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Chem 106, Prof. J.T. Spencer
CHE 106: General Chemistry
CHAPTER
THREE
Copyright © James T. Spencer 1995 - 1999
All Rights Reserved
1
Chem 106, Prof. J.T. Spencer
Stoichiometry
2
Chapter Three
Chem 106, Prof. J.T. Spencer
Stoichiometry
3
• Antoine Lavoisier (1734 - 1794)
– Law of Conservation of Mass - atoms are neither
created nor destroyed in chemical reactions
– total number of atoms = total number of atoms after reaction
before reaction
– Stoichiometry - quantitative study of chemical
formulas and reactions
(Greek; “stoichion”= element, “metron” = measure)
• Chemical Equations - used to describe chemical
reactions in an accurate and convenient fashion
2H2 + O2
reactants
2 H2O
products
Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
To Write and Balance: (Shorthand
Communication for a great deal of
information)
(1) Know Reactants
(2) Know ALL Products
(3) Balance - Same Number and Kinds of
atoms on each side
4
Chem 106, Prof. J.T. Spencer
Chemical Equations
• Chemical Equations
– Must have equal numbers of atoms of
each element on each side of the
equation = BALANCED EQUATION
2 H2 + O2
4 hydrogen
2 oxygen
N2O5(g) + H2O
2 nitrogen
6 oxygen
2 hydrogen
2 H2O
4 hydrogen
2 oxygen
2 HNO3
2 nitrogen
6 oxygen
2 hydrogen
5
NOTE
The coefficients
in front of the
formula for a
compound refers
to the number of
molecules (intact)
involved while a
subscript refers
to the ratio of
atoms within the
molecule
Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
• Chemical Equations
– balancing equations often requires
some trial and error of coefficients
PCl3(l) + 3 H2O(l)
6
3
1
3
hydrogen
oxygen
phosphorus
chloride
C6H12(l) + 9 O2(g)
6 carbon
18 oxygen
12 hydrogen
H3PO3(aq) + 3 HCl
6 hydrogen
3 oxygen
1 phosphorus
3 chlorine
6 CO2(g) + 6 H2O(l)
6 carbon
18 oxygen
12 hydrogen
6
NOTE
Never change
subscripts in
formulas when
balancing
chemical
reactions!
subscripts
change
compounds;
coefficients
change amounts
Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
7
Sample exercise: Balance the following
equations by providing the missing
coefficients:
C2H4 + O2
CO2 + H2O
Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
8
Sample exercise: Balance the following
equations by providing the missing
coefficients:
C2H4 + O2
CO2 + H2O
C
C
H
H
O
O
Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
9
Sample exercise: Balance the following
equations by providing the missing
coefficients:
C2H4 + O2
CO2 + H2O
C 2
C 1
H 4
H 2
O 2
O 3
Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
10
Sample exercise: Balance the following
equations by providing the missing
coefficients:
C2H4 + O2
2 CO2 + H2O
C 2
C (1)2= 2
H 4
H 2
O 2
O 3 5
Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
11
Sample exercise: Balance the following
equations by providing the missing
coefficients:
C2H4 + O2
2 CO2 + 2H2O
C 2
C (1)2= 2
H 4
H (2)2 = 4
O 2
O 3 5 6
Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
12
Sample exercise: Balance the following
equations by providing the missing
coefficients:
C2H4 + 3O2
2 CO2 + 2H2O
C 2
C (1)2= 2
H 4
H (2)2 = 4
O (2)3 = 6 O 3 5 6
Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
13
Sample exercise: Balance the following
equations by providing the missing
coefficients:
Al + HCl
AlCl3 + H2
Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
14
Sample exercise: Balance the following
equations by providing the missing
coefficients:
Al + HCl
AlCl3 + H2
Al
Al
H
H
Cl
Cl
Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
15
Sample exercise: Balance the following
equations by providing the missing
coefficients:
Al + HCl
AlCl3 + H2
Al 1
Al 1
H 1
H 2
Cl 1
Cl 3
Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
16
Sample exercise: Balance the following
equations by providing the missing
coefficients:
Al + 3HCl
AlCl3 + H2
Al 1
Al 1
H (1)3 = 3
H 2
Cl (1)3 = 3
Cl 3
Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
17
Sample exercise: Balance the following
equations by providing the missing
coefficients:
Al + 6HCl
AlCl3 + 3H2
Al 1
Al 1
H (1)6 = 6
H (2)3 = 6
Cl (1)6 = 6
Cl 3
Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
18
Sample exercise: Balance the following
equations by providing the missing
coefficients:
Al + 6HCl
2AlCl3 + 3H2
Al 1
Al (1)2 = 2
H (1)6 = 6
H (2)3 = 6
Cl (1)6 = 6
Cl (3)2 = 6
Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
19
Sample exercise: Balance the following
equations by providing the missing
coefficients:
2Al + 6HCl
2AlCl3 + 3H2
Al (1)2 = 2
Al (1)2 = 2
H (1)6 = 6
H (2)3 = 6
Cl (1)6 = 6
Cl (3)2 = 6
Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
20
• Chemical Reactions
–The course of a chemical reaction can often
be predicted by recognizing general
patterns of reactivity through similar
reactions previously observed. Elements in
same family (column of table) have similar
reactions.
–The periodic table is helpful in predicting
products of reactions. Atoms like to
assume electron configurations of the
Noble Gases.
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
21
• Chemical Reactions
Example, if you know that
2Li + 2H20
2LiOH + H2
then you should be able to predict the products
from the reaction of Na, K and the other members
of group 1 (alkali metals) with water. Thus a
general reaction would be;
2 M + 2 H2O
2 MOH + H2
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
22
•Combustion Reactions
•Combination Reactions
•Decomposition Reactions
•Metathesis Reactions
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
23
• Combustion Reactions
– Reactions with oxygen (usually from the air)
– The complete combustion of hydrocarbons yield
carbon dioxide (CO2) and water (H2O)
CxHy + (2x+y) /2 O2
X CO2 + Y/2 H2O
Generally:
Balance Carbon Atoms First
Balance Hydrogens
Balance Oxygen Atoms
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
24
• Combustion Reactions
– Reactions with oxygen (usually from the air)
– The complete combustion of hydrocarbons yield
carbon dioxide (CO2) and water (H2O)
CxHy + (2x+y) /2 O2
X CO2 + Y/2 H2O
Examples
octane: C8H18 + 25/2 O2
ethanol: C2H5OH + 3 O2
glucose: C6H12O6 + 9 O2
styrene: C8H8 + 10 O2
8 CO2 + 9 H2O
2 CO2 + 3 H2O
6 CO2 + 6 H2O
8 CO2 + 4 H2O
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
25
Sample exercise: Write the balanced
equation for the reaction that occurs when
ethanol, C2H5OH(l) is burned in air.
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
26
Sample exercise: Write the balanced
equation for the reaction that occurs when
ethanol, C2H5OH(l) is burned in air.
C2H5OH + O2
CO2 + H2O
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
27
Sample exercise: Write the balanced
equation for the reaction that occurs when
ethanol, C2H5OH(l) is burned in air.
C2H5OH + O2
C
H
O
CO2 + H2O
C
H
O
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
28
Sample exercise: Write the balanced
equation for the reaction that occurs when
ethanol, C2H5OH(l) is burned in air.
C2H5OH + O2
C 2
H 6
O 3
CO2 + H2O
C 1
H 2
O 3
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
29
Sample exercise: Write the balanced
equation for the reaction that occurs when
ethanol, C2H5OH(l) is burned in air.
C2H5OH + O2
C 2
H 6
O 3
2CO2 + H2O
C (1)2 = 2
H 2
O 3 5
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
30
Sample exercise: Write the balanced
equation for the reaction that occurs when
ethanol, C2H5OH(l) is burned in air.
C2H5OH + O2
C 2
H 6
O 3
2CO2 + 3H2O
C (1)2 = 2
H (2)3 = 6
O 3 5 7
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
31
Sample exercise: Write the balanced
equation for the reaction that occurs when
ethanol, C2H5OH(l) is burned in air.
C2H5OH + 3O2
C 2
H 6
O 3 7
2CO2 + 3H2O
C (1)2 = 2
H (2)3 = 6
O 3 5 7
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
32
• Combination Reactions
– two or more substances react to form a single
product
– especially common in the reactions of pure
elements
A + B
Ni(s) + 4 CO(g)
BF3(g) + NH3(g)
C
Ni(CO)4(g)
BF3NH3(s)
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
33
• Decomposition Reactions
–when one compound reacts to form two
or more products (opposite of
combination reactions)
C
• (often heat required)
2 NaN3(s)
B(OH)3
A + B
Air Bag Inflator (J. Chem. Ed. 1990, 67, 61)
(heat)
2 Na(s) + 3 N2(g)
HBO2 + H2O
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
34
• Metathesis Reactions
–when ionic “partners” switch
AB + CD
AD + BC
• (often in aqueous solutions)
Ag(NO3) + KCl
BaCl2 + Na2SO4
AgCl(s) + KNO3
BaSO4(s) + 2 NaCl
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
35
Sample Exercise: Write balanced chemical
equations for the following reactions:
Solid mercury (II) sulfide decomposes
into its component elements when heated.
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
36
Sample Exercise: Write balanced chemical
equations for the following reactions:
Solid mercury (II) sulfide decomposes
into its component elements when heated.
Hg+2 S-2
HgS  Hg + S
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
37
Sample Exercise: Write balanced chemical
equations for the following reactions:
The surface of aluminum metal
undergoes a combination reaction with
oxygen in air.
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
38
Sample Exercise: Write balanced chemical
equations for the following reactions:
The surface of aluminum metal
undergoes a combination reaction with
oxygen in air.
Al+3 O-2
Al + O2  Al2O3
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
39
Sample Exercise: Write balanced chemical
equations for the following reactions:
The surface of aluminum metal
undergoes a combination reaction with
oxygen in air.
Al+3 O-2
4Al + 3O2  2Al2O3
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
40
Sample Exercise: Write balanced chemical
equations for the following reactions:
Si2H6 burns when exposed to air.
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
41
Sample Exercise: Write balanced chemical
equations for the following reactions:
Si2H6 burns when exposed to air. Hint, Si
is in the same group as C, and therefore reacts similarly.
Si2H6 + O2  SiO2 + H2O
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
42
Sample Exercise: Write balanced chemical
equations for the following reactions:
Si2H6 burns when exposed to air. Hint, Si
is in the same group as C, and therefore reacts similarly.
2Si2H6 + 7O2  4SiO2 + 6H2O
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Atomic and Molecular Weights43
•Chemical equations indicate exactly the
amounts of two reagents which will react to
form an exact amount of products
•Atomic Mass Scale - based upon 12C isotope.
This isotope is assigned a mass of exactly 12
atomic mass units (amu) and the masses of all
other atoms are given relative to this
standard.
•Most elements in nature exist as mixtures of
isotopes.
Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Atomic and Molecular Weights44
• Atomic Mass Scale - given the following;
100 g of water contains 11.1 g of H and 88.9 g of O and
the formula for water is H2O then;
•water has 8 times more O than H by mass
(88.9/11 = 8)
•if water has 2 H for 1 O then O atoms must weigh
16 time more than H atoms
•if H is assigned an atomic mass of 1 amu then O
must weigh 16 amu (using the 12C standard)
•1 amu = 1.66054 x 10-24 g
OR 1 g = 6.02214 x 1023 amu
Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Mass Spectrometer
45
• Direct methods of measuring (separating) mass.
• Sample molecules are ionized by e-beam to cations (+1 by
“knocking off” one electron) which are then deflected by
magnetic field - for ions of the same charge the angle of
deflection in proportional to the ion’s mass
vacuum chamber beam of pos. ions
accelerating grid (-)
N
sample
focusing slits
ionizing e- beam
Hg
S
magnetic field
Mass
Spectrum
Int.
200
mass number (amu)
detector
Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Mass Spectrometer
Mass
Spectrum
Cl
Int.
35
Mass
Spectrum
Int.
C
Mass
Spectrum
P
12
37
mass number (amu)
35Cl:
75% abundant
37Cl: 25% abundant
46
Int.
31
13
mass number (amu)
12C:
98.9% abundant
13C: 1.1% abundant
mass number (amu)
31P:
100% abundant
Chem 106, Prof. J.T. Spencer
Mass Spectrometer
47
Unknown white powdery substance injested by
unconscious patient.
What do you do? Is it Heroin, Cocaine, Caffeine?
Intensity
Mass Spectrum of Unknown Compound
Mass 25
50
75
100
125 150
175
200
225
250
275
300
Chem 106, Prof. J.T. Spencer
Mass Spectrometer
Intensity
MS Library
43
Intensity
Mass 25
50
75
67
55
42
50
other peaks at
327 and 369
268
204
215
94
Mass 25
Heroin
Heroin
48
146
100 125 150 175 200 225 250 275
194
Caffeine
300
109
82
75
MS of Unknown
100 125 150 175 200 225 250 275
300
Chem 106, Prof. J.T. Spencer
Intensity
Mass Spectrometer
MS
Library
82
182
Cocaine
Cocaine
303
42
122
25
Intensity
Mass
49
Mass 25
50
75
67
55
42
50
150
272
100 125 150 175 200 225 250 275
194
Caffeine
300
109
82
75
MS of Unknown
100 125 150 175 200 225 250 275
300
Chem 106, Prof. J.T. Spencer
Mass Spectrometer
Intensity
MS Library
Intensity
Mass 25
Mass 25
67
55
109
75
67
55
42
50
Caffeine
Caffeine
82
42
50
194
50
100 125 150 175 200 225 250 275
194
Caffeine
300
109
82
75
MS of Unknown
100 125 150 175 200 225 250 275
300
Chem 106, Prof. J.T. Spencer
Mass Spectrometer
Unknown white powdery
substance ingested by
unconscious patient.
What do you do?
Mass Spectrum
O
N
N
Mol. Wgt
= 194
N
N
CH3
Caffeine
Intensity
Mass 25
CH3
H3C
O
51
50
75
100
125 150
175
200
225
250
275
300
Chem 106, Prof. J.T. Spencer
Atomic Weights
52
• Average Atomic Mass (AW)- weighted average (by
% natural abundance) of the isotopes of an element.
•Example (1);
10B is 19.78% abundant with a mass of 10.013 amu
11B is 80.22% abundant with a mass of 11.009 amu
therefore the average atomic mass of boron is;
(0.1987)(10.013) + (0.8022)(11.009) = 10.82 amu
Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Atomic Weights
53
• Average Atomic Mass (AW)- weighted average (by
% natural abundance) of the isotopes of an element.
•Example (2):
194Pt
is 33.90% abundant with a mass of 193.963 amu
195Pt is 33.80% abundant with a mass of 194.965 amu
196Pt is 25.30% abundant with a mass of 195.965 amu
198Pt is 7.210% abundant with a mass of 197.968 amu
therefore the average atomic mass of platinum is;
(0.3390)(193.963) + (0.3380)(194.965) + (0.2530)(195.965 ) +
(0.07210)(197.968)= 195.504 amu
Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Atomic and Molecular Weights54
Sample exercise: Three isotopes of silicon
occur in nature: 28Si (92.21%), which has a
mass of 27.97693 amu; 29Si (4.70%), which has
a mass of 28.97659 amu; and 30Si (3.09%),
which has a mass of 29.97376 amu. Calculate
the atomic weight of silicon.
Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Atomic and Molecular Weights55
Sample exercise: Three isotopes of silicon
occur in nature: 28Si (92.21%), which has a
mass of 27.97693 amu; 29Si (4.70%), which has
a mass of 28.97659 amu; and 30Si (3.09%),
which has a mass of 29.97376 amu. Calculate
the atomic weight of silicon.
27.97693(0.9221) + 28.97659(0.0470) + 29.97376(0.0309) =
28.08562 amu
Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Atomic Weights
56
Intensity
• Sample Problem: When a sample of natural copper
is vaporized and injected into a mass spectrometer,
the results shown are obtained. Use these data to
compute the average mass of natural copper.
[masses for 63Cu = 62.93 amu and 65Cu = 64.93 amu]
Given:
•Masses for 63Cu and 65C
69.09%
Mass No. 63
30.91%
65
•Relative abundance of 63Cu
and 65Cu
Find:
•Average Mass of Cu
Chem 106, Prof. J.T. Spencer
Atomic Weights
57
(.6909 atoms)(62.93 amu) + (.3091 atoms)(64.93amu) = 63.55 amu
atom
atom.
average mass per atom is;
6355 amu = 63.55 amu/atom
100 atoms
Intensity
Given:
69.09%
Mass No. 63
30.91%
65
•Masses for 63Cu and 65C
•Relative abundance of 63Cu
and 65Cu
Find:
•Average Mass of Cu
Chem 106, Prof. J.T. Spencer
Molecular Weights
58
• Formula Weights (FW) - sum of the atomic weights of each
atom in its chemical formula. (note AW is atomic weight)
• formula weight of NaN3 = 3(AW of N) + 1(AW of Na)
3(14) + 1(23) = 65 amu for sodium azide
• Molecular Weights (MW) - sum of atomic weights of each
atom in its molecular formula
• molecular weight of B2H6 = 2(AW of B) + 6(AW of H)
2(10.8) + 6(1) = 27.6 amu for diborane
• Difference between Molecular and Formula Weights
• ionic compounds, with extended arrays, have no well
defined molecules (and no molecular formulas) so we
use the formula weights (i.e., NaCl = 58 amu )
Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Molecular Weights
59
• Calculate molecular/formula weights for the following:
– P4O10
– BrCl
– Ca(NO3)2
P =30.97; O = 16.00; Br = 79.90; Cl = 35.45; Ca = 40.08;
N = 14.01
Chem 106, Prof. J.T. Spencer
Molecular Weights
60
– P4O10
4(30.97) + 10(16.00) = 283.88 amu
– BrCl
1(79.90) + 1(35.45) = 115.35 amu
– Ca(NO3)2
1(40.08) + 2(14.01) + 6(16.00) =
164.10 amu
P =30.97; O = 16.00; Br = 79.90; Cl = 35.45; Ca = 40.08;
N = 14.01
Chem 106, Prof. J.T. Spencer
Percentage Composition 61
• Percentage Composition - percentage by mass contributed
by each element in the substance. May be used to verify the
purity or identity of a particular compound.
• 100 [(atoms of an element in formula)(AW)/FW ] = % comp. element
Percentage Composition of C6H12O6
(FW = 180)
% C = 100 (6)(12)/ (180) = 40.0% carbon
% O = 100 (6)(16)/ (180) = 53.3% oxygen
% H = 100 (12)(1)/ (180) = 6.7% hydrogen
2 example calculations follows
Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Percentage Composition 62
Problem: In 1987, the first substance to act as a
superconductor at a temperature above that of liquid
nitrogen (77 K) was discovered. The approximate formula of
the substance is YBa2Cu3O7. Calculate the percent
composition by mass of this material.
M W of YBa2Cu3O7 = (88.9) + 2(137.3) + 3(63.6) + 7(16)
= 666.0 amu
AW: Y = 88.9; Ba =137.3; Cu = 63.6; O = 16.0
Chem 106, Prof. J.T. Spencer
Percentage Composition 63
M W of YBa2Cu3O7 = 666.0 amu
Y
=
1(88.9)
Ba
=
2(137.3)
Cu
=
3(63.5)
O
=
7(16.0)
= 100 (88.9)
666.0
= 100 (274.6)
666.0
= 100 (190.5)
666.0
= 100 (112)
666.0
AW: Y = 88.9; Ba =137.3; Cu = 63.6; O = 16.0
= 13.3 %
= 41.3 %
= 28.6 %
= 16.8 %
Chem 106, Prof. J.T. Spencer
Percentage Composition 64
Sample exercise: Calculate the percentage of
nitrogen, by mass, in Ca(NO3)2.
Chem 106, Prof. J.T. Spencer
Percentage Composition 65
Sample exercise: Calculate the percentage of
nitrogen, by mass, in Ca(NO3)2.
Formula Mass:
1(40.1) = 40.1
2(14.0) = 28.0
6(16.0) = 96.0
164.1 amu
Chem 106, Prof. J.T. Spencer
Percentage Composition 66
Sample exercise: Calculate the percentage of
nitrogen, by mass, in Ca(NO3)2.
Formula Mass:
1(40.1) = 40.1
2(14.0) = 28.0
6(16.0) = 96.0
164.1 amu
part x 100
total
Chem 106, Prof. J.T. Spencer
Percentage Composition 67
Sample exercise: Calculate the percentage of
nitrogen, by mass, in Ca(NO3)2.
Formula Mass:
1(40.1) = 40.1
2(14.0) = 28.0
6(16.0) = 96.0
164.1 amu
part x 100
total
28.0 x 100 =
164.1
Chem 106, Prof. J.T. Spencer
Percentage Composition 68
Sample exercise: Calculate the percentage of
nitrogen, by mass, in Ca(NO3)2.
Formula Mass:
1(40.1) = 40.1
2(14.0) = 28.0
6(16.0) = 96.0
164.1 amu
part x 100
total
28.0 x 100 = 17.1%
164.1
Chem 106, Prof. J.T. Spencer
Percentage Composition 69
• Example ; determine the elemental
percent composition of strychnine
N
•Molecular formula of
strychnine = C21H22N2O2
•Molecular weight (MW)
MW = 21(12) + 22(1) + 2(14) + 2(16) = 334 amu
•Percent composition;
% C = 100 (21)(12)/ (334) = 75.4% carbon
% N = 100 (2)(14)/ (334) = 6.59% nitrogen
% O = 100 (2)(16)/ (334) = 9.59% oxygen
% H = 100 (22)(1)/ (334) = 8.38% hydrogen
N
O
O
Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Mole
70
• Very small macroscopic samples contain VERY many
atoms, molecules, etc... (e.g. 1 tsp. H2O contains 2 x 1023
mol). [Need convenient counting unit]
• known no. of H atoms in 1 g of H = no. of atoms of O in
16 g of O = no. of C atoms in 12 g of C = etc... (based upon
atomic weights)
• Def. = The number of carbon atoms in 12 g of 12C is
called Avogadro’s number. One Mole (latin “mole”
= a mass) is the amount of material that contains
Chapt. 3.4
Avogadro’s number
•Note: a mole refers to a fixed number of any
type of particles!
•Avogadro’s number = 6.023 x 1023
Chem 106, Prof. J.T. Spencer
Avagadro’s Number and the Mole71
•
•
•
•
•
•
1 mole of 12C atoms = 6.02 x 1023 atoms
1 mole of 11B atoms = 6.02 x 1023 atoms
1 mol of PCl3 molecules = 6.02 x 1023 molecules
1 mol of Na+ ions = 6.02 x 1023 Na ions
1 mol of toasters = 6.02 x 1023 toasters
1 mole of students = 6.02 x 1023 students
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Mole
72
•Examples
How many C atoms in 0.5 moles of Carbon?
C atoms = (0.5 moles C)(6.02 x 1023 atoms) = 3.01 x 1023
mole
How many C atoms are in 0.25 mol of C6H12O6?
C ato. = (0.25 mol C6H12O6)(6.02 x 1023 molec) (6 C atoms)
mol
(1 C6H12O6 molec)
C atoms = 1.5 x 1023 atoms
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Mole
73
Sample exercise: How many O atoms are in
a) 0.25 moles Ca(NO3)2
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Mole
74
Sample exercise: How many O atoms are in
a) 0.25 moles Ca(NO3)2
0.25 mol Ca(NO3)2
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Mole
75
Sample exercise: How many O atoms are in
a) 0.25 moles Ca(NO3)2
0.25 mol Ca(NO3)2 6 mole O
1 mole Ca(NO3)2
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Mole
76
Sample exercise: How many O atoms are in
a) 0.25 moles Ca(NO3)2
0.25 mol Ca(NO3)2 6 mole O
1 mole Ca(NO3)2
1.5 mol O
6.02 x 1023 atom O
1 mol O
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Mole
77
Sample exercise: How many O atoms are in
a) 0.25 moles Ca(NO3)2
0.25 mol Ca(NO3)2 6 mole O
1 mole Ca(NO3)2
1.5 mol O
6.02 x 1023 atom O = 9.03 x 1023
1 mol O
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Mole
78
Sample exercise: How many O atoms are in
a) 1.50 moles sodium carbonate
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Mole
79
Sample exercise: How many O atoms are in
a) 1.50 moles sodium carbonate
Na+1 CO3-2  Na2CO3
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Mole
80
Sample exercise: How many O atoms are in
a) 1.50 moles sodium carbonate
Na+1 CO3-2  Na2CO3
1.50 mol Na2CO3
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Mole
81
Sample exercise: How many O atoms are in
a) 1.50 moles sodium carbonate
Na+1 CO3-2  Na2CO3
1.50 mol Na2CO3
3 mol O
1 mol Na2CO3
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Mole
82
Sample exercise: How many O atoms are in
a) 1.50 moles sodium carbonate
Na+1 CO3-2  Na2CO3
1.50 mol Na2CO3
4.50 mol O
3 mol O
1 mol Na2CO3
6.02 x 1023 atom O
1 mol O
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Mole
83
Sample exercise: How many O atoms are in
a) 1.50 moles sodium carbonate
Na+1 CO3-2  Na2CO3
1.50 mol Na2CO3
4.50 mol O
3 mol O
1 mol Na2CO3
6.02 x 1023 atom O = 2.71 x 1024
Chapt. 3.4
1 mol O
Chem 106, Prof. J.T. Spencer
Molar Mass
84
• Example - since one 12C atoms weighs 12 amu
and a 24 Mg atom weighs 24 amu (twice as
massive) and since a mole always contains the
same number of particles, a mole of 24 Mg must
weigh twice as much as a mole of 12C.
1 12C atom weighs 12 amu; 1 mol 12C weigh 12 g
1 24Mg atom weighs 24 amu; 1 mol 24Mg weigh 24 g
1 238U atom weighs 238 amu; 1 mol 238U weighs 238 g
Molar Mass - (in grams) of any substance is
always numerically equal to its formula
weight (in amu).
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Molar Relationships
Name
85
Formula Formula Mass of 1 mol Number and kind
weight of form units of particles in 1 mol
atomic nitrogen
N
14.0
14.0
6.02 X 1023 N atoms
molec. nitrogen
N2
28.0
28.0
6.02 X 1023 N2 molec.
2(6.02 X 1023 ) N atoms
ScCl3
151.5
151.5
6.02 X 1023 ScCl3 units
6.02 X 1023 Sc3+ ions
3(6.02 X 1023) Cl- ions
C6H12O6 180.0
180.0
6.02 X 1023 gluc. molec.
6(6.02 X 1023) C atoms
12(6.02 X 1023 ) H atoms
scandium
chloride
glucose
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Molar Relationships
86
(1) How many moles of phosphorus trichloride, PCl3, are in
50 g of the substance? (MW = 137.4 amu)
Moles of PCl3 = 1mol PCl3 (50 g PCl3) = 0.36 moles PCl3
137.4 g
(2) How many molecules of PCl3 are in 50 g?
molecules of PCl3 = (0.36 moles )(6.023 x 1023 molecules)
1 mole
= 2.2 x 1023 PCl3 molecules
(3) How many grams of PCl3 are in 0.75 moles?
x grams = (0.75 mole)(137.4 g PCl3) = 103 g of PCl3
1 mole PCl3
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Molar Relationships
Grams
use molar mass
87
use molar mass
Moles
use Avagadro’s
number
use Avagadro’s
number
Items
(molecules, atoms, etc...)
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Molar Relationships
88
Sample exercise: How many moles of
NaHCO3 are present in 508 g of this
substance?
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Molar Relationships
89
Sample exercise: How many moles of
NaHCO3 are present in 508 g of this
substance?
508 g
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Molar Relationships
90
Sample exercise: How many moles of
NaHCO3 are present in 508 g of this
substance?
508 g
1 mol
84 g
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Molar Relationships
91
Sample exercise: How many moles of
NaHCO3 are present in 508 g of this
substance?
508 g
1 mol = 6.05 mol NaHCO3
84 g
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Molar Relationships
92
Sample exercise: What is the mass, in
grams, of
a) 6.33 mol NaHCO3
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Molar Relationships
93
Sample exercise: What is the mass, in
grams, of
a) 6.33 mol NaHCO3
6.33 mol NaHCO3
84 g NaHCO3
1 mol NaHCO3
= 532 g NaHCO3
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Molar Relationships
94
Sample exercise: What is the mass, in
grams, of
b) 3.0 x 10-5 mol sulfuric acid
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Molar Relationships
95
Sample exercise: What is the mass, in
grams, of
b) 3.0 x 10-5 mol sulfuric acid
3.0 x 10-5 mol H2SO4
98 g H2SO4
1 mol H2SO4
= 2.9 x 10-3 g
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Molar Relationships
96
Sample exercise: How many nitric acid
molecules are in 4.20 g of HNO3?
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Molar Relationships
97
Sample exercise: How many nitric acid
molecules are in 4.20 g of HNO3?
4.20 g HNO3
6.02 x 1023 molec HNO3
63 g HNO3
= 4.01 x 1022 molec
Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Empirical Formulas
98
• Empirical Formula - Relative number of each element in a
compound.
• Using moles and percent weight (elemental analysis by
chemical means), we can calculate an empirical formula
• Steps;
» assume a 100 g (convenient since working with %
because the elements % can be thought of as g)
» calculate moles of element present in 100g sample
» find ratios of moles (approx) to lead to integral
formula subscripts.
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Empirical Formulas
Mass Percent
of Elements
Empirical
Formula
assume
100 g sample
Grams of
each Element
99
calculate mole
ratio
Use
atomic
weights
Moles of
each Elements
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Empirical Formulas, Examples
100
Determine the empirical formula for a compound which
contains 87.5 % N and 12.5% H by mass.
%
g in 100g
moles
ratio
87.5 % N = 87.5 g N 1 mole = 6.25 moles N = 1
14 g
12.5% H =
12.5 g H 1 mole H = 12.5 moles H = 2
1g
Empirical Formula = NH2
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Empirical Formulas, Examples
101
Sample exercise: A 5.325 g sample of methyl
benzoate, a compound used in the
manufacture of perfumes, is found to contain
3.758 g of carbon, 0.316 g of hydrogen, and
1.251 g of oxygen. What is the empirical
formula of this substance?
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Empirical Formulas, Examples
102
Sample exercise: A 5.325 g sample of methyl
benzoate, a compound used in the
manufacture of perfumes, is found to contain
3.758 g of carbon, 0.316 g of hydrogen, and
1.251 g of oxygen. What is the empirical
formula of this substance?
3.758 g x 100 = 70.58% C -> 70.58 g C 1 mol C = 5.88 mol C
5.325 g
12 g C
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Empirical Formulas, Examples
103
Sample exercise: A 5.325 g sample of methyl
benzoate, a compound used in the
manufacture of perfumes, is found to contain
3.758 g of carbon, 0.316 g of hydrogen, and
1.251 g of oxygen. What is the empirical
formula of this substance?
0.316 g x 100 = 5.93% H -> 5.93 g H 1 mol H = 5.93 mol H
5.325 g
1gH
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Empirical Formulas, Examples
104
Sample exercise: A 5.325 g sample of methyl
benzoate, a compound used in the
manufacture of perfumes, is found to contain
3.758 g of carbon, 0.316 g of hydrogen, and
1.251 g of oxygen. What is the empirical
formula of this substance?
1.251 g x 100 = 23.49% O -> 23.49 g O 1 mol O = 1.47 mol O
5.325 g
16 g O
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Empirical Formulas, Examples
105
Sample exercise: A 5.325 g sample of methyl
benzoate, a compound used in the
manufacture of perfumes, is found to contain
3.758 g of carbon, 0.316 g of hydrogen, and
1.251 g of oxygen. What is the empirical
formula of this substance?
5.88 mol C ; 5.93 mol H ; 1.47 mol O
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Empirical Formulas, Examples
106
Sample exercise: A 5.325 g sample of methyl
benzoate, a compound used in the
manufacture of perfumes, is found to contain
3.758 g of carbon, 0.316 g of hydrogen, and
1.251 g of oxygen. What is the empirical
formula of this substance?
5.88 mol C ; 5.93 mol H ; 1.47 mol O
1.47 mol
1.47 mol
1.47 mol
4
4
1
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Empirical Formulas, Examples
107
Sample exercise: A 5.325 g sample of methyl
benzoate, a compound used in the
manufacture of perfumes, is found to contain
3.758 g of carbon, 0.316 g of hydrogen, and
1.251 g of oxygen. What is the empirical
formula of this substance?
5.88 mol C ; 5.93 mol H ; 1.47 mol O
1.47 mol
1.47 mol
1.47 mol
4
4
1
C 4H4O
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Empirical Formulas, Examples
108
A white unknown substance (mass spec. problem) found on an
unconscious patient is suspected by a forensic chemist of being
either cocaine or caffeine. Combustion of a 50.86 mg sample
yielded 150.0 mg of CO2 and 46.05 mg of water. Further analysis
showed the compound contained 9.39% N by mass. The formula
of cocaine is C17H21NO4. Can the substance be cocaine?
Unknown
% C, H, and O.
Percent Composition
Known
50.86 mg of cmpd gave 150.0 mg of CO2 and
46.05 mg of H2O.
compound contains 9.39% N.
formula of Cocaine is C17H21NO4.
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Empirical Formulas, Examples
109
•Combustion Reactions
–Reactions with oxygen (usually from the air)
–The complete combustion of hydrocarbons yield carbon
dioxide (CO2) and water (H2O)
CxHy + (2x+y) /2 O2
X CO2 + Y/2 H2O
Known
50.86 mg of cmpd gave 150.0 mg of CO2 and
46.05 mg of H2O.
compound contains 9.39% N.
formula of Cocaine is C17H21NO4.
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Empirical Formulas, Examples
110
Compute % C and % H (from combustion).
C:
0.150 g CO2 = 0.00341 mol CO2 = 0.00341 mol C = 40.9 mg C
H:
0.04605 g H2O = 0.00256 mol H2O = 0.00512 mol H = 5.12 mg H
N:
(0.0939)(50.86) = 4.77 mg N = 0.000341 mol N
O:
(50.86 g sample)-(40.9 mg C + 5.12 mg H + 4.77 mg N) =
= 0.08 mg O = 0.000006 mol O
Known
50.86 mg of cmpd gave 150.0 mg of CO2 and
46.05 mg of H2O.
compound contains 9.39% N.
formula of Cocaine is C17H21NO4.
Chem 106, Prof. J.T. Spencer
Empirical Formulas, Examples
111
mg in sample
% in sample calc’n
% sample
C:
40.9 mg C
100 (40.9mg/50.86 mg)
80.5 % C
67.3% C
H:
5.12 mg H
100 (5.11mg/50.86mg)
10.1% H
6.9% H
N:
4.77 mg N
100 (4.77mg/50.86 mg)
9.4% N
4.6% N
O:
0.08 mg O
100 (0.08mg/50.86mg)
0.02% O
21.2% O
Known
50.86 mg of cmpd gave 150.0 mg of CO2 and
46.05 mg of H2O.
compound contains 9.39% N.
formula of Cocaine is C17H21NO4.
% cocaine
Chem 106, Prof. J.T. Spencer
Empirical Formulas, Examples
112
% sample
C:
g in cocaine
% in sample calc’n
(303 grams)
17(12) = 204 g C 100 (204/303)
80.5 % C
67.3% C
H:
21(1) g H
100 (21/303)
10.1% H
6.9% H
N:
1(14) g N
100 (14/303)
9.4% N
4.6% N
O:
4(16) g O
100 (64/303)
0.02% O
21.2% O
Known:
formula of Cocaine is C17H21NO4.
MW = 17(12) + 21(1) + 1(14) + 4(16) = 303 amu
% cocaine
Chem 106, Prof. J.T. Spencer
Empirical Formulas, Examples
113
mg in sample
% in sample calc’n
% sample
% cocaine
C:
40.9 mg C
100 (40.9mg/50.86 mg)
80.5 % C
67.3% C
H:
5.12 mg H
100 (5.11mg/50.86mg)
10.1% H
6.9% H
N:
4.77 mg N
100 (4.77mg/50.86 mg)
9.4% N
4.6% N
O:
0.08 mg O
100 (0.08mg/50.86mg)
0.02% O
21.2% O
Chem 106, Prof. J.T. Spencer
Empirical Formulas, Examples
114
mg in sample
% in sample calc’n
% sample
% cocaine
C:
40.9 mg C
100 (40.9mg/50.86 mg)
80.5 % C
67.3% C
H:
5.12 mg H
100 (5.11mg/50.86mg)
10.1% H
6.9% H
N:
4.77 mg N
100 (4.77mg/50.86 mg)
9.4% N
4.6% N
O:
0.08 mg O
100 (0.08mg/50.86mg)
0.02% O
21.2% O
Compound is Not Cocaine
from analysis
Chem 106, Prof. J.T. Spencer
Combustion Reactions

115
Reaction of Hydrogen with Oxygen
[COMBUSTION] (note precautions)
» 2 H2(g) + O2(g)
2 H2O(g)
H = -232 kJ/mol H2O
» Ignition temperature = 580° - 590° C
» Explosive [“when stuff gets really big really
fast” Beakman’ World]
» The rapid release of energy [-232 kJ/mol
H2O] into the surrounding air causes the air
to very quickly expand. the explosion from
pure H2 sound quiter because the air
expansion is slower.
[Video No. 20-21; 4:42 +1:29 m]
[Video No. 22; 2:50 m]
Chem 106, Prof. J.T. Spencer
Combustion Reactions

Combustion of Alcohol (ethanol):
C2H5OH(g) + 3 O2(g)




116
2CO2(g) + 3 H2O(g)
H =1366.2 kJ mol-1
Tesla coil produces a high voltage electric spark.
The spark is required to initiate this reaction.
Conversion of chemical energy (PE stored in
bonds) to mechanical energy.
Questions for After Demonstration
Are other types of energy are produced besides
mechanical energy?
Why can the reaction not be repeated without
flushing the bottle with air first?
Chem 106, Prof. J.T. Spencer
Combustion Analysis
117
• Empirical Formula from reaction with oxygen
• Organic Compounds - C to CO2 and H to H2O
• Use CO2 and H2O to determine the amount of C
and H in original sample
H2O
furnace
absorbant
(Mg(ClO4)2)
O2 flow
sample
contaminant
catalyst (CuO);
oxidizes traces of
CO and C to CO2
CO2
absorbant
(NaOH)
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Combustion Analysis
118
The combustion of 5.00g of an organic compound containing
C, H, and O yields 9.57 g CO2 and 5.87 g H2O. What is the
empirical formula?
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Combustion Analysis
119
The combustion of 5.00g of an organic compound containing
C, H, and O yields 9.57 g CO2 and 5.87 g H2O. What is the
empirical formula?
9.57 g CO2 = 0.217 mol CO2 = 0.217 mol C = 2.60 g C
5.87 g H2O = 0.326 mol H2O = 0.652 mol H = 0.652 g H
whatever is left over must be the amt. of O originally present;
5.00 g - (2.60 g C + 0.652 g H) = 1.75 g of O = 0.109 mol O
thus;
0.217 mol C
divide each
0.109 mol O by 0.109
0.652 mol H
1.99
1.00
5.98
thus C2H6O
(ethanol)
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Combustion Analysis
120
The combustion of 0.596 g of a compound containing only B
and H yields 1.17 g H2O and all the boron is recovered as B2O3.
What is the empirical formula?
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Combustion Analysis
121
The combustion of 0.596 g of a compound containing only B
and H yields 1.17 g H2O and all the boron is recovered as B2O3.
What is the empirical formula?
(1) Chem Equation; BxHy + O2
H2O + B2O3
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Combustion Analysis
122
The combustion of 0.596 g of a compound containing only B
and H yields 1.17 g H2O and all the boron is recovered as B2O3.
What is the empirical formula?
(1) Chem Equation; BxHy + O2
H2O + B2O3
(2) 1.17 g H2O = 1.17g = 0.065 mol H2O = 0.130 mol H
18 g/mol
(3) g H = (0.065 mol H2O)(2 mol H)
(1 g H) = 0.130 g H
1 mol H2O 1 mol H
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Combustion Analysis
123
The combustion of 0.596 g of a compound containing only B
and H yields 1.17 g H2O and all the boron is recovered as B2O3.
What is the empirical formula?
(1) Chem Equation; BxHy + O2
H2O + B2O3
(2) 1.17 g H2O = 1.17g = 0.065 mol H2O = 0.130 mol H
18 g/mol
(3) g H = (0.065 mol H2O)(2 mol H)
(1 g H) = 0.130 g H
1 mol H2O 1 mol H
(4) (0.596 g tot)-(0.130 g H) = 0.466 g B;
(5) B = 0.0431 mol = 1.00
0.0431
BH3
0.466 g B = 0.043 mol B
10.8 g/ mol
H = 0.130 mol = 3.01
0.0431
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Chemical Problem Solving
124
• Read and UNDERSTAND the problem - determine what
is being given and what is required.
• Identify the Unknown and Given data.
• Set up the problem - determine what kinds of
information bear upon the problem, what solution
pathways may be available, what chemical principles
should give guidance, etc...
• Solve the problem - Use the data given and the
appropriate relationships or equations to work throught
the problem.
• Check your work - not just the mathematical functions
but ask if the answer makes sense and provides what is
being asked for! (sig. figs)
Chapt. 3.5
Chem 106, Prof. J.T. Spencer
Equations and the Mole
125
• Coefficients in a balanced chemical equation refer to
both the relative number of molecules involved in a
reaction and the relative number of moles.
Chapt. 3.6
• Stoichiometric equivalence - from coefficients in a
chemical equation; B2H6 + 3 O2
3 H2O + B2O3
• 1 mol B2H6 equiv. to 3 moles O2 equiv. to 3 mol H20, ...
• Used to calculate quantities involved in a reaction
grams of
compound A
use molar mass of A
moles of
compound A
grams of
compound B
use coeff of A
and B from
balanced eqn.
use molar mass of B
moles of
compound B
Chem 106, Prof. J.T. Spencer
Mole Calculations
126
• Given the reaction for the formation of B2H6 (diborane),
how many grams of diborane can be prepared from 3.0 g
of LiH?
6 LiH + 8BF3
6 LiBF4 + B2H6
[B2H6 MW = 27.6 and LiH MW = 7.9]
3.0 g LiH (1 mol LiH) = 0.38 mol LiH
7.9 g LiH
0.38 mol LiH (1 mol B2H6 ) 27.6 g B2H6 = 1.7 g B2H6
6 mol LiH
1 mol B2H6
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Mole Calculations
127
• Given the reaction for the formation of B2H6 (diborane),
how many grams of BF3 are required to react with 3.0 g of
LiH ?
6 LiH + 8BF3
6 LiBF4 + B2H6
[B2H6 MW = 27.6, BF3 = 67.8 and LiH MW = 7.9]
3.0 g LiH (1 mol LiH) (8 mol BF3) (67.8 g BF3) = 34 g BF3
7.9 g LiH 6 mol LiH 1 mol BF3
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Mole Calculations
128
Sample exercise: A common laboratory
method for preparing small amounts of O2
involves the decomposition of KClO3:
2KClO3  2KCl + 3O2
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Mole Calculations
129
Sample exercise: A common laboratory
method for preparing small amounts of O2
involves the decomposition of KClO3:
2KClO3  2KCl + 3O2
How many grams of oxygen is produced
from 4.50 g KClO3?
4.50 g KClO3
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Mole Calculations
130
Sample exercise: A common laboratory
method for preparing small amounts of O2
involves the decomposition of KClO3:
2KClO3  2KCl + 3O2
How many grams of oxygen is produced
from 4.50 g KClO3?
4.50 g KClO3
1 mol KClO3
122.6 g KClO3
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Mole Calculations
131
Sample exercise: A common laboratory
method for preparing small amounts of O2
involves the decomposition of KClO3:
2KClO3  2KCl + 3O2
How many grams of oxygen is produced
from 4.50 g KClO3?
4.50 g KClO3
1 mol KClO3 3 mol O2
122.6 g KClO3 2 mol KClO3
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Mole Calculations
132
Sample exercise: A common laboratory
method for preparing small amounts of O2
involves the decomposition of KClO3:
2KClO3  2KCl + 3O2
How many grams of oxygen is produced
from 4.50 g KClO3?
4.50 g KClO3
1 mol KClO3 3 mol O2
32 g O2
122.6 g KClO3 2 mol KClO3 1 mol O2
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Mole Calculations
133
Sample exercise: A common laboratory
method for preparing small amounts of O2
involves the decomposition of KClO3:
2KClO3  2KCl + 3O2
How many grams of oxygen is produced
from 4.50 g KClO3?
4.50 g KClO3
1 mol KClO3 3 mol O2
32 g O2
122.6 g KClO3 2 mol KClO3 1 mol O2
= 1.76 g O2
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Mole Calculations
134
Sample exercise: Propane, C3H8, is a common
fuel used for cooking and home heating.
What mass of O2 is consumed in the
combustion of 1.00 g of propane?
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Mole Calculations
135
Sample exercise: Propane, C3H8, is a common
fuel used for cooking and home heating.
What mass of O2 is consumed in the
combustion of 1.00 g of propane?
C3H8 + O2  CO2 + H2O
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Mole Calculations
136
Sample exercise: Propane, C3H8, is a common
fuel used for cooking and home heating.
What mass of O2 is consumed in the
combustion of 1.00 g of propane?
C3H8 + 5O2  3CO2 + 4H2O
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Mole Calculations
137
Sample exercise: Propane, C3H8, is a common
fuel used for cooking and home heating.
What mass of O2 is consumed in the
combustion of 1.00 g of propane?
C3H8 + 5O2  3CO2 + 4H2O
1.00 g C3H8
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Mole Calculations
138
Sample exercise: Propane, C3H8, is a common
fuel used for cooking and home heating.
What mass of O2 is consumed in the
combustion of 1.00 g of propane?
C3H8 + 5O2  3CO2 + 4H2O
1.00 g C3H8
1 mol C3H
44 g C3H8
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Mole Calculations
139
Sample exercise: Propane, C3H8, is a common
fuel used for cooking and home heating.
What mass of O2 is consumed in the
combustion of 1.00 g of propane?
C3H8 + 5O2  3CO2 + 4H2O
1.00 g C3H8
1 mol C3H
44 g C3H8
5 mol O2
1 mol C3H8
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Mole Calculations
140
Sample exercise: Propane, C3H8, is a common
fuel used for cooking and home heating.
What mass of O2 is consumed in the
combustion of 1.00 g of propane?
C3H8 + 5O2  3CO2 + 4H2O
1.00 g C3H8
1 mol C3H
44 g C3H8
5 mol O2
1 mol C3H8
32 g O2
1 mol O2
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Mole Calculations
141
Sample exercise: Propane, C3H8, is a common
fuel used for cooking and home heating.
What mass of O2 is consumed in the
combustion of 1.00 g of propane?
C3H8 + 5O2  3CO2 + 4H2O
1.00 g C3H8
1 mol C3H
44 g C3H8
5 mol O2
1 mol C3H8
32 g O2
1 mol O2
= 3.64 g O2
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Limiting Reagent
142
• Sometimes after one reagent is completely
consumed in the reaction some of another
reagent is left over. The reagent which is
completely consumed limits the extent of the
reaction = LIMITING REAGENT.
Limiting Reagent
+
Chapt. 3.7
Chem 106, Prof. J.T. Spencer
Limiting Reagent Calculations
143
• Given the reaction for the formation of B2H6 (diborane), if
5.0 g of LiH and 5.0 g of BF3 were reacted how much of
which reagent would be left over?
6 LiH + 8BF3
6 LiBF4 + B2H6
[B2H6 MW = 27.6. BF3 = 67.8 and LiH MW = 7.9]
Know:
Quantities (g and moles) of starting materials
Molar ratios between all the starting materials
and products.
Find:
Which reagent is completely consumed (limiting
reagent) and which is left over
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Limiting Reagent Calculations
144
• Given the reaction for the formation of B2H6 (diborane), if
5.0 g of LiH and 5.0 g of BF3 were reacted how much of
which reagent would be left over?
6 LiH + 8BF3
6 LiBF4 + B2H6
[B2H6 MW = 27.6. BF3 = 67.8 and LiH MW = 7.9]
5.0 g LiH (1 mol LiH) = 0.63 mol AND 5.0 g BF3 (1 mol BF3) = 0.074 mol
7.9 g LiH
67.8 g BF3
If all the LiH were consumed, then 0.84 mol BF3 would be required
[(0.63 mol LiH)(8 mol BF3)] = 0.84 mol BF3
6 mol LiH)
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Limiting Reagent Calculations
145
• Given the reaction for the formation of B2H6 (diborane), if
5.0 g of LiH and 5.0 g of BF3 were reacted how much of
which reagent would be left over?
6 LiH + 8BF3
6 LiBF4 + B2H6
[B2H6 MW = 27.6. BF3 = 67.8 and LiH MW = 7.9]
5.0 g LiH (1 mol LiH) = 0.63 mol AND 5.0 g BF3 (1 mol BF3) = 0.074 mol
7.9 g LiH
67.8 g BF3
If all the LiH were consumed, then 0.84 mol BF3 would be required
Since only 0.074 mol of BF3 is available, BF3 is the limiting reagent
(all consumed).
0.074 mol BF3 (6 mol LiH) = 0.056 mol LiH consumed
8 mol BF3
therefore remaining LiH = (0.63 mol - 0.056 mol)(7.9 g/mol) = 4.53 g
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Limiting Reagent Problems
146
• Equal weights (5.00 g) of Zn(s) and I2(s) are mixed
together to form ZnI2. How much ZnI2 is formed? How
much of each reactant remains at the end of the reaction
and which is the limiting reagent?
Zn (AW = 65.4 amu)
I2 (MW = 253.8 amu)
Zn(s) + I2(s)
ZnI2(s)
Chapt. 3.7
Chem 106, Prof. J.T. Spencer
Limiting Reagent Problems
147
• Equal weights (5.00 g) of Zn(s) and I2(s) are mixed
together to form ZnI2. How much ZnI2 is formed? How
much of each reactant remains at the end of the reaction
and which is the limiting reagent?
Zn (AW = 65.4 amu)
I2 (MW = 253.8 amu)
Zn(s) + I2(s)
ZnI2(s)
Zn = 5.0 g Zn (1 mol Zn) = 0.076 mol Zn
65.4 g Zn
I2 = 5.0 g I2 (1 mol I2) = 0.020 mol I2
253.8 g I2
I2 is the limiting reagent.
Zn remaining = (0.076 Zn - 0.020 mol Zn) (64.5 g Zn) = 3.66 g Zn
1 mol Zn
Chapt. 3.7
Chem 106, Prof. J.T. Spencer
Theoretical Yields
148
• Theoretical Yield - quantity of product
calculated to form when all the limiting reagent
is consumed (calculated from molar ratios).
• Actual Yield - the amount of product
experimentally obtained from a reaction
• Percent Yield - describes relationship between
theoretical and actual yields;
percent yield =
actual yield (100)
theoretical yield
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
Percent Yields
149
• Given the reaction of 2.05 g of hydrogen sulfide with 1.84 g
of sodium hydroxide, calculate how the theoretical yield of
Na2S. What is the percent yield if the amt. of Na2S
obtained was 3.65 g. [H2S (MW = 34.1); Na2S (MW = 78.1)]
H2S(g) + 2 NaOH(aq)
Na2S(aq) + 2 H2O
(2.05 g H2S)(1 mol H2S)(1 mol Na2S)(78.1 g Na2S) = 4.70 g Na2S
34.1 g H2S 1 mol H2S 1 mol Na2S
theoretical yield
% yield = 3.65 g (actual yield)
(100) = 77.7 % yield
4.70 g (theoretical yield)
Chapt. 3.6
Chem 106, Prof. J.T. Spencer
End of Chapter 3
150
Major Topics (not exhaustive list):
(1) Chemical Equations
(2) Periodic Table and Reaction
Types
(3) Atomic and Molecular
Weights (formula weights, %
compositions, etc...)
(4) Molar Concepts
(5) Empirical Formulas
(6) Info from Balanced Eqns.
(7) Limiting Reagents
(8) Percent Yields