Transcript 03-Chemical Rxns n Stoichiometryx

Lecture Presentation
Chapter 3
Chemical Reactions
and Reaction
Stoichiometry
© 2015 Pearson Education, Inc.
James F. Kirby
Quinnipiac University
Hamden, CT
LESSON 1
3-1Equations &
3-2 Types of Rxns
Atoms,
Molecules,
and Ions
© 2015 Pearson Education, Inc.
Stoichiometry
• The study of the mass relationships in
chemistry
• Based on the Law of Conservation of
Mass (Antoine Lavoisier, 1789)
“We may lay it down as an
incontestable axiom that, in all the
operations of art and nature, nothing
is created; an equal amount of matter
exists both before and after the
experiment. Upon this principle, the
whole art of performing chemical
experiments depends.”
—Antoine Lavoisier
Chemical Equations
Chemical equations are concise
representations of chemical reactions.
Chemical Reactions & Chemical Equations
• Chemical reaction: a process in which a substance or
substances change into one or more new substances.
Hydrogen gas combusts in the presence of oxygen gas to form water.
• Chemical equation: a description of a chemical
reaction using chemical symbols to show what happens
during the reaction.
BALANCE the
2 H2 + O2  2 H2O
REACTANTS:
Starting compounds;
what you start with
equation to show
conservation of mass
PRODUCTS:
Resulting compounds;
what you end up with
A process in which one or more substances is changed into one
or more new substances is a chemical reaction
A chemical equation uses chemical symbols to show what
happens during a chemical reaction
3 ways of representing the reaction of H2 with O2 to form H2O
reactants
products
What Is in a Chemical Equation?
CH4(g) + 2O2(g)
Reactants appear on the left
side of the equation.
CO2(g) + 2H2O(g)
What Is in a Chemical Equation?
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
Products appear on the right
side of the equation.
What Is in a Chemical Equation?
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
The states of the reactants and products are written in
parentheses to the right of each compound.
(g) = gas; (l) = liquid; (s) = solid;
(aq) = in aqueous solution
What Is in a Chemical Equation?
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
Coefficients are inserted to balance the equation to
follow the law of conservation of mass.
Chemical Equations: Additional Information
2HgO (s)  2Hg (l) + O2 (g)
(g), (l), (s): shows physical state of reactant or
product; i. e., whether reactant or product is a
gas (g), liquid (l), or solid (s)
STOICHIOMETRIC COEFFICIENTS: show
number of moles of reactant or product
Chemical Equations: Additional Information
KBr (aq) + AgNO3 (aq)  KNO3 (aq) + AgBr (aq)
(aq): shows that a reactant or product is
dissolved in H2O; i. e., the reactant or product is
in aqueous solution
How to “Read” Chemical Equations
2 Mg + O2
2 MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
IS NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
Interpretation of a Chemical Equation

2 H2O
Two molecules + one molecule

two molecules
Two moles

two moles

(2)(18.02 g) = 36.04 g
2 H2
+
+
O2
one mole
(2)(2.02 g) = 4.04 g + 32.00 g
36.04 g of reactants
36.04 g of products
Balancing Chemical Equations
Write chemical equations to describe reactions
carried out in the laboratory or other venues.
We know the reactants’ identities, but may need
to establish identities of products via further research, etc.
When identities of reactants and products are established,
we can write their formulas. Still, the number of atoms of
each element must be the same on both sides of the
equation to adhere to the Laws of Conservation
of Matter and Mass.
Thus, we must BALANCE the chemical equation!
Why Do We Add Coefficients Instead of
Changing Subscripts to Balance?
• Hydrogen and oxygen can make water
OR hydrogen peroxide:
 2 H2(g) + O2(g) → 2 H2O(l)
 H2(g) + O2(g) → H2O2(l)
Balancing Chemical Equations
1. Write the correct formula(s) for the reactants on
the left side and the correct formula(s) for the
product(s) on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2
CO2 + H2O
2. Change the numbers in front of the formulas
(coefficients) to make the number of atoms of
each element the same on both sides of the
equation. Do not change the subscripts.
2C2H6
NOT
C4H12
Balancing Chemical Equations
3. Start by balancing those elements that appear in
only one reactant and one product.
C2H6 + O2
2 carbon
on left
C2H6 + O2
6 hydrogen
on left
C2H6 + O2
CO2 + H2O
start with C or H but not O
1 carbon
on right
multiply CO2 by 2
2CO2 + H2O
2 hydrogen
on right
2CO2 + 3H2O
multiply H2O by 3
Balancing Chemical Equations
4. Balance those elements that appear in two or
more reactants or products.
C2H6 + O2
2 oxygen
on left
2CO2 + 3H2O
multiply O2 by 7
2
4 oxygen + 3 oxygen = 7 oxygen
(3x1)
on right
(2x2)
C2H6 + 7 O2
2
2CO2 + 3H2O
2C2H6 + 7O2
4CO2 + 6H2O
remove fraction
multiply both sides by 2
Balancing Chemical Equations
5. Check to make sure that you have the same
number of each type of atom on both sides of the
equation.
2C2H6 + 7O2
4CO2 + 6H2O
12
4 CH
(2
6)
14
O(2
(7xx2)
2)
4(6
C
2)6)
1412
OH
(4
x 2x +
Reactants
4C
12 H
14 O
Products
4C
12 H
14 O
Try to balance these equations…
1.
C6H12O6  C2H5OH + CO2
2.
C4H10 + O2  CO2 + H2O
3.
P4O10 + H2O  H3PO4
4.
NaHCO3 + H2SO4  Na2SO4 + CO2 + H2O
Homework: Worksheet on Balancing
Quiz
Types of Reactions
23
Chemical reactions can be classified as
• combination reactions
• decomposition reactions
• single replacement reactions
• double replacement reactions
• combustion reactions
Combination Reactions
24
In a combination reaction,
• two or more elements form one product
• or simple compounds combine to form one product
2Mg(s) + O2(g)
2MgO(s)
2Na(s) + Cl2(g)
2NaCl(s)
SO3(g) + H2O(l)
H2SO4(aq)
Combination Reaction: MgO
25
Combination Reactions
• In combination
reactions two or
more substances
react to form one
product.
• Examples:
– 2 Mg(s) + O2(g)
– N2(g) + 3 H2(g)
– C3H6(g) + Br2(l)
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2 MgO(s)
2 NH3(g)
C3H6Br2(l)
Stoichiometry
Decomposition Reaction
27
In a decomposition reaction, one substance splits into
two or more simpler substances.
2HgO(s)
2KClO3(s)
2Hg(l) + O2(g)
2KCl(s) + 3O2(g)
Decomposition Reaction: HgO
28
Decomposition Reactions
• In a decomposition
reaction one
substance breaks
down into two or
more substances.
• Examples:
– CaCO3(s)
– 2 KClO3(s)
– 2 NaN3(s)
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CaO(s) + CO2(g)
2 KCl(s) + O2(g)
2 Na(s) + 3 N2(g)
Stoichiometry
Single Replacement Reaction
30
In a single replacement reaction, one element takes the
place of a different element in another reacting a compound.
Zn(s) + 2HCl(aq)
ZnCl2(aq) + H2(g)
Fe(s) + CuSO4(aq)
FeSO4(aq) + Cu(s)
Single Replacement Reaction: ZnCl2
31
The Activity Series for Metals
Displacement Reaction
M + BC
AC + B
M is metal
BC is acid or H2O
B is H2
Ca + 2H2O
Ca(OH)2 + H2
Pb + 2H2O
Pb(OH)2 + H2
4.4
Double Replacement Reaction
33
In a double replacement, two elements in the reactants
exchange places.
AgNO3(aq) + NaCl(aq)
AgCl(s) + NaNO3(aq)
ZnS(s)
ZnCl2(aq) + H2S(g)
+ 2HCl(aq)
Double Replacement Reaction: BaSO4
34
Combustion Reaction
35
In a combustion reaction,
• a carbon-containing compound burns in oxygen gas
to form carbon dioxide (CO2) and water (H2O)
• energy is released as a product in the form of heat
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g) + energy
Combustion Reaction
36
Insert diagram of following balanced reaction with
triangle over the reaction arrow.
C3H8(g) + 5O2(g) Δ 3CO2(g) + 4H2O(g)
Combustion Reactions
• Combustion reactions
are generally rapid
reactions that produce
a flame.
• Combustion reactions
most often involve
oxygen in the air as a
reactant.
• Examples:
– CH4(g) + 2 O2(g)
– C3H8(g) + 5 O2(g)
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CO2(g) + 2 H2O(g)
3 CO2(g) + 4 H2O(g)
Stoichiometry
Summary of Reaction Types
38
Learning Check
39
Classify each of the following reactions as combination,
decomposition, single replacement, double replacement, or
combustion.
A. 2Al(s) + 3H2SO4(aq)
Al2(SO4)3(s) + 3H2(g)
B. Na2SO4(aq) + 2AgNO3(aq)
C. 3C(s) + Fe2O3(s)
D. C2H4(g) + 2O2(g)
Ag2SO4(s) + 2NaNO3(aq)
2Fe(s) + 3CO(g)
2CO2(g) + 2H2O(g)
Solution
40
A. 2Al(s) + 3H2SO4(aq)
Al2(SO4)3(s) + 3H2(g)
Single Replacement
B. Na2SO4(aq) + 2AgNO3(aq)
C. N2(g) + O2(g)
D. C2H4(g) + 2O2(g)
2NO(g)
Ag2SO4(s) + 2NaNO3(aq)
Double Replacement
Combination
2CO2(g) + 2H2O(g)
Combustion
Learning Check
41
Identify each reaction as combination, decomposition,
combustion, single replacement, or double replacement.
A. 3Ba(s) + N2(g)
Ba3N2(s)
B. 2Ag(s) + H2S(aq)
Ag2S(s) + H2(g)
C. 2C2H6(g) + 7O2(g)
4CO2(g) + 6H2O(g)
D. PbCl2(aq) + K2SO4(aq)
E. K2CO3(s)
2KCl(aq) + PbSO4(s)
K2O(aq) + CO2(g)
Solution
42
Identify each reaction as combination, decomposition,
combustion, single replacement, or double replacement.
A. 3Ba(s) + N2(g)
B. 2Ag(s) + H2S(aq)
Ba3N2(s)
Combination
Ag2S(s) + H2(g)
Single Replacement
C. C2H6(g) + 7O2(g)
4CO2(g) + 6H2O(g)
Combustion
Solution
43
Identify each reaction as combination, decomposition,
combustion, single replacement, or double replacement.
D. PbCl2(aq) + K2SO4(aq)
2KCl(aq) + PbSO4(s)
Double Replacement
E. K2CO3(s)
K2O(aq) + CO2(g)
Decomposition
Identify the type of redox reaction for each of the following reactions:
2 CuCl  Cu + CuCl2
Fe + 2 HCl  H2 + FeCl2
S + 3 F2  SF6
2 Ag + PtCl2  2 AgCl + Pt
2 N2O  2 N2 + O2
Homework: Worksheet on Predicting
Products
QUIZ
LESSON 2
3-3 Formula and Molecular
Weights, % Comp &
3-4 Avogadro’s # and the
Mole
Atoms,
Molecules,
and Ions
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Formula Weight (FW)
• A formula weight is the sum of the atomic
weights for the atoms in a chemical formula.
• This is the quantitative significance of a
formula.
• The formula weight of calcium chloride,
CaCl2, would be
Ca: 1(40.08 amu)
+ Cl: 2(35.453 amu)
110.99 amu
Stoichiometry
© 2015 Pearson Education, Inc.
Molecular Weight (MW)
• A molecular weight is the sum of the atomic
weights of the atoms in a molecule.
• For the molecule ethane, C2H6, the molecular
weight would be
C: 2(12.011 amu)
+ H: 6(1.00794 amu)
30.070 amu
Stoichiometry
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Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
1S
SO2
2O
SO2
32.07 amu
+ 2 x 16.00 amu
64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
3.3
Other Names Related to Mass
• Molecular Mass/Molecular Weight: If you have a single
molecule, mass is measured in amu’s instead of grams. But,
the molecular mass/weight is the same numerical value as 1
mole of molecules. Only the units are different. (This is the
beauty of Avogadro’s Number!)
• Formula Mass/Formula Weight: Same goes for
compounds. But again, the numerical value is the same.
Only the units are different.
• THE POINT: You may hear all of these terms
which mean the SAME NUMBER… just different units
Molar Mass:
(a.k.a. molecular mass)
(a.k.a. molecular weight)
The sum of the atomic masses
(in a.m.u.) in a molecule.
Try this:
Calculate the molecular mass of ascorbic acid (i.e., vitamin C),
the chemical formula of which is C6H8O6.
Molar Mass
• A molar mass is the mass of
1 mol of a substance (i.e., g/mol).
• The molar mass of an
element is the atomic
weight for the element
from the periodic table.
If it is diatomic, it is twice
that atomic weight.
• The formula weight (in
amu’s) will be the same
number as the molar mass
(in g/mol).
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Stoichiometry
Ionic Compounds and Formulas
• Remember, ionic compounds exist with
a three-dimensional order of ions.
There is no simple group of atoms to
call a molecule.
• As such, ionic compounds use empirical
formulas and formula weights (not
molecular weights).
Stoichiometry
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Percent Composition
One can find the percentage of the
mass of a compound that comes from
each of the elements in the compound
by using this equation:
(number of atoms)(atomic weight)
% Element =
(FW of the compound)
× 100
Stoichiometry
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Percent Composition of a Chemical Compound
% composition of an element =
(n)(molar mass of element)
------------------------------------ x 100
molar mass of compound
• Useful for verification of purity of compounds for research, etc.
Percent composition of an element in a compound =
n x molar mass of element
x 100%
molar mass of compound
n is the number of moles of the element in 1 mole
of the compound
2 x (12.01 g)
x 100% = 52.14%
46.07 g
6 x (1.008 g)
%H =
x 100% = 13.13%
46.07 g
1 x (16.00 g)
%O =
x 100% = 34.73%
46.07 g
%C =
C2H6O
52.14% + 13.13% + 34.73% = 100.0%
3.5
Example: Percent Composition
of the Elements in Beryl
Aquamarine
(Beryl, Be3Al2Si6O18)
Emerald
(Beryl, Be3Al2Si6O18)
• Moles of atoms:
• Molar masses of:
Beryllium (Be)
Aluminum (Al)
Silicon (Si)
Oxygen (O)
Beryllium (Be)
Aluminum (Al)
Silicon (Si)
Oxygen (O)
Beryl
(Be3Al2Si6O18)
What are the percents of the elements involved?
NEXT SLIDE
Try this one!
Determine the % of each element in the
mineral desmine (CaAl2Si7O18.7H2O).
Find the percentage composition of a compound
that contains 1.94 g of carbon, O.48 g of
hydrogen, and 2.58 g of sulfur in 5.00-g sample
of the compound.
Percent Composition
So the percentage of carbon in ethane is
(2)(12.011 amu)
%C =
=
(30.070 amu)
24.022 amu
30.070 amu
× 100
= 79.887%
Stoichiometry
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We deal with quantities involving more than one object
all the time….
Pair (2 items)
Dozen (12 items)
Gross (144 items)
What would a mole of any substance look like?
WHAT
IS
A
MOLE
?
Avogadro’s Number
• In a lab, we cannot
work with individual
molecules. They are
too small.
• 6.02 × 1023 atoms
or molecules is an
amount that brings
us to lab size. It is
ONE MOLE.
• One mole of 12C has
a mass of 12.000 g.
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Stoichiometry
The mole, as defined by the SI system of units:
A mole is the amount of a substance that
contains as many elementary entities
(i.e., atoms , molecules, or other
particles) as there are atoms in exactly 12
grams of the carbon-12 isotope.
6.022 x
NA
23
10
atoms, molecules, etc.
= Avogadro’s number
Amadeo
Avogadro
The Mole
Q: how long would it take to spend a mole of $1
coins if they were being spent at a rate of 1 billion
per second?
Mollionaire
Q: how long would it take to spend a mole of
$1 coins if they were being spent at a rate of
1 billion per second?
A: $ 6.02 x 1023 / $1 000 000 000
= 6.02 x 1014 payments = 6.02 x 1014 seconds
6.02 x 1014 seconds / 60 = 1.003 x 1013 minutes
1.003 x 1013 minutes / 60 = 1.672 x 1011 hours
1.672 x 1011 hours / 24 = 6.968 x 109 days
6.968 x 109 days / 365.25 = 1.908 x 107 years
A: It would take 19 million years
eggs
Molar mass is the mass of 1 mole of shoes in grams
marbles
atoms
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li
For any element
atomic mass (amu) = molar mass (grams)
One Mole of:
S
C
Hg
Cu
Fe
3.2
Mole Relationships
• One mole of atoms, ions, or molecules contains
Avogadro’s number of those particles.
• One mole of molecules or formula units contains
Avogadro’s number times the number of atoms or
ions of each element in the compound.
Stoichiometry
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1 mole of 12C atoms = 12.00 g of 12C atoms
1 mole of 12C atoms = 6.022 x 1023 atoms of 12C
6.022 x 1023 atoms of 12C = 12.00 g of 12C
What is the mass(in grams) of one ATOM of 12C?
Using Moles
Moles provide a bridge from the molecular
scale to the real-world scale.
Stoichiometry
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1 12C atom
12.00 g
1.66 x 10-24 g
x
=
23
12
12.00 amu
6.022 x 10
C atoms
1 amu
1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu
M = molar mass in g/mol
NA = Avogadro’s number
Molar Mass:
The mass, usually in grams (g),
of 1 mole of atoms or molecules
of a substance.
1 mole of X = molar mass of X (in grams)
1 mole of X = 6.022 x 1023 atoms or molecules of X
Do You Understand Molar Mass?
How many atoms are in 0.551 g of potassium (K) ?
Do You Understand Molar Mass?
How many atoms are in 0.551 g of potassium (K) ?
1 mol K = 39.10 g K
1 mol K = 6.022 x 1023 atoms K
1 mol K
6.022 x 1023 atoms K
0.551 g K x
x
=
1 mol K
39.10 g K
8.49 x 1021 atoms K
3.2
Practice Problem:
(a) How many moles of magnesium (Mg) are
in 88.8 g of Mg?
(b) How many atoms of Mg are contained in
this many moles of Mg?
How many H2O molecules are in a 9.00-g
sample of water?
a. 0.500
b. 3.01 × 1023
c. 2.71 × 1024
d. 1.08 × 1023
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How many H2O molecules are in a 9.00-g
sample of water?
a. 0.500
b. 3.01 × 1023
c. 2.71 × 1024
d. 1.08 × 1023
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What number would you use to convert (a) moles
of CH4 to grams of CH4 grams and (b) number of
molecules of CH4 to moles of CH4?
(a)
a.
b.
c.
d.
Avogadro’s number, 6.02 × 1023 particles/mol
Inverse of molar mass of CH4, 1 mol CH4/16.0 g CH4
Molar mass of CH4, 16.0 g CH4/1 mol CH4
Formula weight of CH4, 16.0 amu
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What number would you use to convert (a) moles
of CH4 to grams of CH4 grams and (b) number of
molecules of CH4 to moles of CH4?
(a)
a.
b.
c.
d.
Avogadro’s number, 6.02 × 1023 particles/mol
Inverse of molar mass of CH4, 1 mol CH4/16.0 g CH4
Molar mass of CH4, 16.0 g CH4/1 mol CH4
Formula weight of CH4, 16.0 amu
© 2015 Pearson Education, Inc.
What number would you use to convert (a) moles
of CH4 to grams of CH4 grams and (b) number of
molecules of CH4 to moles of CH4?
(b)
a. Inverse of Avogadro’s number, 1 mol/6.02 × 1023
molecules
b. Inverse of molar mass of CH4, 1 mol CH4/16.0 g CH4
c. Molar mass of CH4, 16.0 g CH4/1 mol CH4
d. Formula weight of CH4, 16.0 amu
© 2015 Pearson Education, Inc.
What number would you use to convert (a) moles
of CH4 to grams of CH4 grams and (b) number of
molecules of CH4 to moles of CH4?
(b)
a. Inverse of Avogadro’s number, 1 mol/6.02 × 1023
molecules
b. Inverse of molar mass of CH4, 1 mol CH4/16.0 g CH4
c. Molar mass of CH4, 16.0 g CH4/1 mol CH4
d. Formula weight of CH4, 16.0 amu
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Hwk: page 112-120:
23, 27, 29, 35, 37, 39, 44
Quiz to follow
Atoms,
Molecules,
and Ions
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LESSON 3
3-5 Empirical Formulas,
Molecular Formulas,
Combustion Analysis
Atoms,
Molecules,
and Ions
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Determining Empirical Formulas
One can determine the empirical formula
from the percent composition by following
these three steps.
Stoichiometry
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Determining Empirical Formulas—
an Example
The compound para-aminobenzoic acid (you may have
seen it listed as PABA on your bottle of sunscreen) is
composed of carbon (61.31%), hydrogen (5.14%),
nitrogen (10.21%), and oxygen (23.33%). Find the
empirical formula of PABA.
Stoichiometry
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Determining Empirical Formulas—
an Example
Assuming 100.00 g of para-aminobenzoic acid,
C:
H:
N:
O:
1 mol
12.01 g
1 mol
5.14 g ×
1.01 g
1 mol
10.21 g ×
14.01 g
1 mol
23.33 g ×
16.00 g
61.31 g ×
= 5.105 mol C
= 5.09 mol H
= 0.7288 mol N
= 1.456 mol O
Stoichiometry
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Determining Empirical Formulas—
an Example
Calculate the mole ratio by dividing by the smallest number
of moles:
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C:
5.105 mol
0.7288 mol
= 7.005 ≈ 7
H:
5.09 mol
0.7288 mol
= 6.984 ≈ 7
N:
0.7288 mol
0.7288 mol
= 1.000
O:
1.458 mol
0.7288 mol
= 2.001 ≈ 2
Stoichiometry
Determining Empirical Formulas—
an Example
These are the subscripts for the empirical formula:
C7H7NO2
Stoichiometry
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Using percent composition in reverse:
• Can determine the empirical formula of a compound
Example: Ascorbic acid (a.k.a. vitamin C)
Elemental Composition
(mass %):
C
40.92 %
H
4.58 %
O
54.50 %
Goal: Determine the empirical formula of ascorbic acid.
Example: Ascorbic acid (a.k.a. vitamin C)
Step 1: Convert mass percentages to grams of each element.
If a mass of the compound is given, multiply that mass by the
fraction (mass % / 100) of each element to get grams of each
element. If no mass of compound is given, assume its mass
to be 100 g.
Mass of C = 40.92 g C
Mass of H = 4.58 g H
Mass of O = 54.50 g O
Example: Ascorbic acid (a.k.a. vitamin C)
Step 2: Calculate the number of moles of each element
in the compound.
How to do this? Use the molar mass of each element as
follows:
1 mole C
Moles of C = 40.92 g C x ---------------- = 3.407 mole C
12.01 g C
1 mole H
Moles of H = 4.58 g H x ---------------- = 4.54 mole H
1.008 g H
1 mole O
Moles of O = 54.50g O x ---------------- = 3.406 mole O
16.00 g O
Example: Ascorbic acid (a.k.a. vitamin C)
Step 3: Calculate the mole ratio of each element in the
compound to the element with the smallest (or smaller, if 2)
number of moles.
Which element has the smallest number of moles?
oxygen
Thus we obtain the empirical formula
3.407
C: ------------- = 1.000  1
3.406
4.54
H: ----------- = 1.333  1.33
3.406
3.406
O: ------------- = 1.000  1
3.406
CH1.33O
Is this formula
our final result?
Example: Ascorbic acid (a.k.a. vitamin C)
Step 4: By trial and error, convert fractional subscripts in the
formula into whole numbers (integers).
This means changing 1.33 into an integer:
1.33 x 2 = 2.66
1.33 x 3 = 3.99  4
Multiply each subscript
by 3:
Thus the empirical formula
for ascorbic acid is:
C3H4O3
Determining a Molecular Formula
• Remember, the number of atoms in a
molecular formula is a multiple of the
number of atoms in an empirical
formula.
• If we find the empirical formula and
know a molar mass (molecular weight)
for the compound, we can find the
molecular formula.
Stoichiometry
© 2015 Pearson Education, Inc.
Determining a Molecular Formula—
an Example
• The empirical formula of a compound
was found to be CH. It has a molar
mass of 78 g/mol. What is its molecular
formula?
• Solution:
Whole-number multiple = 78/13 = 6
The molecular formula is C6H6.
Stoichiometry
© 2015 Pearson Education, Inc.
Molecular Formula
Calculation of the Molecular Formula
A compound has an empirical formula
of NO2. The colourless liquid, used in
rocket engines has a molar mass of
92.0 g/mole. What is the molecular
formula of this substance?
empirical formula mass: 14.01+2
(16.00) = 46.01 g/mol
n = molar mass
= 92.0 g/mol
emp. f. mass
46.01 g/mol
n = 2
2(NO2) = N2O4
Molecular Formulas from Empirical Formulas:
To get the actual, molecular formula of the compound, we must know:
The approximate molar mass of the compound.
The empirical formula of the compound.
Molecular Formulas from Empirical Formulas:
Example: A sample of a compound containing boron (B) and hydrogen (H)
was analyzed and found to contain 6.4442 g B and 1.8031 g H.
The molar mass of the compound is about 30 g. What is the
molecular formula of this compound?
1 mole B
Moles of B = 6.4442 g B x ---------------- = 0.59608 mole B
10.811 g B
0.59608
B: ------------- = 1.000  1
0.59608
1 mole H
Moles of H = 1.8031 g H x ---------------- = 1.7888 mole H
1.008 g H
1.7888
H: ----------- = 3.0009  3
0.59608
Molecular Formulas from Empirical Formulas:
Example: A sample of a compound containing boron (B) and hydrogen (H)
was analyzed and found to contain 6.4442 g B and 1.8031 g H.
The molar mass of the compound is about 30 g. What is the
molecular formula of this compound?
The empirical molar mass is:
The empirical formula is:
BH3
molar mass
30 g
---------------------------- = ------------ = 2.2  2
empirical molar mass
13.835 g
(10.811 g) + (3)(1.008 g) =
13.835 g
The molecular formula is:
(2)(BH3)  B2H6
Cyclohexane, a commonly used organic
solvent is 85.6% C and 14.4% H by mass
with a molar mass of 84.2 g/mol? What is
the molecular formula?
Stoichiometry
© 2015 Pearson Education, Inc.
Combustion Analysis
• Compounds containing C, H, and O are routinely analyzed
through combustion in a chamber like the one shown in
Figure 3.14.
– C is determined from the mass of CO2 produced.
– H is determined from the mass of H2O produced.
– O is determined by the difference after C and H have been determined.
Stoichiometry
© 2015 Pearson Education, Inc.
Experimental Determination of Empirical Formulas:
O2
Unused O2
Ethanol
Heat
Elemental analyzer for determination
of carbon and hydrogen
H2O
absorber
CO2
absorber
(From Fig. 3.5, p. 81, Chang, 7th ed.)
• Combustion of sample in presence of O2 (Ethanol + O2  CO2 + H2O)
• H2O and CO2 absorber tubes weighed prior to combustion, then weighed
again after combustion is complete.
• Masses of H2O and CO2 obtained by difference, with masses of H and C
in original sample obtained by calculation (see next slide).
Experimental Determination of Empirical Formulas:
O2
Unused O2
Ethanol
Heat
Elemental analyzer for determination
of carbon and hydrogen
H2O
absorber
CO2
absorber
(From Fig. 3.5, p. 81, Chang, 7th ed.)
Consider the following elemental analysis:
An 11.5 g sample of absolute ethanol is combusted in an apparatus
such as the one shown above. Combustion of the 11.5 g ethanol
sample produced 22.0 g of CO2 and 13.5 g of H2O.
Find: (a) The masses of C, H, and O in the sample
(b) The empirical formula for ethanol
Experimental Determination of Empirical Formulas:
O2
Unused O2
Ethanol
Heat
Elemental analyzer for determination
of carbon and hydrogen
H2O
absorber
CO2
absorber
(From Fig. 3.5, p. 81, Chang, 7th ed.)
1 mole CO2
1 mole C
12.01 g C
mass of C = 22.0 g CO2 x ------------------- x ----------------- x ---------------- = 6.00 g C
44.01 g CO2
1 mole CO2
1 mole C
1 mole H2O
2 moles H
1.008 g H
mass of H = 13.5 g H2O x ------------------- x ----------------- x ---------------- = 1.51 g H
18.02 g H2O
1 mole H2O
1 mole H
Experimental Determination of Empirical Formulas:
O2
Unused O2
Ethanol
Heat
Elemental analyzer for determination
of carbon and hydrogen
H2O
absorber
CO2
absorber
(From Fig. 3.5, p. 81, Chang, 7th ed.)
mass of O = 11.5 g ethanol - (6.00 g C + 1.51 g H) = 4.0 g O (obtained by difference)
6.00 g C
1.51 g H
4.0 g O
Calculate moles of C, H, O…
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
g CO2
mol CO2
mol C
gC
6.0 g C = 0.5 mol C
g H2O
mol H2O
mol H
gH
1.5 g H = 1.5 mol H
g of O = g of sample – (g of C + g of H)
4.0 g O = 0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
Isopropyl alcohol, sold as rubbing alcohol,
is composed of C, H, and O. Combustion
of 0.255g of isopropyl alcohol produces
0.561g of CO2 and 0.306g of H2O.
Determine the empirical formula.
Stoichiometry
© 2015 Pearson Education, Inc.
Caproic acid, responsible for the odor of dirty socks, is
composed of C, H, and O atoms. Combustion of a
0.255g sample of this compound produces 0.512g CO2
and 0.209g of H2O. What is the empirical formula? If
caproic acid has a molar mass of 116 g/mol, what is the
molecular formula?
Stoichiometry
© 2015 Pearson Education, Inc.
Hwk: page 112-120:
45, 49, 51, 53, 55, 59
QUIZ
Atoms,
Molecules,
and Ions
© 2015 Pearson Education, Inc.
LESSON 4
3-6 Stoichiometric
Problems & 3-7 Limiting
Reactants, Theoretical
Yields, Percent Yields
Atoms,
Molecules,
and Ions
© 2015 Pearson Education, Inc.
Quantitative Relationships
• The coefficients in the balanced equation show
 relative numbers of molecules of reactants and
products.
 relative numbers of moles of reactants and
products, which can be converted to mass. Stoichiometry
© 2015 Pearson Education, Inc.
Stoichiometry: Amounts of Reactants and Products
Stoichiometry is the quantitative study of reactants and
products in a chemical reaction.
A + 2B  3C
“How much product will I get from specific amounts of starting materials
(reactants)?”
“How much starting material will I need to obtain a specific amount of
product?”
Stoichiometric Calculations
We have already seen in this chapter how to
convert from grams to moles or moles to
grams. The NEW calculation is how to
compare two DIFFERENT materials, using
the MOLE RATIO from the balanced
equation!
Stoichiometry
© 2015 Pearson Education, Inc.
Mole method:
Stoichiometric coefficients in a chemical reaction can be
interpreted as the number of MOLES of each substance.
2 H2 (g) + O2 (g)  2 H2O (l)
“Two moles of hydrogen gas combine with one mole of oxygen gas to form two moles
of liquid water.”
FIVE STEPS TO USING MOLE METHOD:
1. Write correct formulas for all reactants and products, and balance the
resulting equation.
2. Convert the quantities of some or all known substances (usually
reactants) into moles.
3. Use the coefficients in the balanced equation to calculate the number of
moles of the unknown quantities (usually products) in the problem.
4. Using the calculated numbers of moles and the molar masses, convert
the unknown quantities to the required units (usually grams).
5. Check that your answer is reasonable in physical terms.
Moles of
known
Moles of
unknown
Mass of
known
Moles of
known
Moles of
unknown
Mass of
known
Moles of
known
Moles of
unknown
Mass of
unknown
Three types of stoichiometric calculations based on the mole method
Moles of
known
Moles of
unknown
Consider the reaction between methane gas and oxygen gas.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
How many moles of oxygen are required to react with 3.87 mol
of CH4?
Moles of
known
Moles of
unknown
How many moles of hydrogen gas are required to produce 1.76 mol
of iron in the reaction
Fe2O3(s) + 3H2(g)  2Fe(s) + 3H2O(l)
Mass of
known
Moles of
known
Moles of
unknown
Moles of
known
Moles of
unknown
Mass of
unknown
Consider the reaction between potassium and iodine to form
potassium iodide. How many grams of iodine are necessary to
produce 6.97 mol of product?
Mass of
known
Moles of
unknown
Moles of
unknown
Moles of
known
Moles of
unknown
Mass of
unknown
The reaction between dinitrogen pentoxide gas and liquid water yields
an aqueous solution of nitric acid, HNO3(aq). How many grams of nitric
acid would be made from 0.874 mol of dinitrogen pentoxide gas?
How many milliliters of water are needed to react with 0.874 mol of
dinitrogen pentoxide? (Density of water is 1.00 g/mL.)
Stoichiometry: An Example Problem
A well-known precipitation reaction used as a qualitative spot
test for lead ions involves the reaction of lead (II) nitrate with
potassium iodide to produce yellow, insoluble lead (II) iodide
and colorless, soluble potassium nitrate:
Pb(NO3)2 (aq) + KI (aq)  PbI2 (s) + KNO3 (aq)
How many grams of PbI2 will be produced by complete reaction
of excess aqueous KI with 0.8113 g of Pb(NO3)2?
A well-known precipitation reaction used as a qualitative spot
test for lead ions involves the reaction of lead (II) nitrate with
potassium iodide to produce yellow, insoluble lead (II) iodide
and colorless, soluble potassium nitrate:
Pb(NO3)2 (aq) + KI (aq)  PbI2 (s) + KNO3 (aq)
How many grams of PbI2 will be produced by complete reaction
of excess aqueous KI with 0.8113 g of Pb(NO3)2?
Step 1: Balance the chemical equation given in the problem.
Pb(NO3)2 (aq)
Reactants
1 Pb
1K
1I
2N
6O
+ 2 KI (aq)
Products
1 Pb
1K
2I
1N
3O

PbI2 (s)
+ 2 KNO3 (aq)
Pb and K appear to be balanced.
I, N, O are not balanced….
How to balance them?
A well-known precipitation reaction used as a qualitative spot
test for lead ions involves the reaction of lead (II) nitrate with
potassium iodide to produce yellow, insoluble lead (II) iodide
and colorless, soluble potassium nitrate:
Pb(NO3)2 (aq) + KI (aq)  PbI2 (s) + KNO3 (aq)
How many grams of PbI2 will be produced by complete reaction
of excess aqueous KI with 0.8113 g of Pb(NO3)2?
Step 2: We need to convert 0.8113 g Pb(NO3)2 to moles of Pb(NO3)2.
1 mole Pb(NO3)2
Moles of Pb(NO3)2 = 0.8113 g Pb(NO3)2 x -------------------------331.21 g Pb(NO3)2
= 2.450 x 10-3 mole Pb(NO3)2
A well-known precipitation reaction used as a qualitative spot
test for lead ions involves the reaction of lead (II) nitrate with
potassium iodide to produce yellow, insoluble lead (II) iodide
and colorless, soluble potassium nitrate:
Pb(NO3)2 (aq) + KI (aq)  PbI2 (s) + KNO3 (aq)
How many grams of PbI2 will be produced by complete reaction
of excess aqueous KI with 0.8113 g of Pb(NO3)2?
Step 3: From the stoichiometry indicated by the balanced equation, we
know that 1 mole of Pb(NO3)2 = 1 mole of PbI2. Thus, we calculate moles of
PbI2 produced:
1 mole PbI2
Moles of PbI2 = 2.450 x 10-3 mole Pb(NO3)2 x ---------------------1 mole Pb(NO3)2
= 2.450 x 10-3 mole PbI2
A well-known precipitation reaction used as a qualitative spot
test for lead ions involves the reaction of lead (II) nitrate with
potassium iodide to produce yellow, insoluble lead (II) iodide
and colorless, soluble potassium nitrate:
Pb(NO3)2 (aq) + KI (aq)  PbI2 (s) + KNO3 (aq)
How many grams of PbI2 will be produced by complete reaction
of excess aqueous KI with 0.8113 g of Pb(NO3)2?
Step 4: Using the molar mass of PbI2, calculate the mass of PbI2 produced:
460.99 g PbI2
Grams of PbI2 = 2.450 x 10-3 mole PbI2 x ---------------------1 mole PbI2
= 1.129 g PbI2
Step 5: Does this result make sense?
An Example of a Stoichiometric Calculation
• How many grams of water can be
produced from 1.00 g of glucose?
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
• There is 1.00 g of glucose to start.
• The first step is to convert it to moles.
Stoichiometry
© 2015 Pearson Education, Inc.
An Example of a Stoichiometric Calculation
• The NEW calculation is to convert
moles of one substance in the equation
to moles of another substance.
• The MOLE RATIO comes from the
balanced equation.
Stoichiometry
© 2015 Pearson Education, Inc.
An Example of a Stoichiometric Calculation
Stoichiometry
© 2015 Pearson Education, Inc.
Mass of
known
Moles of
known
Moles of
unknown
Mass of
unknown
How many grams of KI are produced by reacting 6.029 g of K?
2K(s) + I2(s)  2KI(s)
Mass of
known
Moles of
known
Moles of
unknown
Mass of
unknown
Methanol burns in air according to the equation
2CH3OH + 3O2
2CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
Hwk: page 112-120:
61, 63, 65, 67, 69
QUIZ
Atoms,
Molecules,
and Ions
© 2015 Pearson Education, Inc.
LESSON 4 cont
3-7 Limiting Reactants,
Theoretical Yields, Percent
Yields
Atoms,
Molecules,
and Ions
© 2015 Pearson Education, Inc.
Limiting Reactants
• The limiting reactant is the reactant present in
the smallest stoichiometric amount.
– In other words, it’s the reactant you’ll run out of first
(in this case, the H2).
Stoichiometry
© 2015 Pearson Education, Inc.
Limiting Reactants
In the example below, the O2 would be the
excess reagent.
Stoichiometry
© 2015 Pearson Education, Inc.
Limiting Reactants
• The limiting reactant is used in all stoichiometry
calculations to determine amounts of products
and amounts of any other reactant(s) used in a
reaction.
Stoichiometry
© 2015 Pearson Education, Inc.
Limiting Reactants:
A limiting reactant (or limiting reagent) is the reactant that is used up
first in a chemical reaction.
• The usual scenario: Reactants are not present in stoichiometric
amounts, or in the amounts indicated by the chemical equation.
• Consequence: One reactant is used up before the other(s).
• The limiting reactant controls how much product is generated by
the reaction, similar to the amount of food prepared for a party. If
too many people attend, the food runs out and folks go away hungry
(and, like snakes, hissed off!). The amount of food limits the number
of people who can eat, and there is thus an excess of people.
Limiting Reagents
66red
green
left used
over up
Limiting Reactants: How to Solve These Problems
Let’s do an example:
The reaction between aluminum and iron (III) oxide
generates temperatures approaching 3000oC and is
used in welding metals:
2 Al (s) + Fe2O3 (s)  Al2O3 (s) + 2 Fe (s)
In one process, 124 g of Al are reacted with 601 g of
Fe2O3. Calculate the grams of Al2O3 formed and how
much of the excess reagent is left at the end of the
reaction.
Limiting Reactants: How to Solve These Problems
Let’s do an example:
Step 1: We can’t tell
by inspection which is
the limiting reactant.
Thus, we must convert
grams of reactants to
available moles of
reactants:
The reaction between aluminum and iron (III) oxide
generates temperatures approaching 3000oC and is
used in welding metals:
2 Al (s) + Fe2O3 (s)  Al2O3 (s) + 2 Fe (s)
In one process, 124 g of Al are reacted with 601 g of
Fe2O3. Calculate the grams of Al2O3 formed and how
much of the excess reagent is left at the end of the
reaction.
Moles of Al = 124 g Al x (1 mole Al / 26.98 g Al) = 4.60 moles Al
Moles of Fe2O3 = 601 g Fe2O3 x (1 mole Fe2O3 / 159.7 g Fe2O3)
= 3.76 moles Fe2O3
These are the AVAILABLE moles of the reactants!
Limiting Reactants: How to Solve These Problems
Let’s do an example:
Step 2: We know from
the balanced equation
that 2 moles Al = 1
mole Fe2O3; thus, we
can calculate the
number of moles of Al
needed to react
stoichiometrically with
3.76 moles of Fe2O3:
The reaction between aluminum and iron (III) oxide
generates temperatures approaching 3000oC and is
used in welding metals:
2 Al (s) + Fe2O3 (s)  Al2O3 (s) + 2 Fe (s)
In one process, 124 g of Al are reacted with 601 g of
Fe2O3. Calculate the grams of Al2O3 formed and how
much of the excess reagent is left at the end of the
reaction.
Moles of Al needed to react stoichiometrically with 3.76 moles Fe2O3
= 3.76 moles Fe2O3 x (2 moles Al / 1 mole Fe2O3)
= 7.52 moles Al (REQUIRED moles of Al)
Limiting Reactants: How to Solve These Problems
Let’s do an example:
Step 2: We can also
calculate the number
of moles of Fe2O3
needed to react
stoichiometrically with
4. 60 moles of Al:
The reaction between aluminum and iron (III) oxide
generates temperatures approaching 3000oC and is
used in welding metals:
2 Al (s) + Fe2O3 (s)  Al2O3 (s) + 2 Fe (s)
In one process, 124 g of Al are reacted with 601 g of
Fe2O3. Calculate the grams of Al2O3 formed and how
much of the excess reagent is left at the end of the
reaction.
Moles of Fe2O3 needed to react stoichiometrically with 4.60 moles Al
= 4.60 moles Al x (1 mole Fe2O3 / 2 moles Al)
= 2.30 moles Fe2O3 (REQUIRED moles of Fe2O3)
Limiting Reactants: How to Solve These Problems
Let’s do an example:
Step 3: Compare
available moles of
reactant with required
moles of reactant for
each reactant. IF
available moles for a
reactant is LESS
THAN required moles
for that reactant, that
reactant is
LIMITING:
The reaction between aluminum and iron (III) oxide
generates temperatures approaching 3000oC and is
used in welding metals:
2 Al (s) + Fe2O3 (s)  Al2O3 (s) + 2 Fe (s)
In one process, 124 g of Al are reacted with 601 g of
Fe2O3. Calculate the grams of Al2O3 formed and how
much of the excess reagent is left at the end of the
reaction.
4.60 moles Al available < 7.60 moles of Al required (LIMITING)
3.76 moles Fe2O3 available > 2.30 moles Fe2O3 required (EXCESS)
Al is the limiting reactant!!!
Fe2O3 is in excess!!!
Limiting Reactants: How to Solve These Problems
Let’s do an example:
Step 4: The amount of
product formed is
based on the available
amount of limiting
reactant. Calculate
the grams of desired
product formed:
The reaction between aluminum and iron (III) oxide
generates temperatures approaching 3000oC and is
used in welding metals:
2 Al (s) + Fe2O3 (s)  Al2O3 (s) + 2 Fe (s)
In one process, 124 g of Al are reacted with 601 g of
Fe2O3. Calculate the grams of Al2O3 formed and how
much of the excess reagent is left at the end of the
reaction.
2 moles of Al = 1 mole of Al2O3, from the balanced equation.
Grams of Al2O3 formed =
4.60 moles Al x (1 mole Al2O3 / 2 moles Al) x (101.96 g Al2O3 / 1 mole Al2O3)
= 235 g Al2O3 formed
Limiting Reactants: How to Solve These Problems
Let’s do an example:
Step 5: Calculate the
amount of excess
reagent left:
The reaction between aluminum and iron (III) oxide
generates temperatures approaching 3000oC and is
used in welding metals:
2 Al (s) + Fe2O3 (s)  Al2O3 (s) + 2 Fe (s)
In one process, 124 g of Al are reacted with 601 g of
Fe2O3. Calculate the grams of Al2O3 formed and how
much of the excess reagent is left at the end of the
reaction.
2 moles of Al = 1 mole of Fe2O3, from the balanced equation.
Grams of Fe2O3 reacted =
4.60 moles Al x (1 mole Fe2O3 / 2 moles Al) x (159.7 g Fe2O3 / 1 mole Fe2O3)
= 367 g Fe2O3 formed
Grams of Fe2O3 left = 601 g - 367 g = 234 g Fe2O3 left
Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3
Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
2NH3(g) + H2S(g)  (NH4)2S(s)
If 6.837 g of NH3 reacts with 4.129 g of H2S, what is the
theoretical yield of (NH4)2S? What is the limiting reactant?
What is in excess and how many grams of that reagent are
left over after the reaction has been completed?
Theoretical Yield
• The theoretical yield is the maximum
amount of product that can be made.
– In other words, it’s the amount of product
possible as calculated through the
stoichiometry problem.
• This is different from the actual yield,
which is the amount one actually
produces and measures.
Stoichiometry
© 2015 Pearson Education, Inc.
Percent Yield
One finds the percent yield by
comparing the amount actually obtained
(actual yield) to the amount it was
possible to make (theoretical yield):
Percent yield =
actual yield
theoretical yield
× 100
Stoichiometry
© 2015 Pearson Education, Inc.
Reaction Yield:
actual yield
% yield = ----------------------- x 100
theoretical yield
Theoretical yield: the amount of product that would
result IF all the limiting reactant reacted
Actual yield: the amount of product actually obtained
from a reaction
A simple example:
In another thermite reaction, 367 g Al2O3
are formed from reaction of a given mass
of Al with a given mass of Fe2O3. What
would be the % yield of that reaction if we
obtained only 124 g of Al2O3?
Actual yield
Solution:
actual yield
124 g Al2O3
% yield = ----------------------- x 100 = ----------------- x 100 = 33.8% yield
theoretical yield
367 g Al2O3
Theoretical yield
When 1.417g of carbon reacts with excess hydrogen,
1.67 g of ethane gas (C2H6) are produced. What is the
theoretical yield? What is the percent yield?
Hwk: page 112-120:
73, 77, 79, 83, 85
Quiz to follow
Atoms,
Molecules,
and Ions
© 2015 Pearson Education, Inc.
Review Questions
Chapter 3
Stoichiometry
© 2015 Pearson Education, Inc.
For the reaction X  Y,
X is referred to as the
a.
b.
c.
d.
yield.
reactant.
product.
coefficient.
© 2015 Pearson Education, Inc.
For the reaction X  Y,
X is referred to as the
a.
b.
c.
d.
yield.
reactant.
product.
coefficient.
© 2015 Pearson Education, Inc.
Hydrocarbons burn to form
a.
b.
c.
d.
H2O and CO2.
charcoal.
methane.
O2 and H2O.
© 2015 Pearson Education, Inc.
Hydrocarbons burn to form
a.
b.
c.
d.
H2O and CO2.
charcoal.
methane.
O2 and H2O.
© 2015 Pearson Education, Inc.
C6H6 + O2  CO2 + H2O
When this equation is correctly
balanced, the coefficients are
a.
b.
c.
d.
1, 7  6, 3.
1, 8  6, 3.
2, 15  12, 6.
2, 16  12, 6.
© 2015 Pearson Education, Inc.
C6H6 + O2  CO2 + H2O
When this equation is correctly
balanced, the coefficients are
a.
b.
c.
d.
1, 7  6, 3.
1, 8  6, 3.
2, 15  12, 6.
2, 16  12, 6.
© 2015 Pearson Education, Inc.
2 NaN3  2 Na + 3 N2
This is an example of a
_______ reaction.
a.
b.
c.
d.
decomposition
combination
combustion
replacement
© 2015 Pearson Education, Inc.
2 NaN3  2 Na + 3 N2
This is an example of a
_______ reaction.
a.
b.
c.
d.
decomposition
combination
combustion
replacement
© 2015 Pearson Education, Inc.
The formula weight of any
substance is also known as
a.
b.
c.
d.
Avogadro’s number.
atomic weight.
density.
molar mass.
© 2015 Pearson Education, Inc.
The formula weight of any
substance is also known as
a.
b.
c.
d.
Avogadro’s number.
atomic weight.
density.
molar mass.
© 2015 Pearson Education, Inc.
The formula weight of Na3PO4
is _______ grams per mole.
a.
b.
c.
d.
70
164
265
116
© 2015 Pearson Education, Inc.
The formula weight of Na3PO4
is _______ grams per mole.
a.
b.
c.
d.
70
164
265
116
© 2015 Pearson Education, Inc.
The percentage by mass of
phosphorus in Na3PO4 is
a.
b.
c.
d.
44.0.
11.7.
26.7.
18.9.
© 2015 Pearson Education, Inc.
The percentage by mass of
phosphorus in Na3PO4 is
a.
b.
c.
d.
44.0.
11.7.
26.7.
18.9.
© 2015 Pearson Education, Inc.
One millionth of one mole of a
noble gas = _______ atoms.
a.
b.
c.
d.
6.02 × 1017
6.02 × 1020
6.02 × 1014
Atoms are too small to count.
© 2015 Pearson Education, Inc.
One millionth of one mole of a
noble gas = _______ atoms.
a.
b.
c.
d.
6.02 × 1017
6.02 × 1020
6.02 × 1014
Atoms are too small to count.
© 2015 Pearson Education, Inc.
Ethanol contains 52.2%
carbon, 13.0% hydrogen, and
34.8% oxygen by mass. The
empirical formula of ethanol is
a.
b.
c.
d.
C 2H 5O 2.
C2H6O.
C 2H 6O 2.
C 3H 4O 2.
© 2015 Pearson Education, Inc.
Ethanol contains 52.2%
carbon, 13.0% hydrogen, and
34.8% oxygen by mass. The
empirical formula of ethanol is
a.
b.
c.
d.
C 2H 5O 2.
C2H6O.
C 2H 6O 2.
C 3H 4O 2.
© 2015 Pearson Education, Inc.
Ribose has a molecular weight
of 150 grams per mole and the
empirical formula CH2O. The
molecular formula of ribose is
a.
b.
c.
d.
C 4H 8O 4.
C5H10O5.
C6H14O4.
C6H12O6.
© 2015 Pearson Education, Inc.
Ribose has a molecular weight
of 150 grams per mole and the
empirical formula CH2O. The
molecular formula of ribose is
a.
b.
c.
d.
C 4H 8O 4.
C5H10O5.
C6H14O4.
C6H12O6.
© 2015 Pearson Education, Inc.
When 3.14 g of Compound X is
completely combusted, 6.91 g
of CO2 and 2.26 g of H2O form.
The molecular formula of
Compound X is
a. C7H16.
c. C5H8O2.
© 2015 Pearson Education, Inc.
b. C6H12O.
d. C4H4O3.
When 3.14 g of Compound X is
completely combusted, 6.91 g
of CO2 and 2.26 g of H2O form.
The molecular formula of
Compound X is
a. C7H16.
c. C5H8O2.
© 2015 Pearson Education, Inc.
b. C6H12O.
d. C4H4O3.
C6H6 + 2 Br2  C6H4Br2 + 2 HBr
When 10.0 g of C6H6 and 30.0 g
of Br2 react as shown above, the
limiting reactant is
a.
b.
c.
d.
Br2.
C 6H 6.
HBr.
C6H4Br2.
© 2015 Pearson Education, Inc.
C6H6 + 2 Br2  C6H4Br2 + 2 HBr
When 10.0 g of C6H6 and 30.0 g
of Br2 react as shown above, the
limiting reactant is
a.
b.
c.
d.
Br2.
C 6H 6.
HBr.
C6H4Br2.
© 2015 Pearson Education, Inc.
2 Fe + 3 Cl2  2 FeCl3
When 10.0 g of iron and 20.0 g
of chlorine react as shown, the
theoretical yield of FeCl3 is
a.
b.
c.
d.
10.0 g.
20.0 g.
29.0 g.
30.0 g.
© 2015 Pearson Education, Inc.
2 Fe + 3 Cl2  2 FeCl3
When 10.0 g of iron and 20.0 g
of chlorine react as shown, the
theoretical yield of FeCl3 is
a.
b.
c.
d.
10.0 g.
20.0 g.
29.0 g.
30.0 g.
© 2015 Pearson Education, Inc.
The percentage yield of a
reaction is 100% × (Z), where
Z is
a.
b.
c.
d.
theoretical yield/actual yield.
calculated yield/actual yield.
calculated yield/theoretical yield.
actual yield/theoretical yield.
© 2015 Pearson Education, Inc.
The percentage yield of a
reaction is 100% × (Z), where
Z is
a.
b.
c.
d.
theoretical yield/actual yield.
calculated yield/actual yield.
calculated yield/theoretical yield.
actual yield/theoretical yield.
© 2015 Pearson Education, Inc.
C3H4O4 + 2 C2H6O 
C7H12O2 + 2 H2O
When 15.0 g of each reactant
was mixed, 15.0 g of C7H12O2
formed. The percentage yield of
this product is
a. 100%.
c. 65%.
© 2015 Pearson Education, Inc.
b. 75%.
d. 50%.
C3H4O4 + 2 C2H6O 
C7H12O2 + 2 H2O
When 15.0 g of each reactant
was mixed, 15.0 g of C7H12O2
formed. The percentage yield of
this product is
a. 100%.
c. 65%.
© 2015 Pearson Education, Inc.
b. 75%.
d. 50%.