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.
Chemical EquationsChemical Equations
Chapter 4 4
2H2(g) + O2(g) 2H2O(g)
• The materials you start with are called Reactants.• The materials you make are called Products.
Chemical EquationsChemical Equations
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.
Chemical EquationsChemical Equations
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.
Chemical EquationsChemical Equations
Chapter 4 7
Balancing Chemical Reactions
___NH4NO3(s) ___N2O(g) + ___H2O(g)
Chemical EquationsChemical Equations
Chapter 4 8
Balancing Chemical Reactions
___NH4NO3(s) ___N2O(g) + ___H2O(g)
Chemical EquationsChemical Equations
Reactants Products
N 2 N 2
H 4 H 2
O 3 O 2
Chapter 4 9
Balancing Chemical Reactions
___NH4NO3(s) ___N2O(g) + _2_H2O(g)
Chemical EquationsChemical Equations
Reactants Products
N 2 N 2
H 4 H 2 4
O 3 O 2 3
Chapter 4 10
Balancing Chemical Reactions
___Mg3N2(s) + ___H2O(l) ___Mg(OH)2(s) + ___NH3(aq)
Chemical EquationsChemical Equations
Reactants Products
Mg 3 Mg 1
N 2 N 1
H 2 H 5
O 1 O 2
Chapter 4 11
Balancing Chemical Reactions
___Mg3N2(s) + ___H2O(l) _3_Mg(OH)2(s) + ___NH3(aq)
Chemical EquationsChemical Equations
Reactants Products
Mg 3 Mg 1 3
N 2 N 1
H 2 H 5 9
O 1 O 2 6
Chapter 4 12
Balancing Chemical Reactions
___Mg3N2(s) + ___H2O(l) _3_Mg(OH)2(s) + _2_NH3(aq)
Chemical EquationsChemical Equations
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
Balancing Chemical Reactions
___Mg3N2(s) + _6_H2O(l) _3_Mg(OH)2(s) + _2_NH3(aq)
Chemical EquationsChemical Equations
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
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?
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)
How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen?
1. Moles of C4H10
F.W. 58.124g
molg
molgHCmoles 0172.0
124.58
100.1104
Chapter 4 19
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?
2. Ratio of C4H10:CO2
2 C4H10 : 8 CO2 or 104
2
2
8
HC
CO
Chapter 4 20
Quantitative InformationQuantitative Information2 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?
3. Set-up ratio and proportion between known and unknown quantities
Chapter 4 21
Quantitative InformationQuantitative Information2 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?
3. Set-up ratio and proportion between known and unknown quantities
104104
2
0172.02
8
HCmol
x
HC
CO
Chapter 4 22
Quantitative InformationQuantitative Information2 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?
3. Set-up ratio and proportion between known and unknown quantities
2
104104
2
0688.0
0172.02
8
COmolx
HCmol
x
HC
CO
Chapter 4 23
Quantitative InformationQuantitative Information2 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?
4. Convert the moles of unknown substance into the desired units
Chapter 4 24
Quantitative InformationQuantitative Information2 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?
4. Convert the moles of unknown substance into the desired unitsFW of CO2: 44.011g/mol molgmolCOg /011.440688.02
Chapter 4 25
Quantitative InformationQuantitative Information2 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?
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
Limiting ReactantsLimiting Reactants
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
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
Limiting ReactantsLimiting Reactants
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
Determine the limiting reagent of this reaction if 10.0 grams of C8H18 and 25.0 grams of O2 are allowed to react.
2. Divide each reagent by its own coefficient
Limiting ReactantsLimiting Reactants
25
781.02
0875.0
2
188
O
HC
Chapter 4 30
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.
2. Divide each reagent by its own coefficient
Limiting ReactantsLimiting Reactants
0313.025
781.0
0438.02
0875.0
2
188
O
HC
Chapter 4 31
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.
3. The substance with the smallest calculated value will be the limiting reagent. In this case, O2 is the limiting reagent.
Limiting ReactantsLimiting Reactants
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).
Limiting ReactantsLimiting Reactants
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
Limiting ReactantsLimiting ReactantsTheoretical YieldTheoretical Yield
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
Limiting ReactantsLimiting ReactantsTheoretical YieldTheoretical Yield
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.
Limiting ReactantsLimiting Reactants
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
Empirical Formulas from AnalysesEmpirical Formulas from Analyses
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
Empirical Formulas from AnalysesEmpirical Formulas from Analyses
2
2
011.44
12829.0
COg
COmolg
Chapter 4 40
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
Empirical Formulas from AnalysesEmpirical Formulas from Analyses
22
2
1
1
011.44
12829.0
COmol
Cmol
COg
COmolg
Chapter 4 41
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
Empirical Formulas from AnalysesEmpirical Formulas from Analyses
Cmol
Cg
COmol
Cmol
COg
COmolg
1
011.12
1
1
011.44
12829.0
22
2
Chapter 4 42
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
Empirical Formulas from AnalysesEmpirical Formulas from Analyses
CgCmol
Cg
COmol
Cmol
COg
COmolg 07721.0
1
011.12
1
1
011.44
12829.0
22
2
Chapter 4 43
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 HydrogenMass of Hydrogenmass Hmass H22O O moles H moles H22O O moles H moles H grams H grams H
Empirical Formulas from AnalysesEmpirical Formulas from Analyses
OHg
OHmolg
2
2
02.18
11159.0
Chapter 4 44
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 HydrogenMass of Hydrogenmass Hmass H22O O moles H moles H22O O moles H moles H grams H grams H
Empirical Formulas from AnalysesEmpirical Formulas from Analyses
OHmol
Hmol
OHg
OHmolg
22
2
1
2
02.18
11159.0
Chapter 4 45
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 HydrogenMass of Hydrogenmass Hmass H22O O moles H moles H22O O moles H moles H grams H grams H
Empirical Formulas from AnalysesEmpirical Formulas from Analyses
Hmol
Hg
OHmol
Hmol
OHg
OHmolg
1
01.1
1
2
02.18
11159.0
22
2
Chapter 4 46
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 HydrogenMass of Hydrogenmass Hmass H22O O moles H moles H22O O moles H moles H grams H grams H
Empirical Formulas from AnalysesEmpirical Formulas from Analyses
gHmol
Hg
OHmol
Hmol
OHg
OHmolg 01299.0
1
01.1
1
2
02.18
11159.0
22
2
Chapter 4 47
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 OxygenMass of Oxygen
mass O = mass of sample – (mass C +mass H)mass O = mass of sample – (mass C +mass H)
Empirical Formulas from AnalysesEmpirical Formulas from Analyses
gggOxygenmass 01299.007721.01005.0
Chapter 4 48
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 OxygenMass of Oxygen
mass O = mass of sample – (mass C +mass H)mass O = mass of sample – (mass C +mass H)
Empirical Formulas from AnalysesEmpirical Formulas from Analyses
g
gggOxygenmass
01030.0
01299.007721.01005.0
Chapter 4 49
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?
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
Empirical Formulas from AnalysesEmpirical Formulas from Analyses
Chapter 4 50
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?
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 =
Empirical Formulas from AnalysesEmpirical Formulas from Analyses
Chapter 4 51
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?
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 = 0.006428 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
Empirical Formulas from AnalysesEmpirical Formulas from Analyses
0006428.0012861.0006428.0 OHC
Chapter 4 52
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?
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 = 0.006428 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
Empirical Formulas from AnalysesEmpirical Formulas from Analyses
0006428.0
0006428.0
0006438.0
012861.0
0006428.0
006428.0 OHC
Chapter 4 53
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?
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 = 0.006428 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
Empirical Formulas from AnalysesEmpirical Formulas from Analyses
12010 OHC
Chapter 4 54
Practice ProblemsPractice Problems
4, 10, 14, 22, 26, 28, 38, 46, 50, 54