Slide

Chapter 1
Chemistry and
Measurement

Example of The law of conservation of
mass: Aluminum powder burns in
? oxygen to produce a substance called
aluminum oxide. A sample of 2.00
grams of aluminum is burned in oxygen
and produces 3.78 grams of aluminum
oxide. How many grams of oxygen
were used in this reaction?
aluminum + oxygen = aluminum oxide
2.00 g + oxygen = 3.78 g
oxygen = 1.78 g

Copyright © Houghton Mifflin Company. All rights reserved. 1|2

Potassium is a soft, silvery-colored
metal that melts at 64 C. It reacts
? vigorously with water, with oxygen, and
with chlorine. Identify all of the physical
properties and chemical properties
given in this description.

Physical Property Chemical Property
Soft Reacts with water
Silvery-colored Reacts with oxygen
Melting point (64 C) Reacts with chlorine


Perform the following calculation and
round your answer to the correct
? number of significant figures:

92.35(0.456 0.421)

Calculator answer:
3.23225000
The answer should be rounded to two significant
figures (92.35 X 0.035):
3.2


In winter, the average low
temperature in interior Alaska is
? -30. F (two significant figures).
What is this temperature in degrees
Celsius and in kelvins?
5ο C
tC tF 32ο F ο
9 F
ο ο 5ο C
tC 30. F 32 F ο
9 F
5ο C
tC 62ο F ο
9 F
tC 34.4444444ο C
tC 34ο C

1K
tK tC 273.15 K
1ο C
1K ο
tK 34 C ο 273.15 K
1 C
tK 34 K 273.15 K

tK 239.15 K

tK 239 K


Oil of wintergreen is a colorless liquid

? used as a flavoring. A 28.1 g sample of
oil of wintergreen has a volume of 23.7
mL. What is the density of oil of
wintergreen?
m 28.1g m
d
V 23.7 m L V
28.1 g
d
23.7 mL
g
d 1.18565491
mL
g
d 1.19
mL

The dimensions of Noah’s ark were
reported as 3.0 102 cubits by 5.0
? 101 cubits. Express this size in units
of feet and meters. (1 cubit = 1.5 ft)
1 cubit 1.5 f t
3 f t 1 yd
1 yd 0.9144 m (exact)
1.5 f t
2 1 1.5 ft
3.0 10 cubits 5.0 10 cubits
1 cubit 1 cubit
4.5000000 102 ft 1
7.5000000 10 ft
4.5 102 ft by 7.5 101ft 75 ft


1 cubit 1.5 ft
3 ft 1 yd
1 yd 0.9144 m (exact)

2 1
4.5 10 ft by 7.5 10 ft 75 ft
1 yd 0.9144 m 1 yd 0.9144 m
4.5 10 ft 2
75 ft
3 ft 1 yd 3 ft 1 yd
1.37160000 102 m 22.8600000 m
1.4 102 m by 23 m
Copyright © Houghton Mifflin Company. All rights reserved. 1 | 10

Chapter 2
Atoms,
Molecules,
and Ions


Write the nuclide symbol for the atom that
has 19 protons and 20 neutrons.
Atomic number: Z = 19
The element is potassium, K.
Mass number: A = 19 + 20 = 39

The nuclide symbol is
39
19 K


An element has four naturally occurring isotopes.
The mass and percentage of each isotope are as
follows:
Percentage Abundance Mass (amu)
1.48 203.973
23.6 205.9745
22.6 206.9759
52.3 207.9766

What is the atomic weight and
name of the element?

To find the portion of the atomic weight due to
each isotope, multiply the fraction by the mass of
the isotope. The atomic weight is the sum of these
products.
Fractional Mass (amu) Mass From
Abundance Isotope
0.0148 203.973 3.01880040
0.236 205.9745 48.6099820
0.226 206.9759 46.7765534
0.523 207.9766 108.771762
Total = 207.177098
The atomic weight is 207 amu; the element is lead.

What is formula of the ionic compound of
Mg2+ and N3-?

The common multiple of the charges is 6, so we
need three Mg2+ and two N3-. The resulting formula
is
Mg3N2


Common Monatomic Ions of the
Main-Group Elements
Period IA IIA IIIA IVA VA VIA VIIA
1 H-
2 Li+ Be2+ N3- O2- F-
3 Na+ Mg2+ Al3+ S2- Cl-
4 K+ Ca2+ Ga3+ Se2- Br-
5 Rb+ Sr2+ In3+ Sn2+ Te2- I-
6 Cs+ Ba2+ Tl3+, Pb2+ Bi3+
Tl+

What are the names of the following ionic
compounds?
– BaO
– Cr2(SO4)3

BaO is barium oxide.
Cr2(SO4)3 is chromium(III) sulfate or chromic
sulfate.


What are the chemical formulas for the
following ionic compounds?
– potassium carbonate
– manganese(II) sulfate

The ions K+ and CO32- form K2CO3

The ions Mn2+ and SO42- form MnSO4


What are the names of the following
compounds?
– OF2
– S4N4
– BCl3

OF2 is oxygen difluoride
S4N4 is tetrasulfur tetranitride
BCl3 is boron trichloride


What are the formulas for the following
binary molecular compounds?
– carbon disulfide
– nitrogen tribromide
– dinitrogen tetrafluoride

The formula for carbon disulfide is CS2.
The formula for nitrogen tribromide is NBr3.
The formula for dinitrogen tetrafluoride is N2F4.


A compound whose common name is green
vitriol has the chemical formula FeSO4·7H2O.
What is the chemical name of this
compound?

FeSO4·7H2O is iron(II) sulfate heptahydrate.


Calcium chloride hexahydrate is used to melt
snow on roads. What is the chemical formula
of the compound?

The chemical formula for calcium chloride
hexahydrate is CaCl2·6H2O.


Balance the following equations:
NH3 + O2  NO + H2O
C2H5OH + O2  CO2 + H2O

4NH3 + 5O2  4NO + 6H2O

C2H5OH + 3O2  2CO2 + 3H2O


Calculate the formula weight of the following
compounds from their formulas. Report your
answers to three significant figures.
– calcium hydroxide, Ca(OH)2
– methylamine, CH3NH2


Calculate the formula weight of the following compounds
from their formulas. calcium hydroxide, Ca(OH)2
Ca(OH)2 methylamine, CH3NH2

1 Ca 1(40.08) = 40.08 amu
2O 2(16.00) = 32.00 amu
2H 2(1.008) = 2.016 amu
Total 74.095 3 significant figures
74.1 amu

CH3NH2
1C 1(12.01) = 12.01 amu
1N 1(14.01) = 14.01 amu
5H 5(1.008) = 5.040 amu
Total 31.060 3 significant figures
31.1 amu

AFirst, find the molar mass of ,HNO3: 0.253 mol
sample of nitric acid, HNO3 contains
HNO3. How many grams is this?
1 H 1(1.008) = 1.008
1 N 1(14.01) = 14.01
3 O 3(16.00) = 48.00
63.018 (2 decimal places)
63.02 g/mol


Next, using the molar mass, find the mass of 0.253
mole:

63.02 g
0.253 mole x
1 mole
= 15.94406 g

15.9 g
(3 significan figures)
t


The average daily requirement of the
essential amino acid leucine, C6H14O2N, is 2.2
g for an adult. What is the average daily
requirement of leucine in moles?

First, find the molar mass of leucine:

6C 6(12.01) = 72.06
2O 2(16.00) = 32.00
1N 1(14.01) = 14.01
14 H 14(1.008) = 14.112 2 decimal places
132.182 132.18 g/mol


Next, find the number of moles in 2.2 g:

1 mole
2.2 g x
132.18g
2
1.6643x 10 mol

2
1.7 x 10 mol or 0.017 mol
t


The daily requirement of chromium in the human diet is 1.0 10-6 g. How many
atoms of chromium does this represent?

First, find the molar mass of Cr:
1 Cr 1(51.996) = 51.996 g/mol

Now, convert 1.0 x 10-6 grams to moles:
23
6 1 mol 6.02 x 10 atoms
1.0 x 10 gx x
51.996g 1 mol
=1.157781368 x 1016 atoms

1.2 x 1016 atoms
(2 significant figures)


Lead(II) chromate, PbCrO4, is used as a paint
pigment (chrome yellow). What is the percentage
composition of lead(II) chromate?

First, find the molar mass of PbCrO4:

1 Pb 1(207.2) = 207.2
1 Cr 1(51.996) = 51.996
4 O 4(16.00) = 64.00
323.196 (1 decimal place)
323.2 g/mol


Now, convert each to percent composition:
207.2 g
Pb : x 100% 64.11%
323.20g
51.996g
Cr : x 100% 16.09%
323.20g
64.00 g
O: x 100% 19.80%
323.20g
Check:
64.11 + 16.09 + 19.80 = 100.00 %


Benzene has the empirical formula CH. Its
molecular weight is 78.1 amu. What is its
molecular formula?

Empirical formula w eight (12.01 1.008) amu
13.02 amu
78.1
6
13.02
Molecular formula
C6H6


Hexamethylene is one of the materials used
to produce a type of nylon. It is composed of
62.1% C, 13.8% H, and 24.1% N. Its
molecular weight is 116 amu. What is its
molecular formula?
1 mol C 5.171
62.1 g C X 5.171 mol C 3
12.01 g C 1.720
1 mol H 13.69
13.8 g H X 13.69 mol H 8
1.008 g H 1.720
1 mol N 1.720
24.1 g N X 1.720 mol N 1
14.01 g N 1.720
Empirical formula
C3H8N

The empirical formula is C3H8N.
Molecular formula = (Empirical formula)n
Find the empirical formula weight:
3(12.01) + 8(1.008) + 1(14.01) = 58.104 amu
116
n 2 Molecular formula: C H N
58.10 6 16 2


Propane, C3H8, is normally a gas, but it is sold
as a fuel compressed as a liquid in steel
cylinders. The gas burns according to the
following equation:

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)

How many grams of CO2 are produced when
20.0 g of propane is burned?


C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)

Molar masses
C3H8: 3(12.01) + 8(1.008) = 44.094 g
CO2: 1(12.01) + 2(16.00) = 44.01 g

1 mol C3H8 3 mol CO 2 44.01 g CO 2
20.0 g C3H8 X X X
44.094 g C3H8 1 mol C3H8 1 mol CO 2
59.8856987 g CO 2
3

59.9 g CO2


Urea, CH4N2O, is used as a nitrogen fertilizer.
It is manufactured from ammonia and carbon
dioxide at high pressure and high temperature:

2NH3 + CO2(g)  CH4N2O + H2O

In a laboratory experiment, 10.0 g NH3 and
10.0 g CO2 were added to a reaction vessel.
What is the maximum quantity (in grams) of
urea that can be obtained? How many grams
of the excess reactant are left at the end of the
reactions?


2NH3 + CO2(g)  CH4N2O + H2O

Molar masses: NH3 1(14.01) + 3(1.008) = 17.02 g
CO2 1(12.01) + 2(16.00) = 44.01 g
CH4N2O 1(12.01) + 4(1.008) + 2(14.01)
+ 1(16.00) = 60.06 g
1 mol NH 3 1 mol CH 4N2 O 60.06 g CH 4N2 O
10.0 g NH 3 X X X
17.02 g NH 3 2 mol NH 3 1 mol CH 4N2O
17.6 g CH 4N2 O
1 mol CO 2 1 mol CH 4N2 O 60.06 g CH 4N2 O
10.0 g CO 2 X X X
44.01 g CO 2 1 mol CO 2 1 mol CH 4N2 O
13.6 g CH 4N2O
CO2 is the limiting reactant.
13.6 g CH4N2O will be produced.

2NH3 + CO2(g)  CH4N2O + H2O

To find the excess NH3, we find how much NH3 reacted:
1 mol CO 2 2 mol NH 3 17.02 g NH 3
10.0 g CO 2 X X X
44.01 g CO 2 1 mol CO 2 1 mol NH 3
7.73460577g NH3
7.73 g NH3 reacted
Now subtract the amount reacted from the
starting amount: 10.0 g at start
-7.73 g reacted
2.27 g remains
2.3 g NH3 is left unreacted.
(1 decimal place)

2NH3 + CO2(g)  CH4N2O + H2O

When 10.0 g NH3 and 10.0 g CO2 are added
to a reaction vessel, the limiting reactant is
CO2. The theoretical yield is 13.6 g of urea.
When this reaction was carried out, 9.3 g of
urea was obtained. What is the percent
yield?

Theoretical yield = 13.6 g 9.3 g
x 100%
Actual yield = 9.3 g 13.6 g
= 68% yield

Chapter 4

Chemical Reactions

Determine whether each of the following
reactions occurs. If it does, write the
molecular, ionic, and net ionic equations.
KBr + MgSO4 
NaOH + MgCl2 
K3PO4 + CaCl2 


Classify the following as strong or weak
acids or bases:
a. KOH
b. H2S
c. CH3NH2
d. HClO4


Write the molecular, ionic, and net ionic
equations for the neutralization of
sulfurous acid, H2SO3, by potassium
hydroxide, KOH.


Write the molecular, ionic, and net ionic
equations for the reaction of copper(II)
carbonate with hydrochloric acid.


Potassium permanganate, KMnO4, is a
purple-colored compound; potassium
manganate, K2MnO4, is a green-
colored compound. Obtain the
oxidation numbers of the manganese in
these compounds.


What is the oxidation number of Cr in
dichromate, Cr2O72-?


Balance the following oxidation-reduction reaction:
FeI3(aq) + Mg(s)  Fe(s) + MgI2(aq)


You place a 1.52 g of potassium dichromate,
K2Cr2O7, into a 50.0 mL volumetric flask. You
then add water to bring the solution up to the
mark on the neck of the flask. What is the
molarity of K2Cr2O7 in the solution?

Molar mass of K2Cr2O7 is 294 g.

1 mol
1.52 g
294 g
0.103 M
50.0 x 10-3 L


A solution of sodium chloride used for
intravenous transfusion (physiological saline
solution) has a concentration of 0.154 M
NaCl. How many moles of NaCl are
contained in 500. mL of physiological saline?
How many grams of NaCl are in the 500. mL
of solution?

mol M V M olarm ass NaCl 58.4 g
0.154 M 0.500 L 58.4 g
0.0770m ol
0.0770 mol NaCl 1 m ol
4.50 g NaCl

A saturated stock solution of NaCl is 6.00 M. How
much of this stock solution is needed to prepare
1.00 L of physiological saline solution (0.154 M)?

M iVi M fVf (0.154 M )(1.00 L)
Vi
M fVf 6.00 M
Vi
Mi Vi 0.0257L or 25.7 mL


A soluble silver compound was analyzed for
the percentage of silver by adding sodium
chloride solution to precipitate the silver ion
as silver chloride. If 1.583 g of silver
compound gave 1.788 g of silver chloride,
what is the mass percent of silver in the
compound?


Molar mass of silver chloride (AgCl) = 143.32 g

1 mol AgCl 1 mol Ag 107.9 g Ag
1.788 g AgCl x x x
143.32g AgCl 1 mol AgCl 1 mol Ag

= 1.346 g Ag in the compound

1.346 g Ag
X 100%
1.583 g silver compound

= 85.03% Ag


Zinc sulfide reacts with hydrochloric acid to
produce hydrogen sulfide gas:

ZnS(s) + 2HCl(aq) ZnCl2(aq) + H2S(g)

How many milliliters of 0.0512 M HCl are
required to react with 0.392 g ZnS?


ZnS(s) + 2HCl(aq) ZnCl2(aq) + H2S(g)

Molar mass of ZnS = 97.47 g

1 mol ZnS 2 mol HCl 1 L solution
0.392 g ZnS x x x
97.47 g ZnS 1 mol ZnS 0.0512mol HCl

= 0.157 L = 157 mL HCl solution


A dilute solution of hydrogen peroxide is sold
in drugstores as a mild antiseptic. A typical
solution was analyzed for the percentage of
hydrogen peroxide by titrating it with
potassium permanganate:

5H2O2(aq) + 2KMnO4(aq) + 6H+(aq) 
8H2O(l) + 5O2(g) + 2K+(aq) + 2Mn2+(aq)

What is the mass percent of H2O2 in a
solution if 57.5 g of solution required 38.9
mL of 0.534 M KMnO4 for its titration?

5H2O2(aq) + 2KMnO4(aq) + 6H+(aq) 
8H2O(l) + 5O2(g) + 2K+(aq) + 2Mn2+(aq)
Molar mass of H2O2 = 34.01 g
0.534 mol KMnO4
3 5 mol H2 O 2 34.01g H2 O 2
38.9 x 10 L x x x
1L 2 mol KMnO4 1 mol H2 O 2

= 1.77 g H2O2

1.77 g H2O 2
X 100%
57.5 g solution

= 3.07% H2O2


A volume of oxygen gas occupies 38.7
mL at 751 mmHg and 21 C. What is the
volume if the pressure changes to 359
mmHg while the temperature remains
constant?
PiVi
Vf
Pf
(38.7 mL)(751mmHg)
Vf
(359 mmHg)

= 81.0 mL

You prepared carbon dioxide by adding
HCl(aq) to marble chips, CaCO3. According
to your calculations, you should obtain 79.4
mL of CO2 at 0 C and 760 mmHg. How
many milliliters of gas would you obtain at
27 C?


Vi = 79.4 mL Vf = ?
Pi = 760 mmHg Pf = 760 mmHg
Ti = 0°C = 273 K Tf = 27°C = 300. K

TfVi
Vf
Ti
(300. K)(79.4 mL)
Vf
(273 K)

= 87.3 mL


Divers working from a North Sea drilling
platform experience pressure of 5.0 101
atm at a depth of 5.0 102 m. If a balloon is
inflated to a volume of 5.0 L (the volume of
the lung) at that depth at a water
temperature of 4 C, what would the volume
of the balloon be on the surface (1.0 atm
pressure) at a temperature of 11 C?


Vi = 5.0 L Vf = ?
Pi = 5.0 101 atm Pf = 1.0 atm
Ti = 4°C = 277 K Tf = 11°C = 284. K

Tf PiVi
Vf
Ti Pf
(284 K)(5.0 x 101 atm)(5.0L)
Vf
(277 K)(1.0 atm)
= 2.6 x 102 L


Ideal Gas Law
The ideal gas law is
given by the equation
PV=nRT

The molar gas
constant, R, is the
constant of
proportionality that
relates the molar
volume of a gas to T/P.


A 50.0 L cylinder of nitrogen, N2, has a pressure
of 17.1 atm at 23 C. What is the mass of
nitrogen in the cylinder?
V = 50.0 L PV
P = 17.1 atm n
RT
T = 23°C = 296 K
(17.1atm)(50.0 L)
n 35.20m ol
L atm
0.08206 (296 K)
mol K
28.01 g
mass 35.20 mol X mass = 986 g
mol (3 significant figures)

Mm = 16.04 g/mol Mm P
d
P = 3.50 atm
What is the density of methaneRT (natural gas),
gas
T = 125°C = 398 K
CH4, at 125 C and 3.50 atm?
g
(16.04 )(3.50 atm) g
mol d 1.72
d L
L atm
0.08206 (398 K) (3 significan figures)
t
mol K


A 500.0-mL flask containing a sample of
octane (a component of gasoline) is placed
in a boiling water bath in Denver, where the
atmospheric pressure is 634 mmHg and
water boils at 95.0 C. The mass of the vapor
required to fill the flask is 1.57 g. What is the
molar mass of octane? (Note: The empirical
formula of octane is C4H9.) What is the
molecular formula of octane?


d = 1.57 g/0.5000 L dRT
= 3.140 g/L Mm
P
P = 634 mmHg
= 0.8342 atm
T = 95.0°C = 368.2 K
g L atm
3.140 0.08206 368.2K
L mol K
Mm
(0.8342atm)
g
M m 114
mol
t


Molar mass = 114 g/mol
Empirical formula: C4H9
Empirical formula molar mass = 57 g/mol

g
114
n mol 2
g
57
mol

Molecular formula: C8H18


When a 2.0-L bottle of concentrated HCl was
spilled, 1.2 kg of CaCO3 was required to
neutralize the spill. What volume of CO2 was
released by the neutralization at 735 mmHg
and 20. C?


First, write the balanced chemical equation:

CaCO3(s) + 2HCl(aq) 
CaCl2(aq) + H2O(l) + CO2(g)

Second, calculate the moles of CO2 produced:

Molar mass of CaCO3 = 100.09 g/mol

1 mol CaCO 3
3 1 mol CO 2
1.2 x 10 g CaCO 3 X X
100.09 g CaCO 3 1 mol CaCO 3

Moles of CO2 produced = 12 mol


nRT
n = 12 mol V
P = 735 mmHg P
= 0.967 atm
T = 20°C = 293 K

L atm
12 mol 0.08206 (293 K)
mol K
V
(0.967 atm)

= 3.0 102 L


A 100.0-mL sample of air exhaled from the
lungs is analyzed and found to contain
0.0830 g N2, 0.0194 g O2, 0.00640 g CO2,
and 0.00441 g water vapor at 35 C. What is
the partial pressure of each component and
the total pressure of the sample?


1 mol N2 L atm
0.0830 g N2 0.08206 308 K
28.01 g N2 mol K
PN 0.749 atm
2 1L
100.0 mL
10 3 mL
1 mol O 2 L atm
0.0194 g O 2 0.08206 308 K
32.00 g O 2 mol K
PO 0.153 atm
2 1L
100.0 mL
10 3 mL
1 mol CO 2 L atm
0.00640 g CO 2 0.08206 308 K
44.01 g CO 2 mol K
PCO 0.0368atm
2 1L
100.0 mL
10 3 mL
1 mol H2 O L atm
0.00441 g H2 O 0.08206 308 K
18.01 g H2 O mol K
PH O 0.0619atm
2 1L
100.0 mL
10 3 mL

PN2 0.749atm
PO2 0.153atm
PCO 2 0.0368atm
PH2O 0.0619atm

P PN2 PO2 PCO 2 PH2O

P = 1.00 atm


The partial pressure of air in the alveoli
(the air sacs in the lungs) is as follows:
nitrogen, 570.0 mmHg; oxygen, 103.0
mmHg; carbon dioxide, 40.0 mmHg;
and water vapor, 47.0 mmHg. What is
the mole fraction of each component of
the alveolar air?
P PN2 PO2 PCO 2 PH2O
570.0 mmHg
103.0 mmHg
40.0 mmHg
47.0 mmHg
P = 760.0 mmHg

Mole fraction of N2 Mole fraction of O2
570.0 mmHg 103.0 mmHg
760.0 mmHg 760.0 mmHg
Mole fraction of CO2 Mole fraction of H2O
40.0 mmHg 47.0 mmHg
760.0 mmHg 760.0 mmHg
Mole fraction N2 = 0.7500

Mole fraction O2 = 0.1355

Mole fraction CO2 = 0.0526

Mole fraction H2O= 0.0618

You prepare nitrogen gas by heating
ammonium nitrite:
NH4NO2(s)  N2(g) + 2H2O(l)
If you collected the nitrogen over water
at 23 C and 727 mmHg, how many
liters of gas would you obtain from 5.68
g NH4NO2?


P = 727 mmHg Molar mass NH4NO2
Pvapor = 21.1 mmHg = 64.04 g/mol
Pgas = 706 mmHg nRT
T = 23°C = 296 K
V
P

1 mol NH4NO 2 1 mol N2
5.68 g NH4NO2 X X
64.04 g NH4NO 2 1 mol NH4NO 2

= 0.08869 mol N2 gas


P = 727 mmHg
Pvapor = 21.1 mmHg
Pgas = 706 mmHg nRT
V
T = 23°C = 296 K P
n = 0.08869 mol
L atm
0.0887 mol 0.08206 (296 K)
mol K
V
1 atm
706 mmHg x
760 mmHg

= 2.32 L of N2

What is the rms speed of carbon dioxide
molecules in a container at 23 C?

3RT kg m 2
u rms s2
Mm 3 8.3145 296 K
mol K
Recall
2 u rms
kg m kg
J 0.04401
s2 mol

5 m2 2 m
urms 1.68x10 2 urms 4.10 x10
s s

Both hydrogen and helium have been used
as the buoyant gas in blimps. If a small leak
were to occur, which gas would effuse more
rapidly and by what factor?

1
Rate H2 2.016 4.002
Rate He 1 2.016
4.002

Hydrogen will diffuse more quickly by a factor of 1.4.


A person weighing 75.0 kg (165 lbs) runs a course
at 1.78 m/s (4.00 mph). What is the person’s
kinetic energy?

m = 75.0 kg EK = ½ mv2
V = 1.78 m/s
2
1 m
EK (75.0 kg) 1.78
2 s
kg m 2
E K 119 2
119 J
s
t

In an endothermic reaction:
The reaction vessel cools.
Heat is absorbed.
Energy is added to the system.
q is positive.

In an exothermic reaction:
The reaction vessel warms.
Heat is evolved.
Energy is subtracted from the system.
q is negative.

Sulfur, S8, burns in air to produce sulfur
dioxide. The reaction evolves 9.31 kJ of
heat per gram of sulfur at constant
pressure. Write the thermochemical
equation for this reaction.
S8(s) + 8O2(g)  8SO2(g)
9.31 kJ 256.52 g S 8
ΔH
1 g S8 1 mol S 8
ΔH 2.39 10 3 kJ

S8(s) + 8O2(g)  8SO2(g); H = –2.39 103 kJ

You burn 15.0 g sulfur in air. How much
heat evolves from this amount of sulfur?
The thermochemical equation is
S8(s) + 8O2(g)  8SO2(g); H = -2.39 x 103 kJ

Molar mass of S8 = 256.52 g
1 mol S 8 2.39 x 103 kJ
q 15.0 g S 8
256.5 g S 8 1 mol S 8

q = –1.40 102 kJ

8.8 x 103 kJ. If all this energy were to be
provided by the combustion of glucose,
C6H12O6, how many grams of glucose would
have to be consumed by the man and the
woman per day?
C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l);
Ho = -2.82 x 103 kJ

The daily energy requirement for a 20-year-old
man weighing 67 kg is 1.3 x 104 kJ. For a 20-year-
old woman weighing 58 kg, the daily requirement
is


C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l);
H = -2.82 x 103 kJ
For a 20-year-old man weighing 67 kg:
1 mol glucose 180.2g glucose
4
mglucose 1.3x10 kJ 3
2.82x10 kJ 1 mol glucose
= 830 g glucose required
For a 20-year-old woman weighing 58 kg:
1 mol glucose 180.2 g glucose
3
mglucose 8.8x10 kJ 3
2.82x10 kJ 1 mol glucose
= 560 g glucose required

A piece of zinc weighing 35.8 g was heated from
20.00 C to 28.00 C. How much heat was
required? The specific heat of zinc is 0.388
J/(g C).
m = 35.8 g
s = 0.388 J/(g C) q=m·s· t
t = 28.00 C – 20.00 C = 8.00 C
0.388 J
q 35.8 g 8.00 C
gC

q = 111 J

CH3NO2(g) + 3/4O2(g) 
CO2(g) organic O(l) + 1burns
Nitromethane, CH3NO2, an
+ 3/2H2 solvent /2N2(g)
You placeaccording of the following reaction:
in oxygen 1.724 g to nitromethane in a
calorimeter with oxygen and ignite it. The
temperature of the calorimeter increases
from 22.23 C to 28.81°C. The heat capacity
of the calorimeter was determined to be
3.044 kJ/°C. Write the thermochemical
equation for the reaction.

We first find the heat evolved for the 1.724 g of
nitromethane, CH3NO2.
q rxn C cal Δt
3.044 kJ
q rxn 28.81 C 22.23 C 20.03kJ
C
Now, covert that to the heat evolved per mole by
using the molar mass of nitromethane, 61.04 g.
- 20.03kJ 61.04 g CH3NO 2
q rxn
1.724 g CH3NO 2 1 mol CH3NO 2
H = –709 kJ
CH3NO2(l) + ¾O2(g)  CO2(g) + 3/2H2O(l) + ½N2(g);
H = –709 kJ

What is the enthalpy of reaction, H, for the
reaction of calcium metal with water?

Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g)
This reaction occurs very slowly, so it is
impractical to measure H directly. However,
the following facts are known:
H+(aq) + OH-(aq)  H2O(l); H = –55.9 kJ
Ca(s) + 2H+(aq)
 Ca2+(aq) + H2(g); H = –543.0 kJ

Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g)

2H2O(l)  2H+(aq) + 2OH-(aq); H = +111.8 kJ

Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g); H = –543.0 kJ

Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g);
H = –431.2 kJ


What is the heat of vaporization of methanol,
CH3OH, at 25 C and 1 atm?
Use standard enthalpies of formation
(Appendix C).
We want H for the reaction:CH3OH(l)  CH3OH(g)
ΔHreaction nΔH f nΔH f
products reactants
kJ
For liquidmethanol: ΔH f 238.7
mol
kJ
For gaseous methanol: ΔH f 200.7
mol
kJ kJ
ΔH vap 1mol 200.7 1mol 238.7 = +38.0 kJ
mol mol

Methyl alcohol, CH3OH, is toxic because
liver enzymes oxidize it to formaldehyde,
HCHO, which can coagulate protein.
Calculate Ho for the following reaction:

2CH3OH(aq) + O2(g)  2HCHO(aq) + 2H2O(l)

Standard enthalpies of formation, ΔH fo:
CH3OH(aq): -245.9 kJ/mol
HCHO(aq): -150.2 kJ/mol
H2O(l): -285.8 kJ/mol


We want H for the reaction:
2CH3OH(aq) + O2(aq)  2HCHO(aq) + 2H2O(l)
ΔHreaction nΔH f nΔH f
products reactants
kJ kJ
ΔH reacton
o
2 mol 150.2 2 mol - 285.8
mol mol
kJ kJ
2 mol 245.9 1mol 0
mol mol
o
ΔHreaction 300.4kJ 571.6kJ 491.8kJ
o
ΔHreaction 872.0kJ 491.8kJ
o
ΔHreaction 380.2kJ

Chapter 7

Quantum Theory of An Atome

What is the wavelength of blue light with a
frequency of 6.4 1014/s?

= 6.4 1014/s c= so
c = 3.00 108 m/s = c/
8 m
3.00 x 10
c s
λ
14 1
6.4 x 10
s

= 4.7 10-7 m


What is the frequency of light having a wavelength
of 681 nm?
= 681 nm = 6.81 10-7 m c= so
c = 3.00 108 m/s = c/

8 m
3.00 x 10
c s
6.81x 10 7 m

v = 4.41 1014 /s


The range of frequencies and wavelengths of
electromagnetic radiation is called the
electromagnetic spectrum.


The blue–green line of the hydrogen atom
spectrum has a wavelength of 486 nm. What is the
energy of a photon of this light?

= 486 nm = 4.86 10-7 m E = h and
c = 3.00 108 m/s c= so
h = 6.63 10-34 J · s E = hc/
34 8 m
6.63x10 Js 3.00 x 10
hc s
E
λ 4.86 x 10 7 m

E = 4.09 10-19 J

What is the wavelength of the light emitted
when the electron in a hydrogen atom
undergoes a transition from n = 6 to n = 3?
1 1
ΔE RH
ni = 6 nf 2 ni 2
nf = 3 hc hc
RH = 2.179 10-18 J ΔE so λ
λ ΔE

18 1 1
ΔE 2.179 x 10 J 2 = -1.816 x 10-19 J
3 62
34 m 8
6.626 x 10 J s 2.998 x 10
s
λ 19 1.094 10-6 m
- 1.816 x 10 J

Compare the wavelengths of
(a) an electron traveling at a speed that is one-
hundredth the speed of light and (b) a baseball of
mass 0.145 kg having a speed of 26.8 m/s (60
mph).


Electron
me = 9.11 10-31 kg; v = 3.00 106 m/s
34
6.63 x 10 J s
λ 2.43 10-10 m
31 m 6
9.11x 10 kg 3.00 x 10
s
Baseball
m = 0.145 kg; v = 26.8 m/s
34
6.63 x 10J s
λ 1.71 10-34 m
m
0.145 kg 26.8 Comment: This is such an
s exceedingly small wavelength that
the wave properties of a baseball
cannot be detected by any existing
measuring device.

Which of the following are permissible sets of
quantum numbers?
n = 4, l = 4, ml = 0, ms = ½
n = 3, l = 2, ml = 1, ms = -½
n = 2, l = 0, ml = 0, ms = ³/²
n = 5, l = 3, ml = -3, ms = ½
(a) Not permitted. When n = 4, the maximum
value of l is 3.
(b) Permitted.
(c) Not permitted; ms can only be +½ or –½.
(b) Permitted.

Chapter 8

Electron Configuration and Periodicity

The building-up principle (or aufbau principle) is
a scheme used to reproduce the ground-state
electron configurations by successively filling
subshells with electrons in a specific order (the
building-up order).

This order generally corresponds to filling the
orbitals from lowest to highest energy. Note that
these energies are the total energy of the atom
rather than the energy of the subshells alone.


1s
2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f
6s 6p 6d
7s 7p
This results in the following order:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p,
6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p


Another way to learn the building-up order is to
correlate each subshell with a position on the
periodic table.

The principal quantum number, n, correlates with
the period number.

Groups IA and IIA correspond to the s subshell;
Groups IIIA through VIIIA correspond to the p
subshell; the ―B‖ groups correspond to the d
subshell; and the bottom two rows correspond to
the f subshell. This is shown on the next slide.

Write the complete electron configuration of
the arsenic atom, As, using the building-up
principle.

For arsenic, As, Z = 33.

1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3


What are the electron configurations for the
valence electrons of arsenic and zinc?

Arsenic is in period 4, Group VA.
Its valence configuration is
4s24p3.

Zinc, Z = 30, is a transition metal in the first transition
series.
Its noble-gas core is Ar, Z = 18.
Its valence configuration is
4s23d10.


Write an orbital diagram for the ground state
of the nickel atom.
For nickel, Z = 28.

1s 2s 2p

3s 3p

4s 3d

Which of the following electron
configurations or orbital diagrams are
allowed and which are not allowed by the
Pauli exclusion principle? If they are not
allowed, explain why?
a. 1s22s12p3 a. Allowed; excited.
b. 1s22s12p8 b. p8 is not allowed.
c. 1s22s22p63s23p63d8 c. Allowed.
d. 1s22s22p63s23p63d11 d. d11 is not allowed.
e. Not allowed; electrons in one
e. orbital must have opposite
spins.

1s 2s


A representation of atomic radii is shown below.


Refer to a periodic table and arrange the
following elements in order of increasing
atomic radius: Br, Se, Te.

Te is larger than Se.
Se is larger than Br.

Br < Se < Te


The size of each sphere indicates the size of the
ionization energy in the figure below.


Refer to a periodic table and arrange the
following elements in order of increasing
ionization energy: As, Br, Sb.

Sb is larger than As.
As is larger than Br.

Ionization energies:
Sb < As < Br


The electron affinity is > 0, so the element must be in Group IIA or VIIIA.

The dramatic difference in ionization energies is at the third ionization.

The element is in Group IIA.


For R2O5 oxides, R must be in Group VA.
R is a metalloid, so R could be As or Sb.
The oxide is acidic, so

R is arsenic, As.


Chapter 9

Ionic and Covalent Bonding

Table 9.1 illustrates the Lewis electron-dot
symbols for second- and third-period atoms.


For chlorine, Cl, Z = 17, so the Cl- ion has 18
electrons. The electron configuration for Cl- is

Give the electron configuration and the Lewis
symbol for the chloride ion, Cl-.

1s2 2s2 2p6 3s2 3p6
-
The Lewis symbol for Cl- is [ Cl ]

Manganese, Z = 25, has 25 electrons;. Its electron
configuration is
Give the electron configurations of Mn and Mn2+.
1s2 2s2 2p6 3s2 3p6 3d5 4s2

Mn2+ has 23 electrons. When ionized, Mn loses
the 4s electrons first; the electron configuration for
Mn2+ is
1s2 2s2 2p6 3s2 3p6 3d5

Using the periodic table only, arrange the following
ions in order of increasing ionic radius: Br-, Se2-,
Sr2+. These ions are isoelectronic,
so their size decreases with increasing
atomic number:

Sr2+ < Br- < Se2-


Using electronegativities, arrange the following
bonds in order by increasing polarity:
C—N
Na—F
O—H
For C—N, the difference is 3.0 (N) – 2.5 (C) = 0.5.
For Na—F, the difference is
4.0 (F) – 0.9 (Na) = 3.1.

For O—H, the difference is 3.5 (O) – 2.1 (H) = 1.4.

Bond polarities: C—N < O—H < Na—F

Writing Lewis Electron-Dot Formulas
Calculate the number of valence electrons.
Write the skeleton structure of the molecule or ion.
The central atom is the one with less
electronegativity.
Distribute electrons to the atoms surrounding the
central atom or atoms to satisfy the octet rule.
Distribute the remaining electrons as pairs to the
central atom or atoms.


Write the electron dot formulas for the
following:
a. OF2
b. NF3
c. NH2OH, hydroxylamine


a. Count the valence electrons in OF2:
O 1(6)
F 2(7) F O F

20 valence electrons

O is the central atom (it is less electronegative).
Now, we distribute the remaining 16 electrons,
beginning with the outer atoms. The last four
electrons go on O.


b. Count the valence electrons in NF3:
N 1(5)
F N F
F 3(7)
26 valence electrons F

N is the central atom (it is less electronegative).
Now, we distribute the remaining 20 electrons,
beginning with the outer atoms. The last two
electrons go on N.


c. Count the electrons in NH2OH:
N 1(5)
H N O H
H 3(1)
O 1(6) H

N is the central atom (it is less electronegative
than O). Now, we distribute the remaining six
electrons, beginning with the outer atoms. The last
two electrons go on N.


Write electron-dot formulas for the following:
a. CO2
b. HCN


a. Count the electrons in CO2:
C 1(4)
O 2(6) C
O O
C is the central atom. Now, we distribute the
remaining 12 electrons, beginning with the outer
atoms.
Carbon does not have an octet, so two of the lone
pairs shift to become a bonding pair, forming
double bonds.


b. Count the electrons in HCN:
H 1(1)
C 1(4)
N 1(5) H C N
10 valence electrons.

C is the central atom. The remaining electrons go
on N.
Carbon does not have an octet, so two of the lone
pairs shift to become a bonding pair, forming a
triple bond.


Phosphorus pentachloride exists in solid
state as the ionic compound [PCl4]+[PCl6]-; it
exists in the gas phase as the PCl5 molecule.
Write the Lewis formula of the PCl4+ ion.


Count the valence electrons in PCl4+:
P 1(5) Cl +
Cl 4(7)
-1 Cl P Cl
32
Cl

P is the central atom. The remaining 24 nonbonding
electrons are placed on Cl atoms. Add square
brackets with the charge around the ion.

Give the Lewis formula of the IF5 molecule.
Count the valence electrons in IF5:
I 1(7)
F
F 5(7)
F
F
I
F
F

I is the central atom. Thirty-two electrons remain;
they first complete F octets. The remaining
electrons go on I.


Compare the formal charges for the following
electron-dot formulas of CO2.

O C O O C O

Formal charge = group number – (number of bond
pairs) – (number of nonbonding electrons)
For the left structure: For the right structure:
C: 4–4–0=0 C: 4–4–0=0
O: 6–2–4=0 O: 6 – 1 – 6 = –1
O: 6 – 3 – 2 = +1
The left structure is better.

The diagrams below illustrate molecular geometry
and the impact of lone pairs on it for linear and
trigonal planar electron-pair arrangements.


Molecular geometries with a tetrahedral electron-
pair arrangement are illustrated below.


Molecular geometries for the trigonal bipyramidal
electron-pair arrangement are shown on the next
slide.


Molecular geometries for the octahedral electron-
pair arrangement are shown below.


Use the VSEPR model to predict the
geometries of the following molecules:
a. AsF3
b. PH4+
c. BCl3

a. Trigonal pyramidal.
b. Tetrahedral.
c. Trigonal planar.

Using the VSEPR model, predict the
geometry of the following species:
a. ICl3
b. ICl4-

a. T-shaped.
b. Square planar.


Which of the following molecules would be
expected to have a zero dipole moment?
a. GeF4
b. SF2
c. XeF2
d. AsF3
a. GeF4 tetrahedral molecular geometry
zero dipole moment
b. SF2 bent molecular geometry
nonzero dipole moment
c. XeF2 linear molecular geometry
zero dipole moment
d. AsF3 trigonal pyramidal molecular geometry
nonzero dipole moment

Hybrid orbitals have definite directional
characteristics, as described in Table 10.2.


F N N F

F F

Use valence bond theory to describe the bonding
about an N atom in N2H4.


The sp3 hybridized N atom is

1s sp3

Consider one N in N2F4: the two N—F bonds are
formed by the overlap of a half-filled sp3 orbital
with a half-filled 2p orbital on F. The N—N bond
forms from the overlap of a half-filled sp3 orbital on
each. The lone pair occupies one sp3 orbital.


Use valence bond theory to describe the bonding
in the ClF2- ion.


The sp3d hybridized orbital diagram for the Cl- ion is

sp3d 3d
Two Cl—F bonds are formed from the overlap of
two half-filled sp3d orbitals with half-filled 2p
orbitals on the F atom. These use the axial
positions of the trigonal bipyramid.

Three lone pairs occupy three sp3d orbitals. These
are in the equatorial position of the trigonal
bipyramid.

Describe the bonding about the C atom
in formaldehyde, CH2O, using valence
bond theory. O

C
H H


After hybridization, the orbital diagram for C is

1s sp2 2p

The C—H bonds are formed from the overlap of two
C sp2 hybrid orbitals with the 1s orbital on the H atoms.

The C—O bond is formed from the overlap of one
sp2 hybrid orbital and one O half-filled p orbital.

The C—O bond is formed from the sideways overlap
of the C 2p orbital and an O 2p orbital.

Give the orbital diagram and electron configuration
of the F2 molecule.
Is the molecular substance diamagnetic or
paramagnetic?
What is the order of the bond in F2?


F2 has 18 electrons. The KK shell holds 4
electrons so 14 remain.

*
2s 2s 2p 2p 2p 2p

The molecular electron configuration is
KK( 2s)2( 2s)2( 2p)4( 2p)2 ( *2p)4
The bond order is ½(8 - 6) = 1.
The molecule is diamagnetic.

For F2 and O2, 2p is lower in energy than 2p. This
order would not affect the determination of bond
order and magnetic properties for these molecules.

A number of compounds of the nitrosonium
ion, NO+, are known, including nitrosonium
hydrogen sulfate, (NO+)(HSO4-). Use the
molecular orbitals similar to those of
homonuclear diatomic molecules and obtain
the orbital diagram, electron configuration,
bond order, and magnetic characteristics of
the NO+ ion.

Note: The stability of the ion results from the
loss of an antibonding electron from NO.


NO+ has 14 electrons. The KK shell holds 4
electrons, leaving 10 electrons for bonding.

*
2s 2s 2p 2p 2p 2p

The molecular electron configuration is
KK( 2s)2( 2s)2( 2p)4( 2p)2

The bond order is ½(8 - 2) = 3.

The ion has a diamagnetic molecular orbital
electron configuration.

The valence-shell electron-pair repulsion (VSEPR)
model predicts the shapes of molecules and ions
by assuming that the valence-shell electron pairs
are arranged about each atom so that electron
pairs are kept as far away from one another as
possible, thereby minimizing electron pair
repulsions.

The diagram on the next slide illustrates this.


Two electron pairs are 180 apart ( a linear
arrangement).
Three electron pairs are 120° apart in one plane
(a trigonal planar arrangement).
Four electron pairs are 109.5° apart in three
dimensions (a tetrahedral arrangement).

Five electron pairs are arranged with three pairs in
a plane 120° apart and two pairs at 90°to the plane
and 180° to each other (a trigonal bipyramidal
arrangement).

Six electron pairs are 90° apart (an octahedral
arrangement).

This is illustrated on the next slide.


These arrangements are illustrated below with
balloons and models of molecules for each.


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