Unit 1: Stoichiometry
Chemistry 2202
5:51 AM
1
Stoichiometry

Stoichiometry deals with quantities
used in OR produced by a chemical
reaction
5:51 AM
2
3 Parts



Mole Calculations (Chp. 2 & 3)
Stoichiometry and Chemical Equations
(Chp. 4)
Solution Stoichiometry (Chp. 6)
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3
PART 1 - Mole Calculations



Isotopes and Atomic Mass (pp. 43 - 46)
Avogadro’s number (pp. 47 – 49)
Mole Conversions (pp. 50 - 74)

M, MV, NA, n, m, v, N
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4
Questions
p.
p.
p.
p.
p.
p.
p.
45 #’s 1 – 4
46 #’s 1 – 6
75 #’s 9 – 12
51-53 #’s 5 – 15
57 #’s 16 – 19
59,60 #’s 20 – 27
63,64 #’s 28 - 37
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p.
p.
p.
p.
54
65
75
76
#’s
#’s
#’s
#’s
5–8
2, 4, 5
13, 14
15, 17–19,
21-23
p. 73 #’s 38 – 43
p. 74 #’s 1 – 4
p. 76 #’s 26, 27
5
PART 1 - Mole Calculations

Percent composition:
- given mass (p. 79 - 82)
 - given the chemical formula (p. 83 - 86)




Empirical Formulas (pp. 87 - 94)
Molecular Formulas (pp. 95 - 98)
Lab: Formula of a Hydrate
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6
Questions
p.
p.
p.
p.
p.
82
85
89
91
97
#’s
#’s
#’s
#’s
#’s
1–4
5–8
9 – 12
13 – 16
17 - 20
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p.
p.
p.
p.
p.
103 #’s 23 – 24
86 #’s 1, 3 – 6
94 #’s 1 - 7
106 #’s 1 - 3, 6, 7
107 – 109
#’s 5 – 23, 25
7
Isotopes and Atomic Mass
atomic number - the number of protons
in an atom or ion
mass number - the sum of the protons
and neutrons in an atom
isotope - atoms which have the same
number of protons and electrons but
different numbers of neutrons
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8
Isotopes and Atomic Mass
eg.
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9
Isotopes and Atomic Mass
35
17
Cl
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37
17
Cl
10
Isotopes and Atomic Mass
24
12
Mg
79 %
25
12
Mg
10 %
26
12
Mg
11 %
not all isotopes are created equal
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11
Isotopes and Atomic Mass
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12
Isotopes and Atomic Mass
atomic mass unit (AMU - p.43)
- a unit used to describe the mass of
individual atoms
- the symbol for the AMU is u
- 1 u is 1/12 of the mass of a carbon-12
atom
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13
Isotopes and Atomic Mass
average atomic mass (AAM)
the AAM is the weighted average of all
the isotopes of an element (p. 45)
p. 14
#5
p. 45
#’s 1 – 4
p. 46
#’s 1 – 6
p. 75
#’s 9 - 12
-
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14
Finding % Abundance
eg. Br has two naturally occurring
isotopes. Br-79 has a mass of 78.92 u
and Br-81 has a mass of 80.92 u. If the
AAM of Br is 79.90 u, determine the
percentage abundance of each isotope.
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15
Finding % Abundance
Let x = fraction of Br-79
Let y = fraction of Br-81
y=1-x
x+y=1
78.92x + 80.92y = 79.90
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16
x+y=1
78.92x + 80.92y = 79.90
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17
Avogadro’s Number (p. 47)
p. 48
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18
Avogadro’s Number



The MOLE is a number used by chemists
to count atoms
The MOLE is the number of atoms
contained in exactly 12 g of carbon-12.
In honor of Amedeo Avogadro, the
number of particles in 1 mol has been
called Avogadro’s number.
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19
How big is Avogadro's number?


An Avogadro's number of soft drink cans
would cover the surface of the earth to a
depth of over 200 miles.
Avogadro's number of unpopped popcorn
kernels spread across the USA, would
cover the entire country to a depth of
over 9 miles.
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20
How big is Avogadro's number?

If we were able to count atoms at the
rate of 10 million per second, it would
take about 2 billion years to count the
atoms in one mole.
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21
Avogadro’s Number
1 mole = 6.02214199 x 1023 particles
1 mol = 6.022 x 1023 particles
NA = 6.022 x 1023 particles/mol
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22
Avogadro’s Number
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23
Avogadro’s Number
Number of moles
Number of atoms
5 mol
3.011 x 1024 atoms
0.01 mol
6.022 x 1021 atoms
7.72 mol
4.65 x 1024 atoms
0.0133 mol
8.01 x 1021 atoms
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24
Avogadro’s Number
Formulas:
N
n
NA
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N = n x NA
n = # of moles
N = # of particles
(atoms, ions, molecules,
or formula units)
NA = Avogadro’s #
25
Avogadro’s Number
a)
How many moles are contained in the
following?
2.56 x 1028 Pb atoms
b)
7.19 x 1021 CO2 molecules

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26
Avogadro’s Number
eg. Calculate the number of moles in
4.98 x 1025 atoms of Al.
eg. How many formula units of Na2SO4
are in 5.69 mol of Na2SO4?
# of Na ions?
# of Oxygen atoms?
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27
Avogadro’s Number
1. How many molecules of glucose are in
0.435 mol of C6H12O6?
How many carbon atoms?
2. Calculate the number of moles in a
sample of glucose that has 3.56 x 1022
hydrogen atoms.
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28
Avogadro’s Number
pp. 51 – 53: #’s 5 – 15
p. 54: #’s 4 - 8
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29
Molar Mass
The mass of one mole of a substance is
called the molar mass of the substance
eg.
1 mole of Pb has a mass of 207.19 g
1 mole of Ag has a mass of 107.87 g

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30
Molar Mass

The symbol for molar mass is M and the
unit is g/mol
MPb = 207.19 g/mol
MAg = 107.87 g/mol
eg.
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31
Molar Mass
The molar mass of a compound is the
sum of the molar masses of the
elements in the compound
eg. Calculate the molar mass of:
a) H2O
b) C6H12O6 c) Ca(OH)2

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32
Molar Mass
H2O has 2 hydrogens and 1 oxygen
2 x 1.01 = 2.02
1 X 16.00 = 16.00
18.02 g/mol
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33
Molar Mass
C6H12O6
6 x 12.01 = 72.06
12 x 1.01 = 12.12
6 x 16.00 = 96.00
180.18 g/mol
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34
Molar Mass
Your calculator
Ca(OH)2
may not show
the zeroes.
1 x 40.08 = 40.08
There should be 2
digits after the
2 x 16.00 = 32.00
decimal
when
2 x 1.01 = 2.02
adding molar
74.10 g/mol
masses
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35
Molar Mass
p. 57: #’s 16 – 19 & Molar Masses Handout
1.
2.
3.
4.
5.
6.
151.92 g/mol
120.38 g/mol
105.99 g/mol
100.40 g/mol
74.44 g/mol
78.01
g/mol
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7.
8.
9.
10.
11.
58.44 g/mol
100.09 g/mol
44.02 g/mol
248.22 g/mol
115.04 g/mol
36
Molar Mass Calculations
Avogadro’s #
N
n
NA
N = n x NA
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m
n
M
mass
molar
mass
m=nxM
37
Molar Mass Calculations
N
n
NA
m
n
M
N = n x NA
m=nxM
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38
Molar Mass Calculations
eg. How many moles
are in 25.3 g of NO2?
m = 25.3 g
MNO2 = 46.01 g/mol
m
n
M
25.3 g

46.01g / mol
 0.550 mol
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39
Molar Mass Calculations
eg. What is the mass of 4.69 mol of water?
n = 4.69 mol
Mwater = 18.02 g/mol
m=nxM
= (4.69 mol)(18.02 g/mol)
= 84.5 g
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40
Molar Mass Calculations
Practice:
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p. 59 #’s 20 - 23
p. 60 #’s 24 - 27
41
The Mole #4
answers
1.a) 0.038 mol 2.a) 17.4 g
b) 3.75 mol
b) 1560. g
c) 0.276 mol
c) 528 g
d) 23 mol
d) 97513 g
e) 0.0575 mol e) 9328 g
4.a) 1.82 mol
b) 13 mol
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3.a) 5631 g
b) 3.73 g
c) 10.02 g
d) 1662.9 g
e) 981.2 g
c) 0.02133 mol e) 0.0000573 mol
d) 4.3423 mol
42
Particle–Mole-Mass Conversions
eg. How many molecules are in 26.9 g of
water?
m = 26.9 g
Mwater= 18.02 g/mol
NA = 6.022 x 1023 molecules/mol
Find N
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43
Particle–Mole-Mass Conversions
m
n
M
26.9

18.02
N = n x NA
= 1.493 X 6.022 x 1023
= 8.99 x 1023 molecules
= 1.493 mol H2O
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44
Particle–Mole-Mass Conversions
eg. A sample of Sn contains 4.69 x 1028
atoms. Calculate its mass.
N = 4.69 x 1028
NA = 6.022 x 1023 molecules/mol
MSn = 118.69 g/mol
Find m
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45
Particle–Mole-Mass Conversions
m=nxM
N
n
NA
= 77881 mol x 118.69 g/mol
28
4.69 x 10

23
6.022 x 10
= 9.24 x 10 6 g
= 77,881 mol
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46
Particle–Mole-Mass Conversions
1. Calculate the mass of 4.80 x 1024 water
molecules.
2. How many grams are in 2.53 x 1026 CH4
molecules?
3. How many atoms are in 68.0 g of Sn?
4. Calculate the number of molecules
in 105 g of C6H12O6.
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47
Particle–Mole-Mass Conversions
Particles (N)
Moles (n)
Mass (m)
4.80 x 1024 H2O
molecules
2.53 x 1026 CH4
molecules
68.0 g of Sn
105 g of C6H12O6
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48
Particle–Mole-Mass Conversions
particles
(N)
Moles
(n)
Mass
(m)
5.98 x 1026 Cu atoms
4.50 g H2O
6.15 mol O3
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49
Particle–Mole-Mass Conversions
particles
(N)
Moles
(n)
Mass
(m)
4.18 x 1022 Al atoms
2.45 g C8H18
0.145 mol S2F4
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50
Particle–Mole-Mass Conversions
particles
(N)
Practice:
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x NA
÷ NA
moles
(n)
xM
÷M
mass
(m)
p. 63 #’s 28 - 33
p. 64 #’s 34 – 37
p. 76 # 15
51
Particle–Mole-Mass Conversions
eg. How many molecules are in 4.78 g of
glucose?
m = 4.78 g
Mwater= 180.18 g/mol
NA = 6.022 x 1023 molecules/mol
Find N
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52
Particle–Mole-Mass Conversions
m
n
M
4.78

180.18
N = n x NA
= 0.02653 X 6.022 x 1023
= 8.99 x 1023 molecules
= 0.02653 mol glucose
5:51 AM
53
Particle–Mole-Mass Conversions
Practice:
5:51 AM
p. 63 #’s 28 - 33
p. 64 #’s 34 – 37
p. 76 # 15
54
Molar Mass Calculations
Practice:
p. 54 #’s
p. 65 #’s
p. 75 #’s
p. 76 #’s
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5-8
2, 4, 5
13, 14,
15, 17 – 19, 21 -23
55
Molar Volume
•The volume of a gas increases when
temperature increases but decreases
when pressure increases .
•The volume of gases is measured
under conditions of Standard
Temperature and Pressure (STP)
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56
Molar Volume
Standard Pressure – 101.3 kPa
 Standard Temperature – 0 °C
 Avogadro hypothesized that equal
volumes of gases at the same
temperature and pressure contain
equal numbers of molecules.

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57
Molar Volume

Experimental evidence shows the
volume of one mole of ANY GAS at
STP is 22.4 L/mol
OR
VSTP = 22.4 L/mol
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58
given volume
in Litres
Molar Mass Calculations
N
n
NA
N = n x NA
m
n
M
v
n
VSTP
v = n x VSTP
m=nxM
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59
Molar Mass Calculations
particles
(N)
x NA
moles
(n)
÷ NA
x VSTP
xM
÷M
mass
(m)
÷ VSTP
volume
(v)
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60
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61
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62
Molar Volume
p. 73 #’s 38 – 43
p. 74 #’s 1 – 4
p. 76 #’s 26, 27
WorkSheet: The Mole #6
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63
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64
Percent Composition (p. 79)

The mass percent of a compound is the
mass of an element in a compound
expressed as a percent of the total mass
of the compound.
mass of element
mass percent 
X 100%
total mass of compound
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65
Percent Composition
eg. 8.50 g of a compound was analyzed
and found to contain 6.00 g of hydrogen
and 2.50 g of carbon. Calculate the mass
percent for each element.
p. 82 #’s 1 - 4
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66
Percent Composition

mass percent may be found using the
formula & the molar mass of a compound.
eg.
Find the percentage composition for CH4
5:51 AM
67
p. 85 #’s 5 - 8
Percent Composition
M = 12.01 g/mol + 4(1.01 g/mol)
= 12.01 g/mol + 4.04 g/mol
= 16.05 g/mol
12.01
%C 
x 100
16.05
 74.8 %
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4.04
%H 
x 100
16.05
 25.2 %
68
5:51 AM
69
p. 85 #’s 5 - 8
p. 86 #’s 1, 3 – 6
p. 107 #’s 6 – 10
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70
Empirical Formulas
An empirical formula gives the simplest
ratio of elements in a compound.
A molecular formula shows the actual
number of atoms in a molecule of a
compound.
Ionic compounds are always written as
empirical formulas
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71
Empirical Formulas
Compound
butane
Molecular
Formula
C4H10
glucose
C6H12O6
water
H2O
H2O
benzene
C6H6
CH
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Empirical
Formula
C2H5
CH2O
72
Empirical Formulas (p.87)
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73
Empirical Formulas
The empirical formula of a compound
may be determined by using the %
composition of a given compound.
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74
Empirical Formulas



Method:
assume you have 100.0 g of the
compound (ie. change % to g)
calculate the moles (n) for each element
divide each n by the smallest n to get the
ratio for the empirical formula
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75
Empirical Formulas
eg. A compound was analyzed and found
to contain 87.4% N and 12.6 % H by
mass. Determine the empirical formula
of the compound.
p. 89 #’s 9 - 12
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76
Empirical Formulas
When finding the EF, the mole ratio may
not be a whole number ratio.
eg. A compound contains 84.73% N and
15.27 % H by mass. Determine the
empirical formula of the compound.

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77
p. 90
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78
Empirical Formulas
eg. A compound contains 89.91% C and
10.08 % H by mass. Determine the
empirical formula of the compound.
p. 91 #’s 13 – 16
p. 94 #’s 2-4, 6, 7
Answers on p. 109
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79
MgO Lab
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80
Molecular Formulas

The molecular formula of a compound is
a multiple of the empirical formula.
See p. 95
5:51 AM
81
Molecular Formulas
To find the molecular formula we need
the empirical formula and the molar
mass of the compound
eg. The empirical formula of hydrazine is
NH2. The molar mass of hydrazine is
32.06 g/mol. What is the molecular
formula for hydrazine?

5:51 AM
82
Molecular Formulas
p. 97 #’s 17 – 20
p. 107, 108 #’s 11 - 14
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83
CHC analyzer (p. 99 – 101)
1. Describe the operation of a carbon
hydrogen combustion analyzer.
2. 22.0 g of carbon dioxide and 10.8 g of
water is collected in a CHC analysis.
Determine the empirical formula of the
hydrocarbon.
p. 101 #’s 21, 22
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84
Formula of a hydrate
To determine the formula of a hydrate:
- calculate the moles of water
- calculate the moles of anhydrous
compound
- determine the simplest ratio
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85
Formula of a hydrate
eg. Use the data below to determine the
value of x in LiCl• xH2O.
mass of crucible = 26.35 g
crucible + hydrate = 42.15 g
crucible + anhydrous compound= 34.94 g
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86
mwater = 42.15 – 34.94 = 7.21 g H2O
mLiCl = 34.94 – 26.35 = 8.59 g LiCl
5:51 AM
87
eg. Na2CO3. xH2O
crucible = 15.96 g
crucible + hydrate = 22.19 g
crucible + anhydrous compound = 19.67 g
5:51 AM
88
eg. CoCl2.xH2O
crucible = 151.96 g
crucible + hydrate = 164.35 g
crucible + anhydrous compound = 158.23 g
p. 103; # 24
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Lab: pp. 104-105
89
Formula of a hydrate
mass of empty beaker
mass of beaker & hydrate
mass of beaker & anhydrous
compound
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90
Review – Chp. 3
p. 86 #’s 1, 3 – 6
p. 94 #’s 1 – 7
p. 103 # 23
pp. 107–109 #’s 5 – 19, 21, 22
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91
Test
p.
p.
p.
p.
p.
p.
p.
45 #’s 1 – 4
46 #’s 1 – 6
75 #’s 9 – 12
51-53 #’s 5 – 15
57 #’s 16 – 19
59,60 #’s 20 – 27
63,64 #’s 28 - 37
5:51 AM
p.
p.
p.
p.
54
65
75
76
#’s
#’s
#’s
#’s
5–8
2, 4, 5
13, 14
15, 17–19,
21-23
p. 73 #’s 38 – 43
p. 74 #’s 1 – 4
p. 76 #’s 26, 27
92
Test
p.
p.
p.
p.
p.
82
85
89
91
97
#’s
#’s
#’s
#’s
#’s
1–4
5–8
9 – 12
13 – 16
17 - 20
5:51 AM
p.
p.
p.
p.
p.
103 #’s 23 – 24
86 #’s 1, 3 – 6
94 #’s 1 - 7
106 #’s 1 - 3, 6, 7
107 – 109
#’s 5 – 23, 25
93
Test Friday – Dec. 4
1.
Formula of a hydrate 3.
- lab activity
- p. 103 # 24
2.
% composition
- given mass
p. 82 #’s 1 – 4
- given formula
p. 85 #’s 5 - 8
p. 86 #’s 3 & 4
p. 103 # 23
p. 107 # 5
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4.
EF and MF
- What is an empirical formula
- Finding EF from % composition
pp. 89, 91 #’s 9 - 16
(watch for .5 or .333)
- Finding MF from EF and molar
mass
p. 97 #’s 17 – 20
p. 108 #13
Mole calculations – Chp. 2
94
Stoichiometry (Chp.4)
Stoichiometry is the determination of
quantities needed for, or produced by,
chemical reactions.
 Ratios from balanced chemical equations
are used to predict quantities.

5:51 AM
95
Stoichiometry – p. 111
Clubhouse sandwich recipe
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96
Clubhouse sandwich recipe
Fill in the missing quantities:
Slices of
Toast
Slices of
Turkey
Strips of
Bacon
# of
Sandwiches
12
27
66
100
97
Mole Ratios

A mole ratio is a mathematical expression
that shows the relative amounts of two
species involved in a chemical change.
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98
Mole Ratios

A mole ratio
Comes from a balanced chemical equation
 Shows the relative amounts of the
reactants/products in moles
 Looks like
coefficient required

coefficient given
5:51 AM
99
N2(g) + 3 H2 (g) → 2 NH3 (g)
20
66
140
81
5:51 AM
100
C3H8 + 5 O2 → 3 CO2 + 4 H2O
n required  n given x
coefficient required
coefficient given
How many moles of CO2 are produced
when 31.5 mol of O2 react?
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101
C3H8 + 5 O2 → 3 CO2 + 4 H2O
How many moles of H2O are produced
when 1.35 mol of O2 react?
5:51 AM
102
C3H8 + 5 O2 → 3 CO2 + 4 H2O
How many moles of C3H8 are needed to
react with 0.369 mol of O2?
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103
C5H12
+
O2 →
CO2 +
H2O
How many moles of CO2 are produced
when 6.35 mol of O2 react?
5:51 AM
104
Al(s) + Br2(l) → AlBr3(s)
How many moles of Br2 are needed to
produce 0.315 mol of AlBr3?
p. 115 #’s 4 – 7
p. 117 #’s 8 - 10
5:51 AM
105
Mole to Mole Stoichiometry
1.
How many moles of copper would be
produced if 20.5 mol of copper (II) oxide
decomposes?
5:51 AM
106
Mass to Mole Stoichiometry
1.
How many moles of water are produced
when 20.6 g of CH4 burns?
5:51 AM
107
Mass to Mole Stoichiometry
2.
How many moles of nitrogen gas are needed
to produce 6.75 g of NH3 in a reaction with
hydrogen gas?
5:51 AM
108
Mass to Mole Stoichiometry
3.
How many moles of silver would be produced
if 10.0 g of silver nitrate reacts with copper
metal?
5:51 AM
109
Mass to Mole Stoichiometry
4.
How many moles of water are produced
when 20.6 g of C6H12 burns?
5:51 AM
110
Mass to Mole Stoichiometry
5.
How many moles of water are produced
when 20.6 g of CH4 burns?
5:51 AM
111
Mass to Mole Stoichiometry
5:51 AM
112
Mole to Mass Stoichiometry
1.
What mass of CaCl2 is produced when
4.38 mol of Ca(NO3)2 reacts with NaCl?
5:51 AM
113
Mole to Mass Stoichiometry
2.
What mass of copper would be produced if
20.5 mol of CuO decomposes?
5:51 AM
114
Four step stoichiometry
1.
2.
3.
4.
Write a balanced chemical equation
Calculate moles given
m
v
N
n
n
n
M
VSTP
NA
Mole ratio – find moles required
Calculate required quantity
m=nxM
v = n x Vstp
N = n x NA
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115
Mass to Mass Stoichiometry
1.
How many grams of water are produced
when 5.45 g of C3H8 burns?
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116
5:51 AM
117
Mass to Mass Stoichiometry
eg. Calculate the mass of HCl needed to
react with 3.56 g of Fe to produce FeCl2.
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118
Stoichiometry (Chp.4)
eg. What mass of CO2 gas is produced
when 45.9 g of CH4 burns ?
Step #1
CH4 + 2 O2 → CO2 + 2 H2O
45.9 g
?g
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119
5:51 AM
120
eg. How many moles of HCl needed to
react with 3.56 g of Fe to produce FeCl2.
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121
Mole Calculations (p. 121 #13)
Fe
+ 2 HCl
3.56 g
Step #2
Step #3
nFe
nHCl
5:51 AM
→ FeCl2 + H2
?g
3.56 g
= 0.06374 mol Fe

55.85g/mol
2 mol HCl
 0.06374 mol Fe x
1 mol Fe
= 0.12748 mol HCl
122
Mole Calculations
p. 122 #15
Given 32.0 g of sulfur (M = 256.56 g/mol)
Find mass of ZnS
#2
n = 0.1247 mol S8
#3
n = 0.9976 mol ZnS
(M = 97.45 g/mol)
m = 97.2 g ZnS
#4
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123
Mole Calculations
p. 123 #18
Given 33.5 g of H3PO4 (M = 98.00 g/mol)
Find mass of MgO
#2
n = 0.3418 mol H3PO4
#3
n = 0.5128 mol MgO
(M = 40.31 g/mol)
m = 20.7 g MgO
#4
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124
Mole Calculations
p. 123 #17
Given 25.0 g of Al4C3 (M = 143.95 g/mol)
Find volume of CH4
#2
n = 0.174 mol Al4C3
#3
n = 0.522 mol CH4
(MV = 22.4 L/mol)
m = 11.7 L CH4
#4
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125
How many moles of aluminum chloride can be
produced from the reaction of chlorine and
10.8 mol of aluminum ?
Cl2(g) + Al(s) → AlCl3(s)
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126
How many moles of magnesium are needed to
react with 27 g of iodine to form magnesium
iodide?
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127
How many grams of nitrogen are needed to
react with 14.0 mol of oxygen to produce
nitrogen dioxide ?
N2(g) + O2(g) → NO2(g)
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128
Mole Calculations (p. 121 #14)
2.34 g
?L
#2 n = 0.05086 mol NO2
#3 n = 0.01272 mol O2
(MV = 22.4 L/mol)
#4 n = (0.01272)(22.4) = 0.285 L O2
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129
Limiting Reactant (p. 128)
10.0 g of Li reacts with 15.0 g of Br2.
Calculate the mass of LiBr produced.
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130
Limiting Reactant (p. 128)
2 Li
Br2
2 LiBr
10.0
15.0
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131
Limiting Reactant (p. 128)


The Limiting Reactant (LR) OR Limiting
Reagent (LR) is the substance that is
completely used in a chemical reaction.
The Excess Reactant is the reactant
that is left over after a reaction is
complete.
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132
Limiting Reactant (p. 128)
eg. 2.00 g of NaI reacts with 2.00 g of
Pb(NO3)2. Determine the LR and
calculate the amount of PbI2 produced.
 write a balanced equation
 find n for each reactant (Step #2)
 find moles produced by each reactant
(Step #3)
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133
Pb(NO3)2 + 2 NaI → 2 NaNO3 + PbI2
2.00 g
2.00 g
nPb(NO)3 = 2.00 g
331.21 g/mol
= 0.006038 mol
nNaI = 2.00 g
149.89 g/mol
= 0.013343 mol
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134
nPbI2 = 0.006038 mol Pb(NO3)2 x 1 mol PbI2
1 mol Pb(NO3)2
= 0.006038 mol PbI2
nPbI2 = 0.013343 mol NaI x 1 mol PbI2
2 mol NaI
= 0.006672 mol PbI2
mPbI2 = 0.006038 mol x 460.99 g/mol
= 2.78 g PbI2
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135
What mass of calcium carbonate will be
produced when 20.0 g of calcium phosphate
reacts with 15.0 g of sodium carbonate?
(14.2 g)
3 Na2CO3 + Ca3(PO4)2 → 3 CaCO3 + 2 Na3PO4
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136
What mass of barium hydroxide will be
produced when 10.0 g of barium nitrate reacts
with 30.0 g of sodium hydroxide? (6.56 g)
Ba(NO3)2 + NaOH → Ba(OH)2 + NaNO3
nBa(NO 3 )2
10.00 g

261.35 g/mol
= 0.03826 mol Ba(NO3)2
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nNaOH
30.00 g

40.00 g/mol
= 0.750 mol NaOH
137
Using Ba(NO3)2
nBa(OH) 2
1 mol Ba(OH) 2
 0.03826 mol Ba(NO 3 )2 x
1 mol Ba(NO 3 )2
= 0.03826 mol Ba(OH)2
Using NaOH
nBa(OH) 2
1 mol Ba(OH) 2
 0.750 mol NaOH x
2 mol NaOH
= 0.375 mol Ba(OH)2
mBa(OH) 2  0.03826 mol x 171.35 g/mol
 6.56 g
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138
What volume of hydrogen gas at STP will be
produced when 10.0 g of zinc metal reacts
with 20.0 g of hydrogen chloride?
Zn + 2 HCl → H2 + ZnCl2
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139
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140
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141
Law of Conservation of Mass (p. 118)
In a chemical reaction, the total mass
of reactants always equals the total
mass of products.
eg. 2 Na3N → 6 Na + N2
When 500.00 g of Na3N decomposes
323.20 g of N2 is produced. How much
Na is produced in this decomposition?
142
Law of Conservation of Mass
( p. 118)
eg. To produce 90.1 g of water, what mass
of hydrogen gas is needed to react with
80.0 g of oxygen?
eg. If 3.55 g of chlorine reacts with exactly
2.29 g of sodium, what mass of NaCl will
be produced?
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143
Percent yield (p. 137)


The theoretical yield is the amount of
product that we calculate using
stoichiometry
The actual yield is the amount of
product obtained from a chemical
reaction
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144
Percent yield (p. 137)
percent yield 
actual yield
x 100
theoretical yield
p. 139 #’s 31, 32, & 33
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145
DEMO: silver nitrate + copper
Equation:
Mass AgNO3 =
Mass Cu =
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146
DEMO: silver nitrate + copper
Mass of filter paper and precipitate =
Mass of empty filter paper =
Mass of precipitate =
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147