Chemical Quantities
A Mrs. J. Kim Production
Version 1.0
The Mole &
Avogadro’s Number
Chemical Quantities 5.1
The Mole
 A counting unit
 Similar to a dozen because it represents a
number
 Dozen = 12
 Baker’s dozen = 13
 Mole =
Pair = 2
Gross = 144
6.02 x 1023
 602 billion trillion
 602,000,000,000,000,000,000,000
Avogadro’s Number
 This number (6.02 x 1023 ) is called Avogadro’s Number
 Italian scientist Amadeo Avogadro di Quaregna (1776-1856)
 The abbreviation for a mole is “mol”
 Definition: A mole is the number of carbon atoms in exactly
12 grams of pure Carbon-12.
How Big is a Mole?
 If you had Avogadro’s number of
 sheets of paper stacked on top of one
another, the pile would reach beyond the sun
 blood cells, that would be more than the
total number of blood cells found in every
human on earth
How Big is a Mole?
 If you had Avogadro’s number of
 popcorn kernels and spread them across
the United States of America, the country
would be covered in popcorn to a depth of
over 9 miles.

marshmallows, they would cover
the surface of the earth with a layer
that is a 12 miles high
How Big is a Mole?
 If you had Avogadro’s number of
 days, that would be a number greater than the
age of the earth

pennies, you could distribute them to all the
people currently in the world so that they could
spend a million dollars per hour every hour (day
and night) for the rest of their lives
Mole Numbers
 1 dozen cookies = 12 cookies
 1 mole of cookies = 6.02 x 1023 cookies
 1 dozen cars = 12 cars
 1 mole of cars = 6.02 x 1023 cars
 1 dozen Al atoms = 12 Al atoms
 1 mole of Al atoms = 6.02 x 1023 atoms
Quick Quiz #1
 (True/False): One mole of carbon and one mole of copper have
the same number of atoms.
 True! One mole of anything means that there are 6.02 x 1023 of that
item.
 In this case, there are 6.02 x 1023 atoms of carbon and 6.02 x 1023
atoms of aluminum.
1 mole
carbon
=
1 mole
copper
Quick Quiz #2
 (True/False): One mole of carbon and one mole of copper weigh
the same amount.
 False! Just because you have the same number of an item does not
mean they weigh the same amount.
 For example, one dozen cookies does not weigh the same as one dozen
cars.
1 mole
copper
1 mole
carbon
A Mole of Representative Particles
 1 mol X
=
A Chemical
Formula
6.02 x 1023 ______ X
Representative Particles
• Atoms
• Ion
• Compounds = 2+ atoms that are
bonded together
• Molecules
• Formula Units
Representative Particles
 Atoms = single element, no subscripts
 ie: X = Fe , Xe , P …
 Ion = atom (or group of atoms) with a charge
 ie: X = Cl– , PO43– , Al3+ , NH4+ …
 Compounds = 2+ atoms that are bonded together
 Molecules = covalent, made up of nonmetals only
 ie: X = Cl2 , H2O , CCl4 , CH3CH2CH3 , SO2…
 Formula Units = ionic, made up of a cation & an anion
 ie: X = CaCl2 , NH4ClO3 , Ba(OH)2 …
A Mole of Representative Particles
= 6.02 x 1023 atoms C
 C is a single element
 1 mol C
6.02 x 1023 molecules H2O
 Hydrogen & oxygen are both nonmetals
 1 mol H2O =
6.02 x 1023 formula units NaCl
 Sodium is a metal (cation) and chloride is a nometal
(anion)
 1 mol NaCl =
Finding the Scientific Notation Button
Using Your Scientific Calculator
6.02 x 1023 / 5.1 x 10-7
6
6
.
.
0
0
2
2
EE
2
EXP
x10x
2

3
3

5
(
.
5
EE
1
.
(-)
EXP
1
7
x10x
(-)
7
)
Example 5.1.A
Use your calculator to find the following values and review SigFigs:
 1)
2.7200 x 1021 =
0.004 517
6.022 x 1023
 2)
 3)
(5.150 x 10–11) x (6.02 x 1023) =
1.5 x 1016
=
2.5 x 10-8
6.02 x 1023
 4)
6.02 x 1023 =
8.140 x 10-9
7.40 x 1031
3.10 x 1013
Example 5.1.B
How many atoms are in 0.500 moles of gold?
 Starting Point (SP) = 0.500 mol Au
 Ending Point (EP) = ? atoms Au

Conversion Factor (CF):
1 mole Au = 6.02 x 1023 atoms Au
0.500 mol Au 6.02 x 1023 atoms Au
1
1 mol Au
3.01 x 1023
=
atoms Au
Example 5.1.C
1) How many representative
particles are in 3.22 x 10–10
moles of CuCl2?
1.94 x 1014 formula units CuCl2
2) How many moles are in
1.8 x 1024 helium atoms?
3.0 mol He
Molar Mass
Chemical Quantities 5.2
Molar Mass
 molar mass
= the mass (g) of 1 mole of a substance
= the average atomic mass
= the atomic weight (From the Periodic Table)
 1 mole of C atoms
 1 mole of Mg atoms
 1 mole of Cu atoms
= 12.01 g C
= 24.31 g Mg
= 63.55 g Cu
Molar Mass Names
 Other names related to molar mass:
 Molecular Mass/Weight (MW) = the mass of all the
atoms in a molecule (nonmetals only, no ions)
 Formula Mass/Weight (FW) = the mass of all the atoms
in an ionic compound (cation & anion)
 Note: You may see any of these terms which
mean the SAME NUMBER… just different units
A Mole of Representative Particles
 1 CaCl2 compound = 1 Ca+2 ions
2 Cl– ions
 3 dozen CaCl2 compounds = 3 dozen Ca+2 ions
6 dozen Cl– ions
 13 mol CaCl2 compounds = 13 mol Ca+2 ions
26 mol Cl– ions
Finding Molar Masses of Compounds
 1) From the chemical formula, count the number of
atoms of each element
 2) Multiply the molar mass of each element (found on
the periodic table) by the number of atoms of the
element present in the compound
 3) Find the sum of the molar masses for each
element’s atom(s)
Finding the Molar Mass of CaCl2
Element
Ca
Cl
# of
Atomic
Multiply
Equals
Atoms
Mass
1
2
x
x
40.08
35.45
Mass of that
Element
=
=
Total:
40.08
70.90
110.98
The molar mass (or FW) of CaCl2 = 110.98 g/mol
Written as a Conversion Factor:
1 mol CaCl2 = 110.98 g CaCl2
Example 5.2.A
1) What is the formula mass of potassium oxide?
1 mol K2O = 94.20 g K2O
2) What is the molar mass of antacid (aluminum hydroxide)?
1 mol Al(OH)3 = 78.01 g Al(OH)3
3) What is the molar mass of Prozac (C17H18F3NO), an
antidepressant that inhibits the uptake of serotonin by the
brain.
1 mol C17H18F3NO = 309.36 g C17H18F3NO
Calculations Involving Molar Mass
Aluminum is often used for the structure of lightweight bicycle
frames. How many grams of aluminum are in 6.15 moles of
aluminum?
SP: 6.15 mol Al
EP: ? g Al
CF: 1 mol Al = 26.98 g Al
6.15 mol Al 26.98 g Al
= 166 g Al
1 mol Al
Example 5.2.B
How many moles of Fe2(SO4)3 are in 8.27 g of
Fe2(SO4)3?
SP: 8.27 g Fe2(SO4)3
EP: ? mol Fe2(SO4)3
CF: 1 mol Fe2(SO4)3 = 399.91 g Fe2(SO4)3
8.27 g Fe2(SO4)3 1 mol Fe2(SO4)3
= 0.0207 mol
399.91 g Fe2(SO4)3 Fe2(SO4)3
Example 5.2.C
The artificial sweetener aspartame (AKA Nutra-Sweet)
(C14H18N2O5) is used to sweeten diet foods, coffee and
soft drinks. How many moles of aspartame are present
in 225 g of aspartame?
225 g C14H18N2O5 1 mol C14H18N2O5
294.34 g C14H18N2O5
=
0.764 mol
C14H18N2O5
Representative Particles  Grams
 You must convert into moles first
 The mole is the ultimate connector!
 There is no single, direct conversion factor to switch between
representative particles  grams
 Analogy: How would you find the mass of 24 cookies, given
that 1 dozen cookies has a mass of 47 g?
 Answer: Convert cookies  dozen  g
Example 5.2.D
How many atoms of Cu are present in 35.4 g Cu?
SP: 35.4 g Cu
EP: ? atoms Cu
CF1: 1 mol Cu = 63.55 g Cu
CF2: 1 mol Cu = 6.02 x 1023 atoms Cu
6.02 x 1023
35.4 g Cu 1 mol Cu
atoms Cu
63.55 g Cu 1 mol Cu
= 3.35 x 1023
atoms Cu
Example 5.2.E
How many molecules of O2 are present in 78.4 g O2?
SP: 78.4 g O2
EP: ? molecules O2
CF1: 1 mol O2 = 32.00 g O2
CF2: 1 mol O2 = 6.02 x 1023 molec O2
78.4 g O2
1 mol O2
32.00 g O2
6.02 x 1023
molec O2
1 mol O2
= 1.47 x 1024
molec O2
Example 5.2.F
What is the mass (in grams) of 1.20 x 1024 molecules of
glucose (C6H12O6)?
1.20 x 1024 molec
C6H12O6
1 mol C6H12O6
6.02 x 1023 molec
C6H12O6
180.18 g
C6H12O6
1 mol
C6H12O6
=
359 g
C6H12O6
Example 5.2.G
How many atoms of O are present in 6.3 grams of
oxygen?
6.3 g O2
1 mol O2
32.00 g O2
6.02 x 1023
molec O2
1 mol O2
2 atoms O
1 molec O2
= 2.4 x 1023
atoms O
Percent Composition
Chemical Quantities 5.3
Percent Composition
Percent
=
Composition
Mass of all atoms of an
element in the compound
Molar mass of the
compound
x 100
Example 5.3.A
What is the mass percent of carbon in C5H8NO4 (the glutamic
acid used to make MSG monosodium glutamate), a compound
used to flavor foods and tenderize meats?
Mass of 5 C
Percent
=
x 100 = ? % C
Composition
MM of C5H8NO4
=
5 (12.01)
146.14
x 100 = 41.09 % C
Empirical & Molecular Formulas
Chemical Quantities 5.4
Chemical Formulas of Compounds
 Chemical Formulas give the relative numbers of atoms or
moles of each element in a formula unit
 Whole number ratios (Law of Definite Proportions)
 If we know (or can determine) the relative number of
moles of each element in a compound, we can determine a
formula for the compound.
Empirical & Molecular Formulas
 Empirical Formula = the lowest whole-number ratio of the atoms of
the elements in a compound
 Ionic compounds are always empirical formulas
 Molecular Formula = the actual formula of a compound (a simple
whole-number multiple of its empirical formula)
Empirical Formula
Molecular Formula
CH2O
x6
C6H12O6
CH4
CH2
x1
x3
CH4
C3H6
CH2
x7
C7H14
The molecular and
empirical formulas
can be the same!
Different molecular
formulas can have the
same empirical formula!
Determining Empirical Formulas
1) Change % of each element into grams (when applicable)
2) Convert grams of each element into mol. (Use molar mass.)
3) Divide each answer from Step 2 by the smallest number of moles.
4) Round each answer from Step 3 to the nearest whole number. The
answers from Step 3 should be very close to a whole number. If not,
multiply all the Step 3 answers by a common number to get whole
numbers.
1) If an answer from Step 3 is closer to __.5, multiply all moles by 2
2) If an answer from Step 3 is closer to __.25, multiply all moles by 4
3) If an answer from Step 3 is closer to __.33, multiply all moles by 3
Example 5.4.A
 A molecule is 40.92% carbon, 4.58% hydrogen, and 54.50%
oxygen by mass. What is the empirical formula of the
molecule?
Element
%
Grams
Moles
Ratio
Mod. Ratio
C
40.92
40.92
3.407
 3.406
1
x3 = 3
H
4.58
4.58
4.535
 3.406
1.33
x3 = 4
O
54.50
54.50
3.406
 3.406
1
x3 = 3
Empirical Formula = C3H4O3
Determining MF from EF
A molecule with the empirical formula C3H4 has a
molecular weight (MW) of 121 g/mol. What is the
molecular formula?
 Step 1) Find the empirical weight (EW)
 C3H4
= 40.07 g/mol
Determining MF from EF
 Step 2) Divide the MW by the EW to see how many
times larger (whole number!) the MW is.

MW
121 g/mol
=
EW
40.07 g/mol
=
3
 Step 3) Determine the MF by multiplying the
subscripts in the EF by your answer from step 2
(C3H4) 3 = C9H12
Example 5.4.B
A colorless liquid used in rocket engines has an empirical
formula of NO2 and a molar mass of 92.0 g/mol. What is the
molecular formula of this substance?
Molecular Formula = N2O4
Example 5.4.C
A substance consisting only of Na, B and H is 60.80%
Na and 28.60% B. What is the empirical formula of the
substance?
Empricial Formula = NaBH4
Stoichiometry
Chemical Quantities 5.5
Representative Particles
(atoms, molecules,
formula units, etc.)
Representative Particles
(atoms, molecules,
formula units, etc.)
Use Avogadro’s Number
1 mol A = 6.02 x 1023 (rp) A
Volume
@ STP
(Liters)
Use Molar
Volume
1 mol A = 22.4 L A
Use Avogadro’s Number
1 mol B = 6.02 x 1023 (rp) B
Use Coefficients from
Balanced Equation
Moles
(mol)
Use Molar Mass
__ mol A = __ mol B
Moles
(mol)
Use Molar
Volume
1 mol B = 22.4 L B
Use Molar Mass
(from periodic table)
1 mol B = (MM) g B
(from periodic table)
1 mol A = (MM) g A
Mass
(grams)
Mass
(grams)
Volume
@ STP
(Liters)
Advanced Stoichiometry
 How many hydroxide ions are in 11.02 mg of calcium
hydroxide?

Solution: 11.02 mg Ca(OH)2  g Ca(OH)2  mol Ca(OH)2 
mol OH –  ions OH–
11.02 mg
Ca(OH)2
1g
1 mol
Ca(OH)2
1000
mg
74.10 g
Ca(OH)2
Molar mass
of Ca(OH)2
2 mol
OH–
6.02 x 1023
ions OH–
1 mol
1 mol OH–
Ca(OH)2
1 Ca(OH)2 has
2 OH– in it
=
1.79 x
1020
ions OH–
Hydrates
Chemical Quantities 5.6
Hydrates
 Hydrate = A compound that has a specific
number of water molecules bound to each
formula unit
 Anhydrate = A hydrate compound that has
had the water removed from it
 How would we remove the water from a hydrate
to make it into an anhydrate?
Hydrate Formulas & Names
 Formulas for hydrates use a dot to connect the formula of the
compound and the number of water molecules per formula unit
 Example: Na2CO3  10 H2O
 This means that there are 10 water molecules for every formula unit
of Na2CO3
 Name: sodium carbonate decahydrate
 Deca = 10
 Hydrate = water
Example 5.6.A
 Calculate the percent by mass of water in Na2CO3  10 H2O (washing
soda, sodium carbonate decahydrate)
mass of water
Percent H2O =
x 100%
mass of hydrate
g 10 H2O
=
x 100%
g Na2CO3  10 H2O
180.20 g
=
x 100% = 62.965 %
105.99 g + 180.20 g
Example 5.6.B
A 34.32 g hydrate with the formula Na2S  x H2O was heated
until all the water evaporated off, leaving an anhydrate with a
mass of 11.15 g. What is the value of x?
 Treat this like an empirical formula problem to find the mole
ratio between the salt and the water. First we need to find
the number of grams of the anhydrate and the water.
Example 5.6.B (Cont.)
A 34.32 g hydrate with the formula Na2S  x H2O was heated
until all the water evaporated off, leaving an anhydrate with a
mass of 11.15 g. What is the value of x?
 Mass of the anhydrate (Na2S) = 11.15 g Na2S

This was given in the problem
 Mass of the water = Mass of hydrate – mass of anhydrate
 34.32
g hydrate (anhydrate + water) – 11.15 g anhydrate
= 23.17 g H2O
Example 5.6.B (Cont.)
 Convert both masses to moles (using molar masses)
g Na2S  0.1429 moles Na2S
 23.17 g H2O  1.286 moles H2O
 11.15
 Then divide both mole values by the smallest number
 0.1429
moles Na2S / 0.1429 = 1 mol Na2S
 1.286 moles H2O / 0.1429 = 8.999 = 9 mol H2O
 The formula is Na2S  9 H2O