Chem 106, Prof. J.T. Spencer 1 CHE 106: General Chemistry u CHAPTER THREE Copyright © James T. Spencer 1995 - 1999 All Rights Reserved

Chem 106, Prof. J.T. Spencer

11CHE 106: General Chemistry

CHAPTER THREE

Copyright © James T. Spencer 1995 - 1999

All Rights Reserved


22StoichiometryStoichiometry

Chapter ThreeChapter Three


33StoichiometryStoichiometry

Chapt. 3.1

2H2 + O2 2 H2Oreactants products

• Antoine Lavoisier (1734 - 1794)Antoine Lavoisier (1734 - 1794)– Law of Conservation of MassLaw of Conservation of Mass - atoms are

neither created nor destroyed in chemical reactions

– total number of atoms = total number of atoms after reaction before reaction

– StoichiometryStoichiometry - quantitative study of chemical formulas and reactions

(Greek; “stoichion”= element, “metron” = measure)

• Chemical EquationsChemical Equations - used to describe - used to describe chemical reactions in an accurate and chemical reactions in an accurate and convenient fashionconvenient fashion


44Chemical Equations

To Write and Balance: (Shorthand Communication for a great deal of information)

(1) Know Reactants(2) Know ALL Products(3) Balance - Same Number and

Kinds of atoms on each side


55

• Chemical EquationsChemical Equations – Must have equal numbers of

atoms of each element on each side of the equation = BALANCED EQUATION

Chemical EquationsChemical Equations

Chapt. 3.1

2 H2 + O2 2 H2O4 hydrogen 4 hydrogen2 oxygen 2 oxygen

N2O5(g) + H2O 2 HNO3

2 nitrogen 2 nitrogen6 oxygen 6 oxygen2 hydrogen 2 hydrogen

The coefficients in front of the

formula for a compound

refers to the number of molecules

(intact) involved while

a subscript refers to the

ratio of atoms within the molecule

NOTE


66

• Chemical EquationsChemical Equations – balancing equations often

requires some trial and error of coefficients

Chapt. 3.1

PCl3(l) + 3 H2O(l) H3PO3(aq) + 3 HCl6 hydrogen 6 hydrogen3 oxygen 3 oxygen1 phosphorus 1 phosphorus3 chloride 3 chlorine

C6H12(l) + 9 O2(g) 6 CO2(g) + 6 H2O(l)

6 carbon 6 carbon18 oxygen 18 oxygen12 hydrogen 12 hydrogen

Never change subscripts in

formulas when

balancing chemical

reactions!subscripts

change compounds; coefficients

change amounts

NOTE



77

Sample exercise: Balance the following equations by providing the missing coefficients:

C2H4 + O2 CO2 + H2O

Chapt. 3.1



88


C2H4 + O2 CO2 + H2O

C C H H O O

Chapt. 3.1



99


C2H4 + O2 CO2 + H2O

C2 C 1 H 4 H 2 O2 O 3

Chapt. 3.1



1010


C2H4 + O2 2 CO2 + H2O

C2 C (1)2= 2 H 4 H 2 O2 O 3 5

Chapt. 3.1



1111


C2H4 + O2 2 CO2 + 2H2O

C2 C (1)2= 2 H 4 H (2)2 = 4 O2 O 3 5 6

Chapt. 3.1



1212


C2H4 + 3O2 2 CO2 + 2H2O

C2 C (1)2= 2 H 4 H (2)2 = 4 O(2)3 = 6 O 3 5 6

Chapt. 3.1



1313


Al + HCl AlCl3 + H2

Chapt. 3.1



1414


Al + HCl AlCl3 + H2

Al Al H H Cl Cl

Chapt. 3.1



1515


Al + HCl AlCl3 + H2

Al1 Al 1 H 1 H 2 Cl1 Cl 3

Chapt. 3.1



1616


Al + 3HCl AlCl3 + H2

Al 1 Al 1H (1)3 = 3 H 2Cl (1)3 = 3 Cl 3

Chapt. 3.1



1717


Al + 6HCl AlCl3 + 3H2

Al 1 Al 1H (1)6 = 6 H (2)3 = 6Cl (1)6 = 6 Cl 3

Chapt. 3.1



1818


Al + 6HCl 2AlCl3 + 3H2

Al 1 Al (1)2 = 2H (1)6 = 6 H (2)3 = 6Cl (1)6 = 6 Cl (3)2 = 6

Chapt. 3.1



1919


2Al + 6HCl 2AlCl3 + 3H2

Al (1)2 = 2 Al (1)2 = 2H (1)6 = 6 H (2)3 = 6Cl (1)6 = 6 Cl (3)2 = 6

Chapt. 3.1



2020

• Chemical ReactionsChemical Reactions

–The course of a chemical reaction can often be predicted by recognizing general patterns of reactivity through similar reactions previously observed. Elements in same family (column of table) have similar reactions.

–The periodic table is helpful in predicting products of reactions. Atoms like to assume electron configurations of the Noble Gases.

Chemical ReactivityChemical Reactivity

Chapt. 3.2


2121

• Chemical ReactionsChemical Reactions


Chapt. 3.2

Example, if you know that

2Li + 2H20 2LiOH + H2

then you should be able to predict the products from the reaction of Na, K and

the other members of group 1 (alkali metals) with water. Thus a general

reaction would be;

2 M + 2 H2O 2 MOH + H2


2222

•Combustion Reactions•Combination Reactions

•Decomposition Reactions

•Metathesis Reactions



2323

• Combustion ReactionsCombustion Reactions – Reactions with oxygen (usually from the

air)– The complete combustion of hydrocarbons

yield carbon dioxide (CO2) and water (H2O)

CxHy + (2x+y) /2 O2 X CO2 + Y/2 H2O


Chapt. 3.2

Generally:Balance Carbon Atoms FirstBalance HydrgoensBalance Oxygen Atoms


2424

octane: C8H18 + 25/2 O2 8 CO2 + 9 H2Oethanol: C2H5OH + 3 O2 2 CO2 + 3 H2Oglucose: C6H12O6 + 9 O2 6 CO2 + 6 H2Ostyrene: C8H8 + 10 O2 8 CO2 + 4 H2O

ExamplesExamples

• Combustion ReactionsCombustion Reactions – Reactions with oxygen (usually from the

air)– The complete combustion of hydrocarbons

yield carbon dioxide (CO2) and water (H2O)

CxHy + (2x+y) /2 O2 X CO2 + Y/2 H2O


Chapt. 3.2


2525

Sample exercise: Write the balanced equation for the reaction that occurs when ethanol, C2H5OH(l) is burned in air.


Chapt. 3.2


2626


C2H5OH + O2 CO2 + H2O


Chapt. 3.2


2727



C CH HO O


Chapt. 3.2


2828



C 2 C 1H 6 H 2O 3 O 3


Chapt. 3.2


2929


C2H5OH + O2 2CO2 + H2O

C 2 C (1)2 = 2

H 6 H 2O 3 O 3 5


Chapt. 3.2


3030


C2H5OH + O2 2CO2 + 3H2O

C 2 C (1)2 = 2

H 6 H (2)3 = 6

O 3 O 3 5 7


Chapt. 3.2


3131


C2H5OH + 3O2 2CO2 + 3H2O

C 2 C (1)2 = 2

H 6 H (2)3 = 6

O 3 7 O 3 5 7


Chapt. 3.2


3232Chemical ReactivityChemical Reactivity

Chapt. 3.2

• Combination ReactionsCombination Reactions

–two or more substances react to form a single product

–especially common in the reactions of pure elements

A + B C

Ni(s) + 4 CO(g) Ni(CO)4(g)

BF3(g) + NH3(g) BF3NH3(s)



Chapt. 3.2

• Decomposition ReactionsDecomposition Reactions

–when one compound reacts to form two or more products (opposite of combination reactions)

C A + B

• (often heat required)

2 NaN3(s) 2 Na(s) + 3 N2(g)

B(OH)3(heat) HBO2 +

H2O

Air Bag Inflator (J. Chem. Ed. 1990, 67, 61)



Chapt. 3.2

• Metathesis ReactionsMetathesis Reactions –when ionic “partners” switch

AB + CD AD + BC

• (often in aqueous solutions)

Ag(NO3) + KCl AgCl(s) + KNO3

BaCl2 + Na2SO4 BaSO4(s) + 2 NaCl



Chapt. 3.2

Sample Exercise: Write balanced Sample Exercise: Write balanced chemical equations for the chemical equations for the following reactions:following reactions:

Solid mercury (II) sulfide Solid mercury (II) sulfide decomposes into its component decomposes into its component elements when heatedelements when heated.



Chapt. 3.2


Solid mercury (II) sulfide Solid mercury (II) sulfide decomposes into its component decomposes into its component elements when heatedelements when heated.

HgHg+2+2 S S-2-2

HgS HgS Hg + S Hg + S



Chapt. 3.2


The surface of aluminum metal The surface of aluminum metal undergoes a combination reaction undergoes a combination reaction with oxygen in air.with oxygen in air.



Chapt. 3.2



AlAl+3+3 O O-2-2

Al + OAl + O22 Al Al22OO33



Chapt. 3.2



AlAl+3+3 O O-2-2

4Al + 3O4Al + 3O22 2Al 2Al22OO33



Chapt. 3.2


SiSi22HH66 burns when exposed to burns when exposed to airair.



Chapt. 3.2


SiSi22HH66 burns when exposed to burns when exposed to airair. Hint, Si is in the same group as C, and therefore reacts similarly.

SiSi22HH66 + O + O22 SiO SiO22 + H + H22OO



Chapt. 3.2


SiSi22HH66 burns when exposed to burns when exposed to airair. Hint, Si is in the same group as C, and therefore reacts similarly.

2Si2Si22HH66 + 7O + 7O22 4SiO 4SiO22 + 6H + 6H22OO


4343

•Chemical equations indicate exactly the amounts of two reagents which will react to form an exact amount of products

•Atomic Mass ScaleAtomic Mass Scale - based upon 12C isotope. This isotope is assigned a mass of exactly 12 atomic mass units (amu) and the masses of all other atoms are given relative to this standard.

•Most elements in nature exist as mixtures of isotopes.

Atomic and Molecular Atomic and Molecular WeightsWeights

Chapt. 3.3


4444

• Atomic Mass ScaleAtomic Mass Scale - given the following;100 g of water contains 11.1 g of H and

88.9 g of O and the formula for water is H2O then;•water has 8 times more O than H by mass (88.9/11 = 8)

•if water has 2 H for 1 O then O atoms must weigh 16 time more than H atoms

•if H is assigned an atomic mass of 1 amu then O must weigh 16 amu (using the 12C standard)

•1 amu = 1.66054 x 10-24 g OROR 1 g = 6.02214 x 1023 amu


Chapt. 3.3


4545

•Direct methods of measuring (separating) mass.

• Sample molecules are ionized by e-beam to cations (+1 by “knocking off” one electron) which are then deflected by magnetic field - for ions of the same charge the angle of deflection in proportional to the ion’s mass

Mass SpectrometerMass Spectrometer

Chapt. 3.3

N

S mass number (amu)

Int.

focusing slits

magnetic fielddetector

accelerating grid (-)

ionizing e- beam

beam of pos. ions

sample

vacuum chamber Mass Spectrum

Hg

200


4646Mass SpectrometerMass Spectrometer

mass number (amu)

Int.

Mass Spectrum

Cl

mass number (amu)

Int.

Mass Spectrum

C

mass number (amu)

Int.

Mass Spectrum

P35

37

35Cl: 75% abundant37Cl: 24% abundant

31

12C: 98.9% abundant13C: 1.11% abundant

31P: 100% abundant

12

13


4747

25 50 75 100 125 150 175 200 225 250 275 300

Inte

nsi

ty

Mass

Mass SpectrometerMass Spectrometer

Unknown white powdery substance injested by unconscious patient.

What do you do? Is it Heroin, Cocaine, Caffeine?

Mass Spectrum of Unknown CompoundMass Spectrum of Unknown Compound



25 50 75 100 125 150 175 200 225 250 275 300

Inte

nsit

y

Mass

Heroin other peaks at 327 and 369

43

94146

204215

268

25 50 75 100 125 150 175 200 225 250 275 300

Inte

nsit

y

Mass

Caffeine

42

55

67

82

109

194

MS of UnknownMS of Unknown

MS LibraryMS Library HeroinHeroin



25 50 75 100 125 150 175 200 225 250 275 300

Inte

nsit

y

Mass

42

82

122 150

182

272

303

Cocaine

25 50 75 100 125 150 175 200 225 250 275 300

Inte

nsit

y

Mass

Caffeine

42

55

67

82

109

194


MS MS LibraryLibrary

CocaineCocaine



25 50 75 100 125 150 175 200 225 250 275 300

Inte

nsit

y

Mass

Caffeine

42

55

67

82

109

194

25 50 75 100 125 150 175 200 225 250 275 300

Inte

nsit

y

Mass

Caffeine

42

55

67

82

109

194


MS LibraryMS Library CaffeineCaffeine



N

NN

N

CH3

CH3

H3C

O

OUnknown white powdery substance injested by unconscious patient. What do you do?

Mass SpectrumMass Spectrum

Mol. Wgt = 194

CaffeineCaffeine

25 50 75 100 125 150 175 200 225 250 275 300

Inte

nsi

ty

Mass


5252

• Average Atomic MassAverage Atomic Mass (AW)- weighted average (by % natural abundance) of the isotopes of an element.

•Example (1);

10B is 19.78% abundant with a mass of 10.013 amu 11B is 80.22% abundant with a mass of 11.009 amu

therefore the average atomic mass of boron is;

(0.1987)(10.013) + (0.8022)(11.009) = 10.82 amu

Atomic WeightsAtomic Weights

Chapt. 3.3


5353

• Average Atomic MassAverage Atomic Mass (AW)- weighted average (by % natural abundance) of the isotopes of an element. •Example (2):

194Pt is 33.90% abundant with a mass of 193.963 amu 195Pt is 33.80% abundant with a mass of 194.965 amu

196Pt is 25.30% abundant with a mass of 195.965 amu 198Pt is 7.210% abundant with a mass of 197.968 amu

therefore the average atomic mass of platinum is;(0.3390)(193.963) + (0.3380)(194.965) + (0.2530)(195.965 ) +

(0.07210)(197.968)= 195.504 amu

Atomic WeightsAtomic Weights

Chapt. 3.3


5454

Sample exercise: Three isotopes of Sample exercise: Three isotopes of silicon occur in nature: silicon occur in nature: 2828Si Si (92.21%), which has a mass of (92.21%), which has a mass of 27.97693 amu; 27.97693 amu; 2929Si (4.70%), which Si (4.70%), which has a mass of 28.97659 amu; and has a mass of 28.97659 amu; and 3030Si (3.09%), which has a mass of Si (3.09%), which has a mass of 29.97376 amu. Calculate the atomic 29.97376 amu. Calculate the atomic weight of silicon.weight of silicon.


Chapt. 3.3


5555


27.97693(0.9221) + 28.97659(0.0470) + 27.97693(0.9221) + 28.97659(0.0470) + 29.97376(0.0309) = 29.97376(0.0309) =

28.0856 amu28.0856 amu


Chapt. 3.3


5656


27.97693(0.9221) + 28.97659(0.0470) + 27.97693(0.9221) + 28.97659(0.0470) + 29.97376(0.0309) = 29.97376(0.0309) =

28.0856 amu28.0856 amu

* 3 sig figs * 3 sig figs 28.1 amu28.1 amu


Chapt. 3.3


5757

• Sample ProblemSample Problem: When a sample of natural copper is vaporized and injected into a mass spectrometer, the results shown are obtained. Use these data to compute the average mass of natural copper. [masses for 63Cu = 62.93 amu and 65Cu = 64.93 amu]

Atomic WeightsAtomic WeightsIn

ten

sit

yIn

ten

sit

y

Mass No. 63 65Mass No. 63 65

69.09%

30.91%

Given:•Masses for 63Cu and 65C•Relative abundance of 63Cu and 65CuFind:•Average Mass of Cu


5858

(.6909 atoms)(62.93 amu) + (.3091 atoms)(.6909 atoms)(62.93 amu) + (.3091 atoms)(64.93amu) = 63.55 amu(64.93amu) = 63.55 amu

atomatom atom. atom.

average mass per atom is;average mass per atom is;

6355 amu = 63.55 amu/atom6355 amu = 63.55 amu/atom

100 atoms100 atoms

Atomic WeightsAtomic WeightsIn

ten

sit

yIn

ten

sit

y

Mass No. 63 65Mass No. 63 65

69.09%

30.91%

Given:•Masses for 63Cu and 65C•Relative abundance of 63Cu and 65CuFind:•Average Mass of Cu


5959

• Formula Weights (FW)Formula Weights (FW) - sum of the atomic weights of each atom in its chemical formula. (note AW is atomic weight)

• formula weight of NaN3 = 3(AW of N) + 1(AW of Na) 3(14) + 1(23) = 65 amu for sodium azide

• Molecular Weights (MW)Molecular Weights (MW) - sum of atomic weights of each atom in its molecular formula

•molecular weight of B2H6 = 2(AW of B) + 6(AW of H) 2(10.8) + 6(1) = 27.6 amu for diborane

• Difference between Molecular and Formula Difference between Molecular and Formula Weights Weights • ionic compounds, with extended arrays, have

no well defined molecules (and no molecular formulas) so we use the formula weights (i.e., NaCl = 58 amu )

Molecular WeightsMolecular Weights

Chapt. 3.3


6060

• Calculate molecular/formula weights for the following:

–P4O10

–BrCl

–Ca(NO3)2

P =31; O = 16; Br = 80; Cl = 35.5; Ca = 40; N = 14



6161

• Calculate molecular weights/formula for the following:

–P4O10

4(31) + 10(16) = 284 amu–BrCl

1(80) + 1(35.5) = 115.5 amu–Ca(NO3)2

1(40) + 2(14) + 6(16) = 164 amu

P =31; O = 16; Br = 80; Cl = 35.5; Ca = 40; N = 14



6262

• Percentage CompositionPercentage Composition - percentage by mass contributed by each element in the substance. May be used to verify the purity or identity of a particular compound.

• 100 [(atoms of an element in formula)(AW)/FW ] = % comp. element

Percentage Percentage CompositionComposition

Chapt. 3.3

Percentage Composition of C6H12O6

(FW = 180)

% C = 100 (6)(12)/ (180) = 40.0% carbon% O = 100 (6)(16)/ (180) = 53.3% oxygen% H = 100 (12)(1)/ (180) = 6.7% hydrogen

2 example calculations follows


6363Percentage Percentage CompositionComposition

ProblemProblem: In 1987, the first substance to act as a : In 1987, the first substance to act as a superconductor at a temperature above that of superconductor at a temperature above that of liquid nitrogen (77 K) was discovered. The liquid nitrogen (77 K) was discovered. The approximate formula of the substance is approximate formula of the substance is YBaYBa22CuCu33OO77. Calculate the percent composition by . Calculate the percent composition by mass of this material.mass of this material.

AW: Y = 88.9; Ba =137.3; Cu = 63.6; O = 16.0

M W of YBaM W of YBa22CuCu33OO77 = (88.9) + 2(137.3) + 3(63.6) + 7(16) = (88.9) + 2(137.3) + 3(63.6) + 7(16) = 666.0 amu= 666.0 amu


6464

Y Y = = 1(88.9) 1(88.9) = 100 (88.9)= 100 (88.9)= 13.3 %= 13.3 % 666.0666.0

Ba Ba == 2(137.3)2(137.3) = 100 (274.6)= 100 (274.6) = 41.3 %= 41.3 % 666.0666.0

CuCu == 3(63.5)3(63.5) = 100 (190.5)= 100 (190.5) = 28.6 %= 28.6 % 666.0666.0

OO == 7(16.0)7(16.0) = 100 (112)= 100 (112) = 16.8 %= 16.8 % 666.0666.0


AW: Y = 88.9; Ba =137.3; Cu = 63.6; O = 16.0

M W of YBaM W of YBa22CuCu33OO77 = 666.0 amu = 666.0 amu


6565

Sample exercise: Calculate the Sample exercise: Calculate the percentage of nitrogen, by mass, in percentage of nitrogen, by mass, in Ca(NOCa(NO33))22..



6666


Formula Mass:Formula Mass:1(40.1) = 40.11(40.1) = 40.12(14.0) = 28.02(14.0) = 28.06(16.0) = 96.06(16.0) = 96.0 164.1 amu164.1 amu



6767


Formula Mass: part x 100Formula Mass: part x 1001(40.1) = 40.1 total1(40.1) = 40.1 total2(14.0) = 28.02(14.0) = 28.06(16.0) = 96.06(16.0) = 96.0 164.1 amu164.1 amu



6868


Formula Mass: part x 100Formula Mass: part x 1001(40.1) = 40.1 total1(40.1) = 40.1 total2(14.0) = 28.02(14.0) = 28.06(16.0) = 96.0 28.0 x 100 6(16.0) = 96.0 28.0 x 100 == 164.1 amu 164.1164.1 amu 164.1



6969


Formula Mass: part x 100Formula Mass: part x 1001(40.1) = 40.1 total1(40.1) = 40.1 total2(14.0) = 28.02(14.0) = 28.06(16.0) = 96.0 28.0 x 100 6(16.0) = 96.0 28.0 x 100 = 17.1%= 17.1% 164.1 amu 164.1164.1 amu 164.1



7070

•Molecular formula of

strychnine = C21H22N2O2

•Molecular weight (MW)

MW = 21(12) + 22(1) + 2(14) + 2(16) = 334 amu

•Percent composition;% C = 100 (21)(12)/ (334) = 75.4% carbon% N = 100 (2)(14)/ (334) = 6.59% nitrogen% O = 100 (2)(16)/ (334) = 9.59% oxygen% H = 100 (22)(1)/ (334) = 8.38% hydrogen

• ExampleExample ; ; determine the elemental percent composition of strychnine


Chapt. 3.3

N

O

N

O


7171MoleMole

Chapt. 3.4

• Very small macroscopic samples contain VERY many atoms, molecules, etc... (e.g. 1 tsp. H2O contains 2 x 1023 mol). [Need convenient counting unit]

• known no. of H atoms in 1 g of H = no. of atoms of O in 16 g of O = no. of C atoms in 12 g of C = etc... (based upon atomic weights)

• Def. = The number of carbon atoms in 12 g of 12C is called Avogadro’s numberAvogadro’s number. One MoleMole (latin “mole” = a mass) is the amount of material that contains Avogadro’s number

•Note: a mole refers to a fixed number of any type of particles!•Avogadro’s number = 6.023 x 1023


7272

• 1 mole of 12C atoms = 6.02 x 1023 atoms• 1 mole of 11B atoms = 6.02 x 1023 atoms

• 1 mol of PCl3 molecules = 6.02 x 1023 molecules

• 1 mol of Na+ ions = 6.02 x 1023 Na ions• 1 mol of toasters = 6.02 x 1023 toasters• 1 mole of students = 6.02 x 1023 students

Avagadro’s Number and the Avagadro’s Number and the MoleMole

Chapt. 3.4


7373MoleMole

•ExamplesExamples

Chapt. 3.4

How many C atoms in 0.5 moles of Carbon?

C atoms = (0.5 moles C)(6.02 x 1023 atoms) = 3.01 x 1023

mole

How many C atoms are in 0.25 mol of C6H12O6?

C ato. = (0.25 mol C6H12O6)(6.02 x 1023 molec) (6 C atoms)

mol (1 C6H12O6 molec)

C atoms = 1.5 x 1023 atoms


7474MoleMole

Chapt. 3.4

Sample exercise: How many O atoms are in a) 0.25 moles Ca(NO3)2


7575MoleMole

Chapt. 3.4


0.25 mol Ca(NO3)2


7676MoleMole

Chapt. 3.4


0.25 mol Ca(NO3)2 6 mole O 1 mole Ca(NO3)2


7777MoleMole

Chapt. 3.4



1.5 mol O 6.02 x 1023 atom O 1 mol O


7878MoleMole

Chapt. 3.4



1.5 mol O 6.02 x 1023 atom O = 9.03 x 1023

1 mol O


7979MoleMole

Chapt. 3.4

Sample exercise: How many O atoms are in a) 1.50 moles sodium carbonate


8080MoleMole

Chapt. 3.4


Na+1 CO3-2 Na2CO3


8181MoleMole

Chapt. 3.4


Na+1 CO3-2 Na2CO3

1.50 mol Na2CO3


8282MoleMole

Chapt. 3.4


Na+1 CO3-2 Na2CO3

1.50 mol Na2CO3 3 mol O1 mol Na2CO3


8383MoleMole

Chapt. 3.4


Na+1 CO3-2 Na2CO3


4.50 mol O 6.02 x 1023 atom O 1 mol O


8484MoleMole

Chapt. 3.4


Na+1 CO3-2 Na2CO3


4.50 mol O 6.02 x 1023 atom O = 2.71 x 1024

1 mol O


8585

• Example - since one 12C atoms weighs 12 amu and a 24 Mg atom weighs 24 amu (twice as massive) and since a mole always contains the same number of particles, a mole of 24 Mg must weigh twice as much as a mole of 12C.

Molar Mass - (in grams) of any substance is always numerically equal to its formula weight (in amu).

Molar MassMolar Mass

Chapt. 3.4

1 12C atom weighs 12 amu; 1 mol 12C weigh 12 g1 24Mg atom weighs 24 amu; 1 mol 24Mg weigh 24 g1 238U atom weighs 238 amu; 1 mol 238U weighs 238 g


8686Molar RelationshipsMolar Relationships

Chapt. 3.4

Name Formula Formula Mass of 1 mol Number and kind

weight of form units of particles in 1 mol

atomic nitrogen N 14.0 14.0 6.02 X 1023 N atoms

molec. nitrogen N2 28.0 28.0 6.02 X 1023 N2 molec.

2(6.02 X 1023 ) N atoms

scandium ScCl3 151.5 151.5 6.02 X 1023 ScCl3 unitschloride 6.02 X 1023 Sc3+ ions

3(6.02 X 1023) Cl- ions

glucose C6H12O6 180.0 180.0 6.02 X 1023 gluc. molec.

6(6.02 X 1023) C atoms 12(6.02 X 1023 ) H atoms


8787

(1) How many moles of phosphorus trichloride, PCl3, are in 50 g of the substance? (MW = 137.4 amu)Moles of PCl3 = 1mol PCl3 (50 g PCl3) = 0.36 moles PCl3

137.4 g

(2) How many molecules of PCl3 are in 50 g?molecules of PCl3 = (0.36 moles )(6.023 x 1023 molecules)

1 mole = 2.2 x 1023 PCl3 molecules

(3) How many grams of PCl3 are in 0.75 moles?x grams = (0.75 mole)(137.4 g PCl3) = 103 g of PCl3

1 mole PCl3

Molar RelationshipsMolar Relationships

Chapt. 3.4


8888

Moles

Items (molecules, atoms, etc...)

Grams use molar mass

use Avagadro’snumber

use molar mass

use Avagadro’snumber


Chapt. 3.4


8989

Sample exercise: How many moles of NaHCO3 are present in 508 g of this substance?


Chapt. 3.4


9090


508 g


Chapt. 3.4


9191


508 g 1 mol 84 g


Chapt. 3.4


9292


508 g 1 mol = 6.05 g NaHCO3

84 g


Chapt. 3.4


9393

Sample exercise: What is the mass, in grams, of

a) 6.33 mol NaHCO3


Chapt. 3.4


9494


a) 6.33 mol NaHCO3

6.33 mol NaHCO3 84 g NaHCO3

1 mol NaHCO3

= 532 g NaHCO3


Chapt. 3.4


9595


b) 3.0 x 10-5 mol sulfuric acid


Chapt. 3.4


9696


b) 3.0 x 10-5 mol sulfuric acid

3.0 x 10-5 mol H2SO4 98 g H2SO4

1 mol H2SO4

= 2.9 x 10-3 g


Chapt. 3.4


9797

Sample exercise: How many nitric acid molecules are in 4.20 g of HNO3?


Chapt. 3.4


9898

Sample exercise: How many nitric acid molecules are in 4.20 g of HNO3?

4.20 g HNO3 6.02 x 1023 molec HNO3

63 g HSO3

= 4.01 x 1022 molec


Chapt. 3.4


9999

• Empirical Formula - Relative number of each element in a compound.

• Using moles and percent weight (elemental analysis by chemical means), we can calculate an empirical formula

• Steps;»assume a 100 g (convenient since working

with % because the elements % can be thought of as g)

»calculate moles of element present in 100g sample

»find ratios of moles (approx) to lead to integral formula subscripts.

Empirical FormulasEmpirical Formulas

Chapt. 3.5


100100

Mass Percentof Elements

Grams of each Element

EmpiricalFormula

Moles of each Elements

assume 100 g sample

Useatomicweights

calculate moleratio

Empirical FormulasEmpirical Formulas

Chapt. 3.5


101101Empirical Formulas, ExamplesEmpirical Formulas, Examples

Chapt. 3.5

Determine the empirical formula for a compound which contains 87.5 % N and 12.5% H by mass.

% g in 100g moles ratio

87.5 % N = 87.5 g N 1 mole = 6.25 moles N = 1

14 g

12.5% H = 12.5 g H 1 mole H = 12.5 moles H = 2

1 g

Empirical Formula = NH2



Chapt. 3.5

Sample exercise: A 5.325 g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.251 g of oxygen. What is the empirical formula of this substance?



Chapt. 3.5


3.758 g x 100 = 70.58% C -> 70.58 g C 1 mol C = 5.88 mol C5.325 g 12 g C



Chapt. 3.5


0.316 g x 100 = 5.93% H -> 5.93 g H 1 mol H = 5.93 mol H5.325 g 1 g H



Chapt. 3.5


1.251 g x 100 = 23.49% O -> 23.49 g O 1 mol O = 1.47 mol O5.325 g 16 g O



Chapt. 3.5


5.88 mol C ; 5.93 mol H ; 1.47 mol O



Chapt. 3.5


5.88 mol C ; 5.93 mol H ; 1.47 mol O1.47 mol 1.47 mol 1.47 mol

4 4 1



Chapt. 3.5


5.88 mol C ; 5.93 mol H ; 1.47 mol O C4H4O1.47 mol 1.47 mol 1.47 mol

4 4 1


109109

A white unknown substance (mass spec. problem) found on an unconscious patient is suspected by a forensic chemist of being either cocaine or caffeine. Combustion of a 50.86 mg sample yielded 150.0 mg of CO2 and 46.05 mg of water. Further analysis showed the compound contained 9.39% N by mass. The formula of cocaine is C17H21NO4. Can the substance be cocaine?

Unknown% C, H, and O.Percent Composition

Empirical Formulas, ExamplesEmpirical Formulas, Examples

Chapt. 3.5

Known50.86 mg of cmpd gave 150.0 mg of CO2 and

46.05 mg of H2O.compound contains 9.39% N.formula of Cocaine is C17H21NO4.



Chapt. 3.5



•Combustion ReactionsCombustion Reactions –Reactions with oxygen (usually from the air)–The complete combustion of hydrocarbons yield carbon dioxide (CO2) and water (H2O)

CxHy + (2x+y) /2 O2 X CO2 + Y/2 H2O


111111




Compute % C and % H (from combustion).

C: 0.150 g CO2 = 0.00341 mol CO2 = 0.00341 mol C = 40.9 mg C

H: 0.04605 g H2O = 0.00256 mol H2O = 0.00512 mol H = 5.12 mg H

N: (0.0939)(50.86) = 4.77 mg N = 0.000341 mol N

O: (50.86 g sample)-(40.9 mg C + 5.12 mg H + 4.77 mg N) = = 0.08 mg O = 0.000006 mol O


112112




mg in samplemg in sample % in sample calc’n% in sample calc’n % sample % cocaine% sample % cocaine

C: 40.9 mg C 100 (40.9mg/50.86 mg) 80.5 % C 67.3% C

H: 5.12 mg H 100 (5.11mg/50.86mg) 10.1% H 6.9% H

N: 4.77 mg N 100 (4.77mg/50.86 mg) 9.4% N 4.6% N

O: 0.08 mg O 100 (0.08mg/50.86mg) 0.02% O 21.2% O


113113

Known:

formula of Cocaine is Cformula of Cocaine is C1717HH2121NONO44..MW = 17(12) + 21(1) + 1(14) + 4(16) = 303 MW = 17(12) + 21(1) + 1(14) + 4(16) = 303

amuamu


g in cocaineg in cocaine % in sample calc’n% in sample calc’n % sample % cocaine% sample % cocaine (303 grams)

C: 17(12) = 204 g C 100 (204/303) 80.5 % C 67.3% C

H: 21(1) g H 100 (21/303) 10.1% H 6.9% H

N: 1(14) g N 100 (14/303) 9.4% N 4.6% N

O: 4(16) g O 100 (64/303) 0.02% O 21.2% O




C: 40.9 mg C 100 (40.9mg/50.86 mg) 80.5 % C 67.3% C

H: 5.12 mg H 100 (5.11mg/50.86mg) 10.1% H 6.9% H

N: 4.77 mg N 100 (4.77mg/50.86 mg) 9.4% N 4.6% N

O: 0.08 mg O 100 (0.08mg/50.86mg) 0.02% O 21.2% O


115115

Compound is Not Compound is Not Cocaine from analysisCocaine from analysis



C: 40.9 mg C 100 (40.9mg/50.86 mg) 80.5 % C 67.3% C

H: 5.12 mg H 100 (5.11mg/50.86mg) 10.1% H 6.9% H

N: 4.77 mg N 100 (4.77mg/50.86 mg) 9.4% N 4.6% N

O: 0.08 mg O 100 (0.08mg/50.86mg) 0.02% O 21.2% O


116116Combustion ReactionsCombustion Reactions

[Video No. 20-21; 4:42 +1:29 m]

Reaction of Hydrogen with OxygenReaction of Hydrogen with Oxygen[COMBUSTION] (note precautions)[COMBUSTION] (note precautions)

» 2 H2 H22(g) + O(g) + O22(g)(g) 2 H2 H22O(g) O(g)

H = -232 H = -232 kJ/mol HkJ/mol H22OO

» Ignition temperature = 580° - 590° CIgnition temperature = 580° - 590° C

» Explosive [“Explosive [“when stuff gets really big really when stuff gets really big really fasfast” Beakman’ World]t” Beakman’ World]

» The rapid release of energy [-232 kJ/mol The rapid release of energy [-232 kJ/mol HH22O] into the surrounding air causes the air O] into the surrounding air causes the air to very quickly expand. the explosion from to very quickly expand. the explosion from pure Hpure H22 sound quiter because the air sound quiter because the air expansion is slower.expansion is slower.


117117

Questions for After DemonstrationQuestions for After Demonstration

Combustion ReactionsCombustion Reactions

[Video No. 22; 2:50 m]

Combustion of Alcohol (ethanol):Combustion of Alcohol (ethanol):

CC22HH55OH(g) + 3 OOH(g) + 3 O22(g)(g)2CO2CO22(g) + 3 H(g) + 3 H22O(g)O(g)

H H =1366.2 kJ mol=1366.2 kJ mol-1-1

Tesla coil produces a high voltage electric spark. Tesla coil produces a high voltage electric spark. The spark is required to initiate this reaction.The spark is required to initiate this reaction.

Conversion of chemical energy (PE stored in Conversion of chemical energy (PE stored in bonds) to mechanical energy.bonds) to mechanical energy.

Are other types of energy are produced besides Are other types of energy are produced besides mechanical energy?mechanical energy?

Why can the reaction not be repeated without Why can the reaction not be repeated without flushing the bottle with air first?flushing the bottle with air first?


118118

• Empirical Formula from reaction with oxygen

• Organic Compounds - C to CO2 and H to H2O

• Use CO2 and H2O to determine the amount of C and H in original sample

Combustion AnalysisCombustion Analysis

Chapt. 3.5

furnace

sample

O2 flow

contaminantcatalyst (CuO); oxidizes traces of CO and C to

CO2

H2O absorbant(Mg(ClO4)2)

CO2 absorbant

(NaOH)


119119

The combustion of 5.00g of an organic compound containing C, H, and O yields 9.57 g CO2 and 5.87 g H2O. What is the empirical formula?


Chapt. 3.5


120120

The combustion of 5.00g of an organic compound containing C, H, and O yields 9.57 g CO2 and 5.87 g H2O. What is the empirical formula?

9.57 g CO2 = 0.217 mol CO2 = 0.217 mol C = 2.60 g C5.87 g H2O = 0.326 mol H2O = 0.652 mol H = 0.652 g H

whatever is left over must be the amt. of O originally present;

5.00 g - (2.60 g C + 0.652 g H) = 1.75 g of O = 0.109 mol O

thus;0.217 mol C 1.990.109 mol O 1.00 thus C2H6O0.652 mol H 5.98 (ethanol)


Chapt. 3.5

divide each by 0.109


121121

The combustion of 0.596 g of a compound containing only B and H yields 1.17 g H2O and all the boron is recovered as B2O3. What is the empirical formula?


Chapt. 3.5


122122


(1) Chem Equation; BxHy + O2 H2O + B2O3


Chapt. 3.5


123123



(2) 1.17 g H2O = 1.17g = 0.065 mol H2O 18 g/mol

(3) g H = (0.065 mol H2O) (1 g H2O) (2 H mol H) = 0.130 mol H

1 mol H 1 mol H2O0.130 mol H = 0.130 g H


Chapt. 3.5


124124



(2) 1.17 g H2O = 1.17g = 0.065 mol H2O 18 g/mol

(3) g H = (0.065 mol H2O) (1 g H2O) (2 H mol H) = 0.130 mol H

1 mol H 1 mol H2O0.130 mol H = 0.130 g H

(4) (0.596 g tot)-(0.130 g H) = 0.466 g B; 0.466 g B = 0.043 mol B

10.8 g/ mol(5) B = 0.0431 mol = 1.00 H = 0.130 mol = 3.01

0.0431 BH3 0.0431


Chapt. 3.5


125125

• Read and UNDERSTAND the problem - determine what is being given and what is required.

• Identify the Unknown and Given data.• Set up the problem - determine what kinds of

information bear upon the problem, what solution pathways may be available, what chemical principles should give guidance, etc...

• Solve the problem - Use the data given and the appropriate relationships or equations to work throught the problem.

• Check your work - not just the mathematical functions but ask if the answer makes sense and provides what is being asked for! (sig. figs)

Chemical Problem SolvingChemical Problem Solving

Chapt. 3.5


126126

• Coefficients in a balanced chemical equation refer to both the relative number of molecules involved in a reaction and the relative number of moles.

• Stoichiometric equivalence - from coefficients in a chemical equation; B2H6 + 3 O2 3 H2O + B2O3

• 1 mol B2H6 equiv. to 3 moles O2 equiv. to 3 mol H20, ...

• Used to calculate quantities involved in a reaction

Equations and the MoleEquations and the Mole

Chapt. 3.6

grams of compound A

moles of compound A

grams of compound B

moles of compound B

use molar mass of A use molar mass of Buse coeff of A and B from

balanced eqn.


127127

• Given the reaction for the formation of B2H6 (diborane), how many grams of diborane can be prepared from 3.0 g of LiH?

6 LiH + 8BF3 6 LiBF4 + B2H6

[B2H6 MW = 27.6 and LiH MW = 7.9]

Mole CalculationsMole Calculations

Chapt. 3.6

3.0 g LiH (1 mol LiH) = 0.38 mol LiH 7.9 g LiH

0.38 mol LiH (1 mol B2H6 ) 27.6 g B2H6 = 1.7 g B2H6

6 mol LiH 1 mol B2H6


128128

• Given the reaction for the formation of B2H6 (diborane), how many grams of BF3 are required to react with 3.0 g of LiH ?


[B2H6 MW = 27.6, BF3 = 67.8 and LiH MW = 7.9]


Chapt. 3.6

3.0 g LiH (1 mol LiH) (8 mol BF3) (67.8 g BF3) = 34 g BF3

7.9 g LiH 6 mol LiH 1 mol BF3


129129

Sample exercise: A common laboratory method for preparing small amounts of O2 involves the decomposition of KClO3:

2KClO3 2KCl + 3O2


Chapt. 3.6


130130


2KClO3 2KCl + 3O2

How many grams of oxygen is produced from 4.50 g KClO3?

4.50 g KClO3


Chapt. 3.6


131131


2KClO3 2KCl + 3O2


4.50 g KClO3 1 mol KClO3

122.6 g KClO3


Chapt. 3.6


132132


2KClO3 2KCl + 3O2


4.50 g KClO3 1 mol KClO3 3 mol O2

122.6 g KClO3 2 mol KClO3


Chapt. 3.6


133133


2KClO3 2KCl + 3O2


4.50 g KClO3 1 mol KClO3 3 mol O2 32 g O2



Chapt. 3.6


134134


2KClO3 2KCl + 3O2


4.50 g KClO3 1 mol KClO3 3 mol O2 32 g O2


= 1.76 g O2


Chapt. 3.6


135135

Sample exercise: Propane, C3H8, is a common fuel used for cooking and home heating. What mass of O2 is consumed in the combustion of 1.00 g of propane?


Chapt. 3.6


136136


C3H8 + O2 CO2 + H2O


Chapt. 3.6


137137


C3H8 + 5O2 3CO2 + 4H2O


Chapt. 3.6


138138


C3H8 + 5O2 3CO2 + 4H2O

1.00 g C3H8


Chapt. 3.6


139139


C3H8 + 5O2 3CO2 + 4H2O

1.00 g C3H8 1 mol C3H

44 g C3H8


Chapt. 3.6


140140


C3H8 + 5O2 3CO2 + 4H2O

1.00 g C3H8 1 mol C3H 5 mol O2

44 g C3H8 1 mol C3H8


Chapt. 3.6


141141


C3H8 + 5O2 3CO2 + 4H2O

1.00 g C3H8 1 mol C3H 5 mol O2 32 g O2

44 g C3H8 1 mol C3H8 1 mol O2


Chapt. 3.6


142142


C3H8 + 5O2 3CO2 + 4H2O

1.00 g C3H8 1 mol C3H 5 mol O2 32 g O2

44 g C3H8 1 mol C3H8 1 mol O2

= 3.64 g O2


Chapt. 3.6


143143

• Sometimes after one reagent is completely consumed in the reaction some of another reagent is left over. The reagent which is completely consumed limits the extent of the reaction = LIMITING REAGENT.

Limiting ReagentLimiting Reagent

Chapt. 3.7

+

Limiting Reagent


144144

• Given the reaction for the formation of B2H6 (diborane), if 5.0 g of LiH and 5.0 g of BF3 were reacted how much of which reagent would be left over?


[B2H6 MW = 27.6. BF3 = 67.8 and LiH MW = 7.9]

Limiting Reagent CalculationsLimiting Reagent Calculations

Chapt. 3.6

KnowKnow::Quantities (g and moles) of starting Quantities (g and moles) of starting

materialsmaterialsMolar ratios between all the starting Molar ratios between all the starting

materials materials and products.and products.FindFind::

Which reagent is completely consumed Which reagent is completely consumed ((limitinglimiting reagentreagent) and which is left over) and which is left over


145145



[B2H6 MW = 27.6. BF3 = 67.8 and LiH MW = 7.9]


Chapt. 3.6

5.0 g LiH (1 mol LiH) = 0.63 mol ANDAND 5.0 g BF3 (1 mol BF3) = 0.074 mol

7.9 g LiH 67.8 g BF3

If all the LiH were consumed, then 0.84 mol BF3 would be required

[(0.63 mol LiH)(8 mol BF3)] = 0.84 mol BF3

6 mol LiH)


146146



[B2H6 MW = 27.6. BF3 = 67.8 and LiH MW = 7.9]


Chapt. 3.6

5.0 g LiH (1 mol LiH) = 0.63 mol AND 5.0 g BF3 (1 mol BF3) = 0.074 mol

7.9 g LiH 67.8 g BF3

If all the LiH were consumed, then 0.84 mol BF3 would be required Since only 0.074 mol of BF3 is available, BF3 is the limiting reagent (all consumed).

0.074 mol BF3 (6 mol LiH) = 0.056 mol LiH consumed

8 mol BF3

therefore remaining LiH = (0.63 mol - 0.056 mol)(7.9 g/mol) = 4.53 g


147147

• Equal weights (5.00 g) of Zn(s) and I2(s) are mixed together to form ZnI2. How much ZnI2 is formed? How much of each reactant remains at the end of the reaction and which is the limiting reagent?

Limiting Reagent ProblemsLimiting Reagent Problems

Chapt. 3.7

Zn (AW = 65.4 amu) Zn(s) + I2(s) ZnI2(s)I2 (MW = 253.8 amu)


148148

• Equal weights (5.00 g) of Zn(s) and I2(s) are mixed together to form ZnI2. How much ZnI2 is formed? How much of each reactant remains at the end of the reaction and which is the limiting reagent?

Limiting Reagent ProblemsLimiting Reagent Problems

Chapt. 3.7

Zn (AW = 65.4 amu) Zn(s) + I2(s) ZnI2(s)I2 (MW = 253.8 amu)

Zn = 5.0 g Zn (1 mol Zn) = 0.076 mol Zn 65.4 g Zn

I2 = 5.0 g I2 (1 mol I2) = 0.020 mol I2

253.8 g I2

I2 is the limiting reagent.

Zn remaining = (0.076 Zn - 0.020 mol Zn) (64.5 g Zn) = 3.66 g Zn

1 mol Zn


149149

• Theoretical Yield - quantity of product calculated to form when all the limiting reagent is consumed (calculated from molar ratios).

• Actual Yield - the amount of product experimentally obtained from a reaction

• Percent Yield - describes relationship between theoretical and actual yields;

percent yield = actual yield (100)

theoretical yield

Theoretical YieldsTheoretical Yields

Chapt. 3.6


150150

• Given the reaction of 2.05 g of hydrogen sulfide with 1.84 g of sodium hydroxide, calculate how the theoretical yield of Na2S. What is the percent yield if the amt. of Na2S obtained was 3.65 g. [H2S (MW = 34.1); Na2S (MW = 78.1)]

Percent YieldsPercent Yields

Chapt. 3.6

H2S(g) + 2 NaOH(aq) Na2S(aq) + 2 H2O

(2.05 g H2S)(1 mol H2S)(1 mol Na2S)(78.1 g Na2S) = 4.70 g Na2S

34.1 g H2S 1 mol H2S 1 mol Na2S theoretical yield

% yield = 3.65 g (actual yield) (100) = 77.7 % yield

4.70 g (theoretical yield)


151151End of Chapter 3End of Chapter 3

Major Topics (not exhaustive list):

(1) Chemical Equations(2) Periodic Table and Reaction Types(3) Atomic and Molecular

Weights (formula weights, % compositions, etc...)(4) Molar Concepts(5) Empirical Formulas(6) Info from Balanced Eqns.(7) Limiting Reagents(8) Percent Yields

Documents

Chem 106, Prof. J.T. Spencer 1 CHE 106: General Chemistry u CHAPTER THREE Copyright © James T. Spencer 1995 - 1999 All Rights Reserved