Download Dear Students, Welcome to AP Chemistry, a little early. We will have

AP Chemistry
Summer Assignment
Dr. Kalish
Dear Students,
Welcome to AP Chemistry, a little early. We will have a fabulous year together in which
you will be taught everything you need for success in the course and on the AP Chemistry Exam
in May 2015. I apologize for the summer assignment, but I want to have at least a month to
review with you prior to the AP Chemistry Exam. The assignment is mostly review and NOT
comprehensive of all material covered in honors chemistry. I would recommend that you start
the assignment one week before school starts, as summer should be reserved for fun.
Summer Assignment:
1. Read the Chapter 1 notes provided. These notes, as well as the two other packets,
correspond with an older textbook and do not match the chapters in your book perfectly.
Complete problems from Hwk 1.1 (the homework for the three chapters has been
packaged as a unit) and Clwk 1.1.
2. Read the Chapter 2 notes provided. Complete problems from Hwk 1.2 and Clwk
1.2A and B (1.2B is NOT available on line). Note: Organic chemistry is no longer
assessed on the APC Exam but is assessed on the SAT Chemistry Subject Test 2 so I
have included some information on this concept.
3. Read the Chapter 3 notes provided. Complete problems from Hwk 1.3. We will
review this section extensively, as stoichiometry is a critical component of APC! The
advanced stoichiometry was not covered at the honors or pre-APC level.
4. Memorize the chemical symbols for elements on the periodic table (PT). For example,
Mg represents Magnesium. You can use a PT on every assessment, but the name of the
element is NOT listed. The periodic table that you will use all year and on the AP
Chemistry Exam is provided in this packet. Also included are formula sheets (p. 160161), which are provided on the APC Exam, polyatomic ions, which are not, and other
pertinent information. Please do not lose these sheets. You must memorize all of the
polyatomic ions that have been provided.
You will take Quiz 1.1 at the beginning of the second week of school. This quiz will assess
material from the summer assignment (i.e. compound naming, formula writing,
stoichiometry, empirical formula, etc.; NOT kinetics, equilibrium, etc.).
If you have any questions or concerns, please email me at [email protected]
Have a wonderful and safe summer. I look forward to working with all of you next year!
Dr. Kalish 
I
:;:.
no NOT DETACH
"CI
FROM nOOK.
n
g
!3
PERIODIC
TABL~E
iii'
'"....
OF THE ELElVlENTS
'<I
n
H
lie
(lOX
::0
CD
M
>:::
:::;
'4
rt
o
::l
10
I
Li
n
C
I N
0.9-1
HU:d
I·tol
16.00
IC).OO l(l,IK
II
I
1
16
Ii' I'
n
Ar
'So·
Na
Mg
AI
22,<)9
::...!.~o
1(1(}X
19
20
21
')~,
24
2:\
I 25
I 16
n
2H
29
Co
Ni
Cu
30
1 .\ 1
7
P
::'-:.(N
.H
~C.97
:n
As
o
:F
s
el
32.06
.q
':'lAS
Yi rt
~
CfQ
~
~
0...
Ca
Sc
Ti
v
- .... 9<1
-I7,<}()
)().t)-I
'i~,O(;
)-l9'"
6.1.55
6.'UtJ
79.90
X3.:':0
3H
Jtj
40
41
.+~
.+3
44
45
.\.6
47
41\
(;9.12
.\.9
711.96
17
52
5.i
54
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Te
I
Xc
X,'i..l7
1-:7.e::
Ml,.91
91,12
tJ2.tJI
95.')-1
l'>X)
101.1
102.91
lOll..!::'
IO?iC
112.-11
11-1.-;2
IIX.71
121.75
127.60
12691
55
56
51
72
73
74
75
76
77
lX
71)
XI)
Xl
X.2
&1
X..J
It')
Cs
B~l I La I Ht'
Au
Hg
TI
Ph
Hi
Po
At
1'291
X7
Fr
1i7
XX
Ra
Cr I Mn I Fe
...:...c....:. :....:....... +...:.. .:. . :. :.. :...............:5....:!!....:...6:.:.9. .....
OS
Ir
Pt
nX.91
17H.-I9
ISO.Y.'i
1K1X'i
IX()21
{\)(1.2
1950X
H<J
IOl
10:"
106
107
lOX
192.2
109
tAc
Rf
Bh
Hs
22()J)21227.()1; (261)
.'is
CD
0
'iLantrmnidi! Scrr:s
g
Cc
"" I Re
Ta
Ob
1~()2)
Sg
I
(266
I
(2b-l11 [277)
I
1 10
Vlt
Os
(26~)
(211)
Zn. Ga
19IJ9~P()()591:[)-l.\X
Ge
Br Kr
Se
"07:l2(H91I!(2()t))
2Jf))
III
Rg
1(272)
59
nO
(, I
Pr
Nd
Pm
().:;
6..t
(],'i
M
I 67
Mi
(,9
71)
}
I
167 .26
16~,93
Lu
162'50
II ~J-l'B
Yb
I ¥1'1I
l) I
17.1()-l
17·1.97
')l\
t}9
l()()
li)1
IU~
1m
Th
Pa
Cf
Es
("m
Md
No
Lr
2J2.11-l
::.H.O-1
(2:;1
L~52)
(257)
(2~X)
(:.'5 Q
I'
1.)0
gs
Dv
I
Ho
Er
Tm
ct
C/]
I' ~9.():'l
36
"'O.CK
r-J
(j
0
~e
K
P>
~
I
.N.IO
51 !:):
:J
s
Q.
6
tv
t...J
Ul(!)
III
-"
22.',
S
g
t Aetinide
SCI' C'i
)
,2()2)
UI
~9
X6 Rn I
(~22)
l".1
~
!3
1:1
~
....
::l
c
STANDARD REDUCTION POTENTIALS IN AQUEOUS SOLUTION AT 25 C
Hal f-reaction
) + 2 e­
+ 1'­
2.87
1.82
Au'''' + 31'­Au(n
2CI­
!K)+2e­
1.50
2 H,O(/)
O:::(K) + 4H'" + 41'­
1.23
2 Br­
Br2 (/) + 2 2
Hg 2 ...
Hg2+
Hg:::+
+ 21'­
+ 2 e­
Hg(/)
Ag(s)
0.80
Hg,:::++2e­
2 Hg(/)
+1'
0.53
curs)
0.52
+ 21'­Cuts)
+ 1'­
0.34
1'-
+ 2 H+ +
0.15
0.15
Sn-+-+2e­
Sis)
0.79
21­
+
Cu 2 +
0.85
0.77
12(~)+2e­
Cu 2+
1.07
0.92
+1'
Cu+
1.36
21'­
H
0.14
2H++2e­
H,(g)
0.00
Pb 2+
PhIs)
0.13
+
21'­
Sn 2-+2e­
Sn(s)
0.14
Ni 2+
+ 21'­Ni(s)
+ 2e­
Cots)
Cd"+ + 21'Cd(s)
Cr2+
Cr'+ + 1'­
Fe"+ + 21'­ Fe(s)
0.25
Co 2+
0.28
Cr 3++3c­
Cr(s)
Zn 2+ + 21'­
0.40
0.41
0.44
0.74
Zn(s)
2H 2 0(/)+ 21'­
H 2 (g)+20H­
0.76
Mn 2+
1.18
+ 21'­
:'vines) AI)+ + 31'­ Al(s) Be 2+ + 2c­
Be(s)
Mg2+ + 21'-
Na+ + e-
+ 2e­
Sr2+ + 2e­
Ba:::'" + 2e­
Rb+ + e­
Ca='-
0.83
1.66
1.70
Mgt.»
Na(s)
-2.37
Cats)
-2.87
2.71
Sr(s)
2.89
Ba(s)
2.90
Rb(s)
-2.92
K+ +e­
K(s)
2.92
+ e­
Li+ + e­
Cs(s)
2.92
Lien
-3.05
Cs+
GO ON TO THE NEXT PAGE.
3
AP Chemistry Course and Exam Description
Appendix B: AP Chemistry Equations and Constants Throughout the test the following symbols have the definitions specified llnles, othemise nOled.
L,mL
g
nm
atm
liter(s). milliliter(s)
gram(s)
nanometer( S)
atmosphere(s)
mm Hg
J. kJ
millimeters or men:ury
jOllle(s). kilojoule(s)
V
\olt{s)
mol
mole(s)
ATOMIC STRUCTURE
E = energy
E
= Ill'
\' = frequency
/.=
Planck's constant. II = 6.626 x 1
Speed of light. c = 2.9l)1{ x
/\\ ogadro' s number
= 6.022
Electron charge. e
=
Js
108 111
xI
1.602
A
s"!
mo\')
10')'1 coulomb
EQUILIBRIUM
[C]'[D]d
K =
. where a A + b B ~ c C + d D
( [A],lrBt K =
(p,
C
l'tP
K, (molar conccntrations)
It!
D
pressure'»
K" (weak acid)
K1, (weak base)
Kit (waler)
Ki'
(P")"(P13/' I'
Equilibrium Constants
K - fOWJlHB
h -
[B]
K"
= IWIIOWj
pH
= -log[WJ, pOH
= 1.0 x 1O,j4 at 25°C
[OWl
14 = pH + pOH
_
[A]
pH - pK" + log rHAl
KINETICS
InlAL
InfAln = -kr
_1_ =kl
[Alo
f ' _
1/2 -
rate constant
r = time
r Vl half,life
k
0.693
-k-
Retllrn to the
Taj~e
0: Contents
Appendix B
GASES, LIQUIDS, AND SOLUTIONS
PV
P == pressure
V = volume
IlRT
PA
Plot,)1
T = temperature
x X..... where X" =
mole, A
n = number of mole,
III
p .... +
Pfl
+ Pc + ...
111
11
K
JJ
== mas"
= molar mass
D =- den,ily
KE ;;; kinetic energy
M
\. = velocity
°c + 273
A == absorbance
= molar absorptivity
h = path length
(I
D=!!.!..
V
KE per molecule
c = concentration
1
,
= -::; 1111"Gas constant. R = 8.314 J mol-I
Molarity, M
A
= moles of solute per liter of ,olution
= abc
= 0.08206 L atm mol-I K- 1
== 62.36 L torr mol-I K
I atm = 760 111m
== 760 tOIl'
I
STP == O.OOcC and 1.000 atm
THERMOCHE:\USTRY/ ELECTROCHEMISTRY
q
lIlcf'::..T
f'::..SC
I
so
MIG
I
!:J.H~ products -
t:,.G
C
products -ISo reactants
I !:J.H! reactants
I t:,.Gi products -I t:,.Gf reactants
q
heat 1/1
mas, c
T
specific heat capacity temperature So
He
qandard enthalpy GO
;,tandard free energy number of moles 11
E' O
-RTln K
1=9..
q
Faraday's constant. F
t
!volt
Returr: to the Table of Corcter:ts
2013 The College Board.
<;tamlan.l entrop: standard reduction potential current (amperes) charge (coulombs) lime (seconds) 96.485 coulombs per mole
of electrons
I joule
[coulomb
Chemistry Reference Tables
Polvatomic Ions
Activitv Series
M~tal
I
.
I
Formula
CH\COO'
Name
Acetatc
Ammonium
lOll
Li
Rb
I
Dr. Kalish
i
~m;
i
I
:,Sr I Arsenate
Ca
Na
i
~Ig
I
Mil
i
Zn
Cr
Fe
Cd
Co
Ni
Sn
Ph
II,
Sb
Bi
Cu Hg Ag I
Azide
Bromate
Carbonate
Chlorate
Chlorite
Chromate
Cyanide
Dichromate
AsO.(
N,
i
BrCh
CO,
CIO,'
CIO,
Cre},"
CN
Cr,O···
l\'ame
Dihvdrogen phosphate
Hydrogen carbonate or
bicarbonate
Hvdrogen sulfate
Hydroxide
Hvpochlorite
Iodate
Iodite
l\1erc ury (l)
Methvlammonilllll
Monohydrogen
phosphate
Nitrate
Nitrite
lOll
Formula
H,P0 4
I-ICO.,
1011 l\'ame
Oxalate
Perchlorate
Formula
C 20 4
U0 4
HSO,j'
OH
ClO
Pennanganate
Peroxide
Phosphate
Phosphite
Sulfate
Sulfite
Thiocvanate
'fhiosull:1te
Mn()j
0:"
P0 4 '
10,'
10::
Hg,·
CH,NH,
HPO.j··
NO.,
NO,
Important Constants
=
0.022 x 10:' atoms
Speed oflight (c):
cc- 2.991\ x ]0'
Planck's constant:
h
111 S
6.6262 x I
J-scc
Pt Au SO.j
S( ),
seN
S::O,"
I
un,
;'\lumber
1
2
~
I mole
Avogadro's constant:
i
I
PO, .
I
Uranvl
Prefix
Mono·
Oi­
...
Tri·
f--..
T etraPentaHexa­
Hepta·
I Octa·
Nona­
Oeca­
I
:1
I
4
I
5
6
7
11
9
10
Electroneaativities
.,.,.
Atomic
Element
[lectroIlegathit~
Atomic
" limber
Element
Electronegativity
-.­
Ni
Cu
1.9
1.6
28
29
30
2.0
34
2.5
3.0
35
Se
Hr
Rb
Ag
Cd
Sn
Sb
I
Cs
Ba
Pt
Au
Hg
Pb
Bi
Po
~umber
It
Li
Be
B
C
N
I
0
F
:-.la
Mg
AI
Si
3.4
:17
47
4.0
4x
0.9
50
1.3
51
53
_.­
16
P
S
17
CI
. ) .')
­
19
20
25
K
O.X
Ca
Mn
Fe
Co
1.0
1.11
1.X
1.9
I
:1
4
5
6
7
1\
9
II
12
13
14
15
26
I
n
I
1.0
1.6
1.9
') I
2.0
'
55
56
78
79
80
fP
,-
x3
84
Zn
1.9
1.6
'.6
:1.0
() .11
1.9
i
1.7
2.0
2.0
2.7
O.X
0.9
_.­
I
i
I
2.4
1.9
1.8
1.9
2.0
!
Reference Tables for 2 nd Semester:
Table I: l!sefu I ('onstants
;\\,'gatiro' -; Numb;:r
l illi\ ;:r,al G'l.-; C"lhtallt
\'
R
R
6J)22 x 10" panicl;:,
0.01121 L-Jtlll ll1(lle-K or
/I ..' 14 L-kPa lllok-K
Planck', ":OlbWnt
Sr;:..:d of light (in \iKUum)
\lolar V"lull1e (STP)
Ion rrodu..:t till' water
°
Ta ble 2 : P ropertles fC,ommon S
~ 0 Ivents
Solvent
ktJ
Til
(0C)
(nC-kg/mole)
I
Acctic acid
Bell/elle
Camphor
Cvelohex<lnc
Cvciohex<1nol
Nitrohcll/ene
Phefwl
Wain
I 17,l)O
/10.100
20742
I
~O.725
2,530
2.53
5,611
2.75
-­
-­
210,/1
5,24
3,MO
0,515
1/l1,~.W
IOO,OO()
Tr
(0C)
16,66
5,533
17 /1, :5
()54
25. i:;
5,7()
40,<)0
kr
(OC-kg/mole)
6J,262, 10 ")-,,-,c
2,')9/1 x I ()' !1l sec
I,,, = 22414 Lnwle
f:.
1,0 X IOIJ
Ii
I
3,90
5,12
~1, 7,7
2()()
39,':;
(,1\52
740
1,1\53
O,()OO
a)l l
e.i S
: ,peCIIC
T
TH eat 0 fC,ommon Substances at 20"('
Substance
Specific
Heat
(Jig 0c)
Table 4- Polvatomic Ions
Air
~
Aluminum
Carh,)ll
(di,1I1Hlllti) Carb"11
il!rarhlte) Cill'holl
dinxide
Copper
Ethyl alcohol
Gold
Granite
Iron
L\.'ad
Parutlin
Sihcr
Stainless steel Water
i
I ,Of)
(J,895
O.5()2
!
CaCO"
CaU".
CI1 1 •. · CJ I",.
('4
i
II I<llel
Formula
CH 1COO
NH.j
Arsenate
Azide
Borate
Bromate
Carbonate
Chlorate
Chlorite
Chromate
AsO.j
N,
80/
BrO,
CO,-'
CIO,
C10,
CrO.j .
Cyanide
Dichromate
eN
(,r::O-·'
() 71 I (U132
O,3l\7
245
(),129
f),803
0,44/1
0,12:-\
2.9
0,233 0,51
4,IX
T a bl c..:; : Ent ha irlles
' 0 f FormatIOn
Substance
~H n (kJ/mole)
-45.9
'\IH, '"
NlljCI,
-314.4
NIl.jFI,
-125
-3(l5,56
NH,NO"
Br,
Ion ;-';ame
Acetate
Ammonium
(),on
-1207,6
-6:'\4.9
-74.9
-104,7 -125.7
-300.9
-393,5
C1"H" "
CO'.
F'::c
n.oo
H>­
HBr:'J
(l,On
-36,29
HC) ·9:U
Ion ~ame
Dihydrogen phosphate
Hydrogen carbonate or
bicarbonate
Hydrogen sulfate
Hydroxide
Hypochlorite
Iodate
Iodite
Mercury (I)
Methylammoniul11
Monohydrogen
phosphate
Nitrate
Nitrite
Substance
Hr', ,
H:O,"
H-Ol\
H,O""
H2SO.j I
FeO"1
Fc:O",.
\lnO-,
NcO,"
'\I:OJ,"
~Ht (kJimole)
0:
n.oo
Na:O"
Na-SO""
so' .
-414.2
-110 I -2%,8
SO"e"
-395,7 I
Formula
HePO,
HC01
Ion :'-lame
Oxalate
Perchlorate
HSO.j
OH'
CIO
10,'
10 2
Hg 2ClhNH,
HP0 1­
Pemlan!!anate
Peroxide
.. Phosphate
Phosphite
Sulfate
Sulfite
ThiocYanate
Thiosu Ifate
NO,
NO,
Lranvl
Formula
C 20 1
ClO.
M1104
(), -
PO+'
PO,"
SO.jSO,"
SCN'
S:O,­
LO, -
-2:3 ..\
Table 6: Heats of Fusion and Vaporization for Some
Common Substances
Substance
lIeat of Fusion
Heat of
(.J/g,
\'aporization (Jig)
472(,
i
205
Coppcr
,>1,79
Ell1\ I ,".koho!
109
()4,5
Gold
157i\
Lead
24.7
X5S
Sil\cr
88
2:'\00
\Vater
334
22<>0
-241.x2
-285,8
-187:-; -/\ J1,.9i\lJ
-~15,5
-111:-\.4 -52(),() -il2.1
'9,16
i
AP Chemi:-;try
II and P: Chapter 1
Dr. Kalish Page I Chemistry: 'latter and Measurement
I. Introduction:
Chemistry enables us to design all kinds of materials: drugs (disease): pesticides: fertilizers:
fuels: fihers (clothing): building materials: plastics: etc.
A. Key Terms:
1. Chemistry: study of the composition. structure. properties. and reactions of matter
a. Matter: anything that has mass and occupies space Examples: wood, sand. water. air. gold. ctc. 1) l\lass: quantity of matter in an ohject
a) use a balance to measure mass: the unit on a balance is the gram. but the
fundamental unit is the kilogram (kg)
2) Volume: space that object occupies
a) use a ruler to measure a regular solid. and a graduated cylinder to measure
an irregular solid (water displacement) or a liquid
b) the units vary: em~. ml. L [1 ml = I em}]
2. Atoms: smallest distinctive units in a sample of matter (huilding blocks)
3. Molecules: larger units in which two or more atoms are joined together
a. The way in which matter hehaves depends on the atoms present and the manner in
which they are comhined
4. Composition: types of atoms and their relative proportions in a sample of matter
B. Properties:
1. Physical property: characteristic displayed hy a sample of matter without it
undergoing any change in its composition: what you can see or measure without
altering the chemical nature of the material
Examples: color. mass. density. state. Tm. Tr, Tb
2. Chemical property: characteristic displayed by a sample of matter as it undergoes a
change in composition
Examples: flammability. ability to react with acids
C. Types of changes:
1. Physical change: change at the macroscopic level but no change in composition: the
same substance must remain after the change
Examples: Phase changes. dissolving
AI' Chemistry
[I and P: Chapter 1
Dr. Kalish
Page ..,
2. Chemical change or chemical reaction: change in composition and/or the structure
of its molecules: one or more substances is altered--new substances arc formed
Examples: cooking and spoiling of foods: burning, digestion. fcnnentation
a. Reactants -7 Products
h. Evidence of a Chemical Change
1) Evolution of a gas
2) Fonnation or a ppt.
3) Evolution or absorption of heat (exo- vs, endothennic reactions)
4) Emission of light
5) Color change
D. Classifying :\:latter
\1aterial
Homogeneous
IVlaterial:­
Substance
Elements
Homogeneous
!vlix tures
(Solutions)
Heterogeneous
Mixtures
Compounds
1. Material: any specific type of matter
a. Homogeneous Materials: unifonn matter
b. Heterogeneous Materials: nonunifo1111 matter
2. Mixture: consists of two or more different atoms or compounds with no fixed
composition; the atoms or compounds are mixed together physically
a. Heterogeneous Mixture: variable composition and/or properties throughout:
two or more distinct phases: interface
Examples: blood, granite: Oil and water
b. Homogenous Mixture or Solution: has the same composition and properties
throughout
Examples: salt water: sugar water
3. Substance: type of matter that has a definite or fixed composition that docs not vary
from one sample to another
Examples: Elements or compounds
Dr. Kalish
AP Chemistry
Hand P: Chapter 1
Page -'
a. Element: substance that cannot be broken down into other simpler substances by
chemical reactions: substances composed of one type of atom; represented by a
chemical symbol
h. Compound: substance made up of atoms of two or more elements that combine
in fixed propoltions or definite ratios: represented by a chemical fonnula
Example: H 20 with 111 and 1 0
4. Key differences behveen mixtures and compounds
a. The properties of a mixture ret1ect the properties of the substances it contains: the
properties of a compound bear no resemblance to the properties of the elements
that comprise the compound.
h. Compounds have a definite composition by mass of their combining elements.
while the components of a mixture may be present in varying propOitions.
E. Scientific !\lethods:
1. Observation
2. Hypothesis: tentative prediction or explanation concerning some phenomenon
3. Experiment: procedure used to test a hypothesis
a. Data
4. Scientific Laws: summary of patterns in a large collection of data
5. Theory: multi-tested and contlnned hypothesis
II. Scientific Measurement:
A. International System of l'nits
1. Length: the SI base unit is the meter '1
Mass: the quantity of matter in an object a. SI base unit is the kilogram
3. Time: SI base unit is the second
4. Temperature: property that tells us in what direction heat will t10w
a. S I base unit is Kelvin (K)
Seven Fundamental Units ill Sf
Qllautity
Length
Mass
Time
Them10dvnamic Temperature
Amount of Substance
Electric Current
Luminous Intensity
I
l"nit
Symholfi)1' qlumti(r
!
III
r
T
11
[
1
I Abbreviation for unit I
i
i
meter
kilogram
second
kelvin
mole
ampere
candela
111
kg
s
K
mol
A
cd
AP Chemistry
Hand P: Chapter I
Dr. Kalish
Page 4
B. Den sity
1. mass per unit volume of a substance
~ - ./
Lowcst
density
d=m/V
Density of water is 1.00 giml
Problems ...
Common Sf prefixes
Prefix
I Exponential:\1 ultiplier
l'nit abbreviation
I
Jfea 11 illJ(
i
GiEaMega­
• KiloHecto­
Deca­
G
lvl
k
h
da
. deci­
centi ­
milli­
microi nano­
, pleo­
femto­
d
10"
10"
10
lO-
i
I 000 000 000 i
t 000 000
1000
tOO .
1
10
10'
10"
10
1cr­
1
1 ]0
1 I U()
H)"
10'('
1 1000
1 1 000000
n
10'
I 1 000 000 000
p
10
10' ,
c
m
Jl
f
,
1
Greatest
density
I 1 000 000 000 000
1 I 000 000 000 000 000
C. Conversions within the same quantity vs. those behYeen different quantities
I. Same Quantity
2. Different Quantities: Factor-Labell\lethod or Dimensional Analysis
Eq utllities and Collversioll Factor.'"
2 cups== 1 pint
2.54 centimeters = I in
12 inches 1 foot
2 pints 1 quart
4 cups 1 quart
3 feet = 1 yard
1 liter = 1.06 quarts
5280 feet = 1 mile
16 ounces = I pound
1 meter = 39.37 inches
I molc= 6.022 x 1O~i awm,
I metcr 1.09 yards
3
1 ml 1 cm
1 km 0.62 miles
Problems....
28.3 grams·=; 1 ounce
1 kilogram = 2.2 pounds
453.6 g = J lb
28.35 g I ounce
3.785 L = 1 gal
29.57 ml I tl oz
'­
Dr. Kalish
AP Chemistry
Hand P: Chapter 1
Page .)
D. Precision and Accuracy in Measurements
I. Precision: how closely individual measurements agree with one another
2. Accuracy: closeness of the average of the set to the "correct" or most probable value
**Precise measurements are 1\OT always accurate.**
Example: Darts--Ifyou hit the same spot outside the bulls-eye five times, you have
precision but not accuracy. You are accurate \vhen you hit thc bulls-eye.
Percent Error
=~(Measured Value -.Accepted VaIUC)}* I 00
1
Accepted Value
Eo Significant Figures
I. All digits known with certainty plus the tirst uncertain one (estimated) arc significant
digits or significant figures
2. Significant figures reflect the precision of the measurement
3. D etermmmg ,S"Igm°fiIcant D""
19lts
. Number
Digits to Count
Example
Number of
Significant Digits
i All
, i\onzero Digits
3279
4
.)
11.2
~
i
Leading Zeroes (zeroes
before an integer) None
Captive Zeroes (zeroes
between two integers) All
Trailing Zeroes (zeroes
, after the last integer)
: 0'()O45 1
5.007
4
7
100
100.
100.0
0.0100
1.7 x 10'-1
1.30 x l((~
,
0.(0) x 10'
All BCT be careful
of "incolTcctly"
written scicntitic
notatilll1
Scientific Notation
O.O()OOO5
6.f)OO.Om~
Counted only if the
number contains a
decimal point
..,
I
3
:4
.,
.)
2
3
1
4. Rules:
a. :vIultiplying and Dividing: Round the calculated result to the same number of
significant figures as the measurement having the least number of significant
figures. [carryall numbers through and then round oft]
Example: 3.45 cm
figures)
*
2
4.5555 cm -7 15.7 cm (Answer is expressed in 3 significant
!\P Chemistry
Hand P: Chapter 1
Dr. Kalish
Page 6
b. Addition and Subtraction: The answer can have NO more digits to the right of
the decimal point than there are in the measurement with the smallest number of
digits to the right of the decimal point.
Example: 3.45 cm + 100,1 em ~ 103,6 em (Express answer to the tenth place)
c, Rounding Fives: If the last significant digit before the five is odd. round up. If
the last significant digit before the five is even (and therc are not any numbers
other than zero after the tive). do NOT round up (leave it alone),
Example: 3.15 ~ For two significant digits or to the tenth place. round to ],2
Example: 3.45 ~ For t\VO significant digits. round to 3.4
Example: 3.451 ~ For two signiticant digits. round to 3,5
28
Chapter 1
Chemistry: Matter and Measurement
Matter is made up of atoms and molecules and can be subdivided into two broad cat­
egories: substances and mixtures. Substances have fixed compositions; they are either ele­
ments or compounds. Compounds can be broken down into their constituent elements
through chemical reactions. but elements cannot be subdivided into simpler substances.
Mixtures are either homogeneous or heterogeneous. Substances can be mixed in varying pro­
portions to produce homogeneous mixtures (also called solutions). The composition and
properties are uniform throughout a solution. The composition and/or properties of a het­
erogeneous mixture vary from one part of the mixture to another.
/
Substances exhibit characteristics called physical properties without undergoing a
change in composition. In displaying a chemical property, a substance undergoes a change
in composition-new substances are formed. A physical change produces a change in the
appearance of a sample of matter but no change in its microscopic structure and composi­
tion. In a chemical change, the composition and/or microscopic structure of matter changes.
Four basic physical quantities of measurement are introduced in this chapter: mass,
length, time, and temperature. In the SI system, measured quantities may be reported in the
base unit or as multiples or submultiples of the base unit. Multiples and submultiples are
based on powers of ten and reflected through prefixes in names and abbreviations.
nanometer(nm)
10- 9 m
micrometer (fLm)
1O-6 m
millimeter (mm)
10- 3 m
meter(m)
m
1
1
kilometer (km)
103 m
The SI base unit of temperature, the kelvin (K), is introduced in Chapter 5, but in this
chapter two other temperature scales, Celsius and Fahrenheit, are considered and compared.
1
tF
= 1.8 tc
+ 32
tc
= -'-'----'­
1.8
To indicate its precision, a measured quantity must be expressed with the proper num­
ber of significant figures. Furthermore, special attention must be paid to the concept of sig­
nificant figures in reporting calculated quantities. Calculations themselves frequently can be
done by the unit-conversion method. The physical property of density also serves as an im­
portant conversion factor. The density of a material is its mass per unit volume: d = m/V.
When the volume of a substance or homogeneous mixture (cm 3) is multiplied by its densi­
ty (glcm\ volume is converted to mass. When the mass of a substance or homogeneous mix­
ture (g) is multiplied by the inverse of density (cm3/g), mass is converted to volume.
In this chapter and throughout the text, Examples and Exercises illustrate the ideas,
methods, and techniques under current discussion. In addition, Estimation Examples and ac­
companying Exercises deal with means of obtaining estimated answers with a minimum of
calculation. Conceptual Examples and accompanying Exercises apply fundamental con­
cepts to answer questions that are often of a qualitative nature.
A
inl
yo
pil
pn
de
tic
PI;
AP Chemistry
Summer Assignment
Dr. Kalish
Page 1
Homework Problems:
Note: Numbers in the left margin correspond with book problems from the PH textbook and your answer key.
2)
11)
12)
13)
14)
15)
32)
36)
42)
46)
48)
50)
2)
8)
10)
12)
19)
Hwk 1.1: Chapter 1
1) Which of the following are examples of matter?
a) Iron
c) The human body
e) Gasoline
b) Air
d) Red light
f) An idea
2) Which of the following is NOT a physical property?
c) Natural gas burns.
a) Solid iron melts at a temperature of 1535 oC.
b) Solid sulfur as a yellow color.
d) Diamond is extremely hard.
3) Which of the following describe a chemical change, and which a physical change?
a) Sheep are sheared and the wool is spun into
c) Milk sours when left out.
yarn.
d) Silkworms feed on mulberry leaves and
b) A cake is baked from a mixture of flour,
produce silk.
baking powder, sugar, eggs, shortening, and
e) An overgrown lawn is mowed.
milk.
4) Which of the following represent elements? Explain.
e) Na
a) C
c) Cl
f) KI
b) CO
d) CaCl2
5) Which of the following are substances, and which are mixtures? Explain.
a) Helium gas used to fill a balloon
c) A premium red wine
b) Juice squeezed from a lemon
d) Salt used to de-ice roads
6) Indicate whether the mixture is homogeneous or heterogeneous.
a) Gasoline
c) Italian salad dressing
b) Raisin pudding
d) Coke
7) Convert the following quantities:
a) 546 mm to meters
c) 181 pm to µm
e) 46.3 m3 to L (careful)
f) 55 mi/h to km/min
b) 87.6 mg to kg
d) 1.00 h to µs
8) How many significant figures are there in each of the following quantities?
e) 1.60 x 10-9 s
a) 4051 m
c) 0.0430 g
9
f) 0.0150 oC
b) 0.0169 s
d) 5.00 x 10 m
9) Perform the indicated operations, and provide answers in the indicated unit with the correct number of
significant digits.
a) 13.25 cm + 26 mm – 7.8 cm + 0.186 m (in cm)
b) 48.834 g + 717 mg – 0.166 g + 1.0251 kg (in kg)
10) Calculate the density of a salt solution if 50.0 ml has a mass of 57.0 g.
11) A glass container has a mass of 48.462 g. A sample of 4.00 ml of antifreeze solution is added, and the container
with the antifreeze has a mass of 54.513 g. Calculate the density of the antifreeze solution expressed in the
correct number of significant figures.
12) A rectangular block of gold-colored material measures 3.00 cm x 1.25 cm x 1.50 cm and has a mass of 28.12 g.
Can the material be gold if the density of Au is 19.3 g/cm3? Calculate the percent error.
Hwk 1.2: Chapter 2
1) When 24.3 g of magnesium is burned in 16.0 g of oxygen, 40.3 g of magnesium oxide is formed. When 24.3 g
of magnesium is burned in 80.0 g of oxygen, (a) What is the total mass of substances present after the reaction?
(b) What mass of magnesium oxide is formed? (c) What law(s) is/are illustrated by this reaction? (d) If 48.6 g of
magnesium is burned in 80.0 g of oxygen, what mass of magnesium oxide is formed? Explain.
2) What is the atomic nucleus? Which subatomic particle(s) is/are found in the nucleus?
3) Which of the following pairs of symbols represent isotopes? Which are isobars?
a)
70
33
E and
70
34
E
d)
7
3
b)
57
28
E and
66
28
E
e)
22
11
c)
186
74
E and 48 E
E and
44
22
E
E and 186
74 E
4) What do atomic mass values represent?
5) What type of information is conveyed by each of the following representations of a molecule?
AP Chemistry
Summer Assignment
20)
24)
25)
26)
38)
47)
48)
50)
52)
54)
56)
58)
60)
62)
64)
68)
Dr. Kalish
Page 2
a) Empirical formula
b) Molecular formula
c) Structural formula
6) A substance has the molecular formula C4H8O2. (a) What is the empirical formula of this substance? (b) Can
you write a structural formula from an empirical formula? Explain.
7) Are hexane and cyclohexane isomers? Explain.
8) For which of the following is the molecular formula alone enough to identify the type of compound? For which
must you have the structural formulas?
a) An organic compound
c) An alcohol
e) A carboxylic acid
b) A hydrocarbon
d) An alkane
9) Explain the difference in meaning between each pair of terms:
a) A group and period on the periodic table
c) An acid and a salt
(P.T.)
d) An isomer and an isotope
b) An ion and ionic substance
10) Indicate the numbers of electrons and neutrons in the following atoms:
a) B-11
c) Kr-81
b) Sm-153
d) Te-121
11) Europium in nature consists of two isotopes, Eu-151, with a mass of 150.92 amu and a fractional abundance of
0.478, and Eu-153, with a mass of 152.92 amu and a fractional abundance of 0.522. Calculate the weighted
average atomic mass of Europium.
12) The two naturally occurring isotopes of nitrogen are N-14, with an atomic mass of 14.003074 amu, and N-15,
with an atomic mass of 15.000108 amu. What are the percent natural abundances of these isotopes? {Hint: set
one at x and the other at 1-x}
13) The two naturally occurring isotopes of rubidium are Rb-85, with an atomic mass of 84.91179 amu, and Rb-87,
with an atomic mass of 86.90919 amu. What are the percent natural abundances of these isotopes? {Hint: set
one at x and the other at 1-x}
14) Identify the elements represented by the following information. Indicate whether the element is a metal or
nonmetal.
a) Group 3A (13), period 4
d) Group 1A (1), period 2
b) Group 1B (3), period 4
e) Group 4A (14), period 2
c) Group 7A (17), period 5
f) Group 1B (3), period 4
15) Write the chemical symbol or a molecular formula for the following, whichever best represents how the element
exists in the natural state.
a) Chlorine
c) Neon
e) Sodium
b) Sulfur
d) Phosphorus
16) Which of the following are binary molecular compounds?
a) Barium iodide
c) Chlorofluorocarbons
e) Sodium cyanide
b) Hydrogen bromide
d) Ammonia
17) Write the chemical formula or name the compound:
d) Phosphorus
f) Dinitrogen pentoxide
a) PF3
b) I2O5
pentachloride
c) P4S10
e) Sulfur hexafluoride
18) Write the chemical symbol or name for the following monatomic ions:
e) Ba2+
a) Calcium ion
c) Sulfide ion
3+
d) Fe
f) Se2b) Cobalt(II) ion
19) Write the chemical formula or name for the following polyatomic ions:
d) CrO42a) HSO4f) Dichromate ion
b) NO3
e) Hydrogen phosphate
g) Perchlorate ion
c) MnO4ion
h) Thiosulfate ion
20) Name the following ionic compounds:
f) KOH
k) K2Cr2O7
a) Li2S
b) FeCl3
l) Ca(ClO2)2
g) NH4CN
c) CaS
h) Cr(NO3)3 9H2O
m) CuI
i) Mg(HCO3)2
d) Cr2O3
n) Mg(H2PO4)2
e) BaSO3
j) Na2S2O3 5H2O
o) CaC2O4 H2O
21) Write the chemical formula for the following ionic compounds:
a) Potassium sulfide
c) Aluminum bromide
d) Potassium sulfite
b) Barium carbonate
hexahydrate
e) Copper(I) sulfide
AP Chemistry
Summer Assignment
f) Magnesium nitride
i) Potassium nitrite
g) Cobalt(II) nitrate
j) Zinc sulfate
h) Magnesium dihydrogen
heptahydrate
phosphate
22) Name the following acids:
a) HClO(aq)
d) HF(aq)
b) HCl(aq)
e) HNO3(aq)
c) HIO4(aq)
f) H2SO4(aq)
23) Write the chemical formula for the following acids.
a) Hydrobromic acid
e)
b) Chlorous acid
f)
c) Perchloric acid
g)
d) Nitrous acid
h)
8)
22)
26)
36)
37)
40)
43)
44)
50)
56)
58)
64)
68)
Dr. Kalish
Page 3
k) Sodium hydrogen
phosphate
l) Iron(III) oxide
g) H2SO3(aq)
h) H2C2O4(aq)
Acetic acid
Phosphorous acid
Hypoiodous acid
Boric acid
Hwk 1.3: Chapter 3
1) What are the empirical formulas of the compounds with the following molecular formulas?
c) C10H8
a) H2O2
b) C6H16
d) C6H16O
2) Calculate the molecular or formula mass of the following.
d) K3[Co(NO2)6]
a) C2H5NO2
b) Na2S2O3
e) Chlorous acid
c) Fe(NO3)3 9H2O
f) Ammonium hydrogen phosphate
3) Calculate the mass, in g, of the following:
c) 0.615 mol chromium(III) oxide
a) 4.61 mol AlCl3
b) 0.314 mol HOCH2(CH2)4CH2OH
4) Calculate the mass percent nitrogen in the compound having the condensed structural formula,
CH3CH2CH(CH3)CONH2.
5) Calculate the mass percent of beryllium in the mineral, Be3Al2Si6O18. Calculate the maximum mass of Be
obtainable from 1.00 kg of Be.
6) The empirical formula of apigenin, a yellow dye for wool, is C3H2O. The molecular mass of the compound is
270 amu. What is the molecular formula?
7) Resorcinol, used in manufacturing resins, drugs, and other products, is 65.44 %C, 5.49 %H, and 29.06 %O by
mass. Its molecular mass is 110. amu. What is the molecular formula?
8) Sodium tetrathionate, an ionic compound formed when sodium thiosulfate reacts with iodine is 17.01 % Na,
47.46 % S, and 35.52 % O by mass. The formula mass is 270 amu. What is its formula?
9) A 0.0989 g sample of an alcohol is burned in oxygen to yield 0.2160 g CO2 and 0.1194 g H2O. Calculate the
mass percent composition and empirical formula of the compound.
10) Balance the following equations:
e) Al2(SO4)3 + NaOH  Al(OH)3 + Na2SO4
a) TiCl4 + H2O  TiO2 + HCl
b) WO3 + H2  W + H2O
f) Ca3P2 + H2O  Ca(OH)2 + PH3
c) C5H12 + O2  CO2 + H2O
g) Cl2O7 + H2O  HClO4
d) Al4C3 + H2O  Al(OH)3 + CH4
h) MnO2 + HCl  MnCl2 + Cl2 + H2O
11) Write a balanced chemical for each of the following:
a) Decomposition of solid potassium chlorate upon heating to generate solid potassium chloride and oxygen
gas
b) Combustion of liquid 2-butanol
c) Reaction of gaseous ammonia (NH3) and oxygen gas to generate nitrogen monoxide gas and water vapor.
d) The reaction of chlorine gas, ammonia vapor, and aqueous sodium hydroxide to generate water and an
aqueous solution containing sodium chloride and hydrazine (N2H4, a chemical used in the synthesis of
pesticides).
12) Toluene and nitric acid are used in the production of trinitrotoluene (TNT), an explosive.
___C7H8 + ___HNO3  ___C7H5N3O6 + ___H2O
a) What mass of nitric acid is required to react with 454 g of C7H8?
b) What mass of TNT can be generated when 829 g of C7H8 reacts with excess nitric acid?
13) Acetaldehyde, CH3CHO (D = 0.789 g/ml), a liquid used in the manufacture of perfumes, flavors, dyes, and
plastics, can be produced by the reaction of ethanol with oxygen gas.
AP Chemistry
Summer Assignment
74)
14)
82)
15)
84)
16)
89)
17)
90)
18)
92)
19)
Dr. Kalish
Page 4
___CH3CH2OH + ___O2  ___ CH3CHO + ___H2O
a) How many liters of liquid ethanol (D = 0.789 g/ml) must be consumed to generate 25.0 L acetaldehyde?
Boron trifluoride reacts with water to produce boric acid and fluoroboric acid.
4BF3 + 3H2O  H3BO3 + 3HBF4
a) If a reaction vessel contains 0.496 mol BF3 and 0.313 mol H2O, identify the limiting reactant.
b) How many moles of HBF4 should be generated?
A student needs 625 g of zinc sulfide, a white pigment, for an art project. He can synthesize it using the
reaction, Na2S(aq) + Zn(NO3)2(aq)  ZnS(s) + 2NaNO3(aq)
a) What mass of zinc nitrate will he need if he can make the zinc sulfide in a 85.0 % yield?
Calculate the molarity of each of the following aqueous solutions:
a) 2.50 mol H2SO4 in 5.00 L solution
b) 0.200 mol C2H5OH in 35.0 ml of solution
c) 44.35 g KOH in 125.0 ml of solution
d) 2.46 g oxalic acid in 750.0 ml of solution
e) 22.00 ml triethylene glycol, (CH2OCH2CH2OH)2 (D = 1.127 g/ml) in 2.125 L of solution
f) 15.0 ml isopropylamine, CH3CH(NH2)CH3, (D = 0.694 g/ml) in 225 ml of solution
A stock bottle of nitric acid indicates that the solution is 67.0 % HNO3 by mass (67.0 g HNO3/100.0 g solution)
and has a density of 1.40 g/ml. Calculate the molarity of the solution.
A stock bottle of potassium hydroxide solution is 50.0 % KOH by mass (50.0 g KOH/100.0 g solution) and has
a density of 1.52 g/ml. Calculate the molarity of the solution.
If 50.00 ml of 19.1 M NaOH is diluted to 2.00 L, calculate the molarity of NaOH in the diluted solution.
AP Chemistry
Clwk 1.1
Dr. Kalish Page 1 Date: Period #:
Matter--Substances vs. 1\1ixtures
All matter can be classified as wither a substance (element or compound) or a mixture (heterogeneous or
homogeneous).
Directions:
Classify each of the following as a s ubstance or a mixture. If it is a substance. write element or a
compound in the substance column. Ifit is a mixture. write heterogeneous or homogeneous in the
mixture column.
Mixture
cream
Physical vs. Chemical Changes
In a physical change. the original substance still exists. It has changed in form only. In contrast. a new substance is
produced when a chemical change occurs. Energy always accompanies chemical changes.
Directions:
Classify each of the follO\ving as a chemical (C) or physical (P) change.
I. Sodium hydroxide dissolves in water.
2. Hydrochloric acid reacts with potassium hydroxide to produce a salt. water. and heat.
3. A pellet of sodium is sliced in two.
4. Water is heated and changed to steam.
5. Potassium chlorate decomposes
to
potassium chloride and oxygen gas.
6. Iron rusts.
7. When placed in water. a sodium pellet catches on tire as hydrogen gas is liberated and sodium
hydroxide forms.
8. Evaporation
9. Ice Melting
10. Milk sours.
AP Chemistry
Clwk 1.1
Dr. Kalish
Page 2
11. Sugar dissolved in water.
12. Wood rotting
13. Pancakes cooking on a griddle
14. Grass growing in a lawn
15. A tire is intlated with air.
16. Food is digested in the stomach.
17. Water is absorbed by a paper towel.
Physical vs. Chemical Properties
A physical property is observed with the senses and can be determined without destroying the object. For example.
color. shape. mass. length. and odor are all examples of physical properties. A chemical property indicates how a
substance reacts with something else. The original substance is altered fundamentally when observing a chemical
property. For example. iron reacts with oxygen to form rust. which is also known as iron oxide.
Directions: Classify each of the following properties as either chemical or physical by denoting with a check
mark.
Physical Property I.
2.
3.
4.
5
-
6.
7.
8.
9.
10.
_.
II.
12.
13.
14.
IS.
Blue color
DeI1~ity
Flammability
Soll:l~ili~y
Reacts with acid to form He ...
?l:lPP?I}:~ combustion ....
Sour taste
~J~I~il1g Point
Reacts with water to form a gas
Reacts with base to form water
Hardness
Boiling Point
Can neutralize a base
Luster Odor Chemical Property
AI' Chemistry
Ii and P: Chapter .2
Dr. Kalish Page 1 Atoms,
'lo1ecules~
and Ions
I. La\\'s and Theories: A Brief Historical Introduction
A. Laws of Chemical Combination
I. Lavosier (1743-1794): The Law of Conservation of 1\1 ass
a. The total mass remains constant during a chemical reaction
Example: HgO ~ IIg -+ O 2 Mass of reactants = Mass of products ')
Proust (1754-1826): The Law of Constant Composition or Definite Proportions
a. All samples of a compound have the same composition or the same proportions
by mass of the elements present
Example:
NaCI is 39.34
l'ia and 60.66 % CI
Example: O:Mg in MgO is 0.6583: I. What mass ofMgO will faTIn when
2.000 g Mg is converted to MgO by buming in pure 02?
2.000 g Mg x 0.6583 0
1 Mg
1.317 gO
2.000 g Mg 1.317 gO
3.317 g MgO
B. John Dalton (1766-1844) and the Atomic Theory of Matter (1803)
1. Law of :Vlultiple Proportions
a. When two or more different compounds of the same two elements are compared.
the masses of one element that combine with a fixed mass of a second element are
in the ratio of small \vhole numbers
Examples: CO vs. CO 2 : S02 \'s. SO,
2. Atomic Theory
a. All matter is composed ofe.\'treme~v small, indivisible particles called atoms
b. All atoms ofa given clement are alike in mass and other properties. but atoms of
one clement differ from the atoms of every other element
c. Compounds are fanned when atoms of different elements unite in fixed
proportions
d. A chcmical reaction involvcs a rearTangemcnt of atoms. l'io atoms are creatcd.
destroyed or broken apart in a chemical reaction.
Examples ...
3. Dalton used the Atomic Theory to restate the Lav. .' of Conservation of Mass: Atoms
can neithcr be created nor destroyed in a chcmical rcaction. and as a consequence. the
total mass remains unchangcd.
AP Chemistry
Hand P: Chapter :::
Dr. Kalish Page ::: C. The Divisible Atom
I. Subatomic Particles
a. Proton
1) Relative mass = 1
2) positive electrical charge
b. Neutron
I) Relative mass 1 (although slightly greater than a proton)
2) no charge == 0
c. Electron
1) mass = I! 1836 of the mass of a proton
2) negative electrical charge -I
Particle
I Symbol
Proton
Neutron
Electron
p
n
e
Approximate
Relative l\lass
1
L
0.000545
Relative
Charge
I
0
1­
Location in
Atom
Inside nucleus
Inside nucleus
Outside nucleus
An atom is neutral (has no net charge) because p = e-. :t The number of protons (Z) detennines the identity of the element. 4. Mass number (A)== protons + neutrons
a. neutrons A - Z
1
Example:
ticr
Detennine the number ofp. e-. and n
5. Isotopes: atoms that have the same number of protons but different numbers of
neutrons Examples: IH. 2H, 3H [or H-J. H-2. H-3J: 32S. "S: 5l)CO. 6!JCO 6. Isobars: atoms with the same mass number but different atomic numbers
1-1
H [
a. Examp Ie: C. N
D. Atomic Masses
1. Dalton arbitrarily assigned a mass number to one atom (H-l) and detennined the
masses of other atoms relative to it.
2. Current atomic mass standard is the pure isotope C-12.
3. Atomic mass unit (amu): l!12themassofC-12.
4. Atomic Mass: weighted average of the masses of the naturally occurring isotopes of
that element
Ne-20: 90.51 %. 19.99244 u a. Example:
Ne-21: 0.27 %. 20.99395 u Ne-22: 9.22 ~O. 21.99138 u AP Chel1li~try
II and P: Chapter :2
Dr. Kalish
Page J
E. The Periodic Table
I. Dmitri Mendeleev's (1869) Periodic Table:
a. Arranged elements in order of increasing atomic mass. from left to right in ro\\'s,
and from top to bottom in groups
b. Elements that most closely resemble each other are in the same vertical group
(more important than increasing mass),
c. The group similarity recurs periodically (once in each row)
d. Gaps for missing clements: predict characteristics of yet to be discovered
clements based on their placement
2. Modern Periodic Table
a. Elements are placed according to increasing atomic number
b. Groups or Families: vel1ical columns
c. Periods: horizontal ro\VS
d. Two series "pulled out" 1) Lanthanide and Actinide Series e. Classes
1) Most elements arc Metals, which are to the len (NOT touching) the stair-step
line
a) luster. good conductors of heat and electricity
b) malleable (hammered into thin sheets or f()il). ductile (drawn into wires)
c) Solids at room temperature (except mercury)
2) Nonmetals are to the right (NOT touching) of the stair-step line
a) poor conductors of heat and electricity
b) many are gases at RT
3) Metalloids: touch the vertical and or horizontal of the stair-step line (except Al
and Po
11. Introduction to :Vlolecular and Ionic Compounds
A. Key Terms
1. Chemical Symbols are used to represent clements
2. Chemical F0l111Ulas are used to represent compounds
a. Subscripts indicate how many atoms of each element are present or the ratio of
Ions
B. Molecules and Molecular Compounds
I. Molecule: group of two or more atoms held together in a definite spatial arrangement
by covalent bonds
') Molecular Compound: molecules arc the smallest entities, and they detennine the
propcI1ies of the substance
3. Empirical Formula: simplest fOl111Ula for a compound
a. indicates the elements present in their smallest integral ratio
Example: CH~O = 1 C: .2 H: 10
4. Molecular Formula: true fonnula for a compound {n = MFmass/EFmassl
a. indicates the elements present and in their actual numbers
Example: C6Hl~06 = Q C: 12 H: Q 0
5. Diatomic Elements: two-atom molecules, which don't exist as single atoms in nature
a. Br~. h, N2, Ch. H> O 2• F~
6. Polyatomic Elements: many-atom molecules
Dr. Kalish
AP Chemi,try
H 3nd P: Chapter 2
Page --l
a. Sg. P-l
7. Structural Formulas: shows the alTangement of atoms
a. lines represent covalent bonds between atoms
C. Writing Formulas and Names of Binary Molecular Compounds
I. Binary Molecular Compounds: comprised of 1\\'0 elements. which are usually
nonmctals
a. The first element symbol is usually the element that lies farthest to the lcft of its
period andlor lowest in its group (exceptions: Hand 0) [Figure 2.7]
b. Molecular compounds contain prefixes far subscripts (exception: mono is not
used far the first element)
Prefix
:".'umber
c. The name consists of two \\'ords:
mono­
1
(prefix) element prefix~ide fonn
I
,
* rule with oxide
!
ditri­
tetra­
pcnta­
hexa­
hepta­
octa­
-
8
!
110113­
9
i
dec a­
10
3
i
4
:'\
f,
7
D. Ions and Ionic Compounds
1. Ion: charged particle due to the loss or gain of one or more electrons
a. ::\1onatomic Ion: a single atom loses or gains one or more e­
1) use the PT to predict charges 2) more than one ion can fann with transition elements b. Cation: positively charged ion [usually a metal]
c. Anion: negatively charged ion [usually a nonmetal]
d. Polyatomic Ion: a group of covalently bonded atoms loses or gains one or more
e
e. Ionic Compounds: comprised of oppositely attracted ions held together by
electrostatic attractions; no identifiable small units
2. Formulas and Names for Binary Ionic Compounds
a. Cation anion (~ide fonn)
b. Cation (Roman Numeral) anion (-ide fann)
3. Polyatomic Ion: charged group of bonded atoms
a. suftixes are often -ite (1 less 0) and ~ate
b. prefixes arc often hypo- (1 less 0 than ~ite fonn) and per-( 1 more 0 than -~atc
fann)
Hypochlorite CIO' c. Example:
CI0 2­
Chlorite
ClO, Chlorate
CIO-l­
Perchlorate
4. Hydrates: ionic compounds in which the tonnula unit includes a fixed number of
water molecules together with cations and anions
a. Example: CaCI 2 ' 6H 2 0
Calcium chloride hexahydrate
AP Chemistry
II and P: Chapter .2
Dr. Kalish
Page"
b. Anhydrous: without water
Acids. Bases, and Salts
1. Basic Characteristics of Acids and Bases when dissolved in water
a. Acids:
I) taste sour
2) sting or prick the skin
3) turn litmus paper red
4) react with many metals to produce ionic compounds and fl:: lgi
5) react with bases
b. Bases
1) taste bitter
2) feel slippery or soapy
3) turn litmus paper blue
4) react with acids
2. The Arrhenius Concept ( 1887)
a. Acid: molecular compound that ionizes in water to form a solution containing If
and anions
b. Base: compound that ionizes in water to t<mn a solution containing OH- and
cations
c. Neutralization: the essential reaction betv,·een and acid and a base. called
neutralization. is the combination of H - and OH- ions to fonn \vater and a salt
1) Example: HCl NaOH -7 i\aCli- H:::O
3. Formulas and ~ames of Acids, Bases, and Salts
a. Arrhenius Bases: cation hydroxide
1) Examples: NaOH = Sodium hydroxide
KOH Potassium hydroxide
Ca(OHb Calcium hydroxide
b. '1olecular Bases: do not contain OH- but produce them when the base reacts with
water
1) Example: NIh = Ammonia
c. Binary Acids: H combincs with a nonmetal
1) Examples: HCl 1g1 = Hydrogen chloride HCl 1a4 ) = Hydrochloric acid
HI 1gi = Hydrogen iodide
HI1<141 = Hydroiodic acid
H:,Slg) Hydrogen sulfide
H:::S (aql Hydrosulfuric acid
d. Ternary Acids: FI combines with two nonmetals
1) oxoacids: H combines with 0 and another nonmetal
a) Examples:
H:ypochlorous Acid HCIO
Chlorous Acid
11CIO:::
Chloric Acid
HCI03
Perchloric Acid
HCIO-1
Sulfurous Acid
H2 S0 3
Sulfuric Acid
H 2S04
b) ate-ic ite-ous
AP Chemistry
1I and P: Chapter :2
Dr. Kalish
Page 6
I II. Introduction to Organic Compounds (Carbon-based Compounds)
A. Alkanes: Saturated Hydrocarbons (contain H and C)
Stem
Number
I. molecules contain a maximum number of H Atoms
ll1eth·
I
2. Formula: C 1H2n-2
eth­
:2
a. Methane: CH 4
: prop
3
b. Ethane: C2H"
but·
4
c. Propane: C.1H~
: pent­
5
(,
hex­
d. Butane: C4H 10 7
hept·
1) Two possible structural fOl11mlas R
oct·
--except methane. ethane, & propane
9
non·
I
2) Compounds with the same molecular formula dec­
lO
but different structural fOl11mlas are known as isomers, and they have di t1crent properties. B. Cyclic Alkanes
1. FOl11mla: CnH::n
2. prefix cyc1o­
C. Alkenes: unsaturated hydrocarbon
1. Formula: CJhn
a. Ethene: C:,H4
b. Propene: C3H6
c. Butene: C4H~
D. Alkynes: unsaturated hydrocarbon
1. Fonnula: C nl1::n.2
a. Ethyne: C2H:~
b. Propyne: C3H4
c. Butvne: C.\H 6
E. Homology
1. a series of compounds \vhose fonnulas and structures vary in a regular manner also
have properties that vary in a predictable manner
a. Example: Both the densities and boiling points of the straight-chain alkanes
increase in a continuous and regular fashion with increasing numbers ofC.
!
!
!
F. Types of Organic Compounds
1. Functional Group: atom or group of atoms attached to or inserted in a hydrocarbon
chain or ring that confers charactcristic properties to the molecule
a. usually where most of the reactions of the molecule occur
2. Alcohols (R-OII) where R represcnt the hydrocarbon
a. Examples:
CH30H methanol CH3CH20H = ethanol CH3CH2CH20H = I-propanol Cf'hCH(OH)CH 3 2-propanol or isopropanol b. Not bases!
3. Ethers (R-O-R') where R' can represent a different hydrocarbon than R
a. Example:
CH,CH 20CH:-.CH, = Diethyl ether
AP Chemistry
If and P: Chapter 2
Dr. Kalish
Page 7
4. Carboxylic Acids (R-COOH)
a. Examples:
HCOOH methanoic or formic acid CH 3COOH ethanoic or acetic acid b. the H of the COOH group is ionizable: the acid is classified as a weak acid
5. Esters (R'-COOR)
a. Flavors and fragrances
CH,COOCH:Crh = ethyl acetate
b. Examples:
CH 1COOCH:::CH 2CH:CH:::CH] pentyl acetate
6. Ketones (R-CO-R ')
7. Aldehydes (R-CO-H)
8. Amines (R-NH:::, R-NHR', R-NR'R")
a. most common organic bases: related to ammonia
b. one or more organic groups are substituted for the H in NH3
c. Examples:
CH,NH::: = methyl amine CH 3CH:::Nlh ethyl amine 68 Chapter 2 Atoms, Molecules. and Ions
TABLE 2.1
Some Classes of Organic Compounds and Their Functional Groups
Class
General
Structural
Formula"
Example
Name of
Example
Cross
Reference
Alkane
R-H
CH3CH2CH2CH2CH2CH3
hexane
Section 2.9, 6.8, Chap. 23
Alkene
"
CH2=CHCH2CH 2CH3
I-pentene
Section 9.10, Chap. 23
Alkyne
-C=C­
CH3C==CCH2CH2CH2CH2CH3
2-octyne
Section 9.10, Chap. 23
Alcohol
R-OH
CH3CH 2CH2CH 2OH
I-butanol
Section 2.10, Chap. 23
Alkyl halide
R_Xb
CH3CH2CH2CH2CH2CH2Br
I-bromohexane
Chap. 23
Ether
R-O-R
CH3-0-CH2CH2CH3
I-methoxypropane
(methyl propyl
ether)"
l-aminopropane
(propylamine)C
Section 2.1
Amine
/
C=C
/
"
R-NH2
CH 3CH2 CH2-NH2
II
CH 3CH2CH 2C-H
0
0
Ketone
R-C-R
0
II
Carboxylic
acid
R-C-OH
0
II
Ester
R-C-OR
0
Amide
Arene
II
R-C-NH z
Ar-Hd
Aryl halide
Ar-Xb
Phenol
Ar-OH
In
or
bo
an
cl~
Sections 2.10, 4.2,
Chap. 15
by
C3
2.1
na
II
R-C-H
II
°
0
0
Aldehyde
H'
butanal
(butyraldehyde)"
Section 4.6, Chap. 23
3-hexanone
(ethyl propyl
ketone)C
Section 4.6, Chap. 23
butanoic acid
(butyric acid)"
Sections 2.10, 4.2,
Chap. 15,23
methyl butanoate
(methyl
butyrate)C
Sections 2.10,
6.8 (fats) Chap. 23,
Chap. 24 (polymers)
butanamide
(butyramide)"
Section 11.6, Chap. 23,
Chap. 24 (polymers)
II
CH3CH2CCH2CH2CH3
0
II
CH 3CH 2CH2C-OH
0
II
CH 3CH 2CH zC-OCH 3
0
II
CH 3CH 2CH2C-NH 2
\(8j-CH 2CH 3
ethylbenzene
o-B'
bromobenzene
Chap. 23
4-chlorophenol
Section 9.10, Chap. 23
CI-Q-OH
Section 10.8, Chap. 23
EI
C(
an
R­
iar
ie.
co
C:
11
"The functional group is shown in red. R stands for an alkyl group.
stands for a halogen atom-F, Cl, Br, or I. C Common name. d Ar- stands for an aromatic (aryl) group such as the benzene ring. bX
su
Summary
71
o
I
R'-C-O-R
or
R'COOR
where R' is the hydrocarbon portion of a carboxylic acid, and R is the hydrocar­
bon group of an alcohol. R and R' may be the same or different.
Esters are named by indicating the part from the alcohol first and then naming
the portion from the carboxylic acid with the name ending in -ate. For instance,
o
II
CH3-C-O-CH2CH3
is ethyl acetate; it is made from ethyl alcohol and acetic acid.
Many esters are noted for their pleasant odors, and some are used in flavors and
fragrances. Pentyl acetate, CH3COOCH2CH2CH2CH2CH3' is responsible for most
of the odor and flavor of ripe bananas. Many esters are used as flavorings in cakes,
candies, and other foods and as ingredients in fragrances, especially those used to
perfume household products. Some esters are also used as solvents. Ethyl acetate,
for example, is used in some fingernail polish removers: It is a solvent for the resins
in the polish.
APPLICATION NOTE
Butyric acid, CH)CH zCH 2COOH,
is one of the most foul-smelling
substances known, but turn it into
the ester methyl butyrate,
CH 3CH2CH 2COOCH 3 , and you
get the aroma of apples.
Amines
The most common organic bases, the amines, are related to ammonia. Amines are
compounds in which one or more organic groups are substituted for H atoms in
NH3. In these two arnines, one of the H atoms has been replaced:
H H
I
I
I
H-C-N-H
H
or
CH3NH Z
H
APPLICATION NOTE
H
I I I
H-C-C-N-H
I I
or
CHFH 2NH 2
H H
H
Methylamine
Ethylamine
The replacement of two and three H atoms, respectively, is seen in dimethylarnine
[(CH3hNH] and trimethylamine [(CH3hN]. In Chapters 4 and 15, we will see that
mUch of what we learn about ammonia as a base applies as well to arnines.
~ummary
The basic laws of chemical combination are the laws of conservation of mass, constant com­
position, and multiple proportions. Each played an important role in Dalton's development
of the atomic theory.
The three components of atoms of most concern to chemists are protons, neutrons, and
electrons. Protons and neutrons make up the nucleus, and their combined number is the
mass number, A, of the atom. The number of protons is the atomic number, Z. Electrons,
found outside the nucleus, have negative charges equal to the positive charges of the pro­
tons. All atoms of an element have the same atomic number, but they may have different mass
numbers, giving rise to isotopes.
A chemical formula indicates the relative numbers of atoms of each type in a com­
POUnd. An empirical formula is the simplest that can be written, and a molecular formula
~fle~ts the actual composition of a molecule. Structural and condensed structural formulas
~. scnbe the arrangement of atoms within molecules. For example, for acetic acid:
Amines with one or two carbon
atoms per molecule smell much
like ammonia. Higher homo logs
smell like rotting fish. In fact, the
foul odors of rotting flesh are due
in large part toamines that are
given off as the flesh decays.
-..- - - -.....
...­
---.--~
72
Chapter 2
Atoms, Molecules, and Ions
Key Terms
H
Acetic acid:
acid (2.8)
alcohol (2.10)
alkane (2.9)
amine (2.10)
anion (2.7)
atomic mass (2.4)
atomic mass unit (2.4)
atomic number (Z) (2.3)
base (2.8)
carboxylic acid (2.10)
cation (2.7)
chemical formula (p. 47)
chemical nomenclature (p. 35)
electron (2.3)
empirical formula (2.6)
ester (2.10)
ether (2.10)
formula unit (2.7)
functional group (2.10)
hydrate (2.7)
ion (2.7)
ionic compound (2.7)
isomer (2.9)
isotope (2.3)
law of conservation of mass
(2.1)
law of constant composition
(2.1)
law of definite proportions
(2.1)
law of multiple proportions
(2.2)
mass number (A) (2.3)
metal (2.5)
metalloid (2.5)
molecular compound (2.6)
molecular formula (2.6)
molecule (2.6)
neutron (2.3)
nonmetal (2.5)
periodic table (2.5)
poly atomic ion (2.7)
proton (2.3)
salt (2.8)
structural formula (2.6)
f
0
I II
H-C-C-O-H
I
H
Empirical
formula
Molecular
formula
Structural
formula
Condensed
structural
formula
The periodic table is an arrangement of the elements by atomic number that places el­
ements with similar properties into the same vertical groups (families). The periodic table
is an important aid in the writing of formulas and names of chemical compounds. A mole­
cular compound consists of molecules; in a binary molecular compound the molecules are
made up of atoms of two different elements. In naming these compounds, the numbers of
atoms in the molecules are denoted by prefixes; the names also feature -ide endings.
Examples:
NI3
nitrogen triiodide
S2F4 = disulfur tetrafluoride
Ions are formed by the loss or gain of electrons by single atoms or groups of atoms. Pos­
itive ions are known as cations and negative ions as anions. An ionic compound is made up
of cations and anions, held together by electrostatic forces of attraction. Formulas of ionic
compounds are based on an electrically neutral combination of cations and anions called a
formula unit. The names of some monatomic cations include Roman numerals to designate
the charge on the ion. The names of monatomic anions are those of the nonm~allic ele­
ments, modified to an -ide ending. For polyatornic anions, the prefixes hypo- and per- and
the endings -ite and -ate are commonly found.
Examples: MgF2 = magnesium fluoride
copper(I) oxide
CU20
Ca(CIOh
calcium hypochlorite
Li 2S = lithium sulfide
CuO = copper (II) oxide
KI04 = potassium periodate
Many compounds are classified as acids, bases, or salts. According to the Arrhenius
theory, an acid produces H+ in aqueous (water) solution, and a base produces OH-. A neu­
tralization reaction between an acid and a base fornls water and an ionic compound called
a salt. Binary acids have hydrogen and a nonmetal as their constituent elements. Their names
feature the prefix hydro- and the ending -ic attached to the stem of the name of the non­
metal. Ternary oxoacids have oxygen as an additional constituent element, and their names
use prefixes (h)po- and per-) and endings (-ous and -ic) to indicate the number of 0 atoms
per molecule.
Examples: HI = hydroiodic acid
HCI0 2 chlorous acid
HI0 3 = iodic acid
HCI0 4 = perchlonc acid
Organic compounds are based on the element carbon. Hydrocarbons contain only the
elements hydrogen and carbon. Alkanes have carbon atoms joined together by single bonds
into chains or rings, with hydrogen atoms attached to the carbon atoms. Alkanes with four
or more carbon atoms can exist as isomers: molecules with the same molecular formula but
different structures and properties.
Functional groups confer distinctive properties to an organic molecule when the groups
are substituted for hydrogen atoms in a hydrocarbon. Alcohols feature the hydroxyl group,
-OH, and ethers have two hydrocarbon groups joined to the same oxygen atom. Carboxylic
acids have a carboxyl group, -COOH. An ester, R'COOR, is derived from a carboxylic acid
(R'COOH) and an alcohol (ROH). Arnines are compounds in which organic groups are sub­
stituted for one or more of the H atoms in anmlonia, NH 3 .
AP Chemistry
Clwk 1.2A
Dr. Kalish
Page 1
Name: ________________________________.. ______.._____
Date: _ _ _ _ ___
Molecular Formula \\-'riting and Naming
Name the following compounds:
1.
..,
SF 4
~.
R1Cl"
3.
PBrs
4.
NcO,
5.
S, L)
6.
SoO.)
\Vrite the chemical formula for cach of the follO\ving compounds:
7
carbon dioxide
R.
sulfur hexafluoride
9.
dinitrogen tetroxide
10. trisulfur heptaiodide
11. disulfur pentachloride
12. triphosphorus monoxide
Ionic Formula Writing and Naming
Directions:
Name the following ionic compounds.
13. MgCh
14. NaF
15. NacO
16. AhOl
17. KI
IR. AIF.
19. Mg 1N2
20. FeCh
21. MnO:
22. erN:,
Compounds that include Polyatomic Ions:
23. Ca(OHh
24. (NH 4 hS
25. Al:(S04h
2A. H 1P0 4
,.,­!.
~
Ca(N01):
AP Chemistry
Clwk 1.2A
Dr. Kalish
Page 2
2R. CaCO,
29. 1\a cSO,
30. Co(CH,COOh
31. Cu c(S03h
32. Pb(OHh
Directions:
Write the correct formula for each of the following compounds.
1.
Magnesium sultide
2.
Calcium phosphide
3.
Barium chloride
4.
Potassium nitride
5.
Aluminum sulfide
6.
Magnesium oxide
7.
Calcium fluoride
R.
Lithium fluoride
9.
Barium iodide
10. Aluminum nitride
II. Silver nitride*
12. Nickel(Il) bromide
.--~-----~-----~--
13. Lead(lV) phosphide
14. Tin(H) sulfide
Compounds that include Polyatomic
Ion~:
15. Aluminum phosphate
IIi. Sodium bromate
17. Aluminum sulfite
18. Ammonium sulfate
19. Ammonium acetate
20. Magnesium chromate
21. Sodium dichromate
22. Zinc hydroxide*
23. Copper(Il) nitrite
24. Manganese(II) hydroxide
25. Iron(II) sulfate
26. lron(III) oxide
---
.•..-
-­
.•..
AI' Chemistry
II and P: Chapter 3 Band S Ch ..\
fk Kabil
Stoichiometr~":
Chemical Calculations
I. Stoichiometry of Chemical Compounds
A. Molecular Masses and Formula Masses
I. Molecular Mass: sum of the masses of the atoms represented in a molecular formula
Example: Mass of CO 2
1 C = 1 x 12.0 amu
l 0 = 2 x 16.0 amu
mass of CO 2 =c 44.0 amu
Fonnula Mass: sum of the masses of the atoms or ions represented in an ionic
fonnula
Example: Mass of BaCb
1Ba= I x D7.3al11u
mass of BaCl 2 20X.3 amu
lCl2x35.5amu
B. The !\:1ole and Avogadro's Number
I. Mole: amount of substance that contains as many elementary entities as there are
atoms in exactly 12 g of the C-12 isotope
a. The elementary entities are atoms in elements. molecules in diatomic elements
and compounds. and t(mnula units in ionic compounds.
b. Avogadro's Number (1'\",) 6'()22 x 1O~' mor l
'I .
I} I mole 6.022 x 10-- atoms. molecules. partIcles. etc.
e. one mole of any element is equal to the mass of that element in grams
I) For the diatomic elements. multiply the mass of the element by two.
2. Molar Mass: mass of one mole of the substance
Example: Mass of BaCb
1moleBa=1 x D7.3g
mass of 1 mole BaCI 2 20X.3 g
I mole CI
2 x 35.5 g
C. Mass Percent Composition from Chemical Fon11ulas
I. Mass Percent Composition describes the prop0l1ions of the constituent elements in a
compound as the number of grams of each element per 100 grams of the compound.
Example: What is the % C in butanc (C..\H1n)?
Mass ofCs
x 100°;;)
4(!2.01g,) x 100°/0
MassofC..\H iIl
5S.14g
D. Chemical Formulas from Mass Percent Composition
I. Steps in the Detennination of Empirical FOI111Ula
a. Change ~!O to grams.
h. Convert mass of each elemcnt to moles.
c. Detennine mole ratios.
d. lfneccssary multiply mole ratios by a t~lctor to obtain positi\'c integers only.
c. Write the empirical fonnula.
/\P Chemistry
II and P: Chapter 3 Band S CI1 4
Dr.
K:lil~h
Example: Cyclohexanol has the mass percent composition:
71.95% C. 12.08% H, and 15.97°0 O. Determine its empirical timl1ula.
A compound has the mass percent composition as 1()llows:
36.33
C. 5.49 H, and 58.18 (~o S. Detcnninc its empirical t(mmtia .
.., Relating Molecular F0ll11ulas to Empirical F0ll11ulas
a. Integral Factor (n)
Molecular Mass Empirical Fonnula Mass Example: Ethylene (M 28.0 u). cyclohexane (M 84.0 u). and I-pcntcnc (70.0 u)
all have the empirical f0ll11ula CH:,. \Vhat is the molecular t(mnula of cach
compound?
II. Stoichiometry of Chemical Reactions
A. Writing and Balancing Equations
1. chemical equation: shorthand description of a chemical reaction using symbols and
formulas to represent clements and compounds. respectivcly.
a. Reactants -7 Products
h. C oefficicnts
c. States: gas (g). liquid (1), solid (s). aqueous (aq)
d. ll: Heat
2. Balancing Equations
a. For an element. the same number of atoms must he on each side of thc equation.
h. only coefticients can be changed
1) Balance the clement that only appears in one compound on each side of the
equation first.
2) Balance any reactants or products that exist as the free clcment last.
3) Polyatomic ions should he treated as a group in most cases.
Example: _SiCI 4 + __ H~O -7
_SiO~ . ~
_He)
B. Stoichiometric Equivalence and Reaction Stoichiometry
1. Mole Ratios or stoichiometric factors
Example: _SiCI.t + lH::O -7 _SiO:: ~·1HCI
2. Problems:
a. Mole-tn-mole
h. Mole-to-gram
c. Gram-to-mole
d. Gram-to-gram
What is the mole ratio ofreactmts'!
Dr. Kalish
AP Chemistry
If and P: Chapter 3 Band S Ch 4
C. Limiting Reactants
I. Limiting reactant (LR) is consumed completely in a reaction and limits the amount of
products t<:mned J
To detenninc the LK compare moles. 3. Usc the LR to detennine theoretical yield.
Example: FeS(s) + 2 HCI(aq)
~
FeCI 2(aq)
H:,S(g)
If 10.2 g HCI is added to 13 ' g FeS. what mass of H:,S can he formed'.'
What is the mass of the excess reactant remaining?
10.2 g HCI x 1 mole HCI
36.46 g HCI
0.280 moles HCI
[3,2 g FeS x 1 mole FeS
0.150 moles FeS
LR
87.92 g FeS
0,280 moles HCI
x 1 mole H2S
:2 mole HCI
0.150 mole FeS - 0.140 mole FeS
=
x
34,()1.) g H,S1 mole H2S
4.n g I \.oS
0.010 mole FcS x 87.92 g FeS
1 mole FcS
= (l~79
g FeS
D. Yields of Chemical Reactions
x 100 I. Percent Yield
Actual Yield
Theoretical '{ield J
Actual Yield may be less than theoretical yield hecause of impurities. errors during
experimentation. side reactions. etc,
Example: If the actual yield of hydrogen sultidc was 3.56 g. calculate lhc
percent yield,
If the percent yield of hydrogen sulfide was 84.7
actual yield?
(~o.
what was the
Solutions and Solution Stoichiometry I, Components of a Solution: a. Solute: substance being dissolving
b. Solvent: substance doing the dissohing
1) water: universal solvent: solutions made \\i(h water as the sol\ent arc called
aqueous solutions
J
Concentration: quantity ofsolutc in a given quantity ofsolwnt or solution
a. ()i1ute: contains relati\cly little solutc with a large amount of solvcnt
,\P Chel1li~try
1I and P: Chapter.3 Rand S Ch -4
Dr. Kalish
['''ic'l' -+
b, Concentrated: contains a relatively large amount of solute in a given quantity
solvent
3. Molarity or Molar Concentration
Molarity moles solute Liters of solution or
Example: Calculate the molarity of solution made by dissolving 2.00 moles
NaCI in enough water to generate 4.00 L of solution.
Molarity
moles solute
l.nO moles NaCI= O.SO() M
4.00 L
Liters of solution
Example: Calculate the molarity of solution made by dissolving 35.1 grams
NaCI in enough water to generate 3.00 L of solution.
35.1 g :\aCI x I mole NaCI
58.44 g NaCI
Molarity 0.60 I moles NaCI
0.601 moles NnCI
3.00 L
moles solute
Liters of solution
f).lOI) M
4. Calculating the :\'lolarity of Ions and Atoms:
Example: Calculate the molarity of Ca::' and
chloride.
Moles:
cr in a 0.600 M solution of Calcium
CaCI:: -7 Ca::- + leT I
I
.,
0.600 M CaCh x I mole Ca 2 '
I mole CaCI::
o.oon \1 Ca2­
0.600 M CaCb x ., mole cr
1.20 M
I mole CaCI::
cr
Example: Calculate the molarity of C and H in 1.50 \1 propane. CJ-lx.
Moles:
C 311~
I
"="
3C
3
8H 8
1.50 M C,Hx x 3 mole C = 4.50 M C
I mole C 3Hx
12.0 M H
1.50 M C;Hx x 8 mole H
I mole C,H~
AI' Ch.:mistry
II and P: Chapt.:r 3 Band S
Ik Kalish
eh 4
5. Dilution: the process by which dilute solutions are made by adding solvent to
concentrated solutions
a. the amount of solute (moles) remains the same. but thc solution concentration is
altered
b. M enllc x VWile = Moil X V cli1
Example:
What is the concentration of a solution made by diluting sO.n 1111 of
1.00 M NaOH with 200. ml of water?
III.Advanced Stoichiometry
A. Allov..·s fiJr the conversion of grams of a compound to grams of an clement and
composition to detem1ine empirical and molecular f0l111ula °
0
Examples:
t. A 0.1204 gram sample of a carboxylic acid is combusted to yield 0.2147 grams of
CO-, and 0.0884 grams of water.
a. Determine the percent composition and empirical ft)Jl11ula of the compound.
Ansl\'cr: C!IIP: (';I/('(),
b. If the molecular mass is .222 gimole. vvhat is the molecular tt)l'lnula'?
c. Write the balanced chemical equation showing combustion of this compound.
Dimethylhydrazine is a C-H-N compound used in rocket fuels. When burned
completely in excess oxygen gas. a 0.312 g sample produces 0.458 g CO:, and 0.374 g
H20. The nitrogen content of a separate 0.525 g sample is cOl1\clied to 0.244 g N
a. What is the empirical t(mTIula of dimethylhydrazinc'?
. 111'\\1 C},,( '//;.\
b. If the molecular mass is 150 gmo)c. what is the molecular ttmnula'?
c. \Vrite the balanced chemical equation showing combustion of this compound.
Empirical fonnulas can be detennined from indirect analyses
In practice, a compound is seldom broken down completely to its elements in a
quantitative analysis. Instead, the compound is changed into other compounds. The
reactions separate the elements by capturing each one entirely (quantitatively) in a
separate compound whose formula is known.
In the following example, we illustrate an indirect analysis of a compound
made entirely of carbon, hydrogen, and oxygen. Such compounds bum completely
in pure oxygen-the reaction is called combustion-and the sole products are car­
bon dioxide and water. (This particular kind of indirect analysis is sometimes called
a combustion analysis.) The complete combustion of methyl alcohol (CH30H), for
example, occurs according to the following equation:
2CH30H + 30 z --- 2CO z + 4HzO
The carbon dioxide and water can be separated and are individually weighed. No­
tice that all of the carbon atoms in the original compound end up among the COz
molecules and all of the hydrogen atoms are in H 20 molecules. In this way at least
two of the original elements, C and H, are entirely separated.
We will calculate the mass of carbon in the CO 2 collected, which equal~ the
mass of carbon in the original sample. Similarly, we will calculate the mass of hy­
drogen in the H 20 collected, which equals the mass of hydrogen in the original
sample. When added together, the mass of C and mass of H are less than the total
mass of the sample because part of the sample is composed of oxygen. By subtract­
ing the sum of the C and H masses from the original sample weight, we can obtain
the mass of oxygen in the sample of the compound.
A 0.5438 g sample of a liquid consisting of only C, H, and 0 was burned in pure oxy­
gen, and 1.039 g of CO 2 and 0.6369 g o( H 20 were obtained. What is the empirical
formula of the compound?
A N A LY SIS: There are two parts to this problem. For the first part, we will find the
number of grams of C in the COz and the number of grams of H in the H 20. (This
kind of calculation was illustrated in Example 4.10.) These values represent the
number of grams of C and H in the original sample. Adding them together and sub­
tracting the sum from the mass of the original sample will give us the mass of oxygen
in the sample. In short, we have the following series of calculations:
grams CO 2 ----l> grams C
grams H 20
----l>
grams H
We find the mass of oxygen by difference.
0.5438 g sample - (g C
+ g H)
g0
In the second half of the solution, we use the masses of C, H, and 0 to calculate
the empirical formula, as in Example 4.14.
SOLUTION: First we find the number of grams of C in the COz and of H ia the H 20. In 1 mol of CO2 (44.009 g) there are 12.011 g of C. Therefore, in 1.039 g of CO . 2
we have
12.011g gCO
C
1.039 g CO2 X 44.009
"" 0.2836 g C
2
In 1 mol of H 20 (18.015 g) there are 2.0158 g of H. For the number of grams of H
in 0.6369 g of H 2 0,
2.0158 g H 0.6369 g H 20 X 18.0]5 g H 0
0.07127 g H 2
The total mass of C and H is therefore the sum of these two quantities. total mass of C and H
=
0.2836 g C
+ 0.07127 g H
-c=
0.3549 g
The difference between this total and the 0.5438 g in the original sample is the mass
of oxygen (the only other element). mass of 0
0.5438 g - 0.3549 g
= 0.1889 g 0 Now we can convert the masses of the elements to an empirical formula. ForC: ForH:
For 0:
1 molC
0.2836 g C X 12.011 g C = 0.02361 mol C
ImolH
0.07127 g H X 1.008 g H = 0.07070 mol H
1 malO
0.1889 g 0 X 15.999 gO = 0.01181 mol 0
Our preliminary empirical formula is thus C1J.02361H0.D707000.01I81' We divide all of
these subscripts by the smallest number, 0.01181.
CQ.02361
O.O1l81
HO.07070 O~ =
0.01181
0.01181
C1.999 H S.987 0 1
.
The results are acceptably close to ~H60, the answer.
Summary
actual yield (3.10) Avogadro's number, NA (3.2) chemical equation (3.7) dilution (3.11) formula mass (3.1) limiting reactant (3.9) mass percent composition (3.4) molar concentration (3.11) molarity, M (3.11) molar mass (3.3) mole (3.2) molecular mass (3.1) percent yield (3.10) product (3.7) reactant (3.7) solute (3.11) solvent (3.11) stoichiometric coefficient (3.7) stoichiometric factor (3.8) stoichiometric proportions (3.9)
stoichiometry (page 82)
theoretical yield (3.10)
Summary Molecular and formula masses relate to the masses of molecules and formula units. Mo­
lecular mass applies to molecular compounds, but only formula mass is appropriate for
ionic compounds.
A mole is an amount of substance containing a number of elementary entities equal to
the number of atoms in exactly 12 g of carbon-12. This number, called Avogadro's number,
is NA
6.022 X 1023 • The mass, in grams, of one mole of substance is called the molar
mass and is numerically equal to an atomic, molecular, or formula mass. Conversions be­
tween number of moles and number of grams of a substance require molar mass as a con­
version factor; conversions between number of grams and number of moles require the
inverse of molar mass. Other calculations involving volume, density, number of atoms or mo­
lecules, and so on may be required prior to or following the gram/mole conversion. That is,
Molar mass
Inverse of
molar mass
Formulas and molar masses can be used to calculate the mass percent compositions of
compounds. And conversely, an empirical formula can be established from the mass percent
composition of a compound;·to establish a molecular formula, we must also know the mo­
lecUlar mass. The mass percents of carbon, hydrogen, and oxygen in organic compounds can
be determined by combustion analysis.
A chemical equation uses symbols and formulas for the elements and/or compounds in­
VolVed in a reaction. Stoichiometric coefficients are used in the equation to reflect that a
chemical reaction obeys the,law of conservation of mass.
Calculations concerning reactions use conversion factors, called stoichiometric fac­
tors, that are based on stoichiometric coefficients in the balanced equation. Also required are
~lar masses and often other quantities such as volume, density, and percent composition.
e general format of a reaction stoichiometry calculation is
~ 123
124 Chapter 3 Stoichiometry: Chemical Calculations
no. mol B
no. mol A
no. mol A
no. molB
The limiting reactant determines the amounts of products in a reaction. The calculat­
ed quantity of a product is the theoretical yield of a reaction. The quantity obtained, called
the actual yield, is often less. It is commonly expressed as a percentage of the theoretical
yield, known as the percent yield. The relationship involving theoretical, actual, and percent
yield is
Percent yield
actual
X I0007
70
theoretical yield
=
The molarity of a solution is the number of moles of solute per liter of solution. Com­
mon calculations include relating an amount of solute to solution volume and molarity. So­
lutions of a desired concentration are often prepared from more concentrated solutions by
dilution. The principle of dilution is that the volume of a solution increases as it is diluted,
but the amount of solute is unchanged. As a consequence, the amount of solute per unit vol­
ume-the concentration-decreases. A useful equation describing the process of dilu­
tion is
tv1conc X
V cone
=
Mdil X
V dil
In addition to other conversion factors, stoichiometric calculations for reactions in solution
use molarity or its inverse as a conversion factor.
Review Questions
1. Explain the difference between the atomic mass of oxy­
gen and the molecular mass of oxygen. Explain how each
is determined from data in the periodic table.
2. \\'hat is Avogadro's number, and how is it related to the
quantity called one mole?
3. How many oxygen molecules and how many oxygen
atoms are in 1.00 mol 0 2?
4. How many calcium ions and how many chloride ions are
in 1.00 mol CaCh?
5. What is the molecular mass, and what is the molar mass
of carbon dioxide? Explain how each is determined from
the formula, CO 2 ,
6. Describe how the mass percent composition of a com­
pound is established from its formula.
7. Describe how the empirical formula of a compound is de­
termined from its mass percent composition.
S. \\'bat are the empirical formulas of the compounds with
the following molecular formulas?
(b) CgHl6
(e) CloHs
(d) C6H 160
(a) HP2
9. Describe how the empirical formula of a compound that
contains carbon, hydrogen, and oxygen is determined by
combustion analysis.
10. \\'bat is the purpose of balancing a chemical equation?
11. Explain the meaning of the equation
at the molecular level. Interpret the equation in terms of
moles. State the mass relationships conveyed by the equation.
12. Translate the following chemical equations into words:
(a) 2 Hig) + 02(g)
(b) 2 KCl0 3 (s)
~
~
2 Hz0(l)
2 KCI(s) + 3 Gig)
(e) 2 AI(s) + 6 HCI(aq) ~ 2 AICI 3(aq) + 3 H2(g)
13. Write balanced chemical equations to represent (a) the
reaction of solid magnesium and gaseous oxygen to form
solid magnesium oxide, (b) the decomposition of solid
ammonium nitrate into dinitrogen monoxide gas and liq­
uid water, and (e) the combustion of liquid heptane,
C 7H 16, in oxygen gas to produce carbon dioxide gas and
liquid water as the sole products.
14. \\'bat is meant by the limiting reactant in a chemical re­
action? Under what circumstances might we say that a
reaction has two limiting reactants? Explain.
15. \\'by are the actual yields of products often less than the
theoretical yields? Can actual yields ever be greater than
theoretical yields? Explain.
16. Define each of the following terms.
(a) solution
(d) molarity
(b) solvent
(e) dilute solution
(e) solute
(0 concentrated solution