### The Mole

```Chapter 10 – The Mole
10.1
10.2
10.3
10.4
10.5
Measuring Matter
Mass and the Mole
Moles of Compounds
Empirical and Molecular Formulas
The Formula for a Hydrate
Section 10.1 Measuring Matter
Chemists use the mole to count atoms,
molecules, ions, and formula units.
• Describe how the mole is defined
• Explain how a mole is used to indirectly count the
number of particles of matter.
• Relate the mole to a common everyday counting unit.
• Explain the relationship between the mole and
• Convert between moles and number of representative
particles and vice-versa using the factor-label process.
Section 10.1 Measuring Matter
Key Concepts
• The mole is a unit used to count particles of matter
indirectly. One mole of a pure substance contains
• Representative particles include atoms, ions,
molecules, formula units, electrons, and other similar
particles.
• One mole of carbon-12 atoms has a mass of exactly
12 g.
• Conversion factors written from Avogadro’s
relationship can be used to convert between moles
and number of representative particles.
Words That Refer to # of Items
Word
Pair
Dozen
Gross
Ream
Number
2
12
144
500
Abbreviation is mol
SI base unit for amount of substance
Defined as number of representative
particles (carbon atoms) in exactly 12 g
of pure carbon-12
Mole of anything contains 6.022 X 1023
representative particles
6.022 X 1023 = Avogadro’s number
The Mole
Examples of representative particles:
Atoms
Ions
Formula Units
Electrons
Molecules
Representative Particle
Molecule
Atom
Formula
Unit
18 mL
H2O
63.5 g
Cu
58.4 g
NaCl
Converting Moles to # of Particles
and # of Particles to Moles
rep. part. = representative particles
Moles to # of rep. part.
# mol  6.02x1023 rep. part./mol
= # of rep. part.
# rep. part. to moles
# rep. part. X 1 mol/6.02x1023 rep. part.
= # mol
Converting Moles to # of Particles
and # of Particles to Moles
4.00 Moles Water  # Molecules of Water
4.00 mol H2O  6.02x1023 molecules H2O
/mol H2O
= ???
2.41x1024 molecules H2O
1.2x1024 Atoms Iron  Moles Iron
1.2x1024 atoms Fe X 1 mol Fe/ 6.02x1023
atoms Fe
==2.0
???mol Fe
Practice
Problems 1- 4, page 323
Problems 5 (a-b), 6 (a-b), p 324
(Put up next slide as reference)
Converting Moles to # of Particles
and # of Particles to Moles
rep. part. = representative particles
Moles to # of rep. part.
# mol  6.02x1023 rep. part./mol
= # of rep. part.
# rep. part. to moles
# rep. part. X 1 mol/6.02x1023 rep. part.
= # mol
Practice
Section Assessment, page 324
Problems 7 - 14
Chapter Assessment, page 358
Problems 90(a-d), 91(a-d), 92(a-d),
93(a-d), 94(a-d), 95(a-c), 96-101
Chapter 10 – The Mole
10.1
10.2
10.3
10.4
10.5
Measuring Matter
Mass and the Mole
Moles of Compounds
Empirical and Molecular Formulas
The Formula for a Hydrate
Section 10.2 Mass and the Mole
A mole always contains the same
number of particles; however, moles of
different substances have different
masses.
• Determine the molar mass of an element given its
atomic mass in amu.
• Convert between number of moles and the mass of
an element and vice-versa using using the factor-label
process.
• Convert between number of moles and number of
atoms of an element and vice-versa using the factorlabel process.
Section 10.2 Mass and the Mole
Key Concepts
• The mass in grams of 1 mol of any pure substance is
called its molar mass.
• The molar mass of an element is numerically equal to
its atomic mass.
• The molar mass of any substance is the mass in
grams of Avogadro’s number of representative
particles of the substance.
• Molar mass is used to convert from moles to mass.
The inverse of molar mass is used to convert from
mass to moles.
Mass vs Number of Objects
mass 1 dozen limes  mass 1 dozen eggs
Mass of a Mole of Atoms
By definition (see page 119), an atomic
mass unit (amu) is exactly 1/12 the
mass of a carbon-12 atom
• Atomic mass carbon-12 = 12 amu
By definition, mol = number of carbon
atoms in 12 g of carbon-12
• Mass of 1 mol carbon-12 = 12 g
• Mass (g) of 1 mol element numerically
same as atomic mass in amu
Molar Mass
Mass in grams of one mole of any
pure substance
Numerically equal to mass of
substance in amu
Has units of g/mol
Molar Mass
Hydrogen (element)
• Atomic mass = 1.01 amu
• Molar mass H = 1.01 g/mol
• 6.02x1023 H atoms has mass of 1.01 g
Manganese
• Atomic mass = 54.94 amu
• Molar mass Mn = 54.94 g/mol
• 6.02x1023 Mn atoms has mass of 54.94 g
Mole to Mass Conversion
3.00 Moles Mn  Mass Mn
3.00 mol Mn  54.94 g Mn/mol Mn
= ???
165 g Mn
0.0450 Moles Cr  Mass Cr
4.50x10-2 mol Cr x 52.00 g Cr/mol Cr
== 2.34
??? g Cr
Mass to Mole Conversion
525 g Ca  Moles Ca
525 g Ca  1 mol Ca/40.08 g Ca
= ???
13.1 mol Ca
Practice
Problems 15 (a-b), 16 (a-b) page 328
Problems 17 (a-b), 18 (a-b) page 329
Mass to # Atoms Conversion
Need two conversion factors
Mass Gold (Au)  # of atoms of Gold
25.0 g Au  1 mol Au/196.97 g Au 
6.02x1023 atoms Au/mol Au
= ???
7.65x1022 atoms Au
Could do in two steps if prefer:
Calculate # moles Au = 0.127 mol
Calculate # atoms from # mol
# Atoms to Mass Conversion
Need two conversion factors
# Atoms of He  Mass of He
5.50x1022 atoms He  1mol He/6.02x1023
atoms He  4.00 g He/ mol He
= ???
0.366 g He
Could do in two steps if prefer:
Calculate # moles He = 0.0914 mol
Calculate mass from # moles
Conversion Pathways
mass
moles
molar
mass
representative
Avaga- particles
dro’s
number
Practice
Problems 19 (a-c), 21 (a-b) page 331
Problems 20 (a-e) page 331
Chapter 10 – The Mole
10.1
10.2
10.3
10.4
10.5
Measuring Matter
Mass and the Mole
Moles of Compounds
Empirical and Molecular Formulas
The Formula for a Hydrate
Section 10.3 Moles of Compounds
The molar mass of a compound can be
calculated from its chemical formula and
can be used to convert from mass to
moles of that compound.
• Recognize the mole relationships shown by a
chemical formula.
• Calculate the molar mass of a compound.
• Convert between the number of moles and mass of a
compound and vice-versa using the factor-label
process.
• Apply conversion factors to determine the number of
atoms or ions in a known mass of a compound.
Section 10.3 Moles of Compounds
Key Concepts
• Subscripts in a chemical formula indicate how many
moles of each element are present in 1 mol of the
compound.
• The molar mass of a compound is calculated from the
molar masses of all of the elements in the compound.
• Conversion factors based on a compound’s molar
mass are used to convert between moles and mass of
a compound.
Mole and Chemical Formulas
Chemical Formula indicates types and
number of each atom contained in
compound
Freon – CCl2F2
Mole and Chemical Formulas
Freon – CCl2F2
In every molecule of freon, there are:
• 1 carbon atom
• 2 chlorine atoms
• 2 fluorine atoms
In one mole
dozenofmolecules,
freon, are there
6.02x10
are2312
times as many atoms as above
Mole and Chemical Formulas
If have 1 mole Freon molecules
(CCl2F2) have:
• 1 mole of C atoms
• 2 moles of Cl atoms
• 2 moles of F atoms
Moles Compound to Moles Atoms
Aluminum oxide (alumina), Al2O3
1.25 moles alumina  moles Al+3
1.25 mol alumina  2 mol Al+3/1 mol
alumina
== 2.50
??? mol Al+3
Practice
Problems 29 - 33 page 335
Molar Mass of Compounds
Mass of one mole of compound
Sum of masses of every particle that
makes up the compound
Molar Mass =  massi
i
massi = mass of particles contained in
one mole of the compound
(mass is conserved)
Molar Mass of Compounds
massi = mass of particles contained in
one mole of the compound
If particles are atoms, then
massi = molar mass of atom  number
of moles of atom contained in one mole
of the compound
Molar Mass of Compounds
Example: K2CrO4 – abbreviate as “KC”
Revised from p. 335 text (credit: Allison Mazur)
elm = element cpd = compound
# mol elm/mol cpd  molar mass elm =
mass elm/mol cpd
2 mol K/mol KC39.10 g K/mol K =78.20 g K/mol KC
1 mol Cr /mol KC52.00 g Cr/mol Cr=52.00 g Cr/mol KC
4 mol O /mol KC16.00 g O/mol O = 64.00 g O /mol KC
Molar mass K2CrO4 = 194.20 g/mol KC
Practice
Problems 34 - 36 (all are a –c) page
335
Mole to Mass Conversion
Allyl sulfide = (C3H5)2S
Mass of 2.50 moles of allyl sulfide?
Step 1 – calculate molar mass
1 mol S  32.07 g S/mol S = 32.07 g S
6 mol C  12.01 g C/mol C = 72.06 g C
10 mol H  1.008 g H/mol H = 10.08 g H
Molar mass = 114.21 g/mol (C3H5)2S
Mole to Mass Conversion
Allyl sulfide = (C3H5)2S
Mass of 2.50 moles of allyl sulfide?
Step 2 – convert moles to mass
Molar mass = 114.21 g/mol (C3H5)2S
2.50 mol (C3H5)2S  114.21 g (C3H5)2S/
mol (C3H5)2S = 286 g (C3H5)2S
Practice
Problems 37-39 page 336
Mass to Mole Conversion
Ca(OH)2
Moles Ca(OH)2 in 325 g?
Step 1 – calculate molar mass
1 mol Ca  40.08 g Ca/mol Ca = 40.08 g Ca
2 mol O  16.00 g O/mol O
= 32.00 g O
2 mol H  1.008 g H/mol H
= 2.016 g H
Molar mass = 74.10 g/mol Ca(OH)2
Mass to Mole Conversion
Ca(OH)2
Moles Ca(OH)2 in 325 g?
Step 2 – convert to moles
Molar mass =
74.10 g/mol Ca(OH)2
325 g Ca(OH)2  1 mol Ca(OH)2/74.10 g
Ca(OH)2 = 4.39 mol Ca(OH)2
Practice
Problems 40(a-c) – 41(a-b) page 337
Mass to # Particles Conversion
Problem 10.9 AlCl3 Sample 35.6 g
# Al+3 ions? # Cl- ions?
Mass of 1 formula unit of AlCl3?
Step 1 – calculate molar mass
1 mol Al  26.98 g Al/mol Al = 26.98 g Al
3 mol Cl  35.45 g Cl/mol Cl = 106.35 g C
Molar mass AlCl3 = 133.33 g/mol AlCl3
Mass to # Particles Conversion
Problem 10.9 AlCl3 Sample 35.6 g
# Al+3 ions? # Cl- ions?
Mass of 1 formula unit of AlCl3?
Step 2 – Convert mass to moles
35.6 g AlCl3  1mol AlCl3/133.33 g AlCl3 =
0.267 mol AlCl3
Mass to # Particles Conversion
Problem 10.9 AlCl3 Sample 35.6 g
# Al+3 ions? # Cl- ions?
Mass of 1 formula unit of AlCl3?
Step 3 – Convert moles to # particles
0.267 mol AlCl3 x 6.02x1023 formula units
AlCl3/mol AlCl3
= 1.61x1023 formula units AlCl3
Mass to # Particles Conversion
Problem 10.9 AlCl3 Sample 35.6 g
# Al+3 ions? # Cl- ions?
Mass of 1 formula unit of AlCl3?
Step 4 – Convert # form. units to # ions
1.61x1023 formula unit AlCl3 x 1 Al+3/formula
unit AlCl3 = 1.61x1023 Al+3 ions
1.61x1023 formula unit AlCl3 x 3 Cl-/formula
unit AlCl3 = 4.83x1023 Cl- ions
Mass to # Particles Conversion
Problem 10.9 AlCl3 Sample 35.6 g
# Al+3 ions? # Cl- ions?
Mass of 1 formula unit of AlCl3?
Step 5 – Convert molar mass to g per
formula unit
133.33 g AlCl3/mol x 1 mol/6.0221x1023
formula unit
= 2.2140x10-22 g AlCl3/ formula unit
Page 339 of book incorrect – need 5 SF
Conversions For Compounds (Cpd)
moles atoms
or ions in cpd
chemical
formula
mass
molar
representmoles
ative
cpd
particles
Practice
Problems 42 (a-c), 43 (a-c), 44 (a-c),
45, 46 (a-d) page 339
Problems 128 – 152 pages 359 – 60
Chapter 10 – The Mole
10.1 Measuring Matter
10.2 Mass and the Mole
10.3 Moles of Compounds
10.4 Empirical and Molecular Formulas +
Laws of Definite Proportions, Multiple
Proportions from section 3.4
10.5 The Formula for a Hydrate
Section 3.4 Elements and Compounds
A compound is a combination of
two or more elements.
• Explain how all compounds obey the laws of definite
and multiple proportions.
• Demonstrate using calculations that these laws apply
to specific compounds.
Section 3.4 Elements and Compounds
Key Concepts
• The law of definite proportions states that a
compound is always composed of the same elements
in the same proportions.
• The law of multiple proportions states that if elements
form more than one compound, those compounds will
have compositions that are whole-number multiples of
each other.
Section 10.4 Empirical & Molecular Formulas
A molecular formula of a compound is a
whole-number multiple of its empirical
formula.
• Explain what is meant by the percent composition of a
compound and calculate it from the formula for the
compound.
• Determine the empirical and molecular formulas for a
compound from mass percent and actual mass data
using calculations.
Section 10.4 Empirical & Molecular Formulas
Key Concepts
• The percent by mass of an element in a compound
gives the percentage of the compound’s total mass
due to that element.
• The subscripts in an empirical formula give the
smallest whole-number ratio of moles of elements in
the compound.
• The molecular formula gives the actual number of
atoms of each element in a molecule or formula unit of
a substance.
• The molecular formula is a whole-number
multiple of the empirical formula.
Percent Composition
Unknown compound’s composition may
be determined by elemental analysis
Results reported as percent by mass
% by mass =
100  mass of element/mass of compound
Percent by mass for all elements in the
compound called percent composition
Laws of Definite and Multiple
Proportions
These laws were developed as result of
careful experimental measurements
prior to the full development of the mole
concept
Law of Definite Proportions
Regardless of amount, compound
composed of same elements in same
proportion by mass
 statement that compound’s formula
doesn’t change with amount of
compound present
•
H2O = formula for water no matter how
much water you have – proportions
always same, mole ratio same, mass
ratios same
Law of Definite
(or Constant)
Proportion
(or Composition)
Both sources of
calcium carbonate
(CaCO3) have same
% composition
Law of Definite Proportions
Focus here is % composition from
some chemical analysis
% by mass = 100  mass element
mass compound
Masses are given rather than derived
from a formula
Law of Definite Proportions
% by mass = 100  mass element
mass compound
Examine analyses of sucrose in table
3.4, page 88
Analysis of 20.00 g sample same as
analysis of 500.0 g sample  have the
same composition
Book says  same compound (sort of)
Practice
Law of Definite Proportions
Problems 19 - 23, page 88
Problems 72, 74-78 , pages 95 - 96
Problem 3, page 977
Law of Multiple Proportions
If elements form more than one
compound, those compounds will have
compositions that are small, wholenumber multiples of each other
Focus: ratio of mass ratios = integer
Water vs Hydrogen Peroxide
H2O (O:H 16:2) vs H2O2 (O:H 32:2)
• Ratio of Mass ratios
O:H (H2O2) / O:H (H2O) = 2:1
Atomic Basis of the Law of Multiple
Proportions
Law of Multiple Proportions
Copper Chloride Compounds
Compound 1 Compound 2
Cpd % Cu
% Cl
Ratio
Cu : Cl
1
64.20 35.80
1.793 : 1
2
47.27 52.73 0.8964 : 1
Ratio of cpd #1
ratio to #2 ratio
2.000
Practice
Law of Multiple Proportions
Problems 73, 79, 80 pages 95-6
Problem 4, page 977
% Composition from Formula
Elemental analysis to determine
percent composition
• Done entirely by experiment
• Don’t need formula
• Easier to do if know elements present
If already know chemical formula, can
compute percent composition
% Composition from Formula
Step 1 – Assume have 1 mol of compound
Step 2 – Look up the molar mass of
element j
Step 3 – Calculate the mass of element j
present using formula
Step 4 – Do steps 2 & 3 until all elements
done.  masses = molar mass
Step 5 – Use mass from step 3 and molar
mass to get % for element j
% Composition from Formula
NaHCO3
Step 1 – assume have 1 mol NaHCO3
Steps 2 & 3 – get masses of each element
1 mol Na  22.99 g Na/mol Na = 22.99 g Na
1 mol H  1.008 g H/mol H
= 1.008 g H
1 mol C  12.01 g C/mol C
= 12.01 g C
3 mol O  16.00 g O/mol O
= 48.00 g O
Step 4 – sum masses to get molar mass
Molar mass = 84.008 g/mol NaHCO3
% Composition from Formula
Step 5 – use molar mass and individual
masses to calculate % composition
% Na = 100  22.99 g Na/84.01 g NaHCO3
= 27.37 %
% H = 100  1.008 g H/ 84.01 g NaHCO3
= 1.200 %
% C = 100  12.01 g C/ 84.01 g NaHCO3
= 14.30 %
% O = 100  48.00 g O/ 84.01 g NaHCO3
= 57.14 %
% Composition from Formula
Cross – check: % should sum to 100
27.37% + 1.200% + 14.30% + 57.14%
= 100.00%
Note: only can keep 2 places after
decimal point even though 1.200% has
3 places after decimal
Practice
Problems 54 - 57, page 344
Problems 162, 168 – 171, 174 pages
360 -1
Empirical Formula
Smallest whole number ratio of
elements in a compound
• If ionic, same as formula unit
• If non-ionic, may or may not be
molecular formula
NaCl – ionic, formula unit
HO - non ionic (H2O2, H3O3, H4O4, etc)
CHO – non ionic (C2H2O2, C3H3O3, etc)
Empirical Formula
Can determine from:
• Percent composition information

Assume 100 g of compound
• Actual masses of elements for given
mass of the compound
Empirical Formula
Step 1 – If given % composition, determine
g of each element in 100 g
Step 2 – Calculate moles of each element
from the mass of the element
Step 3 – Calculate mole ratio of elements
Step 4 – Convert mole ratio to whole
numbers
• Since compounds have whole numbers of
atoms, mole ratio of compound must involve
whole numbers
Empirical Formula Given % Comp
Methyl Acetate
C 48.64% H 8.16% O 43.20%
Steps 1&2 – Assume 100 g and get # moles
48.64 g C  1 mol C/12.01 g C = 4.050 mol C
8.16 g H  1 mol H/1.008 g H = 8.10 mol H
43.20 g O  1 mol O/16.00 g O = 2.700 mol O
Step 3 – Compute mole ratio
4.05 mol C : 8.10 mol H : 2.700 mol O
Empirical Formula Given % Comp
Methyl Acetate
4.05 mol C : 8.10 mol H : 2.700 mol O
Step 4 – Convert mole ratio to whole #
Divide all values by smallest
4.050 mol C  2.700 mol O = 1.500 C/O ratio
8.10 mol H  2.700 mol O = 3.00 H/O ratio
2.700 mol O  2.700 mol O = 1.000 O/O ratio
Find multiplier to make whole number ratios
2 x 1.5 = 3 2 x 3.0 = 6 2 x 1.0 = 2
Empirical Formula Given % Comp
Methyl Acetate
2 x 1.5 = 3 2 x 3.0 = 6 2 x 1.0 = 2
Empirical formula C3H6O2
Practice
Practice includes both percent
composition given and masses given
type problems
Problems 58 - 61, page 346
Problems 161, 165, 167, 171-2 pages
360-1
Molecular Formula
CH – empirical formula
C2H2 – molecular formula for acetylene
C6H6 – molecular formula for benzene
Molecular formula = n x (empirical formula)
Need to know molar mass of actual
compound to determine molecular
formula from molar mass of empirical
formula
Molecular Formula
CH – empirical formula
Molar mass: 13.02 g/mol
1. Molar mass of compound = 26.04 g/mol
 Ratio molar masses = 2.000
 Molecular formula = C2H2
2. Molar mass of compound = 78.12 g/mol
 Ratio molar masses = 6.000
 Molecular formula = C6H6
Empirical and Molecular Formulas
Process Map – Fig 10.15, page 347
Express % by mass in g
Find # moles each element
Examine mole ratio
Write empirical formula
Find integer n relating emp. &
molecular formulas
Multiply subscripts by n
Write molecular formula
Practice
Problems 62-66, page 350
Problem 164, page 361
Problems 29, 30 page 982
Chapter 10 – The Mole
10.1
10.2
10.3
10.4
10.5
Measuring Matter
Mass and the Mole
Moles of Compounds
Empirical and Molecular Formulas
The Formula for a Hydrate
Section 10.5 Formulas of Hydrates
Hydrates are solid ionic compounds in
which water molecules are trapped.
• Explain what a hydrate is and name a hydrate based
on its composition.
• Explain what the term anhydrous means.
• Determine the formula of a hydrate from laboratory
data.
• Explain what a desiccant is and how it is used.
Section 10.5 Formulas of Hydrates
Key Concepts
• The formula of a hydrate consists of the formula of the
ionic compound and the number of water molecules
associated with one formula unit.
• The name of a hydrate consists of the compound
name and the word hydrate with a prefix indicating the
number of water molecules in 1 mol of the compound.
• Anhydrous compounds are formed when hydrates are
heated.
Formulas for Hydrates
Hydrate – compound that has a specific
number of water molecules bound to
the atoms
Number of water molecules associated
with each formula unit is written
following a dot
CaCl22H2O Calcium chloride dihydrate
Formulas for Hydrates – page 351
Hydrates
Single compound can form a number of
different hydrates
• CaCl2 – mono-, di-, and hexa- hydrates
• CaCl2H2O, CaCl22H2O, CaCl26H2O
“Water of hydration” can be driven off
by heating – produces anhydrous form
Mass of water associated with formula
unit must be included in molar mass
calculations
Color Change in Hydrated Compounds
Heating: drives off water and produces
anhydrous form
Hydrated
Anhydrous
Copper (II) sulfate
Determining Hydrate Formula
BaCl2•xH2O - what is value of x ?
Initial mass = 5.00 g
Final mass after heating = 4.26 g
Mass water = 5.00 - 4.26 = 0.74 g H2O
Mol H2O = 0.74 g  1 mol/18.02 g H2O
= 0.041 mol H2O
Mol BaCl2 = 4.26 g BaCl2  1
mol/208.23 g BaCl2 = 0.0205 mol BaCl2
Determining Hydrate Formula
BaCl2•xH2O - what is value of x ?
0.041 mol H2O 0.0205 mol BaCl2
x = moles H2O /moles BaCl2
x = 0.041 mol H2O / 0.0205 mol BaCl2
x = 2.0 mol H2O / 1.00 mol BaCl2
x= 2/1= 2
BaCl2•2H2O
Practice
Problems 74 - 75, page 353
Problems 180 - 187, page 361
Problems 33 - 34, page 982
Hydrates
Anhydrous solids used as drying
agents (desiccants)
End of Chapter
Problems 63-64, page 340
Problems 151-156, page 349
Problems 33-34, page 877
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