1.
a.
Classify each of the following as molecular, ionic or other.
CF2Cl2
b.
CO2
KF
HNCl2
MgSO4
Xe
PF3
HOCl
Show the Lewis structure of each substance you classified as molecular. For
ionic compounds, write charges on the cation and anion.
Give the total number of electrons in each compound.
Draw and name the VSEPR shape for molecular compounds.
Indicate whether the molecule is polar or nonpolar.
Name the dominant intermolecular force in each substance.
Place the compounds in order of increasing predicted melting point.
c.
d.
e.
f.
g.
a.
molecular
molecular
ionic
molecular
ionic
other
1+ 1–
1 N 5 e–
1 H 1 e–
2 Cl 14 e–
20 e–
2+ 2–
Xe
60 e–
54 e–
b., c., d., e.
1 C 4 e–
2 F 14 e–
2 Cl 14 e–
32 e–
1 C 4 e–
2 O 12 e–
16 e–
O
C
28 e–
O
H
C
22 e
F
N
–
58 e–
Cl
C
F
F
C
H
O
N
Cl
Cl
nonpolar
trigonal
pyramid
nonpolar
tetrahedral
C
H
F
F
N
Cl
Cl
P
F
F
trigonal
pyramid
F
Cl
Cl
H
O
P
F
F
Cl
26 e–
42 e–
F
1 e–
6 e–
7 e–
14 e–
F
F
42 e–
O
linear
O
Cl
C
P
Cl
1H
1O
1 Cl
Cl
Cl
O
1 P 5 e–
3 F 21 e–
26 e–
F
F
Cl
molecular molecular
H
O
Cl
bent
(109.5o)
H
O
Cl
polar
polar
polar
polar
f.
dipoledipole
London
ion-ion
H-bond
ion-ion
London
dipoledipole
H-bond
Xe
PF3
CF2Cl4
HOCl
HNCl2
KF
MgSO4
g.
CO2
2.
a.
Draw the best Lewis structure for the appropriate substances or indicate the ions
(with charges):
CF4
b.
a.
1 Se 6 e–
4 F 28 e–
34 e–
F
F
C
F
F
52 e–
1N
3H
1O
5 e–
3 e–
6 e–
14 e–
CaS
Ar
SO3
HOOH
2+ 2–
Ar
36 e–
18 e–
1S
6 e–
3 O 18 e–
24 e–
Se
H
F
N
F
C
Se F
F
F
F
F
F
18 e
nonpolar
f.
London
H
O
O
S
H
1 e–
6 e–
7 e–
14 e–
O
N
HH
H
nonpolar
N trigonal
pyramid
O bent
see-saw
O
Se F
F
F
H
O
S
O
O
N
O
HH
H
both bent
(109.5o)
O
O
O
H
O
H
polar
nonpolar
polar
polar
dipoledipole
ion-ion
H-bond
ion-ion
London
London
London
London
London
H-bond
London
London
g.
CaS
H
O
H
trigonal
planar
S
O
18 e–
40 e–
–
O
F
F
F
O
H
70 e–
F
1H
1O
1 Cl
O
O
tetrahedral
F
1+ 1–
F
42 e–
C
H2NOH
F
F
F
NaBF4
Identify all of the intermolecular interactions that occur in the following (pure)
compounds. Circle the strongest intermolecular force.
Place the compounds in order of decreasing predicted boiling point.
c.
1 C 4 e–
4 F 28 e–
32 e–
SeF4
NaBF4
H2NOH
HOOH
SeF4
CF4
SO3
Ar
3.
Rank the hydrocarbons: pentadecane, pentane, octadecane (all linear structures)
C15H32
C5H12
C18H38
in order of increasing:
4.
a.
viscosity
C5H12 C15H32 C18H38 : since they are all linear, tangling
would be similar. The octadecane, however, has more
electrons (42 vs. 122 vs. 146 ), so the London forces are
stronger.
b.
volatility
C18H38 C15H32 C5H12 : opposite order, because weak forces
means more readily evaporated.
Classify the following solids as molecular, network (covalent), metallic, ionic or atomic.
H
a.
benzene,
H
H
b.
c.
d.
e.
f.
g.
h.
i.
j.
C
C
C
C
diamond
H
krypton
Cd
copper(II) chloride
ice
S8
Mg3(PO4)2
quartz (SiO2)
P4
C
C
H
molecular
H
network
atomic
metallic
ionic
molecular
molecular
ionic
network
molecular
5.
For the two dimensional arrays shown below, draw a set of lattice points, a single
primative unit cell and a centered unit cell.
6.
How many whole atoms are in a primitive cubic unit cell?
1
7.
How many whole atoms are in a body-centered cubic unit cell?
2
8.
How many whole atoms are in a face-centered cubic unit cell?
4
9.
Polonium crystallizes in a sc unit cell with a density of 9.20 g/cm3. Calculate the radius
of a Po atom.
209 g Po
1 mole Po
1 atom Po
1 cm3
x
x
x
 3.773 x 1023cm3  V
1 mole Po 6.022 x 1023 atoms Po 1 unit cell 9.20 g Po
a 
3
V 
a
 r 
2
10.
3
3.773 x 10 23cm 3  3.354 x 10 8 cm
 3.354 x 10 
8
2
 1.68 x 10–8 cm = 1.68 Å
Copper crystallizes in a fcc unit cell, and has an atomic radius of 128 pm. Calculate the
density of copper.
63.546 g Cu
1 mole Cu
4 atoms Cu
x
x
= 4.2209 x 1022 g/unit cell
23
1 mole Cu
6.022 x 10 atoms Cu
1 unit cell
r =
a 2
4r
4(1.28 x 108 cm)
; a =
=
= 3.6204 x 108 cm
4
2
2
V = a3 = (3.6204 x 10–8 cm)3 = 4.7453 x 10–23 cm3
density =
11.
mass of unit cell
4.2209 x 1022 g/unit cell
=
= 8.89 g/cm 3
23
3
volume of unit cell
4.7453 x 10 cm /unit cell
The radius of iron is 124 pm, and it crystallizes in a bcc unit cell, what is the volume of
the unit cell.
r =
a 3
4r
4(124 pm)
; a =
=
= 286.365 pm
4
3
3
V = a3 = (286.365 pm)3 = 2.35 x 107 pm3
12.
For the heating curve below:
a.
(name of the energy) Hfusion
b.
(name of the energy) Hvaporization
c.
Temperature (oC)
d.
Heat (J)
e.
(name of phase change) melting
f.
(name of phase change) boiling
Calculate the energy released when 12.76 g of CH2FCF3 goes from 0.0 oC to –75.0 oC.
normal mp CH2FCF3(s)
normal bp CH2FCF3(ℓ)
Hvaporization CH2FCF3(ℓ)
s CH2FCF3(ℓ)
s CH2FCF3(g)
-4000
-3500
-3000
–101.0 oC
–26.6 oC
22.02 kJ/mole
1.423 J/goC
0.875 J/goC
-2500
-2000
-1500
-1000
-500
0 -5
-15
-25
Temperature (oC)
13.
-35
-45
-55
-65
-75
Heat (J)
1.
q = msT = 12.76 g x 0.875 J/g·oC x (–26.6 – 0.0 oC)
= –296.989 J
2.
H = –22020 J/mole x 12.76 g x 1 mole/102.03 g
= –2753.849 J
3.
q = msT = 12.76 g x 1.423 J/g·oC x (–75.0 – (–26.6 oC))
= –878.822 J
qtotal = (–296.989 J) + (–2753.849 J) + (–878.822 J) = –3929.66 J = –3.930 kJ
14.
Given the phase diagram of xenon, Xe, answer the following:
a.
b.
c.
d.
15.
Estimate the normal boiling point of xenon.
Estimate the normal melting point of xenon.
Estimate the vapor pressure of Xe(ℓ) at –110 oC.
Is Xe(ℓ) or Xe(s) more dense? Explain.
–107 oC
–112 oC
0.8 atm
Xe(s), the mp line slopes to
the right.
Commercial concentrated ammonia is 28.0% NH3 by mass. The density of the solution
is 0.899 g/mL. Calculate:
a.
the molarity of ammonia
28.0 g NH 3
1 mole NH 3
0.899 g sol'n 1000 mL
x
x
x
= 14.8 M NH 3
100 g sol'n
1 mL sol'n
1L
17.031 g NH 3
b.
the molality of ammonia in 100 g soln
28.0 g NH3; 100.0 g sol’n – 28.0 g NH3 = 72.0 g H2O
1 mole NH 3
17.031 g NH 3
= 22.8 m NH 3
0.0720 kg H 2 O
28.0 g NH 3 x
c.
mole fraction of water in 100 g soln
1 mole H 2 O
18.015 g H 2 O
= 0.709

1 mole NH 3
1 mole H 2 O 
+ 72.0 g x
 28.0 g NH 3 x

17.031 g NH 3
18.015 g H 2 O 

72.0 g x
d.
the vapor pressure of the solution at 100 oC (assume ammonia is non-volatile)
vp H2O at 100 oC = 760 torr
Psol’n = waterPowater = (0.709)(760 torr) = 539 torr
f.
the freezing point of this solution
Tf = i Kfm = (1)(1.86 oC/m)(22.8 m) = 42.4 oC lower
Tf = 0.0 oC – 42.4 oC = –42.4 oC
16.
Calculate the difference in solubility of nitrogen (mole fraction in air = 0.78) in blood at
sea level (P = 1.0 atm) versus 10 m underwater (P = 2.0 atm). Assume that blood is
identical to water and kN2 = 7.0 x 10–4 mole/L·atm.
CN2 = kPN2 = 7.0 x 10–4 mole/L·atm x (1.0 atm x 0.78) = 5.5 x 10–4 M N2
CN2 = kPN2 = 7.0 x 10–4 mole/L·atm x (2.0 atm x 0.78) = 1.1 x 10–3 M N2
17.
Calculate the concentration of oxygen gas in water at sea level and 1.00 atm. The partial
pressure of oxygen gas is 0.21 atm and Henry’s constant for oxygen (in water at RT) is
1.3 x 10–3 M·atm–1.
CO2 = kPO2 = 1.3 x 10–3 mole/L·atm x (1.00 atm x 0.21) = 2.7 x 10–4 M O2
18.
Is solid iodine (I2), more soluble in benzene, or chloroform, CH3Cl?
benzene; I2 is nonpolar and so is benzene
19.
Name the intermolecular interaction that occurs between:
a.
ammonia and water
H
H
C
b.
cis-dichloroethylene
Cl
H-bonding
C
Cl
and bromoform (CHBr3) dipole-dipole
c.
d.
e.
20.
dipole-London
dipole-H-bond
ion - dipole
Classify the following as molecular or ionic. How many particles will each form when
dissolved in water?
a.
b.
c.
d.
e.
21.
CH3F and Xe
acetone, CH3COCH3 and water
LiCl and water
copper(II) chloride
CH3F
Mg3(PO4)2
sucrose, C12H22O11
benzene, C6H6
CuCl2, ionic, i
molecular, i =
ionic, i = 5
molecular, i =
molecular, i =
= 3
1
1
1
A solution is prepared by dissolving 5.00 g of glucose, C6H12O6, in 100.0 g of water.
Calculate:
a.
the vapor pressure of water at 90 oC. The vapor pressure of pure water at 90 oC is
525.8 torr.
1 mole H 2 O
100.0 g x
18.015 g H 2 O
= 0.995

1 mole C6 H12 O 6
1 mole H 2 O 
+ 100.0 g x
 5.00 g C6 H12 O 6 x

180.156 g C6 H12 O6
18.015 g H 2 O 

Psol’n = waterPowater = (0.995)(525.8 torr) = 523.2 torr
b.
the boiling point of the solution.
1 mole C6 H12 O6
180.156 g C6 H12 O 6
= 0.2775 m C6 H12 O6
0.100 kg H 2 O
5.00 g C6 H12 O 6 x
Tb = i Kbm = (1)(0.52 oC/m)(0.2775 m) = 0.14 oC higher
Tb = 100.0 oC + 0.14 oC = 100.14 oC
22.
a.
Calculate the freezing point of a solution of 2.00 moles of NaCl in 100.00 mL of
water (density of water = 1.00 g/mL).
2.00 mole NaCl
= 20.0 m NaCl
0.10000 kg H 2 O
Tf = i Kbm = (2)(1.86 oC/m)(20.0 m) = 74.4 oC lower
Tf = 0.0 oC – 74.4 oC = –74.4 oC
b.
Calculate the freezing point of a solution of 2.00 moles of CaCl2 in 100.00 mL of
water.
2.00 mole CaCl2
= 20.0 m CaCl 2
0.10000 kg H 2 O
Tf = i Kbm = (3)(1.86 oC/m)(20.0 m) = 112 oC lower
Tf = 0.0 oC – 112 oC = –112 oC