Post on 24-Nov-2014
Chapter 4Chemical
Quantities and Aqueous Reactions
2008, Prentice Hall
Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro
Roy KennedyMassachusetts Bay Community
CollegeWellesley Hills, MA
Tro, Chemistry: A Molecular Approach 2
Reaction Stoichiometry• the numerical relationships between chemical amounts
in 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 reaction
2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)
2 molecules of C8H18 react with 25 molecules of O2
to form 16 molecules of CO2 and 18 molecules of H2O
2 moles of C8H18 react with 25 moles of O2
to form 16 moles of CO2 and 18 moles of H2O2 mol C8H18 : 25 mol O2 : 16 mol CO2 : 18 mol H2O
Tro, Chemistry: A Molecular Approach 3
Predicting Amounts from Stoichiometry• the amounts of any other substance in a chemical
reaction can be determined from the amount of just one substance
• How much CO2 can be made from 22.0 moles of C8H18 in the combustion of C8H18?
2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)2 moles C8H18 : 16 moles CO2
2188
2188 CO moles 176
HC mol 2CO mol 16HC moles 22.0
Tro, Chemistry: A Molecular Approach 4
Example – Estimate the mass of CO2 produced in 2004 by the combustion of 3.4 x 1015 g gasoline
• assuming that gasoline is octane, C8H18, the equation for the reaction is:
2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)
• the equation for the reaction gives the mole relationship between amount of C8H18 and CO2, but we need to know the mass relationship, so the Concept Plan will be:
g C8H18 mol CO2 g CO2mol C8H18
Example – Estimate the mass of CO2 produced in 2004 by the combustion of 3.4 x 1015 g gasoline
since 8x moles of CO2 as C8H18, but the molar mass of C8H18 is 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 1015 g C8H18
g CO2
Check:
Solution:
Concept Plan:
Relationships:
Given:Find:
g 114.22mol 1
216
2
2
188
2
188
188188
15
CO g 101.0
CO mol 1CO g 44.01
HC mol 2CO mol 16
HC g 114.22HC mol 1HC g 10.43
188
2HC mol 2
CO mol 61
g C8H18 mol CO2 g CO2mol C8H18
mol 1g 44.01
Tro, Chemistry: A Molecular Approach 6
Practice• According to the following equation, how
many milliliters of water are made in the combustion of 9.0 g of glucose?
C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(l)1. convert 9.0 g of glucose into moles (MM 180)2. convert moles of glucose into moles of water3. convert moles of water into grams (MM 18.02)4. convert grams of water into mL
a) How? what is the relationship between mass and volume?
density of water = 1.00 g/mL
Tro, Chemistry: A Molecular Approach 7
Practice
OH mL 5.4
OH g 1.00OH mL 1 x
OH mole 1OH g 18.0 x
OHC mole 1 OH mole 6 x
g 10 x 80.1OHC mole 1 x OHC g 0.9
2
2
2
2
2
6126
22
61266126
According to the following equation, how many milliliters of water are made in the combustion of 9.0 g of glucose?
C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(l)
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 the others
• when this reactant is used up, the reaction stops and no more product is made
• the reactant that limits the amount of product is called the limiting reactantsometimes called the limiting reagent the limiting reactant gets completely consumed
• reactants not completely consumed are called excess reactants
• the amount of product that can be made from the limiting reactant is called the theoretical yield
Tro, Chemistry: A Molecular Approach 9
Things Don’t Always Go as Planned!
• many things can happen during the course of an experiment that cause the loss of product
• the amount of product that is made in a reaction is called the actual yieldgenerally less than the theoretical yield, never more!
• the efficiency of product recovery is generally given as the percent yield
%100yield ltheoretica
yield actualYieldPercent
Tro, Chemistry: A Molecular Approach 10
Limiting and Excess Reactants in the Combustion of Methane
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)• Our balanced equation for the combustion of methane
implies that every 1 molecule of CH4 reacts with 2 molecules of O2
H
HC
H
H+
O
O
C +OO
OO+
OH H
OH H
+
11
Limiting and Excess Reactants in the Combustion of Methane
• If we have 5 molecules of CH4 and 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) + 2 O2(g) CO2(g) + 2 H2O(g)
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
HC
H
H
H
HC
H
H
24
24 CO molecules 16
CH molecules 1CO molecules 2CH molecules 8
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
22
22 CO molecules 10
O molecules 2CO molecules 2O molecules 10
since less CO2 can be made from the O2 than the CH4, the O2 is the limiting reactant
Example 4.4Finding Limiting Reactant,
Theoretical Yield, and Percent Yield
Tro, Chemistry: A Molecular Approach 14
Example:• When 28.6 kg of C are allowed to react with 88.2 kg of
TiO2 in the reaction below, 42.8 kg of Ti are obtained. Find the Limiting Reactant, Theoretical Yield, and Percent Yield.
(g)(s)(s)(s) CO 2 Ti C 2 TiO2
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 C88.2 kg TiO2
42.8 kg Ti produced
Tro, Chemistry: A Molecular Approach 16
• Write down the quantity to find and/or its units.
Find: limiting reactanttheoretical yieldpercent yield
Example:Find the Limiting Reactant, Theoretical Yield, and Percent Yield.
TiO2(s) + 2 C(s) Ti(s) + 2 CO(g)
InformationGiven: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
17
• Write a Concept Plan:
InformationGiven: 28.6 kg C, 88.2 kg TiO2, 42.8 kg
TiFind: 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
kgC
2
2TiO g 9.877
TiO mol 1
C g .0112C mol 1
molC
molTiO2
molTi
molTi
2TiO mol 1Ti mol 1
C mol 2Ti mol 1 }smallest
amount isfrom
limitingreactant
gTiO2
gC
kg 1g 1000
kg 1g 1000
smallestmol Ti g Ti
Ti mol 1g 87.47 % Yield
yield theor.yield act. yield %
kg TiT.Y.g 000 1
kg 1
18
• Collect Needed Relationships:
1000 g = 1 kgMolar Mass TiO2 = 79.87 g/mol
Molar Mass Ti = 47.87 g/molMolar 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)
InformationGiven: 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 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)
Tro, Chemistry: A Molecular Approach 19
Ti mol 104301.1TiO mol 1
Ti mol 1TiO g 79.87TiO mole 1
kg 1g 0001TiO kg 8.28 3
22
22
• Apply the Concept Plan:
InformationGiven: 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 Rel: 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)
Ti mol 100791.1C mol 2Ti mol 1
C g 12.01C mole 1
kg 1g 0001C kg 8.62 3
smallest moles of TiLimiting Reactant
Tro, Chemistry: A Molecular Approach 20
• Apply the Concept Plan:
Ti kg 52.9 g 1000
kg 1mol 1
Ti g 47.87Ti mol 104301.1 3
Theoretical Yield
InformationGiven: 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 Rel: 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)
Tro, Chemistry: A Molecular Approach 21
• Apply the Concept Plan:
YieldPercent 100Yield lTheoretica
Yield Actual %
%9.08%100Ti kg 52.9Ti kg 42.8
InformationGiven: 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 Rel: 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)
Tro, Chemistry: A Molecular Approach 22
• Check the Solutions:
Limiting Reactant = TiO2
Theoretical Yield = 52.9 kgPercent Yield = 80.9%
Since Ti has lower molar mass than TiO2, the T.Y. makes senseThe Percent Yield makes sense as it is less than 100%.
InformationGiven: 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 Rel: 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)
Tro, Chemistry: A Molecular Approach 23
Practice – How many grams of N2(g) can be made from 9.05 g of NH3 reacting 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 N2(g) can be made from 9.05 g of NH3 reacting 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 g2 mol NH3 = 1 mol N2, 3 mol CuO = 1 mol N2
9.05 g NH3, 45.2 g CuOg N2
Concept Plan:
Relationships:
Given:Find:
g 17.03mol 1
3
2
NH mol 2N mol 1
g NH3 mol N2mol NH3
g 28.02mol 1
g CuO mol N2mol CuO
CuO mol 3N mol 1 2
g 79.55mol 1
g N2smallest moles N2
Practice – How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of CuO?
2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)
units are correct, and since there are fewer moles of N2 than CuO in the reaction and N2 has a
smaller mass, the number makes sense
Check:
Solution:
23
2
3
33 N mol .2660
NH mol 2N mol 1
NH g 17.03NH mol 1NH g .059
22 N mol .1890
CuO mol 3N mol 1
CuO g 79.55CuO mol 1CuO g 5.24
22
22 N g .305
N mol 1N g 28.02N mol .1890
Tro, Chemistry: A Molecular Approach 26
Solutions• when table salt is mixed with water, it seems to disappear,
or become a liquid – the 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
Tro, Chemistry: A Molecular Approach 27
Describing Solutions• since solutions are mixtures, the composition can
vary from one sample to anotherpure substances have constant compositionsalt water samples from different seas or lakes have
different 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 describe
them with a single name because all samples identical
Tro, Chemistry: A Molecular Approach 28
Solution Concentration• qualitatively, solutions are often
described as dilute or concentrated
• dilute solutions have a small amount of solute compared to solvent
• concentrated solutions have a large amount of solute compared to solvent
• quantitatively, the relative amount of solute in the solution is called the concentration
Tro, Chemistry: A Molecular Approach 29
Solution ConcentrationMolarity
• moles of solute per 1 liter of solution• used because it describes how many molecules
of solute in each liter of solution
L)(in solution ofamount moles) (in solute ofamount M molarity,
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 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 solutionMolarity, M
Check:• Check
Solution:• Follow the Concept Plan to Solve the problem
Concept Plan:
Relationships:
• Strategize
Given:Find:
• Sort Information
g 119.00mol 1
M 0.122L 1.75
KBr mol 29421.0solution L
KBr moles M molarity,
KBr mol 2940.21 KBr g 119.00
KBr mol 1KBr g 5.52
g KBr mol KBr
L sol’nML
mol M
Tro, Chemistry: A Molecular Approach 32
Using Molarity in Calculations• molarity shows the relationship between the
moles 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 sugar 2 liters = 4.0 moles sugar 0.5 liters = 1.0 mole sugar
• 1 L solution : 2 moles sugar
solution L 1sugar mol 2
sugar mol 2solution L 1
Example 4.6 – How 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 NaOHliters, L
Check:• Check
Solution:• Follow the Concept Plan to Solve the problem
Concept Plan:
Relationships:
• Strategize
Given:Find:
• Sort Information
NaOH mol 0.125solution L 1
solution L 2.04 NaOH mol 0.125
solution L 1NaOH mol 552.0
mol NaOH L sol’n
Tro, Chemistry: A Molecular Approach 34
Dilution• often, solutions are stored as concentrated stock
solutions• to make solutions of lower concentrations from these
stock solutions, more solvent is added the amount of solute doesn’t change, just the volume of
solutionmoles solute in solution 1 = moles solute in solution 2
• the concentrations and volumes of the stock and new solutions are inversely proportional
M1∙V1 = M2∙V2
Example 4.7 – To 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 MV2, L
Check:• Check
Solution:• Follow the Concept Plan to Solve the problem
Concept Plan:
Relationships:
• Strategize
Given:Find:
• Sort Information
22
11 VM
VM
L 1.00
Lmol3.00
L 200.0L
mol15.0
V1, M1, M2 V2
Tro, Chemistry: A Molecular Approach 36
Solution Stoichiometry
• since molarity relates the moles of solute to the liters of solution, it can be used to convert between amount of reactants and/or products in a chemical reaction
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 the Concept Plan to Solve the problem
Concept Plan:
Relationships:
• Strategize
Given:
Find:
• Sort Information
23)Pb(NO L 1mol 0.175
KCl L .3500 mol .1500
KCl L 1)Pb(NO mol 1
KCl mol 2)Pb(NO L 1
mol 0.175)Pb(NO L .15002323
23
23)Pb(NO mol 1KCl mol 2
L Pb(NO3)2 mol KCl L KClmol Pb(NO3)2
mol 0.150KCl L 1
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 the solvent molecules• if the attractions between solute and solvent are strong enough,
the solute will dissolve
Tro, Chemistry: A Molecular Approach 39
Table Salt Dissolving in WaterEach ion is attracted to the surrounding water molecules and pulled off and away from the crystalWhen it enters the solution, the ion is surrounded by water molecules, insulating it from other ions The result is a solution with free moving charged particles able to conduct electricity
Tro, Chemistry: A Molecular Approach 40
Electrolytes and Nonelectrolytes• materials that dissolve
in water to form a solution that will conduct electricity are called electrolytes
• materials that dissolve in water to form a solution that will not conduct electricity are called nonelectrolytes
Tro, Chemistry: A Molecular Approach 41
Molecular View of Electrolytes and Nonelectrolytes
• in order to conduct electricity, a material must have charged 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
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
Tro, Chemistry: A Molecular Approach 43
Acids• acids are molecular compounds that ionize when they
dissolve in water the molecules are pulled apart by their attraction for the waterwhen acids ionize, they form H+ cations and anions
• the percentage of molecules that ionize varies from one acid to another
• acids that ionize virtually 100% are called strong acidsHCl(aq) H+(aq) + Cl-(aq)
• acids that only ionize a small percentage are called weak acids
HF(aq) H+(aq) + F-(aq)
Tro, Chemistry: A Molecular Approach 44
Strong and Weak Electrolytes• strong electrolytes are materials that dissolve
completely as ions ionic compounds and strong acids their solutions conduct electricity well
• weak electrolytes are 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 together
Na2SO4(aq) 2 Na+(aq) + SO42-(aq)
HC2H3O2(aq) H+(aq) + C2H3O2-(aq)
Tro, Chemistry: A Molecular Approach 45
Classes of Dissolved Materials
Tro, Chemistry: A Molecular Approach 46
Solubility of Ionic Compounds• some ionic compounds, like NaCl, dissolve very well in
water at room temperature• other ionic compounds, like AgCl, dissolve hardly at all
in water at room temperature• compounds that dissolve in a solvent are said to be
soluble, while those that do not are said to be insolubleNaCl is soluble in water, AgCl is insoluble in water the degree of solubility depends on the temperatureeven insoluble compounds dissolve, just not enough to be
meaningful
Tro, Chemistry: A Molecular Approach 47
When Will a Salt Dissolve?
• Predicting whether a compound will dissolve in water is not easy
• The best way to do it is to do some experiments to test whether a compound will dissolve in water, then develop some rules based on those experimental resultswe call this method the empirical method
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+
SO42– Ag+, Ca2+, Sr2+, Ba2+, Pb2+
Solubility RulesCompounds that Are Generally Soluble in Water
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 RulesCompounds that Are Generally Insoluble
Tro, Chemistry: A Molecular Approach 50
Precipitation Reactions• reactions between aqueous solutions of ionic
compounds that produce an ionic compound that is insoluble in water are called precipitation reactions and the insoluble product is called a precipitate
51
2 KI(aq) + Pb(NO3)2(aq) PbI2(s) + 2 KNO3(aq)
Tro, Chemistry: A Molecular Approach 52
No Precipitate Formation = No Reaction
KI(aq) + NaCl(aq) KCl(aq) + NaI(aq)all ions still present, no reaction
Tro, Chemistry: A Molecular Approach 53
Process for Predicting the Products ofa 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 each product
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 reaction after the arrow
Tro, Chemistry: A Molecular Approach 54
Process for Predicting the Products ofa Precipitation Reaction
5. If either product is insoluble, write the formulas for the products after the arrow – writing (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 4.10 – Write the equation for the 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+ + CO3
2-) + (Ni2+ + Cl-) (K+ + Cl-) + (Ni2+ + CO32-)
c) Write the formulas of the products cross charges and reduce
K2CO3(aq) + NiCl2(aq) KCl + NiCO3
Tro, Chemistry: A Molecular Approach 56
3. Determine the solubility of each productKCl is soluble
NiCO3 is insoluble
4. If both products soluble, write no reactiondoes not apply since NiCO3 is insoluble
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
Tro, Chemistry: A Molecular Approach 57
5. Write (aq) next to soluble products and (s) next to insoluble products
K2CO3(aq) + NiCl2(aq) KCl(aq) + NiCO3(s)
6. Balance the EquationK2CO3(aq) + NiCl2(aq) 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
Tro, Chemistry: A Molecular Approach 58
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 complete ionic 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) K+1
(aq) + 2NO3-1
(aq) + Mg(OH)2(s)
Tro, Chemistry: A Molecular Approach 59
Ionic Equations• ions that are both reactants and products are called
spectator ions
2K+1(aq) + 2OH-1
(aq) + Mg+2(aq) + 2NO3
-1(aq) K+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)
Tro, Chemistry: A Molecular Approach 60
Acid-Base Reactions• also called neutralization reactions because the
acid and base neutralize each other’s properties2 HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + 2 H2O(l)
• the net ionic equation for an acid-base reaction isH+(aq) + OH(aq) H2O(l)
as long as the salt that forms is soluble in water
Tro, Chemistry: A Molecular Approach 61
Acids and Bases in Solution• acids ionize in water to form H+ ions
more 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 OH ionsbases, like NH3, that do not contain OH ions, produce OH by
pulling H off water molecules• in the reaction of an acid with a base, the H+ from the
acid combines with the OH from the base to make water• the cation from the base combines with the anion from
the acid to make the salt
acid + base salt + water
Common AcidsChemical Name Formula Uses Strength Perchloric Acid HClO4 explosives, catalyst Strong
Nitric Acid HNO3 explosives, fertilizer, dye, glue Strong
Sulfuric Acid H2SO4 explosives, fertilizer, dye, glue,
batteries Strong
Hydrochloric Acid HCl metal cleaning, food prep, ore refining, stomach acid Strong
Phosphoric Acid H3PO4 fertilizer, plastics & rubber,
food preservation Moderate
Chloric Acid HClO3 explosives Moderate
Acetic Acid HC2H3O2 plastics & rubber, food preservation, vinegar Weak
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
Tro, Chemistry: A Molecular Approach 63
Common BasesChemical
Name Formula Common Name Uses Strength
sodium hydroxide
NaOH lye,
caustic soda soap, plastic,
petrol refining Strong
potassium hydroxide
KOH caustic potash soap, cotton, electroplating
Strong
calcium hydroxide
Ca(OH)2 slaked lime cement Strong
sodium bicarbonate
NaHCO3 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
Tro, Chemistry: A Molecular Approach 64
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
Tro, Chemistry: A Molecular Approach 65
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 products
a) Determine the ions present when each reactant dissociates(H+ + NO3
-) + (Ca+2 + OH-) b) Exchange the ions, H+1 combines with OH-1 to 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)
Tro, Chemistry: A Molecular Approach 66
3. Determine the solubility of the saltCa(NO3)2 is 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
Tro, Chemistry: A Molecular Approach 67
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) H2O(l)
H+1(aq) + OH-1(aq) H2O(l)
Tro, Chemistry: A Molecular Approach 68
Titration• often in the lab, a solution’s concentration is
determined by reacting it with another material and using stoichiometry – this process is called titration
• in the titration, the unknown solution is added to a known amount of another reactant until the reaction is just completed, at this point, called the endpoint, the reactants are in their stoichiometric ratio the unknown solution is added slowly from an
instrument called a burettea long glass tube with precise volume markings that
allows small additions of solution
Tro, Chemistry: A Molecular Approach 69
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 that changes 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 of moles of H+ equals the number of moles of OH aka the equivalence point
Tro, Chemistry: A Molecular Approach 70
Titration
Tro, Chemistry: A Molecular Approach 71
TitrationThe base solution is thetitrant in the burette.
As the base is added tothe acid, the H+ reacts withthe OH– to form water. But there is still excess acid present so the colordoes not change.
At the titration’s endpoint,just enough base has been added to neutralize all theacid. At this point the indicator changes color.
Tro, Chemistry: A Molecular Approach 72
Example 4.14: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 end point. 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
Tro, Chemistry: A Molecular Approach 73
• Write down the quantity to find, and/or its units.
Find: concentration HCl, M
InformationGiven: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
Example 4.14: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 end point. What is the concentration of the unknown HCl solution?
Tro, Chemistry: A Molecular Approach 74
• Collect Needed Equations and Conversion Factors:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
1 mole HCl = 1 mole NaOH0.100 M NaOH 0.100 mol NaOH 1 L sol’n
InformationGiven: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
Find: M HCl
Example 4.14: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 end point. What is the concentration of the unknown HCl solution?
solution literssolute molesMolarity
Tro, Chemistry: A Molecular Approach 75
• Write a Concept Plan:
mLNaOH
LNaOH
molNaOH
NaOH L 1NaOH mol 1000.
mL 1L 0010.
molHCl
NaOH mol 1HCl mol 1
InformationGiven: 10.00 mL HCl
12.54 mL of 0.100 M NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L
M = mol/L
Example 4.14: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 end point. What is the concentration of the unknown HCl solution?
mLHCl
LHCl
mL 1L 0010. HCl liters
HCl molesMolarity
Tro, Chemistry: A Molecular Approach 76
NaOH mole 1HCl mol 1
L 1NaOH mol 1000
mL 1L 0.001NaOH mL 2.541
.
• Apply the Solution Map:
= 1.25 x 10-3 mol HCl
InformationGiven: 10.00 mL HCl
12.54 mL of 0.100 M NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L
M = mol/LCP: mL NaOH → L NaOH →
mol NaOH → mol HCl; mL HCl → L HCl & mol M
Example: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 end point. What is the concentration of the unknown HCl solution?
Tro, Chemistry: A Molecular Approach 77
HCl L 010000mL 1
L 0.001NaOH mL 0.001 .
• Apply the Concept Plan:
InformationGiven: 10.00 mL HCl
12.54 mL NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L
M = mol/LCP: mL NaOH → L NaOH →
mol NaOH → mol HCl; mL HCl → L HCl & mol M
Example: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 end point. What is the concentration of the unknown HCl solution?
M 1250HCl L 0.01000
HCl moles 10 x 1.25Molarity-3
.
Tro, Chemistry: A Molecular Approach 78
• 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 thanNaOH solution, so the HCl should be more concentrated.
InformationGiven: 10.00 mL HCl
12.54 mL NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L
M = mol/LCP: mL NaOH → L NaOH →
mol NaOH → mol HCl; mL HCl → L HCl & mol M
Example: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 end point. What is the concentration of the unknown HCl solution?
Tro, Chemistry: A Molecular Approach 79
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 water
K2SO3(aq) + H2SO4(aq) K2SO4(aq) + H2SO3(aq)
H2SO3 H2O(l) + SO2(g)
Tro, Chemistry: A Molecular Approach 80
NaHCO3(aq) + HCl(aq) NaCl(aq) + CO2(g) + H2O(l)
Tro, Chemistry: A Molecular Approach 81
Compounds that UndergoGas Evolving Reactions
ReactantType
ReactingWith
Ion ExchangeProduct
Decom-pose?
GasFormed
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
metal HSO3
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)
Tro, Chemistry: A Molecular Approach 82
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 products
a) 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
Tro, Chemistry: A Molecular Approach 83
3. Check to see either product H2S - No
4. Check to see if either product decomposes – Yes
H2CO3 decomposes 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
Tro, Chemistry: A Molecular Approach 84
5. Determine the solubility of other productNaNO3 is soluble
6. Write an (s) after the insoluble products and a (aq) after the soluble products
Na2CO3(aq) + 2 HNO3(aq) NaNO3(aq) + CO2(g) + H2O(l)
7. Balance the equationNa2CO3(aq) + 2 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
Tro, Chemistry: A Molecular Approach 85
Other Patterns in Reactions• the precipitation, acid-base, and gas evolving
reactions all involved exchanging the ions in the solution
• other kinds of reactions involve transferring electrons from one atom to another – these are called oxidation-reduction reactionsalso known as redox reactionsmany involve the reaction of a substance with O2(g)
4 Fe(s) + 3 O2(g) 2 Fe2O3(s)
Tro, Chemistry: A Molecular Approach 86
Combustion as Redox2 H2(g) + O2(g) 2 H2O(g)
Tro, Chemistry: A Molecular Approach 87
Redox without Combustion2 Na(s) + Cl2(g) 2 NaCl(s)
2 Na 2 Na+ + 2 e
Cl2 + 2 e 2 Cl
Tro, Chemistry: A Molecular Approach 88
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)
Tro, Chemistry: A Molecular Approach 89
Oxidation and Reduction• in order to convert a free element into an ion, the atoms
must gain or lose electronsof course, if one atom loses electrons, another must accept
them• reactions where electrons are transferred from one atom
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– oxidationCl2 + 2 e– → 2 Cl– reduction
Leo
Ger
Tro, Chemistry: A Molecular Approach 90
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 state that allows them to determine the electron flow in the reactioneven though they look like them, oxidation states are
not ion charges!oxidation states are imaginary charges assigned based on a
set of rules ion charges are real, measurable charges
Tro, Chemistry: A Molecular Approach 91
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 equal
to their charge Na = +1 and Cl = -1 in NaCl
3. (a) the sum of the oxidation states of all the atoms in a compound is 0
Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0
Tro, Chemistry: A Molecular Approach 92
Rules for Assigning Oxidation States3. (b) the sum of the oxidation states of all the atoms in
a 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 all their compounds
Na = +1 in NaCl
4. (b) Group II metals have an oxidation state of +2 in all their compounds
Mg = +2 in MgCl2
Tro, Chemistry: A Molecular Approach 93
Rules for Assigning Oxidation States5. in their compounds, nonmetals have oxidation
states according to the table below nonmetals higher on the table take priority
Nonmetal Oxidation State ExampleF -1 CF4
H +1 CH4
O -2 CO2
Group 7A -1 CCl4
Group 6A -2 CS2
Group 5A -3 NH3
Tro, Chemistry: A Molecular Approach 94
Practice – Assign an Oxidation State to Each Element in the following
• Br2
• K+
• LiF
• CO2
• SO42-
• Na2O2
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)
Tro, Chemistry: A Molecular Approach 96
Oxidation and ReductionAnother Definition
• oxidation occurs when an atom’s oxidation state increases during a reaction
• reduction occurs when an atom’s oxidation state decreases during a reaction
CH4 + 2 O2 → CO2 + 2 H2O-4 +1 0 +4 –2 +1 -2
oxidationreduction
Tro, Chemistry: A Molecular Approach 97
Oxidation–Reduction• 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 reactant is 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, Cl2 is the oxidizing agent
Tro, Chemistry: A Molecular Approach 98
Identify the Oxidizing and Reducing Agents in Each of the Following
3 H2S + 2 NO3– + 2 H+ S + 2 NO + 4 H2O
MnO2 + 4 HBr MnBr2 + Br2 + 2 H2O
Tro, Chemistry: A Molecular Approach 99
Identify the Oxidizing and Reducing Agents in Each of the Following
3 H2S + 2 NO3– + 2 H+ S + 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
oxidationreduction
red agox ag
Tro, Chemistry: A Molecular Approach 100
Combustion Reactions• Reactions in which O2(g) is a
reactant are called combustion 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)
Tro, Chemistry: A Molecular Approach 101
Combustion Products• to predict the products of a combustion
reaction, 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)
Tro, Chemistry: A Molecular Approach 102
Practice – Complete the Reactions
• combustion of C3H7OH(l)
• combustion of CH3NH2(g)
Tro, Chemistry: A Molecular Approach 103
Practice – Complete 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)