stoichiometry ii dr. ron rusay spring 2008 © copyright 2008 r.j. rusay

Stoichiometry IIStoichiometry II

Dr. Ron RusayDr. Ron Rusay

Spring 2008Spring 2008

© Copyright 2008 R.J. Rusay© Copyright 2008 R.J. Rusay

Chemical Equations / Chemical Equations / Stoichiometric CalculationsStoichiometric Calculations

•Example:Example:

CC88HH18(l)18(l)+ O+ O2(g)2(g) -----> CO -----> CO2(g)2(g) + H + H22OO(l)(l)

•For the combustion of octane which produces For the combustion of octane which produces

carbon dioxide and water.carbon dioxide and water.•The equation must conserve mass ,i.e. account The equation must conserve mass ,i.e. account

for the mass in product and reactant by having a “balance” for the mass in product and reactant by having a “balance”

which is an equal number of individual atoms in which is an equal number of individual atoms in

reactant and product.reactant and product.

© Copyright 1995-2002 R.J. Rusay© Copyright 1995-2002 R.J. Rusay

Chemical EquationsChemical Equations The equation is “balanced” as before by placing The equation is “balanced” as before by placing

stoichiometric factors before each of the stoichiometric factors before each of the molecules in the reaction so that the atoms equalmolecules in the reaction so that the atoms equal

?? C C88HH18(l) 18(l) + + ?? O O2(g)2(g) -----> -----> ?? CO CO2(g) 2(g) + +

? ? HH22OO(l)(l)

1 25/2 89

Remove fraction:

2 C8H18(l)+ 25 O2(g)-----> 16 CO2(g)+18

H2O(l)

1


Mass CalculationsMass Calculations

The Balanced Equation is a balance on a molar The Balanced Equation is a balance on a molar basis which can be related to mass.basis which can be related to mass.

2 2 CC88HH18(l)18(l)+ + 25 25 OO2(g)2(g)----->----->

16 16 COCO2(g)2(g)++18 18 HH22OO(l)(l)

228 g of octane (228 g of octane (2 moles)* 2 moles)* will react with 800 g of will react with 800 g of oxygen (oxygen (25 moles) 25 moles) to produce 704 g of carbon to produce 704 g of carbon dioxide and 324 g of water. dioxide and 324 g of water.

*(*(2 moles octane x 114 g/mol = 228 g )2 moles octane x 114 g/mol = 228 g )

1


1. Balance the chemical equation.2. Convert mass of reactant or product

to moles.3. Identify mole ratios in balanced equation:

They serve as the “Gatekeeper”.4. Calculate moles of desired product or

reactant.5. Convert moles to grams.

Mass Calculations: Mass Calculations: ProductsProducts Reactants Reactants



Something (S) Another Thing (AT)

Mass (S) Mass (AT)

grams (S) grams (AT)

Chemically Relate:




grams (S) grams (AT)

grams (S)

grams (S)

(Molecular Weight)

1 mol (S)

grams (AT)

grams (AT)

1 mol (AT)

(Molecular Weight)

Avogadro's NumberAtomsMoleculesStoichiometry

?

?"Gatekeeper"

Mass Calculations: Mass Calculations: ReactantsReactants Products Products

� How many grams of salicylic acid are needed to How many grams of salicylic acid are needed to produce 1.80 kg of aspirin?produce 1.80 kg of aspirin?

� Balanced Equation:Balanced Equation:

C7H6O3

MW = 138.12C2H3OClMW = 78.49

C9H8O4

MW = 180.15HClMW= 36.45

O

H

C

O

O

H

H

H

H

H

+ C

H

3

C

O

C

l

O

C

C

H

3

C

O

O

H

H

H

H

H

O

+ Cl

H

Salicylic Acid Aspirin

Mass Calculations: Mass Calculations: ReactantsReactants Products Products


grams (Aspirin) grams (Salicylic Acid)

1800 grams (A)

grams (A)

(Molecular Weight A)

1 mol (A)

grams (SA)

? (SA)

1 mol (SA)

(Molecular Weight SA)

Avogadro's NumberAtomsMoleculesStoichiometry

1 mol SA

1 mol A

"Gatekeeper"

Calculation: Calculation: http://ep.llnl.gov/msds/Stoichiometry/Aspirin-Calc.html

http://ep.llnl.gov/msds/Stoichiometry/Aspirin-Calc.html







Mass Calculations: Mass Calculations: How many grams of salicylic How many grams of salicylic acid are needed to produce acid are needed to produce 1.80 kg1.80 kg of aspirin? of aspirin?


C7H6O3

MW = 138.12C2H3OClMW = 78.49

C9H8O4

MW = 180.15HClMW= 36.45

O

H

C

O

O

H

H

H

H

H

+ C

H

3

C

O

C

l

O

C

C

H

3

C

O

O

H

H

H

H

H

O

+ Cl

H


?g C7 = 1.80 x 103 gC9 x [molC9 /180.15gC9 ] x [1 molC7/ 1 molC9] x 138.12 gC7/molC7

11

?g C7 = 1380 grams

Limiting ReagentLimiting ReagentAn Ice Cream SundaeAn Ice Cream Sundae

??????

Limiting ReagentLimiting ReagentAn Ice Cream SundaeAn Ice Cream Sundae

What’s left?What’s totally consumed?

Limiting ReagentLimiting Reagent

QuickTime™ and aCinepak Codec by Radius decompressor

are needed to see this picture.

QuickTime™ and aSorenson Video decompressorare needed to see this picture.

QuickTime™ and aAnimation decompressor

are needed to see this picture.

QUESTIONQUESTIONThe limiting reactant in a reaction: 1) is the reactant for which there is the least

amount in grams. 2) is the reactant which has the lowest

coefficient in a balanced equation. 3) is the reactant for which there is the most

amount in grams. 4) is the reactant for which there is the fewest

number of moles. 5) none of these

ANSWERANSWER

)

5) none of these

Section 3.10 : Calculations Involving a Limiting Reactant (p. 106

The limiting reactant in a reaction is the reactant that produces the least number of grams of any product.

Mass Applications: Mass Applications: Limiting ReagentLimiting Reagent

How do masses of reactants relate? Is there How do masses of reactants relate? Is there enough mass of each reactant for the reaction to enough mass of each reactant for the reaction to consume all of both of them or will there be some consume all of both of them or will there be some left of one of them? left of one of them?

2 2 CC88HH18(l)18(l)+ + 25 25 OO2(g)2(g)----->----->

16 16 COCO2(g)2(g)++18 18 HH22OO(l)(l)

What would happen if only 600. g of OWhat would happen if only 600. g of O22 were were

available for the reaction of 228 g of octane? available for the reaction of 228 g of octane?

1


Mass Applications: Mass Applications: Determining a Limiting ReagentDetermining a Limiting Reagent

� Does one of the reactants have fewer Does one of the reactants have fewer stoichiometrically adjusted moles than the stoichiometrically adjusted moles than the other reactant? If so, the reactant with the other reactant? If so, the reactant with the smaller value is the limiting reagent.smaller value is the limiting reagent.

Calculation:Calculation:� Divide the mass of each reactant by its Divide the mass of each reactant by its

respective Molar Mass and by its respective Molar Mass and by its Stoichiometric factor from the balanced Stoichiometric factor from the balanced equation; then compare the results.equation; then compare the results.

Limiting Reagent CalculationLimiting Reagent Calculation The The reactant reactant present in the smallest molar present in the smallest molar

amount considering stoichiometry limits the mass amount considering stoichiometry limits the mass basis of any reaction. basis of any reaction.

2 2 CC88HH18(l)18(l)+ + 25 25 OO2(g)2(g)----->----->

16 16 COCO2(g)2(g)++18 18

HH22OO(l)(l)

228 g octane / 114 g/mol = 2 mol octane 228 g octane / 114 g/mol = 2 mol octane

600. g oxygen / 32 g/mol = 18.75 mol oxygen600. g oxygen / 32 g/mol = 18.75 mol oxygen

2 mol octane / 2 mol (stoich.) = 12 mol octane / 2 mol (stoich.) = 1

18.75 mole oxygen / 25 mol (stoich.) = 0.7518.75 mole oxygen / 25 mol (stoich.) = 0.75

1


Mass Effects of theMass Effects of the Limiting ReagentLimiting Reagent

What amount of octane remains unreacted in What amount of octane remains unreacted in the reaction of 600. g of Othe reaction of 600. g of O22 with 228 g of with 228 g of

octane?octane?

600. g O600. g O22 x mol O x mol O22 /32g O /32g O22 x [ x [22mol Cmol C88HH18 18

//2525mol Omol O22] x 114 g / mol ] x 114 g / mol CC88HH18 18 = 171 g = 171 g

CC88HH18 18 are reacted are reacted

228 g - 171 g = 57 g C228 g - 171 g = 57 g C88HH18 18 remain remain

unreactedunreacted

1


Limiting ReagentLimiting Reagent / / Theoretical YieldTheoretical Yield

The limiting reagent governs the theoretical The limiting reagent governs the theoretical yield of products. For the reaction of 228 g yield of products. For the reaction of 228 g of octane with 600. g of oxygen, what is the of octane with 600. g of oxygen, what is the theoretical yield of carbon dioxide?theoretical yield of carbon dioxide?

2 2 CC88HH18(l)18(l)+ + 25 25 OO2(g)2(g)----->----->

16 16 COCO2(g)2(g)++18 18 HH22OO(l)(l)

600. g O600. g O22 x mol O x mol O22 /32g O /32g O22 x x 1616mol COmol CO22 / /2525mol Omol O22 x x

44g / mol CO44g / mol CO22 = = 528 g 528 g COCO22

1


Thoughts to ConsiderThoughts to Consider

How much COHow much CO22 do you produce per gallon of gasoline do you produce per gallon of gasoline

(octane, d= 0.70 g/ml) when gasoline is combusted?(octane, d= 0.70 g/ml) when gasoline is combusted?

How much COHow much CO2 2 do you personally produce from driving do you personally produce from driving

every week?…. every month? …. every year? ….from every week?…. every month? …. every year? ….from other uses and sources?other uses and sources?

Why do people in developed nations, like the U.S., Japan Why do people in developed nations, like the U.S., Japan & in the EU, produce tons more of CO& in the EU, produce tons more of CO2 2 per person per person

than people in under developed or developing nations?than people in under developed or developing nations?

Does the increase in “man-made” CODoes the increase in “man-made” CO2 2 relate to global relate to global

warming?warming?

Greenhouse Gases: The Greenhouse Gases: The Chemistry of WarmingChemistry of Warming

What is a greenhouse gas?The sun’s energy & the molecule’s shape decide.

•Our atmosphere (air) is 78% nitrogen and 21% oxygen. Our atmosphere (air) is 78% nitrogen and 21% oxygen. •Neither are greenhouse gases. They do not absorb infrared radiation (heat). Neither are greenhouse gases. They do not absorb infrared radiation (heat). •However, HHowever, H22O and COO and CO2 2 can absorb infrared energy. Without them earth would can absorb infrared energy. Without them earth would

be very chilly.be very chilly.

http://zebu.uoregon.edu/1998/es202/l13.htmlhttp://zebu.uoregon.edu/1998/es202/l13.html

QUESTIONQUESTIONHow many grams of Ca(NO3)2 can be produced by reacting excess HNO3 with 7.40 g of Ca(OH)2? 1) 10.2 g 2) 16.4 g 3) 32.8 g 4) 8.22 g 5) 7.40 g

ANSWERANSWER

)

2 3

3 2 2

2) 16.4 g

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products (p. 102

The balanced equation is Ca(OH) + 2HNO Ca(NO ) + 2H O. Water is the other product for an acid/base reaction.

Percent YieldPercent Yield In synthesis, the actual yield (g) is measured and In synthesis, the actual yield (g) is measured and

compared to the theoretical yield (g). This is the compared to the theoretical yield (g). This is the

percent yield:percent yield: % % = actual / theoretical x 100= actual / theoretical x 100

If a reaction produced 2.45g of Ibogaine, If a reaction produced 2.45g of Ibogaine, CC2020HH2626NN22O, a natural product with strong promise O, a natural product with strong promise

in treating heroin addiction, and the theoretical in treating heroin addiction, and the theoretical yield was 3.05g, what is the % yield?yield was 3.05g, what is the % yield?

1


% yield = 2.45g / 3.05g x 100= 80.3%

QUESTIONQUESTIONThe amount of tungsten metal, used to make many lightbulb filaments, starting with 25.0 grams of WO3 in the following reaction was 18.0 grams. What percent yield does this represent?

WO3 + 3 H2 W + 3 H2O

1. 25.8%2. 110%3. 90.8%4. I think I may have confused one of the steps in this calculation

ANSWERANSWERChoice 3 has properly converted the initial grams of WO3 to moles and used the mole ratio of the equation to determine the maximum moles of W that would be produced if all the WO3 reacted (theoretical yield – 19.8 g). The division of the actual yield by the theoretical yield 100 produces the answer.

Section 3.8: Stoichiometric Calculations: Amounts of Reactants and Products

Percent YieldPercent Yield

� A reaction was conducted that theoretically A reaction was conducted that theoretically would produce 0.0025 moles of quinine, would produce 0.0025 moles of quinine, CC2020

HH2424 N N22 O O22 . The actual amount of isolated . The actual amount of isolated

quinine was 780 mg. What is the percent yield quinine was 780 mg. What is the percent yield of quinine?of quinine?

� 324 g/mol x 0.0025 mol = 81g = 324 g/mol x 0.0025 mol = 81g = 810mg(theoretical)810mg(theoretical)

� % Yield = 780 mg/ 810 mg x 100% Yield = 780 mg/ 810 mg x 100� % Yield = 96%% Yield = 96%

Mass Calculations: Mass Calculations: How many grams of salicylic How many grams of salicylic acid are needed to produce acid are needed to produce 1.80 kg1.80 kg of aspirin if the process of aspirin if the process

produces an 85.0% yield?produces an 85.0% yield?


C7H6O3

MW = 138.12C2H3OClMW = 78.49

C9H8O4

MW = 180.15HClMW= 36.45

O

H

C

O

O

H

H

H

H

H

+ C

H

3

C

O

C

l

O

C

C

H

3

C

O

O

H

H

H

H

H

O

+ Cl

H


?g C7 = 1.80 x 103 gC9 x [molC9 /180.15gC9 ] x [1 molC7/ 1 molC9] x 138.12 gC7/molC7

11

?g C7 = 1380 grams ?g C7 = 2.12 x 103 gC9 x [molC9 /180.15gC9 ] x [1 molC7/ 1 molC9] x 138.12 gC7/molC7

?g C7 = 1624 grams

1.80 x 103 gC9 /85/100 = 2.12 x 103 gC9

= 1380 grams /85/100

stoichiometry ii dr. ron rusay spring 2008 © copyright 2008 r.j. rusay

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