Chemistry 110 07 8.19.15

General Chemistry 110-07
mon wed 5-6:15 room 209
2015 - 2016
Quinnipiac University
Chemistry 110
Professor Brielmann
Chemistry 110-07 Quinnipiac University Professor Brielmann hbrielmann@quinnipiac.edu chemistryacademy.weebly.com
quick links:
1. Introduction
2. Atoms molecules ions
3. Stoichiometry
4. Reactions in aqueous solutions
5. Gases
6. Energy in reactions
7. Electrons in atoms
8. Periodic table
9. Bonding
10. Bonding 2

date chapter homework
1 M aug 31 begin ch 1
Bring textbook 9/2 ch 1 1-
50 odd due 9/9
2 W sep 2 finish ch 1 ch 1 1-50 odd due 9/9
3 W sep 9 ch 2 ch 2 1-50 odd due 9/16
4 M sep 14 finish 2, review 1 + 2
ch 2 1-50 due 9/16; review
for test
5 W sep 16
test ch 1 and 2
begin ch 3
ch 3 1-91 odd due 9/28
6 M sep 21 complete ch 3 ch 3 1-91 odd due 9/28
7 M sep 28 begin ch 4 ch 4 1-79 odd due 10/5
8 W sep 30 finish ch 4 ch 4 1-79 odd due 10/5;
9 M oct 5 review 3,4 review for test
10 M oct 7
test ch 3 and 4, begin
5
ch 5 1-75 odd due 10/14
11 M oct 12 complete ch 5 ch 5 1-75 odd due 10/14
12 W oct 14 begin ch 6 ch 6 1-63 odd due 10/21
13 M oct 19 finish ch 6 ch 6 1-63 odd due 10/21
14 W oct 21 review 5,6 study for test
15 M oct 26
test ch 5 and 6; begin
ch 7
ch 7 1-91 odd due 11/2
date chapter homework
16 W oct 28 finish ch 7
ch 7 1-91 odd due 11/2
17 M nov 2 begin ch 8
ch 8 1-71 odd due
11/9
18 W nov 4 finish ch 8
ch 8 1-71 odd due
11/9
19 M nov 9 review ch 7 and 8 study for test
20 W nov 11
test ch 7 and 8; begin
ch 9
ch 9 1-69 odd and ch
10 1-59 odd due 10/2
21 M nov 16 ch 9 and 10
10 1-59 odd due 10/2
22 W nov 18 ch 9 and 10
10 1-59 odd due 10/2
23 M nov 30 ch 9 and 10
10 1-59 odd due 10/2
24 W dec 2 review study for test
25 M dec 7
test ch 9 and 10 final
exam review
study for final exam
26 W dec 9
last class
final exam review
study for final exam
27 M dec 14
final exam week
final exam:
Chem 110-07 Schedule
Before class: read each chapter (Chemistry: The essential concepts 7th edition Chang and Goldsby), review lecture online, complete
homework, have questions.
Bring to class room 209 Tator Hall: laptop, scientific calculator, homework, pen, paper
scoring: 5 tests (500 points, 50%), 5 homework assignments (300 points, 30%), final exam (200 points, 20%)
class website: chemistryacademy.weebly.com (blackboard used only for grading)

syllabus
Chemistry 110-07
Fall 2015
General Chemistry I
Tator Hall room 209
Meets Mondays and Wednesdays 5 - 6:15; lab section meets Mondays from 6:30 to 9:30
Instructor: Professor Harry Brielmann
Email: hbrielmann@quinnipiac.edu
Telephone: (203) 738-9348
Office Hours: by appointment
welcome to chemistry!
Course Description and Prerequisites: 3 credits. Students study the atomic theory of matter, nomenclature, chemical formulas and reaction
equations, stoichiometry, the gas laws and the kinetic molecular theory, thermochemistry, atomic structure, periodicity of the elements,
chemical bonding and molecular structure. A math placement score of 3 or higher is required to enroll in CHE 110. Students with scores
below 3 must complete recommended math courses to improve proficiency in algebraic skills before enrolling in CHE 110. Must be taken in
conjunction with CHE 110. Note: This course is designed for science majors.
Course Objectives
By the end of this course, the successful student will be able to:
Learn significant figures, conversions, and scientific notation
Appropriately use chemical terminology
Classify matter as atoms, molecules, ions, elements, compounds and mixtures and describe their structures and properties
Demonstrate an understanding of basic chemical concepts, including chemical reactions, reaction stoichiometry, thermo chemistry,
quantum theory of the atom, chemical bonds, periodicity as shown in the periodic table, atomic and molecular structures, and gases
Develop an ability to analyze and solve problems in general chemistry
Develop an appreciation of the underlying chemistry in the world around him/her
Demonstrate safe laboratory skills
Perform a selection of basic laboratory procedures in general chemistry, collect and analyze data, and clearly express results in a
laboratory report

Learning Strategies
All students are required to purchase and use the textbook and accompanying problem-solving workbook: General Chemistry: The
Essential Concepts, 7th edition, by Raymond Chang and Kenneth Goldsby. Each student should bring it to class on the first day: this
is your first assignment.
This course will faithfully follow units 1-10 of the textbook with some additional lecture material.
Attend each class. You are responsible for all announcements and material covered in class, and your grade will reflect your class
attendance, as lecture and discussion material from class will form the basis of exam questions.
Stay current with the homework assignments
Participate in class by actively listening and asking questions.
Get help when needed through your instructor, other students, or the quinnipiac learning center. .
Methods of Evaluation
5 in-class exams (100 points each 50% of grade)
5 Collected Homework (60 points each; 30% of grade )
Final Exam (200 points 25% of grade)
Required Materials
Bring to class the textbook (General Chemistry: The Essential Concepts, 7th edition, by Raymond Chang and Kenneth Goldsby and
accompanying problem-solving workbook, a scientific calculator, and materials for taking effective notes. Also bring headphones so
you can listen to screencasts without disturbing your neighbors.

+1
alkali
metals
alkaline
earth
metals
+2
transition metals:
+3 -3 -2
-1
noble
gases
halogens
1
H
hydrogen
1.008
11
Na
sodium
22.99
12
Mg
magnesium
24.31
3
Li
lithium
6.94
4
Be
beryllium
9.012
19
K
potassium
39.10
20
Ca
calcium
40.08
37 Rb
rubidium
85.47
21
Sc
scandium
44.96
22
Ti
titanium
47.90
23
V
vanadium
50.94
24
Cr
chromium
25
Mn
manganese
54.94
26
Fe
iron
55.85
38
Sr
strontium
87.62
39
Y
yttrium
88.91
40
Zr
zirconium
91.22
41
niobium
92.91
42
Mo
molybdenum
95.94
43
Tc
technetium
96.91
44
Ru
ruthenium
101.07
55
Cs
cesium
132.91
56
Ba
barium
137.33
72
Hf
hafnium
178.49
73
Ta
tantalum
180.95
74
W
tungsten
183.85
75
Re
rhenium
186.21
76
Os
osmium
190.20
87
Fr
francium
(223.02
88
Ra
radium
(226.03)
71
Lu
lutetium
174.97
103
Lr
lawrencium
262.11
104
Rf
rutherfordium
(267.12)
105
Db
dubnium
(268.13)
106
Sg
seaborgium
(171.13)
107
Bh
bohrium
(270.13)
108
Hs
hassium
(277.15)
27
Co
cobalt
58.93
28
Ni
nickel
58.71
29
Cu
copper
63.55
30
Zn
zinc
65.37
31
Ga
gallium
69.72
32
Ge
germanium
72.59
13
Al
aluminum
26.98
45
Rh
rhodium
102.91
46
Pd
palladium
106.40
47
Ag
silver
107.87
48
Cd
cadmium
112.40
49
In
indium
114.82
50
Sn
tin
118.69
33
As
arsenic
74.92
14
Si
silicon
28.09
15
P
phosphorus
30.97
51
Sb
Antimony)
121.75
77
Ir
iridium
192.22
78
Pt
platinum
195.09
79
Au
gold
196.97
80
Hg
mercury
200.59
81
Tl
thallium
204.37
82
Pb
lead
207.19
83
Bi
bismuth
208.98
109
Mtmeitnerium
(278.16)
110
Ds
darmstadtium
(281.17)
111
Rg
roentgenium
(281.16)
112
Cn
copernicium
(285.18)
113
Uut
ununtrium
(286.19)
114
Fl
flerovium
(289.19)
115
Uup
ununpentium
289.19)
5
B
boron
10.81
6
C
carbon
12.01
7
N
nitrogen
14.01
8
O
oxygen
16.00
9
F
fluorine
19.00
2
He
helium
4.00
10
Ne
neon
20.18
16
S
sulfur
32.07
17
Cl
chlorine
35.45
18
Ar
argon
39.95
34
Se
selenium
78.96
35
Br
79.91
36
Kr
krypton
83.80
52
Te
tellurium
127.60
53
I
iodine
126.90
54 Xe
xenon
131.30
84
Po
polonium
(210)
85
At
astatine
(210)
86
Rn
radon
(220)
116
Lv
livermorium
(293.20)
117
Uus
ununseptium
(294.21)
118
Uuo
ununoctium
(294.21
57
La
lanthanum
138.91
58
Ce
cerium
140.12
59
Pr
praseodymium
140.91
60
Nd
neodymium
144.24
61
Pm
promethium
(144.91)
62
Sm
samarium
150.41
63
Eu
europium
151.96
65
Tb
terbium
158.92
89
Ac
actinium
227.03
90
Th
thorium
232.04
91
Pa
protactinium
231.04
92
U
uranium
238.03
93
Np
neptunium
(237.05)
94
Pu
plutonium
(244.06)
95
Am
americium
(243.06)
96
Cm
curium
(247.07)
Dy
dysprosium
162.50
67
Ho
holmium
164.93
68
Er
erbium
167.26
69
Tm
thulium
168.93
70
Yb
ytterbium
173.04
97
Bk
berkelium
(247.07)
98
Cf
californium
(251.08)
99
Es
einsteinium
(252.08)
100 Fm
fermium
(257.10)
101
Md
mendelevium
(258.10)
102
No
(259.10)
Nb
acetate CH3CO2
- bisulfite HSO3
- chlorite ClO2
- hydroxide OH- nitrite NO2
- phosphide P3-
ammonium NH4
+ bromide Br- chromate CrO4
2- hypochlorite ClO- oxide O2- sulfate SO4
2-
bromide Br- carbonate CO3
2- cyanide CN- iodide I- perchlorate ClO4
- sulfide S2-
bicarbonate HCO3
- chlorate ClO3
- dichromate Cr2O7
2- nitrate NO3
- permanganate MnO4
- sulfite SO3
2-
bisulfate HSO4
- chloride Cl- fluoride F- nitride N3- phosphate PO4
3- thiosulfate S2O3
2-
Gd
gadolinium
157.25
64
nobelium
0
1s
2s
3s
4s
5s
6s
7s
3d
4d
5d
6d
4p
5p
6p
7p
3p
2p
4f
5f
(and NH4
+)
symbol:
solid
liquid
gas
manmade
name
average
atomic mass
(amu)
atomic
number
Sc
scandium
44.96
metal
nonmetalmetalloid
metal
nonmetal
1 valence
electron
2 valence
electrons
valence electrons: 8
4 5 6 73
(H is a nonmetal)
bromine
Group 1
Group 2
group 3 group 4 group 5 group 6 group 7 group 8 group 9 group 10 group 11 group 12
Group 13 Group 14 Group 15 Group 16 Group 17
Group 18
52.00
periodic table of the elements

chemistry formulas chapters 1-5
1. Introduction
2. Atoms, molecules and ions
no formulas
3. stoichiometry
oC = oF −32oF x
5oC
9oF
Fahrenheit (oF) from Celsius (oC)
Celsius (oC) from Fahrenheit (oF)
d =
m
V
d =density (g/mL)
m = mass (g)
v = volume (mL)
density
oF =
9oF
5oC
x oC + 32oF
=
n x molar mass of element
molar mass of compound
x 100%
% composition
example: %
hydrogen in
CH4:
4 x 1.008 g/mol
16.04 g/mol
x 100% =25.13%
=
actual yield
theoretical yield
x 100%
% yield
4. solutions
molarity (M) =
moles of solute
liters of solution
M1V1 = M2V2
5. gases
P1V1=P2V2
boyles law: relates gas pressure to volume at constant temperature
any units may be used for pressure (P) and volume (V)
T1
V1
=
T2
V2
charles law: relates gas temperature to volume at constant pressure
must use Kelvin for temperature (T)
T1
P1
=
T2
P2
gay-lussacs law: relates gas temperature to pressure at constant volume
P1V1
n1T1
=
P2V2
n2T2
combined gas law for adding gas or multiple physical changes
P = pressure
PV = nRT
ideal gas law: relates physical properties of a gas
P = pressure (atm)
V = volume (liters)
n = # of moles (mol)
d =
PM
RT
gas density law (also good for molar mass)
d = density (g/L)
P = pressure (atm)
M = molar mass (g/mol)
urms
=
3RT
M
gas velocity urms = average molecular speed (m/s)
R = 8.314 J/K
T = temp (K)
r1
r2
=
M2
M1
grahams law of diffusion
r1, r2 = diffusion rates (any
units, since they cancel)
M = molar mass (g/mol)
any units may be used for volume (V)
V = volume
T = temperature (K)
M = molarity (mol/L)
v = volume (L)
T = temperature (K)
T = temperature (K)
n = number of moles (mol);
remove if constant
P = pressure
T = temperature (K)
R = 0.0821
L atm
mol K
R = 0.0821 L atm/mol K
M = molar mass (kg/mol)
note that
J
kg
=
m
s

6. Energy
∆U = q + w
1st law of thermodynamics DU = change in the internal energy (J)
q = the heat loss J)
w = work done J)
work of gas expansion and compression at constant external pressure
w = work done (j)
P = external pressure (atm)
DV = change in volume (L)
q = ms∆t
specific heat formula
q = work done (j)
m = mass (g)
s = specific heat (J/gOC)
DT = temperature change (OC)
q = C∆t
heat change in terms of heat capacity
∆Horxn = SnDHo
f products −SmDHo
f reactants
change in standard enthalpy of reaction (∆Horxn)
w = −P∆V
enthalpy change of a gas expansion
∆H = ∆U + P∆V
energy and enthalpy change of a gaseous reaction at constant temperature
∆U = ∆H − RT∆n
heat capacity
C = heat capacity (J/OC)
m = mass (g)
s = specific heat (J/gOC)
C = ms
DH = enthalpy change (j)
DU = internal energy change (j)
note that
101.3 J = 1
L . atmP = pressure (atm)
DV = volume change (L)
note that 101.3 J
= 1 L . atm
DU = change in the internal energy (J)
DH = enthalpy change (J)
R = gas constant = 8.314
J
K
T = temperature (K)
Dn = moles of products – moles of reactants
q = work done (j)
C = heat capacity (J/OC)
DT = temperature change (OC)
S = sum of
n = coefficients for products
m = coefficients for reactants
DHo
f = change in standard enthalpy of formation
Planck’s formula
E = hn
E = energy (j)
h = Planck’s constant
=6.636 x 10-34 j sec
v = frequency (
1
s
)
wavelength-Planck combined
E = h
c
l
E = energy (j)
=6.636 x 10-34 j sec
c = the speed of light
= 3,00 x 108 m/s
l = wavelength (m)
the photoelectric effect
KE = hn - W
KE = kinetic energy of ejected electron (J)
h = Planck’s constant = 6.636 x 10-34 J sec
v = frequency (
1
s
)
W = work function (J)
Rydberg formula: energy of an electron
in the nth state for a hydrogen atom (En)
En = -RH(
1
n2 )
RH = rydberg constant = 2.18 x 10-18 J
n = principal quantum number = shell
number
Balmer formula: Change in energy
(DE)energy of an emitted or absorbed
photon as it travels between shells
∆E = hn = RH(
1
ni
2 ) − (
1
nf
2 )
=6.636 x 10-34 j sec
v = frequency (
1
s
)
RH = rydberg constant = 2.18 x 10-18 J
n = principal quantum number
= shell number
de Broglie relationship between a particle
and a wave
l=
h
mu
l = wavelength (m)
h = Planck’s constant = 6.636 x 10-34 J sec
m = mass (kg)
u = speed (
m
s
) = 3.00 x 108 m/s
u = speed (
m
s
) = 3.00 x 108 m/s
l = wavelength (m)
v = frequency (
1
𝑠
)
wavelength
u = ln
chemistry formulas chapters 6-7
7. The atom

8. the periodic table
Zeff = Z −s
effective nuclear charge (Zeff)
Zeff = effective nuclear charge
Z = actual nuclear charge
s = shielding constant
enthalpy change for a reaction
9. bonding I
∆Ho rxn = SBE reactants − SBE products
10. bonding II: no equations
∆Ho rxn = change in standard enthalpy
for a reaction
S = sum of
BE = bond energy (J)

Introduction to Chemistry
1.1 The study of chemistry
1.2 The Scientific Method
1.3 Classification of matter
1.4 Physical and Chemical Properties of Matter
1.5 Measurement
1.6 Handling Numbers
1.7 Dimensional Analysis in Solving Problems
9
Chapter 1
1. Introduction
oC = oF −32oF x
5oC
9oF
Fahrenheit (oF) from Celsius (oC)
Celsius (oC) from Fahrenheit (oF)
d =
m
V
d =density (g/mL)
m = mass (g)
v = volume (mL)
density
oF =
9oF
5oC
x oC + 32oF

The study of chemistry
why chemistry is awesome:
10
yes, we blow stuff up
provide molecular answers
make a difference: cancer… pain… energy…
gateway to great fields- medicine, engineering,…
chemistry is : a chemical is:
the study of matter
and how it changes
a pure form of
matter (a substance)
core learning- set up so
you can test yourself 
ok…what is matter?
anything that has mass and takes up space.
is it matter?
a perfect vacuum?
a black hole?
religion?energy?
an idea?you? air?yes yes
no
no no theoretically, no
mass: how much matter is in an
object. Location independent.
Weight is gravity dependent.
compare mass and weight:
o
o

organic
biochemical
analytical
physical
inorganic
medicinal
carbon-based:
ex: plastics
not carbon-based
ex: mining
physical change:
ex: reaction rates
analysis:
Medicines:
ex: viagra
forensic
crime solvers:
what do chemists do? they all study matter
the branches of chemistry
chemistry of life:
what kind of chemist
am I?chemists
I make aspirin: I am a
_______________ chemist.medicinal
I make plastics: I am a
____________ chemist.polymer
I analyze: I am an
_____________ chemist
I study physical processes: I
am a ____________ chemist
I study the chemistry
of fruit flies, so I am
a ____chemist
I study gold; this is
____________ chemistry
analytical
bio
inorganic
I solve crimes using
chemistry : I am a
____________ chemist.forensic
where does chemistry fit in?
social science biologyphysics math chemistry
chemistry is the __________science
basic applied
central
physical

the “ohec” scientific method
controls:
standards
for comparison
O
HE
C
example:
pain medication study positive control:
negative control:
ex: aspirin
ex: sugar pill
benchmark
placebo: no effect prevents false positives
oh heck I know that
all experiments need these:
supplemental terms: qualitative (___________),
quantitative (____________),
a tentative explanation for a set of observations.
A law is a concise and permanent statement of a
relationship between phenomena
A theory is a unifying principle that explains a
body of facts and/or laws
hypothesis:
theory:
law
no numbers
numbers

element, compound, or mixture?
element element (Cu) compound (SiO2)mixture mixture
gold ocean milk copper glass
classification of matter
matter
Classification of matter Quinnipiac University Chemistry 110 Professor Brielmann
13
mixture:
nothing is pure in this world. what can we say about mixtures?
looks pure but isn’t doesn’t look pure
homogeneous mixture heterogeneous mixture
one thing visible multiple things visible
“a solution” “a mixture”
classify a drop of blood: classify granite:
either way it’s still a mixture…until it is separated we don’t know much about it.
suggest a solution:
gas-gas:
liquid-liquid:
solid-solid:
solid-liquid:
air
soda
gasoline; vinegar
salt water
brass; steel
gas-liquid:
a substance that cannot be separated into simpler substances by chemical means:
a substance composed of atoms of
two or more elements chemically
united in fixed proportions.
a pure form of matter:
a sample containing more than one substance:
compound
element
substance

• want pure:
• purification: how would you separate these mixtures?
• oil • dry sand
• sugar
• method:
• each pure
oil
• each
pure
solute
• you have:
• oil/water • wet sand • sugar/water • oils • ???
• decant • filter • distill • chromatograph• crystallize
14

physical vs. chemical
chemical property:
physical property:
reacts- new substance(s) formed
no reaction- same substance all along.
burning
rusting
light bending
rising
molding
flattening
stretching observing
boiling
shining
crystallizing
melting
suggest the property responsible and if it is physical or chemical:
boiling point (physical) reactivity (chemical) melting point (physical) density (physical)
flammability (chemical) solubility (physical) malleability (physical)
decomposition
(chemical)
ductility (physical) refractive index (physical) color (physical) luster (physical)

extensive and intensive properties
melting point
density
refractive
index
mass
toxicity
in
in
in
ex
ex
crystalline (shiny) amorphous (dull)
• amount-dependent
• ”extent”
doesn’t matter how much
crystalline or
amorphous?
16

melt
boilcondense
deposit
cold
hot
liquid
solid
gas
phase
changes
freeze
sublime
17
ionizedeionize
plasma

s.i. units: le systeme internationale
18
kilogram
mass
second
time
meter
distance
candela
brightness
mole
amount
ampere
current
Kelvin
temperature
combined Units:
speed acceleration densityvolume density formula:
measurement
m
s
m
s2
cm3
g
mL
density =
m
v

13.5 g of aluminum has a volume of 5.00 mL. Density?
Based on that density, What is the mass of
2.00 mL of aluminum?
19
m
dv
divide
multiply
m = vd
= (2.00 mL)(2.70 g/mL)
= 5.40 g
density calculations

giga (G)billion (109)
million (106)
thousand (103) kilo (k)
centi (c)hundredth 10-2
milli (m)thousandth (10-3)
millionth (10-6)
billionth (10-9)
unit prefixes list the common unit prefixes and their abbreviations
mega (M)
micro (m)
nano (n)
3000 nanograms = 1 ____________
3 x 102 meters = _____________ kilometers
3500 grams = _______________ kilograms
35 seconds = ______________ microseconds
3.42 liters = ______________ milliliters
0.03 seconds = ______________ milliseconds
3 kilograms = ______________ milligrams
3 kilograms = ______ grams
3 milliliters = _____________ liters
3 micrograms = ______________ grams = _______________grams
300 kilometers = ______________ meters
3 milliseconds = ______________ seconds
3 kilograms = 0.003 ___________________
______________ grams = 34.82milligrams
3 x 10-6 liters = 3 __________________liters
3 x 108 bytes = ______________ megabytes
3 x 10-3 seconds = _______________ milliseconds
3000
0.003
3,000,000
300,000
0.03482
0.003
3,000,000
30
3,420
0.000035
microgram
0.3
3000
3000
0.000003
megagrams
3.500
3 x 106

temperature
21
25oC = ?K
Kelvin….why?
0F to 0C:
-40oC = ?0F
-40oF = ?0C
(-40 oC x (9/5) + 32 = -40 0F
-32 = -72, x5 = -360, /9 = -40 = -40 0C
K to 0C:
= 298 K
S.I. unit:
1.5 Measurement Chemistry 110 Professor Brielmann

scientific notation
for big and small numbers
10,000
22
Always 1-10 Always 10x
Draw a line to make it
between 1 and 10; count
to decimal point.
Example:
= 1 x 104
-2860
= 2.12 x 102
.0097 = 9.7x 10-3
602,000,000,000,000,000,000,000
= 6.02 x 1023
= -2.86 x 103
212
= 9.742 x 10-4
.0009742
Try some:

(2 x 101)(1 x 101)= ?
23
(3 x 10-2) x (-4.2 x 10-4) = ?
how do you enter 6.02 x 1023?
2E1x1E1
Enter 3E-2x-4.2E-4
entering scientific data on your calculators: the ee button
find your ee button or equivalent (E, e, ee, exp)
6.02E23
translate: 3E-0.42
3 x 10-0.42
= -1.26 x 10-5 or -.0000126
find change sign button (+/-)
= 200 or 2E2

accurate, precise, or both?
24
accurate:
not accurate
precise:
accurate
precise
not accurate
not precisecorrect
qualitative or quantitative?
several arrows 3 arrows
qualitative: no numbers quantitative: numbers
consistent repeatable
ok…but why are they used interchangeably so often??
accuracy:
precision:
:
The quality, condition, or fact of being exact and accurate.
The quality, of being correct or precise.
1.6 Handling Numbers Chemistry 110 Professor Brielmann

significant figures: measuring using graduations
1.when measuring
include the known digits… plus one estimated digit.
25
volume?
32.0 mL
= 32.0 cm3
2. why is it important to line up level to the meniscus?
to minimize parallax
how does this device minimize parallax??
mL
If reflection is seen there is parallax
1.6 Handling Numbers Chemistry 110 Professor Brielmann

significant figures: how precise is your data?
26
# sig. figs (sf).number why
32 2 “non-zero numbers are
always significant”
0.0323 3 “leading zeroes are never
significant”
3.004 4 “sandwiched zeroes are always
significant”
300 1
“trailing zeroes are only
significant if there is a
decimal place”
300. 3
300.20 5
.030690
dump
keep
keep if decimal present
5sf
Summary:
rounding
+,-
x, /
5 or >: go up
S.f. based on decimal places
Sf based on sf
2.25 to 2 sf:
4.16+ 3.3 =
999 /333 = 3.00
7.5
2.3
infinite sig. figs. for exact
numbers…3
oranges1.7 Dimensional Analysis Chemistry 110 Professor Brielmann
1

significant figures: the hard parts
Which numbers have infinite significant figures?
Defined and counted numbers
12 inches = 1 foot
When should I round?
What if addition and multiplication are in the same problem?
Only at the end whenever possible
Enter entire calculations into a calculator when possible to minimize rounding; use ANS key
If rounding intermediate numbers seems unavoidable, carry along one extra sig fig
Will applying these rules lead to one correct answer?
No, but the questions are written to account for this: choose the closest answer
4 students
Enter it all into you calculator then keep the fewest number of significant figures at the end
And finally, Our general rule
(especially for AP chem):
Answer using 3 sig figs unless
specifically told otherwise

unit conversions
a. 7.25 dollars = _____ quarters
28
x
1. start with
what you are given
3. multiply using
conversion factors2. write the final units
4. cancel your
units.
7.25 dollars 4 quarters
quarters
=
b. 1,285 quarters = _____ dollars
dollar
x1,285 quarters 1 dollar
dollars=
4 quarters
1 mile
1609 meters
x
1 hour
3600 seconds
meters
=
second
. 65 miles/hour = ______meters/second (1609 meters = 1 mile; 3600 seconds = 1 hour
65 miles
hour
x
d. Most gases occupy 24 liters per mole at room temperature. Given that carbon
dioxide has a molar mass of 44 grams per mole, what is the density of carbon dioxide
at room temperature in grams per liter?
1 mole
44 grams
x
1 mole
24 liters
grams
=
liter
29
321.25
29
1.8
1.1 Introduction to Chemistry Chemistry 110 Professor Brielmann 1

chapter 1 solved
(even)problems
29

1.2a Solder is an alloy made of tin and lead that is used in electronic circuits. A certain
solder has a melting point of 224°C. What is its melting point in degrees Fahrenheit?
1.2b. Helium has the lowest boiling point of all the elements at -2452°F. Convert this
temperature to degrees Celsius.
1.2c. Mercury, the only metal that exists as a liquid at room temperature, melts at 238.9°C.
Convert its melting point to Kelvins.
1.2 Temperature conversions

1.4 What is the difference between a physical property and a chemical
property?
A physical property is any property of a substance that can be observed
without transforming the substance into some other substance.
A chemical property is any property of a substance that cannot be studied
without converting the substance into some other substance.

Define these terms:
(a) element,
b) compound.
A compound is a substance composed of
atoms of two or more elements chemically
united in fixed proportions
An element is a substance that cannot be
separated into simple substances by chemical
means.

1.8 Does each of these describe a physical change or a chemical change?
(a) The helium gas inside a balloon tends to leak out after a few hours.
(b) A flashlight beam slowly gets dimmer and finally goes out.
(c) Frozen orange juice is reconstituted by adding water to it.
(d) The growth of plants depends on the sun's energy in a process called
photosynthesis.
(e) A spoonful of table salt dissolves in a bowl of soup.
Physical change. Helium is not changed in any way by leaking
out of the balloon.
Chemical change in the battery.
Physical change. The orange juice concentrate can be
regenerated by evaporation of the water.
Chemical change. Photosynthesis changes water, carbon dioxide,
and so on, into complex organic matter.
Physical change. The salt can be recovered
unchanged by evaporation.

1.10. Which of these properties are intensive and which are extensive?
(a) area,
(b) color,
(c) density.
This is an extensive property- more area leads
to a larger number
This is an intensive property- a small amount of blue is
the same color as a large amount of the same color
This is an intensive property- water has a
density of one gram per milliliter regardless
of whether it is a drop or a gallon.

1.12. Classify each of these as an element or a compound:
(a) sodium chloride (table salt),
(b) helium,
(c) alcohol,
(d) platinum
Compound: contains both the sodium cation
and the chloride anion
Element: contains only helium
Compound: contains O, H, and and more…
Element (Pt)

1.14. Write the numbers for these prefixes:
(a) mega-,
(b) kilo-,
(c) deci-,
(d) centi-,
(e) milli-,
(f) micro-,
(g) nano-,
(h) pico-.
million
hundredth
tenth
thousandth
millionth
billionth
trillionth
thousand

1.16. Write the equations for converting degrees Celsius to degrees Fahrenheit and
degrees Fahrenheit to degrees Celsius.
5 C
? C ( F 32 F)
9 F

 =    

9 F
? F C + 32 F
5 C
 
 =     

1.18. Mercury is the only metal that is a liquid at room temperature. Its density is 13.6
g/mL. How many grams of mercury will occupy a volume of 95.8 mL?
13.6 g
95.8 mL
1 mL
=  = 3
mass of Hg 1.30 10 g

1.20. (a) Convert the following temperatures to kelvin:
(i) 113°C, the melting point of sulfur,
(ii) 37°C, the normal body temperature,
(iii) 357°C, the boiling point of mercury.
(b) Convert the following temperatures to degrees Celsius:
(i) 77 K, the boiling point of liquid nitrogen,
(ii) 4.2 K, the boiling point of liquid helium,
(iii) 601 K, the melting point of lead.
1 K
K ( C 273 C)
1 C
=   

1 K
K ( C 273 C)
1 C
=   

(a)
386 K
310 K
630 K
(add 273)
(subtract 273)
-269 OC
196 OC
328 OC

1.22. Express these numbers in scientific notation:
a) 0.749
(b) 802.6
(c) 0.000000621.
7.49 x 10-1
8.026 x 102
6.21 x 10-7

1.24. Convert these numbers to nonscientific notation:
(a) 3.256 × 10−5
(b) 6.03 × 106
0.00003256
6,030,000

1.26. Express the answers to these operations in scientific notation:
0.0095 + (8.5 × 10
−3
)
653 ÷ (5.75 × 10
−8
)
850,000 − (9.0 × 10
5
)
(3.6 × 10
−4
) × (3.6 × 10
6
)
enter into scientific calculator for ex
0.0095 + (8.5 × 10−3)
= 0.0095 + 8.5E-3
= .018
= 1.8 x 10-35  104
1.3  103
1.14  10
10
1.80 102

1.28. What is the number of significant figures in each of these measured quantities?
(a) 40.2 g/cm3
(b) 0.0000003 cm
(c) 70 min
(d) 4.6 × 1019 atoms.
3
1
1
2

1.30. Carry out these operations as if they were calculations of experimental results, and
express each answer in the correct units and with the correct number of significant
figures:
a. 7.310 km ÷ 5.70 km
b. (3.26 × 10
−3
mg) − (7.88 × 10
−5
mg)
c. (4.02 × 10
6
dm) + (7.74 × 10
7
dm)
d. (7.8 m − 0.34 m)/(1.15 s + 0.82 s)
1.287.310 km
1.28
5.70 km
= 3
0.00318 12 mg to 3 significant figures = 0.00318 mg or 3.18  103 mg
8.14  107dm (note that this has two decimal places
(7.8 m 0.34 m) 7.5 m
(1.15 s 0.82 s) 1.97 s

= =

3.8 m / s

1.32. Carry out these conversions:
(a) 242 lb to milligrams,
(b) 68.3 cm3 to cubic meters.
3
453.6 g 1 mg
242 lb
1 lb 1 10 g
=   =

8
? mg 1.10 10 mg
3
2
3 1 10 m
68.3 cm
1 cm
 
=  = 
 
 
3 5 3
? m 6.83 10 m


1.34. Three students (A, B, and C) are asked to determine the volume of a sample of
methanol. Each student measures the volume three times with a graduated cylinder.
The results in milliliters are A (47.2, 48.2, 47.6); B (46.9, 47.1, 47.2); C (47.8, 47.8,
47.9). The true volume of methanol is 47.0 mL. Which student is the most accurate?
Which student is the most precise?
Calculating the mean for each set of data, we find:
Student A: 47.7 mL
Student B: 47.1 mL
Student C: 47.8 mL
From these calculations, we can conclude that the volume
measurements made by student B were the most accurate of the
three students. The precision in the measurements made by both
students B and C are fairly high, although the measurements by
student C were the most precise.

1.36. A slow jogger runs a mile in 13 min. Calculate the speed in
(a) in/s,
(b) m/min,
(c) km/h. (1 mi = 1609 m; 1 in = 2.54 cm.)
1 mi 5280 ft 12 in 1 min
13 min 1 mi 1 ft 60 s
=    =? in/s 81 in/s
1 mi 1609 m
13 min 1 mi
=  = 2
? m/min 1.2 10 m/min
1 mi 1609 m 1 km 60 min
13 min 1 mi 1000 m 1 h
=    =? km/h 7.4 km/h

1.38. Carry out these conversions:
(a) 1.42 light-years to miles (a light-year is an astronomical measure of
distance—the distance traveled by light in a year, or 365 days)
(b) 32.4 yd to centimeters,
(c) 3.0 × 1010cm/s to ft/s
(d) 47.4°F to degrees Celsius
(e) −273.15°C (the lowest temperature) to degrees Fahrenheit
(f) 71.2 cm3 to m3 (g) 7.2 m3 to liters.
8
365 day 24 h 3600 s 3.00 10 m 1 mi
1.42 yr
1 yr 1 day 1 h 1 s 1609 m

     = 12
8.35 10 mi
36 in 2.54 cm
32.4 yd
1 yd 1 in
  = 3
2.96 10 cm
10
3.0 10 cm 1 in 1 ft
1 s 2.54 cm 12 in

  = 8
9.8 10 ft/s
5 C
(47.4 32.0) F
9 F

=    =

? C 8.6 C 
9 F
273.15 C 32 F
5 C
 
=      =  
? F 459.67 F  
3
3 0.01 m
71.2 cm
1 cm
 
=  = 
 
3 5 3
? m 7.12 10 m

3
3
3
1 cm 1 L
7.2 m
0.01 m 1000 cm
 
=   = 
 
3
? L 7.2 10 L

1.40. The density of ammonia gas under certain conditions is 0.625 g/L. Calculate
its density in g/cm3
3
0.625 g 1 L 1 mL
1 L 1000 mL 1 cm
=   = 4 3
density 6.25 10 g/cm


1.42. In 2010 about 132 billion pounds of sulfuric acid were produced in the United
States. Convert this quantity to tons.
9
3
1 ton
(132 10 lb of sulfuric acid)
2.00 10 lb
  =

7
6.60 10 tons of sulfuric acid

1.44. In the determination of the density of a rectangular metal bar, a student made
the following measurements: length, 8.53 cm; width, 2.4 cm; height, 1.0 cm; mass,
52.7064 g. Calculate the density of the metal to the correct number of significant
figures.
52.7064 g
=
(8.53 cm)(2.4 cm)(1.0 cm)
= = 3
density 2.6 g/cm
m
V

1.46. A cylindrical glass bottle 21.5 cm in length is filled with cooking oil of density
0.953 g/mL. If the mass of the oil needed to fill the bottle is 1360 g, calculate the
inner diameter of the bottle. Volume of cylinder = r2h
mass of oil
volume of oil filling bottle
density of oil
=
3 3 31360 g
volume of oil filling bottle 1.43 10 mL 1.43 10 cm
0.953 g/mL
= =  = 
volume
=
 
r
h
3 3
1.43 10 cm
4.60 cm
21.5 cm
r

= =
 
The inner diameter of the bottle equals 2r.
Bottle diameter = 2r = 2(4.60 cm) = 9.20 cm

1.48. A silver (Ag) object weighing 194.3 g is placed in a graduated cylinder containing
242.0 mL of water. The volume of water now reads 260.5 mL. From these data
calculate the density of silver.
Volume of silver = 260.5 mL  242.0 mL = 18.5 mL = 18.5 cm3
3
194.3 g
18.5 cm
= = 3
density 10.5 g/cm

1.50. How far (in feet) does light travel in 1.00 ns? The speed of light is 3.00 ×
108 m/s.
8
9
1 s 3.00 10 m 100 cm 1 in 1 ft
1.00 ns
1 s 1 m 2.54 cm 12 in10 ns

     = 0.984 ft

chapter 1 assigned
(odd)problems
55

1.1 Define these terms:
(a) matter:
(b) mass:
(c) weight:
(d) substance:
(e) mixture.
1.3 Give an example of
a homogeneous mixture:
and an example of a heterogeneous mixture:
1.5 Give an example of
an intensive property:
and an example of an extensive property:
1.7 Do these statements describe chemical or physical properties?
(a) Oxygen gas supports combustion: ___
(b) Fertilizers help to increase agricultural production:___
(c) Water boils below 100°C on top of a mountain:___
(d) Lead is denser than aluminum: ___
(e) Uranium is a radioactive element: ___
Name:____________________________________________________
Chapter 1 problems. 10 points
1. Please write your answers neatly or type. (20%)
2. Show all your work including cancelled units (40%)
3. Circle your correct answers (40%)

1.9 Which of these properties are intensive and which are extensive?
(a) length:__
(b) volume:___
(c) temperature:___
(d) mass:___
1.11 Classify each of these substances as an element or a compound:
(a) hydrogen:___
(b) water:___
(c) gold: ___
d) sugar: ___
1.13 Give the SI units for expressing these:
(a) length:___
(b) area:___
(c) volume:__
(d) mass:___
(e) time:___
(f) force:___
(g) energy>___
(h) temperature:___
1.15 Define density:
What units do chemists normally use for density?
Is density an intensive or extensive property?
1.17 A lead sphere has a mass of 1.20 × 104 g, and its volume is 1.05 × 103 cm3. Calculate the density of lead. This is the first
calculation: for all calculations be sure to show all work including cancelled units, and circle you answer.

1.19 (a) Normally the human body can endure a temperature of 105°F for only short periods of time without permanent damage to the
brain and other vital organs. What is this temperature in degrees Celsius? Be sure to show your work as for the previous problem…and for
the rest too, thanks.
(b) Ethylene glycol is a liquid organic compound that is used as an antifreeze in car radiators. It freezes at −11.5°C. Calculate its freezing
temperature in degrees Fahrenheit.
(c) The temperature on the surface of the sun is about 6300°C. What is this temperature in degrees Fahrenheit? (d) The ignition
temperature of paper is 451°F. What is the temperature in degrees Celsius?
1.21 Express these numbers in scientific notation:
(a) 0.000000027
(b) 356
(c) 0.096
1.23 Convert these numbers to nonscientific notation:
(a) 1.52 × 104
(b) 7.78 × 10−8
1.25 Express the answers to these operations in scientific notation:
a. 145.75 + (2.3 × 10−1)
b. 79,500 ÷ (2.5 × 102)
c. (7.0 × 10−3) − (8.0 × 10−4)
d. (1.0 × 104) × (9.9 × 106)

1.27 What is the number of significant figures in each of these measured quantities?
(a) 4867 miles: ____
(b) 56 mL:____
(c) 60,104 tons: ____
(d) 2900 g: ____
1.29 Carry out these operations as if they were calculations of experimental results, and express each answer in the correct units and
with the correct number of significant figures:
a. 5.6792 m + 0.6 m + 4.33 m = _______
b. 3.70 g − 2.9133 g = ________
c. 4.51 cm × 3.6666 cm = _______
d. (3 × 104 g + 6.827 g)/(0.043 cm3 − 0.021 cm3) = _____
1.31 Carry out these conversions. As always show your work including cancelled units: No credit for only showing the answer
(a) 22.6 m to decimeters,
(b) 25.4 mg to kilograms.
1.33 The price of gold on a certain day in 2009 was $932 per troy ounce. How much did 1.00 g of gold cost that day? (1 troy ounce =
31.03 g.)

1.35 Three students (X, Y, and Z) are assigned the task of
determining the mass of a sample of iron. Each student makes
three determinations with a balance. The results in grams are X
(61.5, 61.6, 61.4); Y (62.8, 62.2, 62.7); Z (61.9, 62.2, 62.1). The
actual mass of the iron is 62.0 g. Which student is the least
precise? Which student is the most accurate?
1.37 Carry out these conversions:
(a) A 6.0-ft person weighs 168 lb. Express this person's
height in meters and weight in kilograms. (1 lb = 453.6 g; 1
m = 3.28 ft.)
(b) The current speed limit in some states in the United
States is 70 miles per hour. What is the speed limit in
kilometers per hour?
(c) The speed of light is 3.0 × 1010 cm/s. How many
miles does light travel in 1 hour?
(d) Lead is a toxic substance. The “normal” lead content in
human blood is about 0.40 part per million (that is, 0.40 g
of lead per million grams of blood). A value of 0.80 part per
million (ppm) is considered to be dangerous. How many
grams of lead are contained in 6.0 × 103 g of blood (the
amount in an average adult) if the lead content is 0.62 ppm?
1.39 Aluminum is a lightweight metal (density = 2.70 g/cm3) used
in aircraft construction, high-voltage transmission lines, and foils.
What is its density in kg/m3?
?

1.41 Which of these describe physical and which describe chemical properties?
(a) Iron has a tendency to rust.
(b) Rainwater in industrialized regions tends to be acidic.
(c) Hemoglobin molecules have a red color.
(d) When a glass of water is left out in the sun, the water gradually disappears.
(e) Carbon dioxide in air is converted to more complex molecules by plants during photosynthesis.
1.43 Suppose that a new temperature scale has been devised on which the melting point of ethanol (−117.3°C) and the boiling point of
ethanol (78.3°C) are taken as 0°S and 100°S, respectively, where S is the symbol for the new temperature scale. Derive an equation relating
a reading on this scale to a reading on the Celsius scale. What would this thermometer read at 25°C?
1.45 Calculate the mass of each of these:
(a) a sphere of gold of radius 10.0 cm [volume = r3; the density of gold = 19.3 g/cm3]

(b) a cube of platinum of edge length 0.040 mm (the density of platinum = 21.4 g/cm3),
(c) 50.0 mL of ethanol (the density of ethanol = 0.798 g/mL).
1.47 This procedure was carried out to determine the volume of a flask. The flask was weighed dry and then filled with water. If the masses
of the empty flask and the filled flask were 56.12 g and 87.39 g, respectively, and the density of water is 0.9976 g/cm3, calculate the
volume of the flask in cubic centimeters.
1.49 The experiment described in Problem 1.48 is a crude but convenient way to determine the density of some solids. Describe a similar
experiment that would enable you to measure the density of ice. Specifically, what would be the requirements for the liquid used in your
experiment

Atoms, molecules, and ions
2.1 Atomic theory
2.2 The structure of the atom
2.3 Atomic Number, Mass Number, and Isotopes
2.4 The Periodic Table
2.5 Molecules and Ions
2.6 Chemical Formulas
2.7 Naming Compounds
2.8 Introduction to Organic Chemistry
63
2. Atoms, molecules, and ions Chemistry 110 Professor Brielmann
Chapter 2
2. Atoms, molecules and ions
no formulas

a brief history of the atom
400 BC – 1000 AD
symbol idea, source
aristotle
democritus
inventor
earth, air, fire, and water
greek writings
the atom
no written record?
split the atom
the chemistry of happiness
400 BC
400 BC
1000 AD
their evidence: nothing!
ghazali
2.1 Atomic Theory Chemistry 110 Professor Brielmann

Dalton’s Atomic Theory (1808)
65
1. Elements are composed of extremely
small particles called atoms.
2. All atoms of a given element are
identical, having the same size,
mass and chemical properties. The
atoms of one element are different
from the atoms of all other
elements.
3. Compounds are composed of atoms
of more than one element.
combined in small whole numbers.
4. A chemical reaction involves only
the separation, combination, or
rearrangement of atoms; it does
not result in their creation or
destruction.
Describe four aspects:
Describe in one sentence:
The universe is made up of atoms
which are identical for each element
that can combine in small whole
numbers to form compounds

66
C4H10 + O2 2 13 CO28 + 10 H2O
4. A chemical reaction involves only the separation, combination, or
rearrangement of atoms; it does not result in their creation or destruction.
Daltons theory applied: what is a chemical reaction?
balance:

1897: Thomsons cathode ray tube experiments
identify the components
1. Bends “light”
2. Moves a propeller
3. m/c indicates 1000X smaller than a
hydrogen atom (-1.76 x 108 C/g)
proposes: The Electron
thomson and his plum pudding model
battery vacuum tube
light…?
+ pole of magnet
2.2 The structure of the atom Chemistry 110 Professor Brielmann
3 major findings:

68
e- charge = -1.60 x 10-19 C
Thomson’s charge/mass of e- = -1.76 x 108 C/g
e- mass = 9.10 x 10-28 g
1923 Millikan’s Experiment
directly measured electron charge.
how did Millikan determine the mass of an electron?

69
(uranium compound)
Types of Radioactivity

1907:where are the electrons?
rutherfords gold foil experiments
Most particles go right through
the ultra-thin Gold foil.
1. Electrons are outside the
nucleus
2. The atom is mostly space
But..No idea of electron
organizationRutherford
2.2 The structure of the atom Chemistry 110 Professor Brielmann
observation
design and three components:
conclusions

Chadwick’s Experiment (1932)
(1935 Noble Prize in Physics)
71
H atoms: 1 p; He atoms: 2 p
mass He/mass H should = 2
measured mass He/mass H = 4
a + 9Be 1n + 12C + energy
neutron (n) is neutral (charge = 0)
n mass ~ p mass = 1.67 x 10-24 g
what data suggested neutrons might exist?
How did Chadwick discover the missing mass?
What is the charge and approximate mass of a neutron

72
mass p ≈ mass n ≈ 1840 x mass e-
compare the mass and charge of protons, neutrons, and electrons:

atomic bookkeeping: p+, no, and e-
98.9% C-12
1.1 % C-13
“red #”average
atomic mass
U-238:if #no/#p+ >
1.5 run!
band of
stability
7 p+, 10 e- is____#p+ < #e-anion
20 p+, 18 e- is____#p+ > #e-cation
Na+ has __p+, __e-#p+ = #e-ion
He-4 and He-5 are isotopes#no variesisotope
2He or He-5 has __ p+, __no
#p+ + #nomass #
2He# p+; blue #atomic #
examplemeansterm
Fix 32S:
16S
5
2 3
Ca2+
N3-
a.a.m. = (0.989)(12)
+ (0.011)(13) = 12.011
144/92 = 1.56 = run!
11 10
F- and Neon both have ___ electrons: They are _____________.10 isoelectronic
2.3 Atomic Number, Mass Number,, and Isotopes Chemistry 110 Professor Brielmann

organization of the periodic table
74
it’s about the electrons:
valence electrons; orbitals
+1
Alkali
metals
Alkaline
earth
metals
+2
Transition metals: 2 valence electrons
+3
+4,
-4 -3 -2
-1
Noble
gases
halogens
1
H
hydrogen
1.01
11
Na
sodium
22.99
12
Mg
magnesium
24.31
3
Li
lithium
6.94
4
Be
beryllium
9.01
19
K
potassium
39.10
20
Ca
calcium
40.08
37
Rb
rubidium
85.47
21
Sc
scandium
44.96
22
Ti
titanium
47.90
23
V
vanadium
50.94
24
Cr
chromium
52.00
25
Mn
manganese
54.94
26
Fe
iron
55.85
38
Sr
strontium
87.62
39
Y
yttrium
88.91
40
Zr
zirconium
91.22
41
niobium
92.91
42
Mo
molybdenum
95.94
43
Tc
technetium
96.91
44
Ru
ruthenium
101.07
55
Cs
cesium
132.91
56
Ba
barium
137.33
71
Lu
Lutetium
174.97
72
Hf
hafnium
178.49
73
Ta
tantalum
180.95
74
W
tungsten
183.85
75
Re
rhenium
186.21
76
Os
osmium
190.20
87
Fr
francium
223.02
88
Ra
radium
226.02
103
Lr
lawrencium
262.11
104
Rf
rutherfordium
261.11
105
Db
dubnium
262.11
106
Sg
seaborgium
263.12
107 Bh
bohrium
264.12
108
Hs
hassium
265.13
27
Co
cobalt
58.93
28
Ni
nickel
58.71
29
Cu
copper
63.55
30
Zn
zinc
65.37
31
Ga
gallium
69.72
32
Gegermanium
72.59
13
Al
aluminum
26.98
45
Rh
rhodium
102.91
46
Pd
palladium
106.40
47
Ag
silver
107.87
48
Cd
cadmium
112.40
49
In
indium
114.82
50
Sn
tin
118.69
33
As
arsenic
74.9
2
14
Si
silicon
28.09
15
P
phosphorus
30.97
51 Sb
Antimony)
121.75
77
Ir
iridium
192.22
78
Pt
platinum
195.09
79
Au
gold
196.97
80
Hg
mercury
200.59
81
Tl
thallium
204.37
82 Pb
lead
207.19
83
Bi
bismuth
208.98
109
Mt
Meitnerium
(268)
110
Ds
Darmstadtium
(281)
111
Rg
roentgenium
(272)
112 Uub
Ununbium
(285)
113
Uut
ununtrium
(284)
114
Uuq
ununquadium
(289)
115
Uup
ununpentium
(288)
5
B
boron
10.81
6
C
carbon
12.01
7
N
nitrogen
14.01
8
O
oxygen
16.00
9
F
fluorine
19.00
2
He
helium
4.00
10
Ne
neon
20.18
16
S
sulfur
32.07
17
Cl
chlorine
35.45
18
Ar
argon
39.95
34
Se
selenium
78.96
35
Br
79.91
36
Kr
krypton
83.80
52
Te
tellurium
127.60
53
I
iodine
126.90
54 Xe
xenon
131.30
84
Po
polonium
(210)
85
At
astatine
(210)
86
Rn
radon
(220)
116
Uuh
ununhexium
(289)
117
Uusununseptium
(295)
118
Uuo
ununoctium
(293)
57
La
lanthanum
138.91
58
Ce
cerium
140.12
59
Pr
praseodymium
140.91
60
Nd
neodymium
144.24
61
Pm
promethium
144.91
62
Sm
samarium
150.41
63
Eu
europium
151.96
65
Tb
terbium
158.92
89
Ac
actinium
227.03
90
Th
thorium
232.04
91
Pa
protactinium
231.04
92
U
uranium
238.03
93
Np
neptunium
237.05
94
Pu
plutonium
244.06
95
Am
americium
243.06
96
Cm
curium
(247)
66
Dy
dysprosium
162.50
67
Ho
Holmium
164.93
68
Er
erbium
167.26
69
Tm
thulium
168.93
70
Yb
ytterbium
173.04
97
Bk
berkelium
(249)
98
Cf
californium
(251)
99
Es
einsteinium
(254)
100Fm
fermium
257.10
101
Md
mendelevium
(256)
102
No
(254)
Nb
Gd
gadolinium
157.25
64
nobelium
0
1s
2s
3s
4s
5s
6s
7s
3d
4d
5d
6d
4p
5p
6p
7p
3p
2p
4f
5f
Symbol:
Solid
Liquid
Gas
Manmade
name
Atomic mass
Atomic
number
Sc
scandium
44.96
21
metal nonmetal
metalloid
metal
nonmetal
1 valence
electron
2 valence
electrons
Valence electrons: 8
4 5 6 73
(H is a nonmetal)
bromine
Group 1
Group 2
Group 3 Group 4 Group 5 Group 6 Group 7 Group 8 Group 9 Group 10 Group 11 Group 12
Group 13 Group 14 Group 15 Group 16 Group 17
Group 18
to
71
to
103
alkali metals
alkaline earth metals
periods groups or families:
halogens noble gases
metals: nonmetals metalloids
charge
multiple charges
blocks
= columns
left of the
staircase
right of the
staircase
“on” the
staircase
1-8 +4 to -4 1s-7p s p d f = rows
group 1
2 7 8
2.4 The Periodic Table Chemistry 110 Professor Brielmann

Chemical formulas: ionic, covalent, or metallic?
75
NaCl
F2
I2
MgCl2
CO2
NaOH
Fe
Ionic
Covalent
Covalent
Ionic
Covalent
Ionic
metallic
metal present: ionic or metallic
(metallic if alone)
polarity of bonds, molecules
bond type
C-C
(same nonmetals)
nonpolar covalent
C-O
(different
nonmetals)
polar covalent
C-H
(an exception)
nonpolar covalent
2.6 Chemical Formulas Chemistry 110 Professor Brielmann

naming binary covalent molecules
76
H20
NH3
N2H4
• Dihydrogen
monoxide
• (water)
• Nitrogen trihydride
• (ammonia)
• Dinitrogen
tetrahydride
• (hydrazine)
2.7 Naming Compounds Chemistry 110 Professor Brielmann

naming: covalent vs. ionic
77
NaCl
CO
AlCl3
SO2
FeO2
SiO2
Sodium Chloride
Carbon monoxide
Aluminium Chloride
Sulfur dioxide
Iron (IV) oxide
Silicon dioxide
is first
element a
Metal?
no
yes
covalent: use, mono, di, etc.
ionic: don’t use
mono, di, etc…is
it a transition metal?
no
yes
don’t use (I), (II), etc.*
do use (I), (II), etc.
2.7 Naming Compounds Chemistry 110 Professor Brielmann
*note that some non-transition metals are polyvalent
transition metals

78
A molecule is
an aggregate of two or more atoms in a definite arrangement held
together by chemical forces.
H2 H2O NH3 CH4
A diatomic molecule contains
A polyatomic molecule contains
O3, H2O, NH3, CH4
H2, N2, O2, Br2, HCl, CO
some diatomic elements:
some molecules:
some polyatomic molecules:
only two atoms: HBrONClIF
more than two atoms:

79
A monatomic ion contains only one atom:
A polyatomic ion contains more than one atom:
Na+, Cl-, Ca2+, O2-, Al3+, N3-
OH-, CN-, NH4
+, NO3
-

81
A molecular formula shows the exact number of
atoms of each element in the smallest unit of a
substance.
An empirical formula shows the simplest
whole-number ratio of the atoms in a substance.
H2OH2O
molecular empirical
C6H12O6 CH2O
O3 O
N2H4 NH2

Name the following
compounds:
(a) Fe(NO3)2
(b)Na2HPO4
(c) (NH4)2SO3
iron(II) nitrate
sodium hydrogen phosphate
ammonium sulfite
Write chemical formulas for
the following compounds:
(a)mercury(I) nitrite
(a) cesium sulfide
(b)calcium phosphate
Hg2(NO2)2
Cs2S
Ca3(PO4)2
note that mercury(1) is Hg2
2+

85
HI hydrogen iodide
NF3 nitrogen trifluoride
SO2 sulfur dioxide
N2Cl4 dinitrogen tetrachloride
NO2 nitrogen dioxide
N2O dinitrogen monoxide
Name these substances
SiCl4
P4O10
silicon tetrachloride
tetraphosphorus decoxide

86
An acid can be defined as
examples: HCl gas and HCl in water
introduction to acids
a substance that yields hydrogen ions (H+)
when dissolved in water.
Name HF, HCl, HBr, HI, HCN, H2S

87
An oxoacid is
HNO3
nitric acid
H2CO3
carbonic acid
H3PO4
phosphoric acid
an acid that contains hydrogen, oxygen, and another element.
name these:
carbonic acid nitric acid phosphoric acid
name the acid and anion contained in HCl04, HCl03, HCl02, and HClO

Name the following oxoacid and oxoanion:
(a) H2SO3, a very unstable acid formed with SO2(g) reacts with
water
(b)H2AsO4
−, once used to control ticks and lice on livestock
(c) SeO3
2−, used to manufacture colorless glass.
H3AsO4 H2SeO4
sulfurous acid.
dihydrogen arsenate
selenite.
arsenic acid.
selenic acid

89
A base can be defined
NaOH sodium hydroxide
KOH potassium hydroxide
Ba(OH)2 barium hydroxide
formulas:
as a substance that yields
hydroxide ions (OH-) when dissolved in
water.

90
Hydrates are compounds that have a specific number of
water molecules bonded (usually weakly) to
them.
BaCl2•2H2O
LiCl•H2O
MgSO4•7H2O
Sr(NO3)2 •4H2O
barium chloride dihydrate
lithium chloride monohydrate
magnesium sulfate heptahydrate
strontium nitrate tetrahydrate
CuSO4•5H2O CuSO4

91
Organic chemistry is the branch of chemistry that deals
with carbon compounds.
C
H
H
H OH C
H
H
H NH2 C
H
H
H C OH
O
methanol methylamine acetic acid
alcohol, amine, and carboxylic acid functional Groups:

92
names, molecular,
and structural
formulas of the
first ten alkanes

chapter 2 solved
(even)problems
93

2.2 List the types of radiation that are known to be emitted by radioactive
elements.
The most common types of radiation known to be emitted by radioactive
elements are alpha (a) radiation,
beta (b) radiation, and
gamma (g) radiation.

2.4 Describe the contributions of these scientists to our knowledge of atomic
structure:
J. J. Thomson
R. A. Millikan
,
Ernest Rutherford
James Chadwick.
answers will vary
electron
nucleus
neutron
electron charge and mass

2.6 Describe the experimental basis for believing that the nucleus occupies a
very small fraction of the volume of the atom.
Rutherford used a particles to probe the structure of the atom. Most a
particles aimed at thin foils of gold passed through the foil with little or no
deflection. A few a particles were deflected at large angles and occasionally
an a particle bounced back in the direction from which it had come.
Rutherford concluded that most of the atom was empty space with a
small, dense, positively charged core (the nucleus).

2.8 Roughly speaking, the radius of an atom is about 10,000 times greater
than that of its nucleus. If an atom were magnified so that the radius of its
nucleus became 10 cm, what would be the radius of the atom in miles? (1
mi = 1609 m.)
4 4
nucleus
1 m 1 mi
10 10 10 cm
100 cm 1609 m
= =    =atom 0.62 mirr

2.10 Why do all atoms of an element have the same atomic number,
although they may have different mass numbers?
What do we call atoms of the same element with different mass
numbers?
Explain the meaning of each term in the symbol
Z is the atomic number, A is the mass
number, and X represents the symbol of the
element.
The chemical identity of an atom is determined by its number
of protons (atomic number).
Isotopes of an element contain differing numbers of
neutrons, hence the mass numbers of isotopes of an
element will differ.

2.12 Calculate the number of neutrons in 239Pu.
number of neutrons= mass number  number of protons = 239  94 = 145

2.14
Indicate the number of protons, neutrons, and electrons in each of these
species:
Isotope 7
15
N 16
33
S 29
63
Cu 38
84
Sr 56
130
Ba 74
186
W 80
202
Hg
No. Protons 7 16 29 38 56 74 80
No. Neutrons 8 17 34 46 74 112 122
No. Electrons 7 16 29 38 56 74 80

2.16 Write the appropriate symbol for each of these isotopes:
(a) Z = 74, A = 186;
(b) Z = 80, A = 201.
74
186
W
80
201
Hg

2.18 Give two differences between a metal and a nonmetal.
1. Metals are good conductors of heat and electricity, while
nonmetals are usually poor conductors of heat and electricity.
2. Metals, excluding mercury, are solids, whereas many
nonmetals are gases.

2.20 Without consulting a periodic table, name each of the lettered groups in
the following table. Provide two examples from each group.
Column A is the alkali metals. Two examples are sodium (Na) and potassium (K).
Column B is the alkaline earth metals. Two examples are calcium (Ca) and barium (Ba).
Column C is the halogens. Two examples are fluorine (F) and iodine (I).
Column D is the noble gases. Two examples are argon (Ar) and xenon (Xe).

2.22 Describe the changes in properties (from metals to nonmetals or from
nonmetals to metals) as we move
(a) down a periodic group
and (b) across the periodic table.
(a) Metallic character increases as you progress down a group
of the periodic table. For example, moving down Group 4A, the
nonmetal carbon is at the top and the metal lead is at the bottom
of the group.
(b) Metallic character decreases from the left side of the
table (where the metals are located) to the right side of the
table (where the nonmetals are located).

2.24 Group these elements in pairs that you would expect to show similar
chemical properties: K, F, P, Na, Cl, and N.
F and Cl are Group 7A elements; they should have similar chemical properties.
Na and K are both Group 1A elements; they should have similar chemical
properties.
P and N are both Group 5A elements; they should have similar chemical properties.

2.26
What are allotropes?
Give an example.
How are allotropes different from isotopes?
Allotropes of an element differ in structure and properties, whereas isotopes of a
given element contain different numbers of neutrons but have similar chemistries.
Allotropes are two or more forms of the same element that differ significantly in
chemical and physical properties.
Diamond and graphite are allotropes of carbon.

2.28 Give an example of each of the following:
(a) a monatomic cation,
(b) a monatomic anion,
(c) a polyatomic cation,
(d) a polyatomic anion.
SO4
2
Na+
I
NH4
+

2.30 Which of the following diagrams represent diatomic molecules,
polyatomic molecules, molecules that are not compounds, molecules that
are compounds, or an elemental form of the substance?
(a) This is a diatomic molecule that is a compound.
(b) This is a polyatomic molecule that is a compound.
(c) This is a polyatomic molecule that is the elemental form of the
substance. It is not a compound.

2.32 Give two examples of each of the following:
(a) a diatomic molecule containing atoms of the same element,
(b) a diatomic molecule containing atoms of different elements,
(c) a polyatomic molecule containing atoms of the same element,
(d) a polyatomic molecule containing atoms of different elements.
H2 and F2
HCl and CO
S8 and P4
H2O and C12H22O11 (sucrose)

2.34 Give the number of protons and electrons in each of the following
common ions:
K+,
Mg2+,
Fe3+,
Br−,
Mn2+,
C4−,
Cu2+.
19 protons, 18 electrons
12 protons,10 electrons
35 protons, 36electrons

2.36 Define molecular formula and empirical formula.
What are the similarities and differences between the empirical formula
and molecular formula of a compound?
Both give the correct ratio of atoms for each element but not
necessarily the actual number of atoms in a given molecule.
A molecular formula shows the exact number of atoms of each element
in the smallest unit of a substance.
An empirical formula shows the elements present and the simplest whole
number ratio of the atoms

2.38 What does P4 signify?
How does it differ from 4P?
4P represents four atoms of P (phosphorus). Therefore it
differs in both I the number of substances and their identity
P4 signifies one molecule that is composed of four P atoms.

2.40 Explain why the chemical formulas of ionic compounds are usually the
same as their empirical formulas.
Ionic compounds do not consist of discrete molecular
units but are three-dimensional networks of ions. The
formula of ionic compounds represents the simplest
ratio (empirical formula) in which the cation and anion
combine.

2.42 What are the empirical formulas of the following compounds?
(a) Al2Br6,
(b) Na2S2O4,
(c) N2O5,
(d) K2Cr2O7
N2O5
K2Cr2O7
NaSO2
AlBr3

2.44 Write the molecular formula of ethanol. The color codes are: black
(carbon), red (oxygen), and gray (hydrogen).
The molecular formula of ethanol is C2H6O.

2.46
Which of the following compounds are likely to be ionic? Which are
likely to be molecular?
CH4,
NaBr,
BaF2,
CCl4,
ICl,
CsCl,
NF3
molecular
molecular
molecular
molecular
ionic
ionic
ionic

2.48
Name these species:
(a) KClO,
(b) Ag2CO3,
(c) FeCl2,
(d) KMnO4,
(e) CsClO3,
(f) HIO,
(g) FeO,
, (l) Na2O,
(m) Na2O2,
(n) FeCl3 · 6H2O
(h) Fe2O3,
(i) TiCl4,
(j) NaH,
(k) Li3N
potassium hypochlorite
The metals that have only one
charge in ionic compounds are
the alkali metals (+1), the alkaline
earth metals (+2), Ag+, Zn2+,
Cd2+, and Al3+.When naming
acids, binary acids are named
differently than oxoacids. For
binary acids, the name is based
on the nonmetal. For oxoacids,
the name is based on the
polyatomic anion
silver carbonate
iron(II) chloride
potassium permanganate
cesium chlorate
hypoiodous acid
iron(II) oxide
iron(III) oxide
titanium(IV) chloride
sodium hydride
lithium nitride
sodium oxide
sodium peroxide
iron(III) chloride hexahydrate.
arsenate ion.
hydrogen sulfite. ion

2.50 Write the formulas for these compounds:
(a) copper(I) cyanide,
(b) strontium chlorite,
(c) perchloric acid,
(d) hydroiodic acid,
(e) disodium ammonium phosphate,
(f) lead(II) carbonate,
(g) tin(II) fluoride,
(h) tetraphosphorus decasulfide,
(i) mercury(II) oxide,
(j) mercury(I) iodide,
(k) cobalt(II) chloride hexahydrate.
CuCN
Sr(ClO2)2
is HClO4
HI
Na2(NH4)PO4
PbCO3
SnF2
P4S10
HgO
Hg2I2
CoCl2  6H2O

119
add even numbered problems up to “additional”

chapter 2 assigned
(odd) problems
120

(a) α particle,
(b) β particle,
(c) γ ray,
(d) X ray.
2.3 Compare the properties of:
α particles,
cathode rays,
protons, neutrons and electrons.
What is meant by the term
“fundamental particle”?
2.5 A sample of a radioactive element is found to be losing mass gradually. Explain what is happening to the
sample.
chapter 2 problems
name_________________________________

2.7 The diameter of a neutral helium atom is about 1 × 102 pm. Suppose that we could line up helium atoms side by
side in contact with one another. Approximately how many atoms would it take to make the distance from end to
end 1 cm?
(a) atomic number,
(b) mass number.
Why does a knowledge of atomic number enable us
to deduce the number of electrons present in an atom?
2.11 What is the mass number of an iron atom that has 28 neutrons?
2.13 For each of these species, determine the number of protons and the number of neutrons in the nucleus:
2.15 Write the appropriate symbol for each of these isotopes: (a) Z = 11, A = 23; (b) Z = 28, A = 64.
2.17 What is the periodic table, and what is its significance in the study of chemistry? What are groups and periods
in the periodic table?
2.19 Write the names and symbols for four elements in each of these categories:
(a) nonmetal, (b) metal, (c) metalloid.
10
12
0.01 m 1 pm
1 cm 1 10 pm
1 cm 1 10 m
  = 

10
2
1 He atom
(1 10 pm)
1 10 pm
=   =

8
? He atoms 1 10 He atoms
11
23
Na 28
64
Ni

2.21 Elements whose names end with “ium” are usually metals; sodium is one example. Identify a nonmetal whose
name also ends with “ium.”
2.23 find
(a) two metals less dense than water,
(b) two metals more dense than mercury,
(c) the densest known solid metallic element,
(d) the densest known solid nonmetallic element.
2.25 What is the difference between an atom and a molecule?
2.27 Describe the two commonly used molecular models.
2.29 Which of the following diagrams represent diatomic molecules, polyatomic molecules, molecules that are not
compounds, molecules that are compounds, or an elemental form of the substance?

2.31Identify the following as elements or compounds:
NH3,
N2,
S8,
NO,
CO,
CO2,
H2,
SO2.
2.33Give the number of protons and electrons in each of
the following common ions:
Na+,
Ca2+,
Al3+,
Fe2+,
I−,
F
−,
S2−,
O2−,
N3−.
2.35 What does a chemical formula represent?
What is the ratio of the atoms in the following molecular formulas?
2.37 Give an example of a case in which two molecules have
different molecular formulas but the same empirical
formula.
2.39 What is an ionic compound?
How is electrical neutrality maintained in an ionic
compound?
2.41What are the empirical formulas of the following
compounds? (a) C2N2,
(b) C6H6,
(c) C9H20,
(d) P4O10,
(e) B2H6
2.43 Write the molecular formula of glycine, an amino acid present
in proteins. The color codes are: black (carbon),
blue (nitrogen), red (oxygen), and gray (hydrogen).

2.45 Which of the following compounds are likely to be ionic?
Which are likely to be molecular?
SiCl4,
LiF
BaCl2
B2H6
KCl
C2H4
2.47 Name these species:
(a) Na2CrO4
(b) K2HPO4
(c) HBr (gas)
(d) HBr (in water)
(e) Li2CO3, (f) K2Cr2O7
(g) NH4NO2
(h) PF3
.
2.49Write the formulas for these compounds:
(a) rubidium nitrite
(b) potassium sulfide
(c) perbromic acid
(d) magnesium phosphate
(e) calcium hydrogen phosphate
(f) boron trichloride
(g) iodine heptafluoride
(h)ammonium sulfate
(i) silver perchlorate
(j) iron(III) chromate
(k) calcium sulfate dihydrate.
(i) PF5
(j) P4O6
(k) CdI2
(l) SrSO4
(m) Al(OH)3
(n) Na2CO3· 10H2O,

126
add odd numbered problems up to
“additional”

Stoichiometry
127
3.1 Atomic mass
3.2 Avogadro’s Number and Molar Mass
3.3 Molecular mass
3.4 Mass Spectrometry omit
3.5 Percent composition
3.6 Empirical Formula
3.7 Chemical Reactions and Equations
3.8. Amounts of reactants and products
3.9 Limiting reactants
3.10 Yield
Chapter 3
3. stoichiometry
=
n x molar mass of element
molar mass of compound
x 100%
% composition
example: %
hydrogen in
CH4:
4 x 1.008 g/mol
16.04 g/mol
x 100% =25.13%
=
actual yield
theoretical yield
x 100%
% yield

consider water:
An oxygen atom is
≈ times more
massive than a
hydrogen atom
the mole
12 g of C
= ___ mol of C
= _____________ atoms of C
is the SI unit for amount
= 6.02 x 1023
consider the synthesis of water:
__H2 + ___ O2  ___H2O
as written:
___ moles of hydrogen molecules
combine with ___ moles of oxygen
molecules to make ___ moles of
water
2 1 2
2
1
2
stoich.i.om.e.try: element measuring
3 Stoichiometry Chemistry 110 Professor Brielmann
H-O-H
relative masses of atoms
relative atomic masses
atomic mass units (amu) and g
12C exactly 12
H 1.007
O 15.99
every 18 g of H2O contains ___ g H, ___ g O
every molecule of H2O contains ___ H atoms
and ___ O atoms
A carbon-12 atom has a mass of
1.99264664 x 10-23 g. How many carbon
atoms are in 12 g of carbon-12?
1 atom C
1.99264664 x 10−23 g C
x 12 g C = 6.022 x 1023 atoms C
1
6.02 x 1023
2
1
162
carbon has a molar mass of ____________
12
g
mol
16

Let us burn some butane (C4H10):
the mole: balancing equations
C4H10 + O2 2 13 CO28 + 10 H2O
do they really react in that ratio?
write and balance the reaction below. Suggestion: save toughest element (O) for the end

How many moles of CO2 will be produced from 1 mole of butane and excess oxygen?
4 (2 makes 8, so 1 makes 4)
How many moles of oxygen are needed to react with 17.26 moles of butane?
17.26 moles C4H10 x
2 molesC4H10
13 moles O2
== 112.2 moles O2
How many moles of butane react with 0.42 moles of oxygen?
0.42 moles O2
13 moles O2
2 moles C4H10
= 0.065 moles O2
1. mole-mole conversions
C4H10 + O2 2 13 CO28 + 10
2 moles butane: = ___ moles oxygen = ___ moles CO2 = ___ moles water13 8 10
H2O
2. mole-gram conversions
C4H10 + O2 2 13 CO28 + 10 H2O
Example: The combustion of 3 moles of butane (C4H10) with excess oxygen will produce ___ grams of
CO2.
3 moles C4H10 x
2 moles C4H10
8 moles CO2
x
1 moles CO2
44 grams CO2
= 528 grams CO2
For you: The combustion of 4 moles of butane (C4H10) will require____ grams of O2.
4 moles C4H10 x
2 moles C4H10
13 moles O2
x
1 moles O2
32 grams O2
= 832 grams O2

3. gram-gram conversions
C4H10: + O2 2 13 CO28 + 10 H2O
example: The combustion of 100 grams of butane (C4H10) with excess
oxygen will produce ___ grams of CO2.
100 g C4H10 x
58 g C4H10
1 mole C4H10
x
8 moles CO2 = 303 grams CO2
1 mole CO2
44 grams CO2
2 moles C4H10
x
for you: The combustion of 453 grams of butane (C4H10) with excess oxygen
will produce ___ grams of H2O.
453 g C4H10 x
58 g C4H10
1 mole C4H10
x
10 moles H2O
1 mole H2O
18 grams H2O
2 moles C4H10
x = 703 grams H2O
4. molecular and other conversions
a microgram is barely visible. If one microgram of butane is set on fire, how
many molecules of carbon dioxide will be produced?
solution: micrograms grams  moles butane moles CO2  molecules
CO2; note plenty of oxygen.
1 microgram
C4H10 1 x 106
micrograms
C4H10
1 gram C4H10
x
1 mole C4H10 = 4.1 x 1016
grams CO22 mole C4H10
8 moles CO2
58 g C4H10
x
C4H10 + O2 2 13 CO28 + 10 H2O
x
1 mole CO2
6 x 1023 molecules CO2
x
9.4

x
the mole: limiting and excess reactants
C4H10 + O2 2 13 CO28 + 10 H2O
Q. The combustion of 100 grams of butane (C4H10) with 100 g oxygen will produce ___ grams
of CO2.
100 g C4H10 x
58 g C4H10
1 mole C4H10
x
1 mole CO2
44 grams CO2
2 moles C4H10
x
Strategy: find which makes the least product; that limits the reaction.
Answer: This reaction will produce ___ grams of CO2.
100 g O2 x
32 g O2
1 mole O2
x
1 mole CO2
44 grams CO2
13 moles O2
x
85
9.5
How much butane remains?
Easy- find out how much is needed:
(how much butane will the 100 g of oxygen consume? The rest is excess)
The limiting reactant is ____________; the excess reactant is ____________
oxygen butane
100 g O2 x
32 g O2
1 mole O2
2 moles C4H10
1 mole C4H10
58 grams C4H10
13 moles O2
x = ______ grams
butane used; so
_______butane
remains.
27.89
72.11
`

the mole: percent composition
= the mass percent of each element in a substance
H-O-H
what is the % composition by mass of water?
KEY: assume one mole to make it easy.
1 g/mol 16
g/mol
1 g/mol
Water is 16/18 oxygen by mass: 89% O, 11% H
9.6
the mole: empirical and molecular formulas
= the simplest molar ratio for elements in a substance = the actual molar ratio for elements in a substance
CH2 C2H4
KEY: assume 100 grams. Find moles of each element
What is the empirical and molecular formula for a substance that is 86% C
and 14% H, with a molecular mass of 28 g/mol?
= 7.1 mol C86 g C x
12 g C
1 mole C
= 14 mol H14 g H x
1 g H
1 mole H
14 g/mol 28 g/mol

the mole: theoretical and actual yield
maximum product amount
9.7
actual product amount
(100% yield) less
The theoretical yield of CO2 for the combustion of 50 g of butane with
excess oxygen is ____ g. If the reaction produces 25 g CO2, the
yield is ___%.
50 g C4H10
58 g C4H10
1 mole C4H10
x
8 moles CO2
= 151.5 grams
CO2
1 mole CO2
44 grams CO2
2 moles C4H10
x
151.5
% yield =
actual yield
theoretical yield
x 100
% yield =
25 g
151.5 g
x 100 = 16.5 % yield
like a grade on a test.
x
16.5

chapter 3 solved
(even)problems
135

3.2 What is the mass (in amu) of a carbon-12 atom?
Why is the atomic mass of carbon 12.01 amu ?
12.00 amu. On the periodic table, the mass is listed as 12.01 amu
because this is an average mass of the naturally occurring mixture of
isotopes of carbon.

3.4 What information would you need to calculate the average atomic mass of an
element?
You need the mass of each isotope of the element and each isotope’s relative
abundance.

3.6 The atomic masses of Cl-35 and Cl-37 are 6.0151 amu and 7.0160 amu,
respectively. Calculate the natural abundances of these two isotopes. The average
atomic mass of Li is 6.941 amu
Strategy: Each isotope contributes to the average atomic mass based on its relative abundance. Multiplying
the mass of an isotope by its fractional abundance (not percent) will give the contribution to the average
atomic mass of that particular isotope.
It would seem that there are two unknowns in this problem, the fractional abundance of
6
Li and the fractional
abundance of
7
Li. However, these two quantities are not independent of each other; they are related by the
fact that they must sum to 1. Start by letting x be the fractional abundance of
6
Li. Since the sum of the two
abundances must be 1, we can write
Abundance
7
Li = (1 x)
Solution:
Average atomic mass of Li= 6.941 amu =x(6.0151 amu)  (1 x)(7.0160 amu)
6.941 =1.0009x 7.0160
1.0009x= 0.075
x= 0.075
x= 0.075 corresponds to a natural abundance of
6
Li of 7.5%. The natural abundance of
7
Li is (1 x) = 0.925
or 92.5%.

3.8 How many amu are there in 8.4 g?
23
6.022 10 amu
The unit factor required is
1 g
 
 
 
 
23
6.022 10 amu
8.4 g =
1 g

=  24
? amu 5.1 10 amu

3.10 What is the molar mass of an atom?
What are the commonly used units for molar
mass?
The molar mass of an atom is the mass
of one mole, 6.022 × 1023 atoms, of that
element. Units are g/mol

3.12. The thickness of a piece of paper is
0.0036 in. Suppose a certain book has an
Avogadro's number of pages; calculate the
thickness of the book in light-years
The thickness of the book in miles would be:
23 160.0036 in 1 ft 1 mi
(6.022 10 pages) = 3.4 10 mi
1 page 12 in 5280 ft
    
The distance, in miles, traveled by light in one year is:
8
12365 day 24 h 3600 s 3.00 10 m 1 mi
1.00 yr 5.88 10 mi
1 yr 1 day 1 h 1 s 1609 m

     = 
The thickness of the book in light-years is:
16
12
1 light-yr
(3.4 10 mi)
5.88 10 mi
  =

3
5.8 10 light - yr
It will take light 5.8  10
3
years to travel from the first page to the last one!

3.14 How many moles of cobalt (Co) atoms
are there in 6.00 × 109 (6 billion) Co atoms?
9
23
1 mol Co
(6.00 10 Co atoms) =
6.022 10 Co atoms
 

15
9.96 10 mol Co


3.16 How many grams of gold (Au) are
there in 15.3 moles of Au?
Strategy: We are given moles of gold and asked to solve for grams of gold. What conversion factor do we
need to convert between moles and grams? Arrange the appropriate conversion factor so moles cancel and the
unit grams is obtained for the answer.
Solution: The conversion factor needed to covert between moles and grams is the molar mass. In the
periodic table (see inside front cover of the text), we see that the molar mass of Au is 197.0 g. This can be
expressed as
1 mol Au = 197.0 g Au
From this equality, we can write two conversion factors.
1 mol Au 197.0 g Au
and
197.0 g Au 1 mol Au
The conversion factor on the right is the correct one. Moles will cancel, leaving the unit grams for the answer.
We write
197.0 g Au
= 15.3 mol Au =
1 mol Au
 3
? g Au 3.01 10 g Au

3.18 What is the mass in grams of a single
atom of each of the following elements?
(a) As
(b) Ni
(a)
Strategy: We can look up the molar mass of arsenic (As) on the periodic table (74.92 g/mol). We want to
find the mass of a single atom of arsenic (unit of g/atom). Therefore, we need to convert from the unit mole in
the denominator to the unit atom in the denominator. What conversion factor is needed to convert between
moles and atoms? Arrange the appropriate conversion factor so mole in the denominator cancels, and the unit
atom is obtained in the denominator.
Solution: The conversion factor needed is Avogadro’s number. We have
1 mol = 6.022  10
23
particles (atoms)
23
23
1 mol As 6.022 10 As atoms
and
1 mol As6.022 10 As atoms


The conversion factor on the left is the correct one. Moles will cancel, leaving the unit atoms in the
denominator of the answer.
We write
23
74.92 g As 1 mol As
1 mol As 6.022 10 As atoms
=  =

22
? g/As atom 1.244 10 g/As atom

23
58.69 g Ni 1 mol Ni
1 mol Ni 6.022 10 Ni atoms
=  =

23
? g/Ni atom 9.746 10 g/Ni atom


3.20 A modern penny weighs 2.5 g but
contains only 0.063 g of copper (Cu). How
many copper atoms are present in a modern
penny?
Strategy: The question asks for atoms of Cu. We cannot convert directly from grams to atoms of copper. What unit do we need to convert grams of Cu to
in order to convert to atoms? What does Avogadro’s number represent?
Solution: To calculate the number of Cu atoms, we first must convert grams of Cu to moles of Cu. We use the molar mass of copper as a conversion
factor. Once moles of Cu are obtained, we can use Avogadro’s number to convert from moles of copper to atoms of copper.
1 mol Cu = 63.55 g Cu
The conversion factor needed is
1 mol Cu
63.55 g Cu
Avogadro’s number is the key to the second conversion. We have
1 mol = 6.022  10
23
particles (atoms)
23
23
1 mol Cu 6.022 10 Cu atoms
and
1 mol Cu6.022 10 Cu atoms


The conversion factor on the right is the one we need because it has the number of Cu atoms in the numerator, which is the unit we want for the answer.
Let us complete the two conversions in one step.
grams of Cu  moles of Cu  number of Cu atoms
23
1 mol Cu 6.022 10 Cu atoms
0.063 g Cu
63.55 g Cu 1 mol Cu

=   = 20
? atoms of Cu 6.0 10 Cu atoms
Check: Should 0.063 g of Cu contain fewer than Avogadro’s number of atoms? What mass of Cu would contain Avogadro’s number of atoms?

3.22 Which of the following has a greater
mass: 2 atoms of lead or 5.1 × 10−23 mole of
helium?
22
23
1 mol Pb 207.2 g Pb
2 Pb atoms = 6.881 10 g Pb
1 mol Pb6.022 10 Pb atoms

  

23 224.003 g He
(5.1 10 mol He) = 2.0 10 g He
1 mol He
 
  
2 atoms of lead have a greater mass than 5.1  10
23
mol of helium.

3.24 Calculate the molar mass of the following substances:
(a) Li2CO3
(b) CS2
(c) CHCl3 (chloroform)
(d) C6H8O6 (ascorbic acid, or vitamin C)
(e) KNO3
(f) Mg3N2.
molar mass Li2CO3 = 2(6.941 g)  12.01 g  3(16.00 g) = 73.89 g
molar mass CS2 = 12.01 g  2(32.07 g) = 76.15 g
molar mass CHCl3= 12.01 g  1.008 g  3(35.45 g) = 119.4 g
molar mass C6H8O6 = 6(12.01 g)  8(1.008 g)  6(16.00 g) = 176.12 g
molar mass KNO3 = 39.10 g  14.01 g  3(16.00 g) = 101.11 g
molar mass Mg3N2 = 3(24.31 g)  2(14.01 g) = 100.95 g

3.26 How many molecules of acetone,
shown here, are present in 0.435 g of
acetone?
The molar mass of acetone, C3H6O, is 58.08 g. We use the molar mass and Avogadro’s number as conversion factors to convert
from grams to moles to molecules of acetone.
23
3 6 3 6
3 6
3 6 3 6
1 mol C H O 6.022 10 molecules C H O
0.435 g C H O
58.08 g C H O 1 mol C H O

  = 21
3 64.51 10 molecules C H O

3.28 Dimethyl sulfoxide [(CH3)2SO], also called DMSO, is an important solvent that
penetrates the skin, enabling it to be used as a topical drug-delivery agent. Calculate the
number of C, S, H, and O atoms in 7.14 × 103 g of dimethyl sulfoxide.
Strategy: We are asked to solve for the number of C, S, H, and O atoms in 7.14  10
3
g of dimethyl sulfoxide (DMSO). We cannot convert directly from
grams DMSO to atoms. What unit do we need to obtain first before we can convert to atoms? How should Avogadro’s number be used here? How many atoms
of C, S, H, or O are in 1 molecule of DMSO?
Solution: Let us first calculate the number of C atoms in 7.14  10
3
g of dimethyl sulfoxide. First, we must convert grams of DMSO to number of molecules
of DMSO. This calculation is similar to Problem 3.26. The molecular formula of DMSO shows there are two C atoms in one DMSO molecule, which will allow
us to convert to atoms of C. We need to perform three conversions:
grams of DMSO  moles of DMSO  molecules of DMSO  atoms of C
The conversion factors needed for each step are: 1) the molar mass of DMSO, 2) Avogadro’s number, and 3) the number of C atoms in 1 molecule of DMSO.
We complete the three conversions in one calculation.
23
3 1 mol DMSO 6.022 10 DMSO molecules 2 C atoms
7.14 10 g DMSO
78.14 g DMSO 1 mol DMSO 1 molecule DMSO

   
=1.10  10
26
C atoms
The above method utilizes the ratio of molecules (DMSO) to atoms (carbon). We can also solve the problem by reading the formula as the ratio of moles of
DMSO to moles of carbon using the following conversions:
grams of DMSO  moles of DMSO  moles of C  atoms of C
Try it.
Check: Does the answer seem reasonable? We have 7.14  10
3
g DMSO. How many atoms of C would 78.14 g of DMSO contain?
We could calculate the number of atoms of the remaining elements in the same manner or we can use the atom ratios from the molecular formula. The sulfur
atom to carbon atom ratio in a DMSO molecule is 1:2, the hydrogen atom to carbon atom ratio is 6:2 or 3:1, and the oxygen atom to carbon atom ratio is 1:2.
26 1 S atom
(1.10 10 C atoms)
2 C atoms
=   = 25
?atoms of S 5.50 10 S atoms
26 3 H atoms
(1.10 10 C atoms)
1 C atom
=   = 26
?atoms of H 3.30 10 H atoms
26 1 O atom
(1.10 10 C atoms)
2 C atoms
=   = 25
?atoms of O 5.50 10 Oatoms

3.30 The density of water is 1.00 g/mL at
4°C. How many water molecules are present
in 2.56 mL of water at this temperature?
1.00 g
Mass of water = 2.56 mL = 2.56 g
1.00 mL

Molar mass of H2O = (16.00 g)  2(1.008 g) = 18.02 g/mol
23
2 2
2
2 2
1 mol H O 6.022 10 molecules H O
= 2.56 g H O
18.02 g H O 1 mol H O

 2? H O molecules
=8.56  10
22
molecules

3.36 Describe how the knowledge of the
percent composition by mass of an unknown
compound can help us identify the
compound.
If you know the percent composition by
mass of an unknown
compound, you can determine
its empirical formula.

3.38 If we know the empirical formula of a
compound, what additional information do
we need to determine its molecular formula?
The approximate molar mass.

3.40 For many years chloroform (CHCl3) was used as an inhalation anesthetic in spite
of the fact that it is also a toxic substance that may cause severe liver, kidney, and heart
damage. Calculate the percent composition by mass of this compound.
Strategy: Recall the procedure for calculating a percentage. Assume that we have 1 mole of CHCl3. The
percent by mass of each element (C, H, and Cl) is given by the mass of that element in 1 mole of CHCl3
divided by the molar mass of CHCl3, then multiplied by 100 to convert from a fractional number to a
percentage.
Solution: The molar mass of CHCl3= 12.01 g/mol  1.008 g/mol  3(35.45 g/mol) = 119.4 g/mol. The
percent by mass of each of the elements in CHCl3 is calculated as follows:
12.01 g/mol
%C 100%
119.4 g/mol
=  = 10.06%
1.008 g/mol
%H 100%
119.4 g/mol
=  = 0.8442%
3(35.45) g/mol
%Cl 100%
119.4 g/mol
=  = 89.07%
Check: Do the percentages add to 100%? The sum of the percentages is (10.06%  0.8442%  89.07%) =
99.97%. The small discrepancy from 100% is due to the way we rounded off.

3.42 All of the following substances are fertilizers that contribute nitrogen to the
soil. Which of these is the richest source of nitrogen on a mass percentage basis?
Urea, (NH2)2CO
Ammonium nitrate, NH4NO3
Guanidine, HNC(NH2)2
Ammonia, NH3
Compound Molar mass (g) N% by mass
(a) (NH2)2CO 60.06
2(14.01 g)
100% = 46.65%
60.06 g

(b) NH4NO3 80.05
2(14.01 g)
100% = 35.00%
80.05 g

(c) HNC(NH2)2 59.08
3(14.01 g)
100% = 71.14%
59.08 g

(d) NH3 17.03
14.01 g
100% = 82.27%
17.03 g

Ammonia, NH3, is the richest source of nitrogen on a mass percentage basis.

3.44 Peroxyacylnitrate (PAN) is one of the components of smog. It is a compound of C, H, N, and O. Determine
the percent composition of oxygen and the empirical formula from the following percent composition by mass:
19.8 percent C, 2.50 percent H, 11.6 percent N. What is its molecular formula given that its molar mass is about
120 g?
METHOD 1:
Step 1: Assume you have exactly 100 g of substance. 100 g is a convenient amount,
because all the percentages sum to 100%. The percentage of oxygen is found by
difference:
100%  (19.8%  2.50%  11.6%) = 66.1%
In 100 g of PAN, there will be 19.8 g C, 2.50 g H, 11.6 g N, and 66.1 g O.
Step 2: Calculate the number of moles of each element in the compound. Remember, an
empirical formula tells us which elements are present and the simplest whole
number ratio of their atoms. This ratio is also a mole ratio. Use the molar
masses of these elements as conversion factors to convert to moles.
1 mol C
= 19.8 g C = 1.65 mol C
12.01 g C
Cn
1mol H
= 2.50 g H = 2.48 mol H
1.008 g H
Hn
1 mol N
= 11.6 g N = 0.828 mol N
14.01g N
Nn
1 mol O
= 66.1 g O = 4.13 mol O
16.00 g O
On
Step 3: Try to convert to whole numbers by dividing all the subscripts by the smallest
subscript. The formula is C1.65H2.48N0.828O4.13. Dividing the subscripts by
0.828 gives the empirical formula, C2H3NO5.
To determine the molecular formula, remember that the molar mass/empirical mass will
be an integer greater than or equal to one.
molar mass
1 (integer values)
empirical molar mass

In this case,
molar mass 120 g
1
empirical molar mass 121.05 g
= 
Hence, the molecular formula and the empirical formula are the same, C2H3NO5.
METHOD 2:
Step 1: Multiply the mass % (converted to a decimal)
of each element by the molar mass to convert
to grams of each element. Then, use the molar
mass to convert to moles of each element.
1 mol C
(0.198) (120 g) 1.98 mol C
12.01 g C
=   = C 2 mol Cn
1 mol H
(0.0250) (120 g) 2.98 mol H
1.008 g H
=   = H 3 mol Hn
1 mol N
(0.116) (120 g) 0.994 mol N
14.01 g N
=   = N 1 mol Nn
1 mol O
(0.661) (120 g) 4.96 mol O
16.00 g O
=   = O 5 mol On
Step 2: Since we used the molar mass to calculate the
moles of each element present in the
compound, this method directly gives the
molecular formula. The formula is C2H3NO5.
Step 3: Try to reduce the molecular formula to a
simpler whole number ratio to determine the
empirical formula. The formula is already in
its simplest whole number ratio. The
molecular and empirical formulas are the
same. The empirical formula is C2H3NO5.

3.46 How many grams of sulfur (S) are
needed to react completely with 246 g of
mercury (Hg) to form HgS?
Using unit factors we convert:
g of Hg  mol Hg  mol S  g S
1mol Hg 1mol S 32.07 g S
246 g Hg
200.6 g Hg 1mol Hg 1mol S
=    =? g S 39.3 g S

3.48 Tin(II) fluoride (SnF2) is often added to toothpaste as an ingredient to prevent
tooth decay. What is the mass of F in grams in 24.6 g of the compound?
Strategy: Tin(II) fluoride is composed of Sn and F. The mass due to F is based on its percentage by mass in
the compound. How do we calculate mass percent of an element?
Solution: First, we must find the mass % of fluorine in SnF2. Then, we convert this percentage to a fraction
and multiply by the mass of the compound (24.6 g) to find the mass of fluorine in 24.6 g of SnF2.
The percent by mass of fluorine in tin(II) fluoride, is calculated as follows:
2
2
mass of F in 1 mol SnF
mass % F 100%
molar mass of SnF
= 
2(19.00 g)
100% = 24.25% F
156.7 g
= 
Converting this percentage to a fraction, we obtain 24.25/100 = 0.2425.
Next, multiply the fraction by the total mass of the compound.
? g F in 24.6 g SnF2= (0.2425)(24.6 g) =5.97 g F
Check: As a ball-park estimate, note that the mass percent of F is roughly 25 percent, so that a quarter of
the mass should be F. One quarter of approximately 24 g is 6 g, which is close to the answer.
Note: This problem could have been worked in a manner similar to Problem 3.46. You could
complete the following conversions:
g of SnF2 mol of SnF2 mol of F  g of F

3.50 What are the empirical formulas of the compounds with the following compositions? (a) 40.1
percent C, 6.6 percent H, 53.3 percent O, (b) 18.4 percent C, 21.5 percent N, 60.1 percent K..
(a)
Strategy: In a chemical formula, the subscripts represent the ratio of the number of moles of each element that combine to form the compound. Therefore, we
need to convert from mass percent to moles in order to determine the empirical formula. If we assume an exactly 100 g sample of the compound, do we know the
mass of each element in the compound? How do we then convert from grams to moles?
Solution: If we have 100 g of the compound, then each percentage can be converted directly to grams. In this sample, there will be 40.1 g of C, 6.6 g of H, and
53.3 g of O. Because the subscripts in the formula represent a mole ratio, we need to convert the grams of each element to moles. The conversion factor needed is
the molar mass of each element. Let n represent the number of moles of each element so that
C
1 mol C
40.1 g C 3.34 mol C
12.01 g C
=  =n
H
1 mol H
6.6 g H 6.5 mol H
1.008 g H
=  =n
O
1 mol O
53.3 g O 3.33 mol O
16.00 g O
=  =n
Thus, we arrive at the formula C3.34H6.5O3.33, which gives the identity and the mole ratios of atoms present. However, chemical formulas are written with whole
numbers. Try to convert to whole numbers by dividing all the subscripts by the smallest subscript (3.33).
3.34
1
3.33
C:
6.5
2
3.33
H :
3.33
1
3.33
=O :
This gives the empirical formula, CH2O.
Check: Are the subscripts in CH2O reduced to the smallest whole numbers?
(b) Following the same procedure as part (a), we find:
C
1 mol C
18.4 g C 1.53 mol C
12.01 g C
=  =n
N
1 mol N
21.5 g N 1.53 mol N
14.01 g N
=  =n
K
1 mol K
60.1 g K 1.54 mol K
39.10 g K
=  =n
Dividing by the smallest number of moles (1.53 mol) gives the empirical formula, KCN.

3.52 The empirical formula of a compound is CH. If the molar mass of this compound
is about 78 g, what is its molecular formula?
The empirical molar mass of CH is approximately 13.02 g. Let us compare this to the molar mass to determine the
molecular formula.
Recall that the molar mass divided by the empirical mass will be an integer greater than or equal to one.
molar mass
1 (integer values)

In this case,
molar mass 78 g
6
= 
Thus, there are six CH units in each molecule of the compound, so the molecular formula

3.54 Monosodium glutamate (MSG), a food-flavor enhancer, has been blamed for
“Chinese restaurant syndrome,” the symptoms of which are headaches and chest pains.
MSG has the following composition by mass: 35.51 percent C, 4.77 percent H, 37.85
percent O, 8.29 percent N, and 13.60 percent Na. What is its molecular formula if its
molar mass is about 169 g?
METHOD 1:
Step 1: Assume you have exactly 100 g of substance. 100 g is a convenient amount, because all the percentages sum to 100%. In 100 g of MSG there will be 35.51
g C, 4.77 g H, 37.85 g O, 8.29 g N, and 13.60 g Na.
Step 2: Calculate the number of moles of each element in the compound. Remember, an empirical formula tells us which elements are present and the simplest
whole number ratio of their atoms. This ratio is also a mole ratio. Let nC, nH, nO, nN, and nNa be the number of moles of elements present. Use the molar
masses of these elements as conversion factors to convert to moles.
C
1 mol C
35.51 g C 2.957 mol C
12.01 g C
=  =n
H
1 mol H
4.77 g H 4.73 mol H
1.008 g H
=  =n
O
1 mol O
37.85 g O 2.366 mol O
16.00 g O
=  =n
N
1 mol N
8.29 g N 0.592 mol N
14.01 g N
=  =n
Na
1 mol Na
13.60 g Na 0.5916 mol Na
22.99 g Na
=  =n
answer continued on
next page

whole numbers.
Step 3: Try to convert to whole numbers by dividing all the subscripts by the smallest subscript.
2.957
= 4.998 5
0.5916
C:
4.73
= 8.00
0.5916
H :
2.366
= 3.999 4
0.5916
O:
0.592
= 1.00
0.5916
N :
0.5916
= 1
0.5916
Na :
This gives us the empirical formula for MSG, C5H8O4NNa.
To determine the molecular formula, remember that the molar mass/empirical mass will be an integer greater than or equal to one.
molar mass
1 (integer values)

In this case,
molar mass 169 g
1
= 
Hence, the molecular formula and the empirical formula are the same, C5H8O4NNa. It should come as no surprise that the empirical and molecular formulas are the
same since MSG stands
METHOD 2:
Step 1: Multiply the mass % (converted to a decimal) of each element by the molar mass to convert to grams of each element. Then, use the molar
mass to convert to moles of each element.
C
1 mol C
(0.3551) (169 g) 5.00 mol C
12.01 g C
=   =n
H
1 mol H
(0.0477) (169 g) 8.00 mol H
1.008 g H
=   =n
O
1 mol O
(0.3785) (169 g) 4.00 mol O
16.00 g O
=   =n
N
1 mol N
(0.0829) (169 g) 1.00 mol N
14.01 g N
=   =n
Na
1 mol Na
(0.1360) (169 g) 1.00 mol Na
22.99 g Na
=   =n
Step 2: Since we used the molar mass to calculate the moles of each element present in the compound, this method directly gives the molecular
formula. The formula is C5H8O4NNa.

3.56 What is the difference between a
chemical reaction and a chemical equation?
A chemical equation uses chemical symbols to show what happens during a chemical reaction.
A chemical reaction is a process in which a substance (or substances) is changed into
one or more new substances.

3.58 Write the symbols used to represent gas,
liquid, solid, and the aqueous phase in
chemical equations.
(g), (l), (s), (aq).

3.60 Balance the following equations using the method outlined
in Section 3.7:
a. __N2O5 →__ N2O4 + __O2
b. __KNO3 → __KNO2 + __O2
c. __NH4NO3 → __N2O + __H2O
d. __NH4NO2 → __N2 + __H2O
e. __NaHCO3 → __Na2CO3 + __H2O + __CO2
f. __P4O10 + __H2O → __H3PO4
g. __HCl + __CaCO3 → __CaCl2 + __H2O + __CO2
h. __Al + __H2SO4 → __Al2(SO4)3 + __H2
i. __CO2 + __KOH → __K2CO3 + __H2O
j. __CH4 + __O2 → __CO2 + __H2O
k. __Be2C + __H2O → __Be(OH)2 + __CH4
l. Cu + HNO3 → Cu(NO3)2 + NO + H2O
m. __S + __HNO3 → __H2SO4 + __NO2 + __H2O
n. __NH3 + __CuO → __Cu + __N2 + __H2O
2N2O5 2N2O4 O2
NH4NO3 N2O  2H2O
2KNO3 2KNO2 O2
NH4NO2  N2 2H2O
2NaHCO3 Na2CO3 H2O  CO2
P4O10 6H2O  4H3PO4
2HCl  CaCO3 CaCl2 H2O  CO2
2Al  3H2SO4 Al2(SO4)3 3H2
CO2 2KOH  K2CO3 H2O
CH4 2O2 CO2 2H2O
Be2C  4H2O  2Be(OH)2 CH4
3Cu  8HNO3 3Cu(NO3)2 2NO  4H2O
S  6HNO3 H2SO4 6NO2 2H2O
2NH3 3CuO  3Cu  N2 3H2O

3.62 Describe the steps involved in the mole
method.
mol-mol: 1 step
g-mol or mol-g: 2 steps
g-g: 3 steps
g-molecule or atom: 4 or more steps

3.64 Which of the following equations best
represents the reaction shown in the
diagram?A + B → C + D
a. 6A + 4B → C + D
b. A + 2B → 2C + D
c. 3A + 2B → 2C + D
d. 3A + 2B → 4C + 2D
On the reactants side, there are 6 A atoms and 4 B atoms. On the products side, there are 4 C atoms and 2 D atoms.
Writing an equation,
6A  4B  4C  2D
Chemical equations are typically written with the smallest set of whole number coefficients. Dividing the
equation by two gives,
3A  2B  2C  D
The correct answer is choice (d).

3.66 Silicon tetrachloride (SiCl4) can be prepared by heating Si in chlorine gas:
Si + 2 Cl2  SiCl4
In one reaction, 0.507 mole of SiCl4 is produced. How many moles of molecular
chlorine were used in the reaction?
Strategy: Looking at the balanced equation, how do we compare the amounts of Cl2 and SiCl4? We can
compare them based on the mole ratio from the balanced equation.
Solution: Because the balanced equation is given in the problem, the mole ratio between Cl2 and SiCl4 is
known: 2 mole Cl2 1 mole SiCl4. From this relationship, we have two conversion factors.
2 4
4 2
2 mol Cl 1 mol SiCl
and
1 mol SiCl 2 mol Cl
Which conversion factor is needed to convert from moles of SiCl4 to moles of Cl2? The conversion factor on
the left is the correct one. Moles of SiCl4 will cancel, leaving units of “mol Cl2” for the answer. We calculate
moles of Cl2 reacted as follows:
2
4
4
2 mol Cl
0.507 mol SiCl
1 mol SiCl
=  =2 2?mol Cl reacted 1.01mol Cl

3.68 Certain race cars use methanol (CH3OH, also called wood alcohol) as a fuel. The
combustion of methanol occurs according to the following unbalanced equation:
CH4O + O2  CO2 + H2O
In a particular reaction, 9.8 moles of CH3OH are reacted with an excess of O2.
Calculate the number of moles of H2O formed.
Starting with the 9.8 mole of CH3OH, we can use the mole ratio from the balanced equation to calculate the moles of
H2O formed.
2CH3OH(l)  3O2(g)  2CO2(g)  4H2O(l)
2
3 2
3
4 mol H O
9.8 mol CH OH 20 mol H O
2 mol CH OH
=  = = 1
2 2? mol H O 2.0 10 mol H O

3.70 When baking soda (sodium bicarbonate or sodium hydrogen carbonate, NaHCO3)
is heated, it releases carbon dioxide gas, which is responsible for the rising of cookies,
donuts, and bread. (a) Write a balanced equation for the decomposition of the
compound (one of the products is Na2CO3). (b) Calculate the mass of NaHCO3 required
to produce 20.5 g of CO2.
(a) 2NaHCO3 Na2CO3 H2O  CO2
(b) Molar mass NaHCO3= 22.99 g  1.008 g  12.01 g  3(16.00 g) = 84.01 g
Molar mass CO2= 12.01 g  2(16.00 g) = 44.01 g
The balanced equation shows one mole of CO2 formed from two molesof NaHCO3.
3 32
2
2 2 3
2 mol NaHCO 84.01 g NaHCO1 mol CO
= 20.5 g CO
44.01 g CO 1 mol CO 1 mol NaHCO
  3mass NaHCO
=78.3 g NaHCO3

3.72 Fermentation is a complex chemical process of wine making in which glucose is
converted into ethanol and carbon dioxide:
__C6H12O6  __ C2H6O + __CO2
Starting with 500.4 g of glucose, what is the maximum amount of ethanol in grams
and in liters that can be obtained by this process? (Density of ethanol = 0.789 g/mL.)
C6H12O6   2C2H5OH  2CO2
glucose ethanol
Strategy: We compare glucose and ethanol based on the mole ratio in the balanced equation. Before we
can determine moles of ethanol produced, we need to convert to moles of glucose. What conversion factor is
needed to convert from grams of glucose to moles of glucose? Once moles of ethanol are obtained, another
conversion factor is needed to convert from moles of ethanol to grams of ethanol.
Solution: The molar mass of glucose will allow us to convert from grams of glucose to moles of glucose.
The molar mass of glucose = 6(12.01 g)  12(1.008 g)  6(16.00 g) = 180.16 g. The balanced equation is
given, so the mole ratio between glucose and ethanol is known; that is 1 mole glucose  2 mole ethanol.
Finally, the molar mass of ethanol will convert moles of ethanol to grams of ethanol. This sequence of three
conversions is summarized as follows:
grams of glucose  moles of glucose  moles of ethanol  grams of ethanol
6 12 6 2 5 2 5
6 12 6
6 12 6 6 12 6 2 5
1 mol C H O 2 mol C H OH 46.07 g C H OH
500.4 g C H O
180.16 g C H O 1 mol C H O 1 mol C H OH
=   2 5? g C H OH
=255.9 g C2H5OH

Chemistry 110 07 8.19.15

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