Chapter 3Chapter 3

Calculations with

Chemical Formulas

and Equations

HST Mr. Watson

Mr. Watson HST

Molar MassMolar Mass

Sum

atomic masses

represented by formula

atomic masses => gaw

molar mass => MM

Mr. Watson HST

One Mole of each SubstanceOne Mole of each Substance

Clockwise from top left:1-Octanol, C8H17OH;Mercury(II) iodide, HgI2;Methanol, CH3OH; andSulfur, S8.

Mr. Watson HST

ExampleExample

What is the molar mass of ethanol, C2H5OH?

Mr. Watson HST

ExampleExample

What is the molar mass of ethanol, C2H5O1H1?

MM = 2(gaw)C + (5 + 1)(gaw)H + 1(gaw)O

Mr. Watson HST

ExampleExample

What is the molar mass of ethanol, C2H5OH?

MM = 2(gaw)C + (5 + 1)(gaw)H + 1(gaw)O

= 2(12.011)C + 6(1.00794)H + 1(15.9994)O

Mr. Watson HST

ExampleExample

What is the molar mass of ethanol, C2H5OH?

MM = 2(gaw)C + (5 + 1)(gaw)H + 1(gaw)O

= 2(12.011)C + 6(1.00794)H + 1(15.9994)O

= 46.069 g/mol

Mr. Watson HST

The MoleThe Mole

a unit of measurement, quantity of matter present

Avogadro’s Number6.022 x 1023 particles

Latin for “pile”

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ExampleExample

How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide?

Mr. Watson HST

ExampleExample

How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide?

MM = 1(gaw)C + 2(gaw)O = 44.01 g/mol

Mr. Watson HST

ExampleExample

How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide?

MM = 1(gaw)C + 2(gaw)O = 44.01 g/mol

#mol CO2

Mr. Watson HST

ExampleExample

How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide?

MM = 1(gaw)C + 2(gaw)O = 44.01 g/mol

#mol CO2 = (10.00g)

Mr. Watson HST

ExampleExample

How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide?

MM = 1(gaw)C + 2(gaw)O = 44.01 g/mol

#mol CO2 = (10.00g)(1 mol/44.01g)

Mr. Watson HST

ExampleExample

How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide?

MM = 1(gaw)C + 2(gaw)O = 44.01 g/mol

#mol CO2 = (10.00g)(1 mol/44.01g)

Mr. Watson HST

ExampleExample

How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide?

MM = 1(gaw)C + 2(gaw)O = 44.01 g/mol

#mol CO2 = (10.00)(1 mol/44.01)

Mr. Watson HST

ExampleExample

How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide?

MM = 1(gaw)C + 2(gaw)O = 44.01 g/mol

#mol CO2 = (10.00)(1 mol/44.01)

= 0.2272 mol

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Combustion AnalysisCombustion Analysis

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Percentage CompositionPercentage Composition

description of a compound based on the relative amounts of each element in the compound

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EXAMPLE:EXAMPLE: What is the percent What is the percent composition of chloroform, composition of chloroform, CCHHClCl33, a , a substance once used as an anesthetic?substance once used as an anesthetic?

MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl

= (12.011 + 1.00797 + 3*35.453)amu

= 119.377amu

Mr. Watson HST

EXAMPLE:EXAMPLE: What is the percent What is the percent composition of chloroform, composition of chloroform, CCHClHCl33, a , a substance once used as an anesthetic?substance once used as an anesthetic?

MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl

= (12.011 + 1.00797 + 3*35.453)amu

= 119.377amu

1(gaw)%C = ------------ X 100

MM

Mr. Watson HST

EXAMPLE:EXAMPLE: What is the percent What is the percent composition of chloroform, CHClcomposition of chloroform, CHCl33, a , a substance once used as an anesthetic?substance once used as an anesthetic?

MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl

= (12.011 + 1.00797 + 3*35.453)amu

= 119.377amu 1(gaw)

%C = ------------ X 100 MM

1(12.011)%C = -------------- X 100 = 10.061% C

119.377

Mr. Watson HST

EXAMPLE:EXAMPLE: What is the percent What is the percent composition of chloroform, Ccomposition of chloroform, CHHClCl33, a , a substance once used as an anesthetic?substance once used as an anesthetic?

MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl

= (12.011 + 1.00797 + 3*35.453)amu

= 119.377amu

1(1.00797)%H = ---------------- X 100 = 0.844359% H

119.377

Mr. Watson HST

EXAMPLE:EXAMPLE: What is the percent What is the percent composition of chloroform, CHcomposition of chloroform, CHClCl33, a , a substance once used as an anesthetic?substance once used as an anesthetic?

MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl

= (12.011 + 1.00797 + 3*35.453)amu

= 119.377amu

3(35.453)%Cl = -------------- X 100 = 89.095% Cl

119.377

Mr. Watson HST

EXAMPLE:EXAMPLE: What is the percent What is the percent composition of chloroform, CHClcomposition of chloroform, CHCl33, a , a substance once used as an anesthetic?substance once used as an anesthetic?MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl

= 119.377amu

%C = 10.061% C

%H = 0.844359% H

%Cl = 89.095% Cl

Mr. Watson HST

Simplest (Empirical) FormulaSimplest (Empirical) Formula

formula describing a substance based on the smallest set of subscripts

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Acetylene, C2H2, and benzene, C6H6, have the same empirical formula. Is the correct empirical formula:

C2H2

CH

C6H6

Mr. Watson HST

EXAMPLE:EXAMPLE: A white compound is formed A white compound is formed when phosphorus burns in air. Analysis when phosphorus burns in air. Analysis shows that the compound is composed of shows that the compound is composed of 43.7% P43.7% P and and 56.3% O56.3% O by mass. What is by mass. What is the empirical formula of the compound?the empirical formula of the compound?

Relative Number of Atoms

Element % (%/gaw)

P 43.7 43.7/30.97 = 1.41

O 56.3 56.3/15.9994 = 3.52

Mr. Watson HST

EXAMPLE:EXAMPLE: A white compound is formed A white compound is formed when phosphorus burns in air. Analysis when phosphorus burns in air. Analysis shows that the compound is composed of shows that the compound is composed of 43.7% P43.7% P and and 56.3% O56.3% O by mass. What is by mass. What is the empirical formula of the compound?the empirical formula of the compound?

Relative Number of Atoms

Element % (%/gaw) Divide by Smaller

P 43.7 43.7/30.97 = 1.41 1.41/1.41 = 1.00

O 56.3 56.3/15.9994 = 3.52 3.52/1.41 = 2.50

Mr. Watson HST

EXAMPLE:EXAMPLE: A white compound is formed A white compound is formed when phosphorus burns in air. Analysis when phosphorus burns in air. Analysis shows that the compound is composed of shows that the compound is composed of 43.7% P43.7% P and and 56.3% O56.3% O by mass. What is by mass. What is the empirical formula of the compound?the empirical formula of the compound?

Relative Number of Atoms Multiply

% (%/gaw) Divide by Smaller by Integer

P 43.7 43.7/30.97 = 1.41 1.41/1.41 = 1.00 2*1.00 => 2

O 56.3 56.3/15.9994 = 3.52 3.52/1.41 = 2.50 2*2.50 => 5

Mr. Watson HST

EXAMPLE:EXAMPLE: A white compound is formed A white compound is formed when phosphorus burns in air. Analysis when phosphorus burns in air. Analysis shows that the compound is composed of shows that the compound is composed of 43.7% P43.7% P and and 56.3% O56.3% O by mass. What is by mass. What is the empirical formula of the compound?the empirical formula of the compound?

Relative Number of Atoms Multiply

% (%/gaw) Divide by Smaller by Integer

P 43.7 43.7/30.97 = 1.41 1.41/1.41 = 1.00 2*1.00 => 2

O 56.3 56.3/15.9994 = 3.52 3.52/1.41 = 2.50 2*2.50 => 5

Empirical Formula => P2O5

Mr. Watson HST

EXAMPLE:EXAMPLE: A sample of a brown-colored A sample of a brown-colored gas that is a major air pollutant is found to gas that is a major air pollutant is found to contain contain 2.34 g of N2.34 g of N and and 5.34 g of O5.34 g of O. What . What is the empirical formula of the compound?is the empirical formula of the compound?

2.34 %N = ----------------- X 100 = 30.5% N 2.34 + 5.34

5.34%O = ----------------- X 100 = 69.5% O 2.34 + 5.34

Mr. Watson HST

EXAMPLE:EXAMPLE: A sample of a brown-colored A sample of a brown-colored gas that is a major air pollutant is found to gas that is a major air pollutant is found to contain contain 2.34 g of N2.34 g of N and and 5.34 g of O5.34 g of O. What . What is the empirical formula of the compound?is the empirical formula of the compound?%N = 30.5% N%N = 30.5% N %O = 69.5% O%O = 69.5% O

Relative # Atoms

Element % (%/gaw)

N 30.5 30.5/14.0067 = 2.18

O 69.5 69.5/15.9994 = 4.34

Mr. Watson HST

EXAMPLE:EXAMPLE: A sample of a brown-colored A sample of a brown-colored gas that is a major air pollutant is found to gas that is a major air pollutant is found to contain contain 2.34 g of N2.34 g of N and and 5.34 g of O5.34 g of O. What . What is the empirical formula of the compound?is the empirical formula of the compound?

%N = 30.5% N %O = 69.5% O

Relative # Atoms

% (%/gaw) Divide by Smaller

N 30.5 30.5/14.0067 = 2.18 2.18/2.18 = 1.00

O 69.5 69.5/15.9994 = 4.34 4.34/2.18 = 1.99

Mr. Watson HST

EXAMPLE:EXAMPLE: A sample of a brown-colored A sample of a brown-colored gas that is a major air pollutant is found to gas that is a major air pollutant is found to contain contain 2.34 g of N2.34 g of N and and 5.34 g of O5.34 g of O. What . What is the empirical formula of the compound?is the empirical formula of the compound?

%N = 30.5% N %O = 69.5% O Relative # Atoms Multiply

% (%/gaw) Divide by Smaller by Integer

N 30.5 30.5/14.0067 = 2.18 2.18/2.18 = 1.00 1*1.00=>1

O 69.5 69.5/15.9994 = 4.34 4.34/2.18 = 1.99 1*1.99=>2

Mr. Watson HST

EXAMPLE:EXAMPLE: A sample of a brown-colored A sample of a brown-colored gas that is a major air pollutant is found to gas that is a major air pollutant is found to contain contain 2.34 g of N2.34 g of N and and 5.34 g of O5.34 g of O. What . What is the empirical formula of the compound?is the empirical formula of the compound?

%N = 30.5% N %O = 69.5% O Relative # Atoms Multiply

% (%/gaw) Divide by Smaller by Integer

N 30.5 30.5/14.0067 = 2.18 2.18/2.18 = 1.00 1*1.00=>1

O 69.5 69.5/15.9994 = 4.34 4.34/2.18 = 1.99 1*1.99=>2

Empirical Formula => NO2

Mr. Watson HST

Molecular FormulaMolecular Formula

the exact proportions of the elements that are formed in a molecule

Mr. Watson HST

Molecular Formula from Molecular Formula from Simplest FormulaSimplest Formula

empirical formula => EF

molecular formula => MF

MF = X * EF

Mr. Watson HST

Molecular Formula from Molecular Formula from Simplest FormulaSimplest Formula

formula mass => FM

sum of the atomic weights represented by the formula

molar mass = MM = X * FM

Mr. Watson HST

Molecular Formula from Molecular Formula from Simplest FormulaSimplest Formula

first, knowing MM and FM

X = MM/FM

then

MF = X * EF

Mr. Watson HST

EXAMPLE:EXAMPLE: A colorless liquid used in A colorless liquid used in rocket engines, whose empirical formula rocket engines, whose empirical formula is NOis NO22, has a , has a molar mass of 92.0molar mass of 92.0. What is . What is the molecular formula?the molecular formula?

FM = 1(gaw)N + 2(gaw)O = 46.0

MM 92.0X = ------- = -------- = 2 FM 46.0

Mr. Watson HST

EXAMPLE:EXAMPLE: A colorless liquid used in A colorless liquid used in rocket engines, whose empirical formula rocket engines, whose empirical formula is NOis NO22, has a molar mass of 92.0. What is , has a molar mass of 92.0. What is the molecular formula?the molecular formula?

FM = 1(gaw)N + 2(gaw)O = 46.0

MM 92.0X = ------- = -------- = 2 FM 46.0

thus MF = 2 * EF

Mr. Watson HST

What is the correct molecular formula for this colorless liquid rocket fuel?

2NO

NO

N2O4

Mr. Watson HST

StoichiometryStoichiometry

stoi·chi·om·e·try noun1. Calculation of the quantities of

reactants and products in a chemical reaction.

2. The quantitative relationship between reactants and products in a chemical reaction.

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The Mole and Chemical The Mole and Chemical Reactions:Reactions:

The Macro-Nano ConnectionThe Macro-Nano Connection2 H2 + O2 -----> 2 H2O

2 H2 molecules 1 O2 molecule 2 H2O molecules

2 H2 moles molecules 1 O2 mole molecules 2 H2O moles molecules

4 g H2 32 g O2 36 g H2O

Mr. Watson HST

EXAMPLEEXAMPLE

How much H2O, in moles results from burning an excess of H2 in 3.3 moles of O2?

H2 + O2 -----> H2O

2 H2 + O2 -----> 2 H2O

Mr. Watson HST

EXAMPLEEXAMPLE

How much H2O, in moles results from burning an excess of H2 in 3.3 moles of O2?

H2 + O2 -----> H2O

2 H2 + O2 -----> 2 H2O (3.3 mol O2) (2 mol H2O)

#mol H2O = ------------------------------------ (1 mol O2)

Mr. Watson HST

EXAMPLEEXAMPLE

How much H2O, in moles results from burning an excess of H2 in 3.3 moles of O2?

H2 + O2 -----> H2O

2 H2 + O2 -----> 2 H2O (3.3 mol O2) (2 mol H2O)

#mol H2O = ------------------------------------ (1 mol O2)

Mr. Watson HST

EXAMPLEEXAMPLE

How much H2O, in moles results from burning an excess of H2 in 3.3 moles of O2?

H2 + O2 -----> H2O

2 H2 + O2 -----> 2 H2O (3.3) (2 mol H2O)

#mol H2O = ------------------------ = 6.6 mol H2O (1)

Mr. Watson HST

Combination ReactionCombination Reaction

PbNO3(aq) + K2CrO4(aq) PbCrO4(s) + 2 KNO3(aq)

Colorless yellow yellow colorless

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Stoichiometric RoadmapStoichiometric Roadmap

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EXAMPLEEXAMPLE How much iron(III) oxide How much iron(III) oxide and aluminum powder are required to and aluminum powder are required to field weld the ends of two rails together? field weld the ends of two rails together?

Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additional 10% mass of iron.

http://www.cbu.edu/~mcondren/Demos/Thermite-Welding.ppt

Mr. Watson HST

EXAMPLEEXAMPLE How much iron(III) oxide How much iron(III) oxide and aluminum powder are required to and aluminum powder are required to field weld the ends of two rails together? field weld the ends of two rails together?

Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additional 10% mass of iron.

The mass of iron in 1 inch of this rail is:

#g/in = (132 #/yard) (1 yard/36 in) (454 g/#)

= 1.67 X 103 g/in

Mr. Watson HST

EXAMPLEEXAMPLE How much iron(III) oxide How much iron(III) oxide and aluminum powder are required to and aluminum powder are required to field weld the ends of two rails together? field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that

one inch of the rails will be covered by an additional 10% mass of iron.

The mass of iron in 1 inch of this rail is:#g/in = (132) (1/36 in) (454 g)

= 1.67 X 103 g/inThe mass of iron in a weld adding 10% mass:

#g = (1.67 X 103 g) (0.10) = 167 g

Mr. Watson HST

EXAMPLEEXAMPLE How much iron(III) oxide and How much iron(III) oxide and

aluminum powder are required to field weld the ends of aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an and that one inch of the rails will be covered by an additonal additonal 10% mass of iron.10% mass of iron.

The mass of iron in 1 inch of this rail is:

#g/in = (132 #/yard) (1 yard/36 in) (454 g/#)

= 1.67 X 103 g/in

The mass of iron in a weld adding 10% mass:

#g = (1.67 X 103 g) (0.10) = 167 g

Balanced chemical equation:

Fe2O3 + 2 Al ---> 2 Fe + Al2O3

Mr. Watson HST

EXAMPLEEXAMPLE How much iron(III) oxide and How much iron(III) oxide and

aluminum powder are required to field weld the ends of aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an and that one inch of the rails will be covered by an additonal additonal 10% mass of iron.10% mass of iron.

The mass of iron in 1 inch of this rail is:

#g/in = (132 #/yard) (1 yard/36 in) (454 g/#)

= 1.67 X 103 g/in

The mass of iron in a weld adding 10% mass:

#g = (1.67 X 103 g) (0.10) = 167 g

Balanced chemical equation:

Fe2O3 + 2 Al ---> 2 Fe + Al2O3

What mass of Fe2O3 is required for the thermite process?

Mr. Watson HST

EXAMPLEEXAMPLE How much iron(III) oxide and How much iron(III) oxide and

aluminum powder are required to field weld the ends of aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an and that one inch of the rails will be covered by an additonal additonal 10% mass of iron.10% mass of iron.

The mass of iron in a weld adding 10% mass:

#g = (1.67 X 103 g) (0.10) = 167 g

Balanced chemical equation:

Fe2O3 + 2 Al ---> 2 Fe + Al2O3

What mass of Fe2O3 is required for the thermite process?

(1 mol Fe) (1 mol Fe2O3) (159.7 g Fe2O3)#g Fe2O3 = (167 g Fe) * -------------------------------------------------------

(55.85 g Fe) (2 mol Fe) (1 mol Fe2O3)

= 238 g Fe2O3

Mr. Watson HST

EXAMPLEEXAMPLE How much iron(III) oxide and How much iron(III) oxide and

aluminum powder are required to field weld the ends of aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an and that one inch of the rails will be covered by an additonal additonal 10% mass of iron.10% mass of iron.

The mass of iron in a weld adding 10% mass: #g Fe = 167 g Fe

Balanced chemical equation:

Fe2O3 + 2 Al ---> 2 Fe + Al2O3

What mass of Fe2O3 is required for the thermite process?#g Fe2O3 = 238 g Fe2O3

What mass of Al is required for the thermite process?

Mr. Watson HST

EXAMPLEEXAMPLE How much iron(III) oxide and How much iron(III) oxide and

aluminum powder are required to field weld the ends of aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an and that one inch of the rails will be covered by an additonal additonal 10% mass of iron.10% mass of iron.The mass of iron in a weld adding 10% mass: #g Fe = 167 g Fe

Balanced chemical equation:Fe2O3 + 2 Al ---> 2 Fe + Al2O3

What mass of Fe2O3 is required for the thermite process?#g Fe2O3 = 238 g Fe2O3

What mass of Al is required for the thermite process? (1 mol Fe) (2 mol Al) (26.9815 g Al)

#g Al = (167 g Fe) * ----------------------------------------------- (55.85 g Fe) (2 mol Fe) (1 mol Al)= 80.6 g Al

Mr. Watson HST

EXAMPLEEXAMPLE How much iron(III) oxide and How much iron(III) oxide and

aluminum powder are required to field weld the ends of aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an and that one inch of the rails will be covered by an additonal additonal 10% mass of iron.10% mass of iron.

The mass of iron in a weld adding 10% mass:

#g Fe = 167 g Fe

#g Fe2O3 = 238 g Fe2O3

#g Al = 80.6 g Al

Mr. Watson HST

Limiting ReactantLimiting Reactant

reactant that limits the amount of product that can be produced

Mr. Watson HST

EXAMPLEEXAMPLE

What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react?

2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)

Mr. Watson HST

EXAMPLEEXAMPLE

What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react?

2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)

balanced equation relates:

2Fe2S3(S) <=> 6H2O(l) <=> 3O2(g)

Mr. Watson HST

EXAMPLEEXAMPLE

What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react?

2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)

balanced equation relates:

2Fe2S3(S) <=> 6H2O(l) <=> 3O2(g)

have only:

1Fe2S3(S) <=> 2H2O(l) <=> 3O2(g)

Mr. Watson HST

EXAMPLE EXAMPLE What is the number of What is the number of moles of Fe(OH)moles of Fe(OH)3 (S)3 (S) that can be produced that can be produced by allowing 1.0 mol Feby allowing 1.0 mol Fe22SS33, 2.0 mol H, 2.0 mol H22O, O, and 3.0 mol Oand 3.0 mol O22 to react? to react?2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)

balanced equation relates:2Fe2S3(S) <=> 6H2O(l) <=> 3O2(g)

have only:1Fe2S3(S) <=> 2H2O(l) <=> 3O2(g)

not enough H2O to use all Fe2S3

plenty of O2

Mr. Watson HST

EXAMPLE EXAMPLE What is the number of What is the number of moles of Fe(OH)moles of Fe(OH)3 (S)3 (S) that can be produced that can be produced by allowing by allowing 1.0 mol Fe1.0 mol Fe22SS33, 2.0 mol H, 2.0 mol H22O, O, and 3.0 mol Oand 3.0 mol O22 to react? to react?

2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)

if use all Fe2S3:

(1.0 mol Fe2S3) (4 mol Fe(OH)3)

#mol Fe(OH)3 = ------------------------------------------ (2 mol Fe2S3)

= 2.0 mol Fe(OH)3

Mr. Watson HST

EXAMPLE: EXAMPLE: What is the number of What is the number of moles of Fe(OH)moles of Fe(OH)3 (S)3 (S) that can be produced that can be produced by allowing 1.0 mol Feby allowing 1.0 mol Fe22SS33, , 2.0 mol H2.0 mol H22OO, , and 3.0 mol Oand 3.0 mol O22 to react? to react?

2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)

if use all H2O:

(2.0 mol H2O) (4 mol Fe(OH)3) #mol Fe(OH)3 = -----------------------------------------

(6 mol H2O)= 1.3 mol Fe(OH)3

Mr. Watson HST

EXAMPLE: EXAMPLE: What is the number of What is the number of moles of Fe(OH)moles of Fe(OH)3 (S)3 (S) that can be produced that can be produced by allowing 1.0 mol Feby allowing 1.0 mol Fe22SS33, 2.0 mol H, 2.0 mol H22O, O, and and 3.0 mol O3.0 mol O22 to react? to react?2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)

if use all O2

(3.0 mol O2) (4 mol Fe(OH)3) #mol Fe(OH)3 = ---------------------------------------

(3 mol O2)

= 4.0 mol Fe(OH)3

Mr. Watson HST

EXAMPLE: EXAMPLE: What is the number of What is the number of moles of Fe(OH)moles of Fe(OH)3 (S)3 (S) that can be produced that can be produced by allowing by allowing 1.0 mol Fe1.0 mol Fe22SS33, , 2.0 mol H2.0 mol H22OO, , and and 3.0 mol O3.0 mol O22 to react? to react?

2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)

1.0 mol Fe2S3 => 2.0 mol Fe(OH)3

2.0 mol H2O => 1.3 mol Fe(OH)3

3.0 mol O2 => 4.0 mol Fe(OH)3

Mr. Watson HST

EXAMPLE: EXAMPLE: What is the number of What is the number of moles of Fe(OH)moles of Fe(OH)3 (S)3 (S) that can be produced that can be produced by allowing by allowing 1.0 mol Fe1.0 mol Fe22SS33, , 2.0 mol H2.0 mol H22OO, , and and 3.0 mol O3.0 mol O22 to react? to react?

2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)

1.0 mol Fe2S3 => 2.0 mol Fe(OH)3

2.0 mol H2O => 1.3 mol Fe(OH)3

3.0 mol O2 => 4.0 mol Fe(OH)3

Since 2.0 mol H2O will produce only 1.3 mol Fe(OH)3, then H2O is the limiting reactant. Thus the correct number of moles of Fe(OH)3 is 1.33 moles.

Mr. Watson HST

EXAMPLE: EXAMPLE: What is the number of What is the number of moles of Fe(OH)moles of Fe(OH)3 (S)3 (S) that can be produced that can be produced by allowing by allowing 1.0 mol Fe1.0 mol Fe22SS33, , 2.0 mol H2.0 mol H22OO, , and and 3.0 mol O3.0 mol O22 to react? to react?

2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)

1.0 mol Fe2S3 => 2.0 mol Fe(OH)3

2.0 mol H2O => 1.3 mol Fe(OH)3 least amount

3.0 mol O2 => 4.0 mol Fe(OH)3

Since 2.0 mol H2O will produce only 1.3 mol Fe(OH)3, then H2O is the limiting reactant.

Mr. Watson HST

EXAMPLE: EXAMPLE: What is the number of What is the number of moles of Fe(OH)moles of Fe(OH)3 (S)3 (S) that can be produced that can be produced by allowing by allowing 1.0 mol Fe1.0 mol Fe22SS33, , 2.0 mol H2.0 mol H22OO, , and and 3.0 mol O3.0 mol O22 to react? to react?

2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)

1.0 mol Fe2S3 => 2.0 mol Fe(OH)3

2.0 mol H2O => 1.3 mol Fe(OH)3 least amount

3.0 mol O2 => 4.0 mol Fe(OH)3

Since 2.0 mol H2O will produce only 1.3 mol Fe(OH)3, then H2O is the limiting reactant. Thus the maximum number of moles of Fe(OH)3 that can be produced by this reaction is 1.3 moles.

Mr. Watson HST

Theoretical YieldTheoretical Yield

the amount of product produced by a reaction based on the amount of the limiting reactant

Mr. Watson HST

Actual YieldActual Yield

amount of product actually produced in a reaction

Mr. Watson HST

Percent YieldPercent Yield

actual yield% yield = --------------------- * 100 theoretical yield

Mr. Watson HST

EXAMPLE EXAMPLE A chemical plant A chemical plant obtained 0.299 kg of 98.0% Nobtained 0.299 kg of 98.0% N22HH44 for for every 1.00 kg of Clevery 1.00 kg of Cl22 that is reacted with that is reacted with excess NaOH and NHexcess NaOH and NH33. What are the: (a) . What are the: (a) theoretical, (b) actual, and (c) percent theoretical, (b) actual, and (c) percent yield of yield of purepure N N22HH44??

2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O

(a) to calculate the theoretical yield, use the net equation for the overall process

Mr. Watson HST

EXAMPLE EXAMPLE A chemical plant A chemical plant obtained 0.299 kg of 98.0% Nobtained 0.299 kg of 98.0% N22HH44 for for every every 1.00 kg of Cl1.00 kg of Cl22 that is reacted with that is reacted with excess NaOH and NHexcess NaOH and NH33. What are the: (a) . What are the: (a) theoretical, (b) actual, and (c) percent theoretical, (b) actual, and (c) percent yield of yield of purepure N N22HH44??2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation for

the overall process

(1.00 kg Cl2) #kg N2H4 = ---------------------

Mr. Watson HST

EXAMPLE EXAMPLE A chemical plant A chemical plant obtained 0.299 kg of 98.0% Nobtained 0.299 kg of 98.0% N22HH44 for for every 1.00 kg of Clevery 1.00 kg of Cl22 that is reacted with that is reacted with excess NaOH and NHexcess NaOH and NH33. What are the: (a) . What are the: (a) theoretical, (b) actual, and (c) percent theoretical, (b) actual, and (c) percent yield of yield of purepure N N22HH44??2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation for

the overall process

(1.00 kg Cl2) (1000 g Cl2)#kg N2H4 = -----------------------------------

(1 kg Cl2) metric conversion

Mr. Watson HST

EXAMPLE EXAMPLE A chemical plant A chemical plant obtained 0.299 kg of 98.0% Nobtained 0.299 kg of 98.0% N22HH44 for for every 1.00 kg of Clevery 1.00 kg of Cl22 that is reacted with that is reacted with excess NaOH and NHexcess NaOH and NH33. What are the: (a) . What are the: (a) theoretical, (b) actual, and (c) percent theoretical, (b) actual, and (c) percent yield of yield of purepure N N22HH44??2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation for

the overall process

(1.00) (1000 g Cl2) (1 mol Cl2) #kg N2H4 = -----------------------------------------

(1) (70.9 g Cl2) molar mass

Mr. Watson HST

EXAMPLE EXAMPLE A chemical plant A chemical plant obtained 0.299 kg of 98.0% Nobtained 0.299 kg of 98.0% N22HH44 for for every 1.00 kg of Clevery 1.00 kg of Cl22 that is reacted with that is reacted with excess NaOH and NHexcess NaOH and NH33. What are the: (a) . What are the: (a) theoretical, (b) actual, and (c) percent theoretical, (b) actual, and (c) percent yield of yield of purepure N N22HH44??2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation for

the overall process

(1.00)(1000)(1 mol Cl2) #kg N2H4 = -----------------------------------------

(1)(70.9)

Mr. Watson HST

EXAMPLE EXAMPLE A chemical plant A chemical plant obtained 0.299 kg of 98.0% Nobtained 0.299 kg of 98.0% N22HH44 for for every 1.00 kg of Clevery 1.00 kg of Cl22 that is reacted with that is reacted with excess NaOH and NHexcess NaOH and NH33. What are the: (a) . What are the: (a) theoretical, (b) actual, and (c) percent theoretical, (b) actual, and (c) percent yield of yield of purepure N N22HH44??2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation for

the overall process

(1.00)(1000)(1 mol Cl2)(1 mol N2H4) #kg N2H4 = -------------------------------------------------

(1) (70.9) (1 mol Cl2)

Mr. Watson HST

EXAMPLE EXAMPLE A chemical plant A chemical plant obtained 0.299 kg of 98.0% Nobtained 0.299 kg of 98.0% N22HH44 for for every 1.00 kg of Clevery 1.00 kg of Cl22 that is reacted with that is reacted with excess NaOH and NHexcess NaOH and NH33. What are the: (a) . What are the: (a) theoretical, (b) actual, and (c) percent theoretical, (b) actual, and (c) percent yield of yield of purepure N N22HH44??2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation for

the overall process

(1.00)(1000)(1)(1 mol N2H4) #kg N2H4 = -------------------------------------------------

(1) (70.9)(1)

Mr. Watson HST

EXAMPLE EXAMPLE A chemical plant A chemical plant obtained 0.299 kg of 98.0% Nobtained 0.299 kg of 98.0% N22HH44 for for every 1.00 kg of Clevery 1.00 kg of Cl22 that is reacted with that is reacted with excess NaOH and NHexcess NaOH and NH33. What are the: (a) . What are the: (a) theoretical, (b) actual, and (c) percent theoretical, (b) actual, and (c) percent yield of yield of purepure N N22HH44??2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation for

the overall process

(1.00)(1000)(1)(1 mol N2H4) (32.0 g N2H4) #kg N2H4 = --------------------------------------------------------

(1)(70.9) (1) (1 mol N2H4)

molar mass

Mr. Watson HST

EXAMPLE EXAMPLE A chemical plant A chemical plant obtained 0.299 kg of 98.0% Nobtained 0.299 kg of 98.0% N22HH44 for for every 1.00 kg of Clevery 1.00 kg of Cl22 that is reacted with that is reacted with excess NaOH and NHexcess NaOH and NH33. What are the: (a) . What are the: (a) theoretical, (b) actual, and (c) percent theoretical, (b) actual, and (c) percent yield of yield of purepure N N22HH44??2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation for

the overall process

(1.00)(1000)(1)(1) (32.0 g N2H4)(1 kg N2H4)#kg N2H4 = ----------------------------------------------------------

(1)(70.9)(1)(1) (1000 g N2H4)

metric conversion

Mr. Watson HST

EXAMPLE EXAMPLE A chemical plant A chemical plant obtained 0.299 kg of 98.0% Nobtained 0.299 kg of 98.0% N22HH44 for for every 1.00 kg of Clevery 1.00 kg of Cl22 that is reacted with that is reacted with excess NaOH and NHexcess NaOH and NH33. What are the: (a) . What are the: (a) theoretical, (b) actual, and (c) percent theoretical, (b) actual, and (c) percent yield of yield of purepure N N22HH44??2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation for

the overall process(1.00)(1000)(1)(1) (32.0)(1 kg N2H4)

#kg N2H4 = ----------------------------------------------------------(1)(70.9)(1)(1)(1000)

Mr. Watson HST

EXAMPLE EXAMPLE A chemical plant A chemical plant obtained 0.299 kg of 98.0% Nobtained 0.299 kg of 98.0% N22HH44 for for every 1.00 kg of Clevery 1.00 kg of Cl22 that is reacted with that is reacted with excess NaOH and NHexcess NaOH and NH33. What are the: (a) . What are the: (a) theoretical, (b) actual, and (c) percent theoretical, (b) actual, and (c) percent yield of yield of purepure N N22HH44??2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation for

the overall process(1.00)(1000)(1)(1) (32.0)(1 kg N2H4)

#kg N2H4 = ----------------------------------------------------------(1)(70.9)(1)(1)(1000)

= 0.451 kg N2H4

Mr. Watson HST

EXAMPLE EXAMPLE A chemical plant A chemical plant obtained obtained 0.299 kg of 98.0% N0.299 kg of 98.0% N22HH44 for for every 1.00 kg of Clevery 1.00 kg of Cl22 that is reacted with that is reacted with excess NaOH and NHexcess NaOH and NH33. What are the: (a) . What are the: (a) theoretical, (b) actual, and (c) percent theoretical, (b) actual, and (c) percent yield of yield of purepure N N22HH44??2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) theoretical yield #kg N2H4 = 0.451 kg N2H4

(b) actual yield

(0.299 kg product) # kg N2H4 = --------------------------

Mr. Watson HST

EXAMPLE EXAMPLE A chemical plant A chemical plant obtained 0.299 kg of obtained 0.299 kg of 98.0% N98.0% N22HH44 for for every 1.00 kg of Clevery 1.00 kg of Cl22 that is reacted with that is reacted with excess NaOH and NHexcess NaOH and NH33. What are the: (a) . What are the: (a) theoretical, (b) actual, and (c) percent theoretical, (b) actual, and (c) percent yield of yield of purepure N N22HH44??2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) theoretical yield #kg N2H4 = 0.451 kg N2H4

(b) actual yield (0.299 kg product) (98.0 kg N2H4)

# kg N2H4 = -------------------------------------------- (100 kg product)

purity factor

Mr. Watson HST

EXAMPLE EXAMPLE A chemical plant A chemical plant obtained 0.299 kg of 98.0% Nobtained 0.299 kg of 98.0% N22HH44 for for every 1.00 kg of Clevery 1.00 kg of Cl22 that is reacted with that is reacted with excess NaOH and NHexcess NaOH and NH33. What are the: (a) . What are the: (a) theoretical, (b) actual, and (c) percent theoretical, (b) actual, and (c) percent yield of yield of purepure N N22HH44??2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) theoretical yield #kg N2H4 = 0.451 kg N2H4

(b) actual yield (0.299 kg product) (98.0 kg N2H4)

# kg N2H4 = -------------------------------------------- (100 kg product)

= 0.293 kg N2H4

Mr. Watson HST

EXAMPLE EXAMPLE A chemical plant A chemical plant obtained 0.299 kg of 98.0% Nobtained 0.299 kg of 98.0% N22HH44 for for every 1.00 kg of Clevery 1.00 kg of Cl22 that is reacted with that is reacted with excess NaOH and NHexcess NaOH and NH33. What are the: (a) . What are the: (a) theoretical, (b) actual, and (c) percent theoretical, (b) actual, and (c) percent yield of yield of purepure N N22HH44??2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) theoretical yield #kg N2H4 = 0.451 kg N2H4

(b) actual yield # kg N2H4 = 0.293 kg N2H4

Mr. Watson HST

EXAMPLE EXAMPLE A chemical plant A chemical plant obtained 0.299 kg of 98.0% Nobtained 0.299 kg of 98.0% N22HH44 for for every 1.00 kg of Clevery 1.00 kg of Cl22 that is reacted with that is reacted with excess NaOH and NHexcess NaOH and NH33. What are the: (a) . What are the: (a) theoretical, (b) actual, and (c) percent theoretical, (b) actual, and (c) percent yield of yield of purepure N N22HH44??2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) theoretical yield #kg N2H4 = 0.451 kg N2H4

(b) actual yield # kg N2H4 = 0.293 kg N2H4

(c) percent yield 0.293 kg

% yield = -------------- X 100 = 65.0 % yield 0.451kg