Ch. 10 – The Mole
I. Molar Conversions
I
II
III
IV
A. What is the Mole?

A counting number (like a dozen)

Avogadro’s number (NA)

1 mole = 6.022  1023 representative
particles
B. Mole/Particle Conversions
6.022  1023
NA
NUMBER
MOLES
OF
PARTICLES
(particles/mol)
NA atoms/mol
NA molecules/mol
Particles =
atoms,
molecules,
formula units,
ions, etc
B. Mole/Particle Conversion Examples
 How
many molecules are in
2.50 moles of C12H22O11?
2.50 mol
C12H22O11
6.02  1023
molecules
C12H22O11
1 mol
C12H22O11
= 1.51  1024
molecules
C12H22O11
B. Mole/Particle Conversion Examples
you have 2.23 x 1018 atoms of
sodium, how many moles is
that?
 If
2.23  1018 1 mole Na
atoms Na
6.02  1023
atoms Na
3.70 x 10-6
= moles Na
B. Mole/Particle Conversion Examples
 How
many formula units is 3.75
moles of calcium hydroxide?
6.02  1023
3.75
formula units
mol Ca(OH)2 Ca(OH)2
= 2.26  1024
1 mol Ca(OH)2 formula units
Ca(OH)2
C. Molar Mass

Avogadro discovered the
relationship between number of
particles and volume of a gas

This was used to find the
relationship for particles in a mole
Representative Particles & Moles
Substance Chemical Representative Rep Particles
Formula
Particle
in 1.00 mole
Carbon
C
Atom
6.02 x 1023
Nitrogen
gas
N2
Molecule
6.02 x 1023
Calcium ion
Ca2+
Ion
6.02 x 1023
Magnesium
fluoride
MgF2
Formula unit
6.02 x 1023
C. Molar Mass

Mass of 1 mole of an element or
compound

Atomic mass (on the PT) tells the...
 mass of each atom (amu)
 grams per mole (g/mol)
C. Molar Mass Examples

carbon
12.01 g/mol

aluminum
26.98 g/mol

zinc
65.39 g/mol
C. Molar Mass Examples

water
 H2O
 2(1.01) + 15.99 = 18.02 g/mol

sodium chloride
 NaCl
 22.99 + 35.45 = 58.44 g/mol
C. Molar Mass Examples

sodium bicarbonate
 NaHCO3
 22.99 + 1.01 + 12.01 + 3(15.99)
= 84.01 g/mol

sucrose
 C12H22O11
 12(12.01) + 22(1.01) + 11(15.99)
= 342.19 g/mol
D. Molar Conversions
6.02  1023
NA
molar
mass
MASS
NUMBER
MOLES
IN
GRAMS
(g/mol)
OF
(particles/mol)
PARTICLES
Particles = atoms,
molecules, formula
units, ions, etc
NA atoms/mol
NA molecules/mol
D. Molar Conversion Examples
 How
many moles of carbon
atoms are in 26 g of carbon?
26 g C 1 mol C
12.01 g C
= 2.2 mol C
D. Molar Conversion Examples
the mass of 2.1  1024
molecules of NaHCO3.
2.1  1024
84.01 g
Molecules 1 mol
NaHCO3
NaHCO3
NaHCO3
 Find
6.02  1023 1 mol
Molecules
NaHCO3
NaHCO3
= 290 g NaHCO3
D. Molar Conversion Examples
 How
many atoms are in 22.5
grams of potassium?
22.5 g K
1 mol K
6.02  1023
atoms K
39.10 g K 1 mol K
= 3.46 x 1023 atoms K
Mole Ratios






The ratio of the moles of one molecule to the
atoms within that molecule.
molecular formulas give atom-to-atom and moleto-mole ratios
 example: molecular formula C6H12O6
atom-to-atom ratios atom-to-molecule ratios
6 atoms C
12 atoms H
6 atoms C
6 atoms O
12 atoms H
6 atoms O



6 atoms C
1 molecule
12 atoms H
1 molecule
6 atoms O
1 molecule
Mole Ratios
Example: molecular formula C6H12O6
mole-to-mole ratios
(elements)
6 mol C
12 mol H
6 mol C
6 mol O
12 mol H
6 mol O
mole-to-mole ratios
(compound)
6 mol C
1 mol C6H12O6
12 mol H
1 mol C6H12O6
6 mol O
1 mol C6H12O6
problems that ask you to relate one substance to
another require mole-to-mole ratios
Ch. 10 – The Mole
II. Formula
Calculations
I
II
III
IV
A. Percent Composition

the percentage by mass of each
element in a compound
total mass of element
% mass of element 
 100
total mass of compound
mass of element in 1 mol compound
% mass of element 
 100
molar mass of compound
A. Percent Composition
Find the percent composition of a sample
that is 28 g Fe and 8.0 g O.
Known:
Unknown:
 Mass of Fe = 28 g
 % Fe = ?
 Mass of O = 8.0 g
 %O=?
 Total Mass = 28 + 8.0 g = 36 g

%Fe =
%O =
28 g
36 g
8.0 g
36 g
 100 = 78% Fe
 100 = 22% O
Check:
78% + 22%
= 100%

A. Percent Composition


Find the % composition of Cu2S.
Use this formula when finding % composition from a
chemical formula: % mass of element  mass of element in 1 mol compound  100
molar mass of compound
Known:
Unknown:
 Mass of Cu in 1 mol Cu2S =
 % Cu = ?
2(63.55g) = 127.10 g Cu
 %S= ?
 Mass of S in 1 mol Cu2S = 32.07 g S
 Molar Mass = 127.10 g + 32.07 g = 159.17 g/mol
%Cu =
%S =
127.10 g Cu
159.17 g Cu2S
32.07 g S
159.17 g Cu2S
 100 = 79.85% Cu
 100 = 20.15% S
A. Percent Composition
 How
many grams of copper are in
a 38.0-gram sample of Cu2S?
 Use answer from last question as
a conversion factor.
79.85 g Cu
Cu2S is 79.85% Cu =
100 g Cu2S
38.0 g Cu2S 79.85 g Cu
= 30.3 g Cu
100 g Cu2S
A. Percent Composition
 Find
the percent composition of
Cu2SO4.
Known:
 Mass of Cu in 1 mol Cu2SO4 = 2(63.55 g) = 127.10 g Cu
 Mass of S in 1 mol Cu2SO4 = 32.07 g S
 Mass of 0 in 1 mol Cu2SO4 = 4(16.00 g) = 64.00 g O
 Molar Mass = 127.10 g + 32.07 g + 64.00 g = 223.17 g/mol
Unknown:
 % Cu = ?
 %S= ?
 %O= ?
A. Percent Composition
 Find
the percent composition of
Cu2SO4.
%Cu =
%S =
%O =
127.10 g
223.17 g
32.07 g
223.17 g
64.00 g
223.17 g
 100 = 56.95% Cu

 100 = 14.37% S
 100 = 28.68% O
Check:
56.95% +
14.37% +
28.68% =
100%
B. Empirical Formula
 Smallest
whole number ratio of
atoms in a compound
C2H6
reduce subscripts
CH3
B. Empirical Formula
1. Find mass (or %) of each element.
2. Find moles of each element.
3. Divide moles by the smallest # to
find subscripts.
4. When necessary, multiply
subscripts by 2, 3, or 4 to get
whole #’s.
EX 1 Empirical Formula
 Find
the empirical formula for a
sample of 25.9% N and 74.1% O.
25.9 g N 1 mol N = 1.85 mol N
=
1
N
14.01 g N 1.85 mol
N1.85O4.63
74.1 g O 1 mol O = 4.63 mol O
=
2.5
O
16.00 g O 1.85 mol
EX 1 Empirical Formula
N1O2.5
Need to make the subscripts whole
numbers  multiply by 2
N2O5
Mole Ratios that do not end up
in whole numbers




Some mole ratios that are close enough to a
whole number to round off.
(1.04 → 1; 2.98 → 3 )
Other ratios are too far from a whole number to
round. ( 1.25  1.00 1.33  1.00
1.51  2.00 )
Example: It was found that iron and oxygen
combine in the following molar ratios:
Fe 0.195 O 0.291 → Fe1O1.49 → 2 (Fe1O1.49) →
0.195
0.195
»
= Fe2 O3
Mole Ratios that do not end up
in whole numbers



Since we cannot have "fractional" atoms in a
compound, we need to normalize the relative
amount of oxygen to be equal to a whole
number.
1.49 would appear to be 1 and 1/2, so if we
multiply the relative amounts of each atom by
'2', we should be able to get whole number
values for each atom.
A multiple can be used to convert to a whole
number. (1.25 x 4 = 5 ; 1.33 x 3 = 4 ;
1.50 x 2 = 3)
EX 2 Empirical Formula
 Find
the empirical formula for a
sample of 94.1% O and 5.9% H.
94.1 g O 1 mol O = 5.88 mol O
=
1
O
16.00 g O 5.84 mol
5.9 g H 1 mol H = 5.84 mol H
=
1
H
1.01 g H
5.84 mol
EF = OH
Molecular Formulas

“True Formula” - the actual number of atoms
in a compound, either the same as or a
whole-number multiple of the empirical
formula
empirical
formula
CH3
?
molecular
formula
C 2H6
Steps to a Molecular Formula
1. Find the empirical formula.
2. Find the empirical formula mass.
3. Divide the molar mass by the
empirical formula mass.
4. Multiply each subscript in your EF
by the answer from step 3.
molar mass
n
EF mass
EF n
EX 1 Molecular Formula
 The
empirical formula for ethylene
is CH2. Find the molecular formula
if the molar mass is 28.1 g/mol?
empirical formula mass = 14.03 g/mol
n=
28.1 g/mol
14.03 g/mol
= 2.00
(EF)n = (CH2)2  C2H4
Put it all together!!

1,6 - Diaminohexane is 62.1% C, 13.8% H,
and 24.1% N. What is the empirical
formula? If the molar mass is 116.21 g/mol,
what is the molecular formula?
62.1 g C
1 mol C
12.01 g C
13.8 g H 1 mol H
1.01 g H
24.1 g N
1 mol N
14.01 g N
= 5.17 mol C
1.72 mol
=3C
= 13.7 mol H
=8H
1.72 mol
= 1.72 mol N
1.72 mol
=1N
EF = C3H8N
Put it all together!!
C3H8N
empirical formula mass = 58.12 g/mol
n=
116.21 g/mol
58.12 g/mol
(C3H8N)2 
= 2.00
C6H16N2
Combustion Analysis



When a compound containing carbon and
hydrogen is subject to combustion with oxygen in
a special combustion apparatus all the carbon is
converted to CO2 and all the hydrogen to H2O.
The amount of carbon produced can be
determined by measuring the amount of CO2
produced.
The amount of H produced by the amount of H2O
produced.
Example 1





Consider the combustion of isopropyl alcohol. The
sample is known to contain only C, H and O.
The mass of CxHyOz is the total mass of the C + H
+ O. Combustion of 0.255 grams of isopropyl
alcohol produces 0.561 grams of CO2 and 0.306
grams of H2O.
What is the empirical formula of isopropyl alcohol?
The reaction looks like:
CxHyOz + O2 → CO2 + H2O
Determine the amount of C
.561grams
CO2
1 mole
CO2
44.008 g
CO2
= .0128 mole C 12.01 g C
1 mole C
1 mole C
1 mole CO2
= .154 g C
Determine the amount of H
.306 grams
H2O
1 mole
H2O
18.015 g
2 mole H
1 mole H2O
H2O
= .034 mole H
1.008 g H
1 mole H
= .034 g H






If we have .0128 moles of CO2 in our sample,
then we know we have 0.0128 moles of C in
the sample.
How many grams of C is this? 0.154 grams
C (as done above).
If we have 0.017 moles of H2O, then we have
2 x (0.017) = 0.034 moles of hydrogen.
Since hydrogen is about 1.008 gram/mole,
we must have 0.034 grams of hydrogen in
our original sample.
When we add our carbon and hydrogen
together we get:
0.154 grams (C) + 0.034 grams (H) = 0.188 g
What about oxygen?


We know we combusted 0.255 grams of
isopropyl alcohol. The 'missing' mass must be
from the oxygen atoms in the isopropyl
alcohol:
0.255 g - 0.188 g = 0.067 grams oxygen
= .067 grams O
1 mole O
15.999 g O
= .0042 mol O
Find the Empirical Formula
Overall therefore, we have:

C 0.0128 H 0.0340 O 0.0042 →


0.0042
0.0042
0.0042
C 3.05 H 8.1 O 1.0 → C 3 H 8 O
 Within experimental error, the most
likely empirical formula for propanol
would be:

»C3H8O