Chapter 1 Answers to End of Chapter Problems
Section 1.1.
1.1.
Phases of Matter
Identify the following as being either a chemical property or a physical property.
Place a P by all the physical properties and a C by all the chemical properties.
(a) _____ water is clear and colorless
(b) _____ some metals react with water to produce hydrogen gas
(c) _____ water has a density of 1.000 g/cm3 at 4 C
(d) _____ water boils at 100 C
(e) _____ water is the product of a reaction between an acid and a base
(f) _____ water is a polar molecule
Answer: (a) P (b) C
1.2.
(c) P (d) P (e) C
(f) C
Compare solids, liquids, and gases in each given category.
Solids
Liquids
Gases
Definite volume?
Definite shape?
Fixed or changing position
of molecules?
Average distance between
molecules small or large?
Answer:
1.3.
Solids
Liquids
Gases
Definite volume?
Yes
Yes
No
Definite shape?
Yes
No
No
Relative
arrangement of
molecules?
Orderly, fixed
positions
Molecules free to
move
Molecules move
almost
independently
Relative average
distance between
molecules?
Closest of the three
phases
Close, but not as
close as in solids
Not close, but
relatively far apart
Which phases of matter are referred to as “condensed phases”? What is the
justification for the use of this term?
Answer: Solids and liquids are called condensed phases. In both of these phases,
molecules are relatively close together and therefore are likely to have some
interactions. The molecules in the gas phase, by contrast, are spread apart with
significantly less chance for interaction. While the process of condensation
usually refers to the production of liquid from the gas (vapor) phase, both liquid
and solid phases are termed “condensed phases”.
1.4.
(a) Name the phase changes between each of the states of matter indicated by the
arrows in this diagram.
(b) Label each of the four arrows on the diagram to indicate whether energy is
released or absorbed in the process.
Answer: (a)
(b) Energy is released in condensation and freezing. Energy is absorbed in
vaporization and melting.
1.5.
Name a substance other than water that commonly exists as a liquid at STP.
Answer: There are many possible answers, including gasoline, rubbing alcohol,
and cooking oil.
1.6.
Name a substance that commonly exists as a gas at STP.
Answer: There are many possible answers, including nitrogen (N2), oxygen (O2),
argon (Ar) and carbon dioxide (CO2), all found in the Earth’s atmosphere.
1.7.
What happens if a closed glass bottle full of water is kept outside while the
temperature falls below 0 C?
Answer: When water freezes, the closed glass bottle will break because the given
volume of ice is greater than the same volume of water.
1.8.
One method for separating a NaCl(aq), sodium chloride solution, into its
components is to boil the solution of salt water. In this case, water will evaporate
and NaCl(s) will be left behind.
(a) Which property of NaCl(s) and water accounts for this separation?
(b) Design an apparatus that could change the water vapor back into a liquid as
well as recover NaCl(s). Name all phase changes that occur.
Answer: If one boils a solution of NaCl(aq) and water, the water will
evaporate and the salt will be left behind due to the fact that water has a
much lower boiling point than a table salt. The simple apparatus for the
separation of a sodium chloride solution is presented below. Boiling the
solution evaporates the water, which is condensed and collected in the
receiving flask. After all the water has boiled away, NaCl(s) remains in the
boiling flask. The figure below shows the distillation apparatus that is used
to separate this mixture.
1.9.
Given two liquids that do not dissolve in one another (like oil and water), design
an experiment that would allow you to determine which is more dense.
Answer: A very direct experiment is to pour a few drops of either one into the
other. If the added liquid sinks, then it is more dense. If the added liquid remains
on the surface, then it is less dense.
1.10. How could you determine (experimentally) if the solid phase of a particular
substance is more dense or less dense than the liquid phase of the same substance?
Answer: Mixing samples of the solid and liquid will quickly allow for the
determination of relative density. If the solid is more dense, then it will sink in
the liquid. If the solid is less dense, then it will float in the liquid (as is the case
for solid water (ice) in liquid water).
1.11. Consider Figure 1.3 comparing the densities of solid and liquid phases of tbutanol with those of water. What additional information do you need to be able
to predict what will happen if a sample of solid t-butanol were dropped into liquid
water? How would you find that information?
Answer: The figure shows that the density of the solid alcohol is greater than that
of its liquid phase but that the density of the liquid water is greater than that of its
solid phase. What is not known is how the density of solid t-butanol compares
with that of liquid water, an essential piece of information to predict if the solid tbutanol will sink , float, or remain just buoyantly suspended when dropped into
liquid water. The relative density values could be determined experimentally or
by looking up the densities in an appropriate reference book.
1.12.
A block of ice has following dimensions: height = 20 cm, width = 20 cm, and
length = 20 cm. The density of liquid water is 1.000 g·mL–1 at 0 C, and the
density of ice at the same temperature is 0.917 g·mL–1. Calculate the volume of
the puddle of water, at 0 C, that is left behind when the block of ice melts.
Answer: The volume of block of ice:
V = 20cm x 20cm x 20 cm = 8000cm3 = 8000mL
The mass of ice (and water) : d = m/V; m= d x V: m= 8000 mL x 0.917
g/mL = 7336 g
The volume of puddle: V= m/d; V = 7320 g / 1.000 g / mL = 7320 mL
1.13. If an iron bar weighing 100.0 g is heated to a temperature above its melting point
(>1535 °C), it will liquefy. What is the mass of the molten (liquid) iron? Is any
additional information needed in order to answer this question?
Answer: No additional information is necessary. A phase change from solid to
liquid (or from liquid to gas) does not change the amount of matter involved in
the phase change. Mass is conserved in all phase changes, so the mass of the
molten iron must be 100.0 g.
Section 1.2.
1.14.
Atomic Models
Which of the following are chemical elements? How do you decide?
(a) water
(d) iron oxide
(b) salt water
(e) nitrogen
(c) iron
(f) diamond
Answer: Water and iron oxide are compounds. Salt water is a solution of one
compound (salt) in another (water). Nitrogen and diamond are elements; diamond
is pure carbon.
1.15. What is the difference between core electrons and valence electrons?
Answer: Since core electrons are strongly attracted to the nucleus, they are
closest to the nucleus and thus cannot interact with other electrons. Valence
electrons are furthest from the nucleus and interact with other atoms.
1.16. How many valence electrons and core electrons do the following elements
possess?
(a) sodium
(d) phosphorus
(b) bromine
(e) sulfur
(c) barium
Answer: (a) 1 valence electron, 10 core electrons; (b) 7 valence electrons, 28
core electrons; (c) 2 valence electrons, 54 core electrons; (d) 5 valence electrons,
10 core electrons; (e) 6 valence electrons, 10 core electrons.
1.17. Many organisms use these ions in their metabolism: Na+, K+, Mg2+, Ca2+, Cl–, Br–.
(a) Complete the following table concerning these ions.
Ion
# of
protons
# of
electrons
# of valence
electrons
Core
charge
# of core
electrons
Na+
K+
Mg2
+
Ca2+
Cl–
Br–
(b) What patterns do you observe?
Answer:
(a)
Ion
# of
protons
# of
electrons
# of
valence electrons
Core
atomic
charge
# of
core
electrons
Na+
K+
Mg2+
Ca2+
ClBr-
11
19
12
20
17
35
10
18
18
18
18
36
0
0
0
0
0
0
+1
+1
+2
+2
-1
-1
10
18
18
18
18
36
(b) Na+, K+, Mg2+, and Ca2+ have zero valence electrons. Cl- and Br- have eight
valence electrons (a complete octet). To form ions, metals lose electrons while
non-metals gain valence electrons.
1.18. In terms of electronic structure, what is it that elements in the same period (row)
of the periodic table share in common?
Answer: Elements in the same period of the periodic table all have the same
number of core electrons.
1.19. In terms of electronic structure, what is it that elements in the same group
(column) of the periodic table share in common?
Answer: Elements in the same group of the periodic table all have the same
number of valence electrons.
1.20. In general, do elements from the same period or elements from the same group of
the periodic table have similar chemical properties? Justify your answer.
Answer: Recall that chemical reactions and interactions directly involve only the
valence electrons. Therefore, elements in the same group (column) have similar
chemical properties, since they have identical numbers of valence electrons.
1.21. These metal ions, Mn2+, Fe2+, Fe3+, Cu2+, and Zn2+, are available for uptake by
living organisms. How many protons and electrons does each ion have?
Answer: Mn2+ has 25 protons and 23 electrons; Fe2+ has 26 protons and 24
electrons; Fe3+ has 26 protons and 23 electrons; Cu2+ has 29 protons and 27
electrons; Zn2+ has 30 protons and 28 electrons.
Section 1.3.
Molecular Models
1.22. Using equations modeled after Figure 1.9(c), show how to form each of the
following molecules from their constituent atoms. In each case, count electrons in
the products to demonstrate that the octet rule is followed for all second row
atoms.
(a) HF (hydrogen fluoride)
(c) CH3OH (methanol)
(b) NH3 (ammonia)
(d) H2O2 (hydrogen peroxide)
Answer: The octets of electrons have been circled for each of the products. Note
that each covalent bond is composed of two electrons, one formally contributed
by the atoms at each end of the bond.
(a)
H
(b)
H
HF (hydrogen fluoride)
+
H
F
F
NH3 (ammonia)
N
H
H
H
N
H
H
(c)
CH3OH (methanol)
H
H
H
C
O
H
H
C
O
H
O
H
H
H
(d)
H2O2 (hydrogen peroxide)
H
O
O
H
H
O
1.23. Which of these are macroscopic properties? Which are microscopic properties?
(a) the boiling point of water
(b) the HOH bond angle in water
(c) the OH bond length in water
(d) the ability of water to dissolve salt
(e) the density of water
(f) the fact that the oxygen atom of water has two pairs of nonbonding electrons
Answer: The bond angle, bond length, and number of nonbonding pairs of
electrons on oxygen are not directly observable and are therefore microscopic
properties characteristic of individual water molecules. The boiling point and
density of water and its ability to dissolve salt are directly measurable or
observable with bulk samples of water and are therefore macroscopic properties.
These macroscopic properties of are not properties of individual water molecules,
only large (macroscopically observable) collections of water molecules. Of
course, the microscopic structure and macroscopic properties are related and
much of the chemistry we will study is concerned with these relationships.
1.24. Consider these four different models for the ammonia molecule, NH3:
Model 1
NH3
Model 2
H N H
H
Model 3
Model 4
(a) What is the name given to each type of model shown?
(b) What information can be obtained from each type of model?
(c) What other types of models can be used to represent the ammonia molecule?
Answer: (a) Model 1 is a molecular formula, sometimes called a line
formula and Model 2 is an electron-dot model. Model 3 is a ball and stick
representation and Model 4 is a computer-generated model of an ammonia
molecule.
(b) The molecular formula or line formula tells you there are 3 atoms of
hydrogen combined with one atom of nitrogen, forming one molecule of
ammonia. The electron-dot model also shows the connectivity within the
ammonia molecule and locates the valence electrons. The ball and stick
model shows connectivity but also gives some information about the bond
angles within the molecule. Note that this model does not show the pair of
nonbonded electrons. The computer-generated model shows the atoms,
their connectivity and bond angles, and is color-coded to show net
electrical charge on various parts of the molecule.
(c) The “paddle” model can be used to show the location of nonbonded electron
pairs. Space filling models show connectivity, bond angles and give approximate
representations of the space (volume) occupied by electrons around each atomic
core in the ammonia molecule.
1.25. Write the molecular formulas (line formulas) of the compounds represented here:
(a)
(c)
(b)
(d)
Answer: (a) SF4 (b) PCl3 (c) CH4 (d) CH3COOH
1.26. Write the molecular formulas for the following compounds. (If necessary, use a
reference handbook and/or other books to find out the structures.)
(a) glucose, a substance known as "blood sugar"
(b) nitrous oxide, a substance used as an anesthetic and as an aerosol propellant.
It is commonly called "laughing gas."
(c) methanol, an organic solvent and antifreeze
(d) acetylene, a gas that is used in welding torches
Answer: (a) C6H12O6 (b) N2O (c) CH4O (However, the stuand way of writing
the molecular formula for methanol is CH3OH, because it shows how the atoms
are joined in the molecule.) (d) C2H2
Section 1.4.
Valence Electrons in Molecular Models: Lewis Structures
1.27. (a) What information can be obtained from a Lewis structure?
(b) What information cannot be obtained from a Lewis structure?
Answer: (a) From a Lewis structure, you can determine the shape of the
molecule. From the shape of the molecule and relative electronegativities, you
can determine the polarity of the molecule. From the polarity of the molecule,
you can deduce most of the chemical and physical properties such as determining
if it has a high boiling point, melting point, and/or heat of vaporization.
(b) Any properties related to geometry or distribution of electrons are not
conveyed by Lewis structures. These include bond lengths, bond angles, bond
dipoles, and molecular dipoles.
1.28. Consider this electron-dot model for the ammonium ion, NH4+:
H
+
H N H
H
(a) What information can be obtained from this model?
(b) What information cannot be obtained from this model?
(c) Rewrite the ammonium ion using a dash to represent each bonded pair of
electrons. Does this change the information found in the model?
(d) How does this electron-dot model for the ammonium ion compare with that
for methane, CH4, given in Worked Example 1.10?
Answer: (a) There are four atoms of hydrogen combined with one atom
of nitrogen, forming the ammonium ion. There is a positive charge on the
overall ion. The connectivity within the molecule is clearly shown by this
model.
(b) There is no information about bond angles, bond lengths, or
distribution of charge within the ammonium ion.
(c) When rewritten using dashes, the model is:
H
H N H
H
This does not in any way change the information found in the model. It is just a
convenience to replace each pair of bonded electrons with a dash.
(d) The structure for methane is very similar electronically. Both show
four pairs of electrons around the central atom. The central atom nitrogen
has one more valence electron than the central atom carbon and therefore
one electron is lost in forming the ammonium ion, leaving the ion with a
positive charge.
H
Ammonium ion:
+
H N H
H
Methane: H C H
H
H
1.29. Examine Table 1.1. Silicon typically makes four covalent bonds and sulfur
typically makes two covalent bonds. How many covalent bonds do you expect for
each of the following elements? Explain your reasoning in each case.
(a) phosphorus
(c) selenium
(b) chlorine
(d) bromine
Answer: (a) 3 (b) 1 (c) 2 (d) 1
1.30. Neon (Ne) is the element to the right of fluorine (F) in the periodic table.
(a) Examine Table 1.1. How many covalent bonds and nonbonding pairs would
be expected for neon? Explain the reasoning for your answer.
(b) Ne (along with He, Ar, Kr, Xe, and Rn in the same group of the periodic
table) were once known as the “inert gases.” Why were they given this name?
Answer: (a) Neon forms no (“0”) covalent bonds and has four nonbonding pairs of
electrons.
(b) In chemistry, “inert” means incapable of reacting. Chemical reactions are
defined by the making and breaking of chemical bonds. Since neon can not form
covalent bonds, it can be characterized as inert.
1.31. Draw the Lewis structure of ethanol, C2H6O. The formula may also be written
C2H5OH or CH3CH2OH to make the connectivity more apparent.
Answer: Each carbon has 4 valence electrons, each hydrogen has 1
valence electron, and oxygen has 6 valence electrons. The total number of
valence electrons in C2H6O is:
2(4) + 6(1) + 1(6) = 20 valence electrons.
The only possible way to construct ethanol is to connect carbon-to-carbon,
and then carbon-to-oxygen atomic cores. Then the hydrogen atomic cores
are placed around the carbon and oxygen cores so that each carbon has 4
covalent bonds and each oxygen has 2 covalent bonds. The structure is:
H H
H H
H C C O
H C C O
or
H H H
H H H
1.32. Draw the Lewis structures for ozone, O3, sulfur dioxide, SO2, and nitrite ion,
NO2–. What do all three of these structures have in common? Hint: S is the central
atom in SO2 and N is the central atom in NO2–.
Answer:
All three are isoelectronic. This means that the arrangement of their valence
electrons is identical. The fundamental difference amongst them is the number of
balance between the core charges of the three atomic centers and eighteen valence
electrons. When the core charge (positive) is equal to the charge due to the
valence electrons (negative), then the molecule is neutral (ozone and sulfur
dioxide). If they are unequal, then an ion results (nitrite anion).
1.33. Each of the following Lewis structures for NCCN has the correct number of
electrons. Which is the best Lewis structure for NCCN? Explain your reasoning
for rejecting the structures you did not choose. Hint: Multiple bonds between
atoms will be discussed in Chapter 5. For the purposes of this problem, simply
count each stroke (bond) as two electrons shared between the atoms it connects.
(a)
N C C N
(d)
(b)
N C C N
(e)
(c)
N C C N
(f)
N C C N
N C C N
N C C N
Answer:
(d) is the correct Lewis structure and contains an octet for each atom. All other
options lack octets on one or more atoms.
1.34. Which of the following Lewis structures are incorrect? In each case, explain why
the structure is incorrect. Rewrite each of the incorrect structures so it is correct.
Hint: See the hint in Problem 1.33.
O
(a) HOCl
(b) CS2
(c) NH3
H O Cl
S
(d) (HO)2CO
C
S
(e) H2Se
C
H O
O H
H Se H
H N H
H
Answer:
The incorrect structures are (d) and (e). In order to follow the octet rule
they should be drawn as follow:
Section 1.5.
Arranging Electron Pairs in Three Dimensions
1.35. Which of the following molecules or ions has a tetrahedral (or close to
tetrahedral) orientation of bonding and nonbonding electron pairs?
(a) H2S
(d) NH2–
(b) NH4+
(e) CH4
(c) NH3
Answer: With respect to the central atom, tetrahedral orientation requires a total
of four pairs of bonding and non-bonding electrons. For this problem, all have
tetrahedral orientations.
1.36. What is the shape of each of the molecules in Problem 1.35? Recall that the shape
of a molecule describes the position of the atomic cores with respect to one
another.
Answer:
(a) H2S (bent)
(d) NH2– (linear)
(b) NH4+ (tetrahedral)
(e) CH4 (tetrahedral)
(c) NH3 (trigonal pyramidal)
1.37. Figure 1.12 showed one way to stack four balls so that
they are equidistant from a central point. Another way is
to arrange them in a square about the point, as shown in
this picture. Lines drawn between the centers of adjacent
balls form a square of side 2r, where r is the radius of a
ball. The center point of the diagonal of the square is the
center of the square. Use the Pythagorean theorem to find the length of the
diagonal and the distance from the center of any of the balls to the center of the
square.
Answer: Let c = length of diagonal.
2
2
2
4r + 4r = c ; c = 2r 2;
c
distance from center =
= r 2
2
1.38. (a) A cube can be circumscribed about a regular tetrahedron,
as shown in this figure. The four dots represent the centers of
the four balls in Figure 1.12,. The diagonal of one of the cube
faces has a length 2r, where r is the radius of a ball. The center
point of any one of the cube diagonals (one is shown by the
dashed line) is the center of the tetrahedron. Use the
Pythagorean theorem to find the length of the cube edge and
then again to find the length of the cube diagonal. Thus, show that any corner of
6  r from the center of the
the cube, that is, the center of any of the balls, is 
 2 
tetrahedron.
(b) How does the distance you calculated in part (a) compare to the distance of
the center of each ball from the center of the square in Problem 1.37? Do your
results help justify the statement in the text that “the tetrahedral arrangement puts
all the balls as close as possible to the central point”? Explain why or why not.
Answer: (a) Call the length of the cube edges “e”. Apply the Pythagorean
theorem to the right isosceles triangle with hypotenuse 2r: (2r)2 = 2e2; 2r = (2)e;
e=
2r
. Call the length of the cube diagonal, “d”, and apply the Pythagorean
2
theorem to the right triangle with sides 2r and e and hypotenuse d: d2 = (2r)2 +
2
 2r 
2


= 4r2 + 4r /2 = 6r2 ; d = (6)r. Therefore, (1/2)d =  6 2  r.
 2 
(b) The distance calculated in part (a) is less than the distance calculated in
Problem 1.37. Yes, the results help justify the statement that "the tetrahedral
arrangement puts all the balls as close as possible to the central point" because its
distance is shorter than Problem 1.37, indicating that the balls are closer.
1.39. SiCl4, silicon tetrachloride, is used for the production of the very pure silicon
required in many electronic devices such as transistors.
(a) Draw the Lewis structure for SiCl4.
(b) How does this Lewis structure compare to that of methane, CH4?
(c) Predict the shape of the SiCl4 molecule.
Answer: (a) Each silicon has 4 valence electrons and each chlorine has
7 valence electrons. The total number of valence electrons in SiCl4 is:
1(4) + 4(7) = 32 valence electrons.
The way to connect these atom cores is to connect each chlorine atom core to the
central silicon atom core. Silicon has 4 covalent bonds and each chlorine will
have one covalent bond. Each chlorine has 3 pairs of nonbonded electrons. The
structure is:
Cl
Cl Si Cl
Cl
Cl
or
Cl
Si
Cl
Cl
(b) This Lewis structure is similar to that for methane, CH4 in that the
central atom in each case has four covalent bonds. Silicon is in the same
family on the Periodic Table as carbon, so also brings 4 valence electrons
to the structure. Each hydrogen atom cores in methane can only have 1
bonded pair of electrons, but each chlorine atom in silicon tetrachloride
has 1 bonded pair of electrons and 3 nonbonded pairs of electrons.
Cl
H
Methane: H C H
H
Silicon tetrachloride:
Cl Si Cl
Cl
(c) The shape of SiCl4, like that of CH4, will be tetrahedral.
1.40. (a) Draw the Lewis structure for borane, BH3. Compounds like borane are
sometimes called “electron deficient.” How do you think “electron deficiency” is
defined?
(b) What shape do you predict for borane? You might find it useful to try a
modified version of Investigate This 1.14 to help make your prediction. Also see
the Web media, Chapter 1, Section 1.6.
Answer: (a) Borane has only six valence electrons. Because it does not
have a full octet (eight electrons), it is called electron deficient.
(b) As shown in the drawing, borane is trigonal planar with the H–B–H bond
angles all equal to 120°.
Section 1.6.
Polarity of the Water Molecule
1.41. What is electronegativity?
Answer: The force of attraction an atom has for a shared pair of electrons.
1.42. Answer the following questions on electronegativity.
(a) Which element has the highest electronegativity value?
(b) Where on the periodic table do you find the elements with the lowest
electronegativities?
(c) Where on the periodic table do you find the elements with the highest
electronegativities?
(d) What is the general trend for electronegativity as you go from the top to the
bottom within a group?
(e) What is the general trend for electronegativity as you go from the left to right
across a period?
(f) How do the electronegativity trends compare to the atomic size trends? Refer
to Figure 1.7 for information about atomic size.
Answer: (a) Fluorine (b) Bottom lower left (c) Top right (excluding noble
gases) (d) decreases (e) increases (f) Electronegativity increases parallels
atomic size decreases (as the size goes down the electronegativity goes up).
1.43. Without referring to a table of electronegativities, predict which member of each
pair has the greater electronegativity. Explain the basis for your prediction in each
case.
(a) F or S
(c) H or O
(b) C or H
(d) O or C
Answer: (a) F (b) C (c) O (d) O
1.44. You can use the electronegativity difference between two atoms to predict the
polarity of the bonds they make. Choose the pair of atoms in each case that you
predict to make the more polar bond and explain how you make this prediction.
(a) H-F or H-Cl
(c) K-S or Na-S
(b) C-H or N-H
(d) O-O or N-O
Answer: The higher the electronegativity difference between the two atoms, the
more polar the bond will be.
(a) H-F (b) N-H (c) K-S (d) N-O
1.45. What would be the consequences for life if water were a linear molecule?
Consider the effect on the polarity and properties of water if it were linear. You
might find it useful to draw pictures of how the linear molecules might interact
with one another and compare what you get with the various interactions pictured
in this chapter.
Answer: There is no single answer to this problem, but your answer ought
to involve an understanding that a linear water molecule will have no
permanent dipole moment, since the bond moments will be pointing in
opposite directions and cancel one another out. If the hydrogen atoms are
on opposite sides of the oxygen atom, the nonbonding electron pairs are
also likely to be on opposite sides (for symmetry), so the array of bonding
and nonbonding pairs is likely to be square planar. Hydrogen bonds would
still be possible and could easily form planar sheets of molecules, each
hydrogen bonded to four others, as in the left-hand picture below. Such
layers could stack on one another or small arrays of them could jumble
together like potato chips in a bag in the liquid phase. Three-dimensional
structures are also possible, as in the right-hand picture, also with four
hydrogen bonds to each molecule. This three-dimensional structure is
quite open.
Without a permanent dipole moment, the attractions among linear water
molecules would not be as strong and water would probably melt and boil
at lower temperatures. The liquid might exist on Earth at the present
temperature and still be compatible with the development and evolution of
life. The three-dimensional structure shown has large spaces between
molecules, so a solid with this structure could collapse to a more dense
liquid, just as the bent water molecules do. That is, solid linear water
would probably float on the liquid, which is another requirement for life,
as we know it. The extensive hydrogen bonding possible for linear water,
would lead to a higher energy of vaporization and specific heat than if
hydrogen bonding was not possible, but the lack of a permanent dipole
would probably mean lower values than for the bent molecule.
1.46. Consider the two molecules H2O and H2S.
(a) Compare the Lewis structures of these two molecules.
(b) Compare the molecular shape of these two molecules.
(c) Compare the bond dipoles within each molecule. Hint: Use the data in Figure
1.20.
(d) Compare the overall electric dipole of each molecule.
Answer: (a) Lewis structure of H2O is: H O
H
is: H S
H
Lewis structure of H2S
(b)
Both H2O and H2S have bent or angular shapes. Note: Recall that
the nonbonding electrons are not considered when describing the shape of
molecules.
(c) The bond dipoles are different in H2O and H2S. The electronegativity
(EN) of oxygen is 3.44, while that of hydrogen is 2.20. The difference in
EN is therefore 1.24, making each hydrogen to oxygen bond very polar.
The valence electrons are not shared equally between oxygen and each
hydrogen atom, with oxygen having a much greater attraction for the
shared electrons. This is represented in Figure 1.19. The EN of sulfur is
2.58 and that of hydrogen is 2.20. The difference in EN is therefore only
0.38. The rule of thumb is that a significant bond dipole only exists for a
covalent bond between the atoms of two elements that differ by 0.5 or
more EN units, so these bonds are essentially nonpolar.
(d) H2O has a significant electric dipole, as represented in figure 1.18. H2S, on
the other hand, has essentially nonpolar bonds, and therefore even though it is
angular in shape, it cannot exhibit a significant electric dipole.
1.47. Draw the Lewis structures for carbon tetrachloride, CCl4, chloroform, CHCl3, and
dichloromethane, CH2Cl2. Clearly label the bond dipoles for each molecule.
Which molecules are polar and which are non-polar? Explain your answer.
Answer: When the individual bond dipoles cancel out, then there is no net
dipole and the molecule is nonpolar. If the individual bond dipoles do not
cancel out, then there is a net dipole.
1.48. Each of the following molecules has a dipole moment = 0. In each case, explain
why there is no net (molecular) dipole moment. If the molecule has bond dipoles,
draw them and explain how they cancel out.
(a) N2 (molecular nitrogen)
(b) BH3 (borane) Hint: B does not satisfy the octet rule. See Problem 1.40.
(c) SiCl4 (silicon tetrachloride)
(d) BeH2 (beryllium hydride) Hint: Be does not satisfy the octet rule. What
shape must the molecule have, in order not to have a net dipole moment? You
might find it useful to try a modified version of Investigate This 1.14 to help
determine this shape.
Answer: Given: molecular formulas.
Asked for: an explanation why all the molecules have dipole moments = 0
Recall that the dipole moment of any molecule is simply the sum of all the
individual bond dipoles. In order to have no net (molecular) dipole moment, the
bond dipoles must either all equal zero or be symmetrically oriented to exactly
cancel out. [If you have studied vectors in math, we can restate this more simply:
the molecular dipole is the vector sum of the bond dipoles.]
Plan & Analysis: Before attempting an analysis of dipole moments, the threedimensional shape of the molecule must be known. First the Lewis structure must
be written, then the geometric orientation of the bonds must be determined. If
there is any special symmetry, it should be noted.
(a)
N2 (molecular nitrogen) has a triple bond between the two nitrogen atoms.
Since the two nitrogen atoms obviously have the same EN (electronegativity)
value, the bond dipole must be zero. As is true for all diatomic molecules the
bond dipole is equal to the molecular dipole moment.
N
(b)
N
BH3 (borane) has a trigonal planar geometry with all the HBH bond
angles = 120°. Hydrogen is more electronegative than boron, as illustrated. The
cancellation of the bond dipoles is due to the three-fold symmetry of the
molecule.
H
B
H
H
(c)
SiCl4 (silicon tetrachloride) has its four equivalent Si—Cl bonds pointing
to the four corners of a tetrahedron. Chlorine is more electronegative than silicon.
The cancellation of the bond dipoles is due to the tetrahedral symmetry of the
molecule.
Cl
Cl
Si
Cl
(d)
Cl
BeH2 (beryllium hydride) is a linear molecule, with the two Be—H bond
moments 180° apart opposing one another.
Recap: In order to figure the net dipole for any molecule, there are a number of
steps required. Each step is essential in order to arrive at the correct answer.
First, draw a Lewis structure. With only rare exceptions (such as BH3 and BeH2
above), all second row elements should satisfy the octet rule. Next, figure the
three-dimensional shape of the molecule that will minimize the separation of the
negatively charged valence pairs of electrons from the positively charged atomic
core. Finally, the electronegativities of the atoms allows for the relative
magnitudes of individual bond moments to be estimated; from their geometric
sum a molecular dipole direction and magnitude can be qualitatively predicted.
1.49. Which molecule, ammonia, NH3, or phosphine, PH3, has the larger molecular
dipole moment? Explain.
Answer: Since nitrogen is more electronegative than phosphorus, ammonia will
have the greater molecular dipole.
Section 1.7.
Why Is Water Liquid at Room Temperature?
1.50. Describe each of the following types of intermolecular attractions:
(a) induced-dipole attractions
(b) dipole-dipole attractions
(c) hydrogen bond
Answer: (a) The motion of electrons in a nonpolar molecule can creat
partial polar regions in molecules. Polarity in one molecule induces
polarity in a neighboring molecules and two polarized molecules attract
one another. These attractions are called induced-dipole attractions.
(b) A dipole-dipole attracts exist between neutral polar molecules. Polar
molecules attract each other when the positive end of one molecule is near
the negative end of another.
(c) Hydrogen bonding is a psecial trype of intermolecular attractions that exist
between the hydrogen atom in a polar bond (particularly a H-F, H-O, or H-N
bond) and an unshared electron pair on a nearby electronegative atom (usually a
F, N or O atom) on another molecule.
1.51. How are an intramolecular covalent bond and an intermolecular hydrogen bond
similar? How are they different?
Answer: Experiments show that the strength of a hydrogen bond is
approximately five to ten percent the strength of a covalent bond. The relative
weakness of hydrogen bonds means that they can break and form rapidly in
solution. Also, hydrogen bonds are longer than covalent bonds.
1.52. What type(s) of intermolecular attractions are there between
(a) all molecules?
(b) polar molecules?
(c) a hydrogen atom in a water molecule and a nitrogen atom in ammonia (in a
mixture of ammonia and water)?
Answer: a) dispersion forces (b) dipole-dipole attractions (c)
hydrogen bonding
1.53. Astatine, At, element 85, is radioactive and has a half-life of only 8.3 hours (see
Chapter 3). Only minute traces of At have been studied. The hydride, HAt, has
been detected but its physical properties are unknown. Based on the data in Figure
1.24, what would you predict for the boiling point of HAt? Explain how you make
your prediction.
Answer: With the increasing boiling points from HCl through HI, we would
predict that the boiling point for HAt would be approximately 40-50 ºC.
1.54. Methane, CH4, and hydrogen sulfide, H2S, do not form hydrogen bonds. Explain.
Answer: Carbon and sulfur atoms do not have high electronegativity values
required for forming hydrogen bonds. For hydrogen bonds to form, H atoms must
be bonded to a very electronegative atom such as fluorine, nitrogen or oxygen.
1.55. List at least three properties of water that can be attributed to the existence of the
hydrogen bond. Briefly describe how each property would be affected if water did
not form hydrogen bonds.
Answer: 1. Water is in its liquid state at room temperature. At room
temperature, the hydrogen bonds in water are strong enough to hold
molecules together. If the hydrogen bonds are not present, the small
molecules like water would occur in the gas state at room temperature.
2. The lower density of solid ice compared to liquid water. The extended
hydrogen-bonded structure in ice produces more open space between
water molecules than in liquid water. This structure, which extends in all
directions in space, permits the maximum number of hydrogen-bonding
interactions between the H2O molecules. Because the structure has large
hexagonal holes, ice is more open and less dense than the liquid. If the
hydrogen bonds are not present, ice would be more dense than the liquid
as it is the case for most substances.
3. The boiling point of water is much higher than it would be expected on
the basis of its molecular weight. Because of the hydrogen bonds, more
energy is needed to disrupt the intermolecular attractions between the
molecules in the liquid phase.
Other consequences of the presence of hydrogen bonds among
water molecules include:high melting point and high heat capacity.
1.56. (a) How many hydrogen bonds are possible for one water molecule in a sample
of water? Illustrate your answer with a drawing of the structure of water
indicating where the hydrogen bonds are possible.
Answer:
1.57. (a) How many hydrogen bonds are possible for one ammonia molecule in a
sample of ammonia? Illustrate your answer with a drawing of the structure of
ammonia indicating where the hydrogen bonds are possible.
(b) Can all the ammonia molecules in a sample of liquid ammonia have the
maximum number of hydrogen bonds you illustrated in part (a)? If not, what
limits the number and how many hydrogen bonds, on average, can each ammonia
molecule have?
Answer: Compare Figure 1.26 (a), showing hydrogen bonding for water, with the
drawing (below) showing hydrogen bonding for an ammonia molecule. Every
water molecule in a collection of water molecules is capable of forming four
hydrogen bonds because it donates as many nonbonding pairs (2) as hydrogens
(2). The situation for ammonia is quite different. While an individual ammonia
molecule is capable of forming four hydrogen bonds, not all molecules in a
sample of liquid ammonia can do this. The imbalance of hydrogen atoms (3) to
nonbonding pairs of electrons (1) means that, on average, each ammonia molecule
will form a maximum of only one hydrogen bond. There simply aren’t enough
nonbonding pairs of electrons for every available hydrogen.
1.58. Refer to the graph in Consider This 1.23 that relates the boiling points of a series
of hydrocarbons to the number of electrons per molecule. What is the smallest
hydrocarbon in the series that exists as a liquid at room temperature?
Answer: In order to exist as a liquid at room temperature, a substance must have
a boiling point above room temperature. The smallest linear hydrocarbon to fit
this description is pentane (CH3CH2CH2CH2CH3).
1.59. The boiling points of the noble gases are:
element
He
Ne
Ar
Kr
Xe
Rn
bp, C
–269
–246
–186
–152
–107
–62
(a) Plot these data on a graph like the one in Figure 1.24. How are these data
similar to those for the hydrides plotted in Figure 1.24? How are they different?
(b) Noble gas atoms are spherical and Figure 1.22 shows that tetrahedral
hydrides like methane are quite symmetric and almost spherical. What is likely to
be the largest factor responsible for the difference in boiling points between a
noble gas and the corresponding group IV hydride?
Answer: The plot in Figure 1.24 with the noble gases added looks like this:
(a) The periods 2 through 5 group VIII noble gases follow a trend that is
remarkably similar to the group IV hydrides except 50-80 C lower in boiling
point. The trend continues to period 6, but there are no hydrides for comparison.
(The period 6, group IV element is bismuth, which is metallic and does not form a
tetrahydride.) There is, of course, no period 1, group IV element, so no
comparison with helium is possible either.
(b) The symmetry (spherical or near-spherical) of the group VIII noble gases and
the group IV hydrides gives them small surface areas and charge symmetry,
which makes the induced dipole attractions among the atoms or molecules
relatively weak and accounts for their low boiling points. The nuclear charge on
the noble gases is four units higher than the central atom in the corresponding
period hydride, which holds the electrons closer to the nucleus, reduces the
surface area of the atom relative to the hydride and results in even weaker induced
dipole attractions, so the boiling points are substantially lower. This is the trend
we see also in the group VII hydrides, which might have been expected to have
higher boiling points than either the group V or VI (if the trend from IV to V to
VI was followed), but have lower boiling points (at least for periods 4 and 5) than
the group V hydrides.
1.60. There are often different compounds having the same formula. As you will learn
in Chapter 5, these are known as isomers. For C5H12, there are three isomers. The
common names, line formulas, structures, and boiling points of the isomers are
given below. All three have the same molecular mass. Why aren’t their boiling
points closer together? Hint: The more compact a molecule, the less surface area
it has for its electrons to interact with other molecules. Build models of these
molecules to help visualize the surface areas of these molecules.
pentane
isopentane
neopentane
CH3CH2CH2CH2CH3
(CH3)2CHCH2CH3
C(CH3)4
bp = 36 C
bp = 28 C
bp = 10 C
H HH H
H
C
C
C
C H
H C
H H H H
H
C HH H
H
C
C
C
H
H C
H
H
H
H
H
H
H
H H
H
C H
H
C
C H
H C
H
H
H
H
C
Answer: Although the molecular masses (and molecular volumes) are the same
for all three isomers, their surface areas are not. As the degree of branching of the
carbon skeleton increases, the surface area decreases. As the surface area
decreases, the contact with neighboring molecules decreases, thus reducing the
dispersion forces between molecules. Boiling points of hydrocarbons are a
consequence of dispersion forces and those forces “exist across the entire surface
of a molecule (page 34).” You may recall from geometry that a sphere is the
geometric solid with the lowest surface area to volume ratio. Examine your
molecular models of the three isomers and you will find that neopentane most
nearly resembles a sphere. n-Pentane has no skeletal branches and has the
greatest surface area and the greatest boiling point of the three.
Section 1.8.
Further Structural Effects of Hydrogen Bonding in Water
1.61. If an ice cube is dropped into liquid water at 85 C, will it float or sink (before it
melts)? Justify your answer.
Answer: Figure 1.29 shows that the density of liquid water at 85 ˚C is
about 968 g L–1 (= 0.968 g cm–3). The density of ice is 0.917 g cm–3. Ice
is much less dense even than hot water and will float (until it melts) in hot
water. Try the experiment.
1.62. In Investigate This 1.30, you found that the temperature at the bottom of the
container rises to about 4 C and remains almost constant as long as there is ice
left at the top of the container. Thermal energy (heat) must be entering the
container from the warmer room air (the ice does melt). You would expect the
water at the bottom to continue to warm up, but the temperature at the bottom
stays constant. These seem to be contradictory observations.
(a) What is the special property of water at 4 C?
(b) If the water at the bottom warmed a bit above 4 C, how would this property
change? What would the water be likely to do? Draw pictures to illustrate what
you think would happen.
(c) Would the action of the water shown in your drawings resolve the
contradiction suggested above? How could you test your model experimentally?
Answer: (a) This is the temperature range in which liquid water at normal
atmospheric pressure has its maximum density, Figure 1.29, which
explains why it is at the very bottom of the container.
(b) Warming the water will reduce its density, and the warmer water
would be buoyed up and rise. Since the thermal energy is entering the
water at the walls of the container, it is the water next to the walls that
increases in temperature and rises. When water near the top of the
container, which is close to 0 C, is warmed, its density increases and it
sinks in the container. The net effect of these actions is to create “currents”
of sinking water in the temperature range 0-4 or 5 C and rising water with
a temperature above about 5 C. The rising warmer water is cooled as it
mixes with water near the top of the container and, when its temperature
falls below 4 C, it begins to sink. The net effect is a circulation of the
water in the container with rising warmer water near the walls of the
container and sinking colder water near the middle of the container. The
circulation is driven by the thermal energy from the warmer surroundings
of the room entering the ice water mixture.
(c) If the explanation above is correct, the constant temperature measured
at the bottom of the container is not the temperature of the same batch of
water molecules, but of a constantly changing batch of water molecules as
the warmed water is buoyed up and the cooler water sinks. If the
temperatures at the walls and in the center of the container are measured at
the same height in the container, there might be a detectable difference,
with the wall water at temperature above 5 C and the center below 5 C.
If a drop of dye could be added to the water without disturbing whatever
circulation has been set up, the movement of the dye could trace the
circulation over time. More elaborate set ups have been devised to check
whether this circulation occurs. A search of the web could turn up some of
them for you.
1.63. The structure of the water molecule is the molecular basis for the survival of plant
and animal life in a temperate climate lake. The seasonal “turnover” of such a
lake is described in this figure.
These vertical temperature changes are typical of a lake that freezes in winter.
Turnovers, represented by the circling arrows, occur in the spring and fall and mix
nutrients and oxygen into the deeper waters. The turnovers are triggered by winds
at the surface of the water. How would you relate your observations in Investigate
This 1.30 to the changes described in this figure? Explain the connections clearly.
Answer: In Investigate This 1.30, the denser water was at the bottom of the
beaker. In the summer, the sunlight warms the water so the less dense water
(temperature greater than 4 C) is on the surface of the water. As the temperature
decreases in the fall, the nutrients in the lake mix. Now, in the winter, the less
dense water (temperature less than 4 C) freezes over the surface of the lake. As
it melts in the spring, the mixing of nutrients occurs again as the cold water
descends to the bottom of the lake.
1.64. Consider this representation of covalent bonds and
hydrogen bonds in water:
(a) Which bond length is associated with hydrogen bonds
and which with covalent bonds?
(b) Offer a reasonable explanation for why there is a
difference in bond length between these two bonds.
Answer: (a) The shorter bond length, 100 pm, is associated with the
covalent bonds within the water molecule. The longer bond length, 180
pm, is associated with the hydrogen bonds between electropositive and
electronegative regions of different molecules.
(b) Covalent bonds are stronger than hydrogen bonds. The stronger covalent
bonds hold hydrogen to oxygen within the water molecule more closely than
molecules are held together with intermolecular hydrogen bonds.
1.65. The melting points for methane, ammonia, water, and hydrogen fluoride are
shown in the table at the right. You can take the melting
points as an indication of the relative amount of energy
required to disrupt the attractions between molecules in the
solids, so they are free to move about as a liquid. Develop
an explanation for these data that takes into account the
Compound
CH4
NH3
H2O
HF
mp, C
–182
–77.7
0
–83.1
kinds of intermolecular attractions among molecules of each compound. Is water
out of line with the rest of the compounds? Why or why not? Give a molecular
level interpretation of your answer.
Answer: Essentially the same explanation is possible for the melting
points as for the boiling points. The structure of solid water, Figure
1.28(b), shows that each water molecule is hydrogen bonded to four
others. Methane forms no hydrogen bonds and is only held together by
weak induced dipole attractions in the solid (and liquid), so it melts at
quite a low temperature. Ammonia and hydrogen fluoride form hydrogen
bonds, but can’t form as many (per molecule) as water. Any individual
molecule could form four hydrogen bonds, but an extended structure is not
possible. A plot of the melting point data look like this:
One could argue that H2O is “in line” with CH4 and NH3. The melting
points increase as the molar mass and, more importantly, the amount of
hydrogen bonding increases. HF might be seen as “out of line,” but only if
the extent of hydrogen bonding is neglected. On the other hand, the similar
melting points of NH3 and HF can be explained as a result of about the
same amount of hydrogen bonding attraction in the two substances. In this
case, H2O is “out of line” because it has a higher melting point, but, since
it has more hydrogen bonding, it would be “expected” to have more
attractions and a higher melting point. Arguments about what value is “out
of line” are all based on predictions or trends that look at only one
causative parameter (molar mass, number of H bonds, etc.). Since
observable properties are the result of many molecular-level interactions
and phenomena, it isn’t surprising to find that any one of them can’t
explain everything.
Section 1.9.
Hydrogen Bonds in Biomolecules
1.66. The DNA double helix, held together by the hydrogen bonds shown in Figure
1.33, is quite stiff and resistant to movement
through a solution. As shown in this figure, when
a solution of DNA is heated, the absorption of
ultraviolet light at 260 nm rises sharply over a
small temperature range. At the same time the
solution suddenly begins to flow more easily
(more like water than like syrup). The middle of this range is usually labeled Tm
(“melting” temperature). What is happening to the DNA to cause these changes in
the solution properties? Explain your response.
Answer: At the Tm, many of the
hydrogen bonds holding together the two strands of the DNA helix, Figure
1.34, break and the strands begin to come apart to form more flexible
chains that can curl upon themselves to form structures that are more
globular and less resistant to flow. Thus, the viscosity of the solution
decreases. The spectroscopic properties (light absorption) by base pairs
stacked one upon the other in the double helix are quite different than the
“free” bases. The absorption of light (in the ultraviolet) changes markedly
as the double helix comes apart and the bases are no longer stacked.
1.67. This figure shows melting temperature, Tm (see Problem 1.66), data for a number
of different double-helical DNA’s plotted against the fraction of A–T pairs in the
DNA’s. Why are the melting temperatures a function of the fraction of A–T pairs?
Does the direction of the dependence make sense? Clearly explain your reasoning.
Hint: Review Consider This 1.34.
Answer: In Section 1.9, you learn that there are three hydrogen bonds between
G–C base pairs and two between A–T base pairs. Therefore, more energy is
required to break a G–C pairing than an A–T pairing. Higher temperatures
(greater thermal energy in the solution) are required to “melt” DNA with higher
proportions of G–C (or lower proportions of A–T) pairs. Thus the Tm decreases as
the proportion of A–T increases, as shown in the figure.
1.68. In Problem 1.66, the “melting” temperature of DNA in solution was defined as
the temperature at which sharp changes in ultraviolet light absorption and solution
flow occur. Why do you think this is called a “melting” temperature? Is(are) there
any analogy(ies) between what happens to DNA in these solutions and what
happens when ice melts? Explain your reasoning clearly.
Answer: When a solid melts, the temperature of the solid-liquid mixture remains
constant until all the solid has melted. The change from the double helix to
separate strands occurs over a rather short temperature range when the hydrogen
bonds in DNA break at the “melting” temperature. Thus, the relatively sharp
change in properties within a few degree temperature change is the DNA analogy
to “melting.”
1.69. The proteins in most organisms are denatured at temperatures above about 60 C.
Microorganisms that live in hot springs and organisms that live near deep ocean
thermal vents survive at temperatures near or above the boiling point of water,
100 C. Their proteins are made of the same amino acids as all other organisms.
What role do you think hydrogen bonds might play in helping these organisms
survive?
Answer: The larger the number of interactions among the amino acids along the
protein chain, the more energy will be required to disrupt the folded structure of
the protein. These interactions could be hydrogen bonds, ionic attractions between
charged groups (Chapter 2), and/or hydrophobic interactions that stabilize the
folded structure by keeping hydrophobic groups inside the structure away from
water (Chapters 2 and 6). The more such interactions, the higher the temperature
required to provide enough thermal energy to disrupt the protein structure.
1.70. The structure of the DNA of the thermophilic (therme = heat + philos = loving)
organisms discussed in Problem 1.69 also has to be maintained in the high
temperature environments where they live. What kind of A–T versus G–C
composition would you expect to find for the DNA in these organisms? Present
your reasoning clearly. Hint: See Problem 1.67.
Answer: The same kind of reasoning as in Problem 1.49 applies to
thermophilic DNA. The more attraction between strands, the more
thermally stable the DNA (as in Problem 1.47). Thus, you expect the
thermophiles to have a high proportion of G–C (with three H bonds). This
is observed.
1.71. Cellulose is a long-chain molecule made up of glucose molecules bonded together
as shown in this illustration. You will
learn more about cellulose later. Many
chains like these are hydrogen-bonded to
neighboring chains to form the fibers that are used to make paper and cotton and
linen cloth. The hydrogen-bonds make cotton cloth soft and flexible because they
are easily broken and remade, which allows the fibers to change shape.
(a) How do you account for the fact that cotton clothing is easily wrinkled?
Explain your reasoning clearly.
(b) How does ironing wrinkled cotton clothing restore its “press?” Use diagrams
to illustrate your answer.
(c) How might you make “permanent press” cotton cloth? Indicate what you
would try to accomplish; don’t be concerned about the detailed chemical
processes that might be required.
(d) Permanent press cotton clothing is not as soft as regular cotton clothing. Is
this the result you might expect from your response to part (c)? Explain why or
why not.
Answer: (a) Cotton clothing gets wrinkled by being worn and forced to
conform to the shape of its wearer and other forces like chair seats. These
forces on the fibers cause them to be bent and pushed and pulled into
shapes that breaks some H-bonds and makes others. When the forces are
no longer acting, say when the clothing is removed, the new H-bonds hold
the fibers in the positions they had taken during the time the clothing was
worn. Since many of these involve the folding and creasing that goes with
out movements, the fabric is now folded and creased on a small scale, that
is, wrinkled.
(b) Ironing has two effects. The heat of the iron helps to make the Hbonds break more readily. The mass of the iron forces the fabric flat so
that H-bonds that are remade are now holding the fabric flat. Thus the
wrinkles are removed and the “press” restored.
(c) To make the fabric “permanent press”, you want to keep the cellulose
chains in the positions they have in the fiber in the unwrinkled fabric.
Since the H-bonds are too weak to accomplish this, you need to find a way
to bond the chains to one another. You might do this with reagents that
react with —OH groups that are close to one another to form a permanent
bond holding them together.
(d) Since the softness of cotton comes from the ability of the cellulose
chains and fiber made from them to change shape easily, preventing these
easy changes will make the fabric less soft. Thus, bonding the chains
together, as suggested in part (c) would probably make the fabric less soft.
Section 1.10. Phase Changes: Liquid to Gas
1.72. How is energy involved for a substance to change from one state to another?
Explain.
Answer: Energy is required to loosen intermolecular forces that holds molecules
in close proximities when a substance changes from a solid to a liquid or from a
liquid to a gas. Energy is released to condense molecules, strengthening
intermolecular forces, when a substance changes from a liquid to a solid or from a
gas to a liquid.
1.73. Make the following conversions:
(a) 4550 J = ___________ kJ
(b) 250. J = ___________ calories
(c) 500. Cal = ___________ J
Answer: (a) 4.55 kJ
[1 Nutritional Calorie (Cal) = 1000 calories]
(b) 59.8 calories (c) 2.09 X 106 J
1.74. A phase diagram such as this one for water, is a
common way of representing phase changes. Phase
diagrams are pressure-vs.-temperature plots that show
the pressures at which the phase changes of a substance
occur as a function of temperature.
(a) What information does this phase diagram give
you about the phases of water at standard atmospheric pressure of 760 mm Hg (=
1 atm)? Explain your reasoning briefly. Hint: Consider starting at the pressure
axis, moving across the plot (increasing temperature) at a constant 760 mm Hg,
and noting when phase changes occur.
(b) For a change from gaseous water to liquid water at 100 C and 760 mm Hg,
will the sign of E be positive or negative? Explain your reasoning.
(c) For a change from liquid water to solid water at 0 C and 760 mm Hg, will
the sign of E be positive or negative? Explain your reasoning.
Answer: (a) Looking at the normal atmospheric pressure of 760 mmHg,
one can observe as expected that water will be in the solid phase below 0
°C , in the liquid phase up to a temperature of 100 °C, and then in the
gaseous phase. The phase diagram represents a great deal more
information about the phase of water at different temperatures and
pressures.
(b) As water changes from the gaseous phase to the liquid phase, its
energy decreases. This means that E will be less than zero, and its sign
is negative.
(c) As water changes from the liquid phase to the gaseous phase, its
energy increases. This means that E will be greater than zero, and its
sign is positive.
1.75. Sketch the energy diagrams that describe the following processes. Make sure that
you draw an arrow that represents the direction of the change and that you
indicate the sign of E for these processes.
(a) A sample of ice melting, H2O(s)  H2O(l), at 0 C. E > 0 for the sample.
(b) The combustion of hydrogen gas in oxygen gas is one of many chemical
reactions that release considerable quantities of energy. This process is described
by the equation:
2H2(g) + O2(g)  2H2O(l) + energy
(c) The decomposition of mercury oxide (HgO) occurs at high temperatures. For
this process to occur, energy has to be supplied. The process is described by the
equation:
energy + 2HgO(s)  2Hg(l) + O2(g)
(d) The burning (or oxidation) of mercury is the reverse of the decomposition
process in part (c) and is described by the equation:
2Hg(l) + O2(g)  2HgO(s) + energy
(e) Burning a sample of methane in oxygen gas, for which E < 0. This reaction
is described by the equation:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
Answer:
(a)
.
H2O liquid
∆E > 0
H2O solid
(b)
2H2(gas) + O2
(gas)
∆E < 0
2H2O (liquid)
(c)
2 Hg(liquid) + O2
∆E < 0
2 HgO (solid)
(d) reverse of (c)
2 Hg(liquid) + O2
∆E > 0
2 HgO (solid)
(e)
CH4(gas) +
2O2(gas)
∆E < 0
CO2 (gas) + 2H2O
1.76. The amount of energy required to vaporize methanol is 1.22 kJ·g–1. How many
kcal are required to vaporize one gram of methanol?
Answer: See Worked Example 1.38 and Check This 1.39 in which similar
calculations were done for water and hexane. The only additional information
you need is that 1 kJ = 103 J and 1 kcal = 103 cal.
energy of vaporization = 1.22 kJ • g–1 = (1.22 kJ • g–1) (1 kcal / 4.184 kJ) = 0.292
kcal• g–1
Section 1.11. Counting Molecules: The Mole
1.77. Calculate the molar mass of a formula unit of the following substances:
(a) dimethyl ether, CH3OCH3
(b) ethanol, CH3CH2OH
Answer: (a) 46.08 amu (b) 46.08 amu
1.78. How many molecules of water are in exactly one mole of water? How many
grams of water are in exactly one mole of water?
Answer: (a) 6.02 X 1023 molecules (b) 18.02 grams
1.79. How many moles in the following?
(a) 100.0 g of acetone
(b) 100.0 g of methanol
(c) 100.0 g of dimethyl ether
(d) 100.0 g of sucrose
Answer: (a) 100.0 g acetone 
1 mol acetone
 1.721 mol acetone
58.09 g acetone
(b) 3.120 mol methanol
(c) 2.170 mol dimethyl ether
(d) 0.292 mol sucrose
1.80. How many molecules are there in 1 gram of water, methanol, acetone, ethanol,
and dimethyl ether?
Answer: Water:
1 g H 2O 
1 mol H 2 O 6.022  1023 molecules

 3.3  10 22 molecules of
18 g H 2 O
1 mol H 2 O
water
Methanol: 1.9 X 1022 molecules of methanol
Acetone: 1.0 X 1022 molecules of acetone
Ethanol: 1.3 X 1022 molecules of ethanol
Dimethyl ether: 1.3 X 1022 molecules of dimethyl ether
1.81. Argon atoms have a diameter of approximately 100 pm. If a mole of argon atoms
were lined up one after another, how long, in meters, would the line be? The
distance from the Earth to the Sun is 1.5  1010 m. How many round trips will
this line of argon atoms make?
Answer: 100 pm / atom) (6.02 • 1023 atoms / mol) (m / 1012 pm) = 6.02 • 1013 m
To give some concrete meaning to this distance, it can be compared with the
distance to our Sun, 93 million miles (= 1.5 • 1010 m). Our line of argon atoms
would be able to make approximately 2,000 round trips to the Sun!
1.82. (a) How many moles of hydrogen bonds are there in a mole of ice? Explain how
you get your answer. Hint: If each H stopped hydrogen bonding, all the hydrogen
bonds would be gone.
(b) The energy required to melt ice is 6.02 kJ·mol–1. If the model for ice melting
presented in the text is correct, how many moles of hydrogen bonds are broken
when a mole of ice melts? What percentage of the total hydrogen bonds is this?
Clearly explain how you get your answers. Recall that the energy required to
break a hydrogen bond between two water molecules is 20–25 kJ·mol–1.
Answer: Given: 1 mole of ice. The energy required to melt one mole of
ice is 6.02 kJ mol–1.
(a) Asked for: How many moles of H-bonds are there in a mole of ice?
Recall: If each H stopped H-bonding, all the H-bonds would be gone.
Anything that is true of relative numbers of molecules is also true of
relative numbers of moles of those molecules.
Plan: Use the molecule —> mole relationship to find the number of
moles of H-bonds.
Calculations: Each hydrogen atom represents one H-bond; each mole of
hydrogen atoms reprsents one mole of H-bonds. Each molecule of water
has two hydrogen atoms; each mole of water has two moles of hydrogen
atoms. Therefore, there are two moles of H-bonds in one mole of water.
(b) Asked for: How many moles of H-bonds are broken when a mole of
ice melts and what percent of the total moles of H-bonds is this?
Recall: The energy required to break a mole of H-bonds is approximately
21 kJ mol-1,
1 mol H-bonds = 21 kJ

1=
1 mol H - bonds
21 kJ
From part (a), there are two moles of H-bonds in a mole of ice.
Plan: If we know the energy required to melt ice and the energy required
to break a mole of H-bonds we can calculate the number of moles of Hbonds broken when a mole of ice melts. We assume that the energy
required to melt ice is all used to break H-bonds. Convert this energy to
moles of H-bonds broken with the unitary conversion factor above. The
percentage of total H-bonds broken is calculated from this result and the
total number of moles of H-bonds, two moles.
Calculations:
6.02 kJ (melts 1 mol ice) = (6.02 kJ)
1 mol H - bonds


21 kJ
= 0.28 mol (H-bonds broken per mole ice melted)
% H-bonds broken =
mol H - bonds broken
0.28 mol
 100 =
 100
mol H - bonds total
2 mol
= 14% H-bonds broken
Recap: Our approach was to assume that the ratio of the energy required to melt
ice to the energy required to break all the H-bonds in the ice would give the
fraction of H-bonds broken in the melting process. Our result indicates that only
14% of the total number of hydrogen bonds are broken when ice melts. That
means most of the H-bonds are still present even though the water has changed
from a solid to a liquid. This is one of the results that leads to the “ice cluster”
model of liquid water.
1.83. In which of these compounds can the molecules hydrogen bond to themselves?
Draw a diagram of the molecules of each compound hydrogen bonding among
themselves.
(a) water
(d) ethanol
(b) methanol
(e) dimethyl ether
(c) acetone
Answer: “Self hydrogen bonding” requires that the molecule has both (1)
a nonbonding pair of electrons on an oxygen or a nitrogen and (2) an O–H
or N–H. Water, methanol, and ethanol can “self hydrogen bond” since all
three have both the essential features for hydrogen bonding within each
individual molecule. Acetone and diethyl ether molecules lack O–H
bonds. (They are capable of hydrogen bonding with water but not with
themselves.) The answer to Problem 1.41 shows the hydrogen bonded
structure of water. Those of methanol and ethanol are shown below.
(Ethanol would look similar to methanol)
1.84. The boiling points of dimethyl ether (CH3OCH3) and diethyl ether
(CH3CH2OCH2CH3) are –25 °C and 35 °C, respectively. What interaction is
mainly responsible for the observed difference?
Answer: Since these compounds cannot hydrogen bond with themselves,
intermolecular interactions are limited to dipole attractions and dispersion forces.
On a molar basis, both ethers have an equal number of polar C—O bonds.
Therefore, dipolar interactions are not responsible for the 60 °C difference in
boiling points. By process of elimination, we are left with dispersion forces
(induced dipole interactions) as the factor largely responsible for the higher
boiling point of the larger ether.
1.85. Acetone, like dimethyl ether, has no self-hydrogen-bonding capacity, but its
dipole moment is more than double that of the ether. Can this high dipole moment
explain the properties of acetone compared to the other compounds in Table 1.2?
Clearly explain the reasoning for your response and include as many comparisons
as possible.
acetone
Line Formula
Molar
Mass,
g
Energy of
Vaporization,
kJ·mol–1
CH3C(O)CH
58
32
Boiling
Point,
C
56
Dipole
Moment,
Debye
2.88
3
Answer: Dipole-dipole attractions are the major contributor to the forces
holding acetone molecules together in the liquid. Because the acetone
dipole moment (2.88 D) is so much larger than that of the ether (1.30 D),
the attractions are much stronger between acetone molecules and its
boiling point is 81 degrees higher. The energy of vaporization for acetone
is intermediate between dimethyl ether and ethanol. The ethanol dipole
moment is smaller than that of acetone, but the hydrogen bonding in
ethanol raises its boiling point and energy of vaporization above those of
acetone. Although acetone has a molar mass that is lower than that of
hexane, its energy of vaporization and boiling point are comparable to
hexane, because the dipole-dipole attractions in acetone are stronger than
the larger number of induced dipole attractions among hexane molecules.
Section 1.12. Specific Heat of Water: Keeping the Earth's Temperature Stable
1.86. Define specific heat.
Answer: The energy required to raise the temperature of 1 gram of a
substance 1C.
1.87. What is the difference between an intensive and an extensive property?
Answer: An intensive property is independent of the amount of the substance
present, (example: temperature) while an extensive property depends on the
amount of substance present (example: thermal energy).
1.88. Are the following properties intensive or extensive?
(a) the boiling point of water
(b) the density of water
(c) the specific heat of water
(d) the ratio of hydrogen to oxygen atoms in a sample of water
(e) the (maximum) solubility of salt in water
Answer: Recall that intensive properties are independent of the amount of
material and invariant from one sample to another. Extensive properties depend
on the amount of material and may vary from one sample to another. All five
properties the listed in this problem are intensive properties. Oftentimes, an
intensive property can be the ratio of two extensive properties. For example
density (intensive) is the ratio of mass and volume (both extensive properties).
Solubility is typically expressed as a ratio of mass to volume (see Chapter 2).
1.89. Convert 37.0 C to kelvin.
Answer: 37.0 ºC + 273.15 = 310.15 K
1.90. How much thermal energy is required to raise the temperature of 1.0 gram of
water by
(a) 10.0 C?
(c) 25.0 K
(b) 25.0 C?
Answer: (a) 41.8 J (b) and (c) 104.5 J
1.91. How much thermal energy is required to raise the temperature of the following by
10.0 C?
(a) 10.0 grams of water
(b) 25.0 grams of water
Answer: (a) 418 J (b) 1045 J
1.92. How much thermal energy is required to raise the temperature of 20.0 grams of
acetone by 15.0 C?
Answer: 651 J
1.93. In dry parts of the world, “air conditioning” is provided by blowing the hot
outside air through mats soaked in water before it enters the building. On what
scientific principle is this system based? How does it work? What advantages and
disadvantages can you see for the people and things in the building?
Answer: The system is based on the principle of evaporative cooling. As
the hot air flows over the mats, heat is transferred from the air molecules
to the water, the air loses heat energy, and the temperature of the air falls.
The heat energy transferred to the water causes some of the water to
evaporate into the air. The heat required to vaporize the water is quite
substantial (recall the high energy of vaporization of water) so a relatively
small amount of evaporation can remove a lot of heat from the air. In
areas of low humidity this system would have the added advantage of
increasing the humidity in the air in the building. Conversely in regions of
high humidity this method of air conditioning might not be as
advantageous since increasing the humidity could result in increased
levels of mold and mildew and might cause damage to sensitive
instrumentation from the damp. It also wouldn’t work as well in high
humidity;
1.94. Would you expect evaporative cooling of your skin to be more effective on dry
days or humid days? Clearly explain the reasoning for your answer.
Answer: Evaporative cooling should be more important on dry days than
humid days since the amount of water that can evaporate into the
atmosphere depends upon the amount of water already in the atmosphere.
If there’s a lot of water there already, on a humid day for example, then
the amount of sweat that can evaporate will be relatively low. This is
because at a given temperature the atmosphere can only “hold” a certain
amount of water, on humid days the atmosphere is near that limit. (If the
limit is exceeded, water condenses or precipitates out of the air. The
“relative humidity” that your TV weatherperson talks about is the percent
of water the air contains relative to the maximum amount it can contain at
that temperature.) Little water will evaporate and therefore not much
evaporative cooling will occur.
1.95. Table 1.3 gives the heat capacity of liquid water as 75 J·mol–1·C–1 (18 cal·mol–
1
·C–1). The heat capacity of solid water (ice) is 38 J·mol–1·C–1 (9.0 cal·mol–
1
·C–1). Why do you think the heat capacity of the solid is less than that of the
liquid? Are our models of liquid and solid water consistent with your explanation?
Answer: Our model of liquid water says that heat energy added to the
water makes the molecules move faster (increases the temperature) and
causes some H-bonds to be broken. The temperature rise is not as large as
it would be in the absence of the H-bond breaking and the specific heat (or
heat capacity) is high because it takes more heat to make a given change in
temperature. Heat energy added to ice at temperatures below 0 ˚C (so the
solid does not melt), breaks no H-bonds; the energy only causes the
molecules to move faster as they vibrate and jiggle in the position they are
held by their four H-bonds. All the energy added goes to raising the
temperature (increasing molecular motion); it takes less heat to make a
given change in temperature and the heat capacity is lower than for liquid
water.
1.96. The heat capacity of gaseous water (steam) at one atmosphere pressure is 36.7
J·mol–1·C–1 (8.8 cal·mol–1·C–1). What sort of model do you think would
describe gaseous water? Why do you think the heat capacity of the gas is less than
that of the liquid (See Table 1.3 or Problem 1.95)? Is your model of the gas and
the model of the liquid we have discussed in this chapter consistent with your
explanation?
Answer: The high heat capacity of liquid water is reviewed in the solution
to Problem 1.95. Gases occupy so much more space than the same
amount of liquid, that we either have to assume that the molecules get a lot
bigger or that they are far apart and probably not interacting with each
other very much. Gases are all completely miscible with one another, so
the model that has a lot of empty space between molecules seems to fit
this property better. If this model is correct, then we would expect
essentially no H-bonds between water molecules in the gas phase. If there
are no H-bonds all the heat energy we put into the gas goes into increasing
the speed of molecular motion (temperature). It takes less heat energy to
make a given change in temperature and the heat capacity of the gas is less
than that of the liquid. Note that the heat capacities of solid and gaseous
water are similar (about half that of liquid water). This similarity is
fortuitous; the molecular motions involved in the temperature changes in
the solid and the gas are different. In the solid, the molecular motion is
mainly vibration and rocking in place in the crystal. In the gas, the
motions are mainly translation (movement from one place to another) and
rotation.
1.97. Table 1.2 shows that 44 kJ·mol–1 (10.5 kcal·mol–1)
are required to change one mole of liquid water to
water vapor. This value is for water near 25 C.
The energy of vaporization depends on the
temperature of the water, as shown in this figure.
Why does the energy required to vaporize water
vary with temperature this way? Clearly explain
your reasoning.
Answer: At higher temperatures, the water molecules have greater energy
of motion and fewer H-bonds. It takes less energy to vaporize water at a
higher temperature because there are fewer H-bonds to break. For water
molecules to vaporize they must escape from the liquid. The more
hydrogen bonds holding them in the liquid phase, the more energy this
process will take. In addition to understanding the direction of this
phenomenon, you should also note its magnitude. The heat of
vaporization decreases from about 10.7 kcal mol–1 at 0 ˚C to about 9.7
kcal mol–1 at 100 ˚C. This is only a 10% change over the entire liquid
range. The plot might give the impression that the change is much larger.
What choice of scale would you use to show that the change is actually a
relatively small percentage of the total?
Section 1.13. EXTENSION --Liquid Viscosity
1.98. To which physical property of liquids does the expression “slow as molasses in
January” owe its truth? Would molasses be “faster” in June? Why?
Answer: The expression owes its truth to viscosity. Viscosity measures the
resistance of fluids to flow. The greater the viscosity, the more slowly the liquid
flows. The viscosity of a liquid usually decreases as temperature increases.
Therefore, molasses would flow much faster in June than it does in January.
1.99. The viscosity of n-heptane, CH3(CH2)5CH3, as a
function of temperature is plotted here on the same
scale as Figure 1.41, which gives the
corresponding data for the viscosity of water. To
make the plots comparable, the values here are
relative to the viscosity of heptane at 20 ˚C. What
similarities do you observe between these data and
those for water? What differences do you observe?
What explanation can you provide for the similarities and differences?
Answer:
The data for heptane are on the left-hand plot and those for water are on
the right. (The right-hand plot is Figure 1.39; the caption is present but the
graphic is missing from the field test draft of the text.) The axes on both
plots are identical as is the reference point (viscosity at 20 C) for the
relative viscosity. Therefore, the shape and placement of the two lines can
be directly compared. The viscosity of both substances decreases with
temperature. Higher temperature means more average energy per
molecule; it is easier for the molecules to overcome the attractions
between them and therefore move past one another more readily. The low
temperature relative viscosity of heptane is low compared to water and its
decrease with temperature is less pronounced than for water. The
attractions among heptane molecules are induced dipole interactions
(dispersion forces), which are relatively weak to begin with and remain
about the same as the temperature increases the motions of the molecules,
so there is a relatively gradual decrease in viscosity. The relative viscosity
of water undergoes a much steeper decrease with temperature. The
stronger, more directed hydrogen bonds, that are responsible for much of
the high viscosity at low temperatures are disrupted at higher temperatures
and the increased motion of the water molecules makes these directed
bonds less likely to form at higher temperatures. Thus, the drop in relative
viscosity for water is more than twice what is observed for heptane.
1.100. Nature exploits the properties of the
R
hydrogen bond in many ways. Scientists
also work to find ways to use this weak
bond with its strength in numbers to
O
H
N
N
H
H
O
N
create materials with interesting and
N
N
N
H
H
N
N
H
R
O
O
R
O
H
N
N
H
H
O
N
N
N
useful properties. One group of
N
O
H
H
N
N
H
R
researchers has made a compound whose
molecules have “sticky” ends; each end of one molecule forms four hydrogen
bonds to another to produce long chains, as represented in the figure.
(a) For solutions of this compound (in a non-hydrogen-bonding solvent) that
vary in concentration from about 8 to 80 g/L, the
viscosity varies as shown in this logarithmic plot.
As the concentration changes by a factor of 10,
by what factor does the viscosity change? Clearly
explain how you might interpret this result. Hint:
Recall that large molecules can’t move rapidly in
solution, so their solutions resist flow.
(b) The viscosity of these solutions is temperature dependent. Do you predict an
increase or decrease in viscosity as the temperature of a solution is increased?
Explain the reasoning for your prediction.
(c) The researchers also made a compound whose molecules are essentially half
of one of the molecules shown above. These new
molecules have only one “sticky” end. The
viscosities of mixtures of 32 g/L of the original
compound with small amounts of the new compound
are shown in this plot. The horizontal axis shows the
decimal fraction of the mixture that is the new
compound; 0.01, for example, means that 1 in 100 of
the molecules in the solution are the new molecules.
Clearly explain how you might interpret what is going on in the solution to
produce these results.
O
Answer: (a) The scale on both axes is logarithmic. The concentration axis
covers a factor of 10 in concentration, while the viscosity axis covers a
factor of 104 (from 0.1 to 1000). The viscosity increases from about .08 to
about 800 on this scale, which is a factor of 104. As more and more of the
sticky-ended molecules are dissolved, longer and longer chains of them
are formed. The chains attract one another by dipole and induced-dipole
interactions and also can become entangled with one another. All these
attractions makes it harder for the chains to move past one another, so the
viscosity increases.
(b) For the same reasons that the viscosity of water decreases with
increasing temperature, these solutions should show decreasing viscosity
with temperature. The hydrogen bonding is disrupted as the molecules
become more energetic and move faster.
(c) As the fraction of molecules with only one sticky end increases, the
viscosity decreases. There is a competition among the sticky ends for one
another. If molecules with two sticky ends hydrogen bond to one another,
a longer chain is formed that has a higher viscosity and could add more
double-ended molecules to make even longer chains. But, if a molecule
with one sticky end hydrogen bonds to a molecule with two sticky ends,
the chain can no longer continue to grow at the end where the bonding
occurs. As more and more of the molecules present have only one sticky
end, the probability that chain growth will be stunted grows and the chains
are smaller. The smaller chains give the solution a lower viscosity. The
rapid decrease in the viscosity for rather small amounts of the singleended molecule added, show that it doesn’t take much stunting of the
growth of the chains to have a large effect on the viscosity.
General Problems
1.101. Briefly explain why:
(a) You experience a cooling effect after walking out of the ocean onto a warm,
sandy beach, especially on a breezy day.
(b) Liquids can be defined as "disordered" solids. Are there problems with this
definition?
(c) Liquids can be defined as "dense" gases. Are there problems with this
definition?
(d) Solid water (ice) floats on liquid water.
(e) Steam can badly burn you, if it condenses to water on your skin.
(f) Lakes freeze from the top to the bottom.
(g) Water pipes break, if water freezes in them.
Answer:
(a) Water needs heat from your skin to evaporate. While supplying the
heat for water to change from a liquid to a gas, you feel a cooling effect on
your skin.
(b) The molecules in liquids are attracted to one another and stay close
together, similar to solids. However, the molecules in liquids are much
more disordered because they can move around, while still staying close
together.
(c) The fact that the molecules in liquid can move around makes them
similar to gases. The molecules in gases are very far apart, moving
essentially independently of one another. Since ther are many more
molecules in a given volume of a liquid compared to the same volume of a
gas, liquids are sometimes defined as "dense gases".
(d) Ice floats on liquid water because the density of liquid water is lower
than the density of ice.
(e) During condensation, energy is released to the surroundings. This is
why you can be badly burned by steam, if it condenses to water on your
skin.
(f) When lakes freeze during the winter, ice covers the top of the liquid
water, insulating the water below. Since ice is less dense than liquid
water, ice floats and does not fall to the bottom of the lake.
(g) When water freezes in pipes, the ice expands due to the larger volume
of ice compared to liquid water. This causes the pipes to break.
1.102. Chemistry is everywhere. A friend has asked you if the claims are scientifically
accurate: “Keep a bottle of water at your desk and take frequent sips from it. Onethird of water is oxygen, so drinking it will help keep you alert.” Write a response
that will help your friend sort out what is true and what might not be.
Answer: This is not a profound problem, but is designed to suggest being
on the lookout for extraordinary claims like this one. On an atom basis,
one third of the atoms in water are oxygen, so the text is correct in one
way. It seems that the claim is something about the alertness enhancing
qualities of oxygen. You do need oxygen molecules from the air to keep
your metabolism going and you get tired more easily, if the oxygen level
drops somewhat. (That’s why you feel stress when exerting yourself at
high altitude, before your body acclimates to the lower pressure of air and,
hence, of oxygen in the air.) The oxygen atoms in water are, however,
totally unavailable for you to use in metabolism. They are bound in very
stable molecules. The best one can say about this claim is that it can’t hurt
to drink some water and it is free, that is, nothing is being sold here.
1.103. One model of liquid water (an “iceberg” model) is a mixture of molecular-scale,
ice-like structures among other less-ordered, less-hydrogen-bonded molecules
with the molecules continually exchanging between the two forms. In a sample of
liquid water, it is possible to give extra energy of motion only to those molecules
that are pointing (oriented) in the same direction. The natural rocking and jiggling
of these molecules soon changes their orientation (they become more random).
Scientists have measured the time required for the change and find that some of
the water molecules make the change in an average of about 0.7  10–12 s (0.7 ps).
The rest take an average of about 13  10–12 s (13 ps). The scientists concluded
that liquid water acts like it is made up of two species.
(a) Is this conclusion consistent with the “iceberg” model of liquid water? What
might the two species be? Clearly state the reasoning for your answers.
(b) Which of the two species changes orientation rapidly and which more
slowly? Explain your reasoning. Use drawings, if they are helpful.
(c) If there are two species, why aren’t they apparent to our senses in our
everyday contacts with water?
Answer:
(a) The “iceberg” model for liquid water has some of the water molecules
hydrogen bonded in molecular-scale ice-like structures and others that are less
orderly and more mobile. It makes sense that orderly and a disorderly water
structures would behave differently in terms of their ability to randomize the
orientation they have been given by the experimenters.
(b) Since the water molecules “locked” in an ice-like structure cannot move about
independently, their motions are probably more inhibited and take longer to
regain a more random orientation. Thus, we would expect the shorter
randomization time to be characteristic of the less orderly, more mobile molecules
and the longer time characteristic of the molecules in the ice-like structures.
(c) Note that the time scales of the changes observed here are picoseconds. The
molecules can undergo reorientations, which means breaking some hydrogen
bonds and making others in a very short time. Averaged over observation times
on the order of seconds (our everyday contacts with water), such rapid changes
make it impossible for us to detect any differences among the molecules or
possible rapidly interchanging structures in the liquid. Only clever experiments
(and calculations) will enable us to get better pictures of the nature of liquid
water.
1.104. Gases behave much like liquids in terms of things floating and sinking in them.
What can be deduced from each of these facts:
(a) A helium-filled balloon will rise in air.
(b) A hot-air balloon will rise in (colder) air.
(c) A balloon filled with carbon dioxide will sink in air.
Answer: (a) Helium is less dense than air (unstated assumption: when both are at
the same temperature and pressure).
(b) Hot air is less dense than cold air (unstated assumption: when both are at the
same pressure).
(c) Carbon dioxide is more dense than air (unstated assumption: when both are at
the same temperature and pressure).
1.105. What would happen if an astronaut standing on the Moon let go of a helium-filled
balloon she was holding. Recall that the Moon, unlike the Earth, has no
atmosphere. Hint: See Problem 1.104(a).
Answer: It would be pulled by gravity to the surface of the Moon. A helium
balloon on Earth rises because it “floats” on the more dense gases of the Earth’s
atmosphere. Since there is no atmosphere on the Moon, there is nothing for the
helium balloon to float upon.
1.106. Potassium acid fluoride is a salt composed of the ions, K+ and (FHF)–. The
negative ion, (FHF)–, can be thought of as two fluoride ions hydrogen bonded by
a proton with the Lewis structure:
F H F
. The H–F bond length in
hydrogen fluoride, HF, is 93 pm. Each of the H–F bond lengths in (FHF)– is 113
pm. Formation of (FHF)– from HF and F– releases about 155 kJ·mol–1, making it
by far the strongest hydrogen bond known, although much weaker than the twoshared-electron covalent bond in HF which releases about 565 kJ·mol–1 when it
forms.
(a) What, if anything, is peculiar about the Lewis structure shown for (FHF)–?
(b) Discuss the similarities and differences between this hydrogen bond and the
hydrogen bond between two water molecules. Also discuss whether the (FHF)–
example blurs the distinction between covalent bonds and hydrogen bonds.
Answer:
(a) The Lewis structure shows four electrons around the H atom.
(b) The differences are that this hydrogen bond is formed within an ion and with
the most electronegative atom being on the ends of the ion. In water, the oxygen
atom is the central atom. It also looks like this ion would be linear with the two
fluorine atoms trying to keep as far apart from each other is possible, thus
explaining the bond length difference from HF. The similarity is that both form
hydrogen bonds. Yes, this ion does blur the distinction between a hydrogen bond
and a covalent bond.
1.107. (a) About 70% (by mass) of your body is water. How many moles of water does
your body contain? How many molecules of water?
(b) The boxed table on page 1-27 gives the elemental composition of your body.
Assume that the same number of atoms of oxygen and nitrogen are combined in
molecules other than water. How many oxygen atoms (per 100,000 atoms) are
combined with hydrogen atoms to make the water in your body?
(c) Use your results from parts (a) and (b) to calculate the total number of atoms
in your body. How many of these atoms are nitrogen? carbon? How many moles
of nitrogen does your body contain? carbon?
Answer:
(a) Assume a 70 kg (approximately 150 pound) person. The moles and molecules
of water in this person (70% water) are:
mol H2O = (0.7)(70  103 g)/(18 g·mol–1) = 2.72  103 mol
molec H2O = (2.72  103 mol)(6.02  1023 molec·mol–1) = 1.64  1027
molec
(b) If 2440 O atoms are in compounds other than water, then 23230 (= 25670 –
2440) O atoms (out of every 100000 atoms) are present in water.
(c) From part (b) we know that 23230 molecules of O are present as H2O for
every 100000 atoms in the body. From part (a) we know the number of molecules
of H2O in the body, so we can get the total number of atoms in the body as:
total # atoms = [(100000 atoms)/(23230 molecules H2O)](1.64  1027
molec)
= 7.06  1027 atoms
To get the number of atoms of N and C, we take the fraction of atoms that are N
and C times the total number of atoms:
atom N = [(2440 atom C)/(100000 atom)](7.06  1027 atom) = 1.72  1026
atom
mol N = (1.72  1026 atom)/(6.02  1023 atom·mol–1) = 2.86  102 mol
atom N = [(10680 atom C)/(100000 atom)](7.06  1027 atom) = 7.54 
1026 atom
mol N = (7.54  1026 atom)/(6.02  1023 atom·mol–1) = 1.25  103 mol
1.108. This table gives the names, structures, and energies of vaporization for most of
the hydrocarbons (molecules containing only carbon and hydrogen atoms) whose
boiling points are given in the figure in Consider This 1.23.
hydrocarbon
formula
energy of vaporization
kJ·mol–1
ethane
propane
butane
pentane
hexane
heptane
octane
nonane
CH3CH3
CH3CH2CH3
CH3(CH2)2CH3
CH3(CH2)3CH3
CH3(CH2)4CH3
CH3(CH2)5CH3
CH3(CH2)6CH3
CH3(CH2)7CH3
15.65
20.13
24.27
27.61
31.92
35.19
38.58
43.76
(a) Plot these energies of vaporization as a function of the number of electrons in
each molecule. Draw the best possible straight line through the points. (If you use
a graphing calculator or computer graphing program, it can construct the line for
you.) Predict the energy of vaporization for decane, CH3(CH2)8CH3.
(b) Why is there an increase in energy required to vaporize these molecules as CH2- groups are added? Use the slope of your line from part (a) to determine how
much the energy of vaporization increases for each -CH2- group added.
(c) Assume that induced dipole attractions (dispersion forces) are directly
proportional to the number of electrons in molecules with second row elements
connected in a chain, like those in this table. What do you predict for the energy
of vaporization of dimethyl ether? How does your prediction compare to the value
given in Table 1.2? How do you explain any difference?
(d) The energies of vaporization of diethyl ether, CH3CH2OCH2CH3, and
butanol, CH3CH2CH2CH2OH, are 29.1 and 45.9 kJ·mol–1, respectively. Use what
you have learned in the previous parts of this problem, plus the data in Table 1.2,
to predict these energies and compare them with the experimental values. What
attractions among the molecules must you account for in each case?
Answer:
(a) The plot of the data looks like this (a fine straight line as long as methane is
not included):
Since decane has 82 electrons, the equation of the line gives its energy of
vaporization as 47.5 kJ·mol–1.
(b) Each CH2 group adds eight electrons and lengthens the hydrocarbon chain so
that more electrons and more surface area are available for dispersion force
(induced dipole) attractions. The increase of eight electrons adds about 4 kJ·mol–1
(= 8  0.5 kJ·mol–1 -- from the slope of the line).
(c) Dimethyl ether has 26 electrons and is similar in this regard to propane with
an energy of vaporization of 20 kJ·mol–1. We can attribute the extra 3 kJ·mol–1
energy of vaporization of the ether (23 kJ·mol–1) to dipolar attraction due to the
permanent electric dipole of the molecule.
(d) Diethyl ether has 42 electrons and is similar to pentane with an energy of
vaporization of 28 kJ·mol–1. We found in part (c) that an ether might have an
extra 3 kJ·mol–1 energy of vaporization attributable to dipolar attraction, so we
would predict an energy of vaporization of 31 kJ·mol–1. It appears that this is an
over correction for the ether with larger alkyl groups, which might interfere with
the molecule’s ability to orient as readily. The butanol also has 42 electrons and
we would again expect a contribution of 28 kJ·mol–1 to the energy of vaporization
from dispersion forces. In addition, the alcohol can form hydrogen bonds and
these, according to our textual analysis of the data in Table 1.2 might contribute
an additional 22 kJ·mol–1 to the energy of vaporization, for a total of 50 kJ·mol–1.
Again, our prediction is off by about 8%, but this is quite good agreement
considering the simplicity of our assumptions and the models.