The Mole &
Chemical Composition
Section 1
Avogadro’s Number &
Molar Conversions
For Review
Mole
SI unit for amount
# of atoms in 12g of carbon-12
Avogadro’s Number
# of particles in a mole
6.022 x 1023 - # of (?) in 1.000 mole
Used to count any kind of particle
The Mole is a Counting Unit
Ex. – 1 dozen = 12
The mole is used to count out a given
number of particles, whether they
are atoms, molecules, formula units,
ions, or electrons.
Amount in Moles converted to #
of Particles
6.022 x 1023 particles = 1 mol
6.022 10
1 mol
23
1
1 mol
1
23
6.022  10
Choose a conversion factor that cancels the
given units.
Example #1
Find the number of molecules in 2.5 mol
of sulfur dioxide.
6.022 10 23
2.5 mol SO2 
 1.5 10 24 molecules SO 2
1 mol
Homework
Practice A
p. 228
#’s 2 - 4
Example #2
A sample contains 3.01 x 1023 molecules of
sulfur dioxide, SO2. Determine the amount
in moles.
1 mol SO 2
3.0110 molecules SO 2 
6.022 10 23 molecules SO 2
23
 0.500 mol SO 2
Homework
Practice B
p. 229
#’s 2 – 4
#5 c – g
Molar Mass Relates Moles to
Grams
The mass in grams of 1 mole of substance
= atomic mass of monatomic elements &
formula mass of compounds & diatomic
elements
Examples
Carbon = 12g/mol
O2 = 16+16 = 32g/mol
CH4 = 12+1+1+1+1 = 16g/mol
Example #3
Find the mass in grams of 2.44 x 1024 atoms
of carbon, whose molar mass is 12.01
g/mol.
2.44 10
24
 48.7g C
1 mol
12.01g C
atoms 

23
6.022 10
1 mol
Homework
Practice C
p. 231
#’s 2-4
Example #4
Find the number of molecules present
in 47.5 g of glycerol, C3H8O3. The
molar mass of glycerol is 92.11g/mol.
1 mol
6.022 10 23 molecules
47.5 g C3 H 8O 3 

92.11 g C3 H 8O 3
1 mol
 3.1110 23 molecules
Homework
Practice D
p. 232
#’s 2, 3
Quiz.7.1 Answer List
57.41 g
695 g
1.4 x 1024 atm
1.195 mol
0.0206 mol
The Mole &
Chemical Composition
Section 2
Relative Atomic Mass &
Chemical Formulas
For Review
Isotope – atoms of the same element
with different #’s of neutrons
(different mass #’s)
Average Atomic Mass – weighted
average of atomic masses of
elements isotopes
Calculating Average Atomic Mass
Need to know % abundance to calculate
avg. atomic mass
Native copper is a mixture of two isotopes.
Copper-63 contributes 69.17% of the atoms,
and copper-65 the remaining 30.83%.
Example #1
The mass of Cu-63 atom is 62.94 amu,
and that of a Cu-65 atom is 64.93 amu.
Using the data for the previous figure,
find the average atomic mass of Cu.
Isotope
%
Decimal Contribution
Copper-63 69.17% 0.6917
62.94 x 0.6917
Copper-65 30.83% 0.3083
64.93 x 0.3083
(62.94 amu  0.6917)  (64.93 amu  0.3083)  63.55 amu
Practice #1
1. Calculate the average atomic mass for
gallium if 60.00% of its atoms have a
mass of 68.926 amu and 40.00 % have a
mass of 70.925 amu.
2. Calculate the average atomic mass of
oxygen. Its composition is 99.76% of
atoms with a mass of 15.99 amu, 0.038%
with a mass of 17.00 amu, and 0.20%
with a mass 18.00 amu.
To Understand
Chemical formula
Which elements
2. How much of each
1.
Ionic compounds
Show simplest ratio of cations &
anions
KBr – 1:1 – 1 K+ & 1 Br-
Molar Mass of Compound
= sum of masses of all atoms in
g/mol
Ex. H2O - H → 2 x 1.00 = 2.00
- O → 1 x 16.00 = 16.00
18.00 g/mol
Examples
ZnCl2
(NH4)2SO4
Zn
Cl
N
H
S
O
1 65.39  65.39
2  35.45  70.90
136.29 g/mol
2 14  28
8 1  8
1 32  32
4 16  64
132.0 g/mol
Practice – Homework
Practice F
p. 239 - 240
#’s 1 - 4
The Mole &
Chemical Composition
Section 3
Formulas & Percent
Composition
Definitions
Percent composition – percentage by
mass of each element in a compound
Empirical formula – shows simplest
ratio
a chemical formula that shows the
composition of a compound in terms of
relative #’s & kinds of atoms
Example
– Find the empirical formula.
C – 60.0%
H – 13.4%
O – 26.6%
1. Convert
mass to
moles.
Assume you have a
100g sample.
1 mol C
60.0 g 
 5.00 mol C
12g C
1 mol H
13.4g 
 13.4 mol H
1g H
1 mol O
26.6g 
 1.66 mol O
16 g O
Example
– Find the empirical formula.
2. formulas are written using whole #’s
(to convert divide by smallest)
5.00
 3.01
1.66
13.4
 8.07
1.66
3. Write formula
C3H8O
1.66
1
1.66
Practice – In Class
1. A compound is 63.52% Fe and 36.48% S.
2. 26.58% K, 35.35% Cr, 38.07% O
3. 32.37% Na, 22.58% S, 45.05% O
Practice – Homework
Practice G
p. 243
#’s 1 - 4
Definitions
Molecular formulas – whole #
multiple of empirical formula
Ex.
Empirical
– CH2O
Molecular
x 1 = CH2O
formaldehyde
x 2 = C2H4O2 acetic acid
x 6 = C6H12O6 glucose
Example
- Find the molecular formula.
The empirical formula for a
compound is P2O5. Its experimental
molar mass is 284 g/mol. Determine
the molecular formula of the
compound.
P  2  30.97  61.94
O  5  16
 80
141.94 g/mol
284
 2.00
141.94
2(P2O5 )  P4O10
Practice – In Class
What is the molecular formula?
1. Experimental MM = 232.41 g/mol
Empirical formula = OCNCl
2. Experimental MM = 32.06 g/mol
Empirical formula = NH2
Practice - Homework
Practice H
p. 245
#’s 1 - 3
Percent Composition
Calculate the percent composition of
copper (I) sulfide, a copper ore called
chalcocite.
Cu  2  63.55  127.1
S  1 32.07  32.1
Cu 2S  159.2 g/mol
127.1
100  79.8% Cu
159.2
32.1
100  20.2% S
159.2
Check that everything adds up to 100%
Practice – In Class
Determine the % Composition
1. NaClO
2. H2SO3
3. C2H5COOH
Answers to Practice
Na
Cl
1 x 23
1 x 35.5
= 23
= 35.5
O
1 x 16
= 16
NaClO
Na
(23/74.5) x 100
= 30.9%
74.5 g/mol
Cl
O
(35.5/74.5) x 100 (16/74.5) x 100
= 48.0%
= 21.5%
Answers to Practice
H
S
2x1
1 x 32.1
=2
= 32.1
O
3 x 16
= 48
H2SO3
H
(2/82.1) x 100
= 2.5%
82.1 g/mol
S
O
(32.1/82.1) x 100 (48/82.1) x 100
= 39.1%
= 58.5%
Answers to Practice
C
H
3 x 12
6x1
= 36
=6
O
2 x 16
= 32
C2H5COOH
74.0 g/mol
C
H
O
(36/74.0) x 100
= 48.6%
(6/74.0) x 100 =
8.1%
(32/74.0) x 100
= 43.2%
Qz.7.3
Calculate the percent composition
1.
2.
3.
4.
SrBr2
CaSO4
Mg(CN)2
Pb(CH3COO)2
Practice - Homework
Practice I
p. 248
#’s 1- 4