4. Stoichiometry

1. Stoichiometric Equations

2. Limiting Reagent Problems

3. Percent Yield

4. Limiting Reagent Problems

5. Concentrations of Solutes

6. Chemical Analysis

4.1 Stoichiometry

“Chemical Arithmetic”

Objective: Determine Quantitative

Relationships Between Reactants and

Products

Two “Ways” to Quantify Matter

1. Count

2. Measure

Stoichiometry

•Chemical Equations Relate the Numbers of Various

Species to Each Other (Counted Quantities)

•We Observe Measurable Quantities Like Mass

and Volume

Measured

Quantity

Reactants

Counted

Quantity

Reactants

Counted

Quantity

Products

Measured

Quantity

Products

•We Need to be Able to Connect the Two

Chemical

EquationMass, Volume,

Pressure, Temperature)

Mass, Volume,

Pressure, Temperature)

3 Basic Steps

Measured

Quantity

Reactants

Counted

Quantity

Reactants

Counted

Quantity

Products

Measured

Quantity

Products1.

1. Mass to Mole Conversion: Determine # of Moles of all

Species Present (Start with Mass – something you can measure)

2.

2. Mole to Mole Conversion: Use Stoichiometric Ratios to

Relate Various Species to Each Other

3.

3. Mole to Mass Conversion: Multiple Moles Product by Molar

Mass (End with Mass P – something you can measure)

Combustion of Ethanol

Consider the combustion of 16.0 g ethanol

What kind of questions

can we ask?

C2H5OH

Combustion of Ethanol

Consider the combustion of 16.0 g ethanol

C2H5OH + 3O2 2CO2 + 3H2O

4. Quantity of Water Produced (moles or mass)

2. Quantity of Oxygen Consumed (moles or mass) .

3. Quantity of CO2 Produced (moles or mass)

(5. Quantity of Energy Produced or Consumed)

1. Quantity of Ethanol Consumed in Moles

Quantity of Ethanol Consumed

in Moles

C2H5OH + 3O2 2CO2 + 3H2O

2 52 5

2 5

(16.0 )( )46.02

0.348

mol C H OHg C H OH

g

mol C H OH

Quantity of Oxygen Consumed

in Moles or Grams

C2H5OH + 3O2 2CO2 + 3H2O

2 5 22 5 2

2 5

3(16.0 )( )( ) 1.04

46.02

mol C H OH mol Og C H OH mol O

g mol C H OH

2 5 22 5

2 5 2

2

3 32.0(16.0 )( )( )( )

46.02

33.4

mol C H OH mol O gg C H OH

g mol C H OH mol O

g O

Quantity of Carbon Dioxide Produced

in Moles or Grams

C2H5OH + 3O2 2CO2 + 3H2O

2 5 22 5 2

2 5

2(16.0 )( )( ) 0.695

46.02

mol C H OH mol COg C H OH mol CO

g mol C H OH

2 5 22 5

2 5 2

2

2 44.01(16.0 )( )( )( )

46.02

30.6

mol C H OH mol CO gg C H OH

g mol C H OH mol CO

g CO

Quantity of Water Produced

in Moles or Grams

C2H5OH + 3O2 2CO2 + 3H2O

2 5 22 5 2

2 5

3(16.0 )( )( ) 1.04

46.02

mol C H OH mol H Og C H OH mol H O

g mol C H OH

2 5 22 5

2 5 2

2

3 18.015(16.0 )( )( )( )

46.02

18.8

mol C H OH mol H O gg C H OH

g mol C H OH mol H O

g H O

4.2 Limiting Reagent Problems

•Stoichiometric Proportions: Reactants mixed in

Ratios Given by Stoichiometric Coefficients.

Everything is Completely Consumed

•Nonstoichiometric Proportions: Reactants are

Mixed in Ratios Different than Stoichiometric

Coefficients. One species is Completely

Consumed (Limiting Reagent) While Another

is not Completely Consumed (Excess Reagent)

Theoretical Yield

The Maximum Amount of Product Which

Could be Produced by the Complete

Consumption of the Limiting Reagent

Silver Tarnishes in the Presence of Hydrogen

Sulfide and Oxygen due to the Reaction:

4Ag + 2H2S + O2 ----> 2Ag2S + 2H2O

How many grams of Silver Sulfide Would be

Formed from 2.4g Ag, 0.48g H2S & 0.16g O2?

2.4g Ag

0.48g H2S

0.16g O2(1mol O2/32g)

(1mol Ag/107.9g)

(1mol H2S/34g)

(2mol Ag2S/mol O2)=

(2mol Ag2S/4mol Ag)=

(2mol Ag2S/2mol H2S)=

Oxygen is the Limiting Reagent

0.011mol Ag2S

0.014mol Ag2S

0.010mol Ag2S

Mass Silver Sulfide Produced:

(0.010mol Ag2S) (247.9Ag2S/mol)=2.5g Ag2S

4Ag + 2H2S + O2 ----> 2Ag2S + 2H2O

How Do We Quickly Identify Limiting

Reactants in Stoichiometric Problems?

1. Determine Moles for Each Reactant

2.Divide Each by It’s Stoichiometric

Coefficient.

3. Limiting Reagent Has Smallest Value

(Reactants are in Stoichiometric Proportions if

all Values are Equal)

(Now Answer the Question)

A Closer Look at the Last Problem

2.4g Ag(1mol Ag/107.9g) (2mol Ag2S/4mol Ag)= 0.011mol Ag2S

0.48g H2S(1mol H2S/34g) (2mol Ag2S/2mol H2S)=0.014mol Ag2S

0.16g O2 (1mol O2/32g) (2mol Ag2S/mol O2) =0.010mol Ag2S

2.4g Ag(1mol Ag/107.9g) (1/4mol Ag) (2mol Ag2S )= 0.011mol Ag2S

0.48g H2S(1mol H2S/34g) (1/2mol H2S) (2mol Ag2S )= 0.014mol Ag2S

0.16g O2 (1mol O2/32g) (1/mol O2) (2mol Ag2S ) = 0.010mol Ag2S

4Ag + 2H2S + O2 ----> 2Ag2S + 2H2O

Determine Moles

Each Species

Divide by

Coefficient

Note: Next Step Multiples

all by the Same Value

A Closer Look at Short Cut Method

2.4g Ag(1mol Ag/107.9g) /4mol Ag)= 0.00556

0.48g H2S(1mol H2S/34g) /2mol H2S)=0.00706

0.16g O2 (1mol O2/32g) /mol O2) =0.005

4Ag + 2H2S + O2 ----> 2Ag2S + 2H2O

Possible Questions;

1. Theoretical Mass Yield Ag2S

2. Theoretical Mass Yield H2O

3. Excess Ag

4. Excess H2S

Theoretical Mass Yield Ag2S

0.16g O2 (mol O2/32g)(2molAg2S /mol O2)(247.9g/molAg2S )= 2.5g

4Ag + 2H2S + O2 ----> 2Ag2S + 2H2O

or

[0.005](2molAg2S )(247.9g/molAg2S)= 2.5g

Theoretical Mass Yield H2O

0.16g O2 (mol O2/32g)(2molH2S /mol O2)(18g/molH2O )=0.18g

4Ag + 2H2S + O2 ----> 2Ag2S + 2H2O

or

[0.005](2molH2O )(18g/mol H2O)= 0.18g

Excess Mass of Ag

Mass Ag Left Over = Initial Mass -Mass Consumed

4Ag + 2H2S + O2 ----> 2Ag2S + 2H2O

2.4g- 0.16gO2 (mol O2/32g)(4molAg /molO2)(107.9g/molAg ) = 0.2g

or

2.4- [0.005](4 molAg )(107.9g/mol Ag) = 0.2g

Excess Mass of H2S

Mass H2S Left Over = Initial Mass -Mass Consumed

4Ag + 2H2S + O2 ----> 2Ag2S + 2H2O

0.48g- 0.16gO2 (mol O2/32g)(2molH2S /molO2)(34g/molH2S) = 0.24g

or

0.48 - [0.005](2 molH2S)(34g/mol H2S) = 0.24g

How Much Oxygen is Left Over?

None:

It is the limiting reagent

Class Problem

Phosphoric Acid can be synthesized from

phosphorus, oxygen and water according

to the following reactions:

4P + 5O2 -----> P4O10

P4O10 + 6H2O ---> 4H3PO4

Starting with 20.0g P, 30.0g O2 and 15.0 g

H2O, how many grams of Phosphoric

acid can be Produced?

P&O

Combining Equations gives:

eq1. 4P + 5O2 -----> P4O10

eq2 P4O10 + 6H2O ---> 4H3PO4

eq3 4P + 5O2 + 6H2O -----> 4H3PO4

(Note, the P4O10 could be considered as a

reaction intermediate)

4P + 5O2 + 6H2O -----> 4H3PO4

20gP(mol P/31g)(1/4 molP) = 0.1613

30gO2(molO2/32g)(1/ 5mol O2) = 0.1875

15gH2O(molH2O/18g)(1/6mol H2O) = 0.1389

0.1389(4molH3PO4)(98g H3PO4/mol)

= 54.5g H3PO4

(15gH2O)(molH2O/18g)(4molH3PO4/6mol H2O) (98g H3PO4/mol)

= 54.5g H3PO4

OR

4.3 Percent Yields

Often Times the Actual Yield of a

Chemical Reaction is Less Than That

Predicted by the Complete Consumption

of the Limiting Reagent

Percent Yield = Actual Yield (100)

Theoretical Yield

Note: Theoretical Yield Is Based on the Complete

Consumption of the Limiting Reagent

Why Would the Actual Yield Be

Less Than the Theoretical Yield?

1. Equilibrium State (where Reactants exist together

with Products)

2. Existence of Intermediates (two step process like

formation of Phosphoric Acid)

3. Kinetic Factors (the reaction is slow, maybe not all of

the reactants have “diffused” together).

4. Alternative Chemical Reactions (there may be more

than one possible product produced).

Determine Percent Yield if 2.3g of

Silver Sulfide form when 2.4g Ag,

0.28g H2S & 0.16g O2 react.

From Previous Problem,

the Theoretical Yield is

2.5g Ag2S

%Yield (Ag2S) = (2.3g/2.5g)100

= 92%

Class Problem

If, 20.0g P, 30.0g O2 and 15.0 g H2O;

Yields 50.3 g H3PO4, what is the

Percent Yield?

Note in Previous Problem the Theoretical Yield

was determined to be 54.5 g phosphoric acid.

Class Problem:

(50.3g H3PO4/ 54.5g H3PO4)(100)

= 92.3% Yield H3PO4

4.4 Chemical Equations &

Chemical Analysis

Acidum Formicum – Formic AcidWhat is the empirical formula

of Formic Acid?

Obtained through the distillation of Formica rufa

Formic acid contains H, O and C,

combustion of 2.4527 g of formic acid

yields 0.9606g water

and 2.3482g of carbon dioxide.

Empirical Formula of Formic Acid

from Combustion Data

Sample (CxHyOn)

furnace H2O absorber

CO2

absorber

excessoxygen

2.4527g

formic acid0.9606g 2.3482g

mH = .9606g(2gH/18gH2O ) = .1067g H

mC = 2.3482g(12gC/44gCO2 ) = .6404g C

mO = 2.4527g - .1067g - .6404g = 1.7056gO

Step 1: Determine Mass of each element

Step 2: Determine Mole of Each Element1

0.1067 0.10671.008

10.6404 0.0533

12.01

11.7056 0.1066

16.00

molmoles H gH mol H

g

molmoles C gC mol C

g

molmoles O gO mol O

g

Step 3: Divide by Smallest Number

1: 0.1067 2

0.0533

1: 0.0533 1

0.0533

1: 0.1066 2

0.0533

H

C

O

CH2O2

4.5 Concentration of a Solute

1. Mass Concentration: mass solute m

vol solution VTypical Units: g/L

2. Mole Concentration: moles solute n

Mvol solution V

(Molarity) Typical Units: M=mol/L

Conventions

m = mass

n = moles

M = molarity

Molarity

moles soluteM molarity

Liter solution

Why is Molarity so Important in Solution

Stoichiometric Calculations?

Because solutes are typically reactants and we can measure

the solution volume, this gives us the moles solute. ie., we

can “Count moles solute” by measuring the volume of a

solution of known Molarity

Calculate the Molarity of a Solution if 2.00 g NaCl

is diluted to 250.0 ml with water.

1(2.00 )( )( ) .137

58.45 .250

mol NaClg NaCl M

g L

What mass of NaOH(s) would you need to

make 500. mL of .70M NaOH?

.70 40.500 ( )( ) 14

mol NaOH g NaOHL g

L mol

How do You Make 500.0 ml of .1560M

CuSO4?

1. Calculate amount of Copper(II) Sulfate you

will need.

4 40.1560 159.60.5000 ( )( ) 12.45

mol CuSO g CuSOL g

L mol

2. Weight the Copper(II) Sulfate

3. Quantitatively Transfer to 500 ml

Volumetric Flask

4. Dilute to Volume

Concentration of Ions

The concentration of ions are related to the salt

via it’s formula

Al2(SO4)3 2 Al+3 + 3 SO4-2

What are the ion concentrations for a

1M Al2(SO4) solution?

2M in Al+3 and 3M in SO4-2

(5M in Total Ion Concentration)

Dilution Problems

Many Reagents Come as Stock Solutions

Adding Solvent Does Not Change The Moles

of Solute

Moles Before

Dilution

Moles After

Dilution=

MiVi = MfVf

nM n MV

V

ninitial = nfinal

How would we make 100. mL of

4.0 M Sulfuric Acid from Stock

16 M Sulfuric Acid?

1. Determine Amount of 16 M

Sulfuric Acid needed.

4.0( ) ( )100 25

16

f

i f

i

M MV V mL mL

M M

MiVi = MfVf

2. Accurately Measure 25mL

with Volumetric Pipet (TD)

3. Transfer to 100ml Volumetric Flask

4. Dilute to Volume

pH Scale

pH = -log[H3O+]

pX = -log[X]

[H+] = 10-pH = 1

10pH

pH & Significant Figures

What is pH if [H+] = 3.50x10-6?

5.456

Exact # tells

position of

decimal

Shows # of significant

figures

pH of Strong Acids & Bases

Consider 100% Dissociation

What is the pH of a 0.00836M HCl

solution?

pH=-log(0.00836)=2.078

pH of Strong Acids & Bases

What is the hydronium ion

concentration for a solution with

a pH of 4.041?

[H3O+]=10-pH = 10-4.041= 9.099x10-5

= 9.10x10-5

Solution Reaction Stoichiometry

2. Write Balanced Net Ionic Eq.

3. Calculate Moles of Each Reactant

4. Determine Limiting Reagent

5. Calculate Moles of Other Reactants

and Products as Desired

6. Convert Results to Desired Units (Moles,

Molarity, Mass…)

1. Identify species present and possible rxns.

Solution Stoichiometry Problem

Consider mixing 30.0mL of 0.750M

sodium chromate with 40.0 mL of 0.500 M

aluminum chloride.

1. Predict the mass and identity of

any precipitate formed.

2. Predict the concentrations of any

ions remaining in solution.

1. Write Balanced Eq.

3Na2CrO4(aq) + 2AlCl3(aq) 6NaCl + Al2(CrO4)3

2. Use Solubility Rules to Predict

States of Products

3Na2CrO4(aq) + 2AlCl3(aq) 6NaCl(aq) + Al2(CrO4)3(s)

3. Write Net Ionic Equation

3CrO4-2 + 2Al+3

Al2(CrO4)(s)

3. Identify Initial Moles of all

Species

2 4

2 4

.750 20.0300 0.0450

mol Na CrO mol Nal mol Na

l mol Na CrO

222 4 4

4

2 4

.7500.0300 0.0225

mol Na CrO mol CrOl mol CrO

l mol Na CrO

333

3

.50.0400 0.0200

mol AlCl mol All mol Al

l mol AlCl

Na+:

CrO4-2:

Al+3:

Cl-: 3

3

.5 30.0400 0.0600

mol AlCl mol Cll mol Cl

l mol AlCl

4. Determine Precipitate Based

on Complete Consumption of

Limiting Reagent

3CrO4-2 + 2Al+3

Al2(CrO4)(s)

2 4 2 42 3 34 2 42 3

4

401.960.0225 3.01

3

mol Al CrO g Al CrOmol CrO g Al CrO

mol CrO mol

2 4 2 43 3 32 43 3

401.960.0200 4.02

2

mol Al CrO g Al CrOmol Al g Al CrO

mol Al mol

3.01g Al2(CrO4)3

5. Determine Concentration of

Excess Reagent

molesInitial - molesConsumed

volTotal

33 2

4 2

4

20.0200 0.0225

30.0714

0.0400 0.0300

mol Almol Al mol CrO

mol CrOM

l l

0.714M Al+3

6. Determine Concentration of

Spectator Ions

Na+: 0.0450 mol Na+

0.0700 L= 0.643M

Cl-: 0.0600 mol Cl-

0.0700 L0.857M=

Acid-Base Titrations

Analytical Technique to Determine the Quantity

of an Unknown Acid or Base by Neutralizing

With an Acid or Base of Known Concentration.

Titration Procedures

1. Add Acid to

Erlenmeyer Flask,

then add indicator

2. Add Base with

Buret

3. Quit When

Indicator Turns

pink

ab-tit

What is the concentration of an unknown Sulfuric

Acid solution if it takes 35mL of .40 M NaOH to

neutralize 50.0 mL of the acid?

2NaOH + H2SO4 ---> Na2SO4 + 2H2O

4242 14.0)

05.0

1)(

2

1)(

4.(035. SOHM

LNaOHmol

SOHmol

L

NaOHmolNaOHL

-Note, we are using the initial volume of the acid as

we want to know it’s initial strength