Chapter 41 Chemical Equations and Stoichiometry Chapter 4

Chapter 4 1

Chemical Equations and Chemical Equations and StoichiometryStoichiometry

Chapter 4Chapter 4

Chapter 4 2

2H2(g) + O2(g) 2H2O(g)

Chemical EquationsChemical Equations

Chapter 4 3

2H2(g) + O2(g) 2H2O(g)

• The materials you start with are called Reactants.


Chapter 4 4

2H2(g) + O2(g) 2H2O(g)

• The materials you start with are called Reactants.• The materials you make are called Products.


Chapter 4 5

2H2(g) + O2(g) 2H2O(g)

• The materials you start with are called Reactants.• The materials you make are called Products.

• The numbers in front of the compounds (H2 and H2O) are called stoichiometric coefficients.– Coefficients are multipliers, in this equation 2 in front of

the H2 indicates that there are 2 molecules of H2 in the equation.


Chapter 4 6

2H2(g) + O2(g) 2H2O(g)

• Notice that the number of hydrogen atoms and oxygen atoms on the reactant side and the product side is equal.

Law of Conservation of Mass

Matter cannot be created or lost in any chemical reaction.


Chapter 4 7

Balancing Chemical Reactions

___NH4NO3(s) ___N2O(g) + ___H2O(g)


Chapter 4 8


___NH4NO3(s) ___N2O(g) + ___H2O(g)


Reactants Products

N 2 N 2

H 4 H 2

O 3 O 2

Chapter 4 9


___NH4NO3(s) ___N2O(g) + _2_H2O(g)


Reactants Products

N 2 N 2

H 4 H 2 4

O 3 O 2 3

Chapter 4 10


___Mg3N2(s) + ___H2O(l) ___Mg(OH)2(s) + ___NH3(aq)


Reactants Products

Mg 3 Mg 1

N 2 N 1

H 2 H 5

O 1 O 2

Chapter 4 11


___Mg3N2(s) + ___H2O(l) _3_Mg(OH)2(s) + ___NH3(aq)


Reactants Products

Mg 3 Mg 1 3

N 2 N 1

H 2 H 5 9

O 1 O 2 6

Chapter 4 12


___Mg3N2(s) + ___H2O(l) _3_Mg(OH)2(s) + _2_NH3(aq)


Reactants Products

Mg 3 Mg 1 3

N 2 N 1 2

H 2 H 5 9 12

O 1 O 2 6

Chapter 4 13


___Mg3N2(s) + _6_H2O(l) _3_Mg(OH)2(s) + _2_NH3(aq)


Reactants Products

Mg 3 Mg 1 3

N 2 N 1 2

H 2 12 H 5 9 12

O 1 6 O 2 6

Chapter 4 14

2 H2(g) + O2(g) 2 H2O(g)

• The coefficients in a balanced equation represent both the number of molecules and the number of moles in a reaction.

• The coefficients can also be used to derive ratios between any two substances in the chemical reaction.

2 H2 : 1 O2

2 H2 : 2 H2O1 O2 : 2 H2O

Quantitative InformationQuantitative Information

•The ratios can be used to predict•The amount of product formed•The amount of reactant needed

Chapter 4 15

Quantitative InformationQuantitative Information

Chapter 4 16

Quantitative InformationQuantitative Information 2 C4H10(l) + 13 O2(g) 8 CO2(g) + 10 H2O(g)

How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen?

Chapter 4 17



1. Moles of C4H10

F.W. 58.124g

g

molgHCmoles

124.58

100.1104

Chapter 4 18

Quantitative InformationQuantitative Information2 C4H10(l) + 13 O2(g) 8 CO2(g) + 10 H2O(g)


1. Moles of C4H10

F.W. 58.124g

molg

molgHCmoles 0172.0

124.58

100.1104

Chapter 4 19



2. Ratio of C4H10:CO2

2 C4H10 : 8 CO2 or 104

2

2

8

HC

CO

Chapter 4 20



3. Set-up ratio and proportion between known and unknown quantities

Chapter 4 21




104104

2

0172.02

8

HCmol

x

HC

CO

Chapter 4 22




2

104104

2

0688.0

0172.02

8

COmolx

HCmol

x

HC

CO

Chapter 4 23



4. Convert the moles of unknown substance into the desired units

Chapter 4 24



4. Convert the moles of unknown substance into the desired unitsFW of CO2: 44.011g/mol molgmolCOg /011.440688.02

Chapter 4 25



4. Convert the moles of unknown substance into the desired unitsFW of CO2: 44.011g/mol gmolgmolCOg 03.3/011.440688.02

Chapter 4 26

“What runs out first”

2 C8H18 + 25 O2 16 CO2 + 18 H2O

• If you have 2 moles of C8H18 and 20 moles of O2

all the O2 will be used and the reaction will stop• O2 is call the limiting reagent (reactant)

Limiting Reactant – The reagent present in the smallest stoichiometric quantity in a mixture of reactants.

Limiting ReactantsLimiting Reactants

Chapter 4 27

Example2 C8H18 + 25 O2 16 CO2 + 18 H2O

Determine the limiting reagent of this reaction if 10.0 grams of C8H18 and 25.0 grams of O2 are allowed to react.

1. Convert grams to molesFW(C8H18) 114.268g/mol FW(O2) = 32.00g/mol


g

molgOmoles

g

molgHCmoles

00.32

10.25

268.114

10.10

2

188

Chapter 4 28

Example2 C8H18 + 25 O2 16 CO2 + 18 H2O


1. Convert grams to molesFW(C8H18) 114.268g/mol FW(O2) = 32.00g/mol


22

188188

781.000.32

10.25

0875.0268.114

10.10

Omolg

molgOmoles

HCmolg

molgHCmoles

Chapter 4 29

Example2 C8H18 + 25 O2 16 CO2 + 18 H2O


2. Divide each reagent by its own coefficient


25

781.02

0875.0

2

188

O

HC

Chapter 4 30

Example2 C8H18 + 25 O2 16 CO2 + 18 H2O


2. Divide each reagent by its own coefficient


0313.025

781.0

0438.02

0875.0

2

188

O

HC

Chapter 4 31

Example2 C8H18 + 25 O2 16 CO2 + 18 H2O


3. The substance with the smallest calculated value will be the limiting reagent. In this case, O2 is the limiting reagent.


Chapter 4 32

Theoretical YieldTheoretical Yield- The calculated amount of product based on the - The calculated amount of product based on the limiting reactant (Theoretical yield).limiting reactant (Theoretical yield).


Chapter 4 33

2 C8H18 + 25 O2 16 CO2 + 18 H2O

Determine the theoretical yield of CO2 for this reaction if 10.0 grams of C8H18 and 25.0 grams of O2 are allowed to react.

- already know that O2 is the limiting reactant.

Limiting ReactantsLimiting ReactantsTheoretical YieldTheoretical Yield

22 781.000.32

10.25 Omol

g

molgOmoles

Chapter 4 34

2 C8H18 + 25 O2 16 CO2 + 18 H2O

•Calculate moles of oxygen


22 781.000.32

10.25 Omol

g

molgOmoles

•Calculate moles of CO2

2

22

2

500.0

781.025

16

COmolesx

Omol

x

O

CO

Chapter 4 35

2 C8H18 + 25 O2 16 CO2 + 18 H2O

•Calculate moles of CO2


22 0.221

0.44500.0 COg

mol

gmolCOmoles

Chapter 4 36

Percent YieldPercent Yield- Calculation which indicates how much of the - Calculation which indicates how much of the

theoretical yield was obtained.theoretical yield was obtained.


100l yieldTheoretica

ldActual yie% Yield

Chapter 4 37

Combustion AnalysisCombustion Analysis

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Typical example:

2 C2H6(g) + 7 O2 4 CO2(g) + 6 H2O(g)- The combustion of any hydrocarbon produces CO2

and water.- This observation can be used to determine the

empirical formula of the reactant.

Combustion Reaction: The “burning” of any substance in oxygen.

Chapter 4 38

Combustion AnalysisCombustion Analysis


Chapter 4 39

Combustion AnalysisCombustion AnalysisMenthol, the substance we can smell in mentholated cough drops, Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of COcombusted, producing 0.2829g of CO22 and 0.1159g of H and 0.1159g of H22O. What O. What

is the empirical formula for menthol?is the empirical formula for menthol?

Mass of CarbonMass of Carbonmass COmass CO22 moles CO moles CO22 moles C moles C grams C grams C


2

2

011.44

12829.0

COg

COmolg

Chapter 4 40





22

2

1

1

011.44

12829.0

COmol

Cmol

COg

COmolg

Chapter 4 41





Cmol

Cg

COmol

Cmol

COg

COmolg

1

011.12

1

1

011.44

12829.0

22

2

Chapter 4 42





CgCmol

Cg

COmol

Cmol

COg

COmolg 07721.0

1

011.12

1

1

011.44

12829.0

22

2

Chapter 4 43



Mass of HydrogenMass of Hydrogenmass Hmass H22O O moles H moles H22O O moles H moles H grams H grams H


OHg

OHmolg

2

2

02.18

11159.0

Chapter 4 44





OHmol

Hmol

OHg

OHmolg

22

2

1

2

02.18

11159.0

Chapter 4 45





Hmol

Hg

OHmol

Hmol

OHg

OHmolg

1

01.1

1

2

02.18

11159.0

22

2

Chapter 4 46





gHmol

Hg

OHmol

Hmol

OHg

OHmolg 01299.0

1

01.1

1

2

02.18

11159.0

22

2

Chapter 4 47



Mass of OxygenMass of Oxygen

mass O = mass of sample – (mass C +mass H)mass O = mass of sample – (mass C +mass H)


gggOxygenmass 01299.007721.01005.0

Chapter 4 48



Mass of OxygenMass of Oxygen

mass O = mass of sample – (mass C +mass H)mass O = mass of sample – (mass C +mass H)


g

gggOxygenmass

01030.0

01299.007721.01005.0

Chapter 4 49



Now we can determine the empirical formulaNow we can determine the empirical formulaMass of elements:Mass of elements:

C C 0.07721g 0.07721gH H 0.01299g 0.01299gO O 0.01030g 0.01030g


Chapter 4 50



Now we can determine the empirical formulaNow we can determine the empirical formulaMoles of elements:Moles of elements:

C C 0.07721g/12.011g/mol = 0.07721g/12.011g/mol = H H 0.01299g/1.01g/mol = 0.01299g/1.01g/mol =O O 0.01030g/16.00g/mol = 0.01030g/16.00g/mol =


Chapter 4 51




C C 0.07721g/12.011g/mol = 0.006428 mol 0.07721g/12.011g/mol = 0.006428 molH H 0.01299g/1.01g/mol = 0.012861 mol 0.01299g/1.01g/mol = 0.012861 molO O 0.01030g/16.00g/mol = 0.0006428mol 0.01030g/16.00g/mol = 0.0006428mol


0006428.0012861.0006428.0 OHC

Chapter 4 52




C C 0.07721g/12.011g/mol = 0.006428 mol 0.07721g/12.011g/mol = 0.006428 molH H 0.01299g/1.01g/mol = 0.012861 mol 0.01299g/1.01g/mol = 0.012861 molO O 0.01030g/16.00g/mol = 0.0006428 mol 0.01030g/16.00g/mol = 0.0006428 mol


0006428.0

0006428.0

0006438.0

012861.0

0006428.0

006428.0 OHC

Chapter 4 53




C C 0.07721g/12.011g/mol = 0.006428 mol 0.07721g/12.011g/mol = 0.006428 molH H 0.01299g/1.01g/mol = 0.012861 mol 0.01299g/1.01g/mol = 0.012861 molO O 0.01030g/16.00g/mol = 0.0006428 mol 0.01030g/16.00g/mol = 0.0006428 mol


12010 OHC

Chapter 4 54

Practice ProblemsPractice Problems

4, 10, 14, 22, 26, 28, 38, 46, 50, 54

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Chapter 41 Chemical Equations and Stoichiometry Chapter 4