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Notes​ ​by​ ​Aaron​ ​Hui
Chemistry​ ​Lecture​ ​02
...
17
● Diatomic​ ​molecules:
○ Rule​ ​of​ ​7’s
...

● Ion​ ​-​ ​charged​ ​species
○ Cation:​ ​+​ ​charge
■ Metals,​ ​lose​​ ​e○ Anion:​ ​-​ ​charge
■ Nonmetals,​ ​gain​​ ​e● Monatomic​ ​ion​ ​-​ ​only​ ​1​ ​atom​ ​(Ex:​ ​Na​+​,​ ​N3+​
​ )
○ Octet​ ​rule​ ​-​ ​want​ ​8​ ​valence​ ​e-​ ​(be​ ​like​ ​nearest​ ​noble​ ​gas)
■ Valence​ ​e-​ ​=​ ​group​ ​#​ ​(main-group​ ​elements)
● Metals:​ ​charge​ ​=​ ​group​ ​#
● Nonmetals:​ ​charge​ ​=​ ​group​ ​#​ ​-​ ​8
● Polyatomic​ ​ion​ ​-​ ​more​ ​than​ ​1​ ​atom​ ​(Ex:​ ​OH​-​,​ ​NO​3​-​)
● How​ ​many​ ​protons​ ​and​ ​electrons​ ​are​ ​in​ 23​
​ 13​Al​3+
○ Protons:​ ​13​ ​protons
○ Electrons:​ ​10​ ​electrons
● How​ ​many​ ​protons​ ​and​ ​electrons​ ​are​ ​in​ 78​
​ 34​Se​2○ Protons:​ ​34​ ​protons
○ Electrons:​ ​36​ ​electrons
● What​ ​is​ ​the​ ​formula​ ​of​ ​the​ ​phosphorus​ ​ion​ ​(phosphide)?​ ​How​ ​many​ ​electrons
are​ ​present?
○ P​3-​,​ ​(15+3)=​ ​18​ ​electrons
● Ionic​ ​compounds​ ​-​ ​cation(s)​ ​+​ ​anion(s)
○ Sum​ ​of​ ​charges​ ​must​ ​equal​ ​zero
○ The​ ​ionic​ ​compound​ ​NaCl

○ Always​ ​crystalline​ ​structures
● Writing​ ​Formulas​ ​of​ ​Ionic​ ​Compounds
○ Order:​ ​cation​ ​(+)​ ​then​ ​anion​ ​(-)
■ Do​ ​not​ ​write​ ​charges
○ If​ ​equal​ ​+​ ​and​ ​-​ ​charges
■ Ex:​ ​Na+​ ​and​ ​Cl-​ ​=​ ​NaCl
■ Ex:​ ​Al​ ​3+​ ​and​ ​N​ ​3-​ ​=​ ​AlN
○ If​ ​unequal​ ​charges​ ​-​ ​use​ ​crisscross​ ​method
■ Magnitude​ ​of​ ​the​ ​charge​ ​on​ ​one​ ​ion​ ​becomes​ ​the​ ​subscript​ ​for​ ​the
other
...
​ ​ ​It’s​ ​a
diatomic​ ​molecule
...

○ Can​ ​only​ ​change​ ​coefficients​ ​NOT​ ​subscripts
■ Changing​ ​subscripts​ ​change​ ​the​ ​compound
○ Reduce​ ​coefficients​ ​to​ ​the​ ​lowest​ ​whole​​ ​number
...

○ Ex:​ ​Propane​ ​(C​3​H​8​)​ ​reacts​ ​with​ ​oxygen​ ​to​ ​form​ ​carbon​ ​dioxide​ ​and​ ​water
■ C​3​H​8​​ ​+​ ​O​2​​ ​->​ ​CO​2​​ ​+​ ​H​2​O
■ C​3​H​8​​ ​+​ ​5O​2​​ ​->​ ​3CO​2​​ ​+​ ​4H​2​O
● Treat​ ​polyatomic​ ​ions​ ​as​ ​a​ ​“unit”
○ Al​ ​+​ ​FeSO​4​​ ​->​ ​Al​2​(SO​4​)​3​​ ​+​ ​Fe
○ 2Al​ ​+​ ​3FeSO​4​​ ​->​ ​Al​2​(SO​4​)​3​​ ​+​ ​3Fe
● When​ ​an​ ​element​ ​is​ ​in​ ​3​ ​or​ ​more​ ​compounds,​ ​change​ ​the​ ​coefficient​ ​where​ ​it​ ​is
alone​ ​-​ ​you​ ​can​ ​even​ ​use​ ​a​ ​fraction
○ 2NH​3​(g)​ ​+​ ​O​2​(g)​ ​->​ ​2NO(g)​ ​+​ ​3H​2​O(g)
○ 2(2NH​3​(g)​ ​+​ ​(5/2)O​2​(g)​ ​->​ ​2NO(g)​ ​+​ ​3H​2​O(g))
○ 4NH​3​(g)​ ​+​ ​5O​2​(g)​ ​->​ ​4NO(g)​ ​+​ ​6H​2​O(g)
● Synthesis​ ​Reactions
○ Form​ ​1​ ​product
○ Examples:
■ Two​ ​elements​ ​react​ ​to​ ​make​ ​a​ ​compound
● Al​ ​+​ ​S​ ​->​ ​Al​2​​ ​+​ ​S​3
■ Metal​ ​Oxide​ ​plus​ ​water​ ​gives​ ​a​ ​metal​ ​hydroxide
● CaO​ ​+​ ​H​2​O​ ​(not​ ​balanced)​ ​->​ ​Ca​2+​​ ​+​ ​OH​-​ ​->​ ​Ca(OH)​2
■ Non-metal​ ​oxide​ ​plus​ ​water​ ​gives​ ​an​ ​acid
● SO​2​​ ​+​ ​H​2​O​ ​->​ ​H+​​ ​ ​+​ ​SO​3​2-​​ ​->​ ​H​2​SO​3
● Decomposition​ ​Reactions

○ 1​ ​reactant​ ​(opposite​ ​of​ ​synthesis​ ​reactions)
○ Examples​ ​(not​ ​balanced):
■ Compound​ ​breaks​ ​down​ ​into​ ​elements:
● H​2​O​ ​+​ ​electrical​ ​energy​ ​=​ ​H​2​​ ​+​ ​O​2
■ MEMORIZE:​ ​Heating​ ​a​ ​metal​ ​carbonate​ ​gives​ ​carbon​ ​dioxide​ ​and
a​ ​metal​ ​oxide
● CuCO​3​​ ​->​ ​CO​2​​ ​+​ ​CuO
■ MEMORIZE:​ ​Metal​ ​hydroxides​ ​break​ ​down​ ​into​ ​a​ ​metal​ ​oxide​ ​and
water
● AgOH​ ​->​ ​H​2​O​ ​+​ ​Ag​2​O
● Single​ ​Replacement​ ​Reactions
○ One​ ​element​ ​replaces​ ​another
○ Examples​ ​(not​ ​balanced):
■ Na​ ​+​ ​H​2​O​ ​(think​ ​of​ ​water​ ​as​ ​HOH)​ ​->​ ​H​2​​ ​+​ ​NaOH
■ Cl​2​​ ​+​ ​NaI​ ​->​ ​I​2​​ ​+​ ​NaCl
● Double​ ​Replacement​ ​Reactions
○ Ions​ ​switch​ ​partners
○ Examples:
■ HCl(aq)​ ​+​ ​Ca(OH)​2​(aq)​ ​->​ ​H​2​O​ ​+​ ​CaCl​2
● Combustion
○ Compound​ ​+​ ​O​2​​ ​->​ ​CO​2​​ ​+​ ​H​2​O

● Oxidation​ ​Number​ ​(ON):​ ​The​ ​charge​ ​the​ ​atom​ ​would​ ​have​ ​if​ ​electrons​ ​were
completely​ ​transferred
○ Free​ ​elements:​ ​ON​ ​=​ ​0
○ Monatomic​ ​ions:​ ​ON​ ​=​ ​charge
■ Group​ ​1A​ ​=​ ​+2,​ ​2A​ ​=​ ​+2,​ ​Aluminum​ ​=​ ​+3
● Li​+​​ ​=​ ​+1;​ ​Fe​3+​ ​=​ ​+3;​ ​O2-​
​ ​ ​=​ ​-2

○ Oxygen:​ ​ON​ ​usually​ ​-2
○ Hydrogen:​ ​ON​ ​usually​ ​+1
■ Bonded​ ​to​ ​metals​ ​(ex:​ ​LiH,​ ​CaH​2​)
...

○ Oxidizing​ ​agent​ ​is​ ​reduced,​ ​causes​ ​the​ ​other​ ​to​ ​be​ ​oxidized
...
​ ​What​ ​is​ ​the​ ​oxidizing​ ​agent​ ​in​ ​the
reaction?​ ​What​ ​substance​ ​is​ ​being​ ​oxidized?
○ Zn​ ​+​ ​CuSO​4​​ ​->​ ​Cu​ ​+​ ​ZnSO​4
​ ​ ​0​ ​ ​ ​ ​ ​+2​ ​same​ ​ ​ ​ ​0​ ​ ​ ​ ​+2​ ​same
● Cu​ ​went​ ​from​ ​+2​ ​to​ ​0,​ ​so​ ​CuSO​4​​ ​is​ ​the​ ​oxidizing​ ​agent​ ​(Cu​ ​is​ ​being
reduced)
● Zn​ ​went​ ​from​ ​0​ ​to​ ​+2,​ ​so​ ​Zn​ ​is​ ​the​ ​reducing​ ​agent​ ​(Zn​ ​is​ ​being​ ​oxidized)
● Example:​ ​Copper​ ​wire​ ​reacts​ ​with​ ​silver​ ​nitrate​ ​to​ ​form​ ​silver​ ​metal​ ​and
copper(II)​ ​nitrate
...

Lecture​ ​03
...
17
● Average​ ​Atomic​ ​Mass​ ​-​ ​avg
...
42%​ 6​​ Li​ ​(6
...
85%​ 7​​ Li​ ​(7
...
0742​ ​(6
...
446​ ​amu

●
...
016​ ​amu)​ ​=​ ​6
...
941amu
● For​ ​compounds:​ ​use​ ​sum​ ​of​ ​the​ ​atomic​ ​masses​ ​(in​ ​amu)
○ Molecular​ ​Mass​ ​-​ ​for​ ​a​ ​molecule​ ​(covalent​ ​compounds)
■ SO​2
● S​ ​(32
...
07​ ​amu
■ CH​4​O
● C​ ​(12
...
008)​ ​+​ ​O​ ​(16
...
04​ ​amu
■ C​8​H​10​N​4​O​2
● C​ ​(8)(12
...
008)​ ​+​ ​N​ ​(4)(14
...
00)​ ​=
194
...
022​ ​x​ ​10​23
○ Avogadro's​ ​Number
● Molar​ ​Mass​ ​(Molecular​ ​Weight)
○ Mass​ ​of​ ​1​ ​mole​ ​of​ ​*whatever*​ ​in​ ​grams
○ For​ ​any​ ​element,​ ​molecule,​ ​formula​ ​unit:
■ atomic/molecular/formula​ ​mass​ ​(amu)​ ​=​ ​molar​ ​mass​ ​(g)
■ 1​ ​molecule​ ​=​ ​64
...
07​ ​g
● Using​ ​Molar​ ​Mass​ ​as​ ​a​ ​Conversion​ ​Factor:
○ How​ ​many​ ​moles​ ​of​ ​Na​ ​are​ ​in​ ​3
...
45g N a *

mol N a
22
...
150 mol N a

○ Calculate​ ​the​ ​mass​ ​of​ ​1
...

■

1
...
02 g H 2 O
mol H 2 O

= 28
...
2​ ​mol​ ​sample?
■

35
...
022*10 23 atoms N a
mol N a

= 2
...
5​ ​mol​ ​sample​ ​of​ ​copper?
■

6
...
022*10 23 atoms Cu
mol Cu

= 3
...
37​ ​moles​ ​of​ ​ozone?​ ​ ​How​ ​many​ ​oxygen
atoms​ ​are​ ​present
...
37 mol O 3 *

6
...
25 * 10 23 molec
...
25 * 10 23 molec
...
O 3

= 2
...
37 mol O 3 *

3 mol O
mol O 3

*

6
...
48 * 10 24 atoms O

○ How​ ​many​ ​C​6​H​12​O​6​​ ​molecules​ ​are​ ​present​ ​in​ ​a​ ​0
...
225 mol C 6 H

12 O 6 *

6
...
35 * 10 23 molec C 6 H

Part​ ​Two​ ​of​ ​the​ ​problem
1
...
C 6 H 12 O 6

12 O 6

= 8
...

○ “​GO​ ​FOR​ ​THE​ ​MOL”
● Summary:
○ One​ ​mole​ ​of​ ​2​ ​different​ ​substances​ ​contain​ ​same​ ​#​ ​particles​ ​but
different​ ​masses
○ Mass​ ​(g)​ ​→(molar​ ​mass)→​ ​Moles​ ​→(avogadros​ ​#)→​ ​Particles
● How​ ​many​ ​atoms​ ​are​ ​in​ ​0
...
551 g K *

mol K
39
...
022*10 23 atoms K
mol K

= 8
...
3*10​23​​ ​formula​ ​units​ ​of​ ​NaCl
○

1
...
units N aCl *

mol N aCl
6
...
45 g N aCl
mol N aCl

= 13 g N aCl

● How​ ​many​ ​H​ ​atoms​ ​are​ ​in​ ​72
...
5 g C 3 H 8 O *

mol C 3 H 8 O
60
...
022*10 23 atoms H
mol H

= 5
...
4​ ​g​ ​of​ ​urea​ ​(NH​2​)​2​CO?
○

32
...
06 g urea

*

4 mol H
mol urea

*

6
...
30 * 10 24 atoms H

● Percent​ ​Composition​ ​-​ ​percent​ ​by​ ​mass​ ​of​ ​each​ ​element
...
00​ ​g​ ​of​ ​a​ ​compound​ ​made​ ​of​ ​mercury​ ​and​ ​oxygen​ ​decomposes​ ​in​ ​4
...
37​ ​g​ ​of​ ​oxygen
Hg​x​O​y​→Hg(ℓ)​ ​+​ ​O​2​(g)
○ Calculate​ ​the​ ​percent​ ​composition​ ​of​ ​C​2​H​6​O
■ Total​ ​46
...
01 g)
%​ ​C​ ​=​ ​ 2 46
...
14%
g *
(1
...
07
100% = 13
...
00g
%​ ​O​ ​=​ ​ 46
...
73%
g *

○ Molecule​ ​“Y”​ ​(MM​ ​=​ ​166​ ​g/mol)​ ​contains​ ​68
...
​ ​ ​How​ ​many
fluorine​ ​atoms​ ​are​ ​in​ ​each​ ​molecule​ ​“Y?”
atoms (mass F )
%​ ​F​ ​=​ ​ #mass
100%
compound *


...
00 g)
166 g

x = 6 atoms
● Types​ ​of​ ​formulas:
○ Chemical​ ​Formula​ ​(?)↔(?)​ ​Mass​ ​percent​ ​composition
○ Empirical​ ​Formula:
■ Smallest​​ ​whole​ ​number​ ​ratio​ ​of​ ​elements
● H​2​O
● CH​2​O
● O
● NH​2
○ Molecular​ ​Formula:
■ Exact​ ​number​ ​of​ ​atoms​ ​in​ ​a​ ​substance​ ​(true​ ​formula)
● H​2​O
● C​2​H​12​O​6
● O​3
● N​2​H​4

Molecular​ ​Formula

Empirical

H​2​O

H​2​O

C​2​H​12​O​6

CH​2​O

O​3

O

N​2​H​4

NH​2

● Calculating​ ​the​ ​Empirical​ ​Formula​ ​from​ ​Percent​ ​Composition
○ Determine​ ​the​ ​empirical​ ​formula​ ​of​ ​ascorbic​ ​acid​ ​(vitamin​ ​C)​ ​which
contains​ ​40
...
58%​ ​H,​ ​and​ ​54
...

■ Start​ ​with​ ​grams​ ​of​ ​each​ ​element
● If​ ​%​ ​-​ ​assume​ ​100​ ​g
■ g​ ​→​ ​moles​ ​(molar​ ​mass)
40
...
58 g H *

mol C
12
...
008 g H

54
...
407 mol C

= 4
...
00 g O

= 3
...
406
3
...
540 O 3
...
33​O​ ​*​ ​3​ ​=​ ​C​3​H​4​O​3
3
...
406
○ Within​ ​+-​ ​
...
15​ ​<​ ​x<​ ​x+
...
5​ ​g​ ​ethanol
○ Collect​ ​22
...
5​ ​g​ ​H​2​O
C​x​H​y​O​z​​ ​+​ ​O​2​​ ​→​ ​CO​2​​ ​+​ ​H​2​O
​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​11
...
0​ ​g​ ​13
...

To​ ​find​ ​moles​ ​of​ ​C​ ​and​ ​H:

22
...
01 g CO2

*

mol C
mol CO2

= 0
...
5 g H 2 O *

mol H 2 O
18
...
50 mol H

To​ ​find​ ​moles​ ​of​ ​Oxygen:
0
...
50 mol H *

12
...
008 g H
mol H

= 6
...
51 g H

11
...
00g − 1
...
0 g O

4
...
00 g O

= 0
...
500
H 1
...
25
=​ ​C​2​H​6​O
0
...
25
0
...
07
...
407​ ​g​ ​of​ ​the​ ​hydrate​ ​was
heated​ ​to​ ​drive​ ​off​ ​the​ ​water​ ​of​ ​hydration
...
332g​ ​Ba​ ​and​ ​6
...

■ Ba​x​I​y​ ​ ​ ​ ​ZH​2​O​ ​→​ ​Ba​x​I​y​​ ​+​ ​ZH​2​O↑
​ ​Hydrate

anhydrous

10
...
332​ ​g​ ​Ba
6
...
332 g Ba *
6
...
3 g Ba

mol I
126
...
02426 mol Ba

= 0
...
407 g − 3
...
188 g = 0
...
887 g H 2 O *
Ba

B aI


...
02426

2

I


...
02426

mol H 2 O
18
...
0493 mol H 2 O


...
0426 2

2H 2 O

● Determining​ ​Molecular​ ​Formulas
○ Need​ ​empirical​ ​formula​ ​+​ ​~molar​ ​mass
○ (empirical​ ​formula)​n​​ ​=​ ​molecular​ ​formula
○ n​ ​must​ ​be​ ​an​ ​integer
■

n=

molar mass
empirical molar mass

● Stoichiometry​ ​-​ ​Mass​ ​Changes​ ​in​ ​Rxns
○

Conversion​ ​factor​ ​(mole​ ​ratio)​ ​is​ ​derived​ ​from​ ​coefficients​ ​in​ ​balanced​ ​chemical

equation
...
Write​ ​balanced​​ ​equation
2
...
Find​ ​mol​ ​of​ ​desired​ ​substance
4
...
04 g CH 3 OH

*

4 mol H 2 O
2 mol CH 3 OH

*

18
...

■ Guys​ ​are​ ​limiting​ ​reagent
...

● Finding​ ​the​ ​Limiting​ ​Reagent
1
...
Assume​ ​each​ ​reagent​ ​is​ ​completely​ ​used​ ​up​ ​&​ ​calculate​ ​how​ ​much
product​ ​is​ ​formed
...
Limiting​ ​reagent​ ​yields​ ​the​ ​smallest​ ​amount​ ​of​ ​product
...
0​ ​g​ ​of​ ​Al​ ​and​ ​75
...
​ ​Which​ ​reagent​ ​is​ ​limiting​ ​and​ ​how​ ​many​ ​grams​ ​of​ ​H​2​​ ​are​ ​produced?
How​ ​many​ ​grams​ ​of​ ​each​ ​reagent​ ​remains?​ ​ ​If​ ​1
...


■ 2Al​ ​+​ ​6HCl​ ​→​ ​2AlCl​3​​ ​+​ ​3H​2
50
...
0g
50
...
98 g Al

75
...
46 g HCl

*

*

2
...
60 g H 2

2
...
07 g H 2

2
...

Start​ ​with​ ​L
...
​ ​(0​ ​g​ ​HCl​ ​left​ ​since​ ​its​ ​L
...
)
75
...
46 g HCl

*

2 mol Al
6 mol HCl

*

26
...
5 g Al used up

Al : 50
...
5 g = 31
...
92 g
2
...
6%

OTHER​ ​METHOD​​ ​FOR​ ​DETERMINING​ ​L
...

1
...
Compare​ ​to​ ​what​ ​was​ ​given
50
...
98 g Al

*

6 mol HCl
2 mol Al

*

36
...
​ ​ ​Don’t​ ​have​ ​203​ ​g,​ ​only​ ​have​ ​75​ ​g,​ ​thus​ ​HCl​ ​is
L
...

QUIZ​ ​1​ ​TOPICS:
Sig​ ​fig
Measurements
Dim​ ​Analy
Atomic​ ​Theory

Writing​ ​formulas
Nomenclature
Writing​ ​chem​ ​equations
Redox​ ​reactions

Exp​ ​2:
Mix:​ ​CuCO​3​​ ​and​ ​CuO
CuCO​3​​ ​and​ ​CuO​ ​→(heat)​ ​____​ ​(g)​ ​+​ ​____
CuO​ ​does​ ​not​ ​react
——
H​2​​ ​+​ ​O​2​ →​
​H​2​O
​
1 mol O2 *

2 mol O
mol O2

*

mol H 2 O
mol O

mol O

= 2 mol H 2 O ​ ​→​ ​ 2 mol H 2 O
2

● Arrhenius​ ​acid​ ​-​ ​produces​ ​H+​​ ​ ​(H​3​O​+​)​ ​in​ ​water
○ HCl,​ ​HNO​3​,​ ​H​2​SO​4​,​ ​H​3​PO​4
○ Hydronium​ ​ion,​ ​hydrated​ ​proton,​ ​H​3​O​+
● Arrhenius​ ​base​ ​-​ ​produces​ ​OH​-​​ ​in​ ​water
○ NaOH,​ ​KOH,​ ​Ba(OH)​2
● Bronsted​ ​acid​ ​(incorporates​ ​ALL​ ​arrhenius​ ​acids​ ​and​ ​some​ ​additional​ ​ones)​ ​proton​ ​(H​+​)​ ​donor
○ H​ ​in​ ​front

● Bronsted​ ​base​ ​-​ ​proton​ ​acceptor
○ Needs​ ​a​ ​lone​ ​pair,​ ​often​ ​● Monoprotic​ ​acids​ ​-​ ​each​ ​unit​ ​yield​ ​1​ ​H+
○ HCl​ ​→​ ​H+​​ ​ ​+​ ​Cl​● Diprotic​ ​acids​ ​-​ ​each​ ​unit​ ​yields​ ​2​ ​H+
○ H​2​SO​4​​ ​→​ ​2H​+​​ ​+​ ​SO​4​2● Triprotic​ ​acids​ ​-​ ​each​ ​unit​ ​yields​ ​3​ ​H+\
● Acid-Base​ ​Reactions
○ Neutralization​ ​reactions​ ​-​ ​H+​​ ​ ​and​ ​OH​-​​ ​combine​ ​to​ ​form​ ​H​2​O
○ HCl​ ​(aq)​ ​[acid]​ ​+​ ​NaOH​ ​(aq)​ ​[base]​ ​→​ ​NaCl​ ​(aq)​ ​[salt]​ ​+​ ​H​2​O​ ​[water]
■ Salt​ ​-​ ​ionic​ ​compound​ ​without​ ​H+,​ ​OH-,​ ​or​ ​O2​
○ Bronsted​ ​reactions
■ Transfer​ ​only​ ​ONE​ ​proton
● NH​3​​ ​(aq)​ ​+​ ​HF​ ​(aq)​ ​↔​ ​NH​4​+​​ ​(aq)​ ​+​ ​F-​​ ​ ​(aq)
Base ​ ​ ​ ​ ​ ​ ​acid

conj
...
​ ​Base

Lecture​ ​03
...
17
● Reactions​ ​of​ ​Acids
○ React​ ​with​ ​some​ ​metals​ ​to​ ​produce​ ​H​2
○ React​ ​with​ ​carbonates​ ​(CO​3​2-​)​ ​and​ ​bicarbonates​ ​(HCO​3​-​)​ ​to​ ​produce​ ​CO​2​
...


M = molarity =

mol solute
L solution

What​ ​is​ ​the​ ​molarity​ ​of​ ​a​ ​solution​ ​containing​ ​232​ ​g​ ​of​ ​Kl​ ​in​ ​500
...
0 g KI

500
...
40 mol KI
0
...
40 mol KI

L
1000 mL

= 0
...
80 M KI

What​ ​is​ ​the​ ​molarity​ ​of​ ​an​ ​50
...
36​ ​g​ ​of​ ​ethanol?
1
...
0 mL soln *
0
...
0500 L

mol ethanol
46
...
295 mol

= 0
...
590 M ethanol

● Using​ ​molarity​ ​as​ ​a​ ​conversion​ ​factor
○

L​ ​solution​ ​↔​ ​mol​ ​of​ ​solute
What​ ​mass​ ​of​ ​Kl​ ​is​ ​required​ ​to​ ​make​ ​250
...
53​ ​M​ ​Kl​ ​solution?
250
...
53 mol KI
L soln

*

166
...
5 g KI

What​ ​volume​ ​of​ ​HCl​ ​solution​ ​is​ ​required​ ​to​ ​make​ ​a​ ​3
...
2​ ​g​ ​of​ ​HCl
34
...
46 g HCl

*

L soln
3
...
250 L

● Dilution​ ​-​ ​prepare​ ​a​ ​less​ ​concentrated​ ​solution​ ​from​ ​a​ ​more​ ​concentrated​ ​one
...
0​ ​mL​ ​of​ ​0
...
00​ ​M
HNO​3​?​ ​ ​Approximately​ ​how​ ​much​ ​water​ ​would​ ​you​ ​need​ ​to​ ​add?
M 1V 1 = M 2V 2
(4
...
200 M )(60
...
00 mL
Start​ ​with​ ​3
...
00​ ​M​ ​HNO​3
Add​ ​H​2​O​ ​until​ ​final​ ​volume​ ​is​ ​60
...
0 mL − 3
...
0 mL approx how much H 2 O we add

34
...
07​ ​M​ ​NaOH​ ​is​ ​mixed​ ​with​ ​165
...
​ ​ ​What​ ​is​ ​the
final​ ​concentration​ ​of​ ​NaOH​ ​solution​ ​assuming​ ​the​ ​volumes​ ​are​ ​additive?
M 1V 1 = M 2V 2
(5
...
2 mL) = M 2 (165
...
2 mL)
M 2 = 0
...


What​ ​is​ ​the​ ​concentration​ ​of​ ​the​ ​NaOH​ ​solution​ ​if​ ​158​ ​mL​ ​of​ ​NaOH​ ​is​ ​required
titrate​ ​25
...
50​ ​M​ ​H​2​SO​4​​ ​standard​ ​solution?
H 2 SO4 + 2N aOH → 2H 2 O + N a2 SO4
25
...
225 mol N aOH
0
...
225 mol N aOH

= 1
...
2​ ​M​ ​H​2​SO​4​​ ​solution​ ​are​ ​needed​ ​to​ ​neutralize​ ​30
...
256​ ​M​ ​KOH​ ​solution?
H 2 SO4 + 2KOH → 2H 2 O + K 2 SO4
30
...
256 mol KOH
L

*

mol H 2 SO4
2 mol KOH

*

L H 2 SO4
6
...
63 mL H 2 SO4

THERMOCHEMISTRY
● Thermochemistry​ ​-​ ​the​ ​study​ ​of​ ​heat​ ​changes​ ​in​ ​chemical​ ​reactions
● Energy​ ​-​ ​capacity​ ​to​ ​do​ ​work​ ​(force​ ​acts​ ​through​ ​a​ ​distance)​ ​or​ ​transfer​ ​heat
○

SI​ ​unit:​ ​Joules​ ​(J)

○

Calorie

○

1​ ​cal​ ​=​ ​4
...

○

Potential​ ​energy​ ​is​ ​the​ ​same
...


○

Base​ ​sign​ ​on​ ​system:
■ +q​ ​heat​ ​added
■ +w​ ​work​ ​done​ ​on​​ ​system
● Pushing​ ​on​ ​a​ ​gas
■ -q​ ​heat​ ​removed
■ -w​ ​work​ ​done​ ​by​​ ​the​ ​system
● Gas​ ​pushing​ ​out
■ Hot​ ​q<0
■ Cold​ ​q>0

Lecture​ ​3
...
17

● Pressure-Volume​ ​(PV)​ ​Work:
○ Work​ ​done​ ​by​ ​an​ ​expanding​ ​gas​ ​(w​ ​is​ ​-​ ​)
○ Work​ ​done​ ​on​ ​a​ ​gas​ ​compressing​ ​it​ ​(w​ ​is​ ​+​ ​)
● Heat
○ Exothermic​ ​process​ ​-​ ​releases​​ ​heat
■ 2H​2​​ ​(g)​ ​+​ ​O​2​​ ​(g)​ ​→​ ​2H​2​O​ ​(l)​ ​+​ ​energy
■ H​2​O​ ​(g)​ ​→​ ​H​2​O​ ​(l)​ ​+​ ​heat
■ FEELS​ ​HOT
○ Endothermic​ ​process​ ​-​ ​absorbs​​ ​heat
■ Energy​ ​+​ ​2HgO​ ​(s)​ ​→​ ​2Hg​ ​(l)​ ​+​ ​O​2
■ Energy​ ​+​ ​H​2​O​ ​(s)​ ​→​ ​H​2​O​ ​(l)
■ FEELS​ ​COLD
○ Solid​ ​→(endo)→​ ​Liquid​ ​→(endo)→​ ​Gas
○ Solid​ ​←(exo)←​ ​Liquid​ ​←(exo)←​ ​Gas
● Enthalpy​ ​(H)
○ ΔH​ ​=​ ​heat​ ​given​ ​off​ ​or​ ​absorbed​ ​during​ ​a​ ​reaction​ ​at​ ​constant​ ​pressure​
...
01​ ​kJ/mol​ ​(endothermic)
H​2​O​ ​(s)​ ​+​ ​heat​ ​→​ ​H​2​O​ ​(l)
H​2​O​ ​(s)​ ​+​ ​6
...
4​ ​kJ/mol​ ​(exothermic)
(negative​ ​sign​ ​means​ ​put​ ​heat​ ​on​ ​the​ ​product​ ​side)
CH​4​​ ​(g)​ ​+​ ​2O​2​​ ​(g)​ ​→​ ​CO​2​​ ​(g)​ ​+​ ​2H​2​O​ ​(l)​ ​+​ ​heat
CH​4​​ ​(g)​ ​+​ ​2O​2​​ ​(g)​ ​→​ ​CO​2​​ ​(g)​ ​+​ ​2H​2​O​ ​(l)​ ​+​ ​890
...
4​ ​kJ/mol
890
...
01​ ​kJ/mol
H​2​O​ ​(l)​ ​→​ ​H​2​O​ ​(s)
ΔH​ ​ ​=​ ​-6
...

● Practice:
○ How​ ​much​ ​heat​ ​is​ ​evolved​ ​(released/produced)​ ​when​ ​74
...
5​ ​kJ/mol
74
...
99 g N a

*

367
...
47*10​3​​ ​kJ​ ​of​ ​heat​ ​is​ ​produced,​ ​how​ ​many​ ​grams​ ​of​ ​oxygen​ ​must​ ​have
reacted?

P​4​​ ​(s)​ ​+​ ​5O​2​​ ​(g)​ ​→​ ​P​4​O​10​​ ​(s)
ΔH​ ​=​ ​-3013​ ​kJ/mol
(6
...
00 g O2
mol O2

= 344 g O2

● Calorimetry
○ Specific​ ​heat​ ​(s)​ ​-​ ​heat​ ​(q)​ ​required​ ​to​ ​raise​ ​the​ ​temperature​ ​of​ ​1​ ​g​ ​of​ ​the
substance​ ​by​ ​1℃​​ ​(units:​ ​J/g*℃)
○ Heat​ ​capacity​ ​©​ ​-​ ​heat​ ​(q)​ ​required​ ​to​ ​raise​ ​the​ ​temperature​ ​of​ ​a​ ​given
quantity​​ ​of​ ​the​ ​substance​ ​by​ ​1℃​ ​(units:​ ​J/℃)
C​ ​=​ ​m*s​ ​(m=mass)
Heat​ ​(q)​ ​absorbed​ ​or​ ​relased:
q​ ​=​ ​msΔt
Δt​ ​=​ ​t​f​​ ​-​ ​t​i​ =​
​ ​change​ ​in​ ​temp
q​ ​>​ ​0​ ​endo
q​ ​<​ ​0​ ​exo
J
○ MEMORIZE:​ ​ 4
...
25℃​ ​to​ ​26
...
184​ ​J/g​ ​*​ ​℃
Δt​ ​=​ ​26
...
25℃​ ​=​ ​22
...
184
)(22
...
80 * 10 3 J = 9
...
444​ ​J/g​ ​*​ ​℃
Δt​ ​=​ ​5℃​ ​-​ ​94
...
444
)(89℃) = 34 kJ
g*℃

● Calculate​ ​the​ ​amount​ ​of​ ​energy​ ​(in​ ​kJ)​ ​released​ ​when​ ​346​ ​g​ ​of​ ​steam​ ​at​ ​182℃
is​ ​converted​ ​to​ ​liquid​ ​water​ ​at​ ​0℃
...

In​ ​ΔH​ ​of​ ​melting,​ ​it​ ​is​ ​called​ ​Heat​ ​of​ ​Fusion​ ​(q​ ​=​ ​nΔH​fos​,​ ​n​ ​=​ ​mol)
In​ ​ΔH​ ​of​ ​boiling,​ ​it​ ​is​ ​called​ ​Heat​ ​of​ ​Vaporization​ ​(q​ ​=​ ​nΔH​vap​,​ ​n​ ​=​ ​mol)

S​water​​ ​=​ ​4
...
99​ ​J/g℃
ΔH​vap​​ ​=​ ​40
...
01​ ​kJ/mol
q = msg Δt
J
q = (346 g H 2 O)( 1
...
28 J = 56
...
02 g H2 O )( 40
...
5 kJ + 783 kJ + 145 kJ = 985 kJ

EXPERIMENT​ ​3
● Partners
● Calorimeters​ ​-​ ​checkout​ ​1/partner
● Part​ ​1
○ Heat​ ​of​ ​neutralization​ ​of​ ​HCl(aq)​ ​and​ ​NaOH​(aq)
■ Bottles​ ​from​ ​the​ ​sink​ ​(not​ ​on​ ​shelves)
○ 2​ ​trials​ ​for​ ​each​ ​part
○ Stir​ ​until​ ​mixed​ ​and​ ​let​ ​it​ ​sit
● Part​ ​2
○ Heat​ ​of​ ​solution​ ​(ΔH)​ ​of​ ​NaOH(s)
■ ΔH​ ​=​ ​

kJ
mol

○ Keep​ ​all​ ​the​ ​containers​ ​closed
○ Clean​ ​up​ ​immediately
○ Top​ ​loading​ ​balances
○ 10-15​ ​pellets​ ​of​ ​NaOH​ ​=​ ​2​ ​g
○ Only​ ​get​ ​one​ ​sample​ ​of​ ​NaOH​ ​at​ ​a​ ​time
○ Need​ ​3​ ​data​ ​points​ ​after​ ​temp​ ​goes​ ​down
○ Keep​ ​stirring​ ​until​ ​pellets​ ​dissolve

● Part​ ​3
○ Heat​ ​of​ ​reaction​ ​of​ ​HCl(aq)​ ​and​ ​NaOH​(s)​
...

Lecture​ ​3
...
17
● Calorimeter
○ No​ ​heat​ ​enters​ ​or​ ​leaves
○ Assume​ ​no​ ​heat​ ​is​ ​absorbed​ ​by​ ​calorimeter
○ q​sys​ ​=​ ​q​A​​ ​+​ ​q​b​​ ​=​ ​0
■ q​A​ =​
​ ​-q​B
○ A​ ​13
...
36℃​ ​is​ ​placed​ ​in​ ​a​ ​calorimeter​ ​containing
100
...
35℃
...
57℃
...
25 g) s (22
...
36℃) = (100
...
184
)(22
...
35℃)
g℃

s = 2
...
14​ ​g​ ​stainless​ ​steel​ ​ball​ ​(s​ ​=​ ​0
...
82℃​ ​is​ ​placed​ ​in​ ​a
calorimeter​ ​containing​ ​120
...
44℃
...

-q​metal​ =​
​ ​q​h2o
-msΔT​ ​metal​​ ​=​ ​msΔT​​ ​h2o
J
− (30
...
470J
)(tf − 117
...
0 g)( 4
...
44℃)
g℃
g℃

− 14
...
2189 = 502
...
3552
10941
...
366 T f
T f = 21
...

○ 1
...
49​ ​kg​ ​of​ ​water
from​ ​20
...
54℃
...

ΔH​rxn
2CH 3 OH + 3O2 → 2CO2 + 4H 2 O
ΔH(kJ/mol) =

q
mol

​ ​or​ ​ ΔH(kJ/g) =

q
mass

q = msΔT
g 4
...
49 kg)( 1000
)( g℃ )(24
...
34℃)( 1000
)
kg
J

q = 43
...
92 g CH 3 OH *

mol CH 3 OH
32
...
0599 mol CH 3 OH

-q​combustion​​ ​=​ ​q​H2O
ΔH =

43
...
0599 mol

= −

730
...
kJ
mol CH 3 OH

● Problems​ ​-​ ​transferring​ ​heat
○ NaOH​ ​(s)​ ​is​ ​added​ ​to​ ​water​ ​in​ ​a​ ​calorimeter​ ​forming​ ​a​ ​solution​ ​with​ ​a
specific​ ​heat​ ​of​ ​0
...
​ ​mL,​ ​and​ ​a​ ​density​ ​of​ ​1
...
​ ​ ​The​ ​temperature​ ​rises​ ​4
...
7​ ​kcal/mol​ ​(ΔH)
...
98 cal
kcal
q = (110
...
03
)( g℃ )(4
...
477 kcal
0
...
7 kcal

*

40
...
0 g N aOH

● Hess’s​ ​Law
○ Standard​ ​enthalpy​ ​of​ ​formation​ ​ΔH°​f
■ °​ ​means​ ​standard​ ​state:
● gas:​ ​1​ ​atm
● solid​ ​and​ ​liquid:​ ​1​ ​mol,
● Aq:​ ​1​ ​M,
● All​ ​25℃
■

​ ​means​ ​formation​ ​of​ ​1​ ​mol​ ​from​ ​its​ ​elements

f​ ​

■ For​ ​element​ ​in​ ​most​ ​stable​ ​form:​ ​ΔH°​f​ =​
​ ​0
● Examples:​ ​ΔH°​f​ ​(O​2​(g))​ ​=​ ​0​ ​;​ ​ΔH°​f​ (C,​
​(s),​ ​graphite)​ ​=​ ​0
​
Cu(s)​ ​+​ ​S(s)​ ​+​ ​O​2​(g)​ ​→​ ​CuSO​4​(aq)​ ​ ​ΔH°​f​​ ​(CuSO​4​(aq))
C(s,​ ​graphite)​ ​+​ ​ 12 O​2​​ ​→​ ​CO(g)​ ​ΔH°​f​ ​(CO(g))​ ​=​ ​-110
...
19​ ​kJ/mol

● Standard​ ​enthalpy​ ​of​ ​reaction​ ​(ΔH°​rxn​)​ ​-​ ​at​ ​1​ ​atm
○ Hess’s​ ​Law:​ ​enthalpy​ ​change​ ​is​ ​the​ ​same​ ​whether​ ​the​ ​reaction​ ​takes
place​ ​in​ ​one​ ​step​ ​or​ ​a​ ​series​ ​of​ ​steps

ΔH°rxn = ΣnΔH°f (products) − ΣmΔH°f (reactants)
aA​ ​+​ ​bB​ ​→​ ​cC​ ​+​ ​dD
ΔH°rxn = [cΔH°f (C) + dΔH°f (D)] − [aΔH°f (A) + bΔH°f (B)]
Liquid​ ​benzene​ ​C​6​H​6​​ ​burns​ ​in​ ​air​ ​to​ ​produce​ ​carbon​ ​dioxide​ ​and​ ​liquid​ ​water
...
​ ​ ​How​ ​much​ ​heat​ ​(in​ ​kJ)​ ​is​ ​released​ ​per​ ​gram
of​ ​benzene​ ​consumed?
Assume​ ​standard​ ​states
2C 6 H 6 (l) + 15O2 → 12CO2 (g) + 6H 2 O(l) ​ ​ ΔH°rxn
ΔH°rxn = 12ΔH°f (CO2 (g)) + 6ΔH°f (H 2 O(l)) − 2ΔH°f (C 6 H 6 (l)) − 15ΔH°f (O2 (g))
kJ
kJ
0 kJ
ΔH°rxn = 12( −393
...
9
) − 2( mol49CkJH ) − 15( mol
) = − 6535
...
4 kJ
2 mol C 6 H 6

*

mol C 6 H 6
78
...
836 kJ/g C 6 H 6

● 50
...
​ ​ ​Liquid​ ​water​ ​is​ ​formed/
Calculate​ ​the​ ​enthalpy​ ​of​ ​reaction
...
7 kJ
0 kJ
ΔH°rxn = 4( −393
...
9
) − 2( mol
) − 7( mol
) = − 3120
...
0 g C 2 H 6 *

mol C 2 H 6
30
...
0 kJ
2 mol C 2 H 6

2

6

2

= 2
...
5 kJ/mol

S (rhomic) + O2 (g) → S O2 (g)

ΔH°rxn = − 296
...
6 kJ/mol

ΔH° (kJ/mol)

+

C (s, graphite) + O2 → C O2 (g)

-393
...
4)

2SO2 (g) + C O2 (g) → C S 2 (l) + 3O2 (g)

1073
...
5 + 2(− 296
...
6) kJ/mol = 87
...
4 kJ/mol

C (s, graphite) + O2 (g) → C O2 (g)

ΔH°rxn = − 393
...
9 kJ/mol

ΔH° (kJ/mol)
2C(s, graphite) + 2O2 (g) → 2CO2 (g)

2(-393
...
4)

4H 2 (g) + 2O2 (g) → 4H 2 O(l)

4(-285
...
8 kJ/mol
But​ ​we​ ​need​ ​it​ ​to​ ​be​ ​per​ ​1​ ​mol​ ​CH​3​OH(l)​ ​so​ ​divide​ ​by​ ​2
−477
...
9 kJ/mol

● Thermal​ ​Effects​ ​of​ ​Physical​ ​Processes
○ Hydration​ ​-​ ​ions​ ​are​ ​surrounded​ ​by​ ​water
○ Lattice​ ​energy​ ​(U)​ ​-​ ​energy​ ​required​ ​to​ ​completely​ ​separate​ ​1​ ​mol​ ​of​ ​an
ionic​ ​compound​ ​into​ ​gaseous​ ​ions
○ Heat​ ​of​ ​hydration​ ​ ΔH°hydr -​ ​energy​ ​associated​ ​with​ ​the​ ​hydration​ ​process

○ Heat​ ​of​ ​dilution​ ​-​ ​heat​ ​change​ ​associated​ ​with​ ​the​ ​dilution​ ​process



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