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Chapter 4
Chemical
Quantities andAqueous
Reactions
2008, Prentice Hall
Chemistry: A Molecular Approach, 1stEd.
Nivaldo Tro
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
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Tro, Chemistry: A Molecular Approach 2
Reaction Stoichiometry
the numerical relationships between chemical amountsin a reaction is called stoichiometry the coefficients in a balanced chemical equation
specify the relative amounts in moles of each of the
substances involved in the reaction2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)
2 molecules of C8H18react with 25 molecules of O2
to form 16 molecules of CO2
and 18 molecules of H2
O
2 moles of C8H18react with 25 moles of O2to form 16 moles of CO2and 18 moles of H2O
2 mol C8H18: 25 mol O2: 16 mol CO2: 18 mol H2O
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Tro, Chemistry: A Molecular Approach 3
Predicting Amounts from Stoichiometry
the amounts of any other substance in a chemicalreaction can be determined from the amount ofjust one substance
How much CO2 can be made from 22.0 moles ofC8H18in the combustion of C8H18?
2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)
2 moles C8H18: 16 moles CO2
2188
2188 COmoles176
HCmol2
COmol16HCmoles22.0
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Tro, Chemistry: A Molecular Approach 4
ExampleEstimate the mass of CO2produced in
2004 by the combustion of 3.4 x 1015g gasoline
assuming that gasoline is octane, C8H18, the equationfor the reaction is:
2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)
the equation for the reaction gives the mole relationshipbetween amount of C8H18and CO2, but we need to
know the mass relationship, so the Concept Plan will
be:
g C8H18 mol CO2 g CO2mol C8H18
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ExampleEstimate the mass of CO2produced in
2004 by the combustion of 3.4 x 1015g gasoline
since 8x moles of CO2as C8H18, but the molar mass of C8H18is
3x CO2, the number makes sense
1 mol C8H18= 114.22g, 1 mol CO2= 44.01g, 2 mol C8H18= 16 mol CO2
3.4 x 10
15
g C8H18g CO2
Check:
Solution:
Concept Plan:
Relationships:
Given:Find:
g114.22
mol1
216
2
2
188
2
188
188188
15
COg101.0
COmol1
COg44.01
HCmol2
COmol16
HCg114.22
HCmol1HCg10.43
188
2
HCmol2
COmol61
g C8H18 mol CO2 g CO2mol C8H18
mol1
g44.01
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Tro, Chemistry: A Molecular Approach 6
Practice
According to the following equation, howmany milliliters of water are made in thecombustion of 9.0 g of glucose?
C6H
12O
6(s) + 6 O
2(g) 6 CO
2(g) + 6 H
2O(l)
1. convert 9.0 g of glucose into moles (MM 180)
2. convert moles of glucose into moles of water
3. convert moles of water into grams (MM 18.02)
4. convert grams of water into mLa) How? what is the relationship between mass and
volume?
density of water = 1.00 g/mL
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Tro, Chemistry: A Molecular Approach 7
Practice
OHmL5.4
OHg1.00
OHmL1x
OHmole1
OHg18.0x
OHCmole1
OHmole6x
g10x80.1
OHCmole1xOHCg0.9
2
2
2
2
2
6126
2
2
61266126
According to the following equation, how manymilliliters of water are made in the combustion of
9.0 g of glucose?
C6H12O6(s)+ 6 O2(g)6 CO2(g)+ 6 H2O(l)
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Tro, Chemistry: A Molecular Approach 8
Limiting Reactant for reactions with multiple reactants, it is likely that
one of the reactants will be completely used before theothers
when this reactant is used up, the reaction stops and nomore product is made
the reactant that limits the amount of product is calledthe limiting reactant sometimes called the limiting reagent
the limiting reactant gets completely consumed
reactants not completely consumed are called excessreactants the amount of product that can be made from the
limiting reactant is called the theoretical yield
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Tro, Chemistry: A Molecular Approach 9
Things Dont Always Go as Planned!
many things can happen during the course of anexperiment that cause the loss of product
the amount of product that is made in a reactionis called the actual yield
generally less than the theoretical yield, never more!
the efficiency of product recovery is generally
given as the percent yield
%100yieldltheoretica
yieldactualYieldPercent
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Tro, Chemistry: A Molecular Approach 10
Limiting and Excess Reactants in the
Combustion of MethaneCH4(g) + 2O2(g) CO2(g) + 2H2O(g)
Our balanced equation for the combustion of methane
implies that every 1 molecule of CH4reacts with 2molecules of O2
H
H
C
H
H +
O
O
C +
OO
OO
+
OH H
OH H
+
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11
Limiting and Excess Reactants in the
Combustion of Methane
If we have 5 molecules of CH4and 8 molecules
of O2, which is the limiting reactant?
H
HC
H
H
+
OO
OO
OO
OO
OO
OO
OO
OO
?H
HC
H
H
H
HC
H
H
H
HC
H
H H
HC
H
H
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
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12
Limiting and Excess Reactants in the
Combustion of Methane
H
HC
H
H
H
HC
H
H
+
OO
OO
OO
OO
OO
OO
OO
OO
H
HC
H
H
H
HCH
H
H
HC
H
H
24
24 COmolecules16
CHmolecules1
COmolecules2CHmolecules8
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
2
2
22 COmolecules10
Omolecules2
COmolecules2Omolecules10
since less CO2
can be made
from the O2thanthe CH4,the O2
is the limiting
reactant
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Example 4.4
Finding Limiting Reactant,
Theoretical Yield, andPercent Yield
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Tro, Chemistry: A Molecular Approach 14
Example:
When 28.6 kg of C are allowed to react with 88.2 kg ofTiO2in the reaction below, 42.8 kg of Ti are obtained.Find the Limiting Reactant, Theoretical Yield, and
Percent Yield.
(g)(s)(s)(s) CO2TiC2TiO 2
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Tro, Chemistry: A Molecular Approach 15
Example:
When 28.6 kg of C reacts with 88.2
kg of TiO2, 42.8 kg of Ti are
obtained. Find the Limiting
Reactant, Theoretical Yield, and
Percent Yield.
TiO2(s) + 2 C(s) Ti(s) + 2 CO(g)
Write down the given quantity and its units.Given: 28.6 kg C
88.2 kg TiO2
42.8 kg Ti produced
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Tro, Chemistry: A Molecular Approach 16
Write down the quantity to find and/or its units.Find: limiting reactant
theoretical yield
percent yield
Example:
Find the Limiting
Reactant, Theoretical
Yield, and Percent Yield.
TiO2(s) + 2 C(s)
Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
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17
Write a Concept Plan:
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: Lim. Rct., Theor. Yld., % Yld.
Example:
Find the Limiting
Reactant, Theoretical
Yield, and Percent Yield.
TiO2(s) + 2 C(s)
Ti(s) + 2 CO(g)
kgTiO2
kg
C
2
2
TiOg9.877
TiOmol1
Cg.0112
Cmol1
mol
C
molTiO2
mol
Ti
molTi
2TiOmol1
Timol1
Cmol2
Timol1 }smallest
amount is
from
limitingreactant
gTiO2
g
Ckg1
g1000
kg1
g1000
smallest
mol Tig Ti
Timol1
g87.47 % Yieldyieldtheor.
yieldact.
yield%
kg Ti
T.Y.g0001kg1
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18
Collect Needed Relationships:1000 g = 1 kg
Molar Mass TiO2= 79.87 g/mol
Molar Mass Ti = 47.87 g/mol
Molar Mass C = 12.01 g/mol
1 mole TiO2: 1 mol Ti (from the chem. equation)
2 mole C 1 mol Ti (from the chem. equation)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: Lim. Rct., Theor. Yld., % Yld.
CP: kg rct g rct mol rct mol Tipick smallest mol Ti TY kg Ti %Y Ti
Example:
Find the Limiting
Reactant, Theoretical
Yield, and Percent Yield.
TiO2(s) + 2 C(s)
Ti(s) + 2 CO(g)
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Tro, Chemistry: A Molecular Approach 19
Timol104301.1TiOmol1
Timol1TiOg79.87TiOmole1
kg1g0001TiOkg8.28 3
22
22
Apply the Concept Plan:
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg TiFind: Lim. Rct., Theor. Yld., % Yld.
CP: kg rct g rct mol rct mol Ti
pick smallest mol Ti TY kg Ti %Y TiRel: 1 mol C=12.01g; 1 mol Ti =47.87g;1 mol TiO2= 79.87g; 1000g = 1 kg;
1 mol TiO2: 1 mol Ti; 2 mol C : 1 mol Ti
Example:
Find the Limiting
Reactant, Theoretical
Yield, and Percent
Yield.
TiO2(s) + 2 C(s)
Ti(s) + 2 CO(g)
Timol100791.1Cmol2
Timol1
Cg12.01
Cmole1
kg1
g0001Ckg8.62 3
smallest moles of TiLimiting Reactant
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Tro, Chemistry: A Molecular Approach 20
Apply the Concept Plan:
Tikg52.9g1000
kg1
mol1
Tig47.87Timol104301.1 3
Theoretical Yield
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg TiFind: Lim. Rct., Theor. Yld., % Yld.
CP: kg rct g rct mol rct mol Ti
pick smallest mol Ti TY kg Ti %Y TiRel: 1 mol C=12.01g; 1 mol Ti =47.87g;1 mol TiO2= 79.87g; 1000g = 1 kg;
1 mol TiO2: 1 mol Ti; 2 mol C : 1 mol Ti
Example:
Find the Limiting
Reactant, Theoretical
Yield, and Percent
Yield.
TiO2(s) + 2 C(s)
Ti(s) + 2 CO(g)
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Tro, Chemistry: A Molecular Approach 21
Apply the Concept Plan:
YieldPercent100YieldlTheoretica
YieldActual %
%9.08%100Tikg52.9
Tikg42.8
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg TiFind: Lim. Rct., Theor. Yld., % Yld.
CP: kg rct g rct mol rct mol Ti
pick smallest mol Ti TY kg Ti %Y TiRel: 1 mol C=12.01g; 1 mol Ti =47.87g;1 mol TiO2= 79.87g; 1000g = 1 kg;
1 mol TiO2: 1 mol Ti; 2 mol C : 1 mol Ti
Example:
Find the Limiting
Reactant, Theoretical
Yield, and Percent
Yield.
TiO2(s) + 2 C(s)
Ti(s) + 2 CO(g)
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Tro, Chemistry: A Molecular Approach 22
Check the Solutions:
Limiting Reactant = TiO2
Theoretical Yield = 52.9 kg
Percent Yield = 80.9%
Since Ti has lower molar mass than TiO2, the T.Y. makes sense
The Percent Yield makes sense as it is less than 100%.
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg TiFind: Lim. Rct., Theor. Yld., % Yld.
CP: kg rct g rct mol rct mol Ti
pick smallest mol Ti TY kg Ti %Y TiRel: 1 mol C=12.01g; 1 mol Ti =47.87g;1 mol TiO2= 79.87g; 1000g = 1 kg;
1 mol TiO2: 1 mol Ti; 2 mol C : 1 mol Ti
Example:
Find the Limiting
Reactant, Theoretical
Yield, and Percent
Yield.
TiO2(s) + 2 C(s)
Ti(s) + 2 CO(g)
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Tro, Chemistry: A Molecular Approach 23
PracticeHow many grams of N2(g) can be made from
9.05 g of NH3reacting with 45.2 g of CuO?
2 NH3(g) + 3 CuO(s) N2(g) + 3 Cu(s) + 3 H2O(l)
Practice How many grams of N (g) can be made from 9 05 g of
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PracticeHow many grams of N2(g) can be made from 9.05 g of
NH3reacting with 45.2 g of CuO?
2 NH3(g) + 3 CuO(s) N2(g) + 3 Cu(s) + 3 H2O(l)
1 mol NH3= 17.03g, 1 mol CuO = 79.55g, 1 mol N2= 28.02 g
2 mol NH3= 1 mol N2, 3 mol CuO = 1 mol N2
9.05 g NH3
, 45.2 g CuO
g N2
Concept Plan:
Relationships:
Given:
Find:
g17.03
mol1
3
2
NHmol2
Nmol1
g NH3 mol N2mol NH3
g28.02
mol1
g CuO mol N2mol CuO
CuOmol3
Nmol1 2
g79.55
mol1
g N2smallest moles N2
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Tro, Chemistry: A Molecular Approach 26
Solutions
when table salt is mixed with water, it seems to disappear,or become a liquidthe mixture is homogeneous
the salt is still there, as you can tell from the taste, or simply
boiling away the water
homogeneous mixtures are called solutions the component of the solution that changes state is called
the solute
the component that keeps its state is called the solvent if both components start in the same state, the major component
is the solvent
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Tro, Chemistry: A Molecular Approach 27
Describing Solutions
since solutions are mixtures, the composition canvary from one sample to another
pure substances have constant composition
salt water samples from different seas or lakes havedifferent amounts of salt
so to describe solutions accurately, we must
describe how much of each component is presentwe saw that with pure substances, we can describethem with a single name because all samples identical
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Tro, Chemistry: A Molecular Approach 28
Solution Concentration qualitatively, solutions are often
described as dilute orconcentrated
dilute solutionshave a smallamount of solute compared tosolvent
concentrated solutionshave alarge amount of solutecompared to solvent
quantitatively, the relativeamount of solute in the solutionis called the concentration
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Tro, Chemistry: A Molecular Approach 29
Solution Concentration
Molarity moles of solute per 1 liter of solution
used because it describes how many moleculesof solute in each liter of solution
L)(insolutionofamount
moles)(insoluteofamount
Mmolarity,
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Tro, Chemistry: A Molecular Approach 30
Preparing 1 L of a 1.00 M NaCl Solution
Example 4 5 Find the molarity of a solution that
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Example 4.5Find the molarity of a solution that
has 25.5 g KBr dissolved in 1.75 L of solution
since most solutions are between 0 and
18 M, the answer makes sense
1 mol KBr = 119.00 g,
M = moles/L
25.5 g KBr, 1.75 L solution
Molarity, M
Check: Check
Solution: Follow theConcept Plan
to Solvetheproblem
Concept Plan:
Relationships:
Strategize
Given:
Find:
SortInformation
g119.00
mol1
M0.122L1.75
KBrmol29421.0solutionL
KBrmolesMmolarity,
KBrmol2940.21KBrg119.00
KBrmol1KBrg5.52
g KBr mol KBr
L soln
ML
molM
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Tro, Chemistry: A Molecular Approach 32
Using Molarity in Calculations
molarity shows the relationship between themoles of solute and liters of solution
If a sugar solution concentration is 2.0 M, then
1 liter of solution contains 2.0 moles of sugar2 liters = 4.0 moles sugar
0.5 liters = 1.0 mole sugar
1 L solution : 2 moles sugar
solutionL1
sugarmol2
sugarmol2
solutionL1
Example 4 6 How many liters of 0 125 M NaOH
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Example 4.6How many liters of 0.125 M NaOH
contains 0.255 mol NaOH?
since each L has only 0.125 mol NaOH,
it makes sense that 0.255 mol should
require a little more than 2 L
0.125 mol NaOH = 1 L solution
0.125 M NaOH, 0.255 mol NaOH
liters, L
Check: Check
Solution: Follow theConcept Plan
to Solvetheproblem
Concept Plan:
Relationships:
Strategize
Given:
Find:
SortInformation
NaOHmol0.125
solutionL1
solutionL2.04
NaOHmol0.125
solutionL1NaOHmol552.0
mol NaOH L soln
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Tro, Chemistry: A Molecular Approach 34
Dilution
often, solutions are stored as concentrated stocksolutions
to make solutions of lower concentrations from thesestock solutions, more solvent is added
the amount of solute doesnt change, just the volume ofsolution
moles solute in solution 1 = moles solute in solution 2
the concentrations and volumes of the stock and newsolutions are inversely proportional
M1V1= M2V2
Example 4 7 To what volume should you dilute
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Example 4.7To what volume should you dilute
0.200 L of 15.0 M NaOH to make 3.00 M NaOH?
since the solution is diluted by a factor
of 5, the volume should increase by a
factor of 5, and it does
M1V1= M2V2
V1= 0.200L, M1= 15.0 M, M2= 3.00 M
V2, L
Check: Check
Solution: Follow theConcept Plan
to Solvetheproblem
Concept Plan:
Relationships:
Strategize
Given:
Find:
SortInformation
2
2
11 V
M
VM
L1.00
Lmol3.00
L200.0L
mol15.0
V1, M1, M2 V2
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Tro, Chemistry: A Molecular Approach 36
Solution Stoichiometry
since molarity relates the moles of solute to theliters of solution, it can be used to convert
between amount of reactants and/or products ina chemical reaction
Example 4.8 What volume of 0.150 M KCl is required to
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Example 4.8 What volume of 0.150 M KCl is required to
completely react with 0.150 L of 0.175 M Pb(NO3)2 in the
reaction 2 KCl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2 KNO3(aq)
since need 2x moles of KCl as Pb(NO3)2,and
the molarity of Pb(NO3)2> KCl, the volume of
KCl should be more than 2x volume Pb(NO3)2
1 L Pb(NO3)2= 0.175 mol, 1 L KCl = 0.150 mol,
1 mol Pb(NO3)2= 2 mol KCl
0.150 M KCl, 0.150 L of 0.175 M Pb(NO3)2
L KCl
Check: Check
Solution: Follow theConcept
Plan to
Solvetheproblem
Concept Plan:
Relationships:
Strategize
Given:
Find:
SortInformation
23)Pb(NOL1
mol0.175
KClL.3500
mol.1500
KClL1
)Pb(NOmol1
KClmol2
)Pb(NOL1
mol0.175)Pb(NOL.1500
232323
23)Pb(NOmol1
KClmol2
L Pb(NO3)2 mol KCl L KClmol Pb(NO3)2
mol0.150
KClL1
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38
What Happens When a Solute Dissolves? there are attractive forces between the solute particles
holding them together there are also attractive forces between the solvent
molecules
when we mix the solute with the solvent, there are
attractive forces between the solute particles and thesolvent molecules
if the attractions between solute and solvent are strongenough, the solute will dissolve
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Tro, Chemistry: A Molecular Approach 39
Table Salt Dissolving in WaterEach ion is attracted
to the surroundingwater molecules and
pulled off and away
from the crystal
When it enters thesolution, the ion is
surrounded by water
molecules, insulating
it from other ionsThe result is a solution
with free moving
charged particles able
to conduct electricity
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Tro, Chemistry: A Molecular Approach 40
Electrolytes and Nonelectrolytes
materials that dissolvein water to form asolution that willconduct electricity are
called electrolytes
materials that dissolvein water to form a
solution that will notconduct electricity arecalled nonelectrolytes
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Tro, Chemistry: A Molecular Approach 41
Molecular View of
Electrolytes and Nonelectrolytes
in order to conduct electricity, a material must havecharged particles that are able to flow
electrolyte solutions all contain ions dissolved in the
water ionic compounds are electrolytes because they all dissociate
into their ions when they dissolve
nonelectrolyte solutions contain whole molecules
dissolved in the watergenerally, molecular compounds do not ionize when they
dissolve in water
the notable exception being molecular acids
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Tro, Chemistry: A Molecular Approach 42
Salt vs. Sugar Dissolved in Water
ionic compounds dissociate
into ions when they dissolve
molecular compounds do not
dissociate when they dissolve
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Tro, Chemistry: A Molecular Approach 43
Acids
acids are molecular compounds that ionizewhen theydissolve in water the molecules are pulled apart by their attraction for the water
when acids ionize, they form H+cations and anions
the percentage of molecules that ionize varies from oneacid to another acids that ionize virtually 100% are called strong acids
HCl(aq) H+(aq) + Cl-(aq)
acids that only ionize a small percentage are calledweak acids
HF(aq) H+(aq) + F-(aq)
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Tro, Chemistry: A Molecular Approach 44
Strong and Weak Electrolytes strong electrolytesare materials that dissolve
completely as ions ionic compounds and strong acids
their solutions conduct electricity well
weak electrolytesare materials that dissolve mostly as
molecules, but partially as ionsweak acids
their solutions conduct electricity, but not well
when compounds containing a polyatomic ion dissolve,
the polyatomic ion stays togetherNa2SO4(aq) 2 Na+(aq) + SO4
2-(aq)
HC2H3O2(aq) H+(aq) + C2H3O2
-(aq)
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Tro, Chemistry: A Molecular Approach 45
Classes of Dissolved Materials
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Tro, Chemistry: A Molecular Approach 46
Solubility of Ionic Compounds
some ionic compounds, like NaCl, dissolve very well inwater at room temperature
other ionic compounds, like AgCl, dissolve hardly at allin water at room temperature
compounds that dissolve in a solvent are said to besoluble, while those that do not are said to be insoluble
NaCl is soluble in water, AgCl is insoluble in water
the degree of solubility depends on the temperature even insoluble compounds dissolve, just not enough to be
meaningful
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Tro, Chemistry: A Molecular Approach 47
When Will a Salt Dissolve?
Predicting whether a compound will dissolve inwater is not easy
The best way to do it is to do some experimentsto test whether a compound will dissolve in
water, then develop some rules based on those
experimental resultswe call this method the empirical method
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Tro, Chemistry: A Molecular Approach 48
Compounds Containing the
Following Ions are Generally
Soluble
Exceptions
(when combined with ions on the
left the compound is insoluble)
Li+, Na+, K+, NH4+ none
NO3, C2H3O2
none
Cl
, Br
, I
Ag+
, Hg22+
, Pb2+
SO4
2 Ag+, Ca2+, Sr2+, Ba2+, Pb2+
Solubility Rules
Compounds that Are Generally Soluble in Water
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Tro, Chemistry: A Molecular Approach 49
Compounds Containing the
Following Ions are Generally
Insoluble
Exceptions
(when combined with ions on the
left the compound is soluble or
slightly soluble)
OH Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
S2 Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
CO32, PO4
3 Li+, Na+, K+, NH4+
Solubility Rules
Compounds that Are Generally Insoluble
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Tro, Chemistry: A Molecular Approach 50
Precipitation Reactions
reactions between aqueous solutions of ioniccompounds that produce an ionic compound
that is insoluble in water are called
precipitation reactionsand the insoluble
product is called a precipitate
2 KI(aq) + Pb(NO ) (aq) PbI (s) + 2
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51
2 KI(aq)+ Pb(NO3)2(aq)PbI2(s) + 2
KNO3(aq)
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Tro, Chemistry: A Molecular Approach 52
No Precipitate Formation =
No ReactionKI(aq) + NaCl(aq) KCl(aq) + NaI(aq)
all ions still present, no reaction
Process for Predicting the Products of
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Tro, Chemistry: A Molecular Approach 53
Process for Predicting the Products of
a Precipitation Reaction
1. Determine what ions each aqueous reactant has2. Determine formulas of possible products Exchange ions
(+) ion from one reactant with (-) ion from other
Balance charges of combined ions to get formula of eachproduct
3. Determine Solubility of Each Product in Water Use the solubility rules
If product is insoluble or slightly soluble, it will precipitate
4. If neither product will precipitate, write no reactionafter the arrow
Process for Predicting the Products of
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Process for Predicting the Products of
a Precipitation Reaction
5. If either product is insoluble, write the formulasfor the products after the arrowwriting (s)after the product that is insoluble and will
precipitate, and (aq) after products that are
soluble and will not precipitate
6. Balance the equation
Example 4.10Write the equation for the
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p q
precipitation reaction between an aqueous solution
of potassium carbonate and an aqueous solution of
nickel(II) chloride1. Write the formulas of the reactants
K2CO3(aq) + NiCl2(aq)
2. Determine the possible productsa) Determine the ions present
(K++ CO32-) + (Ni2++ Cl-)
b) Exchange the Ions
(K++ CO32-) + (Ni2++ Cl-) (K++ Cl-) + (Ni2++ CO32-)c) Write the formulas of the products
cross charges and reduce
K2CO3(aq) + NiCl2(aq) KCl + NiCO3
Example 4 10 Write the equation for the
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Tro, Chemistry: A Molecular Approach 56
3. Determine the solubility of each productKCl is soluble
NiCO3is insoluble
4. If both products soluble, write no reactiondoes not apply since NiCO3is insoluble
Example 4.10Write the equation for the
precipitation reaction between an aqueous solution
of potassium carbonate and an aqueous solution of
nickel(II) chloride
Example 4 10 Write the equation for the
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5. Write (aq) next to soluble products and (s) nextto insoluble products
K2CO3(aq) + NiCl2(aq) KCl(aq) + NiCO3(s)
6. Balance the Equation
K2CO3(aq) + NiCl2(aq) 2 KCl(aq) + NiCO3(s)
Example 4.10 Write the equation for the
precipitation reaction between an aqueous solution
of potassium carbonate and an aqueous solution of
nickel(II) chloride
I i E ti
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Ionic Equations equations which describe the chemicals put into the water
and the product molecules are called molecular equations2 KOH(aq) + Mg(NO3)2(aq) 2 KNO3(aq) + Mg(OH)2(s)
equations which describe the actual dissolved species are
called completeionic equations aqueous strong electrolytes are written as ions
soluble salts, strong acids, strong bases
insoluble substances, weak electrolytes, and nonelectrolytes
written in molecule formsolids, liquids, and gases are not dissolved, therefore molecule form
2K+1(aq) + 2OH-1
(aq)+ Mg+2
(aq)+ 2NO3-1
(aq) 2K+1
(aq)+ 2NO3-1
(aq)+ Mg(OH)2(s)
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Ionic Equations
ions that are both reactants and products are calledspectator ions
2K+1(aq) + 2OH-1
(aq)+ Mg+2
(aq)+ 2NO3-1
(aq) 2K+1
(aq)+ 2NO3-1
(aq)+ Mg(OH)2(s)
an ionic equation in which the spectator ions are
removed is called a net ionic equation
2OH-1(aq)+ Mg+2(aq) Mg(OH)2(s)
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Acid-Base Reactions
also called neutralization reactionsbecause theacid and base neutralize each others properties
2 HNO3
(aq) + Ca(OH)2
(aq) Ca(NO3
)2
(aq) + 2 H2
O(l)
the net ionic equation for an acid-base reaction is
H+(aq) + OH(aq) H2O(l)
as long as the salt that forms is soluble in water
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Acids and Bases in Solution acids ionize in water to form H+ionsmore precisely, the H from the acid molecule is donated to a
water molecule to form hydronium ion, H3O+
most chemists use H+and H3O+interchangeably
bases dissociate in water to form OHions
bases, like NH3, that do not contain OH
ions, produce OH
bypulling H off water molecules
in the reaction of an acid with a base, the H+from theacid combines with the OHfrom the base to make water
the cation from the base combines with the anion fromthe acid to make the salt
acid + base salt + water
Common Acids
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Chemical Name Formula Uses Strength
Perchloric Acid HClO4 explosives, catalyst Strong
Nitric Acid HNO3 explosives, fertilizer, dye, glue Strong
Sulfuric Acid H2SO4explosives, fertilizer, dye, glue,
batteriesStrong
Hydrochloric Acid HClmetal cleaning, food prep, ore
refining, stomach acidStrong
Phosphoric Acid H3PO4 fertilizer, plastics & rubber,food preservationModerate
Chloric Acid HClO3 explosives Moderate
Acetic Acid HC2H3O2plastics & rubber, food
preservation, vinegarWeak
Hydrofluoric Acid HF metal cleaning, glass etching Weak
Carbonic Acid H2CO3 soda water Weak
Hypochlorous Acid HClO sanitizer Weak
Boric Acid H3BO3 eye wash Weak
Common Bases
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Common BasesChemical
NameFormula
Common
NameUses Strength
sodium
hydroxideNaOH
lye,
caustic soda
soap, plastic,
petrol refiningStrong
potassium
hydroxideKOH caustic potash
soap, cotton,
electroplatingStrong
calciumhydroxide
Ca(OH)2 slaked lime cement Strong
sodium
bicarbonateNaHCO3 baking soda cooking, antacid Weak
magnesium
hydroxide Mg(OH)2
milk of
magnesia antacid Weak
ammonium
hydroxide
NH4OH,
{NH3(aq)}
ammonia
water
detergent,
fertilizer,
explosives, fibers
Weak
HCl( ) + NaOH( ) NaCl( ) + H O(l)
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HCl(aq)+NaOH(aq) NaCl(aq) + H2O(l)
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Example - Write the molecular, ionic, and net-
ionic equation for the reaction of aqueous nitric
acid with aqueous calcium hydroxide1. Write the formulas of the reactants
HNO3(aq) + Ca(OH)2(aq)
2. Determine the possible productsa) Determine the ions present when each reactant dissociates(H++ NO3
-) + (Ca+2+ OH-)
b) Exchange the ions, H+1 combines with OH-1to make H2O(l)
(H
+
+ NO3-
) + (Ca
+2
+ OH
-
)
(Ca
+2
+ NO3-
) + H2O(l)c) Write the formula of the salt
cross the charges
(H++ NO3-) + (Ca+2+ OH-) Ca(NO3)2+ H2O(l)
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3. Determine the solubility of the saltCa(NO3)2is soluble
4. Write an (s) after the insoluble products and a(aq) after the soluble products
HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + H2O(l)
5. Balance the equation2 HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + 2 H2O(l)
Example - Write the molecular, ionic, and net-
ionic equation for the reaction of aqueous nitric
acid with aqueous calcium hydroxide
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Example - Write the molecular, ionic, and net-
ionic equation for the reaction of aqueous nitric
acid with aqueous calcium hydroxide6. Dissociate all aqueous strong electrolytes to
get complete ionic equation
not H2O
2 H+(aq) + 2 NO3-(aq) + Ca+2(aq) + 2 OH-(aq)
Ca+2(aq) + 2 NO3-(aq) + H2O(l)
7. Eliminate spectator ions to get net-ionic
equation2 H+1(aq) + 2 OH-1(aq) 2 H2O(l)
H+1(aq) + OH-1(aq) H2O(l)
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Titration often in the lab, a solutions concentration is
determined by reacting it with another materialand using stoichiometrythis process is calledtitration
in the titration, the unknown solution is addedto a known amount of another reactant untilthe reaction is just completed, at this point,called the endpoint, the reactants are in theirstoichiometric ratio
the unknown solution is added slowly from aninstrument called a burette
a long glass tube with precise volume markings thatallows small additions of solution
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Acid-Base Titrations the difficulty is determining when there has been just
enough titrant added to complete the reaction
the titrant is the solution in the burette
in acid-base titrations, because both the reactant and
product solutions are colorless, a chemical is added thatchanges color when the solution undergoes large
changes in acidity/alkalinity
the chemical is called an indicator
at the endpoint of an acid-base titration, the number ofmoles of H+equals the number of moles of OH
aka the equivalence point
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Titration
Ti i
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TitrationThe base solution is the
titrant in the burette.As the base is added to
the acid, the H+ reacts with
the OHto form water.
But there is still excessacid present so the color
does not change.
At the titrations endpoint,
just enough base has beenadded to neutralize all the
acid. At this point the
indicator changes color.
Example 4.14:
Th i i f 10 00 L f HCl
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The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M
NaOH solution to reach the endpoint. What is the concentration of
the unknown HCl solution?
Write down the given quantity and its units.Given: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
Information
Gi 10 00 L HCl
Example 4.14:
Th i i f 10 00 L f HCl
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Write down the quantity to find, and/or its units.
Find: concentration HCl, M
Given: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M
NaOH solution to reach the endpoint. What is the concentration of
the unknown HCl solution?
Information
Gi 10 00 L HCl
Example 4.14:
Th tit ti f 10 00 L f HCl
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Collect Needed Equations and Conversion Factors:
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
1 mole HCl = 1 mole NaOH
0.100 M NaOH 0.100 mol NaOH 1 L soln
Given: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
Find: M HCl
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M
NaOH solution to reach the endpoint. What is the concentration of
the unknown HCl solution?
solutionliters
solutemolesMolarity
Information
Gi 10 00 L HCl
Example 4.14:
Th tit ti f 10 00 L f HCl
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Write a Concept Plan:
mL
NaOH
L
NaOH
mol
NaOH
NaOHL1
NaOHmol1000.
mL1
L0010.
mol
HCl
NaOHmol1
HClmol1
Given: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
Find: M HCl
CF: 1 mol HCl = 1 mol NaOH0.100 mol NaOH = 1 L
M = mol/L
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M
NaOH solution to reach the endpoint. What is the concentration of
the unknown HCl solution?
mL
HCl
L
HCl
mL1
L0010. HClliters
HClmolesMolarity
Information
Given: 10 00 mL HClExample:
Th tit ti f 10 00 L f HCl
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Tro, Chemistry: A Molecular Approach 76
NaOHmole1
HClmol1
L1
NaOHmol1000
mL1
L0.001NaOHmL2.541
.
Apply the Solution Map:
= 1.25 x 10-3mol HCl
Given: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
Find: M HCl
CF: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH L NaOH
mol NaOH mol HCl;
mL HCl L HCl & mol M
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M
NaOH solution to reach the endpoint. What is the concentration of
the unknown HCl solution?
Information
Given: 10 00 mL HClExample:
Th tit ti f 10 00 L f HCl
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Tro, Chemistry: A Molecular Approach 77
HClL010000mL1
L0.001NaOHmL0.001 .
Apply the Concept Plan:
Given: 10.00 mL HCl
12.54 mL NaOH
Find: M HCl
CF: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH L NaOH
mol NaOH mol HCl;
mL HCl L HCl & mol M
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M
NaOH solution to reach the endpoint. What is the concentration of
the unknown HCl solution?
M1250HClL0.01000
HClmoles10x1.25Molarity
-3
.
Information
Given: 10 00 mL HClExample:
Th tit ti f 10 00 L f HCl
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Check the Solution:HCl solution = 0.125 M
The units of the answer, M, are correct.The magnitude of the answer makes sense since
the neutralization takes less HCl solution than
NaOH solution, so the HCl should be more concentrated.
Given: 10.00 mL HCl
12.54 mL NaOH
Find: M HCl
CF: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH L NaOH
mol NaOH mol HCl;
mL HCl L HCl & mol M
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M
NaOH solution to reach the endpoint. What is the concentration of
the unknown HCl solution?
G E l i R ti
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Gas Evolving Reactions Some reactions form a gas directly from the ion
exchangeK2S(aq) + H2SO4(aq) K2SO4(aq) + H2S(g)
Other reactions form a gas by the decomposition of one
of the ion exchange products into a gas and waterK2SO3(aq) + H2SO4(aq) K2SO4(aq) + H2SO3(aq)
H2SO3 H2O(l) + SO2(g)
N HCO ( ) HCl( ) N Cl( ) CO ( ) H O(l)
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NaHCO3(aq) + HCl(aq) NaCl(aq) + CO2(g) + H2O(l)
Compounds that Undergo
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Compounds that Undergo
Gas Evolving Reactions
Reactant
Type
Reacting
With
Ion
Exchange
Product
Decom-
pose?
Gas
Formed
Example
metalnS,metal HS
acid H2S no H2S K2S(aq) + 2HCl(aq) 2KCl(aq) + H2S(g)
metalnCO3,
metal HCO3
acid H2CO3 yes CO2 K2CO3(aq) + 2HCl(aq)
2KCl(aq) + CO2(g) + H2O(l)
metalnSO3
metalHSO3
acid H2SO3 yes SO2 K2SO3(aq) + 2HCl(aq)
2KCl(aq) + SO2(g) + H2O(l)
(NH4)nanion base NH4OH yes NH3 KOH(aq) + NH4Cl(aq)
KCl(aq) + NH3(g) + H2O(l)
Example 4 15 When an aqueous solution of
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Example 4.15 - When an aqueous solution of
sodium carbonate is added to an aqueous solution
of nitric acid, a gas evolves1. Write the formulas of the reactants
Na2CO3(aq) + HNO3(aq)
2. Determine the possible productsa) Determine the ions present when each reactant dissociates
(Na+1+ CO3-2) + (H+1+ NO3
-1)
b) Exchange the anions
(Na+1+ CO3-2) + (H+1+ NO3
-1) (Na+1+ NO3-1) + (H+1+ CO3
-2)
c) Write the formula of compounds
cross the charges
Na2CO3(aq) + HNO3(aq) NaNO3 + H2CO3
Example 4 15 When an aqueous solution of
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3. Check to see either product H2S -No
4. Check to see if either product decomposes
Yes H2CO3decomposes into CO2(g) + H2O(l)
Na2CO3(aq) + HNO3(aq) NaNO3 + CO2(g) + H2O(l)
Example 4.15 - When an aqueous solution of
sodium carbonate is added to an aqueous solution
of nitric acid, a gas evolves
Example 4 15 When an aqueous solution of
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Tro, Chemistry: A Molecular Approach 84
5. Determine the solubility of other productNaNO3is soluble
6. Write an (s) after the insoluble products and a(aq) after the soluble products
Na2CO3(aq) + 2 HNO3(aq) 2NaNO3(aq) + CO2(g) + H2O(l)
7. Balance the equationNa2CO3(aq) + 2 HNO3(aq) 2NaNO3 + CO2(g) + H2O(l)
Example 4.15 - When an aqueous solution of
sodium carbonate is added to an aqueous solution
of nitric acid, a gas evolves
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Other Patterns in Reactions
the precipitation, acid-base, and gas evolvingreactions all involved exchanging the ions inthe solution
other kinds of reactions involve transferringelectrons from one atom to anotherthese arecalled oxidation-reduction reactions
also known as redox reactions
many involve the reaction of a substance with O2(g)
4 Fe(s) + 3 O2(g) 2 Fe2O3(s)
Combustion as Redox
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2 H2(g) + O2(g)2 H2O(g)
Redox without Combustion
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2 Na(s)+ Cl2(g) 2 NaCl(s)
2 Na2 Na++ 2 e
Cl2+ 2 e2 Cl
R ti f M t l ith N t l
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Reactions of Metals with Nonmetals
consider the following reactions:4 Na(s) + O2(g) 2 Na2O(s)
2 Na(s) + Cl2(g) 2 NaCl(s)
the reaction involves a metal reacting with a nonmetal in addition, both reactions involve the conversion of
free elements into ions
4 Na(s) + O2(g) 2 Na+
2O(s)
2 Na(s) + Cl2(g) 2 Na+Cl(s)
O id i d R d i
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Oxidation and Reduction
in order to convert a free element into an ion, theatoms must gain or lose electrons
of course, if one atom loses electrons, another mustaccept them
reactions where electrons are transferred from oneatom to another are redox reactions atoms that lose electrons are being oxidized, atoms
that gain electrons are being reduced
2 Na(s) + Cl2(g) 2 Na+Cl(s)
Na Na++ 1 e oxidation
Cl2+ 2 e 2 Clreduction
Leo
Ger
El t B kk i
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Electron Bookkeeping for reactions that are not metal + nonmetal, or do
not involve O2, we need a method for determining
how the electrons are transferred
chemists assign a number to each element in a
reaction called an oxidation statethat allows themto determine the electron flow in the reaction
even though they look like them, oxidation states are
not ion charges!
oxidation states are imaginary charges assigned based on aset of rules
ion charges are real, measurable charges
Rules for Assigning Oxidation States
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Rules for Assigning Oxidation States
rules are in order of priority1. free elements have an oxidation state = 0 Na = 0 and Cl2= 0 in 2 Na(s) + Cl2(g)
2. monatomic ions have an oxidation state equalto their charge Na = +1 and Cl = -1 in NaCl
3. (a) the sum of the oxidation states of all theatoms in a compound is 0 Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0
Rules for Assigning Oxidation States
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Rules for Assigning Oxidation States
3. (b) the sum of the oxidation states of all the atoms ina polyatomic ion equals the charge on the ion N = +5 and O = -2 in NO3
, (+5) + 3(-2) = -1
4.(a) Group I metals have an oxidation state of +1 in alltheir compounds
Na = +1 in NaCl
4. (b) Group II metals have an oxidation state of +2 inall their compounds Mg = +2 in MgCl2
Rules for Assigning Oxidation States
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Rules for Assigning Oxidation States
5. in their compounds, nonmetals have oxidationstates according to the table below
nonmetals higher on the table take priority
Nonmetal Oxidation State Example
F -1 CF4
H +1 CH4
O -2 CO2
Group 7A -1 CCl4Group 6A -2 CS2
Group 5A -3 NH3
Practice Assign an Oxidation State to
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Practice Assign an Oxidation State to
Each Element in the following
Br2
K+
LiF
CO2
SO42-
Na2O2
Practice Assign an Oxidation State to
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Tro, Chemistry: A Molecular Approach 95
Practice Assign an Oxidation State to
Each Element in the following
Br2 Br = 0, (Rule 1)
K+ K = +1, (Rule 2)
LiF Li = +1, (Rule 4a) & F = -1, (Rule 5)
CO2 O = -2, (Rule 5) & C = +4, (Rule 3a)
SO42- O = -2, (Rule 5) & S = +6, (Rule 3b)
Na2O2 Na = +1, (Rule 4a) & O = -1, (Rule 3a)
Oxidation and Reduction
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Oxidation and Reduction
Another Definition oxidation occurs when an atoms oxidation stateincreases during a reaction
reduction occurs when an atoms oxidation statedecreases during a reaction
CH4 + 2 O2 CO2+ 2 H2O-4+1 0 +42 +1 -2
oxidation
reduction
OxidationReduction
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oxidation and reduction must occur simultaneously
if an atom loses electrons another atom must take them the reactant that reduces an element in another reactant
is called the reducing agent
the reducing agent contains the element that is oxidized
the reactant that oxidizes an element in another reactantis called the oxidizing agent
the oxidizing agent contains the element that is reduced
2 Na(s) + Cl2(g) 2 Na+Cl(s)Na is oxidized, Cl is reduced
Na is the reducing agent, Cl2is the oxidizing agent
Id tif th O idi i d R d i A t
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Identify the Oxidizing and Reducing Agents
in Each of the Following
3 H2S + 2 NO3 + 2 H+ 3S + 2 NO + 4 H2O
MnO2+ 4 HBr MnBr2+ Br2+ 2 H2O
Id tif th O idi i d R d i A t
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Identify the Oxidizing and Reducing Agents
in Each of the Following
3 H2S + 2 NO3 + 2 H+ 3S + 2 NO + 4 H2O
MnO2+ 4 HBr MnBr2+ Br2+ 2 H2O
+1 -2 +5 -2 +1 0 +2 -2 +1 -2
ox agred ag
+4 -2 +1 -1 +2 -1 0 +1 -2
oxidationreduction
oxidation
reduction
red agox ag
Combustion Reactions
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Combustion Reactions
Reactions in which O2(g) is areactant are calledcombustion reactions
Combustion reactions release
lots of energy Combustion reactions are a
subclass of oxidation-reduction reactions
2 C8H18(g) + 25 O2(g) 16 CO2(g) + 18 H2O(g)
Combustion Products
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Combustion Products
to predict the products of a combustionreaction, combine each element in the other
reactant with oxygen
Reactant Combustion Product
contains C CO2(g)
contains H H2O(g)
contains S SO2(g)
contains N NO(g) or NO2(g)
contains metal M2On(s)
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PracticeComplete the Reactions
combustion of C3H7OH(l)
combustion of CH3NH2(g)
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PracticeComplete the Reactions
C3H7OH(l) + 5 O2(g) 3 CO2(g) + 4 H2O(g)
CH3NH2(g) + 3 O2(g) CO2(g) + 2 H2O(g) + NO2(g)