adv chem chapt 3

Advanced Advanced ChemistryChemistry

Chapter ThreeChapter ThreeStoichiometryStoichiometry

3.1 Counting by 3.1 Counting by WeighingWeighing

Average Mass = total Average Mass = total mass/ number of objectsmass/ number of objects

For purposes of For purposes of counting, objects counting, objects behave as though they behave as though they are identicalare identical

3.2 Atomic Masses3.2 Atomic Masses

Atomic MassesAtomic Masses

The modern system of The modern system of atomic masses is based atomic masses is based on on 1212C, as the C, as the standard.standard.Developed in 1961Developed in 1961

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

Mass SpectrometerMass Spectrometer

An applied electric field An applied electric field accelerates the ions into accelerates the ions into a magnetic field. a magnetic field. The amount of deflection The amount of deflection that occurs with each ion that occurs with each ion depends upon its mass.depends upon its mass.

Mass SpectrometerMass Spectrometer

The most massive ions are The most massive ions are deflected the smallest amount. deflected the smallest amount.

A comparison of the positions A comparison of the positions where the ions hit the where the ions hit the deflector plate provides deflector plate provides accurate values of relative accurate values of relative masses.masses.

A Scientist Injecting a A Scientist Injecting a Sample into a Mass Sample into a Mass

Spectrometer. (right) Spectrometer. (right) Schematic Diagram of a Schematic Diagram of a

Mass SpectrometerMass Spectrometer

Atomic massesAtomic masses

Naturally occurring isotopes Naturally occurring isotopes are averaged to reflect the are averaged to reflect the percent of abundance of those percent of abundance of those isotopes.isotopes.

Counting by averaging the mass Counting by averaging the mass of atoms allows for an of atoms allows for an accurate atomic mass for accurate atomic mass for chemical calculations. chemical calculations.

Mass spec PeaksMass spec Peaks

Mass Spec Bar Mass Spec Bar GraphGraph

Mass Spectrum of Mass Spectrum of Natural CopperNatural Copper

What is the What is the average average mass of mass of natural natural copper?copper?

Mass Spectrum of Mass Spectrum of Natural CopperNatural Copper

What is the What is the average average mass of mass of natural natural copper?copper?

63.55 63.55 amu/atomamu/atom

3.3 The mole3.3 The mole

MoleMoleThe number equal to the The number equal to the number of carbon atoms in number of carbon atoms in exactly 12 grams of pure exactly 12 grams of pure 1212C.C.

6.022 x 106.022 x 102323

A sample of a natural element A sample of a natural element with a mass equal to the with a mass equal to the element’s atomic mass expressed element’s atomic mass expressed in grams. in grams.

Question?Question?

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

AnswerAnswer

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

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12atomsx26.98amu

atom= 323.76amu

AnswerAnswer

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

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12atomsx26.98amu

atom= 323.76amu

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1g = 6.022x1023amu

AnswerAnswer

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

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12atomsx26.98amu

atom= 323.76amu

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1g = 6.022x1023amu€

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3.23x102amux

1g

6.022x1023amu=

Question?Question?

How many moles and number How many moles and number of atoms are in a 10.0g of atoms are in a 10.0g sample of Aluminum?sample of Aluminum?

AnswerAnswer

How many moles and number How many moles and number of atoms are in a 10.0g of atoms are in a 10.0g sample of Aluminum?sample of Aluminum?

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10.0gAlx1molAl

26.98gAl= 0.371molAl

.371molAlx6.022x1023atoms

1mole= 2.23x1023atoms

3.4 Molar Mass3.4 Molar Mass

Molar MassMolar Mass

Is the mass in grams of Is the mass in grams of one mole of the compound.one mole of the compound.

““molecular weight”molecular weight”

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

AnswerAnswerThe formula The formula for juglone, a for juglone, a dye, is dye, is CC1010HH66OO33. What . What is the molar is the molar mass?mass?

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10Cx12.01g =120.1g

6Hx1.008g = 6.048g

3Ox16.00g = 48.00g

120.1g+ 6.048g+ 48.00g =174.1g

Question?Question?

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

AnswerAnswer

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

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1.56x10−2gJuglonex1molJuglone

174.1gJuglone= 8.96x10−5moles

3.5 Learning to 3.5 Learning to Solve ProblemsSolve Problems

Conceptual Conceptual Problem SolvingProblem Solving

1.1.Read the problem and decide Read the problem and decide on final goal. Gather facts on final goal. Gather facts and state the problem as and state the problem as simply as possible.simply as possible.• Where are we going?Where are we going?

Conceptual Conceptual Problem SolvingProblem Solving

2.2. Work backwards from the Work backwards from the final goal to decide where final goal to decide where to start.to start.

• How do we get there?How do we get there?

Conceptual Conceptual Problem SolvingProblem Solving

3.3. Once a solution is obtained, Once a solution is obtained, check to see in answer is check to see in answer is reasonable.reasonable.

• Does it make sense? Does it make sense?

3.6 Percent 3.6 Percent Composition of Composition of

CompoundsCompounds

Mass PercentageMass Percentage

Compare the mass of each Compare the mass of each element in one mole to element in one mole to the molar mass of the the molar mass of the compound.compound.

Question?Question?

What is the mass What is the mass percentage of C, H and O percentage of C, H and O in the following in the following molecule: Cmolecule: C1010HH1414O?O?

AnswerAnswer

What is the mass What is the mass percentage of C, H and O percentage of C, H and O in the following in the following molecule: Cmolecule: C1010HH1414O?O?

3.7 Determining 3.7 Determining the Formula of a the Formula of a

CompoundCompound

Empirical vs. Empirical vs. Molecular FormulaMolecular FormulaEmpirical formula is the Empirical formula is the formula of a molecule in its formula of a molecule in its smallest whole number ratio.smallest whole number ratio.

Molecular formula is the exact Molecular formula is the exact formula of the molecule as it formula of the molecule as it exists.exists.Example: Example: Empirical = CHEmpirical = CH55NNMolecular = (CHMolecular = (CH55N)N)nn

Substances Whose Substances Whose Empirical and Empirical and

Molecular Formulas Molecular Formulas DifferDiffer

Determining the Determining the Empirical Formula Empirical Formula

Use percent composition as a 100g molecule Use percent composition as a 100g molecule sample. sample.

Divide the element mass sample by the molar Divide the element mass sample by the molar mass of each element to determine the molar mass of each element to determine the molar ratios between elements. ratios between elements.

Divide the molar ratios by the smallest Divide the molar ratios by the smallest ratio. ratio.

If needed, multiply all ratios by the same If needed, multiply all ratios by the same number to obtain low whole numbers.number to obtain low whole numbers.

Determining the Determining the Molecular FormulaMolecular FormulaDetermine the empirical formula Determine the empirical formula by using mole ratios between by using mole ratios between elements.elements.

Divide the molar mass of the Divide the molar mass of the molecular formula by the molar molecular formula by the molar mass of the empirical formula. mass of the empirical formula. (n) (n)

Multiply every element in the Multiply every element in the formula by (n). formula by (n).

Question?Question?

Determine the empirical and Determine the empirical and molecular formulas for a molecular formulas for a compound that gives the compound that gives the following percentages on following percentages on analysis (in mass percents):analysis (in mass percents): 71.65% Cl71.65% Cl 24.27% C24.27% C4.07% H4.07% H

The molar mass is 98.96 The molar mass is 98.96 g/molg/mol

AnswerAnswer

71.65% Cl71.65% Cl 24.27% C24.27% C 4.07% 4.07% HH

molar mass = 98.96 molar mass = 98.96 g/molg/mol

Empirical: ClCHEmpirical: ClCH22

Molecular: ClMolecular: Cl22CC22HH44

3.8 Chemical 3.8 Chemical EquationsEquations

Chemical Chemical EquationsEquations

Representation of the Representation of the chemical reaction processchemical reaction processReactants – left sideReactants – left sideProducts – right sideProducts – right side

Chemical Chemical EquationsEquations

Atoms are reorganized. Bonds Atoms are reorganized. Bonds have been broken, and new ones have been broken, and new ones have been formed. have been formed.

Atoms are neither created nor Atoms are neither created nor destroyed therefore all atoms destroyed therefore all atoms present in the reactants must present in the reactants must be accounted for among the be accounted for among the products.products.

Chemical Chemical EquationsEquations

Subscripts apply to an Subscripts apply to an atom or atoms in atom or atoms in parenthesis.parenthesis.

Coefficients apply to Coefficients apply to entire molecule/compound.entire molecule/compound.

Chemical Chemical EquationsEquations

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

3.9 Balancing 3.9 Balancing Chemical Chemical EquationsEquations

Writing and Writing and Balancing Balancing EquationsEquations

Determine what reaction is Determine what reaction is occurring. What are the occurring. What are the reactants, the products, and reactants, the products, and the physical states involved.the physical states involved.

Write the unbalanced equation Write the unbalanced equation that summarizes the reaction.that summarizes the reaction.

Writing and Writing and Balancing Balancing EquationsEquations

Balance the equation by inspection, Balance the equation by inspection, starting with the most complicated starting with the most complicated molecules. Determine what molecules. Determine what coefficients are necessary so that coefficients are necessary so that the same number of each type of the same number of each type of atom appears on both reactant and atom appears on both reactant and product sides. Do not change the product sides. Do not change the identities (formulas) of any of the identities (formulas) of any of the reactants or products. reactants or products.

Question?Question?

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

AnswerAnswer

(NH(NH44))22CrCr22OO7(s) 7(s) CrCr22OO3(s)3(s) + + NN2(g)2(g) + H + H22OO(g)(g)

11 1 + 1 + 4 1 + 1 + 4

Question?Question?

NHNH3(g) 3(g) + O+ O2(g) 2(g) NONO(g) (g) + + HH22OO(g)(g)

AnswerAnswer

NHNH3(g) 3(g) + O+ O2(g) 2(g) NONO(g) (g) + + HH22OO(g)(g)

4 + 5 4 + 5 4 + 64 + 6

3.10 Stoichiometric 3.10 Stoichiometric Calculations: Calculations: Amounts of Amounts of

Reactants and Reactants and ProductsProducts

StoichiometryStoichiometry

Write a balanced equation.Write a balanced equation.

Coefficients in the balanced Coefficients in the balanced equation provide the mole equation provide the mole ratios used in the conversion ratios used in the conversion of mass and other quantities of mass and other quantities from one molecule/compound to from one molecule/compound to another.another.

Calculating Mass Calculating Mass of Reactants and of Reactants and

ProductsProducts

StoichiometryStoichiometry

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mass→molesmole

ratiomoles→mass

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) 2H 2H22OO(l)(l) + MgCl+ MgCl2(aq)2(aq)

Which one of these antacids Which one of these antacids neutralizes more acid with a 1.0g neutralizes more acid with a 1.0g sample?sample?

AnswerAnswer

NaHCONaHCO3(s)3(s) + HCl + HCl(aq) (aq) NaClNaCl(aq)(aq) + H + H22OO(l)(l) + + COCO2(aq)2(aq)

Mg(OH)Mg(OH)2(s)2(s) + 2HCl + 2HCl(aq)(aq) 2H2H22OO(l)(l) + + MgClMgCl2(aq)2(aq)

Which one of these antacids neutralizes Which one of these antacids neutralizes more acid with a 1.0g sample?more acid with a 1.0g sample?

NaHCONaHCO33: 1.19 x 10: 1.19 x 10-2-2 mol HCl neutralized mol HCl neutralized

Mg(OH)Mg(OH)22: 3.42 x 10: 3.42 x 10-2-2 mol HCl neutralized mol HCl neutralized

3.11 The Concept of 3.11 The Concept of Limiting ReagentLimiting Reagent

Stoichiometric Stoichiometric MixturesMixtures

A stoichiometric mixture is A stoichiometric mixture is one that contains the one that contains the relative amounts of reactants relative amounts of reactants that match the numbers in the that match the numbers in the balanced equation. balanced equation. Assuming the reaction goes Assuming the reaction goes to completion, all reactants to completion, all reactants will be consumed to form will be consumed to form products.products.

Limiting ReactantLimiting Reactant

The reactant that runs out first The reactant that runs out first and therefore limits the amount and therefore limits the amount of product that can form.of product that can form.

To determine how much product To determine how much product can be formed from a given can be formed from a given mixture of reactants, the mixture of reactants, the limiting reactant must first be limiting reactant must first be determined.determined.

Solving Solving a a StoichioStoichiometry metry Problem Problem InvolvinInvolving Masses g Masses of of ReactantReactants and s and ProductsProducts

Isopentyl AcetateIsopentyl Acetate

CarvoCarvonene

Carbon DioxideCarbon Dioxide

WaterWater

Figure 3.5 A Figure 3.5 A Schematic Diagram Schematic Diagram of the Combustion of the Combustion Device Used to Device Used to

Analyze Substances Analyze Substances for Carbon and for Carbon and

HydrogenHydrogen

Figure 3.9 Three Different Figure 3.9 Three Different Stoichiometric Mixtures of Stoichiometric Mixtures of

Methane and Water, which React Methane and Water, which React One-to-OneOne-to-One

Figure 3.10 A Figure 3.10 A Mixture of CH4 and Mixture of CH4 and

H20 MoleculesH20 Molecules

Figure 3.11 Figure 3.11 Methane and Water Methane and Water Have Reacted to Have Reacted to Form ProductsForm Products

Figure 3.12 Figure 3.12 Hydrogen and Hydrogen and

Nitrogen React to Nitrogen React to Form AmmoniaForm Ammonia

Jellybeans Can be Jellybeans Can be Counted by Counted by WeighingWeighing

Weighing Hex NutsWeighing Hex Nuts

Copper NuggetCopper Nugget

Figure 3.4 Samples Containing Figure 3.4 Samples Containing One Mole Each of Copper, One Mole Each of Copper,

Aluminum, Iron, Sulfur, Iodine, Aluminum, Iron, Sulfur, Iodine, and Mercuryand Mercury

Pure AluminumPure Aluminum

Bee Stings Cause Bee Stings Cause the Release of the Release of

Isopentyl AcetateIsopentyl Acetate

Penicillin is Isolated from a Mold that Penicillin is Isolated from a Mold that Can be Grown in Large Quantities in Can be Grown in Large Quantities in

Fermentation TanksFermentation Tanks

Figure 3.7 The Figure 3.7 The Two Forms of Two Forms of

DichloroenthaneDichloroenthane

Figure 3.8 The Figure 3.8 The Structure of Structure of

P4O10.P4O10.

Computer-Generated Computer-Generated Molecule of Molecule of CaffeineCaffeine

MethaMethane ne ReactReacts s with with OxygeOxygen to n to ProduProduce ce FlameFlame

HydrocHydrochloric hloric Acid Acid Reacts Reacts with with Solid Solid Sodium Sodium HydrogHydrogen en CarbonCarbonate to ate to ProducProduce e GaseouGaseous s Carbon Carbon DioxidDioxidee

Decomposition of Decomposition of Ammonium Ammonium

DichromateDichromate

Decomposition of Decomposition of Ammonium Ammonium

DichromateDichromate

AstroAstronaut naut SidneSidney M. y M. GutieGutierrezrrez

Milk Milk of of MagneMagnesiasia

Race Cars use Race Cars use Methanol as a FuelMethanol as a Fuel

Table 3.1 Table 3.1 Comparison of 1 Comparison of 1 Mole Samples of Mole Samples of Various ElementsVarious Elements

Table 3.2 Table 3.2 Information Information

Conveyed by the Conveyed by the Balanced Equation Balanced Equation for the Combustion for the Combustion

of Methaneof Methane