Chapter 10 The Mole
“Making Measurements in
Chemistry”
T. Witherup 2006
Chapt. 10 OBJECTIVES
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Define the “mole” & describe its importance.
Identify & use Avogadro’s number.
Define “molar mass” & explain how it relates the
mass of a substance to its number of particles.
Convert among the number of particles, moles &
mass of a substance.
Describe “molar volume” & use it to solve
problems.
Find the percentage composition of a formula.
Use percentage composition to find the formula
of an unknown sample.
Find empirical & molecular formulas.
10-1 Chemical Measurements
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Atomic Mass: the mass of an atom expressed
relative to the mass assigned to the carbon-12
isotope, in amu (atomic mass units).
1 amu = 1/12 of the mass of Carbon-12.
Since C-12 has 6p + 6n = 12 particles, its mass =
12 amu.
Since C-13 has 6p + 7n = 13 particles, its mass =
13 amu.
And C-14 has a mass of ??? amu?
(Why don’t we consider the mass of the electrons
in these atoms?)
Average Atomic Mass
Since “natural” carbon has 1.1% the C-13
isotope and only a trace amount of the C14 isotope, its average atomic mass is
dominated by C-12, or 12.011 amu.
The average atomic mass accounts for all
(natural) isotopes of an element.
The average atomic mass of each element
may be found on the Periodic Table.
NOTE: In our course, when using average atomic
masses, always round to the ‘hundredths’ place.
Average Atomic Mass of Elements
ELEMENT
AVERAGE
ATOMIC MASS
(amu)
AVERAGE
ATOMIC MASS,
rounded (amu)
Hydrogen
1.00794
1.01
Carbon
12.0111
12.01
Oxygen
15.9994
16.00
Chlorine
35.453
35.45
Iron
55.847
55.85
Formula Mass
Formula Mass: the sum of the atomic masses
of all atoms in a compound.
Water (H2O) has 2 Hydrogen atoms and 1
Oxygen atom.
 Formula Mass of H2O is
2 X (1.01) amu + 16.00 amu = 18.02 amu.
 What is the formula mass of methane, CH4?
 Of NaCl? (For ionic compounds we refer to a
“formula unit” rather than a “formula mass,” but
it is essentially the same idea.)
 Of ammonia, NH3?
 Of glucose, C6H12O6?
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But what is a “mole”?
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That furry creature who burrows in the
yard?
The dark pigmentation on our skin?
A massive stone structure used as a
breakwater or pier?
An undercover agent?
NO! A mole is a special chemical term
used to count atoms!
The MOLE (mol)
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A mole of any element is defined as the amount of
the element that contains as many atoms as there
are in exactly 12 g of the carbon-12 isotope.
A mole is found experimentally to be equal to
6.022 X 1023 atoms of C-12, which is called
Avogadro’s Number (NA).
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12 g Carbon-12 = 1 mol Carbon 12 atoms = 6.022 X 1023
Carbon 12 atoms
The molar mass of any substance is the mass of
one mole of that substance.
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Molar mass is numerically equal to the atomic mass,
molecular mass, or formula mass of the substance.
A mole of any substance contains Avogadro’s number of
units of that substance (6.022 X 1023 units).
What’s in a Mole?
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Atoms – one mole of an element contains 6.022 X
1023 atoms of the element.
Molecules – one mole of a molecular (covalent)
compound contains 6.022 X 1023 molecules.
Formula Units – one mole of an ionic compound
contains 6.022 X 1023 formula units.
Gizmos – one mole of gizmos contains 6.022 X 1023
gizmos.
Anything – one mole of anything contains 6.022 X
1023 anythings!
NA (6.022 X 1023) is very practical for counting
small particles, especially things like atoms, ions
and molecules.
EXAMPLES
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How many pens in 1 mole of pens?
How many atoms in 63.546g of Cu?
How many atoms in 6.3546g of Cu?
How many molecules in 1 mole of sugar
(C6H12O6)?
How many molecules in 10 moles of sugar?
How many carbon atoms in 1 mole of sugar?
How many oxygen atoms?
How many formula units in 1 mole of CaCl2?
How many calcium ions in 1 mole of CaCl2?
How many chloride ions in 1 mole of CaCl2?
How many chloride ions in 0.1 mole of CaCl2?
10-2 Mole Conversions by Factor Label
Method (1)
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Mass and Moles
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Use the molar mass of the substance.
1 Mole = n grams of the substance, so
1 = n grams/Mole, but also
1 = Mole/n grams of the substance.
Use these conversions by setting up Factor
Labels to cancel the units!
See examples on next slide.
10-2 Mole Conversions by Factor Label
Method (1) Examples
Example A: How many moles in 75.0 g of iron?
75.0g Fe
X
1 mol Fe
55.85g Fe
= 1.34 mol Fe
Example B: How many grams in 0.250 mol Na?
0.250 mol Na X
22.99g Na
1 mol Na
= 5.75g Na
Solving ‘Mole Problems” (1)
MASS
mol/g
Molar
mass
MOLES
g/mol
10-2 Mole Conversions by Factor Label
Method (2)
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Particles and Moles
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Use Avogadro’s Number (6.022 X 1023) of
particles.
(6.022 X 1023)particles/mol  1 mol/(6.022
X 1023)particles
Again, set up Factor Labels to cancel
units!
See examples on next slide.
10-2 Mole Conversions by Factor Label
Method (2) Examples
Example C: How many atoms in 0.25 mol Na?
6.022 X1023 atoms Na
0.250 mol Na X
= 1.51 X 1023 atoms Na
1 mol Na
Example D: How many moles in 4.20 X 1023 molecules of CO2?
4.20 X
1023
1 mol CO2
Molecules CO2 X
6.022 X 1023 molecules CO2
= 0.697 mol CO2
Solving ‘Mole Problems” (2)
PARTICLES
MOLES
Number of
Particles in
1 mole
(6.02 X 1023)
10-2 Mole Conversions by Factor Label
Method (3)
 Gases and Moles
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Avogadro proposed that equal volumes of
gases contain the same number of gas
particles at a given temperature & pressure.
Therefore one mole of gas #1 would have the
same volume as one mole of gas #2.
It is observed that one mole of any gas
occupies 22.4 liters @ STP (molar
volume).
 STP = Standard Temperature and
Pressure, 0° C and 1 atmosphere.
Once more, set up Factor Labels to cancel
the units!
10-2 Mole Conversions by Factor Label
Method (3) Examples (Gases)
Example E: What is the volume of 13.0 moles of hydrogen
gas at STP?
13.0 mol H2 X
22.4 L H2
1 mol H2
= 291 L H2
Example F: How many moles are in 250. mL of oxygen
at STP?
1 L O2
1 mol O2
250. mL O2 X
X
1000 mL O2
22.4 L O2
= 0.0112 mol O2
Solving ‘Mole Problems” (3)
MOLES
Molar volume
(22.4L/mol @ STP)
VOLUME
of gas @ STP
10-2 Mole Conversions by Factor Label
Method (Summary) (See p 330.)
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Mass and Moles
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Particles and Moles
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Use Avogadro’s Number (6.02 X 1023) of particles.
Gases and Moles
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Use the molar mass of the substance.
Use the Molar Volume (22.4 liters @ STP).
STP = Standard Temperature and Pressure, 0° C and 1
atmosphere.
Set up Factor Labels to cancel units!
 DON’T GET LAZY! Include labels to ensure
that ALL units cancel correctly.
Multi-step conversions are easily done if you
are careful with the labels!
Summary: Solving ‘Mole Problems”
MASS
Molar
mass
PARTICLES
MOLES
Number of
Particles in
1 mole
(6.02 X 1023)
Molar volume
(22.4L/mol @ STP)
VOLUME
of gas @STP
10-3 Empirical & Molecular
Formulas
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Percentage Composition
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The mass of each element in a compound
compared to the entire mass of the compound
and multiplied by 100%.
Example 1
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2.45 g aluminum oxide decomposes into 1.30 g
aluminum & 1.15 g oxygen. What is the percentage
composition?
%O = (1.15g O/2.45g Aluminum Oxide) X 100% =
46.9% Oxygen (O, not O2)
%Al = (1.30g Al/2.45g Aluminum Oxide) X 100% =
53.1% Aluminum
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As a check, note that 46.9% + 53.1% = 100.0%.
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10-3 Empirical & Molecular Formulas
(cont’d)
Percentage Composition (cont’d)
Example 2
 Determine the percent composition of
CaCO3.
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Molar Mass = 40.08 + 12.01 +3(16.00) =
100.09g/mol
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%Ca = (40.08g Ca/100.09g CaCO3) X 100% =
40.04% Ca
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% C = (12.01g C/100.09g CaCO3) X 100% =
12.00% C
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% O = (48.00g O/100.09g CaCO3) X 100% =
47.96% O
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Check: 40.04% + 12.00% + 47.96% = 100.00%
Determining Empirical Formulas
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Empirical Formula
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The formula that gives the simplest whole
number ratio of the atoms of the elements in
the formula.
Example 1
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What is the empirical formula of a compound containing
2.644g of gold and 0.476g of chlorine?
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0.476g Cl X (1 mol Cl/35.45g Cl) = 0.0134 mol Cl
2.644g of Au X (1 mol Au/196.97g Au) = 0.0134 mol Au
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Empirical formula = Au0.0134 Cl0.0134 or simply AuCl
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Determining Empirical Formulas
Example 2
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What is the empirical formula of a compound with
5.75 g Na, 3.50 g N & 12.00 g O?
First, find the mole amounts.
5.75g Na X (1 mol Na/22.99g Na) = 0.250 mol Na
3.50 g N X (1 mol N/14.01g N) = 0.250 mol N
12.00g O X (1 mol O/16.00g O) = 0.750 mol O
Empirical formula = Na0.250 N0.250 O0.750
Divide each mole quantity by the smallest to
get whole numbers. (0.250 in this case)
Empirical formula = NaNO3
Determining Molecular Formulas
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Molecular Formula
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The formula that gives the actual number of
atoms of each element in a molecular compound.
Example 3
 Hydrogen peroxide has a molar mass of 34.00 g/mol and a
chemical composition of 5.90% H & 94.1% O. What is its
molecular formula?
 First, find the empirical formula, assuming the percents are
mass.
 For 5.90% H: 5.90gH X (1 mol H/1.01g H) = 5.84 mol H
 For 94.1% O: 94.1g O X (1 mol O/16.00g O) = 5.88 mol O
 Empirical formula = H5.84O5.88 or HO.
 But molar mass = 34.00g/mol and HO is only 17.01g/mol.
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Therefore, Molecular Formula = 2(HO) or H2O2.
Determining Molecular Formulas
Example 4
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A compound contains 42.56 g Pd and 0.8000 g H. If its
molar mass is 216.8 g/mol, find the molecular formula.
First, find the empirical formula.
42.56g Pd X (1 mol Pd/106.42 g Pd) = 0.4000 mol Pd
0.8000g H X (1 mol H/1.00g H) = 0.800 mol H
Empirical formula: Pd0.4000H0.800 or PdH2
PdH2 has a mass = 108.44 g/mol.
Since molar mass = 216.8 and the empirical
mass = 108.4, the Molecular Formula is twice
the empirical formula [2(PdH2)] or simply
Pd2H4.
Did we cover the OBJECTIVES?
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Define the “mole” & describe its importance.
Identify & use Avogadro’s number.
Define “molar mass” & explain how it relates the
mass of a substance to its number of particles.
Convert among the number of particles, moles &
mass of a substance.
Describe “molar volume” & use it to solve
problems.
Find the percentage composition of a formula.
Use percentage composition to find the formula
of an unknown sample.
Find empirical & molecular formulas.
How to be successful at solving
Mole Problems:
USE FACTOR LABELS!
 PRACTICE!
 PRACTICE!
 PRACTICE!
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