + 16.00 g O = 18.02 g.
29
Practice — How Many CO2 molecules Are
in a tank weighing 3.10 g?
30
Example — What Is the Mass of 4.78 x 1024
NO2 Molecules?
Given: 4.78 x 1024 NO2 molecules
Find: g NO2
Solution Map: molecules
mol NO2
1 mol NO 2
6.02 10 23 molec
Relationships:
46.01 g
N
1 molO
1 mol NO2 = 46.01 g, 1 mol = 6.02 xNO
10223
Solution:
4.78 10 24 molec NO2 
g NO2
 1  14.01 amu
 2  16.00 amu

46.01 amu
1 mol
46.01 g

6.02 10 23 molec 1 mol NO2
 365 g NO2
Check:
Since the given amount is more than Avogadro’s
number, the mass > 46 g makes sense.
31
Practice — How Many Formula Units
Are in 50.0 g of PbO2? (PbO2 = 239.2)
32
Percent
Composition
33
Percent Composition
•
Percentage of each element in a compound.
 By mass.
• Can be determined from:
 The formula of the compound.
 The experimental mass analysis of the
compound.
• The percentages may not always total to 100% due
to rounding.
mass of element
X in 1 mol
part
Percentage 
 100%
Percentage


100%
mass of 1 molwhole
of the compound
34
Example—Find the Percent
Composition of C2Cl4F2.
Given:
Find:
Solution Map:
Relationships:
C2Cl4F2
% Cl by mass
4  molar mass Cl
Mass % Cl 
100%
molar mass C 2Cl4 F2
mass element X in 1 mol
Mass % element X 
100%
mass 1 mol of compound
Solution: 4  molar mass Cl  4(35.45 g/mol)  141.8 g/mol
molar mass C 2Cl4 F2  2(12.01)  4(35.45)  2(19.00)  203.8 g/mol
Mass % Cl 
141.8 g/mol
100%  69.58%
203.8 g/mol
Check: Since the percentage is less than 100 and Cl is much
heavier than the other atoms, the number makes
35
sense.
Empirical Formula
36
Empirical Formulas
• The simplest, whole-number ratio of atoms in a
molecule is called the empirical formula.
Can be determined from percent composition or
combining masses.
• The molecular formula is a multiple of the
empirical formula.
100g
%A
mass A (g)
100g
%B
mass B (g)
MMA
MMB
37
moles A
moles A
moles B
moles B
Empirical Formulas, Continued
Hydrogen Peroxide
Molecular formula = H2O2
Empirical formula = HO
Benzene
Molecular formula = C6H6
Empirical formula = CH
Glucose
Molecular formula = C6H12O6
Empirical formula = CH2O
38
Practice—Determine the Empirical
Formula of Benzopyrene, C20H12.
39
Practice—Determine the Empirical Formula
of Benzopyrene, C20H12, Continued
•
•
Find the greatest common factor (GCF) of
the subscripts.
20 factors = (10 x 2), (5 x 4)
12 factors = (6 x 2), (4 x 3)
GCF = 4
Divide each subscript by the GCF to get the
empirical formula.
C20H12 = (C5H3)4
Empirical formula = C5H3
40
Finding an Empirical
Formula from
Experimental Data
41
Example:
A laboratory analysis of aspirin determined the
following mass percent composition.
C = 60.00%
H = 4.48%
O = 35.53%
Find the empirical formula.
42
Example:
Find the empirical formula
of aspirin with the given
mass percent
composition.
• Write down the given quantity and its units.
Given:
C = 60.00%
H = 4.48%
O = 35.53%
Therefore, in 100 g of aspirin there are 60.00 g C,
4.48 g H, and 35.53 g O.
43
Example:
Find the empirical
formula of aspirin with
the given mass
percent composition.
Information:
Given: 60.00 g C, 4.48 g H, 35.53 g O
• Write down the quantity to find and/or its units.
Find: empirical formula, CxHyOz
44
Example:
Find the empirical
formula of aspirin with
the given mass percent
composition.
Information:
Given: 60.00 g C, 4.48 g H, 35.53 g O
Find: empirical formula, CxHyOz
• Collect needed conversion factors:
1 mole C = 12.01 g C
1 mole H = 1.01 g H
1 mole O = 16.00 g O
45
Example:
Find the empirical
formula of aspirin with
the given mass
percent composition.
Information:
Given: 60.00 g C, 4.48 g H, 35.53 g O
Find: empirical formula, CxHyOz
Conversion Factors:
1 mol C = 12.01 g; 1 mol H = 1.01 g;
1 mol O = 16.00 g
• Write a solution map:
gC
mol C
gH
mol H
gO
mol O
pseudoformula
46
mole
ratio
whole
number
ratio
empirical
formula
Example:
Find the empirical
formula of aspirin with
the given mass percent
composition.
Information:
Given: 60.00 g C, 4.48 g H, 35.53 g O
Find: empirical formula, CxHyOz
Conversion Factors:
1 mol C = 12.01 g;
1 mol H = 1.01 g; 1 mol O = 16.00 g
Solution Map: g C,H,O  mol C,H,O 
mol ratio  empirical formula
• Apply the solution map:
 Calculate the moles of each element.
1 mol C
60.00 g C 
 4.996 mol C
12.01 g C
1 mol H
4.48 g H 
 4.44 mol H
1.01 g H
1 mol O
35.53 g O 
 2.221 mol O
16.00 g O
47
Example:
Find the empirical
formula of aspirin with
the given mass
percent composition.
Information:
Given: 4.996 mol C, 4.44 mol H,
2.221 mol O
Find: empirical formula, CxHyOz
Conversion Factors:
1 mol C = 12.01 g;
1 mol H = 1.01 g; 1 mol O = 16.00 g
Solution Map: g C,H,O  mol C,H,O 
mol ratio  empirical formula
• Apply the solution map:
 Write a pseudoformula.
C4.996H4.44O2.221
48
Example:
Find the empirical
formula of aspirin with
the given mass
percent composition.
Information:
Given: C4.996H4.44O2.221
Find: empirical formula, CxHyOz
Conversion Factors:
1 mol C = 12.01 g;
1 mol H = 1.01 g; 1 mol O = 16.00 g
Solution Map: g C,H,O  mol C,H,O 
mol ratio  empirical formula
• Apply the solution map:
 Find the mole ratio by dividing by the smallest number of
moles.
C 4.996 H 4.44 O 2.221
2.221
2.221
2.221
C 2.25 H 2O1
49
Example:
Find the empirical
formula of aspirin with
the given mass
percent composition.
Information:
Given: C2.25H2O1
Find: empirical formula, CxHyOz
Conversion Factors:
1 mol C = 12.01 g;
1 mol H = 1.01 g; 1 mol O = 16.00 g
Solution Map: g C,H,O  mol C,H,O 
mol ratio  empirical formula
• Apply the solution map:
 Multiply subscripts by factor to give whole number.
{ C2.25H 2O1 } x 4
C9H8O4
50
Practice—Determine the Empirical Formula
of Tin (II) Fluoride, which Contains 75.7%
Sn (118.70) and the Rest Fluorine (19.00).
51
All These Molecules Have the Same
Empirical Formula. How Are the
Molecules Different?
Name
Glyceraldehyde
Molecular
Formula
C3H6O3
Empirical
Formula
CH2O
Erythrose
C4H8O4
CH2O
Arabinose
C5H10O5
CH2O
Glucose
C6H12O6
CH2O
52
All These Molecules Have the Same
Empirical Formula. How Are the
Molecules Different?, Continued
Name
Glyceraldehyde
Molecular
Formula
C3H6O3
Empirical
Formula
CH2O
Molar
Mass, g
90
Erythrose
C4H8O4
CH2O
120
Arabinose
C5H10O5
CH2O
150
Glucose
C6H12O6
CH2O
180
53
Molecular
Formula
54
Molecular Formulas
• The molecular formula is a multiple of the
empirical formula.
• To determine the molecular formula, you
need to know the empirical formula and the
molar mass of the compound.
Molar massreal formula = Factor used to multiply subscripts
Molar massempirical formula
55
Example—Determine the Molecular Formula
of Cadinene if it has a Molar Mass of
204 g and an Empirical Formula of C5H8.
1. Determine the empirical formula.
 May need to calculate it as previous.
C5H8
2. Determine the molar mass of the empirical
formula.
5 C = 60.05, 8 H = 8.064
C5H8 = 68.11 g/mol
56
Example—Determine the Molecular Formula
of Cadinene if it has a Molar Mass of
204 g and an Empirical Formula of C5H8,
Continued.
3. Divide the given molar mass of the
compound by the molar mass of the
empirical formula.
 Round to the nearest whole number.
204 g/mol
3
68.11 g/mol
57
Example—Determine the Molecular Formula
of Cadinene if it has a Molar Mass of
204 g and an Empirical Formula of C5H8,
Continued.
4. Multiply the empirical formula by the
factor above to give the molecular
formula.
(C5H8)3 = C15H24
58
Practice—Benzopyrene has a Molar Mass of
252 g and an Empirical Formula of C5H3.
What is its Molecular Formula? (C = 12.01,
H=1.01)
59
Practice—Benzopyrene has a Molar Mass of 252 g
and an Empirical Formula of C5H3. What is its
Molecular Formula? (C = 12.01, H=1.01),
Continued
C5 = 5(12.01 g) = 60.05 g
H3 = 3(1.01 g) = 3.03 g
C5H3
=
63.08 g
252 g/mol
n
4
63.08 g/mol
Molecular formula = {C5H3} x 4 = C20H12
60
Practice—Determine the Molecular
Formula of Nicotine, which has a Molar
Mass of 162 g and is 74.0% C, 8.7% H, and
the Rest N.
(C=12.01, H=1.01, N=14.01)
61
Practice—Determine the Molecular Formula of
Nicotine, which has a Molar Mass of 162 g and is
74.0% C, 8.7% H, and the Rest N, Continued
Given: 74.0% C, 8.7% H, {100 – (74.0+8.7)} = 17.3% N 
in 100 g nicotine there are 74.0 g C, 8.7 g H, and 17.3 g N.
Find: CxHyNz
Conversion Factors:
1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol N = 14.01 g
Solution Map:
gC
mol C
gH
mol H
gN
mol N
whole
mole number
pseudo- ratio ratio empirical
formula
62
formula
Practice—Determine the Molecular Formula of
Nicotine, which has a Molar Mass of 162 g and is
74.0% C, 8.7% H, and the Rest N, Continued.
Apply solution map:
1 mol C
C5 = 5(12.01 g) = 60.05 g
74.0 g C 
 6.16 mol C
12.01 g
N1 = 1(14.01 g) = 14.01 g
1 mol H
H7 = 7(1.01 g) =
7.07 g
8.7 g H 
 8.6 mol H
C5H7N
=
81.13 g
1.01 g
1 mol N
17.3 g N 
 1.23 mol N
14.01 g
mol. mass nicotine
162 g

2
C6.16H8.6N1.23
mol. mass emp. form. 81.13 g
C 6.16 H 8.6 N1.23  C5H 7 N
1.23
1.23
{C5H7N} x 2 = C10H14N2
1.23
63