CHM141 Chapter4 7e Kotz

Chapter 4Stoichiometry: Quantitative Information About Chemical

Reactions

STOICHIOMETRYSTOICHIOMETRY-- the study of the the study of the

quantitative quantitative aspects of aspects of chemical chemical reactions.reactions.

STOICHIOMETRYSTOICHIOMETRYIt rests on the principle of the It rests on the principle of the

conservation of matterconservation of matter..

2 Al(s) + 3 Br2(liq) → Al2Br6(s)

PROBLEM: If 454 g of NHPROBLEM: If 454 g of NH44NONO33 decomposes, decomposes, how much Nhow much N22O and HO and H22O are formed? What is the O are formed? What is the theoretical yield of products?theoretical yield of products?

STEP 1STEP 1Write the balanced Write the balanced

chemical equationchemical equationNHNH44NONO33 →→

NN22O + 2 HO + 2 H22OO

454 g of NH454 g of NH44NONO33 →→ NN22O + 2 HO + 2 H22OO

STEP 2 STEP 2 Convert mass reactant Convert mass reactant (454 g) (454 g) →→ molesmoles

454 g x 1 mol80.04 g

= 5.68 mol NH4NO3


•• STEP 3 STEP 3 Convert moles reactant Convert moles reactant (5.68 mol) (5.68 mol) →→ moles productmoles product

Relate moles NHRelate moles NH44NONO33 to moles product expected. to moles product expected. 1 mol NH1 mol NH44NONO33 →→ 2 mol H2 mol H22OO

5.68 mol NH4NO3 x 2 mol H2O produced1 mol NH4NO3used

= 11.4 mol


11.4 mol H2O • 18.02 g1 mol

= 204 g H2O

STEP 4 STEP 4 Convert moles product Convert moles product (11.4 mol) (11.4 mol) →→ mass productmass product

Called theCalled the Theoretical YieldTheoretical Yield

ALWAYS FOLLOW THESE STEPS IN ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS!SOLVING STOICHIOMETRY PROBLEMS!

Mass reactant

Molar RatioMolesreactant

Moles product

Mass product

General Steps for Stoichiometric Problems


STEP 5 STEP 5 How much NHow much N22O is formed?O is formed?

Total mass of reactants = Total mass of productsTotal mass of reactants = Total mass of products

454 g NH454 g NH44NONO33 = ___ g N= ___ g N22O + 204 g HO + 204 g H22OO

mass of Nmass of N22O = 250. gO = 250. g

•• CompoundCompound NHNH44NONO33 NN22OO HH22OO

• Initial (g)

• Initial (mol)

• Change (mol)

• Final (mol)

• Final (g)


Amounts Table Amounts Table (page 125)(page 125)

Note that matter is conserved!

•• CompoundCompound NHNH44NONO33 NN22OO HH22OO

• Initial (g) 454 g 0 0

• Initial (mol) 5.68 mol 0 0

• Change (mol) -5.68 mol +5.68 mol +2(5.68) mol

• Final (mol) 0 5.68 mol 11.4 mol

• Final (g) 0 250 g 204g


Amounts Table Amounts Table (page 125)(page 125)


% yield = actual yieldtheoretical yield

• 100%

STEP 6 STEP 6 Calculate the percent yieldCalculate the percent yield

% yield = 131 g250. g

• 100% = 52.4%


PROBLEM: Using 5.00 g of HPROBLEM: Using 5.00 g of H22OO22, what mass of , what mass of OO22 and of Hand of H22O can be obtained?O can be obtained?

2 H2 H22OO22(liq) (liq) →→ 2 H2 H22O(g) + OO(g) + O22(g)(g)(Reaction is catalyzed by MnO(Reaction is catalyzed by MnO22))

Step 1: moles of HStep 1: moles of H22OO22

Step 2: use Molar Ratio to calculate moles of OStep 2: use Molar Ratio to calculate moles of O22

Step 3: mass of OStep 3: mass of O22

Reactions Involving aReactions Involving aLIMITING REACTANTLIMITING REACTANT

•• In a given reaction, there is not In a given reaction, there is not enough of one reagent to use enough of one reagent to use up the other reagent completely.up the other reagent completely.

•• The reagent in short supply The reagent in short supply LIMITSLIMITS the quantity of product the quantity of product that can be formed.that can be formed.

LIMITING REACTANTSLIMITING REACTANTS

Reactantseactants ProductsProducts

2 NO(g) + O2 (g) 2 NO2(g)Limiting reactant = ___________Limiting reactant = ___________Excess reactant = ____________Excess reactant = ____________

LIMITING REACTANTSLIMITING REACTANTS

Rxn 1: Balloon inflates fully, some Zn leftRxn 1: Balloon inflates fully, some Zn left* * More than enough Zn to use up the 0.100 mol HClMore than enough Zn to use up the 0.100 mol HCl

Rxn 2: Balloon inflates fully, no Zn leftRxn 2: Balloon inflates fully, no Zn left* Right amount of each (HCl and Zn)* Right amount of each (HCl and Zn)

Rxn 3: Balloon does not inflate fully, no Zn left.Rxn 3: Balloon does not inflate fully, no Zn left.* Not enough Zn to use up 0.100 mol HCl* Not enough Zn to use up 0.100 mol HCl

React solid Zn with 0.100 React solid Zn with 0.100 mol HCl (aq)mol HCl (aq)Zn + 2 HCl Zn + 2 HCl →→ ZnClZnCl22 + H+ H22

Rxn 1Rxn 1 Rxn 2Rxn 2 Rxn 3Rxn 3mass Zn (g)mass Zn (g) 7.00 g7.00 g 3.27 g3.27 g 1.31 g1.31 gmol Znmol Zn 0.107 mol0.107 mol 0.050 mol0.050 mol 0.020 mol0.020 molmol HClmol HCl 0.100 mol0.100 mol 0.100 mol0.100 mol 0.100 mol0.100 molmol HCl/mol Znmol HCl/mol Zn 0.93/10.93/1 2.00/12.00/1 5.00/15.00/1Lim ReactantLim Reactant LR = HClLR = HCl no LRno LR LR = ZnLR = Zn

LIMITING REACTANTSLIMITING REACTANTSReact solid Zn with 0.100 mol HCl (aq)React solid Zn with 0.100 mol HCl (aq)

Zn + 2 HCl Zn + 2 HCl →→ ZnClZnCl22 + H+ H2 2

0.10 mol HCl [1 mol Zn/2 mol HCl]= 0.050 mol Zn

Reaction to be StudiedReaction to be Studied

2 Al + 3 Cl2 Al + 3 Cl22 →→ AlAl22ClCl66

PROBLEM: PROBLEM: Mix 5.40 g of Al with 8.10 g of ClMix 5.40 g of Al with 8.10 g of Cl22, , What mass of AlWhat mass of Al22ClCl66 can form?can form?

Mass reactant

Molar RatioMolesreactant

Moles product

Mass product


Reactants must be in the mole ratioReactants must be in the mole ratio

Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

mol Cl2mol Al

= 32

Deciding on the Limiting Deciding on the Limiting ReactantReactant

•• IfIf There is not enough Al to There is not enough Al to use up all the Cluse up all the Cl22


mol Cl 2mol Al

> 32

Limiting reagent = AlLimiting reagent = Al

mol Cl 2mol Al

> 32••IfIf There is not enough ClThere is not enough Cl22 to to

use up all the Aluse up all the Al

Limiting reagent = ClLimiting reagent = Cl22

We have 5.40 g of Al and 8.10 g of ClWe have 5.40 g of Al and 8.10 g of Cl22

Step 2 of LR problem: Step 2 of LR problem: Calculate moles of each Calculate moles of each reactantreactant

5.40 g Al • 1 mol27.0 g

= 0.200 mol Al

8.10 g Cl2 • 1 mol70.9 g

= 0.114 mol Cl2

Find mole ratio of reactantsFind mole ratio of reactants

This This should be 3/2 or 1.5/1 if should be 3/2 or 1.5/1 if reactants are present in the reactants are present in the exact stoichiometric ratio.exact stoichiometric ratio.

Limiting reactant is Limiting reactant is ClCl22

mol Cl2mol Al

= 0.114 mol 0.200 mol

= 0.57


Mix 5.40 g of Al with 8.10 g of ClMix 5.40 g of Al with 8.10 g of Cl22, What , What mass of Almass of Al22ClCl66 can form?can form?

Limiting reactant = ClLimiting reactant = Cl22Base all calcs. on ClBase all calcs. on Cl22

molesCl2

moles Al2Cl6

massCl2

mass Al2Cl6

1 mol Al2Cl63 mol Cl2


•• CompoundCompound AlAl ClCl22 AlAl22ClCl66

• Initial (g) 5.40 g 8.10 0

• Initial (mol) 0.200 mol 0.114 mol 0

• Change (mol)

Based on LR -0.0760 mol -0.114 mol +0.0380 mol

• Final (mol) 0.1240 mol 0 mol 0.0380 mol

• Final (g) 3.35 g 0 g 10.1 g


Table of AmountsTable of Amounts


CALCULATIONS: calculate mass ofCALCULATIONS: calculate mass ofAlAl22ClCl66 expected.expected.

Calculate moles of AlCalculate moles of Al22ClCl66 expected based expected based on LR.on LR.

0.114 mol Cl2 • 1 mol Al2Cl63 mol Cl2

= 0.0380 mol Al2Cl6

0.0380 mol Al2Cl6 • 266.4 g Al2Cl6mol

= 10.1 g Al2Cl6

Calculate mass of AlCalculate mass of Al22ClCl66 expected based expected based on LR.on LR.

•• ClCl22 was the limiting reactant. Therefore, was the limiting reactant. Therefore, Al was present in excess. But how Al was present in excess. But how much?much?

•• First find how much Al was required.First find how much Al was required.

•• Then find how much Al Then find how much Al is in excess.is in excess.

How much of which reactant will How much of which reactant will remain when reaction is complete?remain when reaction is complete?

2 Al + 3 2 Al + 3 ClCl22 productsproducts

0.200 mol0.200 mol 0.114 mol = LR0.114 mol = LR

Calculating Excess AlCalculating Excess Al

Excess Al = Al available Excess Al = Al available -- Al requiredAl required

0.114 mol Cl2 • 2 mol Al3 mol Cl2

= 0.0760 mol Al req' d

= 0.200 mol = 0.200 mol -- 0.0760 mol 0.0760 mol = = 0.124 mol Al in excess0.124 mol Al in excess

Percent Yield of Percent Yield of Chemical ReactionChemical Reaction

• Theoretical yield:the amount product predicted by stoichiometry of the chemical reaction.

• Actual yield: the amount of product actually obtained.

• Percent yield = Actual yield / theoretical yield x 100%

Chemical AnalysisChemical Analysis


• An impure sample of the mineral thenardite contains Na2SO4.

• Mass of mineral sample = 0.123 g• The Na2SO4 in the sample is converted to

insoluble BaSO4.• The mass of BaSO4 is 0.177 g• What is the mass percent of Na2SO4 in the

mineral?

•• CompoundCompound Na2SO4(aq) + BaCl2(aq) → 2 NaCl(aq) + BaSO4(s)

• Initial (g) 1.108 g

• Initial (mol) 7.58 x 10-4 mol 7.58 x 10-4 mol

• Change (mol) -7.58 x 10-4 mol +7.58 x 10-4 mol

• Final (mol) 7.58 x 10-4 mol

• Final (g) 0.177 g

Na2SO4(aq) + BaCl2(aq) → 2 NaCl(aq) + BaSO4(s)

Table of AmountsTable of Amounts

•mass percent of Na2SO4 in the mineral:(0.108 g Na2SO4/0.123 g sample)100%

= 87.6% Na2SO4


• Na2SO4(aq) + BaCl2(aq) → 2 NaCl(aq) + BaSO4(s)• 0.177 g BaSO4 (1 mol/233.4 g)

= 7.58 x 10-4 mol BaSO4

• 7.58 x 10-4 mol BaSO4 (1 mol Na2SO4/1 mol BaSO4)= 7.58 x 10-4 mol Na2SO4

• 7.58 x 10-4 mol Na2SO4 (142.0 g/1 mol) = 0.108 g Na2SO4

• (0.108 g Na2SO4/0.123 g sample)100% = 87.6% Na2SO4

Determining the Formula of a Determining the Formula of a Hydrocarbon by CombustionHydrocarbon by Combustion

Active Figure 4.9Active Figure 4.9

Using Stoichiometry to Using Stoichiometry to Determine a FormulaDetermine a Formula

Burn 0.115 g of a hydrocarbon, CBurn 0.115 g of a hydrocarbon, CxxHHyy, and produce , and produce 0.379 g of CO0.379 g of CO22 and and 0.1035 g of H0.1035 g of H22OO..

CCxxHHy y + some oxygen + some oxygen →→ 0.379 g CO0.379 g CO22 + + 0.1035 g H0.1035 g H22OO

What is the empirical formula of CWhat is the empirical formula of CxxHHyy??


First, recognize that all C in COFirst, recognize that all C in CO22 and all H in and all H in HH22O is from CO is from CxxHHyy..

CCxxHHy y + some oxygen + some oxygen →→ 0.379 g CO0.379 g CO22 + 0.1035 g H+ 0.1035 g H22OO

Puddle of CxHy

0.115 g

0.379 g CO0.379 g CO22+O2

+O2 0.1035 g H2O1 H2O molecule forms for each 2 H atoms in CxHy

1 CO2 molecule forms for each C atom in CxHy


First, recognize that all C in COFirst, recognize that all C in CO22 and all H in Hand all H in H22O is O is from Cfrom CxxHHyy..

1. Calculate amount of C in CO1. Calculate amount of C in CO22

8.61 x 108.61 x 10--3 3 mol COmol CO22 →→ 8.61 x 108.61 x 10--3 3 mol Cmol C2. Calculate amount of H in H2. Calculate amount of H in H22OO

5.744 x 105.744 x 10--33 mol Hmol H22O O →→1.149 x 101.149 x 10--2 2 mol Hmol H



Now find ratio of mol H/mol C to find values of x and Now find ratio of mol H/mol C to find values of x and y in Cy in CxxHHyy..

1.149 x 10 1.149 x 10 --2 2 mol Hmol H/ / 8.61 x 108.61 x 10--3 3 mol Cmol C= = 1.33 mol H1.33 mol H / / 1.00 mol C1.00 mol C= = 4 mol H4 mol H / / 3 mol C3 mol CEmpirical formula = CEmpirical formula = C33HH44


Quantitative Quantitative Aspects of Aspects of Reactions in Reactions in

SolutionSolution

TerminologyTerminologyIn solution we need to define In solution we need to define

the the --•• SOLVENTSOLVENTthe component whose the component whose

physical state is physical state is preserved when preserved when solution formssolution forms

•• SOLUTESOLUTEthe other solution componentthe other solution component

Concentration of SoluteConcentration of SoluteThe amount of solute in a solution is The amount of solute in a solution is

given by its given by its concentrationconcentration.

Concentration (M) = [ …]

Molarity (M) =Moles of Solute (mol)Liters of Solution (L)

1.0 L of water 1.0 L of water was used to was used to

make 1.0 L of make 1.0 L of solution. Notice solution. Notice

the water left the water left over.over.

Preparing a SolutionPreparing a Solution

PROBLEM: Dissolve 5.00 g of NiClPROBLEM: Dissolve 5.00 g of NiCl22••6 H6 H22O in enough O in enough water to make 250 mL of solution. Calculate molarity.water to make 250 mL of solution. Calculate molarity.

Step 1: Step 1: Calculate moles Calculate moles of NiClof NiCl22••6H6H22OO

5.00 g • 1 mol237.7 g

= 0.0210 mol

0.0210 mol0.250 L

= 0.0841 M

Step 2: Step 2: Calculate molarityCalculate molarity

[NiClNiCl22••6 H6 H22O O ] = 0.0841 M

The Nature of a CuClThe Nature of a CuCl22 Solution:Solution:Ion ConcentrationsIon Concentrations

CuClCuCl22(aq) (aq) →→

CuCu2+2+(aq) + (aq) + 22 ClCl--(aq)(aq)

If [CuClIf [CuCl22] = 0.30 M, then ] = 0.30 M, then

[Cu[Cu2+2+] = 0.30 M] = 0.30 M

[Cl[Cl--] = 2 x 0.30 M] = 2 x 0.30 M

USING MOLARITYUSING MOLARITY•• What mass of oxalic acid, HWhat mass of oxalic acid, H22CC22OO44, is required , is required

to make 250. mL of a 0.0500 M solution?to make 250. mL of a 0.0500 M solution?

Because Because Conc (M) = moles/volume = mol/VConc (M) = moles/volume = mol/V

this means thatthis means that

mols = M mols = M ·· VV

Step 1: Step 1: Calculate moles of acid required.Calculate moles of acid required.(0.0500 mol/L)(0.250 L) = 0.0125 mol(0.0500 mol/L)(0.250 L) = 0.0125 mol

Step 2: Step 2: Calculate mass of acid required.Calculate mass of acid required.(0.0125 mol )(90.00 g/mol) = (0.0125 mol )(90.00 g/mol) = 1.13 g1.13 g

USING MOLARITYUSING MOLARITY

moles = Mmoles = M••VV

What mass of oxalic acid, What mass of oxalic acid, HH22CC22OO44, is, isrequired to make 250. mL of a 0.0500 Mrequired to make 250. mL of a 0.0500 Msolution?solution?

Preparing SolutionsPreparing Solutions•• Weigh out a solid solute Weigh out a solid solute

and dissolve in a given and dissolve in a given quantity of solvent.quantity of solvent.

•• DiluteDilute a concentrated a concentrated solution to give one that solution to give one that is less concentrated.is less concentrated.

Preparing a Solution by DilutionPreparing a Solution by Dilution

PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you and you want 0.50 M NaOH. What do you do?do?

Add water to the 3.0 M solution to Add water to the 3.0 M solution to lower its concentration to 0.50 M lower its concentration to 0.50 M

Dilute the solution!Dilute the solution!

PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?and you want 0.50 M NaOH. What do you do?

3.0 M NaOH 0.50 M NaOH

H2 O

Concentrated Dilute

But how much water But how much water do we add?do we add?

PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you doand you want 0.50 M NaOH. What do you do??

How much water is added?How much water is added?The important point is that The important point is that →→

moles of NaOH in ORIGINAL solution = moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solutionmoles of NaOH in FINAL solution


Amount of NaOH in original solution = Amount of NaOH in original solution = M M •• VV = =

(3.0 mol/L)(0.050 L) = 0.15 mol NaOH(3.0 mol/L)(0.050 L) = 0.15 mol NaOHAmount of NaOH in final solution must also = 0.15 Amount of NaOH in final solution must also = 0.15

mol NaOHmol NaOHVolume of final solution =Volume of final solution =(0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L(0.15 mol NaOH)(1 L/0.50 mol) = 0.30 Lor or 3.0 x 103.0 x 1022 mLmL


Conclusion:Conclusion:add 250 mL of add 250 mL of waterwater to 50.0 to 50.0 mL of 3.0 M mL of 3.0 M NaOH to make NaOH to make 300 mL of 0.50 300 mL of 0.50 M NaOH. M NaOH.

3.0 M NaOH 0.50 M NaOH

H2 O

Concentrated Dilute

A shortcutA shortcut

MMinitialinitial •• VVinitialinitial = M= Mfinalfinal •• VVfinalfinal

Preparing Solutions by Preparing Solutions by DilutionDilution

pH, a Concentration ScalepH, a Concentration ScalepH: a way to express acidity pH: a way to express acidity ––

the concentration of Hthe concentration of H++ in solution.in solution.

Low pH: high [HLow pH: high [H++]] High pH: low [HHigh pH: low [H++]]

Acidic solutionAcidic solution pH < 7pH < 7Neutral Neutral pH = 7pH = 7Basic solution Basic solution pH > 7pH > 7

The pH ScaleThe pH ScalepH = log (1/ [HpH = log (1/ [H++]) ])

= = -- log [Hlog [H++]]

In a In a neutralneutral solution, solution, [H[H++] = [OH] = [OH--] = 1.00 x 10] = 1.00 x 10--77 M at 25 M at 25 ooCC

pH = pH = -- log [Hlog [H++] = ] = --log (1.00 x 10log (1.00 x 10--77) ) = = -- [0 + ([0 + (--7)] 7)] = 7= 7

[H[H++] and pH] and pHIf the [HIf the [H++] of soda is 1.6 x 10] of soda is 1.6 x 10--33 M, M,

the pH is ____?the pH is ____?Because pH = Because pH = -- log [Hlog [H++] ]

thenthenpH= pH= -- log (1.6 x 10log (1.6 x 10--33) ) pH = pH = --(log (1.6) + log (10(log (1.6) + log (10--33))))pH = pH = --(0.20 (0.20 -- 3.00)3.00)pH = 2.80pH = 2.80

pH and [HpH and [H++]]If the pH of Coke is 3.12, it is ____________.If the pH of Coke is 3.12, it is ____________.

Because pH = Because pH = -- log [Hlog [H++] then] thenlog [Hlog [H++] = ] = -- pHpH

Take antilog and getTake antilog and get[H[H++] = 10] = 10--pHpH

[H[H++] = 10] = 10--3.123.12 = 7.6 x 10= 7.6 x 10--44 MM

•• Zinc reacts with acids to Zinc reacts with acids to produce Hproduce H22 gas. gas.

•• Have 10.0 g of ZnHave 10.0 g of Zn•• What volume of 2.50 M What volume of 2.50 M

HCl is needed to convert HCl is needed to convert the Zn completely?the Zn completely?

SOLUTION STOICHIOMETRYSOLUTION STOICHIOMETRY

GENERAL PLAN FOR STOICHIOMETRY GENERAL PLAN FOR STOICHIOMETRY CALCULATIONSCALCULATIONS

Mass zinc

Molar RatioMoleszinc

Moles HCl

Mass HCl

VolumeHCl

Step 1: Step 1: Write the balanced equationWrite the balanced equation

Zn(s) + 2 HCl(aq) Zn(s) + 2 HCl(aq) →→ ZnClZnCl22(aq) + H(aq) + H22(g)(g)

Step 2: Step 2: Calculate amount of ZnCalculate amount of Zn

10.0 g Zn • 1.00 mol Zn65.39 g Zn

= 0.153 mol Zn

Zinc reacts with acids to produce HZinc reacts with acids to produce H22 gas. If you gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?needed to convert the Zn completely?

Step 3: Step 3: Use the Use the Molar RatioMolar Ratio

0.153 mol Zn • 2 mol HCl1 mol Zn

= 0.306 mol HCl

0.306 mol HCl • 1.00 L2.50 mol

= 0.122 L HCl

Zinc reacts with acids to produce HZinc reacts with acids to produce H22 gas. If you gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?needed to convert the Zn completely?

Step 4: Step 4: Calculate volume of HCl requiredCalculate volume of HCl required

ACIDACID--BASE REACTIONSBASE REACTIONSTitrationsTitrations

HH22CC22OO44(aq) + 2 NaOH(aq) (aq) + 2 NaOH(aq) →→NaNa22CC22OO44(aq) + 2 H(aq) + 2 H22O(liq)O(liq)acidacid basebase

Carry out this reaction using a Carry out this reaction using a TITRATIONTITRATION..

Oxalic acid,Oxalic acid,HH22CC22OO44

Setup for titrating an acid with a baseSetup for titrating an acid with a base

TitrationTitration1. Add solution from the buret.1. Add solution from the buret.2. Reagent (base) reacts with 2. Reagent (base) reacts with

compound (acid) in solution in compound (acid) in solution in the flask.the flask.

3. Indicator shows when exact 3. Indicator shows when exact stoichiometric reaction has stoichiometric reaction has occurred.occurred.

4. Net ionic equation4. Net ionic equationHH++ + OH+ OH-- →→ HH22OO

5. At equivalence point 5. At equivalence point moles Hmoles H++ = moles OH= moles OH--

1.065 g of H1.065 g of H22CC22OO44 (oxalic acid) (oxalic acid) requires 35.62 mL of NaOH for requires 35.62 mL of NaOH for titration to an equivalence point. titration to an equivalence point. What is the concentraWhat is the concentra--tion of the tion of the NaOH?NaOH?

LAB PROBLEM #1: Standardize a solution LAB PROBLEM #1: Standardize a solution of NaOH of NaOH —— i.e., accurately determine its i.e., accurately determine its concentration.concentration.

1.065 g of H1.065 g of H22CC22OO44 (oxalic acid) requires 35.62 mL of NaOH (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the for titration to an equivalence point. What is the concentration of the NaOH?concentration of the NaOH?

Step 1: Step 1: Calculate amount of HCalculate amount of H22CC22OO44

1.065 g • 1 mol90.04 g

= 0.0118 mol

0.0118 mol acid • 2 mol NaOH1 mol acid

= 0.0236 mol NaOH

Step 2: Step 2: Calculate amount of NaOH requiredCalculate amount of NaOH required

1.065 g of H1.065 g of H22CC22OO44 (oxalic acid) requires 35.62 mL of NaOH (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the for titration to an equivalence point. What is the concentration of the NaOH?concentration of the NaOH?

Step 1: Step 1: Calculate amountCalculate amount of Hof H22CC22OO44

= 0.0118 mol acid= 0.0118 mol acidStep 2: Step 2: Calculate amount of NaOH requiredCalculate amount of NaOH required

= 0.0236 mol NaOH= 0.0236 mol NaOHStep 3: Step 3: Calculate concentration of NaOHCalculate concentration of NaOH

0.0236 mol NaOH0.03562 L

= 0.663 M

[NaOH] = 0.663 M[NaOH] = 0.663 M

LAB PROBLEM #2: LAB PROBLEM #2: Use standardized NaOH to determine the Use standardized NaOH to determine the amount of an acid in an unknown.amount of an acid in an unknown.

Apples contain malic acid, CApples contain malic acid, C44HH66OO55..

CC44HH66OO55(aq) + 2 NaOH(aq) (aq) + 2 NaOH(aq) →→

NaNa22CC44HH44OO55(aq) + 2 H(aq) + 2 H22O(liq)O(liq)

76.80 g of apple requires 34.56 mL of 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is 0.663 M NaOH for titration. What is weight % of malic acid?weight % of malic acid?

76.80 g of apple requires 34.56 mL of 0.663 M 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid?NaOH for titration. What is weight % of malic acid?

Step 1: Step 1: Calculate amount of NaOH used.Calculate amount of NaOH used.C C •• V = (0.663 M)(0.03456 L) V = (0.663 M)(0.03456 L)

= 0.0229 mol NaOH= 0.0229 mol NaOHStep 2: Step 2: Calculate amount of acid titrated.Calculate amount of acid titrated.

0.0229 mol NaOH • 1 mol acid2 mol NaOH

= 0.0115 mol acid= 0.0115 mol acid


Step 3: Step 3: Calculate mass of acid titrated.Calculate mass of acid titrated.

0.0115 mol acid • 134 gmol

= 1.54 g

Step 1: Step 1: Calculate amount of NaOH used. Calculate amount of NaOH used. = 0.0229 mol NaOH= 0.0229 mol NaOH

Step 2: Step 2: Calculate amount of acid titratedCalculate amount of acid titrated= 0.0115 mol acid= 0.0115 mol acid

1


Step 1: Step 1: Calculate amount of NaOH used. Calculate amount of NaOH used. = 0.0229 mol NaOH= 0.0229 mol NaOH

Step 2: Step 2: Calculate amount of acid titratedCalculate amount of acid titrated= 0.0115 mol acid= 0.0115 mol acid

Step 3: Step 3: Calculate mass of acid titrated.Calculate mass of acid titrated.= 1.54 g acid= 1.54 g acid

Step 4: Step 4: Calculate % malic acid.Calculate % malic acid.1.54 g76.80 g

• 100% = 2.01%

CHM141 Chapter4 7e Kotz

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Transcript of CHM141 Chapter4 7e Kotz