This document discusses stoichiometry and concepts related to counting atoms and molecules. It introduces atomic masses and how mass spectrometry can be used to determine relative atomic masses. The mole is defined as 6.022x1023 atoms, which allows chemists to convert between the number of atoms or molecules and mass. Empirical and molecular formulas are distinguished, and methods for determining formulas from elemental composition data are presented. Chemical equations are introduced as a way to represent chemical reactions by balancing reactants and products.

1. Advanced ChemistryAdvanced Chemistry
Chapter ThreeChapter Three
StoichiometryStoichiometry

2. 3.1 Counting by3.1 Counting by
WeighingWeighing
Average Mass = total mass/Average Mass = total mass/
number of objectsnumber of objects
For purposes of counting,For purposes of counting,
objects behave as though theyobjects behave as though they
are identicalare identical

3. 3.2 Atomic Masses3.2 Atomic Masses

4. Atomic MassesAtomic Masses
The modern system of atomicThe modern system of atomic
masses is based onmasses is based on 1212
C, as theC, as the
standard.standard.
Developed in 1961Developed in 1961

5. Mass SpectrometerMass Spectrometer
An instrument that passes atoms orAn instrument that passes atoms or
molecules through a beam of high-molecules through a beam of high-
speed electrons, which in turn knockspeed electrons, which in turn knock
electrons off the atoms or moleculeselectrons off the atoms or molecules
being analyzed and change them intobeing analyzed and change them into
positive ions.positive ions.

6. Mass SpectrometerMass Spectrometer
An applied electric fieldAn applied electric field
accelerates the ions into aaccelerates the ions into a
magnetic field.magnetic field.
The amount of deflection thatThe amount of deflection that
occurs with each ion dependsoccurs with each ion depends
upon its mass.upon its mass.

7. Mass SpectrometerMass Spectrometer
The most massive ions areThe most massive ions are
deflected the smallest amount.deflected the smallest amount.
A comparison of the positionsA comparison of the positions
where the ions hit the deflectorwhere the ions hit the deflector
plate provides accurate valuesplate provides accurate values
of relative masses.of relative masses.

8. A Scientist Injecting a Sample into aA Scientist Injecting a Sample into a
Mass Spectrometer. (right)Mass Spectrometer. (right)
Schematic Diagram of a MassSchematic Diagram of a Mass
SpectrometerSpectrometer

9. Atomic massesAtomic masses
Naturally occurring isotopes areNaturally occurring isotopes are
averaged to reflect the percent ofaveraged to reflect the percent of
abundance of those isotopes.abundance of those isotopes.
Counting by averaging the mass ofCounting by averaging the mass of
atoms allows for an accurateatoms allows for an accurate
atomic mass for chemicalatomic mass for chemical
calculations.calculations.

10. Mass spec PeaksMass spec Peaks

11. Mass Spec Bar GraphMass Spec Bar Graph

12. Mass Spectrum ofMass Spectrum of
Natural CopperNatural Copper
What is theWhat is the
average massaverage mass
of naturalof natural
copper?copper?

13. Mass Spectrum ofMass Spectrum of
Natural CopperNatural Copper
What is theWhat is the
average massaverage mass
of naturalof natural
copper?copper?
63.5563.55
amu/atomamu/atom

14. 3.3 The mole3.3 The mole

15. MoleMole
The number equal to the number ofThe number equal to the number of
carbon atoms in exactly 12 grams ofcarbon atoms in exactly 12 grams of
purepure 1212
C.C.
6.022 x 106.022 x 102323
A sample of a natural element with a massA sample of a natural element with a mass
equal to the element’s atomic massequal to the element’s atomic mass
expressed in grams.expressed in grams.

16. Question?Question?
What is the mass, in grams, of 12What is the mass, in grams, of 12
atoms of Aluminum?atoms of Aluminum?

17. AnswerAnswer
What is the mass, in grams, of 12What is the mass, in grams, of 12
atoms of Aluminum?atoms of Aluminum?
12atoms
x
26.98amu
atom
= 323.76amu

18. AnswerAnswer
What is the mass, in grams, of 12What is the mass, in grams, of 12
atoms of Aluminum?atoms of Aluminum?
12atoms
x
26.98amu
atom
= 323.76amu
1g = 6.022x1023
amu

19. AnswerAnswer
What is the mass, in grams, of 12What is the mass, in grams, of 12
atoms of Aluminum?atoms of Aluminum?
12atoms
x
26.98amu
atom
= 323.76amu
1g = 6.022x1023
amu
3.23x102
amu
x
1g
6.022x1023
amu
=

20. Question?Question?
How many moles and number ofHow many moles and number of
atoms are in a 10.0g sample ofatoms are in a 10.0g sample of
Aluminum?Aluminum?

21. AnswerAnswer
How many moles and number ofHow many moles and number of
atoms are in a 10.0g sample ofatoms are in a 10.0g sample of
Aluminum?Aluminum?
10.0gAl
x
1molAl
26.98gAl
= 0.371molAl
.371molAl
x
6.022x1023
atoms
1mole
= 2.23x1023
atoms

22. 3.4 Molar Mass3.4 Molar Mass

23. Molar MassMolar Mass
Is the mass in grams of one mole ofIs the mass in grams of one mole of
the compound.the compound.
““molecular weight”molecular weight”

24. Question?Question?
The formula forThe formula for
juglone, a dye, isjuglone, a dye, is
CC1010HH66OO33. What is the. What is the
molar mass?molar mass?

25. AnswerAnswer
The formula forThe formula for
juglone, a dye, isjuglone, a dye, is
CC1010HH66OO33. What is the. What is the
molar mass?molar mass?
10Cx12.01g =120.1g
6Hx1.008g =6.048g
3Ox16.00g = 48.00g
120.1g+6.048g+ 48.00g =174.1g

26. Question?Question?
If the molar mass of juglone isIf the molar mass of juglone is
174.1g, How many moles of Juglone174.1g, How many moles of Juglone
are in a 1.56 x 10are in a 1.56 x 10-2-2
g sample?g sample?

27. AnswerAnswer
If the molar mass of juglone isIf the molar mass of juglone is
174.1g, How many moles of Juglone174.1g, How many moles of Juglone
are in a 1.56 x 10are in a 1.56 x 10-2-2
g sample?g sample?
1.56x10−2
gJuglone
x
1molJuglone
174.1gJuglone
= 8.96x10−5
moles

28. 3.5 Learning to Solve3.5 Learning to Solve
ProblemsProblems

29. Conceptual ProblemConceptual Problem
SolvingSolving
1.1. Read the problem and decide on finalRead the problem and decide on final
goal. Gather facts and state the problemgoal. Gather facts and state the problem
as simply as possible.as simply as possible.
• Where are we going?Where are we going?

30. Conceptual ProblemConceptual Problem
SolvingSolving
2.2. Work backwards from the final goal toWork backwards from the final goal to
decide where to start.decide where to start.
• How do we get there?How do we get there?

31. Conceptual ProblemConceptual Problem
SolvingSolving
3.3. Once a solution is obtained, check toOnce a solution is obtained, check to
see in answer is reasonable.see in answer is reasonable.
• Does it make sense?Does it make sense?

32. 3.6 Percent Composition of3.6 Percent Composition of
CompoundsCompounds

33. Mass PercentageMass Percentage
Compare the mass of each elementCompare the mass of each element
in one mole to the molar mass ofin one mole to the molar mass of
the compound.the compound.

34. Question?Question?
What is the mass percentage of C,What is the mass percentage of C,
H and O in the following molecule:H and O in the following molecule:
CC1010HH1414O?O?

35. AnswerAnswer
What is the mass percentage of C,What is the mass percentage of C,
H and O in the following molecule:H and O in the following molecule:
CC1010HH1414O?O?

36. 3.7 Determining the3.7 Determining the
Formula of a CompoundFormula of a Compound

37. Empirical vs. MolecularEmpirical vs. Molecular
FormulaFormula
Empirical formula is the formula of aEmpirical formula is the formula of a
molecule in its smallest whole numbermolecule in its smallest whole number
ratio.ratio.
Molecular formula is the exact formulaMolecular formula is the exact formula
of the molecule as it exists.of the molecule as it exists.
Example:Example:
Empirical = CHEmpirical = CH55NN
Molecular = (CHMolecular = (CH55N)N)nn

38. Substances WhoseSubstances Whose
Empirical and MolecularEmpirical and Molecular
Formulas DifferFormulas Differ

39. Determining theDetermining the
Empirical FormulaEmpirical Formula
Use percent composition as a 100g moleculeUse percent composition as a 100g molecule
sample.sample.
Divide the element mass sample by the molarDivide the element mass sample by the molar
mass of each element to determine the molarmass of each element to determine the molar
ratios between elements.ratios between elements.
Divide the molar ratios by the smallest ratio.Divide the molar ratios by the smallest ratio.
If needed, multiply all ratios by the sameIf needed, multiply all ratios by the same
number to obtain low whole numbers.number to obtain low whole numbers.

40. Determining theDetermining the
Molecular FormulaMolecular Formula
Determine the empirical formula byDetermine the empirical formula by
using mole ratios between elements.using mole ratios between elements.
Divide the molar mass of the molecularDivide the molar mass of the molecular
formula by the molar mass of theformula by the molar mass of the
empirical formula. (n)empirical formula. (n)
Multiply every element in the formulaMultiply every element in the formula
by (n).by (n).

41. Question?Question?
Determine the empirical andDetermine the empirical and
molecular formulas for amolecular formulas for a
compound that gives the followingcompound that gives the following
percentages on analysis (in masspercentages on analysis (in mass
percents):percents):
71.65% Cl71.65% Cl 24.27% C24.27% C 4.07% H4.07% H
The molar mass is 98.96 g/molThe molar mass is 98.96 g/mol

42. AnswerAnswer
71.65% Cl71.65% Cl 24.27% C24.27% C 4.07% H4.07% H
molar mass = 98.96 g/molmolar mass = 98.96 g/mol
Empirical: ClCHEmpirical: ClCH22
Molecular: ClMolecular: Cl22CC22HH44

43. 3.8 Chemical Equations3.8 Chemical Equations

44. Chemical EquationsChemical Equations
Representation of the chemicalRepresentation of the chemical
reaction processreaction process
Reactants – left sideReactants – left side
Products – right sideProducts – right side

45. Chemical EquationsChemical Equations
Atoms are reorganized. Bonds haveAtoms are reorganized. Bonds have
been broken, and new ones havebeen broken, and new ones have
been formed.been formed.
Atoms are neither created norAtoms are neither created nor
destroyed therefore all atomsdestroyed therefore all atoms
present in the reactants must bepresent in the reactants must be
accounted for among the products.accounted for among the products.

46. Chemical EquationsChemical Equations
Subscripts apply to an atom orSubscripts apply to an atom or
atoms in parenthesis.atoms in parenthesis.
Coefficients apply to entireCoefficients apply to entire
molecule/compound.molecule/compound.

47. Chemical EquationsChemical Equations
Physical states should be given.Physical states should be given.
Solid – (s)Solid – (s)
Liquid – (l)Liquid – (l)
Gas – (g)Gas – (g)
Dissolved in water – (aq)Dissolved in water – (aq)

48. 3.9 Balancing Chemical3.9 Balancing Chemical
EquationsEquations

49. Writing and BalancingWriting and Balancing
EquationsEquations
Determine what reaction is occurring.Determine what reaction is occurring.
What are the reactants, the products, andWhat are the reactants, the products, and
the physical states involved.the physical states involved.
Write the unbalanced equation thatWrite the unbalanced equation that
summarizes the reaction.summarizes the reaction.

50. Writing and BalancingWriting and Balancing
EquationsEquations
Balance the equation by inspection,Balance the equation by inspection,
starting with the most complicatedstarting with the most complicated
molecules. Determine what coefficientsmolecules. Determine what coefficients
are necessary so that the same number ofare necessary so that the same number of
each type of atom appears on botheach type of atom appears on both
reactant and product sides. Do not changereactant and product sides. Do not change
the identities (formulas) of any of thethe identities (formulas) of any of the
reactants or products.reactants or products.

51. Question?Question?
(NH(NH44))22CrCr22OO7(s)7(s) CrCr22OO3(s)3(s) + N+ N2(g)2(g) + H+ H22OO(g)(g)

52. AnswerAnswer
(NH(NH44))22CrCr22OO7(s)7(s) CrCr22OO3(s)3(s) + N+ N2(g)2(g) + H+ H22OO(g)(g)
11 1 + 1 + 41 + 1 + 4

53. Question?Question?
NHNH3(g)3(g) + O+ O2(g)2(g) NONO(g)(g) + H+ H22OO(g)(g)

54. AnswerAnswer
NHNH3(g)3(g) + O+ O2(g)2(g) NONO(g)(g) + H+ H22OO(g)(g)
4 + 54 + 5 4 + 64 + 6

55. 3.10 Stoichiometric3.10 Stoichiometric
Calculations: Amounts ofCalculations: Amounts of
Reactants and ProductsReactants and Products

56. StoichiometryStoichiometry
Write a balanced equation.Write a balanced equation.
Coefficients in the balanced equationCoefficients in the balanced equation
provide the mole ratios used in theprovide the mole ratios used in the
conversion of mass and other quantitiesconversion of mass and other quantities
from one molecule/compound tofrom one molecule/compound to
another.another.

57. Calculating Mass ofCalculating Mass of
Reactants and ProductsReactants and Products

58. StoichiometryStoichiometry
mass → moles
mole
ratio
moles → mass

59. Question?Question?
NaHCONaHCO3(s)3(s) + HCl+ HCl(aq)(aq) NaClNaCl(aq)(aq) + H+ H22OO(l)(l) + CO+ CO2(aq)2(aq)
Mg(OH)Mg(OH)2(s)2(s) + 2HCl+ 2HCl(aq)(aq) 2H2H22OO(l)(l) + MgCl+ MgCl2(aq)2(aq)
Which one of these antacids neutralizes more acidWhich one of these antacids neutralizes more acid
with a 1.0g sample?with a 1.0g sample?

60. AnswerAnswer
NaHCONaHCO3(s)3(s) + HCl+ HCl(aq)(aq) NaClNaCl(aq)(aq) + H+ H22OO(l)(l) + CO+ CO2(aq)2(aq)
Mg(OH)Mg(OH)2(s)2(s) + 2HCl+ 2HCl(aq)(aq) 2H2H22OO(l)(l) + MgCl+ MgCl2(aq)2(aq)
Which one of these antacids neutralizes more acidWhich one of these antacids neutralizes more acid
with a 1.0g sample?with a 1.0g sample?
NaHCONaHCO33: 1.19 x 10: 1.19 x 10-2-2
mol HCl neutralizedmol HCl neutralized
Mg(OH)Mg(OH)22: 3.42 x 10: 3.42 x 10-2-2
mol HCl neutralizedmol HCl neutralized

61. 3.11 The Concept of3.11 The Concept of
Limiting ReagentLimiting Reagent

62. Stoichiometric MixturesStoichiometric Mixtures
A stoichiometric mixture is one thatA stoichiometric mixture is one that
contains the relative amounts ofcontains the relative amounts of
reactants that match the numbers inreactants that match the numbers in
the balanced equation.the balanced equation.
Assuming the reaction goes toAssuming the reaction goes to
completion, all reactants will becompletion, all reactants will be
consumed to form products.consumed to form products.

63. Limiting ReactantLimiting Reactant
The reactant that runs out first andThe reactant that runs out first and
therefore limits the amount of producttherefore limits the amount of product
that can form.that can form.
To determine how much product can beTo determine how much product can be
formed from a given mixture offormed from a given mixture of
reactants, the limiting reactant mustreactants, the limiting reactant must
first be determined.first be determined.

64. Solving aSolving a
StoichiomeStoichiome
try Problemtry Problem
InvolvingInvolving
Masses ofMasses of
ReactantsReactants
andand
ProductsProducts

66. Isopentyl AcetateIsopentyl Acetate

67. CarvonCarvon
ee

68. Carbon DioxideCarbon Dioxide

69. WaterWater

70. Diagram of theDiagram of the
Combustion Device UsedCombustion Device Used
to Analyze Substances forto Analyze Substances for
Carbon and HydrogenCarbon and Hydrogen

71. Figure 3.9 Three Different StoichiometricFigure 3.9 Three Different Stoichiometric
Mixtures of Methane and Water, whichMixtures of Methane and Water, which
React One-to-OneReact One-to-One

72. Figure 3.10 A Mixture ofFigure 3.10 A Mixture of
CH4 and H20 MoleculesCH4 and H20 Molecules

73. Figure 3.11 Methane andFigure 3.11 Methane and
Water Have Reacted toWater Have Reacted to
Form ProductsForm Products

74. Figure 3.12 Hydrogen andFigure 3.12 Hydrogen and
Nitrogen React to FormNitrogen React to Form
AmmoniaAmmonia

75. Jellybeans Can be CountedJellybeans Can be Counted
by Weighingby Weighing

76. Weighing Hex NutsWeighing Hex Nuts

77. Copper NuggetCopper Nugget

78. Figure 3.4 Samples Containing One MoleFigure 3.4 Samples Containing One Mole
Each of Copper, Aluminum, Iron, Sulfur,Each of Copper, Aluminum, Iron, Sulfur,
Iodine, and MercuryIodine, and Mercury

79. Pure AluminumPure Aluminum

80. Bee Stings Cause theBee Stings Cause the
Release of IsopentylRelease of Isopentyl
AcetateAcetate

81. Penicillin is Isolated from a Mold that Can be Grown inPenicillin is Isolated from a Mold that Can be Grown in
Large Quantities in Fermentation TanksLarge Quantities in Fermentation Tanks

82. Figure 3.7 The Two FormsFigure 3.7 The Two Forms
of Dichloroenthaneof Dichloroenthane

83. Figure 3.8 The StructureFigure 3.8 The Structure
of P4O10.of P4O10.

84. Computer-GeneratedComputer-Generated
Molecule of CaffeineMolecule of Caffeine

85. MethanMethan
e Reactse Reacts
withwith
OxygenOxygen
toto
ProduceProduce
FlameFlame

86. withwith
SolidSolid
SodiumSodium
HydrogeHydroge
nn
CarbonaCarbona
te tote to
ProduceProduce

87. Decomposition ofDecomposition of
Ammonium DichromateAmmonium Dichromate

88. Decomposition ofDecomposition of
Ammonium DichromateAmmonium Dichromate

89. AstronaAstrona
utut
SidneySidney
M.M.
GutierrGutierr
ezez

90. Milk ofMilk of
MagnesMagnes
iaia

91. Race Cars use Methanol asRace Cars use Methanol as
a Fuela Fuel

92. Table 3.1 Comparison of 1Table 3.1 Comparison of 1
Mole Samples of VariousMole Samples of Various
ElementsElements

93. Table 3.2 InformationTable 3.2 Information
Conveyed by the BalancedConveyed by the Balanced
Equation for theEquation for the
Combustion of MethaneCombustion of Methane