•
colourless elements - oxygen
Classification rationale: Clear distinctions can be made among the elements in terms of colour.
(d) Third possible three–group classification:
• malleable solids - aluminum, silicon, magnesium, iron, copper, tin, lead
•
non-malleable solids - carbon, sulfur
•
gases - oxygen.
Classification rationale: Clear distinctions can be made among the solid elements in terms of malleability. Gases (and
liquids) could be classed as “others”.
(e) The description given in (a) of the various samples is fairly comprehensive. At this stage, students may add some of
those attributes to their original descriptions.
(f) The process of classification puts a substance under careful examination and tends to reveal similarities to and differences from other substances. As more and more similarities and differences are identified, a more detailed description
of the substance can be made.
(g) Other ways to classify the substances include: electrical conductivity, chemical reactivity, melting point, boiling point,
freezing point, density, viscosity, etc. To assist in classification, for example, the substances could be connected to a
source of electricity and then a multi-meter could be used to measure voltage and current; the substances could be
reacted with other substances to determine chemical reactivity, etc.
1.1 ELEMENTS AND THE PERIODIC TABLE
PRACTICE
(Page 11)
Understanding Concepts
1. (a) iron; metal
(d) carbon; nonmetal
(b) aluminum; metal
(e) silver; metal
(c) gallium; metal
(f) silicon; metalloid
2. (a) International Union of Pure and Applied Chemistry
(b) IUPAC agrees to and specifies rules for chemical names and symbols. Although the names of elements are
different in different languages, the same symbols are used in all languages. Scientific communication throughout
the world depends on this language of symbols, which is international, precise, logical, and simple.
3. Three sources of names for elements are: Latin names; names based on the country or region in which the element
was discovered; names that pay tribute to a notable scientist.
Making Connections
4. An example: furniture polish contains “isoparaffinic hydrocarbon.” Isoparaffinic hydrocarbon is a compound that
contains hydrogen (a nonmetal) and carbon (a nonmetal).
5. The student is to use the Internet to research the possible link between aluminum and Alzheimer’s disease. The student
is also to comment on whether aluminum is a significant environmental risk. Students may well find evidence to
support either position. The focus of ongoing research is to clarify how aluminum affects the body and whether it is
a factor in Alzheimer’s disease. However, most researchers believe that not enough evidence exists to consider
aluminum a risk factor for Alzheimer’s disease. The strip mining of aluminum ore (bauxite) may be harmful to the
environment. Processing aluminum from ore uses large amounts of electricity. Aluminum is also used in food,
drinking water, cosmetics, and drugs.
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Copyright © 2002 Nelson Thomson Learning
Chapter 1 The Nature of Matter
5
6. An example:
Metal (gold)
Nonmetal (phosphorus)
How Discovered
Gold in small quantities was
probably first found lying on
the ground by prehistoric
humans. It is soft, easy to work,
and retains a shine. It was
probably being used for
ornamentation shortly after
discovery.
A German scientist distilled the
residue from boiled-down,
well-putrefied urine; condensed
the vapours underwater; and
found something that glowed in
the dark.
When Discovered
long before 4000 B.C.
1669
Where Discovered
unknown
Hamburg, Germany
Who Discovered
prehistoric peoples.
Hennig Brand
Common Industrial or
Applications Technological
Gold’s superior electrical
conductivity, its malleability,
and resistance to corrosion have
made it vital to the manufacture
of components of electronic
products and equipment,
including computers and
telephones.
Gold is extraordinarily reflective
and does not tarnish, so is used
to shield spacecraft and satellites
from radiation and to focus
light in industrial and medical
lasers.
Gold is biologically inactive, but is
used in the direct treatment of
rheumatism.
Matches, fireworks displays,
phosphorous acid, phosphoric
acid (used to give a tart taste to
soft drinks, and to make
fertilizer), phosphorous
trichloride, etc.— used mainly
for the further manufacture of
other chemicals.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
Reflecting
7. It would be unwise for each country to choose its own names and symbols for elements because this would seriously
impair scientific communication throughout the world, for example, causing confusion if two countries used the same
name for different elements.
8. Example: “Special Education” as a useful classification:
Intellectual differences, sensory handicaps, communication disorders, physical handicaps, behaviour disorders, and
developmental disabilities have all been used to modify education for students who need something extra to help them
work toward their potential.
“Special Education” as a harmful classification:
Negative connotations are attached to such forms of classification, and educational jurisdictions tend to interpret some
of the “special education” categories in an arbitrary manner. For example, the classification “developmental disabilities” may be interpreted in a variety of different ways, even within a single province like Ontario.
PRACTICE
(Page 17)
Understanding Concepts
9. (a) Mendeleev’s periodic law states that “elements arranged in order of increasing atomic mass show a periodic
recurrence of properties at regular intervals.”
(b) Mendeleev was able to predict the properties of the as yet undiscovered elements, with some accuracy — but not
with complete accuracy. In general, periodic law is inadequate to fully predict properties of elements.
6
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
10. According to the law of triads, the middle element of a triad should have an atomic mass about halfway between the
atomic masses of the other two elements. The halfway value between fluorine, with an atomic mass 19.00, and
bromine, with an atomic mass of 79.90, would be an atomic mass of 49.45. The value for the atomic mass of chlorine
in today’s periodic table is actually 35.45.
11. Johann Döbereiner (1780–1849): Döbereiner was among the first scientists to consider the idea of trends among the
properties of the elements. By 1829, he had noted a similarity among the physical and chemical properties of several
groups of three elements. Döbereiner’s discovery is often referred to as the law of triads.
John Alexander Newlands (1837–1898): In 1864, Newlands arranged all of the known elements in order of
increasing atomic mass. He noticed that similar physical and chemical properties appeared for every eighth element.
He also noticed that some elements shared similar properties with other elements even though they did not follow the
“eighth element” pattern. He identified elements that share similar properties as being in the same family. Newlands
called his discovery “the law of octaves.” However, the law of octaves seemed to be true only for elements up to
calcium.
Julius Lothar Meyer (1830–1895): Meyer also arranged the elements in order of atomic mass. Meyer thought he
saw a repeating pattern in the relative volumes of the individual atoms of known elements. He also observed a change
in length of that repeating pattern. By 1868, Meyer had developed a table of the elements that closely resembles the
modern periodic table.
12. Mendeleev placed sulfur and oxygen in the same family because of their similar chemical properties. For example,
both oxygen and sulfur react with hydrogen according to the formula of H2R.
13. The discovery of noble gases supported Mendeleev’s periodic table in the sense that these new elements had similar
physical properties (they are all gases) and similar chemical properties (they are all unreactive). This added another
family of elements to Mendeleev’s periodic table to illustrate the periodic recurrence of properties.
14. Students will find that science is not always objective. Fear of ridicule (as of Newland’s musical analogy), fear of
risking reputation by making a major blunder (as in Lothar Meyer’s lack of the courage needed to propose new
elements to fill in the gaps in the repeating pattern of atomic volumes), and fear of the power of authority (challenging
an established theory supported by a well-established group of scientists could be hazardous to the career of the
maverick) are all powerful factors that affect the progress of science. Social factors and personal attributes probably
delayed the development and acceptance of the periodic table.
PRACTICE
(Page 19)
Understanding Concepts
15. Element Name
Atomic Symbol
Atomic Number
Group Number
State at SATP
lithium
beryllium
boron
carbon
nitrogen
oxygen
fluorine
neon
Li
Be
B
C
N
O
F
Ne
3
4
5
6
7
8
9
10
1 or lA
2 or llA
13 or lllA
14 or lVA
15 or VA
16 or VlA
17 or VllA
18 or VlllA
solid
solid
solid
solid
gas
gas
gas
gas
16. Metals = 89; nonmetals = 19; metalloids = 7
17.
Representative Elements of Period 2 - Groups l to 18
Element
Name
lithium
beryllium
boron
carbon
nitrogen
oxygen
fluorine
neon
Atomic
Symbol
Li
Be
B
C
N
O
F
Ne
Atomic
Number
3
4
5
6
7
8
9
10
Copyright © 2002 Nelson Thomson Learning
Atomic
Mass
6.94
9.01
10.81
12.01
14.01
16.00
19.00
20.18
State at
SATP
solid
solid
solid
solid
gas
gas
gas
gas
Melting
Point ˚C
181
1278
2300
3550
-210
-218
-220
-249
Boiling
Point ˚C
1342
2970
2550
4827
-196
-183
-188
-246
Chapter 1 The Nature of Matter
7
18.
Alkali Metals
Halogens
Noble Gases
Physical Property
• soft
• silver-coloured
• may be solids, liquids, or gases
at SATP
• not lustrous and
nonconductors of electricity
• gases at SATP
• low melting and boiling points
Chemical Property
• react violently with water,
liberating hydrogen gas
• react with halogens to form
compounds similar to sodium
chloride (NaCl)
• extremely reactive
• react readily with hydrogen
and metals
• extremely unreactive
• heavier gases may form
compounds with fluorine
19. According to the position of phosphorus in the periodic table, the most likely formula for a compound of phosphorus
and hydrogen is PH3(g) — commonly called phosphine, an extremely poisonous gas with a garlic odour. Its chief use
is in the manufacture of plastics used to make compounds that make cotton cloth flame-resistant.
Making Connections
20. Some support for the Age of Silicon: Silicon is a metalloid solid, metallic in appearance, and has a high melting point
(1410°C). However, its electrical conductivity is much less than that of a typical metal. It is this “semiconductor”
property that makes this element extremely useful in the electronics and communication technology industries. Silicon
is the material currently used to make most microchips — integrated circuits — a tiny piece of silicon that contains
thousands of tiny, interconnected electrical circuits that work together to receive and send information. Microchips
can process a great deal of information, and yet take up very little space. However, silicon’s status as indispensable to
computation is being challenged with new methods, including photonics, that will not require silicon. Perhaps, many
years from now, this will be referred to as the Age of Silicon, but a case could be made for a more general description: the Age of Information.
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SECTION 1.1 QUESTIONS
(Page 20)
Understanding Concepts
1. Lithium, sodium, and potassium are elements in the same family that are arranged in order of increasing atomic
number and show a periodic recurrence of properties — all are soft, silvery-coloured elements, all are solids at SATP,
all exhibit metallic properties, and all react violently with water to form basic solutions and liberate hydrogen gas.
2. (a) Percentage approach: If we take each melting point as a percentage of the previous melting point and then average
the percentages, the predicted melting point of Rb would be about 37.6˚C. The actual melting point of Rb is
38.9˚C.
(b) Alkali metals and alkaline earth metals show a periodic recurrence of decreasing melting points as you move
down a group, whereas nonmetals show a periodic recurrence of increasing melting points as you move down a
group. Thus, based on the periodic recurrence of decreasing melting points Li, Na, K, and Rb would be classified
as metals.
(c) As alkali metals, Li, Na, K, and Rb would be expected to be soft, silvery-coloured elements; to be solids at SATP;
to exhibit metallic properties; and to show decreasing boiling points as you move down the group.
3. (a) “Shiny, grey solid at SATP” are physical properties that are in line with elements from the alkaline earth metals
group, or from the transition metals group.
“When heated in the presence of oxygen, a white, powdery solid forms” is a chemical property also in
line with elements from the alkaline earth metals group, or from the transition metals group - elements from either
of these two groups, especially elements from the alkaline earth metals group, react with oxygen to form oxides.
(b) Before 1800, scientists distinguished elements from compounds by heating the substances to find out if they
decomposed. If the products they obtained after cooling had different properties from the starting materials, then
the experimenters concluded that decomposition had occurred, so the original substance was a compound. In this
case, the scientist would have concluded the original grey solid substance to be a compound, and the white
powder (which will not decompose on heating) to be an element.
8
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
4.
Table 4: Elements and Mineral Resources
Mineral resource or use
Element name
radium
Atomic
number
88
Element
symbol
Ra
Group
number
2
Period
number
7
SATP
state
solid
High-quality ores at
Great Bear Lake, NT
Rich ore deposits at
Bernic Lake, MB
Potash deposits in
Saskatchewan
Large deposits in
New Brunswick
Extracted from Alberta
sour natural gas
Radiation source for
cancer treatment
Large ore deposits
in Nova Scotia
World-scale production
in Sudbury, ON
Fuel in CANDU nuclear
reactors from
Saskatchewan
Fluorspar deposits
in Newfoundland
Large smelter in
Trail, BC
cesium
55
Cs
1
6
solid
potassium
19
K
1
4
solid
antimony
51
Sb
15
5
solid
sulfur
16
S
16
3
solid
cobalt
27
Co
9
4
solid
barium
56
Ba
2
6
solid
nickel
28
Ni
10
4
solid
uranium
92
U
_
7
solid
fluorine
9
F
17
2
gas
zinc
30
Zn
12
4
solid
Applying Inquiry Skills
5. The chemical property — “reacts violently with water to form basic solutions and liberate hydrogen gas” — is
associated with the alkali metals. However, the alkaline earth metals can also react with water to liberate hydrogen.
The element could be further investigated to see if the reaction with water produced a basic solution — which is a
chemical property of the alkali metals.
Experimental Design
A small piece of the unknown element is placed in water and the reactivity observed. After the element has reacted
with water, dip litmus paper into the solution. If the litmus paper turns blue, then the solution is basic — this would
indicate that the element is an alkali metal. If the litmus paper does not turn blue, then the solution is either neutral or
possibly acidic. This would indicate that the element is not an alkali metal — other tests could then be carried out to
further narrow down which group the element is from.
6. (a) Radium was discovered by Pierre and Marie Curie in 1898 (France). Radium ore is mined — extracted from the
ground. The element emits alpha particles and gamma rays to form radon, and is used chiefly in luminous
materials and in the treatment of cancer. Uranium is a byproduct of the extraction process.
(b) It is intensely radioactive and poses a serious threat to health — overexposure can cause cancer. Protective measures must be taken in its handling, storage, and disposal.
(c) Advantages — treatment of cancer, use in luminous materials.
Drawbacks — intensely radioactive and can cause cancer.
The main advantage is the use of radium to treat cancer. Providing that adequate protective measures are taken in
its handling, storage, and disposal, it can be argued that the advantages outweigh the drawbacks. Nevertheless, it
would be desirable to discontinue its use in the future, and to replace it with other effective and less health-threatening methods of treating cancer.
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Chapter 1 The Nature of Matter
9
1.2 DEVELOPING A MODEL OF THE ATOM
ACTIVITY 1.2.1
DEVELOPING A MODEL OF A BLACK BOX
(Page 23)
(e) When investigating the contents of an actual black box, there are a number of physical, sensory observations that
can be carried out. For each manipulation of the black box, we can use our senses of touch, vision, hearing, smell,
and taste to give us a clue about the contents, and some ideas for a theoretical description or general statement
that may characterize the nature of the contents of the box. We are unable to carry out such observations of the
nature of the atom, as the atom is too small to touch, see, hear, smell, or taste.
PRACTICE
(Page 26)
Understanding Concepts
1. (a) A theory is based on non-observable ideas. A law is based on observations.
(b) Empirical knowledge is observable, and theoretical knowledge is not.
2. Scientists use models — mental or physical representations of theoretical concepts — to help describe and explore
their ideas.
3. The law of conservation of mass requires that all matter in a reaction be measured, in this case not just the solids (and
liquids) in the unburned wood and the ashes. The missing matter could be accounted for if we were to measure the
mass of the escaped gaseous products of combustion and the particulates in the smoke.
4.
Changing models of the atom:
• Democritus’s model of the atom (third or fourth century B.C.)
Different atoms are of different sizes, have regular geometric shapes, and are in constant motion. There are empty
spaces between atoms.
• Thomson’s model of the atom (1897)
Negatively charged electrons are distributed inside the atom, which is a positively charged sphere consisting
mostly of empty space.
• Rutherford’s model of the atom (1911)
The atom contains a positively charged core, the nucleus, that is surrounded by a predominantly empty space
containing negative electrons.
• Chadwick’s model of the atom (1932)
The atom is composed of a nucleus, containing protons and neutrons, and a number of electrons equal to the
number of protons. An atom is electrically neutral.
5. (a) nucleus: the small, positively charged centre of the atom
(b) proton: a positively charged subatomic particle found in the nucleus of the atom
(c) electron: a negatively charged subatomic particle found around the nucleus
(d) neutron: an uncharged subatomic particle in the nucleus of the atom
6. Cathode ray tubes can be found in some radios, televisions, and computer monitors.
10
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
1.3 UNDERSTANDING ATOMIC MASS
PRACTICE
(Page 29)
Understanding Concepts
1.
Subatomic particle
Electron
Proton
Neutron
Relative atomic mass (u)
5.5 x 10-4
~1
~1
Charge
11+
0
Location
surrounding the atom
nucleus of the atom
nucleus of the atom
2. The proton and the neutron are responsible for most of the mass of an atom.
3. The mass number of an atom is based upon a relative scale that uses the carbon-12 atom as a standard. On this scale
the proton and the neutron both have a mass close to 1 u while the electron has a mass of 0.000 55 u.
4.
Z = 14
N = 13
A=Z+N
= 14 + 13
A = 27
5. (a) Z = 15
(b) A = 31
Z = 15
N=A-Z
N = 31 - 15
N = 16
(c) The element is phosphorus.
6.
A = 37
Z = 17
N=A-Z
N = 37 - 17
N = 20
7. These atoms are of two elements, and so are not isotopes of each other. Each atom has a different atomic number —
a different number of protons. One atom has Z = 15, or 15 protons (phosphorus), and the other atom has Z = 14, or
14 protons (silicon).
8. (a) 24 u, 25 u, and 26 u.
(b) A = 24, 25, and 26, respectively.
Z = 12 for all isotopes.
N=A-Z
For A = 24,
N = 24 - 12 = 12 neutrons
For A = 25,
N = 25 - 12 = 13 neutrons
For A = 26,
N = 26 - 12 = 14 neutrons
(c) A = 24, 80%; A = 25, 10%; A = 26, 10%
Copyright © 2002 Nelson Thomson Learning
Chapter 1 The Nature of Matter
11
ACTIVITY 1.3.1
MODELLING HALF-LIFE
(Page 31)
(a) Table 2
Half-life
0
1
2
3
4
5
Number of disks remaining
30
16
7
4
2
0
The “heads” of the coin were used to represent the original radioactive isotope in this sample.
Analysis
Number of Disks Remaining
30
25
20
15
10
5
0
1
2
3
4
5
6
Half-life
(b) The graph is downward-sloping — very steep slope at first, then becoming more gradual.
(c) If the graph is used to make the prediction, the answer will often be 0 g.
Start with 30 g of radioactive material:
After 2 a (one half-life), 16 g of radioactive material would remain.
After 4 a (two half-lives), 7 g of radioactive material would remain.
After 6 a (three half-lives), 4 g of radioactive material would remain.
(d) Some graphs will reach zero in this activity, others won’t. However, if the activity were continued, all graphs
would reach zero - the last disk(s) will eventually flip.
(e) The removed disks represent atoms that have decayed. This model of radioactive decay is based upon the likelihood that each time the box is shaken, about half of the remaining disks will “shake” to show the original side of
the disk, and the other half will “shake” to show the other side. The model is a good one in the sense that each
shake does result in showing approximately half of one side of the disk and half of the other side of the disk in
each half-life. Thus, the model is helpful in describing the half-life effect for radioactive material. However, the
model is less than perfect because of numbers. Because the number of disks is small, it’s possible to get large
variations from the ideal “half per shake”, and so the graph is not likely to be smooth. In any macroscopic sample
of matter the number of atoms is monstrous. Because of the numbers it is not likely that there will be a wide variation from the half in any given half-life, so it is unlikely that a sample of a radioactive element will decay entirely
in any reasonable time. An activity that counted atoms (if such were possible) would always produce a graph that
smoothly approached, but never actually reached, the x-axis.
PRACTICE
(Page 32)
Understanding Concepts
9. An isotope is a form of an element that has a certain number of neutrons. A radioisotope is an isotope of an element
that emits radiation and in so doing becomes an isotope of another element.
12
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
10.
Table 3: Emission Particles
(a) Another name for this
particle
(b) The symbol for this particle
(c) How the nucleus of a
radioisotope is altered by
emission of this particle
(d) The penetrating ability of
this type of radiation
Alpha particle
nucleus of a helium atom
Beta particle
high-energy electrons
loss of two protons, and two
neutrons (reduction in mass
number by 4)
b
loss of a neutron, and gain of a
proton (no change in mass
but atomic number is
increased by 1)
a few metres in air
a few centimetres in air
11. Many elements have one or more isotopes that are unstable. Atoms of unstable isotopes decay, emitting radiation as
their nucleus changes. Isotopes that decay in this way are known as radioisotopes and are said to be radioactive. Every
radioisotope has a characteristic property called its half-life. The half-life of a radioactive substance is the time taken
for half of the original number of radioactive atoms to decay. Another way of defining the term half-life is the time it
takes for one-half the nuclei in a radioactive sample to decay.
12. (a) After 8.0 d (two half-lives), 1.7 g of radon-222 would remain.
(b) After 16.0 d (four half-lives), 0.42 g of radon-222 would remain.
(c) After 32.0 d (eight half-lives), 0.026 g of radon-222 would remain.
13. Start with 2.0 kg, or 2000 g, of iodine-131:
After 8.0 d (one half-life), 1000 g of iodine-131 would remain.
After 16.0 d (two half-lives), 500 g of iodine-131 would remain.
After 24.0 d (three half-lives), 250 g of iodine-131 would remain.
After 32.0 d (four half-lives), 125 g of iodine-131 would remain.
After 40.0 d (five half-lives), 62.5 g of iodine-131 would remain.
After 48.0 d (six half-lives), 31.2 g of iodine-131 would remain.
After 56.0 d (seven half-lives), 15.6 g of iodine-131 would remain.
After 64.0 d (eight half-lives), 7.81 g of iodine-131 would remain.
Half-Life of Iodine-131
Mass of Iodine-131 (g)
2500
2000
1500
1000
500
0
2
4
6
8
10
Number of Half-Lives (8.0 d)
14. Radioactive decay supports the law of conservation of mass, as matter is not being created or destroyed — the matter
that is emitted continues to exist in the form of protons, neutrons, and electrons.
Making Connections
15. The student is to use the Internet to research and report on a project where scientists are using or have used
carbon-14 dating to find the age of artifacts. The report should cover the scientists’ work and its implications.
The number of possible projects they might discover is huge. Carbon-14 dating is standard in most
archaeological excavations. For example, carbon-14 dating of artifacts from a site in South America implies that
human beings were established in the Americas long before 15 000 years ago. This has generated considerable scientific discussion.
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Chapter 1 The Nature of Matter
13
16. The student is to use the Internet to research and report on the properties and applications of a radioisotope. The
student is to include at least one argument in support of its use, and one argument against.
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17. Some circumstances in which a Geiger counter would be useful:
•
Medical facilities that use radioactive materials
•
Nuclear power facilities
•
Disposal sites for radioactive waste
18. The student is to use the Internet to research and find information related to the production, storage, and practical uses
of tritium. Also, the student is to find out what precautions must be taken when working with this radioactive
substance.
In Ontario, tritium is produced at facilities connected to CANDU reactors and stored on-site.
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PRACTICE
(Page 33)
Making Connections
19. The student is to use the Internet to research and report on a career in nuclear science. The report should include the
following:
• a general description of the work and how radioisotopes are involved;
• current working conditions and a typical salary;
• the education required to work in this field;
• a forecast of employment trends in this field.
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PRACTICE
(Page 35)
Understanding Concepts
20. Two advantages associated with nuclear power:
• Nuclear plants are not associated with immediate environmental pollution such as acid rain, greenhouse gases, or
the emission of toxic gases, all of which result from the burning of fossil fuels.
•
Unlike hydroelectric plants, which must be built where the water is, nuclear plants can be built close to where the
power is needed.
Two disadvantages associated with nuclear power:
Nuclear plants generate long-lived radioactive waste. Currently, there is no socially acceptable solution for the
storage of this waste.
•
Ground water contamination from radioactive tailings at uranium ore mines is a problem.
21. CANDU reactors use the naturally occurring radioisotope of uranium:uranium-235 (U-235).
22. (a) Heavy water is water that contains deuterium instead of hydrogen (D2O).
(b) The nuclear fission reaction produces new neutrons that are travelling too quickly to be used for further fission
reactions. Heavy water serves as a “moderator” to slow down neutrons. It is also used to cool the fuel bundles in
the reactor.
23. As every technology has risks, we need to decide if the benefits of nuclear power are high enough and the risks low
enough to continue using the technology. There can be many positions, generally determined by the values that an
individual or a society holds on a single issue. Which solution is “best” is a matter of opinion; ideally, the solution
that is implemented is the one that is most appropriate for society as a whole. Refer to text Appendix A2, “Decision
Making” and “A Risk – Benefit Analysis Model.”
•
14
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
EXPLORE AN ISSUE
DEBATE: DISPOSING OF NUCLEAR WASTE
(Page 36)
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PRACTICE
(Page 36)
Making Connections
24. The student is to use the Internet to research and present findings on how forensic scientists use the technique of
neutron bombardment to detect small or trace amounts of poisons in human tissue.
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25. The student is to use the Internet to research and report on the use and handling precautions of one of the radioisotopes produced within the core of a nuclear reactor, such as cobalt-60 or iodine-131. These radioisotopes are often
used for medical diagnosis and therapy, or for industrial and research work.
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26. The student is to research and report on the significance of the contribution to atomic theory made by Harriet Brooks,
a Canadian, and Ernest Rutherford, a New Zealander, working at McGill University in Montreal.
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SECTIONS 1.2-1.3 QUESTIONS
(Page 37)
Understanding Concepts
1. (a) An atom is composed of a nucleus, containing protons and neutrons, and anumber of electrons equal to the
number of protons; an atom is electrically neutral.
(b) The number of protons in the nucleus determines the identity of an element and is referred to as that element’s
atomic number (Z). The number of electrons is equal to the number of protons.
The sum of the number of protons and neutrons present in the nucleus of an atom equals the mass number (A).
(c) The element has one or more isotopes. An isotope is a form of an element in which the atoms have the same
number of protons as all other forms of that element, but a different number of neutrons - isotopes have the same
atomic number (Z), but different neutron numbers (N), and so different mass numbers (A).
2.
element Y + e
Z + N1
Z
1–
e
ctiv
ioa β)
rad
(
y
a
dec
×
isotope 1
mass number
(number of protons
plus number of neutrons)
Z + N2
Z
×
isotope 2
Z + N3
atomic number
(number of protons)
Z
element
×
×××.××
×
atomic mass
(weighted average
of mass number of
isotopes)
isotope 3
rad
ioa
dec ctive
ay
(α)
Copyright © 2002 Nelson Thomson Learning
4
element Z + 2 He
Chapter 1 The Nature of Matter
15
3. (a) These two atoms cannot be classified as isotopes of the same element because they do not have the same atomic
number (Z).
(b) This could be a “beta particle” radioisotope and its decay product, where there has been a conversion of two
neutrons into two protons and two electrons. The result is that the mass number stays the same, but the number
of protons — the atomic number — increases.
4. The usual classifying properties of nonradioactive elements are physical and chemical properties. The property of
radioactivity — or radioactive decay — is a property that can be used to classify atoms of an element that can spontaneously change into atoms of another element. The classifying property of radioactivity is different in that it
describes a nuclear change.
Applying Inquiry Skills
5. (a)
Half- Life of Thorium-234
800
Mass of Thorium-234 (g)
700
600
500
400
300
200
100
0
1
2
3
4
5
6
7
8
9
Number of Half-Lives (24.10 d)
(b) The above graph isolates the radioactive decay of thorium-234 through the lower mass values. For 24.0 g of
thorium-234 to remain, approximately 4 half-lives and about 90% of a fifth half-life must pass — a total of
118.2 d.
Making Connections
6. (a) Radioisotopes are useful for diagnostic radiography, radiology, forensic anthropology, nuclear power, for killing
bacteria in food and preventing spoilage, etc.
(b) Safety precautions that may be used when handling radioisotopes include:
• minimize the dose by reducing time of exposure
• minimize the dose by maximizing the distance from the source
• use shielding (common materials are lead, iron, concrete, and water) both for containment and for protective
clothing
• control access to the radioactive material
• conduct frequent surveys of contact
• conduct follow-up bioassays, looking for tissue damage
• wear respiratory protection
• practise good housekeeping; treat any waste from the lab as if it were radioactive
1.4 TOWARD A MODERN ATOMIC THEORY
PRACTICE
(Page 42)
Understanding Concepts
1. Bohr proposed the following explanation for the emission of light when a gas is heated: When energy (heat or electricity) is supplied to hydrogen atoms, electrons gain a certain quantity of energy and become excited (they jump from
16
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
a lower energy level to a higher energy level). When the electrons drop back to a lower energy level they release
energy corresponding to a few precise wavelengths of light.
2. The different colours in a line spectrum correspond to different specific quantities of energy released, as electrons fall
back to lower energy levels. With respect to the line spectrum for hydrogen gas, electrons falling from the sixth energy
level to the second emit violet light; electrons falling from the fifth energy level to the second emit indigo light, etc.
3. Different substances show different spectra because each element has different electron transitions taking place,
releasing energy that corresponds to that substance’s own unique “signature” emission spectrum.
4. Since the sodium vapour emits a yellow light, we can deduce that sodium atoms have an electron transition taking
place in which that frequency of light is emitted.
PRACTICE
(Page 45)
Understanding Concepts
5. The light given off by distant stars results from excited electrons moving from higher to lower energy levels in atoms.
Since each element has a unique emission spectrum, astronomers can use spectroscopy techniques to analyze the
starlight emission and identify the elemental composition of the atmosphere of the star.
6. According to the Bohr theory, when an electron absorbs energy it jumps from a lower energy level to a higher energy
level, and when an electron drops back to lower energy levels it emits energy.
7. Since each element has a unique emission spectrum, the spectral lines act like a “fingerprint” in terms of identifying
the element.
Making Connections
8. (a) Fireworks are composed of many different chemical elements, each element having its own unique electron
energy level transitions, thus producing its own unique emission spectrum.
(b) Students will probably cite the metals they used in their flame tests. Some of the chemicals they used in their tests
are not readily available. Others are very stable and require intense heat to reach vapour state — not convenient
for fireworks. Copper is available as an element, but in that form would be difficult to heat enough to vaporize.
Chemicall Compounds Used For Fireworks
Material
magnesium metal
sodium oxalate
barium chlorate
copper(II) sulfate
strontium carbonate
iron filings and charcoal
potassium benzoate
potassium nitrate and sulfur
potassium perchlorate, sulfur, and aluminum
Special Effect
white flame
yellow flame
green flame
blue flame
red flame
gold sparks
whistle effect
white smoke
flash and bang
Mixing the ingredients is dangerous and should only be attempted by well-trained professionals.
Maintaining the proper balance of ingredients, fuel, and source of oxygen is a difficult challenge.
9. The student is to use the Internet to research and report on the detection and generation of elements in stars and supernovas. The student is to comment on the following hypothesis: All elements are conglomerates of hydrogen.
Students will discover that, using detectors sensitive only to a narrow range of wavelengths, astronomers can
detect specific substances in the atmospheres of stars and in the debris generated by supernovas. Current cosmological theory is that only hydrogen, helium, and a relatively very small amount of lithium were present before the first
star formed. All other elements were built (with hydrogen as a starting point) in the nuclear furnaces of stars.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre
Copyright © 2002 Nelson Thomson Learning
Chapter 1 The Nature of Matter
17
PRACTICE
(Page 47)
Understanding Concepts
10. Valence electrons are those electrons that occupy the highest shell of an atom and form chemical bonds. Elements that
have the same number of valence electrons have similar physical and chemical properties. Valence electrons are significant in that they are a powerful indicator of the relationship between electron arrangement and periodic trends.
11. Electrons within an atom can possess only discrete quantities of energy; electrons fill successive shells.
12.
beryllium
chlorine
krypton
iodine
lead
arsenic
cesium
Number of
Occupied
Energy Levels
2
3
4
5
6
4
6
Number
of Valence
Electrons
2
7
8
7
4
5
1
SECTION 1.4 QUESTIONS
(Page 48)
Understanding Concepts
1. Hydrogen could be placed at the top of Group 17. In terms of atomic structure, hydrogen with its one electron would
be followed by helium with two electrons - this fits with the rule of adding an electron as you move from left to right
on the periodic table. In terms of periodicity, hydrogen, which does sometimes behave like a halogen, would be in
periodic alignment with the other halogens. Like the halogens, hydrogen requires only 1 additional electron to
complete its valence shell.
2. (a)
Table 2: Electron Structure of Selected Elements
Element
oxygen
sulfur
magnesium
sodium
beryllium
calcium
cesium
nitrogen
chlorine
lithium
helium
bromine
phosphorus
fluorine
potassium
18
Number of
Electrons
Number of
Occupied Shells
Number of Valence
Electrons
8
16
12
11
4
20
55
7
17
3
2
35
15
9
19
2
3
3
3
2
4
6
2
3
2
1
4
3
2
4
6
6
2
1
2
2
1
5
7
1
2
7
5
7
1
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
2. (b)
Table 2: Groups According to Number of Valence Electrons
1 Valence
Electron
2 Valence
Electrons
5 Valence
Electrons
6 Valence
Electrons
7 Valence
Electrons
sodium
cesium
lithium
potassium
magnesium
beryllium
calcium
helium
nitrogen
phosphorus
oxygen
sulfur
chlorine
bromine
fluorine
(c) 1 valence electron: physical and chemical properties common to elements of the alkali metals (soft; metallic; react
violently with water to form hydrogen and a basic solution; react strongly with oxygen; react with halogens to
produce a crystalline solid)
2 valence electrons: physical and chemical properties common to elements of the alkaline earth metals (metallic;
light; react with oxygen to form an oxide; react with water to release hydrogen; react with hydrogen to form a
hydride)
5 valence electrons: chemical properties common to elements of Group 15 (nonmetals)
6 valence electrons: chemical properties common to elements of Group 16 (nonmetals)
7 valence electrons: physical and chemical properties common to elements of the halogens (nonmetals; extremely
reactive with hydrogen and most metals)
1.5 TRENDS IN THE PERIODIC TABLE
ACTIVITY 1.5.1
GRAPHING FIRST IONIZATION ENERGY
(Page 54)
Procedure
First Ionization Energies of the First 38 Elements
2500
2250
Ionization Energy (kJ/mol)
2000
1750
1500
1250
1000
750
500
250
0
5
10
15
20
25
30
35
40
Atomic Number
Analysis
(a) •
Group 1 elements have the lowest first ionization energies.
•
Group 8 elements have the highest first ionization energies.
•
First ionization energies of metals are lower than those of nonmetals.
•
First ionization energies generally decrease as you move down a group in the periodic table.
•
First ionization energies generally increase as you move from left to right across a period.
Synthesis
(b) The general decrease in first ionization energies as you move down a group in the periodic table is related to the
greater number of non-valence electrons between the nucleus and the valence electrons (a shielding effect). As a
result, the attraction between the negatively charged electrons and the positive nucleus becomes weaker, so less
energy is required to remove the first valence electron.
The general increase in first ionization energies across a period is related to the increasing size of the charge
in the nucleus. As you move from left to right across a period, the nuclear charge increases while the shielding
Copyright © 2002 Nelson Thomson Learning
Chapter 1 The Nature of Matter
19
effect provided by non-valence electrons remains the same. As a result the valence electrons are more strongly
attracted to the nucleus, resulting in a decrease in atomic radius and an increase in the amount of energy required
to remove an electron.
ACTIVITY 1.5.2
GRAPHING ELECTRONEGATIVITY
(Page 57)
Procedure
Electronegativity Values for the First 38 Elements
Electronegativity Value
5
4
3
2
1
0
5
10
15
20
25
30
35
40
Atomic Number
Analysis
(a) •
•
•
Electronegativity generally increases as you go along a period, from left to right.
Electronegativity decreases as you look down a given group on the periodic table.
The electronegativity values for elements at the bottom of a group are lower than those for elements at the
top of a group.
Synthesis
(b) As the atomic radii tend to decrease along a period, the attraction between the valence electrons and the nucleus
increases as the distance between them decreases. The electronegativity also increases.
(c) As the atomic radii tend to increase down a group, the attraction between the valence electrons and the nucleus
decreases as the distance between them increases. The electronegativity also decreases.
PRACTICE
(Page 58)
Understanding Concepts
1. When an atom gains an electron to form a negative ion, its radius increases. In gaining an electron, repulsion among
the electrons increases but there is no compensating increase in nuclear charge. The result is an enlarged electron
cloud and a greater ionic radius.
When an atom loses an electron to form a positive ion, its radius decreases. In losing an electron, repulsion among
the electrons decreases while nuclear charge remains the same. The result is a reduction in size of the electron cloud
and a smaller ionic radius.
2.
Li
Li +
The atomic radius of the lithium atom is 152 pm. The ionic radius of the lithium ion is 68 pm.
After losing an electron, repulsion among the electrons decreases, and the result is a reduced electron cloud and a
smaller ionic radius.
F
F
The atomic radius of the fluorine atom is 64 pm. The ionic radius of the fluoride ion is 136 pm.
After gaining an electron, repulsion among the electrons increases, and the result is an enlarged electron cloud and a
greater ionic radius.
20
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
3. (a)
•
First ionization energies of metals are lower than non metals.
•
First ionization energies generally decrease as you move down a group in the periodic table.
•
First ionization energies generally increase as you move from left to right across a period.
(b) The general decrease in first ionization energies as you move down a group in the periodic table is related to the
increase in the size of the atoms from top to bottom of a group. As the atomic radius increases, the distance
between the valence electrons and the nucleus also increases. The non-valence electrons between the nucleus and
the valence electrons produce a shielding effect. As a result, the attraction between the negatively charged electrons and the positive nucleus becomes weaker, so less energy is required to remove the first valence electron.
The general increase in first ionization energies across a period is related to the decrease in the size of
the atoms. As you move from left to right across a period, the nuclear charge increases while the shielding effect
provided by the non-valence electrons remains the same. As a result the valence electrons are more strongly
attracted to the nucleus, resulting in a decrease in atomic radius, and more energy is required to remove an electron from the atom.
(c) Ionization energies in kJ/mol of the elements have been observed and measured.
4. (a) The most reactive metal would be cesium (some students will nominate francium, but it has not been discussed).
It has the lowest ionization energy.
(b) The most reactive nonmetal would be fluorine. It has the highest electronegativity value.
5.
Ionization Energies of Calcium
5000
4940
4500
Ionization Energy (kJ/mol)
4000
3500
3000
2500
2000
1500
1000
500
0
1140
590
1st
2nd
3rd
Ionization Energies
6. (a) Reactivity increases with the halogens as you move up the group.
(b) The prediction is supported — the reactivity increases as you look down this group of metals.
(c) Halogens: as you move up the group, valence electrons remain the same, nuclear charge decreases, shielding
decreases, and distance of valence electrons from the nucleus decreases. Electronegativity values increase as you
move up the group, resulting in greater reactivity. Fluorine, at the top of the group, with the highest electronegativity value of all elements, is most reactive.
Alkali Metals: as you look down the group, valence electrons remain the same, nuclear charge increases,
shielding increases, and distance of valence electrons from the nucleus increases. First ionization energy values
decrease as you look down the group, resulting in greater reactivity. Cesium, at the bottom of the group, with an
extremely low first ionization energy value, is most reactive.
(d) Cesium and fluorine.
Applying Inquiry Skills
7. Predictions
(a) The surface of rubidium, if cleaned, should react rapidly with atmospheric oxygen, forming an oxide layer. If
rubidium is reacted with water, there will be a quick and vigorous reaction. The reasoning that supports these
predictions is that as you move down a group, the first ionization energy values decrease, resulting in greater reactivity.
Experimental Design
(b) A small piece of rubidium is placed in water and the reactivity is observed and recorded.
Materials
safety shield, tongs, tweezers, mineral oil, sharp edge (knife or scoop), paper towel, petri dish, large beaker, distilled
water, wire gauze square, sample of rubidium
Copyright © 2002 Nelson Thomson Learning
Chapter 1 The Nature of Matter
21
Procedure
1. Transfer a small piece of rubidium into a petri dish containing mineral oil.
2. Using a knife or similar sharp edge, remove the surface layer from one side of the piece of rubidium. Observe the
fresh surface.
3. Fill the beaker about half full with water.
4. Cut a small piece (about 2 mm cubed) of the element. Remove any oil with paper towel and drop the piece of the
element into the beaker. Immediately cover the beaker with the wire gauze. Record observations of any chemical reaction.
8.(a) Experimental Design
Small pieces of various element solids from the same group in the periodic table are melted, and the melting-point
range of the substance is observed and recorded. The process can be repeated for various element solids from other
groups as well, to further investigate the periodic trends of melting points of elements.
Materials
watch glass, 150-mm test tube, melting-point tube, ruler, rubber band , thermometer, 250-mL beaker, split one-hole
stopper for thermometer, buret clamp, iron ring, wire gauze, ring stand, burner, wire loop stirrer, stirring rod, glass
tube (1-m length), various samples of element solids from the same group
Procedure
1. Place a small piece of an element solid on a clean watch glass and pulverize it if possible, by rubbing it with the
bottom of a clean test tube. Avoid rubbing so hard that you break the test tube or the watch glass.
2. Carefully force a portion of the solid into a melting-point tube by pushing the open end of the tube into the powdered
solid.
3. Allow the melting-point tube to drop (bottom first) through a 1-m length of glass tubing positioned vertically on the
floor or laboratory bench. This knocks the solid into the sealed end of the melting-point tube. Repeat this process until
the tube contains solid to a depth of 5 mm. Do not add too much element to the tube at any one time.
4. Attach the melting-point tube to the thermometer bulb of a thermometer with a rubber band, and support the thermometer in a 250-mL beaker about two-thirds full of water. The open ends of the melting-point tube must be above
the water level.
5. Heat the water very slowly by adjusting the burner flame to a low level and removing the burner under the beaker periodically. Stir the water constantly. Try to obtain a rate of temperature increase of not more than 3˚C/min.
6. Observe the solid in the melting-point tube carefully. Note and record the temperature at which the first crystal begins
to melt and the temperature at which the last crystal disappears.
7. Allow the water to cool. Note and record the temperature at which the first crystal forms and the temperature at which
all of the element has become solid.
8. Repeat the procedure for each of the other element solids.
Reflecting
9. Elements that are used for structural support of buildings, roads, vehicles, furniture, and so on should be
unreactive — for example, metals that are resistant to the “rusting” reaction. Containers should be made of unreactive
materials, especially in a chemistry lab or a kitchen.
A match that does not “light” is useless, as is a fuel that will not burn, or a food that will not “cook”.
SECTION 1.5 QUESTIONS
(Page 59)
Understanding Concepts
1. The valence electrons are the ones that usually participate in chemical bonding. The number of valence electrons
possessed by an atom determines the number of other atoms with which that atom can combine. The most chemically
reactive metals are found in Groups 1, 2, and, to a lesser extent, 13 of the periodic table. These metals have low ionization energies. The most active nonmetals are found in Groups 16 and 17. These nonmetals have relatively large electron affinities.
2. (a) Ionization energy is the amount of energy required to remove an electron from an atom or ion in the gaseous state.
Electron affinity is the energy change that occurs when an electron is accepted by an atom in the gaseous state.
Both represent energy changes related to either the removal or the gaining of an electron.
(b) Electronegativity describes the relative ability of an atom to attract electrons, whereas electron affinity is the
energy change that occurs when an electron is accepted by an atom in the gaseous state. The electron affinity of
22
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
3.
4.
5.
6.
7.
an atom affects the ability of an atom to attract electrons and is one of the components considered in the determination of an atom’s electronegativity. Both terms are related to the gaining of an electron by an atom.
(a) Order of increasing ionization energy: students might predict K, Sr, Mg, Al, S (see below)
Order of increasing atomic radius: S, Al, Mg, Sr, K
Order of increasing electron affinity: K, Sr, Mg, Al, S
Order of increasing electronegativity: K, Sr, Mg, Al, S
(b) Ionization energy: Elements that have an electron in the valence shell farthest from the nucleus, where it is most
shielded from the attractive forces of the positive nuclear charge, have lower ionization energies. (Ionization energies generally increase as you move left to right across the periodic table. The exception in this case is Al.
Although it is to the right of Mg it has a lower ionization energy.)
Atomic radius: The atomic radii decrease from left to right across each period. As you move left to right across
a period there is no increase in the electron shielding effect. Yet as a result of an increasing nuclear charge, there
is an increase in nuclear attraction, which pulls the electrons closer to the nucleus, thus decreasing the size of the
atom. The atomic radii increase from top to bottom in a given group because of the increasing number of energy
levels that are filled with electrons. This has the effect of increasing the shielding effect, decreasing nuclear attraction, and increasing the size of the atom.
Electron affinity: When an electron is added to an atom of smaller radius, there is a greater attractive force
between the nucleus of the atom and the new electron. Electron affinity and atomic radius are therefore inversely
related. Smaller atoms should generally have higher electron affinities. Since the size of atoms generally
decreases across a period, electron affinity will increase as we move from left to right within a period. And since
the size of atoms generally decreases as we move up a group, electron affinity will also increase as we move from
the bottom to the top of a group.
Electronegativity: An atom of smaller radius has a greater attractive force between the nucleus of the atom and a
new electron. Electronegativity and atomic radius are therefore inversely related. Smaller atoms should generally
have higher electronegativities. Since the size of atoms generally decreases across a period, electronegativity will
increase as we move from left to right within a period. And since the size of atoms generally decreases as we
move up a group, electronegativity will also increase as we move from the bottom to the top of a group.
First ionization energies for two isotopes of the same element would be expected to be the same. It is the protons and
electrons in atoms that are responsible for determining the element’s chemical behaviour. Consequently, isotopes of
the same element share chemical properties despite their slight difference in mass.
(a) Na+; Mg2+; Al3+. The valence electrons are the ones that usually participate in chemical bonding. Na, Mg, and Al
are metals that have relatively low ionization energies and lose their valence electrons during chemical bonding
to become positive charged ions.
(b) From larger ionic radius to smaller ionic radius: Na+, Mg2+, Al3+. Each ion contains the same number of
electrons; however, for each successive positive ion, the nuclear charge is greater by 1+. Thus, each electron in
an aluminum ion experiences greater attraction than an electron in a magnesium ion, and each electron in a
magnesium ion experiences greater attraction than each electron in a sodium ion. The attraction of the nucleus
for the electrons in positive ions increases across the period, and the ionic radius gets smaller.
(a) 2 valence electrons
(b) Group 2
(c) It should exhibit the properties of an alkaline earth metal.
(d) The first two electrons are valence electrons in the outer shell, and so ionization energies for these electrons is
low. The third electron is in a lower shell, and so the ionization energy is very high.
As the atomic number increases, atomic radii decrease while ionization energies increase as you move across a period
from left to right on the periodic table. Atomic radii increase while ionization energies decrease as you move down a
group from top to bottom on the periodic table. Graph shapes for atomic radii as compared to ionization energy would
be reversed (slope and direction of graph line), similar to a mirror image.
Copyright © 2002 Nelson Thomson Learning
Chapter 1 The Nature of Matter
23
Applying Inquiry Skills
8. (a)
Atomic Radius of the First 38 Elements
Atomic Radius (pm)
300
250
200
150
100
50
0
5
10
15
20
25
30
35
40
Atomic Number
High Points: Li, Na, K, Rb
Low Points: H, Ne, Ar, Kr
(b) Group 1 corresponds to the elements indicated by the peaks on the graph (excluding hydrogen). Group 18 corresponds to the elements indicated by the lowest points on the graph.
(c) The atomic radii decrease, as the atomic number increases across a period. There are exceptions — on this graph,
elements that fall between atomic number 26 and atomic number 32 fluctuate slightly from this decreasing
pattern.
(d) By taking the individual differences in atomic radius between all of the elements in the same group that precede
cesium and barium, and then calculating an averaged difference, and then adding this averaged difference to the
value of the element that directly precedes cesium and barium, the predicted atomic radii would be:
cesium 280 pm
barium 250 pm
These predicted values for atomic radii are higher than the actual atomic radii indicated on the periodic table,
as shown below:
cesium 265 pm (actual atomic radius)
barium 217 pm (actual atomic radius)
However, lower values for cesium and barium would most likely be predicted if the trend of progressively
smaller changes in atomic radii in the two elements directly preceding both cesium and barium was noted, and
weighed into the calculations.
CHAPTER 1 SUMMARY
(Page 61)
Make a Summary
The student is to summarize the concepts learned so far, by creating a concept map that begins with the word
“element” and uses as many words as possible from the list of key words. Refer to page 61 of the text for the list ofkey
words.
Reflect on your Learning
(Page 61)
By the end of the chapter the student should have developed a more in-depth understanding of the constituents of the
atom, how these constituents are arranged, how atoms of various elements differ from each other, how the elements
in the periodic table are organized, how there are patterns in properties among the elements, and how these patterns
can be explained using the model of the atom. The student should also have developed skill in locating and interpreting
information on a periodic table.
24
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
CHAPTER 1 REVIEW
(Page 62)
Understanding Concepts
1. Metals are malleable, ductile, and conductors of electricity. They are also described as lustrous, or shiny. Nonmetals
are generally nonconductors of electricity in their solid form; at SATP the nonmetals are mostly gases or solids. Solid
nonmetals are brittle and lack the lustre of metals.
2. The symbol CA is not an acceptable international symbol for calcium. The International Union of Pure and Applied
Chemistry (IUPAC) specifies rules for chemical names and symbols — for elements, the first letter (only) of the
symbol is to be an uppercase letter. The acceptable international symbol for calcium is Ca.
3. The periodic law came to be accepted because of Mendeleev’s detailed predictions of the properties of undiscovered
elements, using his knowledge of periodic trends.
4. (a) liquids at SATP: bromine, mercury; gases at SATP: hydrogen, helium, nitrogen, oxygen, fluorine, neon, chlorine,
argon, krypton, xenon, radon
(b) The “staircase line” is a zigzag line that separates metals (to the left) from nonmetals (to the right).
(c) The noble gases remained undiscovered until the late 1800s because they are so stable and unreactive and their
boiling points are so low. It is difficult to detect such an element — they would not form compounds under conditions of a typical early lab.
(d) The atomic number of hydrogen is 1, of oxygen is 8, of aluminum is 13, of silicon is 14, of chlorine is 17, and
of copper is 29.
(e) Oxygen has 6 electrons in its outer energy level.
(f) Carbon has 4 of its 8 electrons in its second energy level.
(g) Beryllium would tend to lose 2 electrons.
(h) Fluorine would tend to gain 1 electron.
5. The order of the elements did change, but only slightly. Cobalt and nickel would exchange places, as would tellurium
and iodine. Scientists also added the newly discovered group, the “noble gases.”
6. (a)
metals
nonmetals
transition metals
noble gases
halogens
alkaline earth metals
alkali metals
staircase
lanthanides (rare earths)
actinides, including transuranic elements
(b) The number of valence electrons for alkali metals is 1, for alkaline earth metals is 2, and for halogens is 7. From
his study of the periodic table, Bohr concluded that the maximum number of electrons that can populate a given
shell can be calculated from the equation 2n2 (where n is the principal quantum number). The electrons in the
highest occupied shell are called the valence electrons. According to Bohr’s theory, there is a pattern linking electron arrangements to the periodic table for the representative elements.
7. (a) The most reactive metal is francium (accept cesium). The most reactive nonmetal is fluorine.
(b) With the alkali metals, reactivity increases moving down the group, due to increasing atomic radii and decreasing
first ionization energies. With the halogens, reactivity increases moving up the group, due to decreasing atomic
radii and increasing electron affinity.
(c) The Bohr model
Copyright © 2002 Nelson Thomson Learning
Chapter 1 The Nature of Matter
25
8. Hydrogen has only one electron in the first energy level. All other elements have two electrons in the first energy level.
9. (a) Dalton
(b) Nagaoka
(c) Rutherford
(d) Bohr
a positive core (nucleus)
surrounded by a ring of
electrons
–
–
–
–
–
–
indivisible
atoms
–
–
–
–
–
–
–
– – + ++ – –
– + ++ –
–
–
–
–
a small positive core
(nucleus) surrounded
by a much larger cloud
of electrons
a nucleus containing
protons and neutrons
surrounded by electrons
in shells
10. While each atomic theory made significant contributions to the current understanding of atomic structure, each had
shortcomings and limitations. These shortcomings resulted in ongoing intense investigation that led to the development of the succeeding theory.
11.
Subatomic particle
Electron
Proton
Neutron
Relative mass
1
1836.12
1838.65
Charge
11+
0
Location
space surrounding the atom
nucleus of the atom
nucleus of the atom
12. Bohr suggested that the properties of the elements could be explained by the arrangement of electrons in orbits around
the nucleus. From his study of the periodic table, Bohr concluded that there was a restriction on the number of electrons that can populate a given shell, with the valence electrons occupying the highest energy level. According to Bohr,
elements in the same chemical group shared not only properties but also the number of valence electrons — a powerful
indicator of the relationship between electron arrangement and periodic trends.
13. Use the representative element chlorine as an example:
(a) The atomic number of chlorine is 17. Therefore, there are 17 protons.
(b) The number of protons equals the number of electrons. Therefore, chlorine has 17 electrons.
(c) In Groups 13–18 the number of valence electrons corresponds to the second digit of the group number; as
chlorine is in Group 17, the number of valence electrons is 7.
(d) Chlorine is in Period 3, so it has three occupied energy levels.
14. (a) magnesium: 12 protons, 12 electrons, 2 valence electrons
(b) aluminum: 13 protons, 13 electrons, 3 valence electrons
(c) iodine: 53 protons, 53 electrons, 7 valence electrons
15. (a) 11 protons and 10 electrons: sodium ion, Na+
(b) 18 electrons and a net charge of 3- phosphorus ion, P3(c) 16 protons and 18 electrons: sulfur ion, S216. No. Both alpha and beta decay alter the number of protons in the nucleus.
17. (a) calcium-42: 20 protons, 20 electrons, 22 neutrons
(b) strontium-90: 38 protons, 38 electrons, 52 neutrons
(c) cesium-137: 55 protons, 55 electrons, 82 neutrons
(d) iron-59: 26 protons, 26 electrons, 33 neutrons
(e) sodium-24: 11 protons, 11 electrons, 13 neutrons
18. (a) The mass number for iodine in the periodic table is 126.90. Iodine-123 has an atomic mass of 3.90 u less than the
average iodine atom. The atomic mass of iodine given in the table is a weighted average of the atomic masses of
each of the isotopes of the element that are normally present in any sample of the element.
(b) As the atomic mass of 126.90 is not very close to 123, this would indicate that iodine-123 is not in great abundance in a typical sample of iodine atoms.
19. (a) Within Group 1 and 2 elements, reactivity increases moving down the group.
(b) Within Group 16 and 17 elements, reactivity decreases moving down the group.
(c) Within a period, reactivity tends to be high in Group 1 metals, lower in elements toward the middle of the table,
and increase to a maximum in Group 17 nonmetals.
(d) Within Group 18, the elements are stable and extremely unreactive. However, reactivity does increase from the
top to the bottom of the group.
20. (a) Atomic radii decrease as you move from left to right across each period. From left to right across a period, the
nuclear charge increases while the shielding effect provided by the non-valence electrons remains the same. As a
26
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
result, of the increasing nuclear charge, the electrons are more strongly attracted to the nucleus, pulling them
closer to the nucleus and decreasing the size of the atom.
The atomic radii increase as you move down a given group. As you move down a group, there is an
increasing number of energy levels that are filled with electrons. This has the effect of increasing the shielding
effect, decreasing nuclear attraction, and increasing the size of the atom.
(b) First ionization energies generally increase as you move from left to right across a period. From left to right across
a period, the nuclear charge increases while the shielding effect provided by the non-valence electrons remains
the same. As a result the valence electrons are more strongly attracted to the nucleus, and more energy is required
to remove an electron from the atom.
First ionization energies generally decrease as you move down a group. As you move down a group, there
is an increase in the size of the atoms. As the atomic radius increases, the distance between the valence electrons
and the nucleus also increases. The non-valence electrons between the nucleus and the valence electrons produce
a shielding effect. As a result, the attraction between the valence electrons and the nucleus becomes weaker, so
less energy is required to remove an electron from the atom.
(c) Electronegativity generally increases as you move from left to right across a period, as the attraction of the
nucleus for any new electrons increases.
Electronegativity decreases as you move down a group, as new electrons are shielded from the nucleus by
increasing numbers of shells of electrons.
(d) Electron affinity increases as you move from left to right across a period. Smaller atoms with larger nuclei have
a stronger attraction to new electrons.
Electron affinity decreases as you move down a group. Larger atoms with more distant and more shielded
nuclei have a smaller attraction for new electrons.
21. When chlorine gains an electron to form a negative ion, the repulsion forces among the electrons increase while
nuclear charge remains the same. The result is an enlarged electron cloud and a greater ionic radius. When sodium
loses an electron to form a positive ion, the repulsion forces among the electrons decrease while the nuclear charge
remains the same. The result is a reduced electron cloud and a smaller ionic radius.
22. (a) Group 17 nonmetals from most reactive to the least reactive: fluorine, chlorine, bromine, iodine, astatine.
(b) During chemical reactions, the nonmetals of Group 17 have a tendency to gain an electron. However, electron
affinity declines as you move down Group 17.
23. The noble gas group is especially interesting to chemists because elements in this group are so stable and unreactive.
Bohr suggested that the properties of the elements could be explained by the arrangement of electrons around the
nucleus. Bohr hypothesized that these elements have filled outermost energy levels.
Applying Inquiry Skills
24. (a)
Electron Affinity vs. Atomic Number
Electron Affinity (kJ/mol)
400
300
200
100
0
–100
0
5
10
15
20
Atomic Number
Note:Where the electron affinity value was given as less than zero, an arbitrary -50 kJ/mol has been assigned.
This has been done so that the graph will show a connected line through all points, and to show that the values
fluctuate above and below zero as we move through these 18 elements.
(b) Similarities: Both graphs show a general increase in energy as you move left to right across a period. With respect
to electron affinity, we know that the atomic radii tend to decrease across a period, so the attractive force between
the nucleus of the atom and the new electron is stronger. And with respect to first ionization energy, it is again
the decreasing atomic radii across a period and the increasing nuclear attraction that makes it harder to remove
an electron.
Differences: The electron affinity graph fluctuates dramatically up and down as you move left to right across a
period and shows some negative values. The negative values occur when the electron repulsive force is greater
than the nuclear attractive force, and energy must be injected, instead of energy being released, in ion formation.
Copyright © 2002 Nelson Thomson Learning
Chapter 1 The Nature of Matter
27
(c)
25. (a)
(b)
26. (a)
(b)
Unlike the graph for first ionization energy, the graph of electron affinity for these 18 elements does not show a
consistent decrease in energy as you move down the groups.
If the trend shown on the graph continues, the electron affinity value for rubidium would be approximately
45 kJ/mol.
The unknown substance is a radioisotope undergoing radioactive decay.
The half-life of the radioisotope is one hour — the time taken for half of the original number of radioactive atoms
to decay.
With the alkaline earth metals, reactivity increases moving down the group, due to increasing atomic radii and
decreasing first ionization energies. Barium has a greater reactivity than magnesium because it has a larger atomic
radius and a lower first ionization energy.
The product of the reaction should be hydrogen gas. Hydrogen gas is indicated if a flaming splint at the mouth
of a collecting test tube causes a “pop” or small explosion in the gas.
Making Connections
27. It would not be appropriate to store the radioisotope in a cardboard box. Due to the high speed (speed of light) and
penetration characteristics of gamma rays, an effective barrier would be 1 m of lead or concrete.
28. (a) Calcium and strontium are both alkaline earth metals, with similar chemical properties. It would not be unreasonable to predict that the human body would treat strontium ions much as it does calcium ions — and so they
would be incorporated into bone tissue. (This does, in fact, happen.) The effect of an unusually high concentration of strontium in the diet is greatest among those who are building bone the most rapidly — infants and children in growth spurts. If the strontium is radioactive, it will irradiate neighbouring tissue. Leukemia is often the
result.
(b) The student is to use the Internet to research and report on how radioactive strontium-90 is produced and the effect
it has when ingested.
(c) The student is to use the information gathered from the Internet research on strontium-90 to prepare a brief
cost/benefit analysis of the testing of nuclear warheads.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre
29. The student is to use the Internet to investigate Group 1 element bottle labels, MSDS information, and other sources
to create a leaflet advising high-school laboratory technicians on safe ways to handle and store each of these elements.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre
Exploring
30. The student is to use the Internet to research and report on the CANDU nuclear power process. The report is to include
a simple diagram of a CANDU reactor and a description of the function of each part.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre
28
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
(g) Two more classification schemes that might be used are: models that have the same elements bonding (O2), and
models that have different elements bonding (HCl); models with a carbon atom (CH4), and models without a carbon
atom (H2O). (Students are likely to discover quickly the variety of compounds that can be formed using carbon
atoms.) The rationale for this classification is that once again, clear distinctions can be made between such bonding
arrangements.
2.1 CLASSIFYING COMPOUNDS
PRACTICE
(Page 68)
Understanding Concepts
1. (a) A metal element and a nonmetal element combine to form an ionic compound.
(b) A nonmetal element and a nonmetal element combine to form a molecular compound.
2. Electrical conductivity can be used as a diagnostic test for an ionic compound. Ionic compounds (many of which
dissolve readily in water) form solutions that conduct electricity.
Applying Inquiry Skills
3. Experimental Design
Solubility: Obtain a small amount of the unknown substance. Observe and record its state at the ambient temperature.
Add a small quantity of the substance to about 10 mL of distilled water. Stir the mixture with a stirring rod and note
whether the chemical dissolves. Many ionic compounds readily dissolve in water.
Conductivity: Obtain a small sample of distilled water in a beaker. Use a low-voltage conductivity apparatus to test
the electrical conductivity of the sample. The apparatus should indicate a reading of zero. Test the electrical conductivity of the mixture from the above solubility procedure and record observations. Ionic compounds form solutions
that conduct electricity.
4. • Compound A is ionic — its solution conducts electricity.
•
Compound B is molecular — it is a liquid at SATP, and its solution does not conduct electricity.
•
Compound C is molecular — it is a gas at SATP.
•
Compound D is ionic — its solution conducts electricity.
•
Compound E is molecular — its solution does not conduct electricity.
2.2 IONIC BONDING
PRACTICE
(Page 71)
Understanding Concepts
1. The properties of ionic compounds that suggest ionic bonds are strong are: they are solids at SATP, they have hard
surfaces, and they have high melting and boiling points.
2. Metal elements and nonmetal elements form ionic bonds with each other.
3. Groups 1, 2, and 3 (13), tend to lose electrons to become positive ions. Groups 15, 16, and 17 tend to gain electrons
to form negative ions.
4. The minimum number of different ions in the formula of an ionic compound is 2. This is because the smallest unit of
an ionic compound that would still have the properties of the compound is a 1:1 ratio of the different ions, with the
general formula MX.
5. (a) S2–
(f) K+
2+
(b) Ba
(g) P3–
–
(c) Br
(h) Rb+
–
(d) Cl
(i) Be2+
2+
(e) Ca
30
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
Applying Inquiry Skills
Experimental Design
Conductivity of an Ionic Solid: Obtain a small sample of ionic solid such as a piece of chalk (calcium carbonate,
CaCo3(s). Use a low-voltage conductivity apparatus to test the electrical conductivity of the sample. Record your
observations.
Alternatively, fill a 100-mL beaker about one-third full with crystals of sodium chloride. Use a low-voltage
conductivity apparatus to test the electrical conductivity of the sample. Record your observations.
(b) The student is to research the conductivity of molten (liquid) ionic compounds. The student will discover that in
liquid form, ionic compounds conduct electricity.
(c) Ionic compounds are solids at room temperature. As solids they are nonconductors of electricity. However, as
liquids they conduct electricity quite well. Some examples of molten ionic compounds are sodium chloride (melts
at 801˚C), and aluminum oxide, Al2O3 (melts at 2000˚C).
(d) In the solid state, ionic compounds do not conduct electricity. In ionic solids, the ions are tightly held in the crystal
structure, so they are not free to move and carry an electrical current. When an ionic compound is melted, the
attractive forces are overcome and the crystal collapses. The ions are now free to move and carry an electrical
current, as they are in solution.
6. (a)
PRACTICE
(Page 72)
Making Connections
7. The student is to use the Internet to research and report upon the importance of one of the ions that make up the human
body, and comment on whether supplements of the ion are recommended. There are many ions from which to choose.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
PRACTICE
(Page 73)
Understanding Concepts
8. (a) The electron dot diagrams of metal ions differ from those of nonmetal ions in that the diagrams of metal ions have
vacant valence orbitals, and no lone pairs, and show a positive charge outside the square bracket, while the
diagrams of nonmetal ions have no vacant valence orbitals, have lone pairs, and show a negative charge outside
the square bracket.
(b) Electron dot diagrams of metal ions are similar to those of nonmetal ions in that both diagrams show a total charge
that is equal in value but opposite in sign, and both diagrams represent an ion that has full outer orbits of electrons — a configuration exactly the same as that of the nearest noble gas.
9. (a) lithium iodide
Li + I → [Li]+ [I](b) barium chloride
(c) potassium oxide
(d) calcium fluoride
10. (a) nitrogen
(e) lithium
(b) sulfur
(f) cesium
(c) argon
(g) calcium
(d) iodine
(h) sodium
Copyright © 2002 Nelson Thomson Learning
Chapter 2 Chemical Bonding
31
11. (a) Ca + O → [Ca]2+ [O]2–
The two elements will combine in a ratio of 1:1. The formula is CaO.
(b) 2 Rb + O → [Rb]1+ [O]2– [Rb]1+
The two elements will combine in a ratio of 2:1. The formula is Rb2O.
(c) Sr + O → [Sr]2+ [O]2–
The two elements will combine in a ratio of 1:1. The formula is SrO.
(d) 2 Al + 3 O → [O]2– [Al]3+ [O]2– [Al]3+ [O]2–
The two elements will combine in a ratio of 2:3. The formula is Al2O3.
12.
The electron dot diagrams for the five halogens are identical. Elements in the same chemical family have the same
number of valence electrons, and will thus be represented by the same electron dot diagram.
13. (a) A magnesium atom has two valence electrons. By transferring the two electrons to chlorine atoms, the resulting
magnesium ion will have a stable octet with the same electron configuration as neon.
A chlorine atom has seven valence electrons. By attracting an electron from a magnesium atom, the
resulting chlorine ion will have a stable octet with the same electron configuration as argon. Two chlorine atoms
will attract one electron each from a single magnesium atom.
Mg + 2 Cl → [Cl]– [Mg]2+ [Cl]–
The formula is MgCl2.
(b) A sodium atom has one valence electron. By transferring this electron to a sulfur atom, the resulting sodium ion
will have a stable octet with the same electron configuration as neon.
A sulfur atom has six valence electrons. By attracting two electrons from sodium atoms, the resulting
sulfur ion will have a stable octet with the same electron configuration as argon. One sulfur atom will attract one
electron each, from two separate sodium atoms.
2 Na + S → [Na]+ [S]2– [Na]+
The formula is Na2S.
(c) An aluminum atom has three valence electrons. By transferring the three electrons to oxygen atoms, the resulting
aluminum ion will have a stable octet with the same electron configuration as neon.
An oxygen atom has six valence electrons. By attracting two electrons from aluminum atoms, the resulting
oxygen ion will have a stable octet, also with the same electron configuration as neon. Three oxygen atoms will
attract two electrons each, from two separate aluminum atoms.
2 Al + 3 O → [O]2– [Al]3+ [O]2– [Al]3+ [O]2–
The formula is Al2O3.
(d) A barium atom has two valence electrons. By transferring the two electrons to chlorine atoms, the resulting
barium ion will have a stable octet with the same electron configuration as xenon.
A chlorine atom has seven valence electrons. By attracting an electron from a barium atom, the resulting
chlorine ion will have a stable octet with the same electron configuration as argon. Two chlorine atoms will attract
one electron each from a single barium atom.
Ba + 2 Cl → [Cl]– [Ba]2+ [Cl]–
The formula is BaCl2.
(e) A calcium atom has two valence electrons. By transferring the two electrons to fluorine atoms, the resulting
calcium ion will have a stable octet with the same electron configuration as argon.
A fluorine atom has seven valence electrons. By attracting an electron from a calcium atom, the resulting
fluorine ion will have a stable octet with the same electron configuration as neon. Two fluorine atoms will attract
one electron each from a single calcium atom.
Ca + 2 F → [F]– [Ca]2+ [F]–
The formula is CaF2.
32
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
(f) A sodium atom has one valence electron. By transferring this electron to an iodine atom, the resulting sodium ion
will have a stable octet with the same electron configuration as neon.
An iodine atom has seven valence electrons. By attracting an electron from a sodium atom, the resulting
iodine ion will have a stable octet with the same electron configuration as xenon.
Na + F → [Na]+ [F]–
The formula is NaF.
(g) A potassium atom has one valence electron. By transferring this electron to a chlorine atom, the resulting
potassium ion will have a stable octet with the same electron configuration as argon.
A chlorine atom has seven valence electrons. By attracting an electron from a potassium atom, the resulting
chlorine ion will have a stable octet with the same electron configuration as argon.
K + Cl → [K]+ [Cl]–
14. (a)
(b)
(c)
(d)
The formula is KCl.
baking soda
table salt
limestone or chalk
slaked lime — used to make mortar and plaster
SECTION 2.1- 2.2 QUESTIONS
(Page 74)
Understanding Concepts
1. With respect to periodic trends, elements within a chemical family (group) tend to participate in similar chemical reactions, producing ionic compounds with the same general formula. With respect to electronegativity, Group 1 and 2
metals, which have low electronegativities, will readily react with the elements in Group 17, which have high electronegativities, to form ionic compounds. As most Group 1 and 2 metals and Group 17 elements are relatively abundant in nature, it makes sense that the ionic compounds that these elements form would also be abundant in nature.
2. To reach a stable state with a full outer orbit of electrons, oxygen must gain two electrons. To reach a stable state with
a full outer orbit of electrons, lithium must lose one electron. Thus, for lithium and oxygen to combine, there must be
two lithium atoms for every one oxygen atom, with each lithium atom donating one electron to a single oxygen
atom — a ratio of two lithium atoms for every one oxygen atom.
3. Ion formation shows periodic trends. Elements within a chemical family (group) tend to produce similar ions, and to
participate in similar chemical reactions, producing ionic compounds with the same general formula.
4. (a) calcium carbonate
(b) calcium hydroxide
(c) sodium chloride
(d) sodium bicarbonate
Cs + F → [Cs]+ [F]–
CsF
An ionic compound (an ionic halide).
Solid at SATP, with a hard surface, is brittle, has a high melting point, and forms a solution that conducts
electricity.
(e) The properties of the compound are due to the strong, simultaneous forces of attraction between the positive and
negative ions, which hold the ions firmly in a rigid structure. The solid state, hardness, brittleness, and the high
melting point result from the strong attractions, which occur in the crystal structure. And because the ionic bonds
break down in water, the resulting ions are free to move in solution and conduct electricity.
(f) Cs has a low first ionization energy, and F has a high electron affinity, so it makes sense that the reaction would
be a vigorous one. With the alkali metals, reactivity increases moving down the group, due to increasing atomic
radii and decreasing first ionization energies. With the halogens, reactivity increases moving up the group, due to
decreasing atomic radii and increasing electron affinity.
5. (a)
(b)
(c)
(d)
Copyright © 2002 Nelson Thomson Learning
Chapter 2 Chemical Bonding
33
2.3 COVALENT BONDING
PRACTICE
(Page 77)
Understanding Concepts
1. (a) F2(g)
(d)
PCl3(s)
(b) H2O(l)
(e)
H2S(g)
(c) CH4(g)
(f)
SiO2(s)
2. (a) H2(g)
(d)
NF3(g)
(b) O3(g)
(e)
N2H2(g)
(c) OF2(g)
(f)
P2H4(g)
PRACTICE
(Page 79)
Understanding Concepts
34
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
3. (a) PO43–(aq)
(b) OH
(aq)
(c) BrO
3(aq)
(d) ClO
4(aq)
4. Ozone, question 2(b), can be considered to have a coordinate covalent bond, with one of the oxygens starting as [O]2–.
Question 3 (a), (c), and (d) have coordinate covalent bonds; (b) would have a coordinate covalent bond if the ions
involved are [O]2– and [H]+.
PRACTICE
(Page 81)
Understanding Concepts
5. The term “bonding electrons” describes a pair of electrons that are shared by two atoms forming a covalent bond,
whereas the term “lone pairs” describes a pair of electrons not involved in bonding.
6. (a) covalent
(b) covalent
(c) ionic
(d) ionic
7. (a) Molecular elements and compounds: N2(g), O2(g), F2(g), C12H22 O11(s) (sugar), C6H6(l) (liquid benzene), NH3(g)
(ammonia).
Ionic compounds: NaCl(s) , MgO(s) , Ca(OH)2(s) , CaF2(s) , NaHCO3(s) , CaCO3(s).
(b) Molecular elements and compounds are formed by covalent bonds. Molecular compounds may be solids, liquids,
or gases at SATP and tend to be soft or waxy. Covalent bonds between the atoms are strong. However, the intermolecular forces in molecular compounds are weaker in comparison — adding a relatively small amount of heat
will cause a solid molecular compound to change state from a solid to a liquid, and then to a gas. Ionic compounds
are formed by ionic bonds. Ionic compounds are solids at SATP and are hard and brittle. The properties of ionic
compounds are due to the strong, simultaneous forces of attraction between the positive and negative ions, which
hold the ions firmly in a rigid structure.
8. (a) The bonding capacity of nitrogen is three, and the bonding capacity of chlorine is one.
Copyright © 2002 Nelson Thomson Learning
Chapter 2 Chemical Bonding
35
(b) The number of covalent bonds (shared electron pairs) that an atom can form is known as its bonding capacity.
Each atom of nitrogen shares three electron pairs, so it has a bonding capacity of three. Each atom of chlorine
shares one electron pair, so it has a bonding capacity of one.
9. Coordinate covalent bonds are similar to covalent bonds in that once the bond is formed, there is no way to tell the
difference from a covalent bond. Coordinate covalent bonds are different from covalent bonds in that instead of having
the shared electrons come from both atoms, both of the shared electrons come from the same atom.
10. (a) Two nitrogen atoms require three electrons each to form stable octets. The two atoms form a triple bond by
sharing three electron pairs.
(b)
(c) The strength of a bond between two atoms increases as the number of electron pairs in the bond increases. The
triple bond in a nitrogen molecule is very strong, preventing the atoms from forming a bond with another atom.
11. (a) HCl
(c)
H2S
(b) NH3
(d)
CO2
12. Bonding Capacities of Some Common Atoms
Atom
carbon
nitrogen
oxygen
hydrogen
fluorine
chlorine
bromine
odine
Number of valence
electrons
4
5
6
1
7
7
7
7
13. (a) O2
36
Number of bonding
electrons
4
3
2
1
1
1
1
1
(e)
HCN
(b) Br2
(f)
C2H5OH
(c) H2O2
(g)
CH3OCH3
(d) C2H4
(h)
CH3NH2
Unit 1 Matter and Chemical Bonding
Bonding capacity
4
3
2
1
1
1
1
1
Copyright © 2002 Nelson Thomson Learning
14. (a)
(b) There are covalent and coordinate covalent bonds within the hydronium ion.
(c) Ionic bonds. (Note that it will also form hydrogen bonds.)
15. It is incorrect to show the structural formula of H2S as H—H—S. The central H atom is shown as bonding to two
other atoms — yet the bonding capacity of hydrogen is one. The correct structural formula is H–S–H and its formation is shown below:
SECTION 2.3 QUESTIONS
(Page 82)
Understanding Concepts
1. (a) Properties of compound A: a solid at SATP, is hard and brittle, has a high melting point, and its solution conducts
electricity.
Properties of compound B: may be a solid, a liquid, or a gas at SATP — if a solid it is soft, waxy or flexible, has
a low melting point, and its solution does not conduct electricity.
(b) Compound A is an ionic compound involving the bonding of a metal with a nonmetal. Sodium, a metal, has
formed an ionic bond with fluorine, a nonmetal.
Compound B is a molecular compound involving the bonding of a nonmetal with another nonmetal. Nitrogen, a
nonmetal, has formed covalent bonds with fluorine, also a nonmetal.
(c)
2. The difference in electronegativities between aluminum and oxygen is strong enough to transfer electrons from the
metal atom to the nonmetal atom, producing ionic bonds and an ionic compound. Metals and nonmetals show a periodic trend to form ionic compounds, and the metals of Group 13 show a periodic trend to form ionic oxides
(compounds composed of a metal and oxygen) when burned in air.
The difference in electronegativities between carbon and oxygen is not strong enough to transfer electrons, and
instead, the electrons are shared in a covalent bond, resulting in the formation of a molecular compound. Nonmetals
show a periodic trend to combine with other nonmetals to form molecular compounds.
3. (a) Group 1 metals readily react with the nonmetal elements in Group 17 to form ionic compounds, with the general
formula MX.
(b) The compound MX is a solid at SATP, is hard and brittle, has a high melting point, and its aqueous solution
conducts electricity.
4. (a) NaCl is an ionic compound; Cl2 is a covalent compound.
(b) NaCl is a solid at SATP because of the strong ionic bonds. The simultaneous forces of attraction between the
positive and negative ions hold the ions firmly in a rigid structure.
Cl2 is a gas at SATP because of the weak intermolecular forces between the Cl2 molecules. The covalent bonds
between the Cl atoms are strong, but with the addition of a relatively small amount of heat, the intermolecular
forces between Cl2 molecules are easily overcome, causing the molecular compound to be a gas at SATP.
Copyright © 2002 Nelson Thomson Learning
Chapter 2 Chemical Bonding
37
5. Some of the molecular compounds that can be created using oxygen and sulfur include SO, SO2 , SO3.
From the examples above, SO contains only multiple bonds.
Applying Inquiry Skills
6. Experimental Design
Solubility: Obtain a small amount of compound A. Observe and record its state at the ambient temperature. Add a
small quantity of the substance to about 10 mL of distilled water. Stir the mixture with a stirring rod and note whether
the chemical dissolves. Many ionic compounds readily dissolve in water.
Conductivity: Obtain a small sample of distilled water in a beaker. Use a low-voltage conductivity apparatus to test
the electrical conductivity of the sample. The apparatus should indicate a reading of zero. Test the electrical
conductivity of the mixture from the above solubility procedure and record observations. Ionic compounds (many of
which dissolve readily in water) form solutions that conduct electricity.
Repeat the procedure for compound B. Molecular compounds (some of which dissolve in water) form solutions that
do not conduct electricity.
7. Experimental Design
State, Hardness and Brittleness: Obtain a small-sized piece of NaCl(s) in road salt form. Observe and, in a table, record
its state at SATP, its hardness, and brittleness.
Solubility: Pour about 10 mL of distilled water into a 50-mL beaker. Add a small quantity of the road salt to the water.
Use a stirring rod to stir the mixture. Note whether the road salt dissolves.
Conductivity: Obtain a small sample of distilled water in a beaker. Test the electrical conductivity of the sample. The
apparatus should indicate a reading of zero. Test the conductivity of the mixture of sodium chloride and water from
the Solubility procedure. Record observations.
2.4 ELECTRONEGATIVITY, POLAR BONDS, AND POLAR MOLECULES
PRACTICE
(Page 84)
Understanding Concepts
1. To predict whether a chemical bond between two atoms will be ionic, polar covalent, or covalent, we must consider
the electronegativities of the elements involved. The absolute value of the difference in electronegativities of two
bonded atoms provides a measure of the polarity in the bond: the greater the difference, the more polar the bond. By
convention, a difference in electronegativity greater than 1.7 indicates an ionic bond.
2. (a) covalent
(f) covalent
(b) covalent
(g) covalent
(c) ionic
(h) ionic
(d) ionic
(i) covalent
(e) covalent
Si and O would be the most polar of the covalent bonds.
3. (a) H—F
(e) N—H
(b) C—O
(f) P—O
(c) O—H
(g) C—N
(d) P—Cl
38
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
4. (a) H2O
(b) Br2
(d)
PCl3
(e)
OF2
(c) HBr
PRACTICE
(Page 85)
Understanding Concepts
5. If a molecule that contains polar covalent bonds is quite symmetrical, it lacks oppositely charged ends and is not polar.
or
6. While the difference in electronegativity values between carbon and hydrogen results in polar covalent bonds
(the difference is 0.4 for each carbon — hydrogen bond), the molecule of methane, CH4(g,) is quite symmetrical, and
lacks oppositely charged ends. Therefore, it is not polar.
Try This Activity : Molecular Models
(Page 87)
(a)
(b)
(c) Carbon dioxide required a double bond in order to complete its octet. Nitrogen required a triple bond in order to
complete its octet.
(d) Water, hydrogen sulfide, hydrogen fluoride, and ammonia would be expected to be polar.
Copyright © 2002 Nelson Thomson Learning
Chapter 2 Chemical Bonding
39
(e) (c)
Similar shape to water:
Similar shape to ammonia:
Similar shape to hydrogen fluoride:
PRACTICE
(Page 88)
Understanding Concepts
7. In ice, the hydrogen bonds between the molecules result in a regular hexagonal crystal structure that forms an open
lattice with a great deal of empty space between the molecules. This causes ice to be less dense than liquid water. Thus,
as water freezes, the ice floats to the top of the water surface, and as time passes, the lake freezes from the top down.
8. (a) polar
(d) nonpolar
(b) nonpolar
(e) nonpolar
(c) polar
(f) nonpolar
9. PCl3(s) , HC2H3O2(aq), CCl4(l) were classified as nonpolar but contain polar covalent bonds. Due to the symmetrical
shape of these molecules, they lack oppositely charged ends and are therefore not polar.
10. A dipole–dipole force describes in general the attractive force acting between polar molecules. A hydrogen bond is a
relatively strong dipole–dipole force acting specifically between a positive hydrogen atom of one molecule and a
highly electronegative atom (F, O, or N) in another molecule.
11. (a) London dispersion force
(b) Hydrogen bond
(c) Hydrogen bond
12. (a) I2(s) is a nonpolar molecule. Thus it is the London dispersion force that exists between all molecules — both polar
and nonpolar — that is responsible for the intermolecular attraction between molecules of I2(s).
(b) H2O(l) is a highly polar molecule containing an O–H bond. Thus it is the hydrogen bond force — a relatively
strong dipole–dipole force between a positive hydrogen atom of one molecule and a highly electronegative atom
(F, O, or N) in another molecule — that is responsible for the intermolecular attraction between molecules of
H2O(l).
(c) NH3(g) is a highly polar molecule containing an N–H bond, and so forms hydrogen bonds with other molecules.
Applying Inquiry Skills
13.Prediction
(a) The molecules that will be affected by the charged object are NCl3 , H2O, CH3OH, and H2O2. The charged object
will have an effect on polar molecules (only) because the polar molecules are slightly charged at each end.
Evidence
(b) Samples 1 to 3 are Br2, CCl4, and vegetable oil. Samples 4 to 7 are NCl3, H2O, CH3OH, and H2O2.
Analysis
(c) The evidence shows that a charged object will have an effect on a thin stream of liquids that are composed of
polar molecules, but will have no effect on liquids composed of nonpolar molecules. The thin stream of liquids
composed of polar molecules will be attracted to the charged object. Of the seven sample liquids, only NCl3, H2O,
CH3OH, and H2O2 are polar, and therefore affected by the charged object.
Synthesis
(d) The polar molecules — NCl3, H2O, CH3OH, and H2O2 — are positively charged at one end, and negatively
charged at the other because of electronegativity differences. Thus, the end of the polar molecule that has the
opposite charge of the charged object will be attracted to the charged object. This is similar to dipole–dipole
forces.
(e) The liquids were affected by both positive and negative charges because the polar molecules of the liquid are
positively charged at one end, and negatively charged at the other. If a positively charged object is used, the negatively charged end of the polar molecule will be attracted to the charged object, and if a negatively charged object
is used, the positively charged end of the polar molecule will be attracted to the charged object.
40
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
SECTION 2.4 QUESTIONS
(Page 89)
Understanding Concepts
1. Covalent bonds and ionic bonds are the forces that bond atoms and ions together within a compound — the
intramolecular forces. These forces are sufficient to explain the existence of molecular and ionic compounds, and to
explain many of the properties of ionic compounds, but they aren’t sufficient to explain the physical state of molecular compounds. If covalent bonds were the only forces at work, molecular compounds would all be gases, as there
would be no attraction between the molecules strong enough to order the molecules into solids or liquids. The
concepts of the polar molecule and small charges on atoms that result in intermolecular forces help to explain why
these molecular compounds are not all gases at SATP.
2. When the atoms are identical, such as in a chlorine molecule, the electrons are shared equally. However, this is not
the case for a compound like hydrogen chloride, where electrons are shared between two different elements. In this
situation, the sharing is unequal, as the bonding electrons in the H—Cl bond spend more time near the chlorine atom
than near the hydrogen atom. This is because of chlorine’s greater attraction for electrons.
3. (a) Both BCl3 and NCl3 are molecular compounds. By convention, compounds with bonds that have electronegativity
differences less than or equal to 1.7, have covalent-type bonds and are classified as molecular compounds.
(b) The bonds between B—Cl and N—Cl are similar in that they are both covalent-type bonds and involve the
sharing of a pair of electrons. The bonds are different in that the B—Cl bonds are polar covalent bonds due to
differences in electronegativities. In this situation, the sharing of electrons is unequal, as the bonding electrons in
the B—Cl bond spend more time near the chlorine atom than near the boron atom. This is because of chlorine’s
greater attraction for electrons. Another difference is that the boron atom in BCl3 does not have an octet in its
valence shell.
(c) BCl3 is quite symmetrical and lacks oppositely charged ends. Thus it is not a polar molecule. Intermolecular
attractions would be due to the weaker London dispersion forces. NCl3 is a polar molecule due to having nitrogen
at one end. Intermolecular attractions would be due to the stronger dipole-dipole forces.
4. Since a molecule of carbon tetrachloride, CCl4(l)n, is quite symmetrical, it lacks oppositely charged ends and is not
polar.
2.5 THE NAMES AND FORMULAS OF COMPOUNDS
PRACTICE
(Page 93)
Understanding Concepts
1. Substances were named in a variety of ways. In some cases, the name referred to the use of the compound; in other
cases, it incorporated an obvious property, or perhaps referred to the sources of the substance.
2. (a) muriatic acid
(b) baking soda
(c) laughing gas
(d) grain alcohol
3. A binary compound is composed of two kinds of elements.
4. In the formula of a binary ionic compound, the metal cation is always written first, followed by the nonmetal anion.
The name of the metal is stated in full and the name of the nonmetal ion has an -ide suffix; for example, NaCl(s) is
sodium chloride.
Copyright © 2002 Nelson Thomson Learning
Chapter 2 Chemical Bonding
41
5. Most transition metals and some representative metals can form more than one kind of ion. Metals that can have more
than one valence, or charge, are classified as multivalent. For example, iron can form an Fe2+ ion or an Fe3+ ion.
6. (a) CaF2
(l) Hg2O
(b) Na2S
(m) NiBr2
(c) AlN
(n) ZnO
(d) AlCl3
(o) CoCl3
(e) K2O
(p) SrBr2
(f) CaCl2
(q) AuF
(g) CuS
(r) LiCl
(h) PbBr2
(s) Sr3N2
(i) AgI
(t) BaBr2
(j) Ba3N2
(u) SnI4
(k) FeF2
7. (a) sodium chloride
(h) copper(I) sulfide
(b) calcium oxide
(i) lead(IV) sulfide
(c) calcium chloride
(j) iron(III) oxide
(d) magnesium oxide
(k) molybdenum oxide
(e) aluminum oxide
(l) silver sulfide
(f) zinc sulfide
(m) zinc oxide
(g) tin(IV) oxide
8. (a) sodium oxide
(g) nickel(III) oxide
(b) tin(IV) chloride
(h) silver sulfide
(c) zinc iodide
(i) iron(II) chloride
(d) strontium chloride
(j) potassium bromide
(e) aluminum bromide
(k) copper(II) iodide
(f) lead(IV) chloride
(l) nickel(II) sulfide
9. (a) SrO - strontium oxide
(b) Na2S - sodium sulfide
(c) AgI - silver iodide
(d) BaF2 - barium fluoride
(e) CaBr2 - calcium bromide
(f) LiCl - lithium chloride
10. (a) HgS
(d) NiBr2
(b) MoS2
(e) CuCl2
(c) MnO2
(f) FeI3
11. (a) iron(II) sulfide
(b) lead(IV) bromide
(c) tin(II) chloride
Reflecting
12. Because old systems die hard. For example, the -ous and -ic suffixes are still used extensively in industry.
PRACTICE
(Page 96)
Understanding Concepts
13. Tertiary compounds are composed of three different elements.
14. (a) Ionic compounds can be composed of a single ion and a polyatomic ion.
(b) Some ionic compounds are composed of a single ion and a polyatomic ion that includes oxygen. A polyatomic
ion that includes oxygen is called an oxyanion.
(c) Ionic compounds that form crystals that contain molecules of water within the crystal structure are referred to as
hydrates.
15. (a) sodium nitrate
(f) calcium hydroxide
(b) sodium nitrite
(g) lead(II) carbonate
(c) copper(II) nitrate
(h) tin(II) phosphate
(d) copper(I) nitrate
(i) iron(III) sulfate
(e) aluminum sulfite
42
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
16. (a) CaCO3
(f) (NH4)3PO4
(b) NaHCO3
(g) CuSO4
(c) NaClO
(h) NaOH
(d) CaSO4
(i) KmNO4
(e) NH4NO3
17. (a) lithium chlorate
(n) silver sulfate
(b) barium sulfate
(o) mercury(II) bromate
(c) mercury(I) carbonate
(p) iron(III) carbonate
(d) magnesium nitrate
(q) ammonium hypochlorate
(e) iron(III) bromate
(r) gold(III) nitrate
(f) sodium phosphate
(s) magnesium bromate
(g) ammonium iodate
(t) sodium iodate
(h) gold(I) acetate
(u) zinc chlorite
(i) zinc phosphate
(v) tin(II) carbonate
(j) antimony(V) chlorate
(w) strontium sulfite
(k) manganese(II) sulfite
(x) nickel(III) phosphate
(l) potassium hypobromite
(y) copper(II) acetate
(m) aluminum perphosphate (z) barium perphosphate
18. (a) copper(I) hypophosphite
CuPO2
(b) tin(IV) chlorite
Sn(ClO2)4
(c) iron(II) bromate
Fe(BrO3)2
(d) iron(III) chlorite
Fe(ClO2)3
(e) lead(IV) sulfate
Pb(SO4)2
19. (a) copper(II) pentahydrate
(b) sodium sulfate decahydrate
(c) magnesium sulfate heptahydrate
20. (a) Fe2O3 3 H2O
(d) Cd(NO3)2 4 H2O
(b) AlCl3 6 H2O
(e) LiCl 4 H2O
(c) NaS2O3 5 H2O
(f) CaCl2 2 H2O
21. When heat is applied to a hydrate, it will decompose to produce water vapour and an associated ionic compound.
When this water, called water of hydration, is removed, the product is referred to as anhydrous.
PRACTICE
(Page 98)
Understanding Concepts
22. (a) N2
(b) CO2
(c) CO
(d) NO2
(e) NO
(f) N2O
(g) N2O4
(h) SO2
(i) I2O5
(j) SiF4
(k) BF3
(l) PI3
(m) P2O5
23. (a) sulfur hexafluoride
(b) dinitrogen trioxide
(c) nitrogen dioxide
(d) phosphorus trichloride
(e) phosphorus pentachloride
(n)
(o)
(p)
(q)
(r)
(s)
(t)
(u)
(v)
(w)
(x)
(y)
(z)
(f)
(g)
(h)
(i)
Copyright © 2002 Nelson Thomson Learning
SF4
PCl5
S2Cl2
CCl4
SO3
SF6
ClO2
N2O5
PCl3
SiCl4
CS2
PBr5
CF4
iodine heptafluoride
boron trifluoride
diphosphorus pentasulfide
diphosphorus pentoxide
Chapter 2 Chemical Bonding
43
PRACTICE
(Page 101)
Understanding Concepts
24. (a) HCl(aq)
(b) HCl(aq)
(c) H2SO4(aq)
(d) H2SO4(aq)
(e) HC2H3O2(aq)
(f) HC2H3O2(aq)
25.
Classical
(a) sulfurous acid
(b) phosphoric acid
(c) hydrocyanic acid
(d) carbonic acid
(e) hydrosulfuric acid
(f) hydrochloric acid
(g) hydrocyanic acid
(h) sulfuric acid
(i) phosphoric acid
26. (a) potassium hydroxide
(b) calcium hydroxide
27. (a) Mg(OH)2(aq)
(b) NaOH(aq)
(c) Al(OH)3(aq)
(g)
(h)
(i)
(j)
(k)
(l)
HNO2(aq)
HNO3(aq)
HBr(aq)
H2SO2(aq)
HI(aq)
HClO4(aq)
IUPAC
aqueous hydrogen sulfite
aqueous hydrogen phosphate
aqueous hydrogen cyanide
aqueous hydrogen carbonate
aqueous hydrogen sulfide
aqueous hydrogen chloride
aqueous hydrogen cyanide
aqueous hydrogen sulfate
aqueous hydrogen phosphate
SECTION 2.5 QUESTIONS
(Page 101)
Understanding Concepts
1. (a) NH3, HCN
(b) Ammonia and hydrogen cyanide are classified as covalent molecules.
(c) Hydrogen cyanide is a polar covalent molecule that ionizes in water to form H+, and CN. The ionic nature of
the compound could be verified by dissolving the substance in water and testing for electrical conductivity. The
covalent nature of the compound could be verified by calculating the electronegativity difference between H and
C, and between C and N — the differences are not greater than 1.7.
(d) If hydrogen cyanide is added to water, aqueous hydrogen cyanide, also known as hydrocyanic acid, is formed.
The substance might cause blue litmus paper to turn red, indicating an acidic solution. (Actually, hydrocyanic
acid is a very weak acid and may not turn blue litmus paper red.)
2. (a) KOH is an ionic hydroxide, and its aqueous solution is a base. The bond consists of a K+ ion, and an OH ion.
Chemists have discovered that all aqueous solutions of ionic hydroxides are bases. HCl as a gas is covalent, and
its aqueous solution is an acid. The bond consists of an unequal sharing of a pair of electrons. When dissolved in
water, the resulting aqueous solution displays a set of specific properties called acidic.
(b)
[ K] + [ O — H ] – → K OH
H — Cl → H — Cl
(c) aqueous potassium hydroxide, aqueous hydrogen chloride
(d) Bases are reactive — for example, bases react with proteins to break them down into smaller molecules.
Precautions must be taken when handling bases. Acids are reactive and can combine with many other substances.
Acids must be treated with care, as they can be very corrosive and can cause serious damage to the environment.
3. The student is to write the IUPAC names and formulas for as many compounds as possible, using only the following
elements: K, C, H, F, Mg, O, Cl, and Na.
Use K as an example:
(a) KCl, KOH, KClO3
44
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
(b) KCl — potassium chloride, KOH — potassium hydroxide, KClO3 — potassium chlorate.
(c) KCl — ionic, KOH — ionic, KClO3 — ionic.
[ K] + [ Cl ] – → KCl
[ K] + [ O — H ] – → K OH
[ K] +
O — Cl — O
→ K ClO3
:O:
(d) KCl — binary, KOH — tertiary and basic, KClO3 — tertiary.
(e) KCl — ionic bonds only, KOH — ionic and covalent bonds, KClO3 — ionic and covalent bonds.
CHAPTER 2 SUMMARY
Make a Summary
(Page 102)
The examples of NaCl and H2O are used in the table below. The student is to include as many examples of compounds
as possible for each type of intramolecular bond.
Table 1: Summarizing Bonds and Forces
Compound
Properties
NaCl
Solid at SATP,
hard and
brittle, high
melting point,
its solution
conducts
electricity.
H 2O
Liquid at
SATP, low
boiling point.
Electron dot
diagram/ Lewis
structure
[Na]+ [Cl]–
Intramolecular
bond type
Polarity
Intermolecular
forces
Ionic
Ionic
Locked in
a regular
structure, held
by the balance
of attractive
bonds and
electrical repulsion.
H–O–H
Covalent
Polar
Hydrogen bonds.
Reflect on your Learning
(Page 102)
By the end of the chapter the student should have developed a more in-depth understanding of why atoms form
compounds, an awareness of the many different compounds that are possible, the types of forces present between
atoms in compounds, and how the forces that hold atoms together in a compound determine the chemical properties
of the compound.
CHAPTER 2 REVIEW
(Page 103)
Understanding Concepts
1. When elements that are found in the “metals” position in the periodic table react with elements that are found in the
“nonmetals” position in the periodic table, they form ionic compounds that have ionic bonds. When elements that are
found in the “nonmetals” position in the periodic table react with elements also found in the “nonmetals” position in
the periodic table, they form molecular compounds that have covalent bonds.
Copyright © 2002 Nelson Thomson Learning
Chapter 2 Chemical Bonding
45
2. An ionic bond occurs when one or more valence electrons are transferred from a metal atom to a nonmetal atom. This
leaves the metal atom as a positive ion, or cation, and the nonmetal atom as a negative ion, or anion. An ionic bond is
the electrostatic attraction between positive and negative ions in a compound.
A covalent bond arises from the simultaneous attraction of two nuclei for a shared pair of electrons. The result is a
covalent bond — a shared pair of electrons held between two nonmetal atoms that hold the atoms together in a molecule.
3. (a) The properties of ionic compounds: Are solid at SATP, with hard surfaces, are brittle, have high melting points,
and form solutions that conduct electricity. The properties are due to the strong ionic bonds, simultaneous forces
of attraction between the positive and negative ions, which hold the ions firmly in a rigid structure. The solid state,
hardness, brittleness, and the high melting point result from the strong attractions, which occur in the crystal
structure. And because the ionic bonds break down in water, the resulting ions are free to move in solution and
conduct electricity.
The properties of molecular compounds: May be solids, liquids, or gases at SATP, and are soft, waxy, or
flexible. Covalent bonds between the atoms are strong. However, the intermolecular forces in molecular
compounds are weaker in comparison — adding a relatively small amount of heat will cause a solid molecular
compound to change state from a solid to a liquid, and then to a gas.
(b) Ionic compounds (many of which dissolve readily in water) form solutions that conduct electricity. Because the
ionic bonds often break down in water, the resulting ions are free to move in solution and conduct electricity.
Molecular compounds form solutions that do not generally conduct electricity.
4. (a) Intramolecular bonding is the force that bonds atoms and ions together in a compound. One main type of
intramolecular bonding is ionic bonding. An example of ionic bonding is the electrostatic attraction that occurs
between the [Na] cation and the [Cl] anion to form the ionic compound NaCl. Another main type of
intramolecular bonding is covalent bonding. An example of covalent bonding is the sharing of a pair of electrons
that occurs between hydrogen and chlorine to form the molecular compound HCl.
(b) Ionic bonding results in compounds that are solid at SATP with hard surfaces, are brittle, have high melting
points, and form solutions that conduct electricity. Covalent bonding results in compounds that may be solids,
liquids, or gases at SATP, and are soft, waxy, or flexible. Covalently bonded compounds form solutions that do
not generally conduct electricity.
5. (a) two nonmetal atoms that are sharing a pair of electrons
(b) oppositely charged ends of polar molecules
(c) a positive hydrogen atom of one molecule and a highly electronegative atom (F, O, or N) in another molecule
(d) a positively charged ion (cation) of a metal and a negatively charged ion (anion) of a nonmetal
6. The chemical formulas of ionic compounds consist of a metal joined to a nonmetal. Examples are NaCl, CuSO4, and
NaHCO3. The chemical formulas of molecular compounds consist of nonmetals combined with other nonmetals.
Examples are SO2, CO2, and NH3.
7. Halogens tend to form diatomic molecules because they have only one bonding electron, and thus a capacity to bond
with only one other atom.
8. Students will reproduce the bonding continuum of Figure 3 on p. 84. Cl2, with difference in electronegativities of 0,
will be placed at the far right (covelent). NaCl, with an electronegativity difference of 2.1, should be placed left of
centre, in the “ionic” area. Na–Cl involves an electron transfer, resulting in the formation of cations and anions that
are attracted to each other. Cl–Cl involves equal sharing of a pair of electrons.
Ca
(e) S
S
(b) Al
Al
(f) Br
Br
(c) K
K
(g) Ne
Ne
(d) N
N
9. (a) Ca
10. (a)
(b)
(c)
(d)
(e)
(f)
11. (a)
46
covalent
polar covalent
polar covalent
ionic
ionic
ionic
Na2O(s) is ionic, MgO(s) is ionic, Al2O3(s) is ionic, SiO2(s) is molecular, P2O5(s) is molecular, SO2(g) is molecular,
and Cl2O(g) is molecular.
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
(b)
(c) The ionic compounds are solid at SATP with hard surfaces, are brittle, have high melting points, and form solutions that conduct electricity. The molecular compounds may be solids, liquids, or gases at SATP, are soft, waxy,
or flexible, and form solutions that do not generally conduct electricity.
(d) The difference in electronegativity of the constituent elements determines whether the compound has ionic bonds
or covalent-type bonds. An electronegativity difference greater than 1.7 indicates an ionic bond. Ionic bonds result
in ionic properties, and covalent-type bonds result in molecular properties.
12.
Table 1: Structures of Covalent Compounds
Compound
Lewis Structure
HF(g)
H—F
: : Cl
: :
BCl3(g)
:
: : ::
:
:
: B:
: :
: : Cl : Cl
:: ::
:
: :: :
:
:
:
H
:
:: :
: :
:
:
:
:
H — Si — H
:
:
H
:
:
SiH4(g)
CCl4(l)
NCl3(g)
H2O2(l)
:
:
:
HCN(g)
H—C
Copyright © 2002 Nelson Thomson Learning
N:
Cl
Polar covalent
Cl
H
:
:
Polar covalent
H — Si — H
:
H
:
: : : :: :
H—O—O—H
:
:
O = C =: O
:
:
:
Polar covalent
Cl
:
: :: :
: :
:
: : : :
: Cl : : : :
: :
: :: : ::
: :: :
: : : :: ::
:
: Cl :—: C
::—: Cl
:
:
:
::
:
:
: : : : :: ::
: : Cl :
: :
:
: :
: : :: :: : : : : :: : :
:
:
: :: :: :
: :
: :
: ::
: : : :N
: ::: ::: : ::: ::: :
: : : :
:
: : : ::
: Cl : Cl : :Cl : :
: :
:
: : : : : :: :: : :
:
: : : :: : : : :
CO2(g)
H—F
B
:
: :
:
Types of Bonds
:
:
:
: : ::
: : :
:
: :
: :: :
Structural Formula
: :
:
:
:
Cl
Polar covalent
Cl — C — Cl
Cl
Cl — N — Cl
Polar covalent
Cl
H—O—O—H
Polar covalent and covalent
O=C=O
Polar covalent
H—C
Polar covalent
:
:
:
:
:
:
N
Chapter 2 Chemical Bonding
47
13. (a) Cl is most electronegative and will have a δ– charge.
2.5
3.0
C
C
δ+
δ–
(b) F is most electronegative and will have a δ– charge.
1.0
4.0
+
Ca + F → [ Ca] [ F
]–
(c) Cl is most: electronegative and will have a δ– charge.
1.5
3.0
+
–
Al + Cl: → [ Al] [ Cl ]
:
(d) O is most: electronegative and will have a δ– charge.
1.8
2.5
Si + O : → Si
δ+
δ–
O
:
(e) O is most: electronegative and will have a δ– charge.
2.5
3.5
C
O
C
δ+
O
δ–
:
14. (a)
+
H
Cl
H—N—H
:
H
: :
(b)
(c)
NH4Cl(s) contains a coordinate covalent bond. A hydrogen ion, which has no electrons of its own, has bonded to
an NH
:
3(g): molecule by sharing the unbonded, lone pair of electrons in NH3(g).
:
:F :
: :
: :
B
F
B
:F
F:
:
:
:
:
H—N—H
F
H—N—H
H: :
(d)
(e)
H
H
H
: :
H
:
H
: :
: :
: F:
:
:
H — N —: B: — F:
H
F
H
H
F
H—N—B—F
: :
: F:
H
F
NH3BF3(g) contains a coordinate covalent bond. The boron in BF3(g),which does not have an octet in its valence shell,
has bonded to NH3 by sharing the unbonded, lone pair of electrons in NH3.
15. (a)
(b)
(c)
(d)
(e)
(f)
(g)
16. (a)
(b)
(c)
(d)
(e)
48
NaHSO4(s)
NaOH(s)
CO2(g)
HC2H3O2(aq)
NaS2O3 5 H2O(s)
NaClO(s)
S8(s)
MgBr2
CS2
Hg(NO2)2
HCl(aq)
LiOH
(h)
(i)
(j)
(k)
(l)
(m)
NaNO3(s)
H3PO4(aq)
I2(s)
Al2O3(s)
KOH(s)
HCO3(aq)
(n)
(o)
(p)
(q)
(r)
H2CO3(aq)
Ca(OH)2
Zn(ClO)2
Pb(ClO4)4
PBr5
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
17. (a)
(b)
(c)
(d)
(e)
(f)
18. (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
19. (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
20. (a)
(b)
(c)
(d)
(e)
21. (a)
Ag2CO3
(s) AsCl5
Al(ClO4)3
(t) Bi(NO3)3
CuSO4
(u) NaClO
SO3
(v) OCl2
NiPO4
(w) SnBr2
MgO
(x) H2SO4(aq)
N2O
(y) KOH
FeSO5
(z) BaCO3
NH4H2PO3
LiHSO3
KHSO4
BaCl2 3 H2O
NaH2PO4
NaHCO3
calcium carbonate
diphosphorus pentoxide
magnesium sulfate heptahydrate
dinitrogen monoxide
sodium silicate
calcium hydrogen carbonate
hydrochloric acid
copper(II) sulfate pentahydrate
sulphuric acid
calcium hydroxide
sulfur trioxide
sodium fluoride
sodium chloride
diphosphorus trioxide
nitric acid
lead(II) acetate
ammonium hypochlorite
tin(IV) bromate
antimony(III) oxide
zinc iodate
iron(II) pernitrate
calcium hydroxide
potassium iodide
sulfur difluoride
hydrobromic acid
calcium hydrogen phosphate
copper(II) sulfate heptahydrate
sodium hydrogen phosphate
lithium hydrogen carbonate
potassium hydrogen sulfate
potassium bromide
KBr(s)
[ K] + [ Br
(n)
(o)
(p)
(q)
(r)
(s)
(t)
(u)
(v)
(w)
(x)
(y)
]–
(b) silver iodide
AgI (s)
[ Ag] + [ I ] –
(c) lead(II) oxide
PbO(s)
[ Pb] 2+ [ O ] 2–
(d) zinc sulfide
ZnS(s)
(e) copper(II) oxide
CuO(s)
[ Zn] 2+ [ S ] 2–
[ Cu] 2+ [ O ] 2–
(f) lithium nitride
Li3N(s)
[ Li] 2+ [ N ] 3 –
Copyright © 2002 Nelson Thomson Learning
copper(II) carbonate
aluminum sulfite
ammonium hydroxide
barium acetate
iodine monochloride
gold(III) chloride
magnesium sulfide
dinitrogen difluoride
nickel(II) sulfate
hydrosulfuric acid
silver bromate
lithium perchlorate
Chapter 2 Chemical Bonding
49
Applying Inquiry Skills
22. (a) Analysis
Substance 1 is an ionic compound: KCl(aq)
Substance 2 is an acid: HCl(aq)
Substance 3 is soluble but not ionic: C2H5OH(aq)
Substance 4 is a base: Ba(OH)2(aq)
Synthesis
(b) The water is used as a control. The result of the conductivity and litmus tests is the dependent variable. The different
substances that are dissolved in water are the independent variables. Since water is used to prepare the solutions for
each of the different substances, any change in the dependent variable from the control results can be attributed to the
changing independent variable.
(c) Solutions 1, 2, and 4 all have high conductivities, and could have been involved in somebody getting electrocuted;
however, the most likely to have been involved is KCl. If the solutions were of high concentration, it seems possible
that the lawsuit would have been about corrosion rather than electrocution if the substance involved were the acid or
the base. When KCl, HCl, and Ba(OH)2 are dissolved in water, the positive and negative ions dissociate. The ions can
then carry an electrical current. The electronegativity differences between the constituent atoms of C2H5OH result in
polar covalent bonds — C2H5OH is a polar molecule. With C2H5OH there are no positive and negative ions to dissociate, so the solution does not conduct electricity and could not have been involved in an electrocution.
[ K] + [ Cl ] –
[ H] + [ Cl ] –
[ Ba] 2+
H
2 [OH ]
–
H
H — C — C — OH
H
H
Making Connections
23. (a) The student is to use the Internet to determine the IUPAC name and chemical formula for sal volatile. Ammonium
carbonate: (NH4)2CO3
(b) The student is to predict the properties of the substance based upon the nature of the bonds. It is an ionic
compound of the NH+4 and CO23 ions. Students would predict that it is a solid, white, hard, and is highly soluble
in water (hydrogen bonding). It is actually volatile at room temperature and even more so upon slight heating,
releasing ammonia, water, and carbon dioxide.
(c) The student is to explain how the properties of the substance relate to the strong, sharp, ammonia smell. The
release of ammonia (NH3(g)) is responsible for the odour.
(d) The student is to use the Internet to find out how smelling salts used to be administered, how they worked, and
comment on the safety concerns that this use would raise today. A bottle containing the salt, or the salt in an
ammonia–water solution, was waved under the nose of the unconscious person. Human noses are extremely
sensitive to ammonia, because it is relatively common in nature and also toxic. If our noses detect the smell of
toxic chemicals, this information, vitally important for survival, can bypass the temporary shutdown that is characteristic of some forms of unconsciousness and “shock” the victim into consciousness. Ammonia dissolves
readily in water, forming a basic solution that could damage the lining of the respiratory tract and the lungs.
(Unconsciousness should also be respected. It is a defence of the body — shutting down consciousness prevents
the brain, in an addled state caused by oxygen or nutrient deprivation or severe bodily trauma, from instructing
the body to do something that endangers survival (like moving, for example). It also reduces the metabolic
requirements of the brain in a time of shortage, preventing cell death.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
24. The student is to use the Internet to research the structure of boron nitride. Crystals with a hexagonal, graphite lattice
are the most common form of boron nitride.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
50
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
25. The student is to use the Internet to research the foods that use glycerol, the physical and chemical properties of
glycerol, the chemical formula and molecular shape of glycerol, and to use the knowledge of the molecular shape and
intra- and intermolecular bonding to explain at least three of the properties of glycerol. The student is also to write a
short report, outlining why glycerol is so useful to the food industry.
(a) In addition to the uses mentioned in the question, it is added to tobacco to keep it moist, and also to fruits, for the
same purpose.
(b) Glycerol is also known as propanetriol (IUPAC), glycyl alcohol, and glycerine. At SATP it is a viscous liquid with
a sweet taste. It dissolves readily in water (antifreeze). It reacts with a rather large range of organic substances,
and with acids to produce esters.
(c) CH2OHCHOHCH2OH; the molecule is polar and forms hydrogen bonds (explaining its solubility in water and
its usefulness as a moistening agent, and also its viscosity in pure form).
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
Exploring
26. The student is to use the Internet to research Fritz London’s work on intermolecular forces, and to write a brief report
on his major findings. Dr. London (1900–1954) was born in what is now Wroclaw in Poland but was at that time
Breslau in Germany. He emigrated to the United States in 1939 and became a citizen in 1945. A physicist and theoretical chemist, he worked on superconductivity, superfluids, and quantum chemistry (and is, of course, the London
after whom London forces were named).
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
Copyright © 2002 Nelson Thomson Learning
Chapter 2 Chemical Bonding
51
CHAPTER 3
CHEMICAL REACTIONS
Reflect on your Learning
(Page 106)
1. Clues that indicate that a chemical reaction has taken place include: a change in colour, a change in odour, formation
of a gas/solid, release/absorption of heat.
2. Combustion, synthesis, decomposition, single displacement and double displacement reactions.
3. Whether or not one element displaces another in a compound depends on the relative reactivities of the two elements.
The collision–reaction theory suggests that particles must collide with the correct orientation and at sufficient speed
to react.
Try This Activity: Observing Chemical Change
(Page 107)
(a) A change in colour is evidence that a chemical change has taken place.
(b) Iron(II) sulfate and elemental copper
3.1 RECOGNIZING AND UNDERSTANDING CHEMICAL CHANGES
Try This Activity: The KMT in Action
(Page 109)
(a) The food colouring starts as a small drop, but soon begins to spread throughout the beaker of tap water.
(b) According to kinetic molecular theory, the motion of all particles results in random collisions. The food colouring
particles collide with each other and with molecules of water, bounce off in different directions, and thus spread
throughout the beaker of tap water.
Try This Activity: A Model for the Collision–Reaction Theory
(Page 110)
(a) When the box is gently shaken, the particles exhibit the continuous random motion of all particles of matter.
(b) The student is to count the number of free and combined particles, and to indicate the proportion of particles that
have formed a “compound.”
(c) The vigorous shaking of the box represents the effect of higher temperature — greater speeds of the motion of
particles.
(d) The student is to again count the number of free and combined particles, and to indicate the proportion of particles that have formed a compound. It is expected that a greater number of compounds will have formed.
(e) All of the particles may, or may not, have reacted.
(f) The collision–reaction theory suggests that particles must collide with the correct orientation and at sufficient
speed to react. It is possible that there may not have been the correct orientation, and/or sufficient speed, for all
particles to react.
(g) A lower number of particles within the same space should result in fewer collisions and a lower proportion of
particles forming a compound.
PRACTICE
(Page 111)
Understanding Concepts
1. The formation of gas is evidence that a chemical reaction has occurred.
2. When gasoline evaporates (a physical change) it absorbs heat from its surroundings. A visual observation of this
absorption of heat might lead to a conclusion that a chemical change has taken place. Thus it is important not to rely
solely on visual observation when drawing conclusions about chemical change.
3. According to kinetic molecular theory, the motion of all particles results in random collisions. At lower temperatures,
the colliding molecules of liquid gasoline and oxygen simply bounce off one another unchanged. However, at higher
Copyright © 2002 Nelson Thomson Learning
Chapter 3 Chemical Reactions
53
temperatures, the molecules of gaseous gasoline and oxygen collide at greater speeds and with greater energy. In some
collisions, the valence shells of the reactants overlap, and their electrons can be rearranged to form new bonds.
4. According to the collision–reaction theory, the occurrence of a chemical reaction is dependent on the energy and
orientation of the collisions.
ACTIVITY 3.1.1
UNDERSTANDING CHEMICAL REACTIONS
(Page 112)
Analysis
(a) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
2 H2O(l) → 2 H2(g) + O2(g)
C(s) + O2(g) → CO2(g)
Synthesis
(b) The law of conservation of mass.
EXPLORE AN ISSUE
TAKE A STAND: CATALYTIC CONVERTERS
(Page 113)
The student is to use the Internet to carry out research on catalytic converters, to assemble arguments for and against
making catalytic converters compulsory in all vehicles, and to put together a presentation — pamphlet or video — aimed
at people who are about to purchase a used vehicle.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre
PRACTICE
(Page 113)
Understanding Concepts
5. (a) Change in colour, change in odour, formation of gas/solid, release/absorption of heat.
(b) Change in colour: iron nails that rust turn brownish in colour.
Change in odour: food that decomposes gives off an odour.
Formation of gas/solid: seltzer tablets added to water give off a gas.
Release/absorption of heat: wood that burns gives off heat.
6. (a) Reactant: the substances that combine in a chemical reaction.
(b) Product: the substances that are formed in a chemical reaction.
(c) Coefficient: a whole number indicating the ratio of formula units of each substance involved in a chemical
reaction.
(d) Balanced: the reactants and products contain equal numbers of atoms of each type.
7. (a) 2 H2(g) + O2(g) → 2 H2O(g)
(c) Pb(s) + 2 AgNO3(aq) → 2 Ag(s) + Pb(NO3)2(aq)
(b) and (d) are balanced.
SECTION 3.1 QUESTIONS
(Page 113)
Understanding Concepts
1. (a) sodium chloride and water → chlorine and hydrogen and sodium hydroxide
(b) 2 NaCl(aq) + 2 H2O(l) → Cl2(s) + H2(s) + 2 NaOH(aq)
(c) The reaction takes place in a sealed container because chlorine is a poisonous gas at SATP.
(d) The formation of gases — Cl2(s) + H2(s)
(e) NaCl(aq) and NaOH(aq) are ionic compounds, H2O(l) is a molecular compound, Cl2(s) + H2(s) are molecular
elements.
(f) H2O(l) can be classified as a polar molecule.
(g) Chlorine compounds have different chemical properties than does chlorine gas.
54
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
3.2 COMBUSTION, SYNTHESIS, AND DECOMPOSITION REACTIONS
PRACTICE
(Page 117)
Understanding Concepts
1. Reactants: a substance for fuel, and oxygen; conditions: three things must be present - fuel, oxygen, and heat; products: common oxides of the elements making up the substance that is burned.
2. CO2(g); H2O(g); SO2(g); NO2(g); Fe2O3(s)
3. Carbon dioxide and water are products of the combustion of carbon compounds that have had a significant effect on
the atmosphere, including making it warmer than it would otherwise be. This phenomenon is known as the “greenhouse effect.”
Gaseous oxides of nitrogen and sulfur are released from sources such as automobiles and coal-burning power plants.
These oxides join the naturally produced oxides in the atmosphere, react with water vapour to form acids, and are
responsible for the increased acidity of precipitation known as “acid rain.”
PRACTICE
(Page 121)
Understanding Concepts
4. (a) Decomposition — the product is a simpler elemental compound and an element.
(b) Synthesis — the product is a more complex compound.
(c) Synthesis — the product is a more complex compound.
5. (a) 2 Al(s) + 3 F2(g) → 2 AlF3(s)
synthesis
6.
(b) 2 KCl(s) → 2 K(s) + Cl2(g)
decomposition
(c) S8(s) + 8 O2(g) → 8 SO2(g)
synthesis (combustion)
(d) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
combustion
(e) 2 Al2O3(s) → 4 Al(s) + 3 O2(g)
decomposition
(f) 2 Hg(l) + O2(g) → 2 HgO(s)
synthesis (combustion)
(g) 2 FeBr3(s) → 2 Fe(s) + 3 Br2(g)
decomposition
Reaction Type
Reactants
Synthesis
element + element
element + compound
compound + compound
Decomposition
binary compound
complex compound
Combustion
metal + oxygen
nonmetal + oxygen
fossil fuel
7. (a) 2 Na(s) + Cl2(g) → 2 NaCl(s)
synthesis
(b) 2 CuO(s) → 2 Cu(s) + O2(g)
decomposition
(c) Cu(s) + Cl2(aq) → CuCl2(s)
synthesis
8. (a) 2 Li2O(s) → 4 Li(s) + O2(g)
2 MgO(s) → 2 Mg(s) + O2(g)
ZnCl2(s) → Zn(s) + Cl2(g)
Mg(OH)2(s) → MgO(s) + H2O(l)
Copyright © 2002 Nelson Thomson Learning
Chapter 3 Chemical Reactions
55
9. Evaluation
(a) The experimental design is not sufficient to identify all of the elements of the white powder. However, there is
sufficient evidence to indicate that the white powder could be a metal carbonate of some sort. Carbonates will
decompose when heated to produce a metal oxide and carbon dioxide gas.
To identify the elements of the metal oxide, the melting point of the white powder could be determined and
matched against known melting points of ionic solids.
To identify the gas, the gas could be bubbled into a limewater solution. Carbon dioxide gas is indicated if
limewater solution turns milky. The limewater undergoes a chemical change to form an insoluble white precipitate. Carbon dioxide gas is also indicated if a flaming splint, held at the mouth of the test tube, is extinguished.
Synthesis
(b) A thermal decomposition reaction of the general form AB → A + B
(c) Calcium carbonate → calcium oxide + carbon dioxide
CaCO3(s) → CaO(s) + CO2(g)
10. The student is to use the Internet to investigate a vehicle exhaust pollutant, and to describe the chemical reactions that
produce it and its harmful effects on living organisms. Also, the student is to develop a list of suggestions for reducing
the production of this pollutant, and to discuss the pros and cons of implementing these suggestions from the perspective of a car manufacturer and from the perspective of an environmentalist.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre
11. The student is to use the Internet to investigate the properties of diesel fuel and gasoline, and the products of their
combustion. Also, the student is to determine which fuel poses the greater threat to the environment.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre
12. The student is to use the Internet to research the chemical compositions of natural gas and propane, for use as alternative fuels for vehicles. Also, the student is to compare the products of combustion with those of gasoline.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre
SECTION 3.2 QUESTIONS
(Page 123)
Understanding Concepts
1. (a) The warning signs at gas stations are there to insure that people refrain from any activity that could result in the
occurrence of a gasoline-related chemical reaction, such as a fire or an explosion.
(b) A combustion reaction is the concern.
(c) The necessary conditions for this reaction to proceed are fuel, oxygen, and heat.
2. (a) Butane — combustion reaction. Butane is a fuel.
Butane + oxygen → carbon dioxide + water
2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)
Calcium carbonate — thermal decomposition reaction. Carbonates will decompose when heated to produce a metal
oxide and carbon dioxide gas.
Calcium carbonate → calcium oxide + carbon dioxide
CaCO3(s) → CaO(s) + CO2(g)
Li and Br — synthesis reaction. Two substances combine to form a single product.
Lithium + bromine → lithium bromide
Li(s) + Br(l) → LiBr(s)
Bluestone — decomposition reaction. Compounds consisting of more than two elements often decompose to
form simpler compounds. When a hydrated salt is heated, the products are the anhydrous salt and water.
Copper(II) sulfate pentahydrate → copper(II) sulfate + water
CuSO4· 5 H2O(s) → CuSO4(s) + 5 H2O(g)
Making Connections
3. (a) 2 C8H18(g) + 25 O2(g) → 16 CO2(g) + 18 H2O(g) — a combustion reaction.
Gasoline often contains trace amounts of elemental sulfur, and when the gas is burned, the sulfur combines with
oxygen to produce sulfur dioxide.
56
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
S(s) + O2(g) → SO2(g)
synthesis (combustion)
2 SO2(g) + O2(g) → 2 SO3(g)
synthesis (combustion)
SO3(g) + H2O(l) → H2SO4(aq)
synthesis
(b) Sulfur trioxide is a byproduct of the combustion of gasoline in car engines. In the atmosphere it reacts with
condensed water on dust particles, producing sulfuric acid. Atmospheric sulfuric acid is one of the acids that is
responsible for the increased acidity of precipitation known as acid rain.
4. (a) Nitrogen monoxide decomposes into nitrogen and oxygen.
Pt/Pd
2 NO(g) → N2(g) + O2(g)
(b) The platinum/palladium catalytic converter built into today’s automobiles catalyzes the decomposition of
nitrogen monoxide — which is a combustion engine exhaust pollutant — into harmless nitrogen and oxygen.
3.3 SINGLE DISPLACEMENT REACTIONS
PRACTICE
(Page 128)
Understanding Concepts
1. In single displacement reactions, like displaces like – a metallic element takes the place of a metal in a compound; a
nonmetallic element takes the place of a nonmetal in a compound.
2. An “activity series” is a list of elements arranged in order of their reactivity, based upon empirical evidence gathered
from single displacement reactions.
3. (a) Zn(s) + CuCl2(aq) → Cu(s) + ZnCl2(aq)
(b) Br2(aq) + CaCl2(aq) → NR
(c) Pb(s) + 2 HCl(aq) → H2(g) + PbCl2(aq)
(d) Cl2(aq) + 2 NaI(aq) → 2 NaCl(aq) + I2(s)
(e) Ca(s) + 2 H2O(l) → H2(g) + Ca(OH)2(aq)
(f) Au(s) + ZnSO4(aq) → NR
(g) Sn(s) + 2 AgNO3(aq) → 2 Ag(s) + Sn(NO3)2(aq)
(h) 2 Al(s) + 3 H2O(l) → 3 H2(g) + Al2O3(aq)
(i) Br2(aq) + MgI2(aq) → MgBr2(aq) + I2(s)
(j) 2 Al(s) + 3 ZnSO4(aq) → 3 Zn(s) + Al2(SO4)3(aq)
4. Generally speaking, the more reactive elements will replace the less reactive elements. Thus, within the metal group,
the more reactive metal elements are the ones with low electronegativity values and they will replace metal elements
with higher electronegativity values. For example, lithium has an electronegativity value of 1.0 and will replace potassium, which has an electronegativity value of 0.8. (It should be noted that there are a number of exceptions to this
generalization.)
Within the nonmetal group, the more reactive nonmetal elements are the ones with high electronegativity values
and they will replace nonmetal elements with lower electronegativity values. For example, fluorine has an electronegativity value of 4.0 and will replace chlorine, which has an electronegativity value of 3.0. (Again, it should be
noted that there are a number of exceptions to this generalization.)
5. As you move from left to right within the same period, the elements of the periodic table show a general increase in
electronegativity values. The most reactive metals are the ones with lower electronegativity values and are positioned
at the left of a period. Thus, as you move from left to right within the same period, the metals become more electronegative and therefore less reactive. However, the most reactive nonmetals are the ones with high electronegativity
values and are positioned at the right of a period. Thus, as you move from left to right within the same period, the
nonmetals become more electronegative and therefore more reactive.
Copyright © 2002 Nelson Thomson Learning
Chapter 3 Chemical Reactions
57
6. (a) The Goldschmidt process is a single displacement reaction.
(b) CrO(s) or Cr2O3(s), MnO(s), BaO(s)
(c) 2 Al(s) + Cr2O3(s) → 2 Cr(s) + Al2O3(s)
2 Al(s) + 3 MnO(s) → 3 Mn(s) + Al2O3(s)
2 Al(s) + 3 BaO(s) → NR
(According to the activity series, aluminum is less reactive than barium and should not displace it from barium oxide.
There should be no reaction.)
(d) The more reactive aluminum displaces the chromium and manganese from their oxides, producing the pure
metals of chromium and manganese, along with aluminum oxide.
7. (a) titanium dioxide + carbon + chlorine → titanium tetrachloride + carbon dioxide
magnesium + titanium tetrachloride → titanium + magnesium chloride
(b) 2 Mg(s) + TiCl4(s) → Ti(s) + 2 MgCl2(s)
single displacement reaction
(c) For the metals, the lower the electronegativity, the more reactive the metal should be.
Magnesium has a lower electronegativity value than titanium and is more reactive than titanium. Therefore
magnesium displaces titanium, producing pure titanium along with magnesium chloride.
PRACTICE
(Page 130)
Understanding Concepts
8. As a result of their greater reactivity, calcium and sodium are most commonly found in compounds, ores, rather than
as elements. In order to extract these two elements from their ores, they must be reacted with metals that are higher
in the activity series. However, according to the activity series, calcium and sodium are high in the activity series and
are two of the most reactive elements, so they would tend not to be displaced by other metals in an aqueous
environment.
Making Connections
9. (a) The student is to use the Internet to carry out research on a metallic element that is mined in Canada, with respect
to how and where it is mined, extracted, and purified, and the uses of the metal.
(b) The findings are to be compiled in a table that lists the positive and negative aspects. The table is to be used to
help the student decide on changes that could be made in the way we use the metal or the way we obtain it. The
student is to report on the findings.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre
PRACTICE
(Page 134)
Understanding Concepts
10. Yellow brass is an alloy that comprises 70% copper by mass, and 30% zinc by mass. It is harder and more resistant
to corrosion than copper alone.
Stainless steel is an alloy that comprises a number of metals, but mainly comprises 79.06 – 81.06% iron by mass,
and 16–18% chromium by mass. It is more resistant to corrosion than iron alone.
Making Connections
11. (a) The higher acidic nature of women’s skin causes some of the copper in the gold ring to dissolve, leaving deposits
of dissolved copper. Women wear 18K gold rings because there is less copper content by mass in the 18K gold,
thus reducing the amount of copper available to dissolve.
(b) The 14K gold alloy is a solid solution that consists of gold, silver, copper, and zinc. The greenish stain that
develops beneath a 14K gold bracelet is the result of acidic skin that has caused some of the copper in the gold
bracelet to dissolve back into the liquid phase. The green copper stain becomes visible when the wearer’s body
dissolves the copper faster than it absorbs it. This happens when there is profuse perspiration, or when our sweat
becomes more acidic.
58
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
12. The student is to use the Internet to research the composition of various steels and to choose one alloy and list its properties and its applications. The student is then to write a short “infomercial” advertising the benefits of this material
to potential users and to include any precautions necessary for its safe use.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre
13. The student is to use the Internet to research the applications of aluminum and its alloys and the environmental issues
surrounding aluminum production. The student is to use the findings to comment on the following statement: Risks
to the environment posed by mining and refining aluminum are outweighed by the technological benefits of aluminum
alloys.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre
PRACTICE
(Page 135)
Making Connections
14. The student is to research one of the careers listed on page 135 or a related career, and write a report that:
(a) provides a general description of the nature of the work and how chemical reactions are involved;
(b) describes the educational background and the length of study required to obtain employment in this field;
(c) gives examples of programs offered by educational institutions leading to this career;
(d) forecasts employment trends for this field; and
(e) describes working conditions and salary.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre
3.4 DOUBLE DISPLACEMENT REACTIONS
PRACTICE
(Page 138)
Understanding Concepts
1. The following monatomic ions form compounds that have high solubility in water:
Group 1 monatomic ions that form compounds with Cl—, Br—, and I—.
–
Group 1 and Group 2 monatomic ions that form compounds with S2 .
+
2. NH4 is the positive polyatomic ion that forms compounds that all have high solubility in water.
3. (a) KCl(aq)
(i) Fe(OH)3(s)
(b) Ca(NO3)2(aq)
(j) PbSO4(s)
(c) Na2SO4(aq)
(k) Ca3(PO4)2(s)
(d) AgC2H3O2(s)
(l) KMnO4(aq)
(e) NH4Br(aq)
(m) NH4NO3(aq)
(f) BaS(aq)
(n) CoCl2(aq)
(g) PbI2(s)
(o) CaCO3(aq)
(h) Ca(OH)2(s)
PRACTICE
(Page 141)
Understanding Concepts
4. The reaction involves two ionic compounds as reactants.
5. (a) Cu(NO3)2(aq) + MgCl2(aq) → CuCl2(s) + Mg(NO3)2(aq)
(b) 3 Ba(OH)2(aq) + Fe2(SO4)3(aq) → 3 BaSO4(s) + 2 Fe(OH)3(s)
(c) Mg(OH)2(s) + H2SO4(aq) → MgSO4(aq) + 2 H2O(l)
(d) (NH4)2S(aq) + FeSO4(aq) → (NH4)2SO4(aq) + FeS(s)
Copyright © 2002 Nelson Thomson Learning
Chapter 3 Chemical Reactions
59
6. AB + CD → AD + CB
As the equation above shows, the ions “change partners” to form products. This type of reaction commonly occurs in
aqueous solutions.
7. (a) KNO3(aq)
(b) CaCl2(aq)
(c) Mg(OH)2(s)
(d) Al2(SO4)3(aq)
(e) PbI2(s)
(f) Ca3(PO4)2(s)
(g) (NH4)2CO3(aq)
8. (a) 2 KCl(aq) + Cu(NO3)2 (aq) → 2 KNO3(aq) + CuCl2(s)
(b) MgCl2(aq) + Ca(OH)2(aq) → Mg(OH)2(s) + CaCl2(aq)
(c) MgCl2(aq) + Ca(OH)2(aq) → CaCl2(aq) + Mg(OH)2(s)
(d) 3 K2SO4(aq) + 2 AlCl3(aq) → 6 KCl(aq) + Al2(SO4)3(aq)
(e) CuI2(s) + PbSO4(s) → CuSO4(aq) + PbI2(s)
(f) 3 CaI2(aq) + Pb3(PO4)2(s) → 3 PbI2(s) + Ca3(PO4)2(s)
(g) (NH4)2S(aq) + CaCO3(s) → CaS(aq) + (NH4)2CO3(aq)
9. (a) single displacement
(b) double displacement
10. (a) Cl2(g) + 2 NaBr(aq) → 2 NaCl(aq) + Br2(g)
single displacement
(b) H2SO4(aq) + 2 NaOH(aq) → 2 H2O(l) + Na2SO4(aq)
double displacement
(c) 3 Ca(NO3)2(aq) + 2 Na3PO4(aq) → Ca3(PO4)2(s) + 6 NaNO3(aq)
double displacement
11. (a) Al(s) + 3 AgNO3(aq) → 3 Ag(s) + Al(NO)3(aq)
single displacement
(b) Cl2(g) + 2 NaBr(aq) → 2 NaCl(aq) + Br2(g)
single displacement
(c) Zn(s) + H2SO4(aq) → H2(g) + ZnSO4(aq)
single displacement
(d) 2 AgNO3(aq) + MgCl2(aq) → 2 AgCl(s) + Mg(NO3)2(aq)
double displacement
(e) Na2C2O4(aq) + CaCl2(aq) → CaC2O4(aq) + 2 NaCl(aq)
double displacement
(f) 2 Na(s) + 2 H2O(l) → H2(g) + 2 NaOH(aq)
single displacement
(g) 3 KOH(aq) + FeCl3(aq) → 3 KCl(aq) + Fe(OH)3(s)
double displacement
12. (a) Ca3(PO4)2(s)
calcium phosphate
(b) 3 H2SO4(aq) + Ca3(PO4)2(s) → 2 H3PO4(aq) + 3 CaSO4(s)
(c) The simple procedure of “filtering” could be employed to isolate aqueous phosphoric acid from the solid calcium
phosphate.
(d) If sodium phosphate is used in place of calcium phosphate as a reactant, the products of the reaction would be
aqueous phosphoric acid and aqueous sodium sulfate. Sodium sulfate is highly soluble at SATP, and would pass
through a simple filter. Thus the simple procedure of filtering would not be sufficient to isolate the phosphoric
acid. The fertilizer manufacturer would prefer to avoid the additional costs that would be associated with a more
involved process of isolating the phosphoric acid.
60
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
Copyright © 2002 Nelson Thomson Learning
Pb2+
Cu+
Al3+
Ca2+
K+
lead(II)
chloride,
binary ionic,
sdr with zinc
will occur
PbBr2(s)
lead(II)
bromide,
binary ionic,
sdr with zinc
will occur
copper(I)
bromide,
binary ionic,
sdr with zinc
will occur
PbCl2(s)
CuBr
CuCl
AlBr3
aluminum
bromide,
binary ionic,
sdr with zinc
would not
occur
CaBr2
calcium
bromide,
binary ionic,
sdr with zinc
would not
occur
copper(I)
chloride,
binary ionic,
sdr with zinc
will occur
aluminum
chloride,
binary ionic,
sdr with zinc
would not
occur
AlCl3
calcium
chloride,
binary ionic,
sdr with zinc
would not
occur
CaCl2
KBr
potassium
bromide,
binary ionic,
sdr with zinc
would not
occur
potassium
chloride,
binary ionic,
sdr with zinc
would not
occur
Br-
KCl
Cl-
Pb(NO3)2
lead(II)
nitrate,
tertiary
ionic,
sdr with zinc
will occur
CuNO3
copper(I)
nitrate,
tertiary
ionic,
sdr with zinc
will occur
Al(NO3)3
aluminum
nitrate,
tertiary
ionic,
sdr with zinc
would not
occur
Ca(NO3)2
calcium
nitrate,
tertiary
ionic,
sdr with zinc
would not
occur
KNO3
potassium
nitrate,
tertiary
ionic,
sdr with zinc
would not
occur
NO3
Pb(C2H3O2)2
lead(II)
acetate,
quaternary
ionic,
sdr with zinc
will occur
CuC2H3O2
copper(I)
acetate,
quaternary
ionic,
sdr with zinc
will occur
Al(C2H3O2)3
aluminum
acetate,
quaternary
ionic,
sdr with zinc
would not
occur
Ca(C2H3O2)2
calcium
acetate,
quaternary
ionic,
sdr with zinc
would not
occur
KC2H3O2
potassium
acetate,
quaternary
ionic,
sdr with zinc
would not
occur
C2H3O2
Pb3(PO4)2(s)
lead(II)
phosphate,
tertiary
ionic,
sdr with zinc
will occur
Cu3PO4(s)
copper(I)
phosphate,
tertiary
ionic,
sdr with zinc
will occur
AlPO4(s)
aluminum
phosphate,
tertiary
ionic,
sdr with zinc
would not
occur
Ca3(PO4)2(s)
calcium
phosphate,
tertiary
ionic,
sdr with zinc
would not
occur
K3PO4
potassium
phosphate,
tertiary
ionic,
sdr with zinc
would not
occur
PO 3
4
PbSO4(s)
lead(II)
sulfate,
tertiary
ionic,
sdr with zinc
will occur
Cu2SO4
copper(I)
sulfate,
tertiary
ionic,
sdr with zinc
will occur
Al2(SO4)3
aluminum
sulfate,
tertiary
ionic,
sdr with zinc
would not
occur
CaSO4(s)
calcium
sulfate,
tertiary
ionic,
sdr with zinc
would not
occur
K2SO4
potassium
sulfate,
tertiary
ionic,
sdr with zinc
would not
occur
SO 2
4
Pb(OH)2(s)
lead(II)
hydroxide,
tertiary
ionic,
sdr with zinc
will occur
CuOH(s)
copper(I)
hydroxide,
tertiary
ionic,
sdr with zinc
will occur
Al(OH)3(s)
aluminum
hydroxide,
tertiary
ionic,
sdr with zinc
would not
occur
Ca(OH)2(s)
calcium
hydroxide,
tertiary
ionic,
sdr with zinc
would not
occur
potassium
hydroxide,
tertiary
ionic,
sdr with zinc
would not
occur
KOH
OH-
PbSO3(s)
lead(II)
sulfite,
tertiary
ionic,
sdr with zinc
will occur
Cu2SO3(s)
copper(I)
sulfite,
tertiary
ionic,
sdr with zinc
will occur
Al2(SO3)3(s)
aluminum
sulfite,
tertiary
ionic,
sdr with zinc
would not
occur
CaSO3(s)
calcium
sulfite,
tertiary
ionic,
sdr with zinc
would not
occur
K2SO3
potassium
sulfite,
tertiary
ionic,
sdr with zinc
would not
occur
SO 23
SECTIONS 3.3-3.4 QUESTIONS
(Page 143)
Understanding Concepts
1. (a)-(e)
Chapter 3 Chemical Reactions
61
Word and chemical equation examples:
zinc + potassium chloride → NR
Zn(s) + KCl(aq) → NR
zinc + iron(III) acetate → iron + zinc acetate
3 Zn(s) + 2 Fe(C2H3O2)3(aq) → 2 Fe(s) + 3 Zn(C2H3O2)2(aq)
62
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
NOTE:
+
NH4
Mg2+
Fe3+
NH4Br
ammonium
bromide,
tertiary ionic,
sdr with zinc
will occur
MgBr2
magnesium
bromide,
binary ionic,
sdr with zinc
will occur
FeBr3
iron(III)
bromide,
binary ionic,
sdr with zinc
will occur
NH4NO3
ammonium
nitrate,
tertiary
ionic,
sdr with zinc
will occur
Mg(NO3)2
magnesium
nitrate,
tertiary
ionic,
sdr with zinc
will occur
Fe(NO3)3
iron(III)
nitrate,
tertiary
ionic,
sdr with zinc
will occur
All substances aqueous (aq), unless noted otherwise.
sdr stands for “single displacement reaction.”
ammonium
chloride,
tertiary ionic,
sdr with zinc
will occur
NH4Cl
magnesium
chloride,
binary ionic,
sdr with zinc
will occur
MgCl2
FeCl3
iron(III)
chloride,
binary ionic,
sdr with zinc
will occur
NH4C2H3O2
ammonium
acetate,
quaternary
ionic,
sdr with zinc
will occur
Mg(C2H3O2)2
magnesium
acetate,
quaternary
ionic,
sdr with zinc
will occur
Fe(C2H3O2)3
iron(III)
acetate,
quaternary
ionic,
sdr with zinc
will occur
(NH4)3PO4
ammonium
phosphate,
quaternary
ionic,
sdr with zinc
will occur
Mg3(PO4)2(s)
magnesium
phosphate,
tertiary
ionic,
sdr with zinc
will occur
FePO4(s)
iron(III)
phosphate,
tertiary
ionic,
sdr with zinc
will occur
(NH4)2SO4
ammonium
sulfate,
quaternary
ionic,
sdr with zinc
will occur
MgSO4
magnesium
sulfate,
tertiary
ionic,
sdr with zinc
will occur
Fe2(SO4)3
iron(III)
sulfate,
tertiary
ionic,
sdr with zinc
will occur
NH4OH
ammonium
hydroxide,
tertiary
ionic,
sdr with zinc
will occur
Mg(OH)2(s)
magnesium
hydroxide,
tertiary
ionic,
sdr with zinc
will occur
Fe(OH)3(s)
iron(III)
hydroxide,
tertiary
ionic,
sdr with zinc
will occur
(NH4)2SO3
ammonium
sulfite,
quaternary
ionic,
sdr with zinc
will occur
MgSO3(s)
magnesium
sulfite,
tertiary
ionic,
sdr with zinc
will occur
Fe2(SO3)3(s)
iron(III)
sulfite,
tertiary
ionic,
sdr with zinc
will occur
2.
1. octane + oxygen → carbon dioxide + water
2 C8H18(g) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)
combustion
Rain becomes slightly acidic due to the presence of carbon dioxide, which dissolves in atmospheric moisture and
reacts to form very dilute carbonic acid, H2CO3(aq). Carbon dioxide gas also contributes to the greenhouse effect,
which may lead to global warming and dramatic climate changes.
2. sulfur + oxygen → sulfur dioxide
synthesis (combustion)
S(s) + O2(g) → SO2(g)
Sulfur dioxide joins the naturally produced oxides in the atmosphere, and reacts with water vapour to form acids, and
is responsible for the increased acidity of precipitation known as acid rain.
3. nitrogen + oxygen → nitrogen monoxide
synthesis (combustion)
N2(s) + O2(g) → 2 NO(g)
Nitrogen monoxide joins the naturally produced oxides in the atmosphere, and reacts with water vapour to form acids,
and is responsible for the increased acidity of precipitation known as acid rain.
4. nitrogen monoxide + oxygen → nitrogen dioxide
synthesis
Pt
2 NO(s) + O2(g) → 2 NO(g)
Nitrogen dioxide joins the naturally produced oxides in the atmosphere, and reacts with water vapour to form acids,
and is responsible for the increased acidity of precipitation known as acid rain.
5. sulfur trioxide + water → sulfuric acid
synthesis
SO3(s) + H2O(l) → H2SO4(aq)
Sulfur trioxide reacts with water vapour to form sulfuric acid, and is responsible for the increased acidity of precipitation known as acid rain.
6. nitrogen monoxide → nitrogen + oxygen
decomposition
Pt/Pd
2 NO(g) → N2(g) + O2(g)
The platinum/palladium catalytic converter built into today’s automobiles catalyzes the decomposition of nitrogen
monoxide — which is a combustion engine exhaust pollutant — into harmless nitrogen and oxygen.
Applying Inquiry Skills
3. Analysis
(a) Y would be highest in the activity series, followed by Z, and then X, as indicated below:
Y
Z
X
(b) Y and Z would be located higher in the activity series than hydrogen. X would be located lower in the activity
series than hydrogen, unless in fact it is actually hydrogen.
Using empirical evidence gathered in many experiments, scientists have been able to list the elements in order
of their reactivity. The empirical evidence has shown that the more reactive element will replace the less reactive
element. Since Y and Z showed evidence of reaction in water and acid - bubbles were observed - it is evident that
Y and Z have reacted with the water and the acid to displace hydrogen and produce hydrogen gas. And since Y
and Z displaced hydrogen, they cannot be hydrogen. The only possible choice for hydrogen would be X.
Synthesis
(c) X would appear higher in order than Y, as you descend the group.
For metals, the lower the electronegativity value, the more reactive the metal should be. As the evidence
showed Y to be more reactive than X, it would be expected to have a lower electronegativity value than X. And
since electronegativity values decrease as you move down a group, it makes sense that X would appear higher in
order than Y, as you descend the group.
Y would appear to be further to the left in order than Z, as you move from left to right within the same
period.
For metals, the lower the electronegativity value, the more reactive the metal should be. As the evidence
showed Y to be more reactive than Z, it would be expected to have a lower electronegativity value than Z. And
since electronegativity values increase as you move from left to right within the same period, it makes sense that
Y would appear further to the left in order than Z, as you move from left to right within the same period.
Copyright © 2002 Nelson Thomson Learning
Chapter 3 Chemical Reactions
63
(d) Assign the following element identities:
X = Li
Y = Na
Z = Mg
2 Li(s) + 2 H2O(l) → H2(g) + 2 LiOH(aq)
2 Li(s) + 2 HCl(aq) → H2(g) + 2 LiCl(aq)
2 Na(s) + 2 H2O(l) → H2(g) + 2 NaOH(aq)
2 Na(s) + 2 HCl(aq) → H2(g) + 2 NaCl(aq)
Mg(s) + 2 H2O(l) → H2(g) + Mg(OH)2(s)
Mg(s) + 2 HCl(aq) → H2(g) + MgCl2(aq)
(e) All reactions are single displacement reactions.
With respect to the reaction with water:
Use a lighted splint to ignite the gas produced. A “pop” sound when ignited indicates hydrogen gas. The aqueous
solution could be tested with pH paper to determine if it is basic. These tests would indicate that the reaction was a
single displacement type, with the metal displacing hydrogen in the water (H2O(l)) to produce hydrogen gas and a
metal hydroxide.
With respect to the reaction with acid:
Use a lighted splint to ignite the gas produced. A “pop” sound when ignited indicates hydrogen gas. The aqueous
solution could be evaporated and the remaining compound analyzed to determine if it is a metal chloride. These tests
would indicate that the reaction was a single displacement type, with the metal displacing hydrogen in the acid
(HCl(aq)) to produce hydrogen gas and a metal chloride.
Making Connections
Pt
4. (a) 2K(s) + BeCl2(s) → Be(s) + 2 KCl(l)
single displacement
(b) Beryllium chloride should be stored dry within air and water vapour tight sealed containers. The substance should
only be handled in environmentally enclosed and controlled areas, and all required protective equipment such as
gas masks, protective clothing, and protective gloves should be worn.
5. (a) C(s) + SiO2(l) → Si(l) + CO2(g) single displacement
Si(s) + 2 Cl2(g) → SiCl4(l)
synthesis
2 Mg(s) + SiCl4(l) → Si(s) + 2 MgCl2(aq)
single displacement
(b) Emissions of carbon dioxide gas (a product of the first reaction) cause rain to become slightly acidic. Carbon
dioxide gas also contributes to the greenhouse effect, which may lead to global warming and dramatic climate
changes.
CHAPTER 3 REVIEW
(Page 146)
Understanding Concepts
1. The kinetic molecular theory states that all matter is made up of particles in continuous random motion.
A gas has widely separated molecules in constant, chaotic motion. The average kinetic energy of the molecules
is much larger than the energy associated with the attractive forces between them.
With liquids, the attractive forces between molecules have energies comparable to the kinetic energies of the
molecules. The attractive forces are able to hold the molecules close to each other. However, the attractive forces are
not strong enough to hold the molecules rigidly in place. In fact, molecules within a liquid are able to move in a more
or less chaotic fashion, allowing liquids to be poured, and to flow to take the shape of their container.
With solids, the intermolecular attractions are sufficiently strong enough to hold the molecules rigidly in place.
The average kinetic energy of the molecules is much smaller than the energy associated with the attractive forces
between them. The particles of a solid are not free to move. However, the molecules within a solid may undergo vibrational motion.
64
Unit 1 Matter and Chemical Bonding
Copyright © 2002 Nelson Thomson Learning
2. The collision–reaction theory suggests that particles must collide with the correct orientation and at sufficient speed
to react. Chemical reactions will not occur when the collision orientation is not correct, and/or when the speed of
collision is not sufficient.
3. (a) S8 + 8 O2 → 8 SO2
synthesis (combustion)
(b) HBr + NaOH → NaBr + H2O
double displacement
(c) N2 + 3 H2 → 2 NH3
synthesis
(d) PtCl4 → Pt + 2 Cl
decomposition
(e) 2 MgO + Si → 2 Mg + SiO2
single displacement
(f) Na2S + 2 HCl → 2 NaCl + H2S
double displacement
(g) P4 + 5 O2 → P4 O10
synthesis (combustion)
(h) Zn + 2 HCl → ZnCl2 + H2
single displacement
4. (a) potassium chlorate → potassium chloride + oxygen
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
(decomposition)
(b) sodium + water (HOH) → hydrogen + sodium hydroxide
2 Na(s) + 2 H2O(l) → H2(g) + 2 NaOH
(single displacement)
(c) carbon + oxygen → carbon dioxide
C(s) + O2(g) → CO2(g)
(combustion)
(d) zinc + sulfuric acid → zinc sulfate + hydrogen
Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)
(single displacement)
(e) silver nitrate + potassium iodide → silver iodide + potassium nitrate
AgNO3(aq) + KI(aq) → AgI(s) + KNO3(aq)
(double displacement)
(f) sodium sulfate + barium chloride → barium sulfate + sodium chloride
Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2 NaCl(aq)
(double displacement)
(g) iron + oxygen → iron(III) oxide
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
(combustion/synthesis)
(h) sulfur trioxide + water → sulfuric acid
SO3(g) + H2O(g) → H2SO4(aq)
(synthesis)
5. (c) calcium sulfate; (h) silver iodide; (i) copper(I) chloride; (j) lead(II) sulfate are all of low solubility
6. Na2SO4(aq) + Ca(NO3)2(aq) → 2 NaNO3(aq) + CaSO4(s)
KI(aq) + AgNO3(aq) → KNO3(aq) + AgI(s)
NaCl(aq) + CuNO3(aq) → NaNO3(aq) + CuCl(s)
Na2SO4(aq) + Pb(NO3)2(aq) → 2 NaNO3(aq) + PbSO4(s)
7. By referring to the relevant activity series. A metal will be displaced by a metal above it in the series; a nonmetal will
be similarly displaced by a nonmetal.
8. (a) 2 Li(s) + 2 H2O(l) → H2(g) + 2 LiOH(aq)
(b) 2 K(s) + 2 H2O(l) → H2(g) + 2 KOH(aq)
(c) Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)
(d) Fe(s) + NaCl(aq) → NR
(e) Mg(s) + Ca(NO3)2(aq) → NR
(f) 2 Al(s) + 6 HCl(aq) → 3 H2(g) + 2 AlCl3(aq)
(g) Pb(s) + Cu(NO3)2(aq) → Cu(s) + Pb(NO3)2(aq)
(h) F2(g) + 2 HCl(aq) → Cl2(g) + 2 HF(aq)
(i) I2(s) + NaBr(aq) → NR
9. (a) single displacement
(b) solid iodine and potassium chloride
Copyright © 2002 Nelson Thomson Learning
Chapter 3 Chemical Reactions
65
(c) Cl2(g) + 2 KI(aq) → I2(s) + 2 KCl(aq)
(d) Chlorine, being more electronegative, displaces the iodide ion.
10. (a) Assume their elemental states: fluorine, bromine, nitrogen. Nitrogen gas is unreactive; fluorine is more reactive
than bromine as it has a higher electronegativity.
(b) rubidium, potassium, magnesium. Rubidium is less electronegative than potassium, which is more reactive than
magnesium.
Applying Inquiry Skills
11. (a) 2 Li(s) + 2 H2O(l) → H2(g) + 2 LiOH(aq)
Reasons:
• Group 1 metals are soft and silvery
•
water is colourless
•
the gas test indicates hydrogen
•
the litmus test indicates one of the products is a base
•
the flame test (bright red) indicates that lithium could be a constituent of one of the products
(b) Reasonably sure. We don’t know if the metal floated when it was put in the liquid (lithium would). If we assume
the liquid was pure water, the metal in the reaction could be strontium (flame test).
Sr(s) + 2 H2O(l) → H2(g) + Sr(OH)2(aq)
To be sure that the metal wasn’t strontium, another sample of the metal could be observed when it is put in water. If it
floats (as lithium would), it can’t be strontium (density 2.6 g/cm3). A chemical test would involve adding sulfuric acid to
the product solution. Strontium sulfate has low solubility, but lithium sulfate has high solubility. If sulfuric acid were
added to the solution, and strontium ions were present, strontium sulfate should precipitate in a double displacement reaction:
Sr(OH)2(aq) + H2SO4(aq) → 2 H2O(l) + SrSO4(aq)
It is also possible the liquid was water, but containing one or more solutes, for example, an acid. A test with pH paper or
litmus paper would test for this kind of solute.
12.
part 1: step 1, Ni(OH)2(s) should form and be filtered
step 2, adding the barium chloride should result in the precipitation of BaSO4(s)
NiSO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + Ni(OH)2(s)
Na2SO4(aq) + BaCl2(aq) → 2 NaCl(aq) + BaSO4(s)
both are double displacement reactions
part 2: step 5, the mass of the crucible and contents should increase due to the synthesis of MgO(s) as the
magnesium combusts
2 Mg + O2 → 2 MgO(s)
part 3: step 6, small bubbles may form on the zinc as the zinc slowly replaces hydrogen in water in a single
displacement reaction; after the addition of sulfuric acid, gas should be generated more vigorously in
another single displacement reaction. The small amount of zinc hydroxide produced in the first reaction
should also react with the sulfuric acid in a double displacement reaction.
step 7, the gas will pop, indicating hydrogen
Zn(s) + 2 H2O(l) → H2(g) + Zn(OH)2(s)
Zn + H2SO4(aq) → H2(g) + ZnSO4(aq)
Zn(OH)2(s) + H2SO4(aq) → ZnSO4(aq) + 2 H2O(l)
2 H2(g) + O2(g) → 2 H2O(g)
Making Connections
13. Barium sulfate is a low-solubility solid. Because it does not dissolve, no barium ions are freed into solution within the
patient and so there are no toxic effects.
14. Students are to investigate the refining process for a major industrial element, including any chemical reactions.
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Unit 1 Matter and Chemical Bonding
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15. Students are to research the composition and design of safety matches and produce a poster.
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UNIT 1 REVIEW
(Page 150)
Understanding Concepts
1. (a) iodine-127: p = 53; e = 53; n = 74 (127 53)
(b) phosphorus-32: p = 15; e = 15; n = 17 (32 15)
(c) Cu-64 (copper-64): p = 29; e = 29; n = 35 (64 29)
(d) Hg-203 (mercury-203): p = 80; e = 80; n = 123 (203 80)
2. Comparison of Radiation
alpha particles
beta particles
gamma rays
mass
4u
very small
none (for now)
speed
relatively low
high
speed of light
charge
2+
10
3.
200
175
150
Mass (g)
125
100
75
61
50
25
0
10
20
30
40
50
60
Time (d)
4. (a) Bromine and fluorine each have seven electrons in their outer shell.
(b) After the removal of the first electron, the nuclear charge has not been reduced, but the number of electrons has
been. The pull of the nucleus on the remaining valence electron should be stronger. As a result it will take more
energy to remove the second electron.
(c) In order to achieve a full octet (full outer shell of electrons), oxygen must gain two electrons, producing the O2
ion.
(d) Argon has a full outer shell of electrons, which is a very stable arrangement.
(e) Fluorine requires one electron to reach a stable octet and it has few electrons shielding the charge of the nucleus.
As a result it has a high electron affinity.
5. The phenomenon of radioactivity indicated that atoms could change, i.e., that Dalton’s model of an indivisible, balllike atom was flawed. That radiation often consisted of charged particles (alpha and beta) indicated that the atom had
positive and negative components.
6. (a) mercury; transition metal
(b) halogen
(c) alkali metal
(d) alkaline earth metal
(e) halogen
(f) noble gas
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Unit 1 Matter and Chemical Bonding
67
4.1 PROPORTIONS IN COMPOUNDS
PRACTICE
(Page 161)
Understanding Concepts
1. Students will largely reproduce Table 1 on page 160 of the text, adding information on the chemical properties of the
two compounds, e.g.:
Comparing CO and CO2
physical properties
melting point (°C)
boiling point (°C)
density (g/L)
colour
solubility (g/100mL H2O)
chemical properties
odour
toxicity
taste
blood
photosynthesis
applications
CO
CO2
–199.0
–191.5
1.250
none
0.004
–78.5
–78.5
1.977
none
0.339
none
high
none
binds to hemoglobin
not used
?
none
low
slightly acid
affects acidity
reactant
dry ice;
carbonation;
fire extinguishers
2. Carbon monoxide is toxic because it bonds to hemoglobin in the bloodstream, preventing blood cells from carrying
oxygen to body tissues.
3. Carbonated beverages are made with carbon dioxide gas. Its physical property of dissolving under pressure and
escaping when pressure is low makes the drink “fizzy.” Its chemical properties make the drink taste “tangy.”
Carbon dioxide is useful in fire extinguishers. It can be liquified and stored under pressure, and will propel itself out
of its container when opened. It is chemically unreactive and doesn’t support combustion of most substances.
PRACTICE
(Page 162)
Understanding Concepts
4. When compounds form from elements, the ratio of masses of those elements in the compound is always precisely the
same.
5. Stoichiometry refers to the study of quantities of substances involved in chemical reactions.
6. (a) Nonmetals can often form many bonds, and bond to each other in several different ways, resulting in numerous
different compounds.
(b) Sulfur and phosphorus should bond to oxygen in more than one way, since they are nonmetals.
7. (a) CuCl(s) and CuCl2(s)
(b) FeO(s) and Fe2O3(s)
(c) PbS(s) and PbS2(s)
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Chapter 4 Quantities in Chemical Formulas
75
Making Connections
8. Examples might include medical lab technologist, food chemistry, cosmetics chemistry, plastics chemistry, or detergent/cleanser chemistry analyst, and so on.
SECTION 4.1 QUESTIONS
(Page 162)
Understanding Concepts
1. Statements (a), (b), and (c) illustrate the law of definite proportions, and (d) does not.
Applying Inquiry Skills
2. Experimental Design
Aqueous silver ions and chromate ions will be mixed indifferent ratios to see how they combine. The solutions of ions
will each contain the same number of ions per unit volume.
Materials
• lab apron
• eye protection
• 5 small test tubes of equal size
• test-tube rack
• eyedropper
• 10 mL of silver nitrate solution (0.10 mol/L)
• 10 mL of sodium chromate solution (0.10 mol/L)
• distilled water
Procedure
Safety: Chromate compounds are poisonous if ingested. Wash hands after use.
1. Number the test tubes from 1 to 5, and place in the test-tube rack.
2. Using the dropper, add drops of silver nitrate solution to each test tube. Add 2, 4, 6, 8, and 10 drops respectively, to
tubes 1, 2, 3, 4, and 5.
3. Wash the dropper thoroughly with distilled water and use the same dropper to add drops of sodium chromate solution
to each test tube. Add 10, 8, 6, 4, and 2 drops respectively, to tubes 1, 2, 3, 4, and 5. After you have finished putting
drops in each tube, the test tubes should be filled to equal depth since they contain the same number of drops
(12 drops total).
4. Swirl each test tube gently to mix the contents. Allow the precipitates to settle for about 5 minutes.
5. Wash your hands thoroughly.
Analysis
(a) The test tube with the most precipitate indicates the ratio of combination of silver and chromate ions — which will
be proportional to the ratio of drops used in that tube.
(b) This experiment can be done with any desired ratio of drops, but the result will always be that silver and chromate
combine in a 2:1 ratio, showing that the proportions of the precipitate compound are fixed.
Making Connections
3.
Compound
Colour
Hazard
Use
N2O(g)
colourless
slight
anesthetic
NO(g)
colourless
toxic/irritant
bleaching
NO2(g)
red–brown
very toxic
rocket fuels
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Unit 2 Quantities in Chemical Reactions
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4.2 RELATIVE ATOMIC MASS AND ISOTOPIC ABUNDANCE
PRACTICE
(Page 164)
Understanding Concepts
1. Atomic mass, like any other quantity, has to have a reference value with which it can be compared, for accuracy.
2. The atomic mass unit is exactly 1/12 the mass of a single atom of carbon-12. The unit symbol is u.
3. Hydrogen, oxygen, and carbon atoms have been used as reference elements for atomic mass.
4. 1.92 12 u = 23.0 u is the relative atomic mass of sodium.
Making Connections
5. A communication system must be convenient, simple, and practical. To be international it must also be accepted
universally. The SI and IUPAC systems are both designed around these criteria.
PRACTICE
(Page 166)
Understanding Concepts
6. (a) Average atomic mass is the average value of the mass of atoms in a sample of a naturally occurring element.
(b) Isotopes of an element are atoms with the same number of protons, but different numbers of neutrons, and so
different masses.
(c) Isotopic abundance refers to the proportion of atoms of an element that are specific isotopes, usually given as a
percentage.
7. The isotope C-12 refers to atoms with 6 protons and 6 neutrons.
8. The atomic mass of carbon is not exactly 12 u because a small percentage of carbon atoms are C-13 atoms, which
increase the average value of the mass.
9. Since atoms exist as indivisible objects for chemistry purposes, we avoid talking about fractions of atoms to avoid
confusion.
10. Assume 10 000 K atoms, for convenience
93.10% or 9310 atoms are K-39
6.90% or 690 atoms are K-41
mtot
= (9310 39 u) + (690 41 u)
mtot
mav
mav
= 391 380 u
391 380 u
= 10 000
= 39.14 u
The actual average atomic mass value for potassium is 39.10 u, which is very close to the value calculated by this
method.
11. Assume 10 000 Ar atoms, for convenience
99.60% or 9960 atoms are Ar-40
0.34% or 34 atoms are Ar-36
0.06% or 6 atoms are Ar-38
= (9960 40 u) + (34 36 u) + (6 38 u)
mtot
mtot
= 399 852 u
mav
399 852 u
= 10 000
mav
= 39.99 u
The actual average atomic mass value for argon is 39.95 u, which is very close to the value calculated by this method.
Copyright © 2002 Nelson Thomson Learning
Chapter 4 Quantities in Chemical Formulas
77
SECTION 4.2 QUESTIONS
(Page 167)
Understanding Concepts
1. 9 12 u = 108 u (assume both numbers are exact), which is very close to the average value for silver, Ag, which is
actually 107.87 u.
2. The atomic mass unit is defined as 1/12 the mass of a C-12 atom. Atomic mass is the actual average mass of the atoms
of a given element.
3. Because the elements are compared to each other based on the way they combine when they react, the reaction proportions must be known.
4. Chlorine atoms have an average atomic mass of 35.45 u, because chlorine is made up of about 75% Cl-35 atoms, and
about 25% Cl-37 atoms.
5. Assume 1000 B atoms, for convenience
19.8% or 198 atoms are B-10
80.2% or 802 atoms are B-11
mtot
= (198 10 u) + (802 11 u)
mtot
= 10 802 u
mav
10 802 u
= 1000
mav
= 10.8 u
The average atomic mass for boron is 10.8 u, calculated by this method.
4.3 THE MOLE AND MOLAR MASS
PRACTICE
(Page 168)
Understanding Concepts
1. A mole of anything is the same number of things as the number of C atoms there would be in exactly 12 g of the
isotope C-12.
2. Avogadro’s constant is 6.02 1023 (rounded to three digits ...).
Note:
Since the concepts of “pure” C-12 and “exactly” 12 g are imaginary, there is no pretense in the scientific community that we will ever know the “exact” value for Avogadro’s constant. The mole is a purely theoretical
definition. As technology improves, we are, of course, able to determine the value to greater precision. Rounded to
six digits, the precision routinely stated in postsecondary level work, the currently accepted value is 6.022 14 1023.
The Canadian Metric Practice Guide lists 8 digits — 6.022 136 7 1023. The most precise recent reported value,
obtained from ion X-ray diffraction evidence, is 6.022 141 99 1023 . This constant, like many others that are
frequently used, is usually rounded to three digits for high-school calculations.
3. 602 000 000 000 000 000 000 000 (only the first 3 digits are significant)
Note:
It is useful for students to extend this number to see how large it is, being mindful that all those trailing
zeros represent numbers they don’t know. It is also thought-provoking for students to consider that when written in
scientific notation to the usual high-school text precision, the error of the number is 2 1020, meaning we blithely
ignore some two hundred millions of millions of millions, as being too small to make a noticeable difference.
6.02 × 1023 molecules
1 mol
molecules
4. NCO = 3.00 mol 2
NCO = 1.81 × 1024
2
There are 1.81 1024 molecules of carbon dioxide in the sample.
78
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
6.02 × 1023 atoms
1 mol
23
= 3.01 × 10 atoms
5. NAr = 0.500 mol NAr
There are 3.01 × 1023 atoms of argon in the sample.
1.43 kg
6. (a) m = (12 is an exact value — counted)
12 oranges
m = 0.119 kg/orange = 119 g/orange
The average mass of one of these oranges is 119 g.
1.01 g
(b) m = 1m
ol
1 mol
6.02 × 1023 H atoms
m = 1.68 × 10–24 g/H atom (average value)
The average mass of an atom in a hydrogen sample is 1.68 × 10–24 g.
Note: No actual hydrogen atom, of course, really has this mass value, since the relative atomic molar
mass used to derive it is an average value, itself derived from the natural isotopic composition of the element.
PRACTICE
(Page 170)
Understanding Concepts
7. Molar mass is the mass of one mole (the Avogadro number) of any entity, usually atoms, molecules, ions, or ionic
compound formula units. The SI unit is technically kg/mol, but is commonly used and stated as g/mol.
8. 1 mol Ca(OH)2 = 1 mol Ca + 2 mol O + 2 mol H
MCa(OH) = [(40.08 × 1) + (16.00 × 2) + (1.01 × 2)] g/mol
2
MCa(OH) = 74.10 g/mol
2
The molar mass of calcium hydroxide is 74.10 g/mol.
9. 1 mol Cl2 = 2 mol Cl
MCl = (35.45 × 2) g/mol
2
MCl = 70.90 g/mol
2
The molar mass of chlorine is 70.90 g/mol
10. 1 mol OH– = 1 mol O + 1 mol H + 1 mol e–
MOH– = [(16.00 × 1) + (1.01 × 1) + (negligible mass)] g/mol
MOH– = 17.01 g/mol
The molar mass of hydroxide ions is 17.01 g/mol.
Note:
Students are taught routinely to simply ignore the electrical charges of ions when calculating molar
masses, because the tiny variation in mass due to electron increase/decrease is negligible, never enough to affect the
value.
11. The element is most likely gold, Au(s), which has a molar mass listed on the periodic table of elements of
196.97 g/mol.
67.2 g
12. m = 8.0 mol mol
2
m = 5.4 × 10 g = 0.54 kg
The mass of the substance is 0.54 kg.
13. (a) Diatomic means composed of two-atom molecules.
(b) Hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, and iodine form diatomic molecules.
(c) Avogadro’s number of molecules (6.02 × 1023 molecules) are present.
Copyright © 2002 Nelson Thomson Learning
Chapter 4 Quantities in Chemical Formulas
79
SECTION 4.3 QUESTIONS
(Page 171)
Understanding Concepts
1. Avogadro’s constant is the number of atoms of carbon in exactly 12 g of the isotope C-12, also called one mole. It is
useful in chemistry because a mole is defined so that the mass of this amount in grams will equal numerically the relative atomic mass of any atom.
2. Atomic mass is the average value of the mass of atoms of a particular element. Molar mass is the mass of one mole
of any entity.
3. 1 mol C12H22O11 = 12 mol C + 22 mol H + 11 mol O
MC
= [(12.01 × 12) + (1.01 × 22) + (16.00 × 11)] g/mol
MC
= 342.34 g/mol
12H22O11
12H22O11
NA = 6.02 × 1023 entities/mol
mC
12H22O11
=?
1 mol
342.34 g
m = 23 molecules
6.02
×
10
1
mol
m = 5.69 × 10–22 g/molecule (on average)
The average mass of a molecule in a sucrose sample is 5.69 × 10–22 g.
Note:
No actual single sucrose molecule, of course, has this mass value. It is an average value derived from
the natural isotope mixtures of elements in this compound.
4. 1 mol C8H18 = 8 mol C + 18 mol H
MC H = [(12.01 × 8) + (1.01 × 18)] g/mol
8 18
MC H = 114.26 g/mol
8 18
The molar mass of octane is 114.26 g/mol.
Note:
You may find it convenient, from this point on in the course, to assume that students will calculate molar
masses accurately, i.e., not requiring them to show the addition of periodic table values. Since all of the periodic table
(average atomic molar mass) values are given to two decimal places, and the precision rule for addition always applies,
any molar mass calculation will automatically be precise to hundredths of a gram per mole.
Making Connections
5. Molar masses:
H2(g)
2.02 g/mol
He(g)
4.00 g/mol
N2(g)
28.02 g/mol
O2(g)
16.00 g/mol
CO2(g) 44.01 g/mol
The density of each gas is found as follows; assume the volume of a mole of each is approximately 22.4 L.
2.02 g
1
mol
densityH = 2
1
mol
22.4 L
densityH = 0.0902 g/L
2
The density of hydrogen gas at STP is 0.0902 g/L.
4.00 g
1
mol
densityHe = 1
mol
22.4 L
80
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
densityHe = 0.179 g/L
The density of helium gas at STP is 0.179 g/L.
28.02 g
1
mol
densityN = 2
1m
ol
22.4 L
densityN = 1.25 g/L
2
The density of nitrogen gas at STP is 1.25 g/L.
32.00 g
1
mol
densityO = 2
1m
ol
22.4 L
densityO = 1.43 g/L
2
The density of oxygen gas at STP is 1.43 g/L.
44.01 g
1
mol
densityCO = 2
1
mol
22.4 L
densityCO = 1.96 g/L
2
The density of carbon dioxide gas at STP is 1.96 g/L.
Note : The molar volume of 22.4 L/mol at STP for gases is an approximation derived from a theoretical “ideal”
gas system, rounded to three significant digits. See page 469 of the text (Did You Know?) for more precise values for
some real gases.
Reflecting
6. Molar mass is a conversion factor that can be used to convert measured masses of substances into numerical amounts,
in moles. This will be useful because the numbers in chemical reaction equations represent numerical values.
4.4 CALCULATIONS INVOLVING THE MOLE CONCEPT
PRACTICE
(Page 172)
Understanding Concepts
1. mNaCl = 2.5 g
MNaCl = 58.44 g/mol
nNaCl = ?
1 mol
nNaCl = 2.5 g 58.44 g
nNaCl = 0.043 mol = 43 mmol
The amount of sodium chloride is 43 mmol.
2. mC H
6 12O6
= 1.0 kg
MC H
= 180.18 g/mol
nC H
= ?
6 12O6
6 12O6
nC H
6
12O6
nC H
6 12O6
1 mol
= 1.0 kg 180.18 g
= 0.056 kmol = 5.6 mol
The amount of glucose is 5.6 mol.
Copyright © 2002 Nelson Thomson Learning
Chapter 4 Quantities in Chemical Formulas
81
3. mO = 25.0 g
2
MO = 32.00 g/mol
2
nO
= ?
2
1 mol
nO = 25.0 g 2
32.00 g
nO = 0.781 mol
2
The amount of oxygen is 0.781 mol (or 781 mmol).
4. (a) mC = 24.0 g
MC = 12.01 g/mol
nC = ?
1 mol
nC = 24.0 g 12.01 g
nC = 2.00 mol
The amount of carbon atoms in the compound is 2.00 mol.
mH = 6.0 g
MH = 1.01 g/mol
nH = ?
1 mol
nH = 6.0 g 1.01 g
nH = 5.9 mol
The amount of hydrogen atoms in the compound is 5.9 mol.
mO = 16.0 g
MO = 16.00 g/mol
nO = ?
1 mol
nO = 16.0 g 16.00 g
nO = 1.00 mol
The amount of oxygen atoms in the compound is 1.00 mol.
(b) The mole ratio of C:H:O is approximately 2:6:1.
(c) The likely formula for the compound then, is C2H6O(l).
PRACTICE
(Page 173)
Understanding Concepts
5. nMg(OH) = 0.45 mol
2
MMg(OH) = 58.33 g/mol
2
mMg(OH)
2
= ?
mMg(OH)
2
mMg(OH)
2
58.33 g
= 0.45 mol 1
mol
= 26 g
The mass of magnesium hydroxide is 26 g.
6. nNH
= 87 mmol
3
MNH
= 17.04 g/mol
3
mNH = ?
3
82
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
17.04 g
mNH = 87 mmol
3
1m
ol
mNH = 1.5 × 103 mg = 1.5 g
3
The mass of ammonia is 1.5 g.
7. nC H O = 63.28 mol
8 6 4
MC H O = 166.14 g/mol
8 6 4
mC H O
= ?
8 6 4
166.14 g
mC H O = 63.28 mol 8 6 4
1
mol
mC H O = 10.51 × 103 g = 10.51 kg
8 6 4
The mass of 1,4-benzenediotic acid is 10.51 kg.
8. nHC H O = 1.0 mmol
9 7 4
MHC H O = 180.17 g/mol
9 7 4
mHC H O = ?
9 7 4
180.17 g
mHC H O = 1.0 mmol
9 7 4
1
mol
mHC H O = 1.8 × 102 mg = 0.18 g
9 7 4
The mass of acetylsalicylic acid (Aspirin) is 0.18 g.
Try This Activity : Counting Atoms, Molecules, and Other Entities
(Page 176)
Calculate the evaporation rate (in molecules) of a drop of water:
mass of 50 drops of water = 1.95 g
mass of 1 drop of water = 1.95 g/50 = 0.039 g
time of 1 drop of water to evaporate = 65 min = 3.9 103 s
1 mol
nH O = 0.039 g 2
18.02 g
= 0.00216 mol
There is 0.00216 mol of H2O molecules in every drop.
6.02 1023 molecules
1 mol
= 1.30 1021 molecules
NH O = 0.00216 mol 2
There are 1.30 1021 molecules of H2O in every drop.
1.30 1021 molecules
3.9 10 s
evaporation rate = 3
evaporation rate = 3.34 1017 molecules/s
Water molecules evaporate at the average rate of 3.34 1017 molecules/s.
Calculate the number of atoms and the value of each atom in a copper penny:
mass of 1 penny = 2.44 g
1 mol
nCu = 2.44 g 63.55 g
= .0384 mol
6.02 1023 atoms
NCu = .0384 mol 1
mol
NCu = 2.31 1022 atoms
Copyright © 2002 Nelson Thomson Learning
Chapter 4 Quantities in Chemical Formulas
83
There are 2.31 1022 atoms in a copper penny.
1 cent
value of a copper atom = 2.31 1022 atoms
value of a copper atom = 4.33 10–23 cents
Each copper atom in a penny has a value of 4.33 10–23 cents.
Measure into a graduated cylinder half a mole of sucrose molecules:
MC12H22O11 = 337.34 g/mol
337.34 g
1
mol
mass of half a mole of sucrose = 1
mol
2
= 168.67 g
When poured into a graduated cylinder, the volume of sucrose = 185 mL.
Measure into a graduated cylinder the quantity of sugar that contains two moles of carbon atoms:
One mole of sucrose, C12H22O11(s), contains twelve moles of carbon atoms. Therefore, you would need one-sixth of a
mole of sucrose to get two moles of carbon atoms.
337.34 g
1
mol
required mass of sucrose = 1
mol
6
= 56.22 g
When poured into a graduated cylinder, the volume of sucrose = 60 mL.
Calculate the number of atoms needed to write your name in chalk:
initial mass of chalk = 14.23 g
final mass of chalk = 14.15 g
change in mass of chalk = 0.08 g
MCaCO3 = 100.09 g/mol
number of atoms in each formula unit = 5
1 mol
nCaCO
= 0.08 g 3
100.09 g
= 0.000799 mol
6.02 1023 formula units
3
1 mol
= 4.81 1020 formula units
5 atoms
number of atoms = 4.81 1020 form
unit 1 fo
rm u nit
= 2.40 1021 atoms
NCaCO
= 0.000799 mol The number of atoms needed to write my name in chalk is 2.40 1021.
Calculate the number of sodium ions in a solution of 3.00 g of NaCl in 200 mL of water:
mNaCl = 3.00 g
MNaCl = 58.44 g/mol
1
mol
6.02 1023 formula units
NNaCl = 3.00 g 58.44 g
1 mol
= 3.09 1022 formula units
Since there is one sodium ion in every NaCl formula unit, the number of sodium ions is 3.09 1022.
Calculate the number of iron atoms in a nail:
mnail = 4.16 g
MFe
= 55.85 g/mol
NFe
1m
ol
6.02 1023 atoms
= 4.16 g 55.85 g
1
mol
= 4.48 1022
84
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
There are 4.48 1022 atoms in the iron nail.
Calculate the number of years to span a mole of seconds:
60 m
in
24 h
365 d
60 s
Number of seconds in one year = 1 h
1 d
1a
1
min
= 3.15 107 s/a
Years in one mole of seconds
6.02 1023 s
1a
= 1 mol
3.15 107 s
= 1.91 1016 a/mol
It takes 1.91 1016 years to span a mole of seconds.
PRACTICE
(Page 176)
Understanding Concepts
9. (a) The molar mass of H2O(l) is 18.01 g/mol.
(b) The molar mass of CO2(g) is 44.01 g/mol.
(c) The molar mass of NaCl(s) is 58.44 g/mol.
(d) The molar mass of C12H22O11(s) is 342.34 g/mol.
(e) The molar mass of (NH4)2Cr2O7(s) is 252.10 g/mol.
Note: Solutions for the rest of this course assume that students can correctly total a molar mass for
a substance without showing work for the operation, unless specifically requested to do so. As a final example of
the full work:
1 mol (NH4)2Cr2O7 = 2 mol N + 8 mol H + 2 mol Cr + 7 mol O
M(NH ) Cr O = [(14.01 × 2) + (1.01 × 8) + (52.00 × 2) + (16.00 × 7)] g/mol
4 2
2 7
M(NH ) Cr O = 252.10 g/mol
4 2
2 7
10. (a) mC
12H22O11
= 10.00 kg
MC
= 342.34 g/mol
nC
=?
12H22O11
12H22O11
nC
12H22O11
nC
12H22O11
1 mol
= 10.00 kg 342.34 g
= 0.02921 kmol = 29.21 mol
The amount of sucrose is 29.21 mol.
(b) mNaCl = 500 g
MNaCl = 58.44 g/mol
nNaCl = ?
1 mol
nNaCl = 500 g 58.44 g
nNaCl = 8.56 mol
The amount of pickling salt (sodium chloride) is 8.56 mol.
(c) mC H = 40.0 g
3 8
MC H = 44.11 g/mol
3 8
nC H = ?
3 8
1 mol
nC H = 40.0 g 3 8
44.11 g
nC H = 0.907 mol
3 8
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Chapter 4 Quantities in Chemical Formulas
85
The amount of propane is 0.907 mol (or 907 mmol).
(d) mHC H O = 325 mg
9 7 4
MHC H O = 180.17 g/mol
9 7 4
nHC H O = ?
9 7 4
1 mol
nHC H O = 325 mg 9 7 4
180.17 g
nHC H O = 1.80 mmol
9 7 4
The amount of acetylsalicylic acid (Aspirin) is 1.80 mmol.
(e) Write CH3CHOHCH3 condensed to C3H8O for convenience.
mC H O
= 150 g
3 8
MC H O = 60.11 g/mol
3 8
nC H O
= ?
3 8
1 m ol
nC H O = 150 g 3 8
60.11 g
nC H O = 2.50 mol
3 8
The amount of 2-propanol (rubbing alcohol) is 2.50 mol.
Note: The first printing of Chemistry 11 incorrectly gives the formula of 2-propanol as
CH3CH2OHCH3(l). Using that formula would yield an answer of 2.45 mol.
11. (a) nNH = 4.22 mol
3
MNH = 17.04 g/mol
3
mNH = ?
3
mNH
3
mNH
3
17.04 g
= 4.22 mol 1
mol
= 71.9 g
The mass of ammonia is 71.9 g.
(b) nNaOH = 0.224 mol
MNaOH = 40.00 g/mol
mNaOH = ?
40.00 g
mNaOH = 0.224 mol 1
mol
mNaOH = 8.96 g
The mass of sodium hydroxide is 8.96 g.
(c) nH O = 57.3 mmol
2
MH O = 18.02 g/mol
2
mH O = ?
2
mH O
2
mH O
2
18.02 g
= 57.3 mmol
1
mol
= 1.03 × 103 mg = 1.03 g
The mass of water is 1.03 g.
(d) nKMnO = 9.44 kmol
4
MKMnO = 158.04 g/mol
4
mKMnO = ?
4
158.04 g
mKMnO = 9.44 kmol
4
1
mol
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Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
mKMnO = 1.49 × 103 kg = 1.49 Mg
4
The mass of potassium permanganate is 1.49 Mg (or 1.49 t).
(e) n(NH ) SO = 0.77 mol
4 2
4
M(NH ) SO = 132.16 g/mol
4 2
4
m(NH ) SO = ?
4 2
4
132.16 g
m(NH ) SO = 0.77 mol 4 2
4
1
mol
m(NH ) SO = 1.0 × 102 g = 0.10 kg
4 2
4
The mass of ammonium sulfate is 0.10 kg.
12. (a) nCO
= 15 mol
2
= 6.02 × 1023 entities/mol
NA
NCO = ?
2
NCO
2
NCO
2
6.02 × 1023 molecules
1 mol
molecules
= 15 mol = 9.0 × 1024
There are 9.0 × 1024 molecules of carbon dioxide in the sample.
(b) mNH = 15 g
3
MNH = 17.04 g/mol
3
= 6.02 × 1023 entities/mol
NA
NNH = ?
3
nNH
1 mol
= 15 g 17.04 g
= 0.88 mol
NNH
= 0.88 mol NNH
= 5.3 × 1023
NNH
6.02 × 10 molecules
1
mol
= 15 g 1 mol
17.04 g
= 5.3 × 1023 molecules
nNH
3
3
3
3
or
6.02 × 1023 molecules
1 mol
molecules
23
3
NNH
3
There are 5.3 × 1023 molecules of ammonia in the sample.
(c) mHCl = 15 g
MHCl = 36.46 g/mol
NA
= 6.02 × 1023 entities/mol
NHCl = ?
nHCl
1 mol
= 15 g 36.46 g
nHCl
= 0.41 mol
NHCl
= 0.41 mol 1 mol
NHCl
6.02 × 1023 molecules
= 2.5 × 1023 molecules
or
Copyright © 2002 Nelson Thomson Learning
Chapter 4 Quantities in Chemical Formulas
87
NHCl
6.02 × 1023 molecules
1 m
ol
= 15 g 1 mol
36.46 g
NHCl
= 2.5 × 1023 molecules
There are 2.5 × 1023 molecules of hydrogen chloride in the sample.
(d) mNaCl = 15 g
MNaCl = 58.44 g/mol
= 6.02 × 1023 entities/mol
NA
NNaCl = ?
nNaCl
1 mol
= 15 g 58.44 g
= 0.26 mol
NNaCl
= 0.26 mol NNaCl
= 1.5 × 1023
NNaCl
6.02 × 1023 molecules
1
mol
= 15 g 1 mol
58.44 g
= 1.5 × 1023 formula units
nNaCl
or
NNaCl
6.02 × 1023 molecules
1 mol
formula units
There are 1.5 × 1023 formula units of sodium chloride in the sample.
13. (a) MCO = 44.01 g/mol
2
= 6.02 × 1023 entities/mol
NA
mCO = ?
2
mCO
2
mCO
2
1 mol
44.01 g
= 23 molecules
6.02
×
10
1
mol
= 7.31 × 10–23 g/molecule (on average)
The average mass of a molecule in a sample of carbon dioxide from respiration is 7.31 × 10–23 g.
(b) MC H
6 12O6
= 180.18 g/mol
= 6.02 × 1023 entities/mol
NA
mC H
6 12O6
=?
mC H
180.18 g
1 mol
= 1
mol
6.02 × 1023 molecules
mC H
= 2.99 × 10–22 g/molecule (on average)
6 12O6
6 12O6
The average mass of a molecule in a sample of glucose from photosynthesis is 2.99 × 10–22 g.
(c) MO = 32.00 g/mol
2
NA = 6.02 × 1023 entities/mol
mO = ?
2
mO
2
1 mol
32.00 g
= 23
6.02 × 10 molecules
1
mol
mO
= 5.32 × 10–23 g/molecule (on average)
2
The average mass of a molecule in a sample of oxygen from photosynthesis is 5.32 × 10–23 g.
88
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
14. mH O = 1.000 L
1.00 kg/L
= 1.00 kg
2
MH O = 18.02 g/mol
2
= 6.02 × 1023 entities/mol
NA
NH O = ?
2
Note: It is assumed that students by this point will automatically apply the 1.00 g/mL, 1.00 kg/L, and
1.00 Mg/m3 (1.00 t/m3) conversions for the mass–volume relationship of pure water.
1 mol
nH O
= 1.00 kg 2
18.02 g
nH O
= 0.0555 kmol = 55.5 mol
2
6.02 × 1023 molecules
1 mol
molecules
NH O
= 55.5 mol NH O
= 3.34 × 1025
NH O
6.02 × 1023 molecules
1m
ol
= 1.00 kg 1 mol
18.02 g
NH O
= 3.34 × 1025 molecules
2
2
or
2
2
There are 3.34 × 1025 molecules of water in 1.000 L.
SECTION 4.4 QUESTIONS
(Page 177)
Understanding Concepts
1. (a) nO = 1.5 mol
2
NA
= 6.02 × 1023 entities/mol
NO
= ?
2
6.02 × 1023 molecules
1 mol
molecules
NO
= 1.5 mol NO
= 9.0 × 1023
2
2
There are 9.0 × 1023 molecules of oxygen in the sample.
(b) 2 atoms/molecule 9.0 × 1023 molecules = 1.8 × 1024 atoms.
There are 1.8 × 1024 atoms of oxygen in the sample.
2. mH C H O = 90 mg
2 6 6 6
MH C H O = 176.14 g/mol
2 6 6 6
NA
= 6.02 × 1023 entities/mol
NH C H O = ?
2 6 6 6
nH C H O
1 mol
= 90 mg 176.14 g
= 0.51 mmol
NH C H O
= 0.51 mmol
NH C H O
= 3.1 × 1020
nH C H O
2 6 6 6
2 6 6 6
2 6 6 6
2 6
6 6
Copyright © 2002 Nelson Thomson Learning
6.02 × 1023 molecules
1 mol
molecules
Chapter 4 Quantities in Chemical Formulas
89
or
NH C H O
2 6 6 6
NH C H O
2 6 6 6
6.02 × 1023 molecules
1
mol
= 90 mg 1 mol
176.14 g
= 3.1 × 1020 molecules
There are 3.1 × 1020 molecules of ascorbic acid in the tablet.
3. (a) nCa(IO ) = 1.00 × 10–2 mol
3 2
NA
= 6.02 × 1023 entities/mol
NI
= ?
6.02 × 1023 molecules
NCa(IO ) = 1.00 × 10–2 mol 3 2
1 mol
NCa(IO ) = 6.02 × 1021 formula units
3 2
2 ato ms
NI
= 6.02 × 1021 formula units formula u nit
NI
= 1.20 × 1022 atoms
There are 1.20 × 1022 atoms of iodine in the sample.
(b) nCa(IO ) = 1.00 × 10–2 mol
3 2
MCa(IO ) = 389.88 g/mol
3 2
mCa(IO ) = ?
3 2
389.88 g
mCa(IO ) = 1.00 × 10–2 mol 3 2
1
mol
mCa(IO ) = 3.90 g
3 2
The mass of calcium iodate is 3.90 g.
4. mH O = 500 g
2
MH O = 18.02 g/mol
2
nH O = ?
2
1 mol
nH O = 500 g 2
18.02 g
nH O = 27.7 mol
2
mC
12H22O11
= 200 g
MC
= 342.34 g/mol
nC
=?
12H22O11
12H22O11
nC
12H22O11
nC
12H22O11
1 mol
= 200 g 342.34 g
= 0.584 mol (or 584 mmol)
mHC H O = 25 g
2 3 2
MHC H O = 60.06 g/mol
2 3 2
nHC H O = ?
2 3 2
mH C H O
3 6 5 7
1 mol
nHC H O = 25 g 2 3 2
60.06 g
nHC H O = 0.42 mol
2 3 2
= 15 g
MH C H O = 192.14 g/mol
3 6 5 7
nH C H O = ?
3 6 5 7
90
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
1 mol
nH C H O = 15 g 3 6 5 7
192.14 g
nH C H O = 0.078 mol (or 78 mmol )
3 6 5 7
mNaCl = 5 g
MNaCl = 58.44 g/mol
nNaCl = ?
nNaCl
1 mol
= 5 g 58.44 g
nNaCl
= 0.09 mol
Converted to amounts, the recipe is: 27.7 mol water, 0.584 mol sugar, 0.42 mol vinegar, 0.078 mol citric acid, and
0.09 mol salt.
5. (a) vC H OH = 17 mL
2 5
dC H OH = 0.789 g/mL
2 5
(density)
MC H OH = 46.08 g/mol
2 5
NA
= 6.02 × 1023 entities/mol
NC H OH = ?
2 5
0.789 g
mC H OH = 17 mL
2 5
1
mL
mC H OH = 13 g
2 5
1 mol
nC H OH = 13 g 2 5
46.08 g
nC H OH = 0.29 mol
2 5
6.02 × 1023 molecules
1 mol
molecules
NC H OH = 0.29 mol 2 5
NC H OH = 1.8 × 1023
2 5
or
6.02 × 1023 molecules
0.789 g
1m
ol
NC H OH = 17 mL
2 5
1 mol
1
mL
46.08 g
NC H OH = 1.8 × 1023 molecules
2 5
There are 1.8 × 1023 molecules of ethanol in the beer.
(b) vNi = 0.72 cm3
dNi = 8.90 g/cm3 (density)
MNi = 58.69 g/mol
NA = 6.02 × 1023 entities/mol
NNi = ?
mNi
mNi
nNi
nNi
8.90 g
= 0.72 cm3 cm3
= 6.4 g
1 mol
= 6.4 g 58.69 g
= 0.11 mol
Copyright © 2002 Nelson Thomson Learning
Chapter 4 Quantities in Chemical Formulas
91
= 0.11 mol NNi
= 6.6 × 1022
NNi
6.02 × 1023 atoms
8.90 g
1
mol
= 0.72 cm3 1 mol
c
m3
58.69 g
22
= 6.6 × 10 atoms
or
NNi
There are 6.6 ×
6.02 × 1023 atoms
1 mol
atoms
NNi
1022
atoms of nickel in the quarter.
(c) mH O = 100 mL 1.00 g/mL
= 100 g
2
MH O = 18.02 g/mol
2
NA
= 6.02 × 1023 entities/mol
NH O = ?
2
nH O
1 mol
= 100 g 18.02 g
= 5.55 mol
NH O
= 5.55 mol NH O
= 3.34 × 1024
NH O
6.02 × 1023 molecules
1m
ol
= 100 g 1 mol
18.02 g
NH O
= 3.34 × 1024 molecules
nH O
2
2
2
2
or
2
2
6.02 × 1023 molecules
1 mol
molecules
There are 3.34 × 1024 molecules of water in 100 mL.
Applying Inquiry Skills
6. Experimental Design
A sample of silver nitrate is dissolved and completely reacted with copper metal. The ratio of amounts of silver
product and copper reactant are determined from mass measurements.
Materials
•
silver nitrate
•
copper wire
•
pure water
•
250-mL beaker
•
stirring rod
•
centigram balance
•
paper towels
Procedure
1. Measure the mass of a piece of coiled copper wire to 0.01 g.
2. Dissolve the silver nitrate in about 150 mL of water in the beaker.
3. Place the copper wire in the solution and let stand until the reaction is complete.
4. Remove the silver from the surface of the copper wire, wipe the wire dry, and measure its final mass to 0.01 g.
5. Rinse the silver crystals in the beaker with water, and drain (decant) as much water as possible.
6. Place the wet silver on a piece of paper towel to dry.
7. When dry, measure the mass of the silver to 0.01 g.
8. Dispose of materials as instructed.
92
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
Evidence
mass of copper (initial)
_______ g
mass of copper (final)
_______ g
mass of silver
_______ g
Analysis
The masses of silver and copper are divided by their molar masses to convert the quantities from masses to amounts.
The ratio, amount of copper: amount of silver, is then calculated.
Making Connections
7. The primary point in any report should be that the concept of the mole is essential to converting the easily measurable quantity of substances (mass) into numerical amounts that are the numerical quantities represented by formulas
and equations. Actual numbers are far too large for convenience, so the mole is defined so as to make these conversions quick and easy. It is, in fact, not necessary to know what the value of a mole is, numerically, to do predictive
and descriptive work in chemistry — any more than one needs to know how many salt grains are in a shaker.
4.5 PERCENTAGE COMPOSITION
PRACTICE
(Page 179)
Understanding Concepts
1. mC = 7.2 g
mH
= 2.2 g
mO
= 17.6 g
mtotal = 27.0 g
%C
7.2 g
= 100%
27.0 g
%C
= 27%
%H
2.2 g
= 100%
27.0 g
%H
= 8.1%
%O
17.6 g
= 100%
27.0 g
%O
= 65.2%
The percentage composition of the compound is 27% carbon atoms, 8.1% hydrogen atoms, and 65.2% oxygen atoms
by mass.
2. (a) mO = (30.80 – 8.40) g = 22.40 g
(b) mC
= 8.40 g
mO = 22.40 g
mtotal = 30.80 g (CO2(g))
%C
%C
%O
8.40 g
= 100%
30.80 g
= 27.3%
22.40 g
= 100%
30.80 g
%O
= 72.7%
The percentage composition of carbon dioxide is 27.3% carbon atoms and 72.7% oxygen atoms by mass.
Copyright © 2002 Nelson Thomson Learning
Chapter 4 Quantities in Chemical Formulas
93
3. a)
m1 = 3.12 g
mCu = 2.50 g
mO = (3.12 – 2.50) g = 0.62 g
% Cu
2.50 g
= 100%
3.12 g
% Cu
= 80.1%
%O
0.62 g
= 100%
3.12 g
%O
= 20%
The percentage composition of compound 1 is 80.1% copper atoms and 20% oxygen atoms by mass.
m2 = 1.62 g
mCu = 1.44 g
mO = (1.62 – 1.44) g = 0.18 g
% Cu
1.44 g
= 100%
1.62 g
% Cu
= 88.9%
%O
0.18 g
= 100%
1.62 g
%O
= 11%
The percentage composition of compound 2 is 88.9% copper atoms and 11% oxygen atoms by mass.
(b) The two compounds cannot be the same substance, because the proportions of the elements are different.
PRACTICE
(Page 182)
Making Connections
4. (a) The total mass is 9.7 kg. Mass units cancel in percentage calculations.
natural rubber is 1.8/9.7 100%
= 19%
carbon black is 2.3/9.7 100%
= 24%
steel cord is 0.5/9.7 100%
= 5%
polyester and nylon is 0.5/9.7 100%
= 5%
steel bead wire is 0.5/9.7 100%
= 5%
chemicals are 1.4/9.7 100%
= 14%
synthetic rubber is 2.7/9.7 100%
= 28%
(b) From a Goodyear web site, the synthetic:natural rubber ratios of:
Passenger car tires
55%:45%
Light truck tires
50%:50%
Racing car tires
65%:35%
Earthmover tires
20%:80%
(c) Synthetic rubber gives more flexibility and traction, while natural rubber gives durability. Racing car tires must
have superior traction, for example, but often don’t even last for one race.
(d) Charles Goodyear discovered the process of vulcanizing rubber, which made it stable at higher temperatures. This
made the rubber tire for vehicles possible, which has shaped the entire evolution of technological society.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
94
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
Try This Activity : What Makes Popcorn Pop?
(Page 182)
Popcorn Initial
Mass (g)
whole
18.16
Final Mass (g)
Change in Mass (g)
Percentage Water (%)
16.30
1.86
10.2
Percentage of
Popped Corn (%)
~90
split crosswise 15.34
13.90
1.45
9.45
~20
split lengthwise 15.04
13.54
1.50
9.97
<1
(a) In general, the results seem to confirm that popcorn pops because the moisture inside the large part (endosperm) of
the seed vaporizes, builds up the internal pressure, and suddenly breaks the hard outer coating (pericarp). The evidence
of the popping percentage of the whole seeds versus the lengthwise-split seeds clearly illustrates this. The results for
the crosswise-split seeds are inconclusive, perhaps because this splitting will, in some cases, include some of the
endosperm and, in other cases, be restricted only to the bottom, germinating part of the seed.
PRACTICE
(Page 184)
Understanding Concepts
5. mH 1.01 u 2
2.02 u
mS 32.06 u 1
32.06 u
mO 16.00 u 4
64.00 u
mH SO
98.08 u
2
4(aq)
2.02 u
% H 100%
98.08 u
% H 2.06%
32.06 u
% S 100%
98.08 u
% S 32.69%
64.00 u
% O 100%
98.08 u
% O 65.25%
The percentage composition of H2SO4(aq) is 2.06% hydrogen atoms, 32.69% sulfur atoms, and 65.25% oxygen atoms
by mass.
6. mMg2+ 24.31 u 1 24.31 u
58.33 u
mMg(OH)
2(s)
24.31 u
% Mg2+ 100%
58.33 u
% Mg2+ 41.68%
The percentage, by mass, of magnesium ions in Mg(OH)2(s) is 41.68%.
7. mFe2+ 55.85 u 1
55.85 u
mO2– 16.00 u 1 16.00 u
mFeO
(s)
71.85 u
55.85 u
% Fe2+ 100%
71.85 u
% Fe2+ 77.73%
Copyright © 2002 Nelson Thomson Learning
Chapter 4 Quantities in Chemical Formulas
95
% O2–
% O2–
16.00 u
100%
71.85 u
22.27%
mFe3+ 55.85 u 2 111.70 u
mO2– 16.00 u 3 48.00 u
159.70 u
mFe O
2 3(s)
111.70 u
% Fe3+ 100%
15 9.70 u
% Fe3+ 69.94%
% O2–
% O2–
48.00 u
100%
159.70 u
30.06%
The percentage composition of FeO(s) is 77.73% iron(II) ions, 22.27% oxide ions by mass. The percentage
composition of Fe2O3(s) is 69.94% iron(III) ions, 30.06% oxide ions by mass.
8. mN 14.01 u 3
m(NH ) PO
4 3
4
42.03 u
149.12 u
42.03 u
% N 100%
149.12 u
% N 28.19%
The percentage, by mass, of nitrogen atoms in (NH4)3PO4(s) is 28.19%.
SECTION 4.5 QUESTIONS
(Page 184)
Understanding Concepts
1. The percentage composition of a new compound can be used to establish the correct formula.
2. The total mass of the compound sample is (33.5 + 30.4) g = 63.9 g.
33.5 g
% K+ 100% 52.4%
63.9 g
30.4 g
% Cl– 100% 47.6%
63.9 g
The percentage composition of the compound is 52.4% potassium ions and 47.6% chloride ions by mass.
3. The total mass of the compound sample is (23.0 + 16.0 + 32.0) g = 71.0 g.
23.0 g
% Na+ 100% 32.4%
71.0 g
16.0 g
% S 100% 22.5%
71.0 g
32.0 g
% O 100% 45.1%
71.0 g
The percentage composition of the compound is 32.4% sodium ions, 22.5% sulfur atoms, and 45.1% oxygen atoms
by mass.
28.02 u
4. mN 14.01 u 2
mNH NO
4
3(s)
80.06 u
28.02 u
% N 100%
80.06 u
96
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
% N 35.00%
mN 14.01 u 2 28.02 u
m(NH ) SO
4 2
4(s)
132.16 u
28.02 u
% N 100%
132.16 u
% N 21.20%
The percentage by mass of nitrogen atoms in ammonium nitrate is 35.00%. In ammonium sulfate this percentage is
21.20%.
5. m2H O 2[(1.01 u 2) + 16.00 u]
2
172.18 u
mCaSO 2H O
4
2 (s)
36.04 u
36.04 u
% H2O 100%
172.18 u
% H2O 20.93%
mH O [(1.01 u 2) + 16.00 u]
18.02 u
mCaSO H O
154.16 u
2
4
2 (s)
18.02 u
% H2O 100%
154.16 u
% H2O 11.69%
The percentage by mass of water molecules in calcium sulfate dihydrate is 20.93%. In calcium sulfate monohydrate
this percentage is 11.69%.
Applying Inquiry Skills
6. (a) Procedure
1. Measure and record the mass of the empty crucible to 0.01 g.
2. Place the copper in the crucible and measure and record the total mass to 0.01 g.
3. Add excess sulfur to the crucible, and heat strongly in a fume hood until all the copper has reacted, and all the
excess sulfur has burned off.
4. Allow the crucible and contents to cool, and measure and record the total mass to 0.01 g.
(b) Analysis
The masses of copper and of copper sulfide can be obtained by subtraction, and used to calculate the percent (by
mass) of copper ions in the compound. Subtracting from 100 will give the mass percent of sulfide ions in this
compound.
(c) Evaluation
The experimental design is judged to be adequate. It should easily provide dependable evidence from which to
calculate the required answer to the question.
Making Connections
7. (a) Percentage by mass is routinely used for home products like medications (0.05% oxymetazoline hydrochloride
in a nasal spray) or cleaners (3% sodium hypochlorite when packed, in bleach). Foodstuffs almost never have
mass percent data. Instead they list mass of each component per “serving size.”
(b) Typical products that use a percentage other than mass include vinegar, which is typically 5 or 7% acetic acid by
volume, and alcoholic beverages, which list alcohol (ethanol) as volume percent as well.
Copyright © 2002 Nelson Thomson Learning
Chapter 4 Quantities in Chemical Formulas
97
4.6 EMPIRICAL AND MOLECULAR FORMULAS
PRACTICE
(Page 186)
Understanding Concepts
1. Empirical means derived from observation and experimentation.
2. A molecular formula gives the number of each kind of atom or ion, as opposed to an empirical formula which gives
the simplest numerical ratio of the component atoms and/or ions.
3. Different compounds can exist because the same number and kind of atoms are bonded together differently, like
ethanol, CH3CH2OH, and dimethyl ether, CH3OCH3. These two different compounds have very different properties,
but would have the same percentage composition.
4. Possible molecular formulas could be C2H6, C3H9, C4H12, or indeed, any compound with the general formula CnH2n,
where n is any integer.
5. (a) NO2
(b) CO2
(c) CH2O
(d) C3H2Cl
6. Sodium chloride does not exist as molecules, but as a three-dimensional lattice of ions; so there is no such concept as
a molecular formula for this, or any other, ionic compound. The same rule applies to network solid elements and
compounds — the formulas we use are always the simplest ratio of component ions or atoms.
Try This Activity : Distinguish Between Empirical and Molecular Formulas
(Page 187)
Name
Empirical Formula
Molecular Formula
ethane
CH3
C 2H 6
butane
C 2H 5
C4H10
hexane
C 3H 7
C6H14
ethene
CH2
C 2H 4
butene
CH2
C 4H 8
hexene
CH2
C6H12
4.7 CALCULATING CHEMICAL FORMULAS
PRACTICE
(Page 188)
Understanding Concepts
1. Empirical formulas can be determined from mass percent information.
2. Molecular formulas can be determined if one also has molar mass information.
3. Assume a 100 g sample, for convenience.
mK+
= 28.9 g
MK+ = 39.10 g/mol
mS
= 23.7 g
MS = 32.06 g/mol
mO
= 47.4 g
MO = 16.00 g/mol
nK+
nK+
98
1 mol
= 28.9 g 39.10 g
= 0.739 mol
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
nS
nS
nO
nO
1 mol
= 23.7 g 32.06 g
= 0.739 mol
1 mol
= 47.4 g 16.00 g
= 2.96 mol
The mole ratio, K+ : S : O, is 0.739 : 0.739 : 2.96.
Simplifying (dividing each value by the lowest), we obtain 1.00 : 1.00 : 4.01, or almost exactly 1 : 1 : 4, making the
empirical formula KSO4(s).
4. (a) Assume a 100.0 g sample, for convenience.
mC
= 40.87 g
MC = 12.01 g/mol
mH
= 3.72 g
MH = 1.01 g/mol
mN
= 8.67 g
MN = 14.01 g/mol
mO
= 24.77 g
MO = 16.00 g/mol
mCl
= 21.98 g
MCl = 35.45 g/mol
nC
nC
nH
nH
nN
nN
nO
nO
nCl
nCl
1 mol
= 40.87 g 12.01 g
= 3.403 mol
1 mol
= 3.72 g 1.01 g
= 3.68 mol
1 mol
= 8.67 g 14.01 g
= 0.619 mol
1 mol
= 24.77 g 16.00 g
= 1.548 mol
1 mol
= 21.98 g 35.45 g
= 0.6200 mol
The mole ratio, C:H:N:O:Cl, is 3.403 : 3.68 : 0.619 : 1.548 : 0.6200.
Simplifying (dividing each by 0.619), we obtain 5.50 : 5.95 : 1.00 : 2.50 : 1.00. This is obviously not an
integral ratio, but two more of the values become integers upon doubling, which gives 11.0 : 11.9 : 2.00 : 5.00 :
2.00, or 11: 12 : 2 : 5 : 2, making the empirical formula for chloromycetin C11H12N2O5Cl2(s).
(b) Assume a 100.0 g sample, for convenience.
mC
= 41.86 g
MC = 12.01 g/mol
mH
= 4.65 g
MH = 1.01 g/mol
mN
= 16.28 g
MN = 14.01 g/mol
mO
= 18.60 g
MO = 16.00 g/mol
mS
= 18.60 g
MS = 32.06 g/mol
Copyright © 2002 Nelson Thomson Learning
Chapter 4 Quantities in Chemical Formulas
99
nC
nC
nH
nH
nN
nN
nO
nO
nS
nS
1 mol
= 41.86 g 12.01 g
= 3.485 mol
1 mol
= 4.65 g 1.01 g
= 4.60 mol
1 mol
= 16.28 g 14.01 g
= 1.162 mol
1 mol
= 18.60 g 16.00 g
= 1.163 mol
1 mol
= 18.60 g 32.06 g
= 0.5802 mol
The mole ratio, C:H:N:O:S, is 3.485 : 4.60 : 1.162 : 1.163 : 0.5802.
Simplifying, we obtain 6.007 : 7.93 : 2.003 : 2.005 : 1.000. This is obviously nearly an integral ratio, or 6 : 8 : 2 : 2 : 1,
making the empirical formula for sulfanilamide C6H8N2O2S(s).
5. (a) Assume a 100 g sample, for convenience.
mP
= (100 – 43.6) g = 56.4 g
MP = 30.97 g/mol
mO
= 43.6 g
MO = 16.00 g/mol
nP
nP
nO
nO
1 mol
= 56.4 g 30.97 g
= 1.82 mol
1 mol
= 43.6 g 16.00 g
= 2.73 mol
The mole ratio, P : O is 1.82 : 2.73.
Simplifying (dividing each value by the lowest), we obtain 1.00 : 1.50, which, when doubled, is 2 : 3, making
the empirical formula P2O3(s).
(b) Assume a 100 g sample, for convenience.
mP
= (100 – 56.6) g = 43.4 g
MP = 30.97 g/mol
mO
= 56.6 g
MO = 16.00 g/mol
nP
nP
nO
nO
1 mol
= 43.4 g 30.97 g
= 1.40 mol
1 mol
= 56.6 g 16.00 g
= 3.54 mol
The mole ratio, P : O, is 1.40 : 3.54.
Simplifying (dividing each value by the lowest), we obtain 1.00 : 2.53, which, when doubled, is 2.00 : 5.06, or
2:5, making the empirical formula P2O5(s).
100
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
6. Assume a 100.0 g sample, for convenience.
mC
= (26.80 – 4.90) g = 21.90 g
MC = 12.01 g/mol
mH
= 4.90 g
MH = 1.01 g/mol
nC
nC
nH
nH
1 mol
= 21.90 g 12.01 g
= 1.823 mol
1 mol
= 4.90 g 1.01 g
= 4.85 mol
The mole ratio, C:H, is 1.823 : 4.85.
Simplifying, we obtain 1.000 : 2.66. This is obviously not an integral ratio, nor will doubling the ratio help.
Multiplying the ratio through by three gives 3.000 : 7.98, or 3 : 8, making the empirical formula for propane
C3H8(g).
Applying Inquiry Skills
7. (a) Experimental Design
A sample of copper oxide will be reacted with carbon to burn away the oxygen and leave copper. The empirical
formula of the oxide will be determined from the proportions of copper and oxygen in the oxide.
Procedure
1. Measure and record the mass of the empty crucible to 0.01 g.
2. Add the copper oxide sample to the crucible and measure and record the total mass to 0.01 g.
3. Add excess carbon to the crucible, and heat strongly until all the carbon has burned away, and only copper metal
remains.
4. Allow the crucible and contents to cool, and measure and record the total mass to 0.01 g.
Evidence
mass of crucible
______ g
mass of crucible plus copper oxide
______ g
mass of crucible plus copper
______ g
Analysis
The masses of copper and oxygen in the sample are found by subtraction, if we assume the sample contained only
copper and oxide ions. These masses are converted to moles, to allow determination of the mole ratio, and thus
the formula.
(b) If the oxide is Cu2O(s), then the evidence would show:
mCu+
63.55 u 2 127.10 u
mO2–
16.00 u 1
mCu O
2 (s)
16.00 u
143.10 u
127.10 u
% Cu+ 100%
143.10 u
+
% Cu 88.819%
16.00 u
% O2– 100%
143.10 u
% O2– 11.18%
The percentage composition of Cu2O(s) is approximately 89% copper ions and 11% oxygen ions by mass. If the
sample were this compound, the evidence would show these proportions.
Copyright © 2002 Nelson Thomson Learning
Chapter 4 Quantities in Chemical Formulas
101
PRACTICE
(Page 193)
Understanding Concepts
8. Assume one mole of compound, 58.1 g.
mass % C = 62.0%
MC = 12.01 g/mol
mass % H = 10.4%
MH = 1.01 g/mol
mass % O = 27.5%
MO = 16.00 g/mol
62.0
1 mol
nC = 58.1 g 100
12.01 g
nC = 3.00 mol
10.4
1 mol
nH = 58.1 g 100
1.01 g
nH = 5.98 mol
27.5
1 mol
nO = 58.1 g 100
16.00 g
nO = 0.999 mol
The integral mole ratio is 3 : 6 : 1, so the molecular formula is C3H6O.
9. Assume one mole of compound, 92.0 g.
mass % N = 30.4%
MN = 14.01 g/mol
mass % O = 69.6%
MO = 16.00 g/mol
30.4
1 mol
nN = 92.0 g 100
14.01 g
nN = 2.00 mol
69.6
1 mol
nO = 92.0 g 100
16.00 g
nO = 4.00 mol
The integral mole ratio is 2 : 4, so the molecular formula is N2O4, and the name is dinitrogen tetraoxide.
Applying Inquiry Skills
10. (a) Analysis
Assume one mole of compound, 180.2 g.
mass % C = 40.0%
MC = 12.01 g/mol
mass % H = 6.8%
MH = 1.01 g/mol
mass % O = 53.2%
MO = 16.00 g/mol
40.0
1 mol
nC = 180.2 g 100
12.01 g
nC = 6.00 mol
6.8
1 mol
nH = 180.2 g 100
1.01 g
nH = 12 mol
53.2
1 mol
nO = 180.2 g 100
16.00 g
nO = 5.99 mol
The integral mole ratio is 6 : 12 : 6, so the molecular formula of this carbohydrate is C6H12O6.
102
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
PRACTICE
(Page 194)
Understanding Concepts
11. Natural source products usually contain traces of many other substances, and will normally vary somewhat in composition.
Making Connections
12. Vitamin D is produced naturally beneath the human skin surface by reaction of the body to exposure to sunlight. It is
classed as an essential substance (a vitamin) because a lack of it causes a wide variety of deficiency diseases, the most
notable of which is rickets. Rickets is characterized by a softening of the bones, causing limbs to grow bent, especially in children; and by a whole host of other symptoms. In fact, vitamin D is a steroid hormone that affects a great
number of bodily functions.
In 1921 Sir Edward Mellanby reported that dogs raised in the absence of sunlight could be kept healthy with a
proper diet and that cod-liver oil contained some trace substance that prevented rickets. Cod-liver oil became a
common tonic for people for the next several decades. Vitamin D (which actually has several forms) has since been
synthesized, and is commonly available without prescription in tablet form in pharmacies. One of vitamin D’s many
aspects is that it aids bones to use calcium, so patients who take calcium supplements usually take vitamin D as well.
Adding vitamin D to milk automatically ensures that the calcium in the milk is more useful and that small children
receive an adequate amount of the vitamin itself.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
Explore an Issue
Debate: Are Natural Vitamins Better for Your Health?
(Page 195)
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
PRACTICE
(Page 196)
Understanding Concepts
13. A mass spectrometer can provide data that allows the determination of the molar mass of a substance.
14. Since CH would be 13.02 g/mol, and the molar mass is known to be 26 g/mol, the molecular formula must be C2H2(g).
15. Assume one mole of the compound ethane, 30.1 g.
mass % C = 79.8%
MC = 12.01 g/mol
mass % H = 20.2%
MH = 1.01 g/mol
79.8
1 mol
nC = 30.1 g 100
12.01 g
nC = 2.00 mol
20.2
1 mol
nH = 30.1 g 100
1.01 g
nH = 6.02 mol
The integral mole ratio is 2 : 6, so the molecular formula of ethane from this evidence is C2H6(g).
Applying Inquiry Skills
16. Experimental Design
A sample of compound will be strongly heated to cause it to decompose to CaO(s) and CO2(g). The proportions of these
two constituents will be determined from mass measurements.
Copyright © 2002 Nelson Thomson Learning
Chapter 4 Quantities in Chemical Formulas
103
Procedure
1. Measure and record the mass of the empty crucible to 0.01 g.
2. Add the compound sample to the crucible and measure and record the total mass to 0.01 g.
3. Heat strongly until all the sample has decomposed, and only CaO(s) remains.
4. Allow the crucible and contents to cool, and measure and record the total mass to 0.01 g.
Evidence
mass of crucible
mass of crucible plus compound
mass of crucible plus calcium oxide
______ g
______ g
______ g
Analysis
The masses of CaO and CO2 in the sample are found by subtraction, if we assume the sample contained only these
two constituents. These masses are converted to moles, to allow determination of the mole ratio, and thus the proportions.
Reflecting
17. The compound can be analyzed with a combustion analyzer and with a mass spectrometer to provide evidence to use
in calculating its molecular formula.
SECTIONS 4.6 – 4.7 QUESTIONS
(Page 197)
Understanding Concepts
1. An empirical formula gives the simplest integral ratio of its constituent atoms, while a molecular formula also gives
the correct actual number of atoms.
2. To calculate a molecular formula, the molar mass is needed.
3. To determine the empirical formula of a compound from the percentage composition, find the mass of each element
in 100 g of the compound, using percentage composition; then find the amount in moles of each element by using
the molar mass of the element; and then find the lowest integral ratio of atoms to determine the empirical formula.
4. The empirical formulas are, respectively:
C5H4
C2H6O
NH2
O
C5H7
C6H8N2O2S
5. Ionic compounds do not exist as molecules, but as three-dimensional lattices of ions; so there is no such concept as a
molecular formula for an ionic compound. The formulas we use are always the simplest ratio of component ions.
6. Assume a 100.00 g sample, for percentage conversion convenience.
mC
= 40.00 g
MC = 12.01 g/mol
mH
= 6.71 g
MH = 1.01 g/mol
mO
= (100.00 – 40.00 – 6.71) g = 53.29 g
MO = 16.00 g/mol
nC
nC
nH
nH
nO
nO
1 mol
= 40.00 g 12.01 g
= 3.331 mol
1 mo l
= 6.71 g 1.01 g
= 6.64 mol
1 m ol
= 53.29 g 16.00 g
= 3.331 mol
The mole ratio, C : H : O, is 3.331 : 6.64 : 3.331.
104
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
Simplifying (dividing each value by the lowest), we obtain a ratio of 1.000 : 1.99 : 1.000, or almost exactly 1 : 2 : 1,
making the empirical formula for the lactic acid CH2O.
(b) Given a molar mass for lactic acid of 90 g/mol, and a molar mass for CH2O (the empirical formula) of
30.03 g/mol, the molecular formula must be triple the empirical ratio, or C3H6O3.
Applying Inquiry Skills
7. (a) Procedure
1. Use a centigram balance to measure the masses of a penny and of a quarter, to 0.01 g.
2. Use a decigram balance to measure the total masses of the pennies and of the quarters, to 0.1 g.
3. Divide the total mass of the pennies by the mass of one penny to calculate the number of pennies.
4. Divide the total mass of the quarters by the mass of one quarter to calculate the number of quarters.
5. Express the numbers of the two types of coins as a ratio.
(b) A parallel procedure a scientist might use:
1. Use a reference to find the molar masses of each kind of atom in the compound, to 0.01 g/mol.
2. Use a combustion analyzer to measure the total masses of each kind of atom in the compound, to 0.01 g.
3. Divide the total mass of each type of atom by the molar mass, to calculate the amount of atoms.
4. Express the numbers of the different types of atoms as a simplest integral ratio.
CHAPTER 4 REVIEW
(Page 199)
Understanding Concepts
1. Every compound has a specific proportion of constituent substances. This led scientists to the concept of specific kinds
of atoms for each element, and to comparing their combining masses to obtain a relative scale of masses of atoms.
2. The relative atomic mass would be (exactly) 24 u.
3. A hydrogen atom would be 18/12 of its current mass, or 1.5 × 1.01 u, or 1.52 u.
4. Relative atomic masses cannot be assigned correctly unless the combining ratio is correctly known; so the correct
molecular formula is necessary.
5. Elements consist of atoms with identical numbers of protons and electrons. The nuclei of atoms of elements, however,
may vary in numbers of neutrons, which has negligible effect on chemical properties, but does change the atom’s mass
significantly. For nearly every element there are several of these isotopes, and the average mass of atoms of such an
element is a value that depends on the mass of these isotopes and also their proportion in nature. The classic example
is chlorine, where roughly 3/4 of any sample will consist of chlorine-35 atoms (molar mass 35.00 g/mol), and roughly
1/4 of the sample will be chlorine-37 atoms (molar mass 37.00 g/mol). The molar mass of chlorine, then, is the
average molar mass of all the chlorine atoms in a sample, which works out to 35.45 g/mol.
6. For silicon, with significant amounts of three isotopes, and if we assume the molar mass of isotopes is the same as the
mass number, to two decimals (which is approximately valid, although the last digit is uncertain ...)
M = [(0.9221 × 28.00) + ( 0.0470 × 29.00) + ( 0.0309 × 30.00)] g/mol
M = 28.11 g/mol
(The actual value is 28.09 g/mol.)
7. (a) 12 u exactly (by definition)
(b) 12 g exactly (by definition)
(c) Avogadro’s constant is (by definition) the number of entities in exactly 12 g of pure carbon-12. This number is
used to define the mole, which is the (numerical) amount of any substance that is this number of entities (atoms,
ions, molecules, or formula units).
(d) The symbol M represents the quantity, molar mass, in g/mol units.
8. (a) Formula: CaCO3(s)
M = [(40.08) + (12.01) + (16.00 × 3)]
M = 100.09 g/mol
Copyright © 2002 Nelson Thomson Learning
Chapter 4 Quantities in Chemical Formulas
105
(b) Formula: N2O4(g)
M = [(14.01 × 2) + (16.00 × 4)]
M = 92.02 g/mol
(c) Formula: Na2CO310H2O(s)
M = [(22.99 × 2) + (12.01) + (16.00 × 3) + (18.02 × 10)]
M = 286.19 g/mol
Note: The last solution above assumes that students will have memorized the molar mass of water — an
extremely useful quantity to commit to memory. For future study, memorization of the molar mass of carbon
dioxide, 44.01 g/mol, will be almost as useful, because it is a product in so many reactions.
9. (a) mNaCl = 1.000 kg
MNaCl = 58.44 g/mol
1 mol
nNaCl = 1.000 kg 58.44 g
nNaCl = 0.01711 kmol = 17.11 mol
1.000 kg of table salt is 17.11 mol.
(b) mCO = 1.000 kg
2
MCO = 44.01 g/mol
2
1 mol
nCO = 1.000 kg 2
44.01 g
nCO = 0.02272 kmol = 22.72 mol
2
1.000 kg of dry ice is 22.72 mol.
(c) mH O = 1.000 kg
2
MH O = 18.02 g/mol
2
1 mol
nH O = 1.000 kg 2
18.02 g
nH O = 0.05549 kmol = 55.49 mol
2
1.000 kg of water is 55.49 mol.
10. (a) nO = 1.50 mol
2
MO = 32.00 g/mol
2
32.00 g
mO = 1.50 mol 2
1
mol
mO = 48.0 g
2
1.50 mol of liquid oxygen is 48.0 g.
(b) nHg = 1.50 mmol
MHg = 200.59 g/mol
200.59 g
mHg = 1.50 mmol
1
mol
mHg = 301 mg (or 0.301 g)
1.50 mmol of liquid mercury is 301 mg.
(c) nBr = 1.50 kmol
2
MBr = 159.80 g/mol
2
159.80 g
mBr = 1.50 kmol
2
1
mol
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Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
mBr = 240 kg
2
1.50 kmol of liquid bromine is 240 kg.
11.(a) nHC H O = 0.42 mol
2 3 2
= 6.02 × 1023 (entities)/mol
6.02 × 1023
NHC H O = 0.42 mol 2 3 2
1
mol
NHC H O = 2.5 × 1023
NA
2 3 2
0.42 mol of acetic acid is 2.5 × 1023 molecules.
= 7.6 × 10–4 mol
(b) nCO
= 6.02 × 1023 (entities)/mol
NA
6.02 × 1023
= 7.6 × 10–4 mol 1m
ol
20
= 4.6 × 10
NCO
NCO
7.6 × 10–4 mol of carbon monoxide is 4.6 × 1020 molecules.
(c) mCCl
4
MCCl
4
= 100 g
= 153.81 g/mol
= 6.02 × 1023 (entities)/mol
1 mol
nCCl
= 100 g 4
153.81 g
nCCl
= 0.650 mol
4
6.02 × 1023
NCCl
= 0.650 mol 4
1
mol
NCCl
= 3.91 × 1023
NA
4
or
NCCl
4
NCCl
4
1
mol
6.02 × 1023
= 100 g 153.81 g
1m
ol
= 3.91 × 1023
100 g of carbon tetrachloride is 3.91 × 1023 molecules.
(d) mH S
2
= 100 g
MH S
= 34.08 g/mol
NA
= 6.02 × 1023 (entities)/mol
1 mol
nH S
= 100 g 2
34.08 g
nH S
= 2.93 mol
2
6.02 × 1023
NH S
= 2.93 mol 2
1m
ol
NH S
= 1.77 × 1024
2
2
or
NH S
1
mol
6.02 × 1023
= 100 g 34.08 g
1m
ol
NH S
= 1.77 × 1024
2
2
100 g of hydrogen sulfide is 1.77 × 1024 molecules.
Copyright © 2002 Nelson Thomson Learning
Chapter 4 Quantities in Chemical Formulas
107
12. (a) MC
14H18N2O5
MC
14H18N2O5
= [(12.01 × 14) + (1.01 × 18) + (14.01 × 2) + (16.00 × 5)] g/mol
= 294.34 g/mol
The molar mass of Aspartame is 294.34 g/mol
(b) mC
14H18N2O5
= 35 mg
MC
= 294.34 g/mol
1 mol
nC H N O = 35 mg 14 18 2 5
294.34 g
nC H N O = 0.12 mmol
14H18N2O5
14 18 2 5
35 mg of Aspartame is 0.12 mmol.
= 0.12 mmol = 1.2 × 10–4 mol
(c) nC
14H18N2O5
= 6.02 × 1023 (entities)/mol
NA
NC
14H18N2O5
NC
14H18N2O5
6.02 × 1023
= 1.2 × 10–4 mol 1m
ol
= 7.2 × 1019
NH
= 7.2 × 1019 × 18
NH
= 1.3 × 1021
35 mg of Aspartame contains 1.3 × 1021 hydrogen atoms.
13. (a) Assume one mole of compound, for convenience.
g
mNa+
22.99 1 mol 22.99 g
m
ol
mNaN
65.02 g
3(s)
`
22.99 g
% Na+ 100%
65.02 g
+
% Na 35.36%
The percentage, by mass, of sodium ions in NaN3(s) is 35.36%.
(b) Assume one mole of compound, for convenience.
g
mAl3+
26.98 2 mol 53.96 g
m
ol
mAl O
101.96 g
2 3(s)
53.96 g
% Al3+ 100%
101.96 g
% Al3+ 52.92%
The percentage, by mass, of aluminum ions in Al2O3(s) is 52.92%.
(c) Assume one mole of compound, for convenience.
g
mN 14.01 1 mol 14.01 g
m
ol
mC H
8 11NO2(s)
153.20 g
14.01 g
% N 100%
153.20 g
% N 9.145%
The percentage, by mass, of nitrogen atoms in C8H11NO2(aq) is 9.145%.
14. (a) An empirical formula gives the simplest integral ratio of the constituent atoms of a compound, while a molecular
formula also gives the correct actual number of atoms in one molecule.
108
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
(b) An empirical formula is just a proportion of atoms, and cannot give a correct number of atoms per molecule without
evidence of the mass of the molecule — it is like saying Tom is twice as old as Harry, when, without the age of one,
the other’s age cannot be determined.
15. (a) Assume a 100.00 g sample, for percentage conversion convenience.
mC
= 49.5 g
MC = 12.01 g/mol
mH
= 5.15 g
MH = 1.01 g/mol
mN
= 28.9 g
MN = 14.01 g/mol
mO
= (by subtraction) = 16.5 g
MO = 16.00 g/mol
nC
nC
nH
nH
nN
nN
nO
nO
1 mol
= 49.5 g 12.01 g
= 4.12 mol
1 mol
= 5.15 g 1.01 g
= 5.10 mol
1 mol
= 28.9 g 14.01 g
= 2.06 mol
1 mol
= 16.5 g 16.00 g
= 1.03 mol
The mole ratio, C : H : N : O, is 4.12 : 5.10 : 2.06 : 1.03.
Simplifying (dividing each value by the lowest), we obtain a ratio of 4.00 : 4.95 : 2.00 : 1.00, or almost exactly
4 : 5 : 2 : 1, making the empirical formula for the caffeine C4H5N2O.
(b) Since the molar mass of the caffeine is 195 g/mol, about double the value of 97.11g/mol that we get for
C4H5N2O, the molecular formula must be C8H10N4O2.
16. (a) mC
= 3.161 g
MC = 12.01 g/mol
mH
= 0.266 g
MH = 1.01 g/mol
mO
= 1.052 g
MO = 16.00 g/mol
nC
1 mol
= 3.161 g 12.01 g
nC
= 0.2632 mol
nH
nH
nO
nO
1 mol
= 0.266 g 1.01 g
= 0.263 mol
1 mol
= 1.052 g 16.00 g
= 0.06575 mol
The mole ratio, C : H : O, is 0.2632 : 0.263 : 0.06575.
Simplifying (dividing each value by the lowest), we obtain a ratio of 4.00 : 4.00 : 1.00, making the empirical
formula for the ester C4H4O.
(b) Since the molar mass of the ester is 136 g/mol, about double the value of 68.08 g/mol that we get for C4H4O,
the molecular formula must be C8H8O2.
17. Assume a 100.00 g sample, for percentage conversion convenience.
mC = 63.2 g
MC = 12.01 g/mol
Copyright © 2002 Nelson Thomson Learning
Chapter 4 Quantities in Chemical Formulas
109
mH = 5.26 g
MH = 1.01 g/mol
mO = 31.6 g
MO = 16.00 g/mol
nC
1 mol
= 63.2 g 12.01 g
nC
= 5.26 mol
nH
1 mol
= 5.26 g 1.01 g
nH
= 5.21 mol
nO
1 mol
= 31.6 g 16.00 g
nO
= 1.98 mol
Simplifying (dividing each value by the lowest), we obtain a ratio of 2.66 : 2.63 : 1.00, which is not an integral ratio.
Tripling this ratio gives 7.97 : 7.89 : 3.00, or almost exactly 8 : 8 : 3, making the empirical formula for the vanillin
C8H8O3.
18. mNa+ 22.99 u 3
68.97 u
mP 30.97 u 1
30.97 u
mO 16.00 u 4
64.00 u
mNa PO
163.94 u
3
4(aq)
68.97 u
% Na+ 100%
163.94 u
% Na+ 42.07%
%P
30.97 u
100%
163.94 u
%P
18.89%
%O
64.00 u
100%
163.94 u
%O
39.04%
The percentage composition of Na3PO4(aq) is 42.07% sodium ions, 18.89% phosphorus atoms, and 39.04% oxygen
atoms, by mass.
19. mNa+ 22.99 u 3 68.97 u
mAs 30.97 u 1
74.92 u
mO 16.00 u 4
64.00 u
mNa AsO
207.89 u
3
4(aq)
68.97 u
% Na+ 100%
207.89 u
% Na+ 33.18%
110
% As
74.92 u
100%
207.89 u
% As
36.04%
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
64.00 u
% O 100%
207.89 u
% O 30.79%
The percentage composition of Na3AsO4(aq) is 33.18% sodium ions, 36.04% arsenic atoms, and 30.79% oxygen
atoms, by mass.
Applying Inquiry Skills
20. (a) Analysis
mZn
= (36.244 – 35.603) g
= 0.641 g
mcompound = (36.933 – 35.603) g
= 1.330 g
mCl
= (1.330 – 0.641) g
nZn2+
nZn2+
nCl–
nCl–
= 0.689 g
1 mol
0.641 g 65.38 g
0.00980 mol
1 mol
0.689 g 35.45 g
0.0194 mol
MZn = 65.38 g/mol
MCl = 35.45 g/mol
Simplifying (dividing each value by the lowest), we obtain a ratio of 1.00 : 1.98, or almost exactly 1 : 2, making
the empirical formula for the compound ZnCl2.
(b) Evaluation
The design is simple and straightforward, and is likely to provide reliable evidence, from which the answer can
be easily calculated.
21. (a) If the mass of the nail is measured before and after reaction, the mass difference will be copper, and can be used
to determine the number of atoms.
(b) Besides the evidence from (a), the molar mass of copper and the value of Avogadro’s constant will be needed.
(c) If the mass of copper(II) sulfate is initially measured, and the reaction is continued until all of the copper(II) ions
react, then masses can be obtained to allow the determination of percentage by mass of copper in the compound.
22. (a) The crucible and lid are preheated to drive off any combustible or vaporizable material, so the mass of the crucible
will not change later when heated.
(b) Handling hot equipment means wearing proper clothing, using heat-resistant gloves or tongs, and being careful
not to set hot materials down on unprotected bench tops.
23. Subtracting the mass of the element from the mass of the oxide formed should give the mass of oxygen reacted.
Making Connections
24. Carbon monoxide and carbon dioxide are formed from the same elements; but one of these compounds is highly toxic
to humans, and the other is not.
25. Moles are used for medical materials because the reaction quantities are the principal concern, and reaction equations
are dependent on numerical amounts. For foods, masses are common because that is the easy and convenient way to
measure how much food there is in a package or container.
26. Typical answers might include ...
(a) Food testing
Water treatment
Cement manufacturing
Plastics manufacturing
Prescription lens grinding
(b) Natural gas treatment labs
Water quality test labs
Steel composition analysis
Lubricant contamination analysis
Air quality test facilities
Copyright © 2002 Nelson Thomson Learning
Chapter 4 Quantities in Chemical Formulas
111
Exploring
27. Typical information ...
Salicylate compounds were known to be painkillers by the 5th century B.C. In an attempt to find a relief from arthritic
pain for his father, Felix Hoffmann first synthesized acetylsalicylic acid in 1893. The compound was named Aspirin
and was being marketed by the Bayer corporation in 1897. By 1899 Heinrich Dreser was routinely using this drug to
treat arthritis.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
28. Typical information ...
The most recent value for Avogadro’s constant is obtained by precise measuring of atomic (ionic) sizes in metallic
crystals, yielding a value with 9 significant digits — 6.022 141 99 1023 entities.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
29. Typical information ...
Cheese has been made for at least 5500 years. The protein casein, in milk, clumps into curds upon the addition of
rennin. Cottage cheese is sold in this initial stage. These curds may be heated, pressed, or strained to remove liquid
whey. Cheddaring is the process of compressing curds to remove moisture. Many cheeses are also “ripened” by treatment with bacteria and/or moulds, which may be added into the cheese (internal treatment, like Roquefort) or rubbed
on the surface (surface treatment, like Brie or Camembert). The texture and flavour of cheeses depends primarily on
their moisture content and the agents used to ripen them. The holes in Swiss cheese are formed during ripening by
gases produced by the bacteria used to create the characteristic flavour.
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112
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
(f) Mass measurements are in general more accurate than volume measurements for solids. The volume depends on the
size of the particles and how tightly the solid is packed into a measuring spoon or cup. However, for small quantities
it is much easier, faster, and requires less expensive equipment to measure volumes than masses. (People who bake
use specific techniques and spoons to minimize the problem of varying density of solids.)
5.1 QUANTITATIVE ANALYSIS
PRACTICE
(Page 205)
Understanding Concepts
1. In a quantitative (complete) reaction of two reactants, both cannot be in excess. If both are present in stoichiometric
ratio quantity, neither would, strictly speaking, be the limiting reagent (or both would ...).
2. A would have to be limiting. B would have to be reacted in excess, to ensure that all of A reacts — so that subsequent
calculations would correctly indicate how much A was originally present.
PRACTICE
(Page 208)
Making Connections
3. A typical report might deal with the recently developed highly complex blood/urine analysis for recombinant human
erythropoietin. EPO is a substance that can be used (illegally) by athletes to artificially increase production of red
blood cells, which improves the oxygen-carrying capacity of the blood, and thus improves performance in endurance
sports. The correlated career would be that of a sports-related laboratory analyst, or of a research scientist (in one
example an Australian, Canadian, Chinese, and Norwegian team, working for Bayer corporation at a research centre
in New York state).
4. A typical career might be that of a process and quality control analyst for any chemical industry. Such a career usually
requires certification from a postsecondary technical institution.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
PRACTICE
(Page 209)
Applying Inquiry Skills
5. A standard curve is a graph of the relationship between quantities of any two substances involved in a quantitative
stoichiometric chemical reaction. Such a curve is used to quickly predict the quantity of one of the substances when
the quantity of the other is known.
6. (a) Successive 10-mL samples (aliquots) of sodium sulfide solution should be added to the lead(II) nitrate solution,
allowing the precipitate to settle each time, until addition of the next aliquot causes no reaction. The total volume
of sodium sulfide solution used is then in excess.
(b) The mass of lead(II) nitrate should be the dependent variable (y-axis) and the mass of lead(II) sulfide should be
the independent variable (x-axis).
Making Connections
7. The postal data could be used to make a bar graph but not a standard curve, because the postal service charges the
same cost for a range of masses.
114
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
SECTION 5.1 QUESTIONS
(Page 209)
Understanding Concepts
1. (a) B is the limiting reagent.
(b) 1 A +
1B → 2C
2.1 mol n
2 mol C
nC = 2.1 mol
B 1
mol B
nC = 4.2 mol
The amount of C formed will be 4.2 mol.
2. CH4(g)
2 O2(g)
→ CO2(g) 2 H2O(g)
32 g
m
16.05 g/mol
32.00 g/mol
1 mol
nCH
= 32 g 4
16.05 g
nCH
= 2.0 mol
4
2
= 2.0 mol (mole ratio O2 : CH4 is 2 : 1)
nO
2
1
nO
= 4.0 mol
2
32.00 g
mO
= 4.0 mol 2
1
mol
mO
= 1.3 × 102 g = 0.13 kg
mO
1
mol C
H4
2
mol O2
32.00 g O
= 32 g CH4 2
16.05 g CH4
1
mol CH4
1
mol O
2
mO
= 1.3 × 102 g = 0.13 kg
2
or
2
2
The minimium mass of oxygen required for complete combustion would be 0.13 kg. A 10% excess would be
0.13 kg 110% 0.14 kg.
3. A standard curve is a graph of the relationship between quantities of any two substances involved in a quantitative
stoichiometric chemical reaction. If the curve is plotted empirically, from repeated mass measurements, the reaction
equation becomes irrelevant.
Applying Inquiry Skills
4. The limiting reagent must be measured accurately, because the predictive calculation is made from this value.
5. Successive samples (aliquots) of excess reagent solution should be added to the limiting reagent solution, allowing the
precipitate to settle each time, until addition of the next aliquot causes no reaction. The total volume of excess reagent
solution used is then empirically in excess.
6. Several reactions must be done with samples of aluminum of varying masses, and excess copper(II) sulfate solution
each time. From the measured mass of copper produced each time, an average copper/aluminum mass ratio can be
calculated, and used to draw a standard “curve” for this reaction.
Making Connections
7. Almost all everyday and workplace situations that involve a quantity ratio are now performed by calculators, rather
than by use of tables or graphs. Everyday examples include cooking quantities and automobile mileage; while work
examples might be quantities used in photofinishing, or mixing herbicide solutions on a farm, or a whole host of other
examples.
Copyright © 2002 Nelson Thomson Learning
Chapter 5 Quantities in Chemical Equations
115
5.2 BALANCING CHEMICAL EQUATIONS
PRACTICE
(Page 211)
Understanding Concepts
1. Conservation of mass requires that a reaction equation somehow represent the fact that total reactant mass equals total
product mass.
2. (a) Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
(b) 2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)
(c) CaCO3(s) + 2 HCl(aq) → CaCl2(s) + H2O(l) + CO2(g)
(d) Cu(s) + 4 HNO3(aq) → Cu(NO3)2(aq) + 2 NO(g) + 2 H2O(l)
(e) 2 C3H6(g) + 9 O2(g) → 6 CO2(g) + 6 H2O(g)
3. (a) 2 H2(g) + O2(g) → 2 H2O(g)
(b) Correct
(c) Pb(s) + 2 AgNO3(aq) → 2 Ag(s) + Pb(NO3)2(aq)
(d) Correct
4. Fe(NO3)3(aq) 1 3 LiOH(aq) → 3 LiNO3(aq) 1 Fe(OH)3(s)
PRACTICE
(Page 213)
Understanding Concepts
5. The formula coefficients of a chemical equation represent the mole ratio of substances in the reaction.
6. Consider:
C(s) + O2(g) → CO2(g)
This reaction equation shows clearly that the number of moles of substances in reactions is not conserved: two moles
of reactants become one mole of product.
7. 3 NO2(g) H2O(l) → 2 HNO3(aq) NO(g)
The mole ratio is 3:1:2:1 for the reaction equation as written here.
8. In a chemical industry, the amounts produced and consumed in reactions determine the economics of the process —
so the mole ratio is essential knowledge for determining whether any process is practical.
PRACTICE
(Page 214)
Understanding Concepts
9. Fertilizers increase crop production. Without them there would not be enough food produced to support the current
population of the Earth.
10. Fertilizer excess that enters the environment can cause serious changes in watersheds, affecting all living things that
depend on that water supply.
Making Connections
11. Typical reports might select groups like Ducks Unlimited, Canada, which is dedicated to actively working to preserve
Canada’s wetlands. DU emphasize that not only do wetlands provide habitat for a complex ecosystem of living things;
they act as natural filters for our water supply.
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116
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
Explore an Issue
Role Play: Controlling the Use of Fertilizers
(Page 215)
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SECTION 5.2 QUESTIONS
(Page 215)
Understanding Concepts
1. Balancing by inspection uses a trial-and-error approach to determining the coefficients of a reaction equation.
2. (a) 2 Al(s) + 3 CuCl2(aq) → 3 Cu(s) + 2 AlCl3(aq)
(b) Cu(s) + 2 HCl(aq) → H2(g) + CuCl2(aq)
(c) 2 HgO(s) → 2 Hg(l) + O2(g)
(d) CH4(g) + H2O(g) → CO(g) + 3 H2(g)
(step i)
CO(g) + 2 H2(g) → CH3OH(g)
(step ii)
3. (a) C(s) + O2(g) → CO2(g)
(b) S8(s) + 8 O2(g) → 8 SO2(g)
(c) Cu(OH)2(s) + H2SO4(aq) → 2 H2O(l) + CuSO4(aq)
(d) CaSiO3(s) + H2SO3(aq) → H2SiO3(aq) + CaSO3(s)
(e) CaCO3(s) + 2 HNO3 → H2CO3(aq) + Ca(NO3)2(aq)
(f) 2 Al(s) + 3 H2SO4(aq) → 3 H2(g) + Al2(SO4)3(aq)
(g) SO2(g) + H2O(l) → H2SO3(aq)
(h) 2 Fe(s) + 3 H2SO3(aq) → 3 H2(g) + Fe2(SO3)3(s)
(i) N2(g) + O2(g) → 2 NO(g)
(j) CO2(g) + H2O(l) → H2CO3(aq)
(k) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
(l) 2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)
(m) 2 FeS(s) + 3 O2(g) → 2 FeO(s) + 2 SO2(g)
(n) 2 H2S(g) + 3 O2(g) → 2 H2O(g) + 2 SO2(g)
(o) 2 CaCO3(s) + 2 SO2(g) + O2(g) → 2 CaSO4(s) + 2 CO2(g)
4. (a) 4 CH3NO2(l) + 7 O2(g) → 4 CO2(g) + 6 H2O(g) + 4 NO2(g)
The mole ratio is 4:7:4:6:4 for the reaction equation as written here.
(b) 2 Al(s) + 3 CuCl2(aq) → 3 Cu(s) + 2 AlCl3(aq)
The mole ratio is 2:3:3:2 for the reaction equation as written here.
Making Connections
5. Fertilizers are sold with information about what area can be treated per kilogram. Farmers would have to do area
calculations, cost calculations, and predictive calculations about the value of projected increased crop yield.
Copyright © 2002 Nelson Thomson Learning
Chapter 5 Quantities in Chemical Equations
117
→ 187
76Os +
(c)
187 Re
75
(d)
11 B+ 1 p
5
1
(e)
98
42 Mo
0
–1e
→ 3 42He
1
+ 21H → 99
43Tc + 0n
7. The mass number drops by 28, and only alpha decay drops the mass number, and each alpha particle is 4, so 7 alpha
particles are emitted. The atomic number drops by 10, which is 4 less than the drop of 14 caused by the alpha decays,
so there must be 4 beta particles emitted, because each one raises the atomic number by 1.
Making Connections
8. A typical report would include information on directing a particle beam through the body to cause maximum damage
at a tumour location — and probably on the development of the “cobalt bomb” cobalt-60 cancer therapy machines by
AECL, and the extensive worldwide use of this treatment program.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
5.4 CALCULATING MASSES OF REACTANTS AND PRODUCTS
PRACTICE
(Page 227)
Understanding Concepts
1. 2 Al2O3(s)
→
3 O2(g) 125 g
101.96 g/mol
1 mol
nAl O
= 125 g 2 3
101.96 g
nAl O
= 1.23 mol
2 3
4
nAl
= 1.23 mol 2
nAl
= 2.45 mol
26.98 g
mAl
= 2.45 mol 1
mol
mAl
= 66.2 g
or
mAl
mAl
4 Al(s)
m
26.98 g/mol
1
mol A
l2O3
l
26.98 g Al
4
mol A
= 125 g Al
2O3 l2O3
1
mol A
101.96 g A
l2O3 2 mol A
l
= 66.2 g
The (maximum) mass of aluminum that can be produced is 66.2 g.
2. C3H8(g) 5 O2(g) → 3 CO2(g) 4 H2O(g)
10.0 g
m
44.11 g/mol 32.00 g/mol
1 mol
nC H
= 10.0 g 3 8
44.11 g
nC H
= 0.227 mol
3 8
5
nO
= 0.227 mol 2
1
nO
= 1.13 mol
2
32.00 g
mO
= 1.13 mol 2
1
mol
120
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
mO
= 36.3 g
mO
1
mol C
5
mol O2
32.00 g O
3H8
= 10.0 g C
2
3H8 44.11 g C
H
1
m
o
l
C
H
1
mol O
2
3 8
3 8
= 36.3 g
2
or
2
mO
2
The (minimum) mass of oxygen required is 36.3 g.
3. 2 NaCl(aq) Pb(NO3)2(aq) → 2 NaNO3(aq) PbCl2(s)
2.57 g
m
58.44 g/mol
278.10 g/mol
1 mol
nNaCl
= 2.57 g 58.44 g
nNaCl
= 0.0440 mol
1
nPbCl
= 0.0440 mol 2
2
nPbCl
= 0.0220 mol
2
278.10 g
mPbCl
= 0.0220 mol 2
1
mol
mPbCl
= 6.11 g
2
or
1
mol P
bCl
278.10 g PbCl2
1m
ol N
aCl
mPbCl = 2.57 g NaCl
2 2
58.44 g N
aCl
2
mol N
aCl
1
mol P
bCl2
mPbCl
= 6.11 g
2
The (maximum) mass of lead(II) chloride produced would be 6.11 g.
4. 2 Al(s)
3 H2SO4(aq)
→ 3 H2(g)
2 Al2(SO4)3(aq)
2.73 g
m
26.98 g/mol
2.02 g/mol
1 mol
nAl
= 2.73 g 26.98 g
nAl
= 0.101 mol
3
nH
= 0.101 mol 2
2
nH
= 0.152 mol
2
2.02 g
mH
= 0.152 mol 2
1
mol
mH
= 0.307 g
2
or
mH
2
mH
2
3
mol H
2.02 g H
l
1
mol A
= 2.73 g Al
2 2
2
mol A
1
mol 26.98 g A
l
l
H2
= 0.307 g
The (predicted) mass of hydrogen would be 0.307 g, or 307 mg.
5. 2 KOH(aq) Cu(NO3)2(aq)
→ 2 KNO3(aq)
Cu(OH)2(s)
2.67 g
m
56.11 g/mol
97.57 g/mol
1 mol
nKOH
= 2.67 g 56.11 g
nKOH
= 0.0476 mol
1
nCu(OH) = 0.0476 mol 2
2
nCu(OH) = 0.0238 mol
2
97.57 g
mCu(OH) = 0.0238 mol 2
1
mol
Copyright © 2002 Nelson Thomson Learning
Chapter 5 Quantities in Chemical Equations
121
mCu(OH) = 2.32 g
2
or
1
mol Cu(
OH)
1m
ol K
OH
mCu(OH) = 2.67 g KOH
2
2
56.11 g K
OH
2
mol K
OH
mCu(OH) = 2.32 g
97. 57 g Cu(OH)
2
1
mol C
u(OH)2
2
The mass of copper(II) hydroxide produced would be 2.32 g.
6. 2 H2O(l) → 2 H2(g)
O2(g)
25.0 g
N
18.02 g/mol
6.02 × 1023 molecules/mol
1 mol
nH O
= 25.0 g 2
18.02 g
nH O
= 1.39 mol
2
1
nO
= 1.39 mol 2
2
nO
= 0.694 mol
6.02 × 1023 molecules
1 mol
= 4.18 × 1023 molecules
2
= 0.694 mol NO
2
NO
2
or
NO
6.02 × 1023 molecules O2
1
mol H
1m
ol O2
2O
= 25.0 g H2O 18.02 g H2O
2
mol H2O
1 mol
O2
NO
= 4.18 × 1023 molecules
2
2
In this reaction, 4.18 × 1023 molecules of oxygen would be produced.
PRACTICE
(Page 228)
Understanding Concepts
7. The fundamental test for a scientific concept is its ability to predict.
8. The reaction equation coefficients show the ratio of substances in moles, so measurements must be changed to
amounts before the ratio can be applied.
9. (a) The mole ratio CO2 : C8H18 (from the equation) is 16 : 2.
(b) 2 C8H18(g) 25 O2(g)
→
16 CO2(g) 18 H2O(g)
22.8 g
m
114.26 g/mol
44.01 g/mol
1 mol
nC H
= 22.8 g 8 18
114.26 g
nC H
= 0.200 mol
8 18
16
nCO
= 0.200 mol 2
2
nCO
= 1.60 mol
2
44.01 g
mCO
= 1.60 mol 2
1
mol
mCO
= 70.3 g
2
or
mCO
2
mCO
2
1
mol C
16 mol C
O2
44.01 g CO2
8H18
= 22.8 g C
8H18 114.26 g C
2
mol C
1m
ol C
O2
8H18
8H18
= 70.3 g
The mass of carbon dioxide produced will be 70.3 g.
(c) The mole ratio O2 : C8H18 (from the equation) is 25 : 2.
122
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
(d) 2 C8H18(g)
22.8 g
114.26 g/mol
nC H
8 18
nC H
8 18
nO
2
nO
2
mO
2
mO
2
or
mO
2
mO
2
25 O2(g)
→
16 CO2(g)
m
32.00 g/mol
1 mol
= 22.8 g 114.26 g
= 0.200 mol
25
= 0.200 mol 2
= 2.49 mol
32.00 g
= 2.49 mol 1
mol
= 79.8 g
18 H2O(g)
1
mol C
25 mol O2
32.00 g O
8H18
= 22.8 g C
2
8H18 114.26 g C
2
mol C
1
mol O
2
8H18
8H18
= 79.8 g
The mass of oxygen consumed will be 79.8 g.
(e) If oxygen is the limiting reagent, some carbon will not react completely to CO2(g), but instead will react incompletely to form CO(g), which is very toxic.
10. TiO2(s) 2 Cl2(g) 2 C(s) → TiCl4(l) 2 CO(g)
1.00 kg
79.88 g/mol
TiCl4(l)
2 Mg(s)
→
Ti(s) 2 MgCl2(s)
m
47.88 g/mol
Note: This question requires the students to see that the mole ratio TiCl4 : TiO2 from the first equation can be
combined with the mole ratio Ti : TiCl4 from the second equation.
nTiO
2
nTiO
2
nTiCl
4
nTiCl
4
nTi
nTi
mTi
mTi
or
1 mol
= 1.00 kg 79.88 g
= 0.0125 kmol
1
= 0.125 kmol (mole ratio TiCl4 : TiO2 is 1 : 1 — step 1)
1
= 0.125 kmol
1
= 0.125 kmol (mole ratio Ti : TiCl4 is 1 : 1 — step 2)
1
= 0.125 kmol
47.88 g
= 0.125 kmol
1
mol
= 0.599 kg or 599 g
mTi
1
mol T
iO 2
1
mol T
i Cl
i
47.88 g Ti
1
mol T
= 1.00 kg TiO2 4 1
mol T
79.88 g T
iO2
1
mol T
iO2
1
mol T
iC l4
i
mTi
= 0.599 kg or 599 g
The (maximum) mass of titanium produced would be 599 g.
Copyright © 2002 Nelson Thomson Learning
Chapter 5 Quantities in Chemical Equations
123
SECTION 5.4 QUESTIONS
(Page 229)
Understanding Concepts
1. Reaction equations read in amounts, not masses, so combining reagents by mass according to equation values will not
work.
2. (a) Fe2O3(s) 3 H2(g)
→
2 Fe(s) 3 H2O(g)
500 kg
m
159.70 g/mol
55.85 g/mol
1 mol
nFe O
= 500 kg 2 3
159.70 g
nFe O
= 3.13 kmol
2 3
2
nFe
= 3.13 kmol 1
nFe
= 6.26 kmol
55.85 g
mFe
= 6.26 kmol
1
mol
mFe
= 350 kg
or
mFe
mFe
1
mol Fe
2
mol Fe
55.85 g Fe
2O3
= 500 kg Fe
2O3 159.70 g F
e2O3 1 mol F
e2O3
1
mol F
e
= 350 kg
The mass of iron produced would be 350 kg.
(b) Fe2O3(s)
3 H2(g) →
2 Fe(s) 1.0 t = 1.0 Mg
m
159.70 g/mol
2.02 g/mol
1 mol
nFe O
= 1.0 Mg 2 3
159.70 g
nFe O
= 0.0063 Mmol = 6.3 kmol
2 3
3
nH
= 0.0063 Mmol 2
1
nH
= 0.019 Mmol
2
2.02 g
mH
= 0.019 Mmol
2
1
mol
mH
= 0.038 Mg = 38 kg
3 H2O(g)
2
or
mH
2
mH
2
1
mol Fe
3m
ol H2
2.02 g H
2O3
= 1.0 Mg Fe
2
2O3 159.70 g F
e2O3 1 mol F
e2O3 1 mol H2
= 0.038 Mg = 38 kg
The mass of hydrogen required would be 38 kg.
(c) Fe2O3(s) 3 H2(g)
→
2 Fe(s)
m
159.70 g/mol
1 mol
nH O
= 220 kg 2
18.02 g
nH O
= 12.2 kmol
2
1
nFe O
= 12.2 kmol 2 3
3
nFe O
= 4.07 kmol
2 3
159.70 g
mFe O = 4.07 kmol
2 3
1
mol
124
Unit 2 Quantities in Chemical Reactions
3 H2O(g)
220 kg
18.02 g/mol
Copyright © 2002 Nelson Thomson Learning
mFe O
= 650 kg
mFe O
1
mol H
1
mol F
e2O3 159.70 g Fe2O3
2O
= 220 kg H2O 18.02 g H2O
3m
ol H2O
1
mol F
e2O3
= 650 kg
2 3
or
2 3
mFe O
2 3
The mass of iron(III) oxide consumed would be 650 kg.
3. (a) 2 C6H12(l) 5 O2(g)
→
2 C6H10O4(s) m
280 kg
84.18 g/mol
146.16 g/mol
1 mol
nC H O = 280 kg 6 10 4
146.16 g
nC H O = 1.92 kmol
6 10 4
2
nC H
= 1.92 kmol 6 12
2
nC H
= 1.92 kmol
6 12
84.18 g
mC H
= 1.92 kmol
6 12
1
mol
mC H
= 161 kg
2 H2O(g)
6 12
or
mC H
6 12
mC H
6 12
1
mol C6
H10O4
2m
ol C
84.18 g C6H12
6H12
= 280 kg C6
H10O4 146.16 g C6
H10O4
2
mol C6
H10O4
1
mol C
6H12
= 161 kg
The mass of cyclohexane required is 161 kg.
(b) 2 C6H12(l) 5 O2(g)
→
2 C6H10O4(s)
125 kg
m
84.18 g/mol
32.00 g/mol
1 mol
nC H
= 125 kg 6 12
84.18 g
nC H
= 1.48 kmol
6 12
5
nO
= 1.48 kmol 2
2
nO
= 3.71 kmol
2
32.00 g
mO
= 3.71 kmol
2
1
mol
mO
= 119 kg
2 H2O(g)
2
or
mO
2
mO
2
1m
ol C
5
mol O2
32.00 g O
6H12
= 125 kg C
2
6H12 84.18 g C
H
2
m
ol
C
H
1
mol O
2
6 12
6 12
= 119 kg
The mass of oxygen required is 119 kg.
(c) 2 C6H12(l) 5 O2(g)
→
2 C6H10O4(s)
m
200 g
84.18 g/mol
32.00 g/mol
1 mol
nO
= 200 g 2
32.00 g
nO
= 6.25 mol
2
2
nC H
= 6.25 mol 6 12
5
nC H
= 2.50 mol
2 H2O(g)
6 12
Copyright © 2002 Nelson Thomson Learning
Chapter 5 Quantities in Chemical Equations
125
mC H
6 12
mC H
6 12
or
mC H
6 12
mC H
6 12
84.18 g
= 2.50 mol 1
mol
= 210 g
1
mol O
2
2
mol C
84.18 g C6H12
6H12
= 200 g O2 32.00 g O2
5
mol O2
1
mol C
6H12
= 210 g
The maximum mass of cyclohexane that can be reacted is 210 g.
4. 2 Fe(s)
O2(g)
→
2 FeO(s)
56.8 g
m
55.85 g/mol
71.85 g/mol
1 mol
nFe
= 56.8 g 55.85 g
nFe
= 1.02 mol
2
nFeO
= 1.02 mol 2
nFeO
= 1.02 mol
71.85 g
mFeO
= 1.02 mol 1
mol
mFeO
= 73.1 g
or
mFeO
mFeO
e
2
mol FeO
71.85 g FeO
1
mol F
= 56.8 g Fe
2
mol Fe
1
mol F
eO
55.85 g Fe
= 73.1 g
The mass of iron(II) oxide that can be produced is 73.1 g.
Applying Inquiry Skills
5. (a) Prediction
From the balanced reaction equation, one mole of calcium ions produces one mole of solid calcium oxalate
precipitate, so the amount of calcium ions in a sample will be the same as the amount of calcium oxalate precipitate that forms.
(b) Experimental Design
Excess sodium oxalate solution will be added to a measured sample (independent variable) of hard water. The
precipitate formed will be dried and the mass measured (dependent variable). The stoichiometric method will be
used to calculate an answer to the question.
(c) The mass of calcium oxalate precipitate will be divided by its molar mass, 128.10 g/mol, to determine the amount
of precipitate. The amount of calcium ions is the same, since the mole ratio is 1 : 1.
(d) Materials
•
hard-water sample
126
•
sodium oxalate solution
•
wash bottle with pure water
•
two 100-mL graduated cylinders
•
250-mL beaker
•
filtration apparatus
•
filter paper
•
centigram balance
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
(e) Procedure
1. Use a graduated cylinder to add 10.0 mL of hard water to a 250-mL beaker.
2. Use another graduated cylinder to add about 10 mL of sodium oxalate solution to the beaker.
3. Allow the precipitate to settle.
4. Add another 10-mL aliquot of sodium oxalate solution to the beaker, observing whether or not more precipitate forms as the added solution mixes with the clear upper portion of the solution in the beaker.
5. Repeat step 4 until no more precipitate forms.
6. Measure and record the mass of a piece of filter paper.
7. Filter, wash, and dry the precipitate.
8. Measure and record the mass of the filter paper plus dry precipitate.
9. Dispose of all materials as instructed.
Making Connections
6. (a) Raw materials, like aluminum oxide, are purchased according to the amount needed for reaction.
(b) Products, like aluminum, are priced so that the amount produced will be profitable.
(c) Pollutant amounts will be calculated and decisions on processes to control them will be made accordingly.
(d) Amounts of all materials must be calculated to allow costing of things like packaging, disposal, and shipping.
5.5 CALCULATING LIMITING AND EXCESS REAGENTS
PRACTICE
(Page 231)
Understanding Concepts
1. (a) 2 CH4(g) 3 O2(g)
→
m
16.05 g/mol
3.0 mol
nCH
4
nCH
4
mCH
4
mCH
4
mCH
4
mCH
4
or
2 CO(g)
4 H2O(g)
2
= 3.0 mol 3
= 2.0 mol
16.05 g
= 2.0 mol 1
mol
= 32 g
2
mol CH4
16.05 g CH4
= 3 mol
O2 3
mol O2
1
mol C
H4
= 32 g
The mass of methane that will react is 32 g.
(b) CH4(g)
2 O2(g)
→
CO2(g)
m
16.05 g/mol
3.0 mol
1
nCH
= 3.0 mol 4
2
nCH
= 1.5 mol
4
16.05 g
mCH
= 1.5 mol 4
1
mol
mCH
= 24 g
2 H2O(g)
4
or
mCH
4
1
mol CH4
16.05 g CH4
= 3 mol
O2 2
mol O2
1
mol C
H4
Copyright © 2002 Nelson Thomson Learning
Chapter 5 Quantities in Chemical Equations
127
mCH
4
= 24 g
The mass of methane that will react is 24 g.
(c) In a closed garage the oxygen in the air will be the limiting reagent.
2. 8 Zn(s) 5.00 g
65.38 g/mol
nZn
nZn
nS
8
nS
8
mS
8
mS
8
or
S8(s)
→ 8 ZnS(s)
m
256.48 g/mol
1 mol
= 5.00 g 65.38 g
= 0.0765 mol
1
= 0.0765 mol 8
= 0.00956 mol
256.48 g
= 0.00956 mol 1
mol
= 2.45 g
1
mol S8
256.48 g S
1
mol Z
n
= 5.00 g Zn 8
65.38 g Zn
8
mol Zn
1
mol S
8
= 2.45 g
mS
8
mS
8
excess mS = 2.45 g 110% = 2.70 g
8
The mass of sulfur, to be in reasonable excess, should be 2.70 g.
3. 2 Na(s) Cl2(g)
→
2 NaCl(s)
0.75 g
m
22.99 g/mol
70.90 g/mol
1 mol
nNa
= 0.75 g 22.99 g
nNa
= 0.033 mol
1
nCl
= 0.033 mol 2
2
nCl
= 0.016 mol
2
70.90 g
mCl
= 0.016 mol 2
1
mol
mCl
= 1.2 g
2
or
mCl
2
mCl
2
1
mol Cl2
70.90 g Cl2
a
1
mol N
= 0.75 g Na
2
mol N
1
mol C
l2
22.99 g Na
a
= 1.2 g
excess mCl = 1.2 g 110% = 1.3 g
2
The mass of chlorine, to be in reasonable excess, should be 1.3 g.
4. Ca(OH)2(s)
2 HCl(aq) → CaCl2(aq) 2 H2O(l)
m
2.17 g
74.10 g/mol
36.46 g/mol
1 mol
nHCl
= 2.17 g 36.46 g
nHCl
= 0.0595 mol
1
nCa(OH) = 0.0595 mol 2
2
nCa(OH) = 0.0298 mol
2
74.10 g
mCa(OH) = 0.0298 mol 2
1
mol
128
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
mCa(OH) = 2.21 g
2
or
1
mol Ca(
OH)
74.10 g Ca(OH)2
1
mol H
Cl
mCa(OH) = 2.17 g HCl 2 2
36.46 g HCl
2m
ol HCl
1
mol Ca
(OH)2
mCa(OH) = 2.21 g
2
excess mCa(OH)
2
= 2.21 g 110% = 2.43 g
The mass of calcium hydroxide, to be in reasonable excess, should be 2.43 g.
PRACTICE
(Page 235)
Understanding Concepts
5. Prediction cannot be made from the amount of reagent in excess because not all of the amount will react.
6. (a) Zn(s)
CuSO4(aq)
→ Cu(s)
ZnSO4(aq)
0.42 mol
0.22 mol
Since this reactant mole ratio is 1:1, it is obvious by inspection that zinc is in excess, by an amount of
(0.42 – 0.22) mol = 0.20 mol.
(b) Cl2(g)
2 NaI(aq)
→ I2(s)
2 NaCl(aq)
10 mmol
10 mmol
Since this reactant mole ratio is 1:2, 10 mmol of NaI(aq) will require 10 mmol 1/2 5.0 mmol of Cl2(g) for reaction. The Cl2(g) is in excess by an amount of (10 – 5.0) mmol = 5 mmol.
(c) AlCl3(aq) 20 g
133.33 g/mol
nAlCl
3
nAlCl
3
nNaOH
nNaOH
3 NaOH(aq)
→ Al(OH)3(s)
19 g
40.00 g/mol
1 mol
= 20 g 133.33 g
= 0.150 mol
1 mol
= 19 g 40.00 g
= 0.475 mol
3 NaCl(aq)
Since this reactant mole ratio is 1:3, 0.150 mol of AlCl3(aq) will require 0.150 mol 3/1 0.450 mol of NaOH(aq)
for reaction.
The NaOH(aq) is in excess, by an amount of (0.475 0.450) mol = 0.025 mol, or 25 mmol.
7. (a) 8 Zn(s)
S8(s)
→
8 ZnS(s)
5.00 g
3.00 g
65.38 g/mol
256.48 g/mol
1 mol
nZn
= 5.00 g 65.38 g
nZn
= 0.0765 mol
1 mol
nS
= 3.00 g 8
256.48 g
nS
= 0.0117 mol
8
Since this reactant mole ratio is 8:1, 0.0765 mol of Zn(s) will require 0.0765 mol 1/8 0.00956 mol of S8(s) for
reaction. The S8(s) is in excess; so the Zn(s) is the limiting reagent.
(b) 8 Zn(s)
S8(s)
→
8 ZnS(s)
5.00 g
m
65.38 g/mol
97.44 g/mol
Copyright © 2002 Nelson Thomson Learning
Chapter 5 Quantities in Chemical Equations
129
nZn
nZn
nZnS
nZnS
mZnS
mZnS
or
mZnS
mZnS
1 mol
= 5.00 g 65.38 g
= 0.0765 mol
8
= 0.0765 mol 8
= 0.0765 mol
97.44 g
= 0.0765 mol 1
mol
= 7.45 g
8
mol S8
97.44 g S
1
mol Z
n
= 5.00 g Zn 8
65.38 g Zn
8
mol Zn
1
mol S
8
= 7.45 g
The mass of zinc sulfide produced would be 7.45 g.
5 O2(g)
→
367.4 g
32.00 g/mol
1 mol
= 100.7 g 44.11 g
= 2.28 mol
1 mol
= 367.4 g 32.00 g
= 11.5 mol
8. (a) C3H8(g) 100.7 g
44.11 g/mol
nC H
3 8
nC H
3 8
nO
2
nO
2
3 CO2(g)
4 H2O(g)
Since this reactant mole ratio is 1:5, 2.28 mol of C3H8(g) will require 2.28 mol 5/1 11.4 mol of O2(g) for reaction.
The O2(g) is in excess; so the C3H8(g) is the limiting reagent.
(b) C3H8(g) 5 O2(g)
100.7 g
44.11 g/mol
nC H
3 8
nC H
3 8
nH O
2
nH O
2
mH O
2
mH O
2
or
mH O
2
mH O
2
→
3 CO2(g)
4 H2O(g)
m
18.02 g/mol
1 mol
= 100.7 g 44.11 g
= 2.283 mol
4
= 2.283 mol 1
= 9.132 mol
18.02 g
= 9.132 mol 1
mol
= 164.6 g
1
mol C
4
mol H
18.02 g H2O
3H8
2O
= 100.7 g C
3H8 44.11 g C
1m
ol C
1
mol H
3H8
3H8
2O
= 164.6 g
The mass of water formed is 164.6 g.
9. (a) 2 SO2(g) O2(g)
→
2 SO3(g)
4.55 kg
2.88 kg
64.06 g/mol
32.00 g/mol
1 mol
nSO
= 4.55 kg 2
64.06 g
nSO
= 0.0710 kmol
2
1 mol
nO
= 2.88 kg 2
32.00 g
130
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
nO
= 0.0900 kmol
2
Since this reactant mole ratio is 2:1, 0.0710 kmol of SO2(g) will require 0.0710 kmol 1/2 0.0355 kmol of
O2(g) for reaction. The O2(g) is in excess; so the SO2(g) is the limiting reagent.
(b) 2 SO2(g)
4.55 kg
64.06 g/mol
nSO
2
nSO
2
nSO
3
nSO
3
mSO
3
mSO
3
or
mSO
3
mSO
3
→
O2(g)
2 SO3(g)
m
80.06 g/mol
1 mol
= 4.55 kg 64.06 g
= 0.0710 kmol
2
= 0.0710 kmol 2
= 0.0710 kmol
80.06 g
= 0.0710 kmol
1
mol
= 5.69 kg
1
mol S
O2
2
mol SO
80.06 g SO
= 4.55 kg SO
2 3 3
64.06 g SO2
2
mol SO2
1
mol S
O3
= 5.69 kg
The mass of sulfur trioxide produced is 5.69 kg.
10. (a) Fe2O3(s) 3 CO(g) → 2 Fe(s) 74.2 kg
40.3 kg
159.70 g/mol
28.01 g/mol
1 mol
nFe O
= 74.2 kg 2 3
159.70 g
nFe O
= 0.465 kmol
2 3
1 mol
nCO
= 40.3 kg 28.01 g
nCO
= 1.44 kmol
3 CO2(g)
Since this reactant mole ratio is 1:3, 0.465 kmol of Fe2O3(s) will require 0.465 kmol 3/1 1.39 kmol of CO(g) for
reaction. The CO(g) is in excess; so the Fe2O3(s) is the limiting reagent.
(b) Fe2O3(s) 74.2 kg
159.70 g/mol
nFe O
2 3
nFe O
2 3
nFe
nFe
mFe
mFe
or
mFe
mFe
→
2 Fe(s) m
55.85 g/mol
1 mol
= 74.2 kg 159.70 g
= 0.465 kmol
2
= 0.465 kmol 1
= 0.929 kmol
55.85 g
= 0.929 kmol
1
mol
= 51.9 kg
3 CO(g)
3 CO2(g)
1
mol Fe
2
mol F
e
55.85 g Fe
2O3
= 74.2 kg Fe
2O3 159.70 g F
e2O3 1 mol F
e2O3
1
mol F
e
= 51.9 kg
The mass of iron produced would be 51.9 kg.
Copyright © 2002 Nelson Thomson Learning
Chapter 5 Quantities in Chemical Equations
131
Applying Inquiry Skills
11. Experimental Design
A solid reaction product can be removed (by filtration, for instance) from the reaction system. If some more of the
(presumed) excess reagent is added, and no reaction occurs, then the limiting reagent is completely reacted and the
other reagent was in excess.
PRACTICE
(Page 236)
Applying Inquiry Skills
12. (a) If aluminum is the limiting reagent, the silver metal solid will have completely disappeared when the reaction
stops.
(b) If copper(II) sulfate is the limiting reagent, the silver metal solid will not have completely disappeared when the
reaction stops.
13. (a) If magnesium is the limiting reagent, the silver metal solid will have completely disappeared when the reaction
stops.
(b) If hydrochloric acid is the limiting reagent, the silver metal solid will not have completely disappeared when the
reaction stops.
SECTION 5.5 QUESTIONS
(Page 237)
Understanding Concepts
1. The limiting reagent is determined by the reaction mole ratio; so the mass may actually be much greater than the
excess reagent, as long as the amount is such that it is completely reacted.
2. The reaction only occurs until the limiting reagent is consumed, so the amount of the limiting reagent determines all
other amounts involved in the reaction.
3. One (or more) reactants must be in excess to ensure that the limiting reagent will be completely reacted.
4. (a) Zn(s)
2 HCl(aq) → H2(g) ZnCl2(aq)
0.35 mol
0.60 mol
Since this reactant mole ratio is 1:2, 0.60 mol of HCl(aq) will require 0.60 mol 1/2 0.30 mol of Zn(s) for reaction. The Zn(s) is in excess; so the HCl(aq) is the limiting reagent, and will be completely consumed.
(b) The Zn(s) is in excess, by an amount of (0.35 0.30) mol = 0.05 mol.
excess mZn = 0.05 mol 65.38 g/1 mol = 3 g
The mass of zinc remaining after the reaction ends will be 3 g.
5. (a) 2 KBr(aq) Pb(NO3)2(aq) → 2 KNO3(aq) PbBr2(s)
m
3.50 g
119.00 g/mol
331.22 g/mol
1 mol
nPb(NO )
= 3.50 g 3 2
331.22 g
nPb(NO )
= 0.0106 mol
3 2
1
nKBr
= 0.0106 mol 2
nKBr
= 0.00528 mol
119.00 g
mKBr
= 0.00528 mol 1
mol
mKBr
= 0.629 g
or
mKBr
mKBr
132
1
mol Pb
(NO3)2
1
mol KBr
119.00 g KBr
= 3.50 g Pb(NO
3)2 331.22 g Pb
(NO3)2 2 m
ol Pb
(NO3)2
1
mol K
Br
= 0.629 g
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
excess mKBr = 0.629 g 110% = 0.692 g
The mass of potassium bromide, to be in reasonable excess, should be 692 mg.
(b) 2 KBr(aq)
→ 2 KNO3(aq) m
367.00 g/mol
1 mol
= 3.50 g 331.22 g
= 0.0106 mol
1
= 0.0106 mol 1
= 0.0106 mol
119.00 g
= 0.0106 mol 1
mol
= 0.629 g
Pb(NO3)2(aq)
3.50 g
331.22 g/mol
nPb(NO )
3 2
nPb(NO )
3 2
nPbBr
2
nPbBr
2
mPbBr
2
mPbBr
2
or
mPbBr
2
mPbBr
PbBr2(s)
1
mol Pb
(NO3)2
1
mol P
bBr2
119.00 g PbBr
= 3.50 g Pb(NO
3)2 2
1m
ol P
bBr2
331.22 g Pb
(NO3)2 1 mol Pb(N
O3)2
= 0.629 g
2
The mass of lead(II) bromide produced would be 0.629 g.
6. (a) Cu(s) 10.0 g
63.55 g/mol
nCu
nCu
nAgNO
3
nAgNO
3
2 AgNO3(aq) → 2 Ag(s) 20.0 g
169.88 g/mol
1 mol
= 10.0 g 63.55 g
= 0.157 mol
1 mol
= 20.0 g 169.88 g
= 0.118 mol
Cu(NO3)2(aq)
Since this reactant mole ratio is 1:2, 0.118 mol of AgNO3(aq) will require 0.118 mol 1/2 0.0589 mol of Cu(s) for
reaction. The Cu(s) is in excess; so the AgNO3(aq) is the limiting reagent.
(b) Cu(s)
2 AgNO3(aq)
20.0 g
169.88 g/mol
nAgNO
3
nAgNO
3
nAg
nAg
mAg
mAg
or
mAg
mAg
→
2 Ag(s) Cu(NO3)2(aq)
m
107.87 g/mol
1 mol
= 20.0 g 169.88 g
= 0.118 mol
2
= 0.118 mol 2
= 0.118 mol
107.87 g
= 0.118 mol 1
mol
= 12.7 g
1
mol Ag
NO3
2
mol A
g
107.87 g Ag
= 20.0 g AgNO
3 169.88 g A
gNO3 2 mol A
gNO3
1
mol A
g
= 12.7 g
The mass of silver crystals produced would be 12.7 g.
Copyright © 2002 Nelson Thomson Learning
Chapter 5 Quantities in Chemical Equations
133
7. (a) FeCl3(aq) 26.8 g
162.20 g/mol
3 NaOH(aq)
21.5 g
40.00 g/mol
nNaOH
3
nFeCl
3
Fe(OH)3(s)
3 NaCl(aq)
1 mol
= 21.5 g 40.00 g
= 0.538 mol
1 mol
= 26.8 g 162.20 g
= 0.165 mol
nNaOH
nFeCl
→
Since this reactant mole ratio is 1:3, 0.165 mol of FeCl3(aq) will require 0.165 mol 3/1 0.496 mol of NaOH(aq)
for reaction. The NaOH(aq) is in excess; so the FeCl3(aq) is the limiting reagent.
(b) FeCl3(aq) 3 NaOH(aq) →
Fe(OH)3(s) 3 NaCl(aq)
26.8 g
m
m
162.20 g/mol
106.88 g/mol
58.44 g/mol
1 mol
nFeCl
= 26.8 g 3
162.20 g
nFeCl
= 0.165 mol
3
1
= 0.165 mol nFe(OH)
3
1
nFe(OH)
= 0.165 mol
3
106.88 g
mFe(OH) = 0.165 mol 3
1
mol
mFe(OH) = 17.7 g
3
or
mFe(OH)
mFe(OH)
3
1
mol Fe
Cl3
1
mol Fe
(OH)
106.88 g Fe(OH)
= 26.8 g FeCl
3 3 3
162.20 g F
eCl3
1m
ol F
eCl3
1
mol Fe
(OH)3
= 17.7 g
3
The mass of iron(III) hydroxide produced would be 17.7 g
3
and nNaCl
= 0.165 mol 1
nNaCl
= 0.496 mol
58.44 g
mNaCl = 0.496 mol 1
mol
mNaCl = 29.0 g
or
mNaCl
mNaCl
1
mol Fe
Cl3
3
mol N
aCl
58.44 g NaCl
= 26.8 g FeCl
3 162.20 g F
eCl3 1 m
ol F
eCl3
1
mol N
aCl
= 29.0 g
The mass of sodium chloride produced would be 29.0 g.
Applying Inquiry Skills
8. Determining if enough excess reagent has been added is the same as determining if all of the limiting reagent has
reacted. This can be done in any situation where a product can be removed from the reaction. For example, in a reaction of zinc and hydrochloric acid, where hydrogen bubbles escape, if more zinc is added and no bubbles form, then
all of the hydrochloric acid must have reacted.
134
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
5.6 THE YIELD OF A CHEMICAL REACTION
PRACTICE
(Page 239)
Understanding Concepts
1. The quantity of product predicted by stoichiometric calculation is the theoretical yield. When the reaction is carried
out, the measured quantity of product obtained is the actual yield.
2. An actual yield is a measured quantity, and normal measurement error as well as experimental error can result in this
value being greater than the theoretical yield.
3. (a) NaBr(aq) AgNO3(aq) → NaNO3(aq) AgBr(s)
(b) NaBr(aq)
AgNO3(aq)
→
NaNO3(aq)
5.00 g
169.88 g/mol
1 mol
nAgNO = 5.00 g 3
169.88 g
nAgNO = 0.0294 mol
3
1
nAgBr = 0.0294 mol 1
nAgBr = 0.0294 mol
187.77 g
mAgBr = 0.0294 mol 1
mol
mAgBr = 5.53 g
or
mAgBr
mAgBr
AgBr(s)
m (5.03 g actual)
187.77 g/mol
1
mol A
gNO3
1
mol A
gBr
187.77 g AgBr
= 5.00 g AgNO
3 169.88 g A
gNO3 1 mol A
gNO3
1
mol A
gBr
= 5.53 g
The mass of silver bromide produced should be 5.53 g.
(c) The mass of silver bromide actually produced is 5.03 g.
actual yield
(d) % yield = 100%
theoretical yield
5.03 g
% yield = 100% 91.0%
5.53 g
The percentage yield of silver bromide in this reaction is 91.0%.
4. (a) 4 FeS(s)
7 O2(g) →
2 Fe2O3(s) 4 SO2(g)
16.1 g
10.8 g
87.91 g/mol
32.00 g/mol
1 mol
nFeS
= 16.1 g 87.91 g
nFeS
= 0.183 mol
1 mol
nO
= 10.8 g 2
32.00 g
nO
= 0.338 mol
2
Since this reactant mole ratio is 4:7, 0.183 mol of FeS(s) will require 0.183 mol 7/4 0.320 mol of O2(g) for reaction. The O2(g) is in excess; so the FeS(s) is the limiting reagent.
(b) 4 FeS(s)
7 O2(g) →
2 Fe2O3(s) 4 SO2(g)
16.1 g
m (14.1 g actual)
87.91 g/mol
159.70 g/mol
1 mol
= 16.1 g nFeS
87.91 g
Copyright © 2002 Nelson Thomson Learning
Chapter 5 Quantities in Chemical Equations
135
nFeS
nFe O
2 3
nFe O
2 3
mFe O
2 3
mFe O
2 3
or
mFe O
2 3
mFe O
2 3
= 0.183 mol
2
= 0.183 mol 4
= 0.0916 mol
159.70 g
= 0.0916 mol 1
mol
= 14.6 g
2m
ol F
e2O3
159.70 g Fe2O3
1
mol F
eS
= 16.1 g FeS
87.91 g FeS
4m
ol F
eS
1
mol F
e2O3
= 14.6 g
The theoretical yield of iron(III) oxide would be 14.6 g.
actual yield
(c) % yield = 100%
theoretical yield
14.1 g
% yield = 100% 96.6%
14.6 g
The percentage yield of iron(III) oxide in this reaction is 96.6%.
5. Fe2O3(s) 3 CO(g)
→
2 Fe(s) 3 CO2(g)
1000 kg
m (635 kg actual)
159.70 g/mol
55.85 g/mol
1 mol
nFe O
= 1000 kg 2 3
159.70 g
nFe O
= 6.26 kmol
2 3
2
nFe
= 6.26 kmol 1
nFe
= 12.5 kmol
55.85 g
mFe
= 12.5 kmol
1
mol
mFe
= 699 kg
or
1
mol Fe
ol F
e
55.85 g Fe
2m
2O3
= 1000 kg Fe
2O3 1
mol F
159.70 g F
e2O3
1
mol F
e2O3
e
mFe
= 699 kg
actual yield
= 100%
theoretical yield
mFe
% yield
635 k
g
= 100% 90.8%
699 kg
The percentage yield of iron in this reaction is 90.8%.
CH3OH(l)
→
C8H8O3(l)
+
6. C7H6O3(s) +
2.00 g
m (1.65 g actual)
138.13 g/mol
152.16 g/mol
1 mol
nC H O = 2.00 g 7 6 3
138.13 g
nC H O = 0.0145 mol
7 6 3
1
nC H O = 0.0145 mol 8 8 3
1
nC H O = 0.0145 mol
8 8 3
152.16 g
mC H O = 0.0145 mol 8 8 3
1
mol
mC H O = 2.20 g
% yield
H2O(l)
8 8 3
or
136
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
mC
H O
8 8 3
mC
H O
8 8 3
1
mol C
1
mol C
152.16 g C8H8O3
7H6O3
8H8O3
= 2.00 g C
7H6O3 138.13 g C
ol C
1
mol C
7H6O3
7H6O3 1 m
8H8O3
= 2.20 g
actual yield
% yield = 100%
theoretical yield
1.65 g
% yield = 100% 74.9%.
2.20 g
The percentage yield of methyl salicylate in this reaction is 74.9%
Applying Inquiry Skills
7. The procedure listed is fatally flawed if the reaction produces any other product (besides the precipitate) that does not
vaporize upon heating, because any such product will also be mixed with the precipitate in the evaporating dish. The
normal way to efficiently recover a precipitate is to filter it using a piece of filter paper of known (measured) mass,
wash and dry it, and measure the mass of the paper plus precipitate.
Making Connections
8. (a) Student research should indicate how large-scale industry practices and procedures result in a more efficient
extraction of the red dye, carmine, from the bodies of female Dactylopus coccus insects (cochineal insects) than
the original hand process of crushing the insects and simmering them in water to extract the dye. Since so many
insects are needed to produce a reasonable amount (150 000 insects per kilogram of carmine), the product is
costly and the process efficiency is, therefore, very important.
(b) As the insects live on desert cactus plants, cochineal carmine can be produced in arid areas where no other profitable crop can be harvested. This means that in Peru a significantly profitable industry and source of income for
the residents of desert areas depends on these tiny bugs.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
PRACTICE
(Page 243)
Understanding Concepts
9. Yield less than predicted in a reaction may be due to experimental error inherent in the procedure; to impurities in the
reagents; to unwanted side reactions; and to reactions that are not quantitative — that do not “go to completion.”
10. (a) 2 Cu2O(s) Cu2S(s) →
6 Cu(s) SO2(g)
250 kg
129 kg
143.10 g/mol
159.16 g/mol
1 mol
nCu O = 250 kg 2
143.10 g
nCu O = 1.75 kmol
2
1 mol
nCu S
= 129 kg 2
159.16 g
nCu S
= 0.811 kmol
2
Since this reactant mole ratio is 2:1, 0.811 kmol of Cu2S(s) will require 0.811 kmol 2/1 1.62 kmol of Cu2O(s)
for reaction. The Cu2O(s) is in excess; so the Cu2S(s) is the limiting reagent.
(b) 2 Cu2O(s) Cu2S(s) →
6 Cu(s) SO2(g)
129 kg
m (285 kg actual)
159.16 g/mol
63.55 g/mol
1 mol
= 129 kg nCu S
2
159.16 g
nCu S
= 0.811 kmol
2
Copyright © 2002 Nelson Thomson Learning
Chapter 5 Quantities in Chemical Equations
137
6
= 0.811 kmol 1
= 4.86 kmol
63.55 g
= 4.86 kmol
1
mol
= 309 kg
nCu
nCu
mCu
mCu
or
1
mol C
u2S
6
mol Cu
63.55 g Cu
= 129 kg Cu
2S 159.16 g C
u2S
1
mol C
u2S
1m
ol C
u
= 309 kg
mCu
mCu
The theoretical yield of copper would be 309 kg.
actual yield
(c) % yield = 100%
theoretical yield
285 kg
% yield = 100% 92.2%
309 kg
The percentage yield of copper in this reaction is 92.2%.
11. (a) C(s)
2 H2(g)
→
CH4(g)
10.0 kg
m (4.20 kg actual)
12.01 g/mol
16.05 g/mol
1 mol
nC
= 10.0 kg 12.01 g
nC
= 0.833 kmol
1
nCH
= 0.833 kmol 4
1
nCH
= 0.833 kmol
4
16.05 g
mCH
= 0.833 kmol
4
1
mol
mCH
= 13.4 kg
4
or
mCH
4
mCH
4
1
mol CH4
16.05 g CH4
1
mol C
= 10.0 kg C
12.01 g C
1
mol C
1
mol C
H4
= 13.4 kg
actual yield
% yield = 100%
theoretical yield
4.20 kg
% yield = 100% 31.4%
13.4 kg
The percentage yield of methane in this reaction is 31.4%.
(b) If the coal is only 40% carbon, then the percentage yield of methane is increased by a factor of 100/40. The
percentage yield of methane in this reaction becomes 31.4% 100/40 = 78.6%.
Making Connections
12. (a) Water is a preferable (nonpolluting) solvent.
(b) Room temperature reactions don’t require energy input.
(c) Drying agents are less acceptable — they add an extra chemical.
(d) Purification is less preferred — it will take energy and maybe more chemicals.
(e) Biomass is preferable because it is a renewable resource.
138
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
SECTION 5.6 QUESTIONS
(Page 244)
Understanding Concepts
1. (a) Yield in school experiments can be improved mostly by being careful (e.g., using only clean equipment),
following procedure, and using good technique.
(b) Yields in industrial processes can be improved by adjusting reaction conditions, and sometimes by following a
different reaction sequence.
2. (a) C7H6O3(s) +
C4H6O3(s) → C9H8O4(s) +
C2H4O2(s)
2.00 g
4.00 g
138.13 g/mol
102.10 g/mol
1 mol
nC H O = 2.00 g 7 6 3
138.13 g
nC H O = 0.0145 mol
7 6 3
1 mol
nC H O = 4.00 g 4 6 3
102.10 g
nC H O = 0.0392 mol
4 6 3
Since the reactant mole ratio is 1:1, the C7H6O3(s) is obviously the limiting reagent for this reaction.
C7H6O3(s) +
2.00 g
138.13 g/mol
C4H6O3(s)
nC H O
7 6 3
nC H O
7 6 3
nC H O
9 8 4
nC H O
9 8 4
mC H O
9 8 4
mC H O
9 8
→
C9H8O4(s) +
C2H4O2(s)
m (2.09 g actual)
180.17 g/mol
1 mol
= 2.00 g 138.13 g
= 0.0145 mol
1
= 0.0145 mol 1
= 0.0145 mol
180.17 g
= 0.0145 mol 1
mol
= 2.61 g
4
or
mC
1
mol C
1
mol C
180.17 g C9H8O4
7H6O3
9H8O4
2.00 g C
7H6O3 138.13 g C
mol C
1
mol C
7H6O3 1 7H6O3
9H8O4
= 2.61 g
H O =
9 8 4
nC
H O
9 8 4
The theoretical yield of Aspirin would be 2.61 g.
actual yield
(b) % yield = 100%
theoretical yield
2.09 g
% yield = 100% 80.1%
2.61 g
The percentage yield of Aspirin in this reaction is 80.1%.
Applying Inquiry Skills
3. (a) Experimental Design
A measured sample of sodium silicate is dissolved and reacted with an excess of iron(III) nitrate in solution. The
resulting precipitate is filtered, washed, and dried to allow measurement of its mass.
Procedure
A typical student procedure will be very similar to the one created for Investigation 5.5.1. It should be preceded
by a calculation to determine what mass of iron(III) nitrate (per gram of sodium silicate) is required to ensure an
excess.
Copyright © 2002 Nelson Thomson Learning
Chapter 5 Quantities in Chemical Equations
139
(b) Evaluation
A percentage yield of 80% is very low for a precipitation reaction. This would probably indicate that the precipitate remained at least partly dissolved in the original solution, or dissolved to some extent in the wash water.
Obviously, a different process must be tried to improve the yield.
Making Connections
4. There are many “green” projects for students to research — one example might be the work on fuel cells to allow cars
to run on hydrogen, thus reducing pollutant levels significantly.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
5. Student reports will be specific to the educational institution they choose and especially to the person they choose to
interview. The report should concentrate on educational requirements, and the nature of the workday.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
5.7 CHEMISTRY IN TECHNOLOGY
PRACTICE
(Page 246)
Understanding Concepts
1. Inspection of the reaction equations shows that all mole ratios in each step are 1:1, so two moles (in total) of
hydrochloric acid are produced for each one mole of sodium carbonate produced.
2. The LeBlanc process was expensive because it required using a large amount of fuel for heat.
3. Air pollutants were a matter of personal concern to people affected directly by them in these centuries, but no government had, as yet, considered that controlling pollutants was its responsibility. People in general were not aware of longterm health hazards or, indeed, of specific pollution hazards, other than odours and skin or lung irritation. Also, because
industries were smaller and relatively few in number, pollutant effects tended to be local rather than widespread.
PRACTICE
(Page 247)
Understanding Concepts
4. (a) CaCO3(s) → CaO(s) CO2(g)
(b) CO2(g) NH3(aq) H2O(l) → NH4HCO3(aq)
(c) NH4HCO3(aq) NaCl(aq) → NH4Cl(aq) NaHCO3(s)
(d) 2 NaHCO3(s) → Na2CO3(s) H2O(g) CO2(g)
(e) CaO(s) H2O(l) → Ca(OH)2(s)
(f) Ca(OH)2(s) 2 NH4Cl(aq) → 2 NH3(g) CaCl2(aq) 2 H2O(l)
(g) CaCO3(s) 2 NaCl(aq) → Na2CO3(s) CaCl2(aq) (from text, p. 247)
5. (i) CaCO3(s) → CaO(s) CO
2(g)
(ii) CO2(g) NH3(aq) H2O(l) → NH
4HCO3 (aq)
(iii) NH
3(s)
4HCO3(aq) NaCl(aq) → NH
4Cl (aq) NaHCO
(iv) 2 NaHCO
3 (s) → Na2CO3(s) H2O(g) CO
2(g)
(v) CaO(s) H2O(l) → Ca(OH)
2 (s)
140
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
(vi) Ca(OH)
2 (s) 2 NH
NH3(g) CaCl2(aq) 2 H2O(l)
4Cl (aq) → 2 Totalling the above, we get:
CaCO3(s) 2 NaCl(aq) → Na2CO3(s) CaCl2(aq)
6. The raw materials are CaCO3(s) and NaCl(s), both of which are mined easily in large quantity.
7. The primary product is sodium carbonate (washing soda), Na2CO3(s). The byproduct is calcium chloride, CaCl2(s).
8. Sodium hydrogen carbonate (sodium bicarbonate or baking soda), NaHCO3(s), is a marketable intermediate.
Removing this intermediate from the reaction sequence means less sodium carbonate, Na2CO3(s), will be produced,
and some water and carbon dioxide must be added back into the reaction sequence.
9. Water and energy will also be required for this process.
10. Larger reaction scale is more efficient, and thus more economic. It does not cost twice as much to build a reaction
vessel twice as large, nor are twice as many employees required to operate it, but it will produce twice as much
product, and therefore, twice the value.
PRACTICE
(Page 249)
Making Connections
11. Industries may resist changing processes because of high restructuring costs (new manufacturing plants, new delivery
methods, perhaps a need for new resources) and employee change/training/dislocation costs. The risk is loss of profitability; the benefit may not be sufficient, or may not be sufficiently well defined.
12. Consumers have input through consumer action groups, by personal complaints, and through government, but their
primary input is the effect they have on the profitability of the process, through their decision to purchase or refuse to
purchase the product.
13. Student discussion should include examples of percentage composition, stoichiometric reaction quantities, and
percentage yield for the industrial chemical process chosen.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
SECTION 5.7 QUESTIONS
(Page 250)
Understanding Concepts
1. Science is the study of the natural world to describe, predict, and explain changes and substances; technology encompasses the skills, processes, and equipment required to make useful products or to perform useful tasks.
2. (a) Technological question.
(b) Scientific question.
(c) Scientific question.
(d) Technological question.
(e) Scientific question.
(f) Technological question.
3. The LeBlanc process produced sodium carbonate (washing soda), which was in demand for glassmaking, among other
things.
4. The LeBlanc process required a great quantity of costly fuel for energy, and produced a large quantity of undesirable
pollutants.
5. Some (only a few) of the uses of Solvay products and a byproduct:
(a) Na2CO3(s) — cleaning compounds, pH control, food additive, glassmaking, chemical analysis.
(b) NaHCO3(s) — baking powder, pharmaceuticals, sponge rubber, gold and platinum plating, fire extinguishers,
cleaners, and antacids.
(c) CaCl2(s) — de-icing, dust control on roads, fungicides, use in the paper industry, drying agent, and pharmaceuticals.
Copyright © 2002 Nelson Thomson Learning
Chapter 5 Quantities in Chemical Equations
141
CHAPTER 5
REVIEW
(Page 252)
Understanding Concepts
1. (a) In a reagent mix, the one consumed first, causing the reaction to cease, is the limiting reagent. Some of the other
reagent will remain, so it is said to be in excess.
(b) A chemical reaction involves change in the electrons of an entity; a nuclear reaction involves change in the atomic
nuclei.
(c) Alpha decay involves the emission of an alpha particle (helium-4 nucleus) from an atomic nucleus, while beta
decay involves the emission of a beta particle (electron) from a nucleus.
(d) The quantity of product predicted by stoichiometric calculation is the theoretical yield. When the reaction is
carried out, the measured quantity of product obtained is the actual yield.
(e) An empirical formula shows the simplest integral ratio of component entities. A molecular formula shows the
actual numerical ratio of atoms in a molecule of the substance.
2. (a) 2 SO2(g) O2(g) → 2 SO3(g)
(b) SO3(g) H2O(l) → H2SO4(aq)
(c) CaO(s) SO2(g) → O2(g) CaSO4(s)
(d) CaO(s) H2SO3(aq) → H2O(l) CaSO3(s)
(e) Al2(SiO3)3(s) 3 H2SO4(aq) → 3 H2SiO3(aq) Al2(SO4)3(s)
3. (a)
233Th
90
→ -10e 233 Pa
91
(b)
233 Pa
91
→ -10e 233 U
92
131 I
53
→ -10e 4.
131 Xe
54
122 I
53
0
→ 122
54Xe + -1e
(b)
59 Fe
26
→
(c)
222 Rn
86
4
→ 218
84Po + 2He
(d)
252 Cf
98
1
+ 105B → 259
103 Lr + 3 0n
(e)
239 Pu
94
1
+ 42He → 242
96Cm + 0n
5. (a)
59 Co
27
+ –10e
6. The mass number drops by 28, which means 28/4 = 7 alpha decays. The atomic number only drops by 10, which
means 14 10 = 4 beta decays.
0
190
7. 190
75Re → 76Os + -1e
9Li
3
→ 83Li + 10n
214 Bi
83
4
→ 210
79Au + 2He
162 Tm
69
120 In
49
0
→ 162
70Yb + -1e
0
→ 120
50Sn + -1e
8. 2 C8H18(l) 692 g
114.26 g/mol
25 O2(g)
nC H
8 18
nC H
8 18
nCO
142
2
→
16 CO2(g) 18 H2O(g)
m
44.01 g/mol
1 mol
= 692 g 114.26 g
= 6.06 mol
16
= 6.06 mol 2
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
nCO
= 48.5 mol
2
mCO
2
mCO
2
or
mCO
2
mCO
2
44.01 g
= 48.5 mol 1
mol
= 2.13 × 103 g = 2.13 kg
1
mol C
16 mol C
O2
44.01 g CO2
8H18
= 692 g C
8H18 114.26 g C
2
mol C
1
mol C
O2
8H18
8H18
= 2.13 × 103 g = 2.13 kg
The mass of carbon dioxide formed is 2.13 kg.
9. (a) WO3(s) 3 H2(g) →
m
231.85 g/mol
nW
nW
nWO
3
nWO
3
mWO
3
mWO
3
or
mWO
3
mWO
3
W(s) 3 H2O(g)
5.00 g
183.85 g/mol
1 mol
= 5.00 g 183.85 g
= 0.0272 mol
1
= 0.0272 mol 1
= 0.0272 mol
231.85 g
= 0.0272 mol 1
mol
= 6.31 g
1
mol W
O
231.85 g WO3
1
mol W
= 5.00 g W
3 183.85 g W
1
mol W
1
mol W
O3
= 6.31 g
The mass of tungsten(VI) oxide needed is 6.31 g.
(b) WO3(s) 3 H2(g) →
W(s)
3 H2O(g)
5.00 g
m
183.85 g/mol
18.02 g/mol
1 mol
nW
= 5.00 g 183.85 g
nW
= 0.0272 mol
3
nH O
= 0.0272 mol 2
1
nH O
= 0.0816 mol
2
18.02 g
mH O = 0.0816 mol 2
1
mol
mH O = 1.47 g
2
or
mH O
2
mH O
2
3
mol H2O
18.02 g H2O
1
mol W
= 5.00 g W
183.85 g W
1
mol W
1m
ol H
2O
= 1.47 g
The mass of water vapour produced is 1.47 g.
10. (a) HC4H7O2(l) + C2H5OH(l) → C4H7O2C2H5(l)
30.0 g
18.0 g
88.12 g/mol
46.08 g/mol
1 mol
nHC H O = 30.0 g 4 7 2
88.12 g
nHC H O = 0.340 mol
+
H2O(l)
4 7 2
Copyright © 2002 Nelson Thomson Learning
Chapter 5 Quantities in Chemical Equations
143
1 mol
nC H OH = 18.0 g 2 5
46.08 g
nC H OH = 0.391 mol
2 5
Since the reactant mole ratio is 1:1, the HC4H7O2(l) is obviously the limiting reagent for this reaction.
(b) HC4H7O2(l) + C2H5OH(l) → C4H7O2C2H5(l) + H2O(l)
30.0 g
m
88.12 g/mol
116.18 g/mol
1 mol
nHC H O
= 30.0 g 4 7 2
88.12 g
nHC H O
= 0.340 mol
4 7 2
1
nC H O C H
= 0.340 mol 4 7 2 2 5
1
nC H O C H
= 0.340 mol
4 7 2 2 5
116.18 g
mC H O C H
= 0.340 mol 4 7 2 2 5
1
mol
mC H O C H
= 39.6 g
4 7 2 2 5
or
116.18 g C4H7O2C2H5
1 mol
HC
1 mol
C4H7O2C2H5
4H7O2
mC H O C H = 30.0 g HC
4H7O2 4 7
2 5
2
mC H O C H = 39.6 g
4 7 2 2 5
The mass of ethylbutanoate produced is 39.6 g.
11. Use the equation mole ratios in sequence: 1 mol NH3(g) reacts to form 1 mol NO(g) ; and 1 mol NO(g) reacts to form
1 mol NO2(g) ; and 1 mol NO2(g) reacts to form 2/3 mol HNO3(aq) . This simplifies to: 1 mol NH3(g) reacts to form
2/3 mol HNO3(aq) , or integrally, 3 mol NH3(g) react to form 2 mol HNO3(aq) .
since
3 NH3(g) react to form
4.00 mol
nHNO
3
nHNO
3
mHNO
3
mHNO
3
or
mHNO
3
mHNO
3
2 HNO3(aq)
m
63.02 g/mol
2
= 4.00 mol 3
= 2.67 mol
63.02 g
= 2.67 mol 1
mol
= 168 g
2
mol H
NO
63.02 g HNO
= 4.00 mol NH3 3 3
3
mol NH3
1m
ol H
NO3
= 168 g
The mass of nitric acid produced is 168 g.
12. (a) C3H5(OH)3(l) 3 HNO3(aq) →
C3H5(NO3)3(s) H2O(l)
(b) C3H5(OH)3(l) 3 HNO3(aq) →
10.4 g
19.2 g
92.11 g/mol
63.02 g/mol
nC H (OH)
3 5
3
nC H (OH)
3 5
nHNO
3
nHNO
3
144
3
C3H5(NO3)3(s) H2O(l)
1 mol
= 10.4 g 92.11 g
= 0.113 mol
1 mol
= 19.2 g 63.02 g
= 0.305 mol
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
Since the reactant mole ratio is 1:3, 0.305 mol of HNO3(aq) would require 0.305 mol 1/3 = 0.102 mol of
C3H5(OH)3(l) to react completely. The amount of C3H5(OH)3(l) is in excess, so the HNO3(aq) is the limiting reagent
for this reaction.
(c) C3H5(OH)3(l) 3 HNO3(aq) →
19.2 g
63.02 g/mol
nHNO
3
nHNO
3
nC H (NO )
3 5
3 3
nC H (NO )
3 5
3 3
mC H (NO )
3 5
3 3
mC H (NO )
3 5
3 3
or
mC H (NO )
3 5
3 3
mC H (NO )
3 5
3 3
C3H5(NO3)3(s) H2O(l)
m (22.6 g actual)
227.11 g/mol
1 mol
= 19.2 g 63.02 g
= 0.305 mol
1
= 0.305 mol 3
= 0.102 mol
227.11 g
= 0.102 mol 1
mol
= 23.1 g
227.11 g C3H5(NO3)3
1
mol H
NO3
1
mol C3H5(NO3)3
= 19.2 g HNO
3 3
63.02 g H
NO3
mol H
NO3
= 23.1 g
The theoretical yield of nitroglycerin should be 23.1 g.
actual yield
(d) % yield = 100%
theoretical yield
22.6 g
% yield = 100% 97.8%
23.1 g
The percentage yield of nitroglycerin in this reaction is 97.8%
13. Yield less than predicted in a reaction may be due to experimental error inherent in the procedure; to impurities in the
reagents; to unwanted side reactions; and to reactions that are not quantitative — that do not “go to completion.”
14. (a) SiO2(s) 4 HF(aq) →
SiF4(g) 2 H2O(l)
6.80 g
m
60.09 g/mol
104.09 g/mol
1 mol
nSiO
= 6.80 g 2
60.09 g
nSiO
= 0.113 mol
2
1
nSiF
= 0.113 mol 4
1
nSiF
= 0.113 mol
4
104.09 g
mSiF
= 0.113 mol 4
1
mol
mSiF
= 11.8 g
4
or
mSiF
4
mSiF
4
1m
ol S
iO2
1
mol S
iF4
104.09 g SiF
= 6.80 g SiO
2 4
60.09 g S
iO2
1
mol S
iO2
1
mol Si
F4
= 11.8 g
The mass of silicon tetrafluoride produced is 11.8 g.
(b) SiO2(s) 4 HF(aq) →
SiF4(g) 53.2 g
m
60.09 g/mol
20.01 g/mol
1 mol
nSiO
= 53.2 g 2
60.09 g
nSiO
= 0.885 mol
2 H2O(l)
2
Copyright © 2002 Nelson Thomson Learning
Chapter 5 Quantities in Chemical Equations
145
4
= 0.885 mol 1
= 3.54 mol
20.01 g
= 3.54 mol 1
mol
= 70.9 g
nHF
nHF
mHF
mHF
or
1m
ol S
iO2
4
mol HF
20.01 g HF
= 53.2 g SiO
2 60.09 g S
iO2
1
mol SiO2
1m
ol H
F
= 70.9 g
mHF
mHF
The mass of hydrofluoric acid required is 70.9 g.
(c) SiO2(s) 10.6 g
60.09 g/mol
4 HF(aq)
nSiO
2
nSiO
2
nH O
2
nH O
2
mH O
2
mH O
2
or
mH O
2
mH O
2
→
SiF4(g)
2 H2O(l)
m
18.02 g/mol
1 mol
= 10.6 g 60.09 g
= 0.176 mol
2
= 0.176 mol 1
= 0.353 mol
18.02 g
= 0.353 mol 1
mol
= 6.36 g
1m
ol S
iO2
2
mol H
18.02 g H2O
2O
= 10.6 g SiO
2 60.09 g S
iO2
1
mol SiO2
1
mol H
2O
= 6.36 g
The mass of water produced is 6.36 g.
15. (a) C2H5OH(l) 3 O2(g)
(b) C2H5OH(l) 450.0 g
46.08 g/mol
3 O2(g)
nC H OH
2 5
nC H OH
2 5
nCO
2
nCO
2
mCO
2
mCO
2
or
mCO
mCO
2
→
3 H2O(g)
2 CO2(g) m
44.01 g/mol
1 mol
= 450.0 g 46.08 g
= 9.766 mol
2
= 9.766 mol 1
= 19.53 mol
44.01 g
= 19.53 mol 1
mol
= 859.6 g
3 H2O(g)
2 CO2(g)
→
= 450.0 g C2
H5OH 2m
ol C
O2
44.01 g CO2
1
mol C2
H5OH
1
mol C2
H5OH
1
mol C
O2
= 859.6 g
2
The mass of carbon dioxide produced is 859.6 g.
(c) C2H5OH(l) 450.0 g
146
3 O2(g)
→
Unit 2 Quantities in Chemical Reactions
2 CO2(g)
3 H2O(g)
m
Copyright © 2002 Nelson Thomson Learning
46.08 g/mol
18.02 g/mol
1 mol
nC H OH = 450.0 g 2 5
46.08 g
nC H OH = 9.766 mol
2 5
3
nH O
= 9.766 mol 2
1
nH O
= 29.30 mol
2
18.02 g
mH O = 29.30 mol 2
1
mol
mH O = 527.9 g
2
or
mH
2
mH
O
O
1
mol C2
H5OH
3
mol H2O
18.02 g H2O
= 450.0 g C2
H5OH 46.08 g C2
H5OH
1m
ol C2
H5OH
1
mol H
2O
= 527.9 g
2
The mass of water produced is 527.9 g.
(d) C2H5OH(l) 450.0 g
46.08 g/mol
3 O2(g)
→
m
32.00 g/mol
2 CO2(g)
3 H2O(g)
1 mol
nC H OH = 450.0 g 2 5
46.08 g
nC H OH = 9.766 mol
2 5
3
nO
= 9.766 mol 2
1
nO
= 29.30 mol
2
32.00 g
mO
= 29.30 mol 2
1
mol
mO
= 937.5 g
2
or
mO
2
mO
2
1
mol C2
H5OH
3m
ol O2
32.00 g O
= 450.0 g C2
H5OH 2
46.08 g C2
H5OH
1m
ol C2
H5OH
1
mol O
2
= 937.5 g
The mass of oxygen required is 937.5 g.
(e) C2H5OH(l) 3 O2(g)
→
2 CO2(g)
3 H2O(g)
The masses involved, respectively, in this reaction are:
450.0 g
937.5 g
859.6 g
527.9 g
Addition shows that (450.0 + 937.5) = (859.6 + 527.9) = 1387.5, so the result agrees with the law of conservation of
mass.
C(s)
→
2 Cu(s)
m
2.01 g/mol
1 mol
= 50.0 kg 79.55 g
= 0.629 kmol
1
= 0.629 kmol 2
= 0.314 kmol
12.01 g
= 0.314 kmol
1
mol
16. 2 CuO(s) 50.0 kg
79.55 g/mol 1
nCuO
nCuO
nC
nC
mC
Copyright © 2002 Nelson Thomson Learning
CO2(g)
Chapter 5 Quantities in Chemical Equations
147
mC
= 3.77 kg
mC
1
mol C
1
mol C
uO
12.01 g C
= 50.0 kg CuO 79.55 g CuO
2
mol CuO
1
mol C
= 3.77 kg
or
mC
The mass of carbon required is 3.77 kg.
Applying Inquiry Skills
17. A solid reaction product can be removed (by filtration, for instance) from the reaction system. If some more of the
(presumed) excess reagent is added, and no reaction occurs, then the limiting reagent is completely reacted and the
other reagent is present in excess.
18. (a) Prediction
Na2SO4(aq) BaCl2(aq)
→
142.04 g/mol
2 NaCl(aq) BaSO4(s)
233.39 g/mol
Since the substances are in a 1 : 1 mole ratio, the mass ratio of BaSO4(s) : Na2SO4(aq) will be 233.39 : 142.04 or 1.6431
: 1.0000. This means that for each 1.0000 g of sodium sulfate reacted, 1.6431 g of barium sulfate should be produced.
(b) Experimental Design
A measured mass of sodium sulfate is dissolved and reacted with excess barium chloride solution. The precipitate is filtered, dried, and weighed.
(c) Procedure
1. Use a clean, dry 250-mL beaker to obtain a 2.00 g sample of Na2SO4(aq).
2. Use a 100-mL graduated cylinder to add approximately 50 mL of BaCl2(aq) to the beaker.
3. Allow the precipitate to settle, and test the clear liquid above the precipitate (the supernatant liquid) with a
small amount of the BaCl2(aq) from a medicine dropper to see if further precipitation occurs.
4. If the test in step 3 indicates the reaction is not yet complete, repeat step 3 until no further precipitation
occurs.
5. Measure and record the mass of a piece of filter paper to 0.01 g.
6. Filter, wash, and dry the BaSO4(s) precipitate.
7. Measure and record the mass of the filter paper plus dry precipitate to 0.01 g.
8. Dispose of all waste materials according to instructions.
(d) Analysis
To determine the purity of the sample, the percentage yield of the BaSO4(s) precipitate is calculated.
actual yield
% yield = 100%
theoretical yield
The BaSO4(s) percentage yield represents the purity of the sodium sulfate reagent as a percentage.
(e) Evaluation
This analysis assumes: that the other reagent is pure; that the precipitate is negligibly soluble; and that the experimental error is negligible. If any of these assumptions are incorrect, the actual yield value will be affected.
Making Connections
19. The report for this question should emphasize primarily the social perspective (the advantage of increased food
supply) and the environmental perspective (the disadvantage of water pollution) as a typical tradeoff involving technology.
148
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
3. The dissolving of both salt and sugar involves the solid separating into particles too small to see. The salt solution
contains ions of sodium and chlorine and will conduct a current, while the sugar dissolves to release sugar molecules,
so its solution will not conduct electricity.
Try This Activity: Substances in Water
(Page 265)
(a) The potassium permanganate, sugar, and ethanol dissolve.
(b) Sugar and ethanol are certainly soluble, as they disappear completely. The solubility of potassium permanganate is
less certain, as some of it remains undissolved.
Note: Students may be uncertain about any substance that does not “disappear” completely upon dissolving, because
they rarely encounter this. Expect discussion about potassum permanganate, if they had some remain in solid state.
Students may be uncertain if they speculate about whether some calcium carbonate or vegetable oil dissolves, even
though there is no visible reduction of the original phase. They have only visible evidence of sample “shrinking” to
go on, where no colour change is involved.
(c) The calcium carbonate and vegetable oil do not dissolve.
(d) We cannot be entirely certain, as a small amount may have dissolved.
(e) Properties are different: solutions are visibly homogeneous. Some other properties that might differ include electrical
conductivity, acidity, melting/freezing points, viscosity, and so forth.
(f) Acidity could be tested with pH paper or conductivity with a multimeter.
Note: Tests listed by students should be consistent with their answers to (e).
6.1 DEFINING A SOLUTION
PRACTICE
(Page 269)
Understanding Concepts
1. (a) Heterogeneous: different substances are visible.
(b) Homogeneous: only one phase is visible.
(c) Homogeneous if it has been decanted; if not, there may be sediment in the bottle and the red wine would then be
considered heterogeneous.
(d) Heterogeneous if corroded; if clean, bronze appears homogeneous.
(e) Homogeneous: the metal looks all the same throughout.
(f) Heterogeneous if corroded; otherwise it is homogeneous.
(g) Humid air is usually homogeneous; however, when cloud, fog, or rain forms, the solution is heterogeneous.
(h) Heterogeneous: the suspended droplets of water make it opaque.
(i) Heterogeneous: the water is not clear.
2. The solutions are (b), (d), (h) and (i).
(a), (c), (e), (f) and (g) are not solutions.
3. Solutions may be classified by type of solvent, by electrical conductivity, by acidity, by colour, or by physical state at
room conditions. Even categories such as viscosity, volatility, etc., can be used to classify substances.
4. (a) An aqueous solution is one in which the solvent is water.
(b) Aqueous solutions found around the home will be substances such as shampoo, vinegar, syrup, clear fruit juices,
tea, bleach, drain cleaners.
5. Methanol is a nonelectrolyte (it is a nonacidic molecular substance); sodium chloride is an electrolyte (it dissolves to
release ions); hydrochloric acid is an electrolyte (acids are the only molecular substances to conduct electricity); and
potassium hydroxide is an electrolyte (it is ionic).
6. (a) Electrolyte solutes include soluble ionic compounds (including ionic hydroxides) and acids.
(b) electrolyte: a substance that dissolves in water to form a conducting solution
7. (a) Acidic solutions have acid solutes.
(b) Basic soutions have ionic hydroxide solutes.
(c) Neutral solutions have molecular solutes (other than acids) or ionic solutes (other than ionic hydroxides).
8. (a) Electrolytes: citric acid, salt (assume sodium chloride), sodium citrate, and monosodium phosphate (4 of the 11
substances listed).
Copyright © 2002 Nelson Thomson Learning
Chapter 6 The Nature and Properties of Solutions
165
Nonelectrolytes: water, liquid sugar, glucose-fructose, and brominated vetgetable oil (4 of the 11 substances
listed).
The remaining ingredients (natural flavour, colour, and ester gum) cannot be classified at this stage.
Although there are the same number of ingredients that are electrolytes and nonelectrolytes, there is a greater
quantity of nonelectrolytes than electrolytes.
(b) In this product salt (sodium chloride) and sodium citrate contain sodium ions; only the monopotassium phosphate
contains potassium ions. The drink therefore contains more sodium ions than potassium ions.
(c) All of the food energy in the drink comes from carbohydrates—essentially from sugars.
(d) The drink attempts to replace water, energy, and alkali metal ions that are lost from the body during physical
activity.
SECTION 6.1 QUESTIONS
(Page 271)
Understanding Concepts
1. (a) An appropriate table might be:
Solute
Solvent
Solid
Liquid
Gas
Solid
brass or solder
sugar syrup
oxygen in ice
Liquid
Hg amalgams
vinegar
humid air
Gas
lead in air
soda pop
air
(b) Solutions of gaseous, liquid, and solid solutes in liquid solvents
Note: Some students might correctly suggest gas in gas as being a very common solution.
2.
gas
liquid
solute
aqueous
solution
water
solvent
solid
solutions
homogenous
mixture
electrolyte
nonelectrolyte
acid
base
neutral
3. Examples include gasoline, lubricating oils, Varsol, mineral spirits, and turpentine.
4. The solvent in alkyd paints is a mixture of oils. Any thinner/cleaner for alkyd paints must be an oil, such as mineral
oil or Varsol. Nonpolar alkyd paint components dissolve in nonpolar solvents. Water is used as a thinner/cleaner
(solvent) or diluent for latex paints because water is a component of the paint mixture. Polar latex paint components
dissolve in polar water molecules. Since water and oil will not dissolve in each other (polar does not dissolve in
nonpolar), it is critical not to mix the two types of solvent.
5. (a) electrolyte
(b) nonelectrolyte
(c) electrolyte
(d) nonelectrolyte
6. (a) HCl(aq): acidic solution; blue litmus turns pink.
(b) NaOH(aq): basic solution; pink litmus turns blue.
166
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
(c) Methanol: neutral solution; neither colour of litmus changes from its original colour.
(d) Sodium hydrogen carbonate: neutral solution; neither colour of litmus changes from its original colour.
Applying Inquiry Skills
7. (a)
Predicting Properties of Compounds
Substance
Acidic/Basic/Neutral
Electrolyte/Nonelectrolyte
neutral
nonelectrolyte
calcium hydroxide (slaked lime)
basic
electrolyte
H3PO4(aq) (for manufacturing fertilizer)
acidic
electrolyte
glucose (a product of photosynthesis)
neutral
nonelectrolyte
sodium fluoride (in toothpaste)
neutral
electrolyte
C3H7OH(l) (a rubbing alcohol)
(b) Each compound is dissolved in pure water, and the resulting solutions are tested with litmus. They are also tested
for electrical conductivity.
Making Connections
8. Using gasoline as a cleaner in a basement is unsafe because the liquid is very volatile (evaporates very readily), and
the vapours are very flammable. Gasoline should only be used in a very well-ventilated area, with no sources of ignition anywhere nearby — preferably outdoors.
Reflecting
9. If water really were a universal solvent, everything (and everyone) on Earth would be part of a huge unchanging
sphere of homogeneous solution.
6.2 EXPLAINING SOLUTIONS
PRACTICE
(Page 273)
Understanding Concepts
1. Most molecules are polar. The four categories of polar molecules are:
AB
(e.g., HCl(g), CO(g))
NxAy
(e.g., NH3(g), NF3(g))
OxAy
(e.g., H2O(l), OCl2(g))
CxAyBz
(e.g., CHCl3(l), C2H5OH(l))
2. Nonpolar molecules are those in which no part of the molecule is significantly more (or less) electronegative than any
other part. The two categories of nonpolar molecules are
molecular elements (e.g., N2(g) or P4(s));
compounds consisting only of carbon and one other type of atom, with a general formula CxAy (e.g., CH4(g), CO2(g),
and C8H18(l)).
PRACTICE
(Page 277)
Understanding Concepts
3. Intramolecular forces act between atoms within a molecule; intermolecular forces act between molecules.
4. (a) No, gasoline will not dissolve in water. The film floating on puddles of water at gas stations is evidence that
supports this statement.
(b) the “like dissolves like” rule
(c) Since gasoline is nonpolar and water is highly polar, the two liquids would not be miscible.
Copyright © 2002 Nelson Thomson Learning
Chapter 6 The Nature and Properties of Solutions
167
5. (a) The —OH groups of both methanol and water molecules allow hydrogen bonds to form among them, so these
liquids will dissolve well in each other.
(b)
H
H
O
H
H
—C—O
H
H
O
H
H
O
H
(c) The more hydrogen bonds that can form, the higher the solubility should be. This occurs because the molecules
will attract each other better if they can form many intermolecular bonds.
PRACTICE
(Page 278)
Understanding Concepts
6. (a) Observations might include: the solids disappear; the change of refraction of water near each solid as it dissolves;
and the solutions are colourless. Depending on the type of salt used, the rate of dissolving will vary. If table salt
is used, the solution remains cloudy because virtually all brands of table salt have insoluble additives that help to
keep the crystals from bonding together.
(b) In both cases, the solids are separated into particles too small to see, theoretically, by the attraction of the water
molecules.
(c) In theory, the solutions are different because the particles in the sugar solution are neutral molecules, and those
in the salt solution are ions.
(d) Theoretically, the high polarity of water molecules explains why they strongly attract ions in ionic compounds,
and the presence of many —OH groups on a sugar molecule allows a lot of hydrogen bonding (hence high solubility) with water.
+
+ F (aq)–
7. (a) NaF(s) → Na(aq)
+
(b) Na3PO4(s) → 3 Na(aq)
+ PO3–
4(aq)
(c) KNO3(s) →
+
K(aq)
+ NO–3(aq)
3+
(d) Al2(SO4)3(s) → 2 Al(aq)
+ 3 SO2–
4(aq)
+
(e) (NH4)2HPO4(s) → 2 NH4(aq)
+ HPO2–
4(aq)
2+ + 2 Cl–
(f) CoCl26H2O(s) → Co(aq)
(aq)
PRACTICE
(Page 279)
Understanding Concepts
8. Answers will vary but may include the following.
Empirical properties: melting and boiling points; heat capacity; colour and clarity; excellence as a solvent; expansion
upon solidification, formation of hexagonal crystals from vapour state; necessity for life; other observable,
measurable properties.
Theoretical properties: formation of hydrogen bonds; dissociation of dissolved ionic compounds, explaining the
expansion of ice and hexagonal-form snowflakes by theorizing V-shaped molecules that hydrogen bond to each other
in 6-sided patterns; other imagined, explanatory properties.
9. A nonpolar solvent such as mineral spirits (Varsol or turpentine) should dissolve grease, according to the “like
dissolves like” rule, because grease is a nonpolar substance.
10. (a) C6H12O6(s), because it is polar.
168
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
(b) C2H5OH(l), because it has hydrogen bonding.
(c) Na2CO3(s), because it is an ionic compound containing sodium ions.
11. (a) CH3Cl(l), CH2Cl2(l), and CHCl3(l) are most likely to be water soluble because they are polar molecules.
(b) CH4(g) and CCl4(l) are most likely to be soluble in nonpolar solvents because they are nonpolar.
12. A logical hypothesis would be that butanol has more C and H atoms, which reduce its polarity, making it more soluble
in nonpolar solvents.
Applying Inquiry Skills
13. Explanations are accepted by the scientific community only if they are logical, relatively simple, and are consistent
with evidence and other related explanations.
Making Connections
14. Dry cleaning solvents are chosen for their ability to dissolve nonpolar “dirt and grease” from clothing and to evaporate completely from the fabric. Health regulations apply to all such chemicals — most are toxic/noxious substances.
One such solvent that was used for years both domestically and commercially is carbon tetrachloride, CCl4(l). Carbon
tetrachloride was subsequently found to cause liver damage. Its first substitute, trichloroethene, C2HCl3(l), was also
found to produce liver damage. Eventually trichloroethane, C2H3Cl3(l), was found to be a safe replacement that neither
caused liver damage nor was carcinogenic.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
SECTION 6.2 QUESTIONS
(Page 280)
Understanding Concepts
1. Water is an effective solvent because its molecules are very polar. This polarity enables it to separate the particles of
the solute, surrounding each of them with water molecules. The polar —OH bonds result in polar molecules that
strongly attract any charged ion or other polar molecules.
A second reason that water is an effective solvent for a few solutes is hydrogen bonding. Although the number of solutes
that can hydrogen bond to water in solution is limited, solutes that have an unpaired electron on a N, O, or F atom, or
have a hydrogen bonded to an N, O, or F atom, have much higher than expected (from polarity) solubility in water.
2. (a) Soluble combinations are (iii), (iv), and (v).
(b) Empirical: answers based on observations are by definition empirical.
3. (a) Not soluble: one is polar and the other is nonpolar.
(b) Soluble: both are polar.
(c) Not soluble: one is nonpolar and the other is polar.
(d) Soluble: both have hydrogen bonding.
(e) Not soluble: one is nonpolar and the other is polar.
4. Methane, methanol, ammonia. Methane is nonpolar, methanol has hydrogen bonding from one —OH bond, and
ammonia has hydrogen bonding from three —NH bonds.
Applying Inquiry Skills
5. The experimental design is unacceptable because it involves too many variables; the presence of water will make it
impossible to decide anything about the mutual solubility of the other two substances.
6. Question
Which of C6H6(l) and C6H5OH(l) is more soluble in water?
Prediction
C6H5OH(l) is predicted to be more soluble.
Experimental Design
Each liquid will be added, in 1-mL increments, to 100 mL of water (in separate beakers) while stirring, until no more
will dissolve.
•
•
•
Materials
C6H6(l) and C6H5OH(l)
two 250-mL beakers
pure water
Copyright © 2002 Nelson Thomson Learning
Chapter 6 The Nature and Properties of Solutions
169
•
10-mL graduated cylinder
•
100-mL graduated cylinder
•
stirring rod
Making Connections
7. Children’s glue must be nontoxic and water washable. This means the chemical substance(s) must be unreactive and
must have very polar molecules.
6.3 SOLUTION CONCENTRATION
PRACTICE
(Page 284)
Understanding Concepts
1. W/W (weight to weight), W/V (weight to volume), or V/V (volume to volume) ratios
2. v ethanol
4.1 L
55 L
4.1 L
c ethanol 100%
55 L
c ethanol 7.5% V/V
v gasohol
The ethanol concentration in fuel solution is 7.5% V/V (by volume).
3. m zinc chloride 16 g
50 mL
16g
c zinc chloride 100%
50 m L
c zinc chloride 32% W/V
v solution
The concentration of zinc chloride in the flux solution is 32% W/V.
4. m zinc 1.7 g
m brass 35.0 g
1.7 g
c zinc 100%
35.0 g
c zinc 4.9% W/W
The zinc concentration in the brass is 4.9% W/W (by mass).
5. 8 ppm 8 mg/L
8 mg
m oxygen 1 L
1
L
m oxygen 8 mg
The mass of oxygen in each litre of water is 8 mg.
6. m formaldehyde 3.2 mg
m air
0.59 kg
3.2 mg
c formaldehyde 0.59 kg
c formaldehyde 5.4 mg/kg 5.4 ppm
The concentration of formaldehyde in air is 5.4 ppm.
7. (a) 1 ppb is 1/1000 of 1 ppm, or 0.001 ppm.
(b) 1 ppb
1 mg/106 mL (1 mg/kL)
1 mg/1000 L (1 mg/m3)
170
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
1 g/L
1 g/kg
1 ng/g
(c) 30 ppb 30 g/kg
8. n CaCl 0.11 mol
2
v CaCl 60 mL 0.060 L
2
0.11 mol
C CaCl 2
0.060 L
C CaCl 1.8 mol/L
2
The molar concentration of calcium chloride is 1.8 mol/L.
Making Connections
9. LD50 refers to a toxic substance dosage in mass per kilogram of the recipient (e.g., milligrams per kilogram of body
mass) that will prove lethal to half of the organisms in a test sample. Descriptors of toxicity vary, so students may well
find differing definitions of “extremely” toxic and “slightly” toxic in their research. However, a common definition
for extremely toxic is less than 5 mg/kg, while slightly toxic is 5–15 g/kg. Above 15 g/kg is considered to be almost
nontoxic.
The quantity of toxin ingested depends on the concentration of the chemical to which one may be exposed.
However, the LD50 levels express the concentration of toxin in the body (e.g., 15 mg/kg). Cumulative toxins may be
ingested at very low concentrations for long periods of time and become toxic over time. Examples of cumulative
toxins include mercury and lead.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
Reflecting
10. A report mark is a ratio of marks achieved/possible marks. Other common ratios are prices of bulk goods ($/kg),
speeds (km/h), nutritional information (energy per serving, in kJ/g) and cooking times (min/kg).
PRACTICE
(Page 287)
Understanding Concepts
11. c C H OH 70.0% V/V 70.0 mL/100 mL
3 7
v C H OH 500 mL
3 7
70.0 mL
v C H OH 500 mL
3 7
100 mL
v C H OH 350 mL
3 7
The volume of rubbing alcohol present is 350 mL.
12. c H O 3.0% W/V 3.0 g/100 mL 3.0 g/0.100 L
2 2
0.250 L
v H O 1000 bottles
250 L
2 2
1 b
ottle
3.0 g
m H O 250 L
2 2
0.100 L
m H O 7.5 kg
2 2
The mass of hydrogen peroxide needed is 7.5 kg.
13. c F– 1.5 ppm 1.5 mg/L
v F– 0.250 L
Copyright © 2002 Nelson Thomson Learning
Chapter 6 The Nature and Properties of Solutions
171
1.5 mg
m F– 0.250 L
1
L
m F– 0.38 mg
The mass of fluoride ions present in the glass of water is 0.38 mg.
14. C MgCl 0.055 mol/L
2
v MgCl 75 L
2
0.055 mol
n MgCl 75 L
2
1L
n MgCl 4.1 mol
2
The amount of magnesium chloride present is 4.1 mol.
15. C HCl 5.0 mol/L
v HCl 50 mL 0.050 L
5.0 mol
n HCl 0.050 L
1L
n HCl 0.25 mol
The amount of hydrogen chloride in the beaker is 0.25 mol.
16. C NH 1.24 mol/L
3
n NH
3
0.500 mol
1L
v NH 0.500 mol 3
1.24 mol
v NH 0.403 L
3
The volume of aqueous ammonia solution needed is 403 mL.
17. C Na SO
2
n Na SO
2
2.6 mol/L
4
0.14 mol
4
v Na SO
4
v Na SO
4
2
2
1L
0.14 mol 2.6 mol
0.054 L 54 mL
The volume of aqueous sodium sulfate solution needed is 54 mL.
Making Connections
18. Prediluted commercial solutions are usually more expensive for equal quantities of solute. If cost is the concern, you
would calculate cost/quantity for the solute. On the other hand, prediluted solutions are both more convenient to use
and less dangerous to store.
PRACTICE
(Page 290)
Understanding Concepts
19. C NaOH
0.125 mol/L
v NaOH
3.00 L
M NaOH
40.00 g/mol
0.125 mol
3.00 L
1L
0.375 mol
40.00 g
0.375 mol 1
mol
15.0 g
n NaOH
m NaOH
m NaOH
172
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
or
0.125 mol
40.00 g
m NaOH 3.00 L
1L
1
mol
m NaOH 15.0 g
The mass of solid sodium hydroxide required is 15.0 g.
20. C NaCl 0.56 mol/L
v NaCl 5.0 L
M NaCl 58.44 g/mol
n NaCl
m NaCl
m NaCl
or
0.56 mol
5.0 L
1L
2.8 mol
58.44 g
2.8 mol 1
mol
0.16 kg
0.56 mol
m NaCl 5.0 L
1L
m NaCl 0.16 kg
58.44 g
1
mol
The mass of sodium chloride in the sample is 0.16 kg.
21. (a) c CO 355 ppm 355 mg/L
2
The mass of carbon dioxide present in a litre of acid rain is 355 mg.
(b) MCO
mCO
vCO
2
2
2
44.01 g/mol
0.355 g
1.00 L
nCO
2
CCO
2
CCO
2
1 mol
0.355 g 44.01 g
8.07 mmol
8.07 mmol
1.00 L
8.07 mmol/L
The molar concentration of aqueous carbon dioxide is 8.07 mmol/L.
22. (a) m NaCl
v NaCl
235 g
3.00 L 3.00 103 mL
c NaCl
235 g
100%
3.00 103 mL
c NaCl
7.83% W/V
The percent concentration of sodium chloride in aqueous solution is 7.83% W/V.
(b) M NaCl
58.44 g/mol
m NaCl
235 g
v NaCl
3.00 L
CNaCl
1 mol
235 g 58.44 g
4.02 mol
4.02 mol
3.00 L
CNaCl
1.34 mol/L
nNaCl
The molar concentration of aqueous sodium chloride is 1.34 mol/L.
Copyright © 2002 Nelson Thomson Learning
Chapter 6 The Nature and Properties of Solutions
173
SECTION 6.3 QUESTIONS
(Page 290)
Understanding Concepts
1. Consumer products often have percentage concentrations given on the label because these units are more familiar to
most consumers.
2. (a) cdextrose 5.0% W/V 5.0 g/100 mL
vsolution 500.0 mL
5.0 g
mdextrose 500.0 mL
100 mL
mdextrose 25 g
The mass of dextrose in the bag is 25 g.
(b) cdextrose 5.0 g/100 mL 50 g/L 5.0 104 ppm
The concentration of dextrose in the solution is 5.0 104 ppm.
3. c NaCl
v NaCl
m NaCl
m NaCl
31.6 g/100 mL
250 mL
31.6 g
250 mL
100 mL
79.0 g
The mass of sodium chloride that will dissolve is 79.0 g.
4. The nandrolone concentration of 1000 times 2 mg/L is 2 103 mg/L, so the test result concentration is 2 103 ppm
or 2 ppt.
5. c PCBs 18.9 mg/kg
m chick 0.60 kg
18.9 mg
m PCBs 0.60 kg
1 k
g
m PCBs 11 mg
The chick would contain 11 mg of PCBs.
6. c PCBs 4.00 ppm 4.00 mg/kg
m person 64.0 kg
4.00 mg
m PCBs 64.0 kg
1 k
g
m PCBs 256 mg
The mass of PCBs in the person is 256 mg.
7. Assuming the 5 mL volume is a theoretical (exact) value, then the substance concentrations in g/L are, respectively
153 m
g
ammonium carbonate 30.6 g/L
5
mL
267 m
g
potassium bicarbonate 53.4 g/L
5
mL
174
menthol
22 m
g
4.4 g/L
5
mL
camphor
2.2 m
g
0.44 g/L
5
mL
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
8. c each sol’n 0.10 mol/L
v each sol’n 100 mL 0.100 L
0.10 mol
n each compound 0.100 L
0.010 mol
1L
M NaCl
58.44 g/mol
74.55 g/mol
M KCl
110.98 g/mol
M CaCl
2
58.44 g
0.010 mol 1
mol
0.58 g
(a) m NaCl
m NaCl
or
0.10 mol
58.44 g
0.100 L
1L
1m
ol
0.58 g
m NaCl
m NaCl
The mass of sodium chloride required is 0.58 g.
74.55 g
(b) m KCl 0.010 mol 1
mol
m KCl
0.75 g
or
0.10 mol
74.55 g
0.100 L
1L
1m
ol
0.75 g
m KCl
m KCl
The mass of potassium chloride required is 0.75 g.
110.98 g
(c) m CaCl 0.010 mol 2
1
mol
m CaCl 1.1 g
2
or
0.10 mol
110.98 g
0.100 L
1L
1
mol
m CaCl
2
m CaCl 1.1 g
2
The mass of calcium chloride required is 1.1 g.
25 g/100 mL 25 g/0.100 L
9. c NaCl
M NaCl
C NaCl
C NaCl
58.44 g/mol
25 g
1 mol
0.100 L
58.44 g
4.3 mol/L
The molar concentration of sodium chloride in the brine is 4.3 mol/L.
10. m C H O
6 12 6
MCH O
6 12 6
CCH O
6 12 6
nCH
6 12O6
nCH
6 12O6
vCH
6 12O6
vCH
6 12O6
or
vCH
6 12O6
vCH
6 12O6
2.0 g
180.18 g/mol
0.055 mol/L
1 mol
2.0 g 180.18 g
0.011 mol
1L
0.011 mol 0.055 mol
0.20 L
1
mol
1L
2.0 g 180.18 g
0.055 mol
0.20 L
The volume of solution that contains 2.0 g of glucose is 0.20 L.
Copyright © 2002 Nelson Thomson Learning
Chapter 6 The Nature and Properties of Solutions
175
Making Connections
11. Points in favour: patients would have to swallow less of the medicine; the medicines would be more compact and
could be sold in smaller bottles, which might result in a lower cost (although the saving is likely to be minimal).
Points against: selling medicines in highly concentrated solutions would almost certainly increase the frequency of
patients accidentally taking the wrong dose (an extra 5 mL would contain considerably more of the active ingredient
in a concentrated solution than in a dilute solution); more precise equipment would be required for measuring the
appropriate quantity for the prescription and the dose.
12. A common system of communication is crucial for clarity, so there is no confusion about concentrations of blood test
results or drug dosages. There are currently several systems for communicating the concentration of medicines,
although all are metric. That all systems are metric is important for international communication and for labelling of
bottles of medicine.
• Percent composition is very common for medicines dispensed in solution (e.g., salicylic acid in acne preparations
and D5W intravenous (5% dextrose in water)).
• Percent concentration may also be used for blood-test results such as percent of alcohol (e.g., a maximum limit
of 0.080% for drinking and driving).
• Molar concentrations are used for blood sugar analysis where units of millimoles per litre (e.g., 5.1 mmol/L) are
employed.
• Parts per million and/or milligrams per kilogram of body mass are used when very small concentrations are
involved (e.g., testing for toxins or drug-enhanced sports performances). The effective concentration of medicines
are determined by extensive research with animals and humans. Once the concentration is determined through
research, the effective mass of medicine that needs to be ingested every so-many hours is calculated. Taking one
capsule of a medicine every 8 h or at every meal maintains an effective concentration of the chemical in the body
systems. Time-release medication and patches are newer technologies that release the medication slowly in order
to keep the concentration of the medication fairly constant over time. Sometimes, such as in chemotherapy, the
concentration of chemicals is kept near the toxic level for the patient. Very careful monitoring of the concentrations of the chemicals in the blood system is required.
Miscommunication of concentration levels can, of course, result in death. Conventions of communication are not just
convenient and money saving; they also save lives.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
6.4 DRINKING WATER
PRACTICE
(Page 294)
Understanding Concepts
1. Contaminants are biological (e.g., viruses, bacteria, protozoa), chemical (e.g., mercury, lead, PCBs) and physical (e.g.,
sand, mud, suspended particles of organic matter).
2. Some ways contaminants enter water are by leaching from landfills; from inadequate septic/sewage systems; from
overuse of fertilizers, pesticides and herbicides; from livestock wastes; and from road salt runoff.
3. Leaking sewer pipes are environmental hazards because untreated sewage contains nitrates and phosphates (which can
act as fertilizers in the environment) and bacteria (which can cause a variety of infections resulting in illness). Raw
sewage also has a very unpleasant odour.
4. Water leaching from a landfill site may seep through the surrounding earth and rock, and into wells or streams, thus
ending up in a drinking water supply. Such leachate could contain heavy metal ions (e.g., mercury, lead, and
cadmium); bacteria; acids; and organic compounds (e.g., benzene and tetrachloromethylene). If not removed, these
pollutants could have a negative influence on health: heavy metal ions interfere with brain and nerve development;
bacteria can cause infections; acids can damage pipes; and organic compounds may be poisonous or carcinogenic.
176
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
5.
lead
pesticide
may
come
from
may
come
from
gasoline
may
come
from
forest or
crop spraying
landfill
leachate
leaky tanks
or pipelines
and flow into
ground
water
6. c tetrachloroethylene 0.03 ppm 0.03 mg/L
v water
250 L
0.03 mg
m tetrachloroethylene 250 L
1
L
m tetrachloroethylene 8 mg
The mass of tetrachloroethylene in the bath water is 8 mg.
7. c cadmium 0.005 ppm 0.005 mg/L
0.010 ppm 0.010 mg/L
c lead
c mercury 0.001 ppm 0.001 mg/L
v water
1.5 L
365.25 d
1 d
1a
5.5 102 L/a
5.5 102 L
0.005 mg
m cadmium 1a
1
L
m cadmium 3 mg/a
The mass of cadmium consumed per year would be 3 mg.
5.5 102 L
0.010 mg
m lead 1a
1
L
m lead 5.5 mg/a
The mass of lead consumed per year would be 5.5 mg/a.
5.5 102 L
0.001 mg
m mercury 1a
1
L
m mercury 0.5 mg/a
The mass of mercury consumed per year would be 0.5 mg.
10.00 mL 0.01000 L
8. v water
m NO –
3
5.4 mg
5.4 mg
c NO – 3
0.01000 L
c NO – 0.54 g/L
3
This concentration of nitrate ion, 0.54 g/L, is much higher than the MAC level, about 540/45, or 12 times the acceptable
maximum.
Making Connections
9. Student reports will vary widely. Suitable starting points for research include the Ontario Groundwater Management
program, the Ontario Ministry of Agriculture, Food and Rural Affairs, and reports from the Walkerton Inquiry. Locally
relevant issues should also be investigated.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
Copyright © 2002 Nelson Thomson Learning
Chapter 6 The Nature and Properties of Solutions
177
10. Student reports will vary. The chosen source of contamination could be any of those listed in Table 1, page 293 of the
text. A proposed solution could involve preventing the contamination at source, preventing the contaminant from
reaching the well, relocating the well, or removing the contaminant from well water before it reaches our taps.
Try This Activity: Simulated Water Treatment
(Page 297)
(a) In glass 1, a control, adding a teaspoon (5 mL) of ammonia to the alum solution causes a white flocculent precipitate
to form. The nature of the precipitate is observed. The precipitate gradually settles to the bottom of the glass.
In glass 2, the components are added in the order water, soil, alum, and ammonia, and the contents clear as the flocculent precipitate settles. The removal of the suspended silt is obvious. The soil/silt adheres to the precipitate. The
process is much more complete and quicker than in glass 3.
In glass 3, a control with water and soil only, the soil gradually settles but leaves some of the silt in suspension for a
longer period of time than in glass 2.
(b) The precipitate, being gelatinous, will quickly clog a filter paper, making separation by filtration impractical. Allowing
the precipitate to fall to the bottom of the container is slower, but simple and effective.
PRACTICE
(Page 299)
Understanding Concepts
11. Disinfection — killing disease-causing organisms — is the most important step in water treatment.
12. Usual (all areas)
Optional (some areas)
collection
aeration
coagulation/sedimentation
softening
filtration
fluoridation
disinfection
postchlorination
ammoniation
Making Connections
13. Physical treatments might include a variety of filters, such as ceramic, ultrafine, and carbon, plus reverse osmosis. The
pore sizes in the most common filters sold today are 0.1-4 µm. Cysts and bacteria are removed by the filter, but viruses
are too small. Boiling the water is effective for disinfecting the water completely.
Chemical treatments might include the use of iodine crystals, iodine-complex tablets, chlorine bleach, calcium
hypochlorite crystals, and halazone tablets. The iodine, for example, is added as a measured volume of a saturated
solution. The dilution procedure adds 4 mg of iodine to 1 L of water to produce a 4 ppm solution that is treated for
20 – 30 min before drinking.
Note that information on purifying water while hiking is also readily available at outdoor activity stores.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
14. In favour of bottled water: we can be fairly certain that it is free of microorganisms and toxic chemicals; distilled or
reverse osmosis-treated water is much more pure; bottled water is convenient, if tap water is contaminated.
Against bottled water: most bottled water is not significantly different from that available through most municipal
systems; bottled water may, in fact, be less safe than local municipally treated water; it is more expensive than tap
water, and distilled or reverse osmosis-treated water is particularly expensive; it is less convenient (e.g. it has to be
moved from storage to display areas, and it is heavy).
The note to the school cafeteria should outline a position and give reasons. It could also request some action on the
part of the cafeteria staff, such as making free tap water available to all students.
178
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
15. Water testing, at anything other than a very basic level, is really a branch of analytical chemistry. A person trained as
a laboratory technologist can analyze water for chemicals that may be present. However, the analysis also needs to
include screening for possible biological components. The water must be tested for bacteria (e.g., E-coli), intestinal
parasites (e.g., Giardia cysts and cryptosporidium), and viruses. Special equipment and training are required for the
bioanalysis. The water-treatment plant operators must also be trained in the interpretation of analytical reports and in
how to respond to these reports (how to adjust the process to remedy any problems).
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
Explore an Issue
Take a Stand: Safe to Drink?
(Page 299)
(a) Aspects chosen may include: the choice of ground or surface water as a source; the location of wells or intake pipes;
the age and condition of the pipes; the treatment the water receives; water metering; or the privatization of water
delivery. The letter should clearly state the current situation (found through research) and make reasonable suggestions for improvements.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
SECTION 6.4 QUESTIONS
(Page 300)
Understanding Concepts
1. A typical list might be taken from Section 6.4; Tables 1, 2, and 3:
Pollutant
Classification
acid
chemical
bacteria
biological
cysts
biological
viruses
biological
lead
chemical
mercury
chemical
cadmium
chemical
mineral solids
physical
nitrates
chemical
phosphates
chemical
organic compounds
chemical
benzene
chemical
gasoline
chemical
pesticides
chemical
salt
chemical
2. The first action should be to notify the community to stop using the water. Next, the water supply system must be
disinfected and flushed.
3. Exceding the MAC for chemicals in drinking water is a potential health hazard. For example, high lead levels can
cause brain damage; organic compounds may be carcinogenic; salt makes water taste bad and may contribute to circulatory problems.
4. Contaminants may be any of those listed in Tables 2 or 3. If benzene is chosen, the presentation should include its
source (e.g., leaked gasoline, industrial effluent, or landfill leachate); its MAC (0.005 ppm); and its effects (it is a
suspected carcinogen and, because it floats on water, it interferes with water’s ability to exchange gases with the air).
Research information is plentiful on water contaminants. Students might be encouraged to report on a contaminant
that poses a problem locally.
5. A lead concentration of 0.01 g/L is 1000 times higher than the MAC level of 0.010 mg/L (0.000010 g/L).
Copyright © 2002 Nelson Thomson Learning
Chapter 6 The Nature and Properties of Solutions
179
Making Connections
6. Water treatment is designed to remove physical, biological, and chemical contaminants. Physical contaminants are
removed through coagulation, flocculation, sedimentation, and filtration; biological contaminants are removed
through disinfection and postchlorination; chemical contaminants are removed through aeration and softening.
Water treatment on a large scale is usually a continuous process (rather than a batch process) in order to provide a
continuous flow of treated water into the water system. Continuous-flow designs are much more difficult to create and
monitor because they must be timed correctly. The size of the container, the mixing of the fluids, and the time that a
sample of water remains in the container must all be pre-engineered in order for the process to be effective. In a batch
process, there are fewer variables to control and manipulate — the water is relatively static (still) while being treated.
7. In support of the statement: Almost any contamination problem eventually becomes a human (personal) problem
because of the interconnectedness of the entire ecosystem, which, of course, includes us. Most ground water contamination is caused by people, so it is our responsibility to clean it up.
In opposition to the statement: Contamination is a problem regardless of whether it affects people and should be
avoided if at all possible. Humans are not the only organisms damaged by contamination.
Presentations should include suppporting information and reasoned arguments.
8. Precise equipment is required for water testing because “safe” levels of toxic and noxious materials are so low that
they are hard to measure with imprecise equipment.
6.5 SOLUTION PREPARATION
PRACTICE
(Page 302)
Understanding Concepts
1. In solid form, ammonium oxalate is a monohydrate, so the calculation of molar mass must take this into account.
C (NH ) C O 0.250 mol/L
4 2 2 4
v (NH ) C O
4 2 2 4
100.0 mL 0.1000 L
M (NH ) C O H O 142.14 g/mol
4 2 2 4
2
n (NH ) C O
4 2 2 4
m (NH ) C O H O
4 2 2 4
2
m (NH ) C O H O
4 2 2 4
or
2
0.250 mol
0.1000 L
1L
0.0250 mol
142.14 g
0.0250 mol 1
mol
3.55 g
0.250 mol
142.14 g
m (NH ) C O H O 0.1000 L
4 2 2 4 2
1 L
1
mol
m (NH ) C O H O 3.55 g
4 2 2 4
2
The mass of ammonium oxalate monohydrate required is 3.55 g.
2. C NaOH
v NaOH
500 mL 0.500 L
M NaOH
40.00 g/mol
n NaOH
m NaOH
m NaOH
or
180
10.0 mol/L
10.0 mol
0.500 L
1L
5.00 mol
40.00 g
5.00 mol 1
mol
200 g
10.0 mol
40.00 g
m NaOH 0.500 L
1L
1m
ol
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
Making Connections
6. Water treatment is designed to remove physical, biological, and chemical contaminants. Physical contaminants are
removed through coagulation, flocculation, sedimentation, and filtration; biological contaminants are removed
through disinfection and postchlorination; chemical contaminants are removed through aeration and softening.
Water treatment on a large scale is usually a continuous process (rather than a batch process) in order to provide a
continuous flow of treated water into the water system. Continuous-flow designs are much more difficult to create and
monitor because they must be timed correctly. The size of the container, the mixing of the fluids, and the time that a
sample of water remains in the container must all be pre-engineered in order for the process to be effective. In a batch
process, there are fewer variables to control and manipulate — the water is relatively static (still) while being treated.
7. In support of the statement: Almost any contamination problem eventually becomes a human (personal) problem
because of the interconnectedness of the entire ecosystem, which, of course, includes us. Most ground water contamination is caused by people, so it is our responsibility to clean it up.
In opposition to the statement: Contamination is a problem regardless of whether it affects people and should be
avoided if at all possible. Humans are not the only organisms damaged by contamination.
Presentations should include suppporting information and reasoned arguments.
8. Precise equipment is required for water testing because “safe” levels of toxic and noxious materials are so low that
they are hard to measure with imprecise equipment.
6.5 SOLUTION PREPARATION
PRACTICE
(Page 302)
Understanding Concepts
1. In solid form, ammonium oxalate is a monohydrate, so the calculation of molar mass must take this into account.
C (NH ) C O 0.250 mol/L
4 2 2 4
v (NH ) C O
4 2 2 4
100.0 mL 0.1000 L
M (NH ) C O H O 142.14 g/mol
4 2 2 4
2
n (NH ) C O
4 2 2 4
m (NH ) C O H O
4 2 2 4
2
m (NH ) C O H O
4 2 2 4
or
2
0.250 mol
0.1000 L
1L
0.0250 mol
142.14 g
0.0250 mol 1
mol
3.55 g
0.250 mol
142.14 g
m (NH ) C O H O 0.1000 L
4 2 2 4 2
1 L
1
mol
m (NH ) C O H O 3.55 g
4 2 2 4
2
The mass of ammonium oxalate monohydrate required is 3.55 g.
2. C NaOH
v NaOH
500 mL 0.500 L
M NaOH
40.00 g/mol
n NaOH
m NaOH
m NaOH
or
180
10.0 mol/L
10.0 mol
0.500 L
1L
5.00 mol
40.00 g
5.00 mol 1
mol
200 g
10.0 mol
40.00 g
m NaOH 0.500 L
1L
1m
ol
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
m NaOH 200 g
The mass of sodium hydroxide required is 200 g.
3. Typical answers may include sugar (in coffee, tea, lemonade), salt (in cooking water), fruit drink crystals, and soap
solutions.
Applying Inquiry Skills
4. (a) In solid form cobalt (II) chloride is a dihydrate, so the calculation of molar mass must take this into account.
C CoCl
0.100 mol/L
2
v CoCl
2.00 L
2
M CoCl 2H O 168.57 g/mol
2
2
0.100 mol
2.00 L
1L
0.200 mol
165.87 g
m CoCl 2H O 0.200 mol 2
2
1
mol
m CoCl 2H O 33.2 g
n CoCl
2
2
or
2
0.100 mol
m CoCl 2H O 2.00 L
2
2
1L
m CoCl 2H O 33.2 g
2
165.87 g
1
mol
2
The mass of cobalt (II) chloride dihydrate required is 33.2 g.
(b) Procedure
1. Wear eye protection and a laboratory apron.
2. Calculate the mass of CoCl22H2O(s) required to prepare 2.00 L of 0.100 mol/L solution.
3. Obtain the calculated mass of CoCl22H2O(s) in a clean dry 400-mL beaker.
4. Dissolve the solid in about 200 mL of water.
5. Transfer the solution into a clean 2-L volumetric flask, making sure to rinse the beaker and funnel
several times. Transfer the rinsings into the flask.
6. Add pure water to make the final volume 2.00 L.
7. Stopper the flask and mix the contents thoroughly by inverting the flask repeatedly.
5. (a) C KMnO 75.0 mmol/L 0.0750 mol/L
4
v KMnO
500.0 mL 0.5000 L
4
M KMnO 158.04 g/mol
4
0.0750 mol
n KMnO 0.5000 L
4
1
L
0.0375 mol
158.04 g
mol mKMnO 0.0375 4
1
mol
mKMnO 5.93 g
4
or
0.0750 mol
158.04 g
mKMnO 0.5000 L
4
1L
1
mol
mKMnO 5.93 g
4
The mass of potassium permanganate required is 5.93 g.
(b) Procedure
1. Wear eye protection and a laboratory apron.
2. Calculate the mass of KMnO4(s) required to prepare 500.0 mL of 75.0 mmol/L solution.
3. Obtain the calculated mass of KMnO4(s) in a clean, dry 250-mL beaker.
4. Dissolve the solid in about 100 mL of water.
Copyright © 2002 Nelson Thomson Learning
Chapter 6 The Nature and Properties of Solutions
181
5. Transfer the solution into a clean 500-mL volumetric flask, being sure to rinse the beaker and funnel several
times. Transfer the rinsings into the flask.
6. Add pure water to make the final volume 500.0 mL.
7. Stopper the flask and mix the contents thoroughly by inverting the flask repeatedly.
ACTIVITY 6.5.2: A STANDARD SOLUTION BY DILUTION
(Page 305)
Analysis
(a) Some water is always placed in the volumetric flask initially for safety reasons, so that when concentrated acids are
diluted their heat of solution will be dissipated through this water, and will not cause the flask contents to boil and
spurt out. The rule is "Add concentrated acid to water; NEVER add water to concentrated acid.” When the concentrated solution is not an acid, the rule is followed anyway, for consistency and to build the habit.
(b) If 100 mL of water had initially been placed in the flask, adding the concentrated solution and rinse water during
transfer would make the final volume too great.
PRACTICE
(Page 306)
Understanding Concepts
6. vi ?
vf 2.00 L
Ci 17.8 mol/L
Cf 0.200 mol/L
viCi vfCf
vi
vC
f f
Ci
vi
vf
0.0225 L 22.5 mL
vi
0.200 m
ol/L
2.00 L 17.8 mol/L
0.0225 L 22.5 mL
or
vi
2.00 L 0.200 mol/L
17.8 mol/L
The initial volume of 17.8 mol/L hydrochloric acid required is 22.5 mL.
7. (a) vi
5.00 mL
vf
100.0 mL
Ci
0.05000 mol/L
Cf
?
viCi vfCf
Cf
vC
i i
vf
Cf
182
5.00 mL
0.05000 mol/L
0.00250 mol/L 2.50 mmol/L
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
or
Cf
Cf
0.05000 mol
5.00 mL
L
100.0 mL
0.00250 mol/L 2.50 mmol/L
The final concentration of copper(II) sulfate solution is 2.50 mmol/L.
(b) C CuSO
v CuSO
4
2.50 mmol/L 0.00250 mol/L
10.0 mL 0.0100 L
4
M CuSO 159.61 g/mol
4
2.50 mmol
n CuSO 0.0100 L
4
1L
0.0250 mmol
159.61 g
m CuSO 0.0250 mmol
4
1
mol
m CuSO 3.99 mg
4
or
2.50 mmol
m CuSO 0.0100 L
4
1L
m CuSO 3.99 mg
159.61 g
1
mol
4
The mass of copper(II) sulfate in the solution sample is 3.99 mg.
(c) This final dilute solution would be quite difficult to prepare directly, since the entire 100.0 mL volume would
only contain 39.9 mg of solute. This mass could only be measured accurately by using a balance with a
precision of at least tenths of milligrams (ten-thousandths of grams), which is on the order of a hundred times
more precise than standard school lab (centigram) balances. Even more difficult to measure would be the 3.99
mg needed to prepare the 10.0 mL of the solution. The technique of dilution solves this problem.
Furthermore, mass measurements this precise must be done with the balance pan in an enclosed space in
order to prevent error caused by air currents. The sample must be enclosed in a solid container so that a correction for the buoyant force of air will not be required.
8. (a) The volume increase is 250.0 mL/10.00 mL or 25.00 times.
Concentration decrease is thus 1/25.00 or 0.04000 times or 4.000%.
(b) The reacting volume would now be 25.00 times as much.
(c) We would expect the speed of the reaction to be slower with the diluted solution, since the particles that react
would be spread out more in the solvent and would not collide as often.
9. Typical examples of dilutions may include liquid soap solutions (hand cleaner), fruit juice concentrates, chocolate
syrup in milk, and bleach in laundry loads.
Reflecting
10. The statement will not be true if the instrument available for measuring mass is very precise, and the equipment
available for measuring liquid volumes is not very precise.
SECTION 6.5 QUESTIONS
(Page 306)
Understanding Concepts
1. Scientists make solutions for chemical analysis, as well as for precise control of reactions. Solutions make substances
easy to handle and measure and safer to use, and many reactions only occur in solution.
2. (a) Solutions may be made by dissolving a known mass of solute up to a known volume, or by diluting an existing
solution of higher concentration.
(b) The method used normally depends on the form (pure substance or concentrated solution) in which the desired
reagent is available. A solution is prepared from a solid solute by measuring the mass of the solute and adding
water (e.g., the preparation of a sodium carbonate solution). A solution is prepared from an available (more
concentrated) solution by diluting the available solution (e.g., hydrochloric acid solutions are prepared by diluting
concentrated hydrochloric acid).
Copyright © 2002 Nelson Thomson Learning
Chapter 6 The Nature and Properties of Solutions
183
0.125 mol/L
3. C Ba(NO )
3 2
v Ba(NO )
100 mL 0.100 L
M Ba(NO )
261.35 g/mol
3 2
3 2
0.125 mol
0.100 L
1L
0.0125 mol
261.35 g
0.0125 mol 1
mol
3.27 g
n Ba(NO )
3 2
m Ba(NO )
3 2
m Ba(NO )
3 2
or
0.125 mol
261.35 g
0.100 L
1L
1
mol
3.27 g
m Ba(NO )
3 2
m Ba(NO )
3 2
The mass of pure barium nitrate required is 3.27 g.
4. vi 1.00 L
vf ?
Ci 17.4 mol/L
Cf 0.400 mol/L
viCi vfCf
vf
vC
i i
Cf
vf
vf
43.5 L
vf
17.4 mol/L
1.00 L 0.400 m
ol/L
43.5 L
or
vf
1.00 L 17.4 mol/L
0.400 mol/L
The final volume of diluted acetic acid is 43.5 L
5. vi 10.00 mL
vf 250.0 mL
ci ?
cf 0.274 g/L
vici vfcf
vc
ci ff
vi
250.0 mL
0.274 g/L
ci 10.00 mL
ci 6.85 g/L
or
0.274 g
250.0 mL
ci 1L
10.00 mL
184
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
ci 6.85 g/L
The initial concentration of the solution was 6.85 g/L.
Applying Inquiry Skills
6. C KSCN 0.155 mol/L
v KSCN 100 mL 0.100 L
M KSCN 97.18 g/mol
n KSCN
m KSCN
m KSCN
or
0.155 mol
0.100 L
1L
0.0155 mol
97.18 g
0.0155 mol 1
mol
1.51 g
0.155 mol
97.18 g
m KSCN 0.100 L
1L
1
mol
m KSCN 1.51 g
The mass of potassium thiocyanate required is 1.51 g.
Procedure
1. Wear eye protection and a laboratory apron.
2. Calculate the mass of KSCN(s) (1.51 g) required to prepare 100 mL of 0.155 mol/L solution.
3. Obtain the calculated mass of KSCN(s) in a clean, dry, 250-mL beaker.
4. Dissolve the solid in about 50 mL of water.
5. Transfer the solution into a clean 100-mL volumetric flask, being sure to rinse the beaker and funnel several times.
Transfer the rinsings into the flask.
6. Add pure water to make the final volume 100.0 mL.
7. Stopper the flask and mix the contents thoroughly by inverting the flask repeatedly.
7. vi ?
vf 1.00 L
Ci 5.00 mol/L
Cf 0.125 mol/L
viCi vfCf
vC
vi f f
Ci
1.00 L 0.125 mol/L
5.00 mol/L
vi
vf
0.0250 L 25.0 mL
vi
0.125 m
ol/L
1.00 L 5.00 mol/L
vi
0.0250 L 25.0 mL
or
The initial volume of 5.00 mol/L sulfuric acid required is 25.0 mL.
Procedure
1. Wear eye protection and a laboratory apron.
2. Calculate the initial volume of 5.00 mol/L H2SO4(aq) (25.0 mL) that is required to prepare 1.00 L of 0.125 mol/L
solution.
Copyright © 2002 Nelson Thomson Learning
Chapter 6 The Nature and Properties of Solutions
185
3.
4.
5.
6.
7.
8.
Add 400–500 mL of pure water to a 1-L volumetric flask.
Use a 25-mL volumetric pipet to transfer 25.00 mL of 5.00 mol/L acid to the flask.
Add pure water to make the final volume 1.000 L (to the mark on the flask).
Stopper the flask and mix the contents thoroughly by inverting the flask repeatedly.
This solution may be safely disposed of down the sink, provided it is washed down with plenty of water.
(a) In solid form, the sample of cobalt (II) chloride is a hexahydrate, so the calculation of molar mass must take this
into account.
C CoCl
0.100 mol/L
2
v CoCl
100.0 mL 0.1000 L
2
M CoCl 6H O 237.95 g/mol
2
2
n CoCl
2
m CoCl 6H O
2
2
m CoCl 6H O
2
2
or
m CoCl 6H O
2
2
m CoCl 6H O
2
2
0.100 mol
0.1000 L
1L
0.0100 mol
237.95 g
0.0100 mol 1
mol
2.38 g
0.100 mol
0.1000 L
1L
2.38 g
237.95 g
1
mol
The mass of solid cobalt (II) chloride hexahydrate required is 2.38 g.
(b) vi ?
vf 100.0 mL
Ci 0.100 mol/L
Cf 0.0100 mol/L
viCi
vfCf
vi
vC
f f
Ci
vi
0.0100 mol/L
100.0 mL 0.100 mol/L
vi
10.0 mL
The initial volume of 0.100 mol/L CoCl2(aq) required is 10.0 mL.
(c) Materials
• lab apron
• eye protection
• CoCl26H2O(s)
• wash bottle with pure water
• centigram balance
• 250-mL beaker
• stirring rod
• funnel
• 2 100-mL volumetric flasks with stoppers
• 10-mL volumetric pipet with pipet bulb
• medicine dropper
• meniscus finder
Procedure
1. Wear eye protection and a laboratory apron.
2. Calculate the mass of CoCl26H2O(s) required to prepare 100.0 mL of 0.100 mol/L solution.
3. Obtain the calculated mass (2.38 g) of CoCl26H2O(s) in a clean, dry 250-mL beaker.
186
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
4. Dissolve the solid in about 50 mL of water.
5. Transfer the solution into a clean 100-mL volumetric flask, being sure to rinse the beaker and funnel several times.
Transfer the rinsings into the flask.
6. Add pure water to make the final volume 100.0 mL.
7. Stopper the flask and mix the contents thoroughly by inverting the flask repeatedly.
8. Calculate the volume of 0.100 mol/L CoCl2(aq) required to make 100.0 mL of 0.0100 mol/L solution by dilution
(according to the above calculations: 10.0 mL).
9. Use a 10-mL volumetric pipet to obtain and transfer 10.0 mL of 0.100 mol/L CoCl2(aq) solution to a clean 100-mL
volumetric flask.
10. Add pure water to the flask until the final 100.0 mL volume is reached.
11. Stopper and invert the flask to thoroughly mix the contents.
12. Solutions may be disposed of down the sink with plenty of water.
Making Connections
9. Arguments for the position: There will be a fuel saving from transporting concentrated reagents (less mass and
volume); fewer trucks will be required for their transportation, resulting in less traffic on the roads; loading and
unloading will be quicker.
Arguments against the position: Any spills or careless handling would be much more dangerous; the recipient will
likely have the inconvenience of diluting the product upon its arrival.
10. The common system that concentrates pollutants (bioaccumulation) is the “web of life” or “food chain.” Organisms
that ingest pollutants can concentrate them in their bodies and pass this concentration on to predators. An example of
such a chain is aquatic microorganisms → plankton → fish → seabird chicks → eagles.
Reflecting
11. The procedure should show serial dilution, that is, begin by diluting stock HCl(aq) solution and then use samples of
each new dilution to produce a further dilution of, say, one-tenth the concentration each time. These solutions can then
easily be compared according to how rapidly they react with equal samples of zinc.
Make a Summary
(Page 308)
1
molar mass
(n = m /M)
amount = mass ×
molecular
substance
amount = volume × concentration
(n = v × c)
molecules
in solution
does not
conduct electricity
amount of
solution (n)
ionic
compound
nonelectrolytes
conducts
electricity
concentration
of solution
electrolytes
acids
ions in
water
acid
compound
determine
mass
prepared
by dilution
molecular
substance
concentrated
acid
refined
water
lake
determine
volume
other
concentrated
solution
Copyright © 2002 Nelson Thomson Learning
graduated
cylinder
volume
of solution
or solute (v)
prepared
from solid
ionic
compound
pipet
solution
volumetric
flask
water
purification
process
water
agriculture
runoff
pollution
landfill
leachate
aquifer
river
human
waste
industrial
waste
Chapter 6 The Nature and Properties of Solutions
187
CHAPTER 6
REVIEW
(Page 309)
Understanding Concepts
1. All concentration units express the ratio of the quantity of solute to the quantity of solution.
2. (a) Highly polar solvents
(b) Polar solvents
(c) Nonpolar solvents
3. (a) Water dissolves very many substances because it is liquid over a wide temperature range and because it is made
up of highly polar molecules that strongly attract (electrostatically) both ions and polar molecules. Water also
hydrogen bonds to substances with N, O and F atoms.
(b) Substances are dissolved in water to make them easier to store, easier to manipulate, and to control when and how
rapidly they react. Examples:
• Ammonia gas is so corrosive and toxic in pure form that it is very dangerous to handle, but when dissolved
in water can be stored, shipped, and easily handled in ordinary plastic bottles, such as in window cleaner solutions, or “household” ammonia cleaner.
• Baking soda (sodium hydrogen carbonate) is an example of a substance that doesn’t react (in cooking) until
dissolved. Solid plant fertilizers are another good example of this point.
• All-purpose cleansers are examples of solutions that react more rapidly and to greater extent when they are
made with a higher solute/solvent ratio (more concentrated).
(c) The ability of water to dissolve so many substances is a problem when it comes to the purification of water for
drinking, because often a large number of toxic or noxious solutes must be removed.
4. (a) solute: sodium chloride
solvent: water
(b) solute: acetic acid
solvent: water
(c) solute: sodium carbonate decahydrate
solvent: water
(d) solute: sucrose
solvent: water
(e) solute: ethanol
solvent: water
5. c MF 2 g/100 mL
v milk 250 mL
mfat
mfat
2.0 g
250 mL
100 mL
5.0 g
The mass of fat in one glass of milk is 5.0 g.
6. cMF 5.9%, 2.0%, and 1.2% , therefore,
cMF 5.9 g/100 mL, 2.0 g/100 mL, and 1.2 g/100 mL , respectively.
m fat 3.0 g maximum in each serving
100 g
mserving 3.0 g
fat 5.9 g fat
mserving 51 g or 0.051 kg
100 g serving
mserving 3.0 g
fat 2.0 g fat
mserving 0.15 kg
100 g serving
mserving 3.0 g
fat 1.2 g fat
mserving 0.25 kg
The serving sizes containing 3.0 g of fat for the three choices of yogurt are as follows: for 5.9% MF yogurt, a 51 g or
0.051 kg serving; for 2.0% MF yogurt, a 0.15 kg serving; and for 1.2% MF yogurt, a 0.25 kg serving.
188
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
7. (White) vinegar is most commonly sold in stores as a 5% (by volume) solution of acetic acid, HC2H3O2(aq), by
volume. Assume a minimum of 5 mL of acetic acid in 100 mL of solution. (Commercial labelling is always a guaranteed legal minimum. Significant digits do not apply to this value, because the 5% is not a measurement.)
cHC H O 5% 5 mL/100 mL
2 3 2
vHC H O 15 mL
2 3 2
100 m
L
vHC H O 15 mL 2 3 2
5
mL
vHC H O 0.30 L
2 3 2
The volume of vinegar containing 15 mL of acetic acid is 0.30 L.
8. c nitrate 55 ppm 55 mg/L
v water 200 mL 0.200 L
55 mg
m nitrate 0.200 L
1
L
m nitrate 11 mg
The mass of nitrate ion in the water is 11 mg.
9. (a)
n Cu(NO )
= 0.35 mol
v Cu(NO )
C Cu(NO )
= 500 mL 0.500 L
0.35 mol
= 0.500 L
C Cu(NO )
= 0.70 mol/L
3 2
3 2
3 2
3 2
The molar concentration of the solution is 0.70 mol/L.
(b) M NaOH = 40.00 g/mol
m NaOH = 10.0 g
v NaOH = 2.00 L
1 mol
n NaOH = 10.0 g 40.00 g
= 0.250 mol
C NaOH
0.250 mol
= 2.00 L
C NaOH
= 0.125 mol/L
C NaOH
1 mol
1
= 10.0 g 40.00 g
2.00 L
= 0.125 mol/L
or
C NaOH
The molar concentration of the solution is 0.125 mol/L.
(c)
vi
25 mL
vf
145 mL
Ci
11.6 mol/L
Cf
?
viCi vfCf
vC
Cf i i
vf
25 mL
11.6 mol/L
145 mL
Cf Copyright © 2002 Nelson Thomson Learning
Chapter 6 The Nature and Properties of Solutions
189
Cf
= 2.0 mol/L
or
11.6 mol/L
Cf = 25 mL
145 m
L
Cf = 2.0 mol/L
The final molar concentration of the solution is 2.0 mol/L.
(d) 16 ppm of Mg2+(aq) = 16 mg/L
MMg2+
= 24.31 g/mol
mMg2+
= 16 mg
v
= 1.00 L (assume one litre of solution, with a certainty that does not limit the certainty of
the answer)
n Mg2+
1 mol
= 16 mg 24.31 g
= 0.66 mmol
CMg2+
0.66 mmol
= 1.00 L
CMg2+
= 0.66 mmol/L
CMg2+
1 mol
1
= 16 mg 21.43g
1.00L
= 0.66 mmol/L
or
CMg2+
The molar concentration of the solution is 0.66 mmol/L.
10. (a) The labelled compounds are (by formula and class of compound):
water
H2O(l)
neutral molecular
glucose
C6H6O11(l)
neutral molecular
citric acid
C3H4OH(COOH)3(s)
acid molecular
potassium citrate
K3C3H4OH(COO)3(s)
ionic
sodium chloride
NaCl(s)
ionic
potassium phosphate
K3PO4(s)
ionic
(b) Dissolved glucose would make the drink taste sweet, and dissolved citric acid would make it taste tangy.
(c) Note that the ion mass values in this question are from a commercial label. They thus represent required legal
minimum content, and are not measured values. Significant digits can not be assigned to such values, so the certainty
for this calculation must be taken from (related to) the given serving volume value, 400 mL (3 significant digits).
50 mg
mNa+
mK+
55 mg
v solution 400 mL = 0.400 L
50 mg
cNa+ = 0.400 L
cNa+ = 125 mg/L = 125 ppm
The concentration of sodium ions is 125 ppm.
55 mg
cK+ = 0.400 L
cK+ = 138 mg/L = 138 ppm
The concentration of potassium ions is 138 ppm.
190
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
11. M Na C O = 134.00 g/mol
2 2 4
v Na C O
= 250.0 mL 0.2500 L
C Na C O
= 0.375 mol/L
2 2 4
2 2 4
0.375 mol
n Na C O 0.2500 L
2 2 4
1L
0.0938 mol
134.00 g
m Na C O 0.0938 mol 2 2 4
1
mol
m Na C O 12.6 g
2 2 4
or
0.375 mol
134.00 g
m Na C O 0.2500 L
2 2 4
1L
1
mol
m Na C O 12.6 g
2 2 4
The mass of sodium oxalate required is 12.6 g.
12. vi ?
vf 500 mL
Ci 14.6 mol/L
Cf 1.25 mol/L
viCi
vfCf
vi
vC
f f
Ci
vi
=
500 mL 1.25 mol/L
14.6 mol/L
= 42.8 mL
vi
or
1.25 mol/L
= 500 mL 14.6 mol/L
= 42.8 mL
vi
vi
The required volume of the initial phosphoric acid solution is 42.8 mL.
13. c HCl
36% W/V = 0.36 kg/L
M HCl
= 36.46 g/mol
CHCl
0.36 kg
1 mol
1L
36.46 g
0.0099 kmol/L 9.9 mol/L
vi ?
vf 5.00 L
Ci 9.9 mol/L
Cf 0.12 mol/L
viCi vfCf
vi
vC
f f
Ci
Copyright © 2002 Nelson Thomson Learning
Chapter 6 The Nature and Properties of Solutions
191
5.00 L 0.12 mol/L
9.9 mol/L
vi
= vi
= 0.061 L 61 mL
vi
0.12 mol/L
= 5.00 L 9.9 mol/L
vi
= 0.061 L 61 mL
or
The required initial volume of the hydrochloric acid is 61 mL.
Applying Inquiry Skills
14. Standard solutions are prepared either by dissolving a measured mass of a solid solute to make a known volume of
solution; or by diluting an existing (more concentrated) solution of known concentration to decrease its concentration.
15. A standard solution is one for which the concentration is accurately known. It is necessary for accurate chemical
analysis or for precise control of chemical reactions.
16. (a) Electrolytes can be distinguished from nonelectrolytes by testing solutions of the compound with an ohmmeter,
or conductivity meter. Electrolytes form conducting solutions; nonelectrolytes form nonconducting solutions.
(b) Acids, bases, and neutral compounds can be distinguished by testing a solution of each compound with blue and
pink litmus. In an acidic solution, blue litmus will change to pink (but pink litmus will be unchanged); in a basic
solution, pink litmus will change to blue (but blue litmus will be unchanged); and in a neutral solution, neither
pink nor blue litmus will change colour.
17. (a) M KHC H O
4 4 6
= 188.19 g/mol
v KHC H O
= 100.0 mL 0.1000 L
C KHC H O
= 0.150 mol/L
4 4 6
4 4 6
n KHC H O
4 4 6
m KHC H O
4 4 6
m KHC H O
4 4 6
or
m KHC H O
4 4 6
m KHC H O
4 4 6
0.150 mol
0.1000 L
1L
0.0150 mol
188.19 g
0.0150 mol 1
mol
2.82 g
0.150 mol
188.19 g
0.1000 L
1L
1
mol
2.82 g
The mass of potassium hydrogen tartrate measured is 2.82 g.
(b) Procedure
1. Put on eye protection and a lab apron.
2. Calculate the mass of solid potassium hydrogen tartrate needed to prepare 100.0 mL of a 0.150 mol/L solution. (2.82
g, as shown above.)
3. Obtain the calculated mass of potassium hydrogen tartrate in a clean, dry 150 mL beaker.
4. Dissolve the solid in 40 to 50 mL of pure water.
5. Transfer the solution into a 100 mL volumetric flask. Rinse the beaker and funnel two or three times with small
quantities of pure water, transferring the rinsings into the volumetric flask.
6. Add pure water to the flask until the volume is 100.0 mL.
7. Stopper the flask and mix the contents thoroughly by repeatedly inverting the flask.
18. (a) vi ?
vf 100.0 mL
Ci 0.400 mol/L
Cf 0.100 mol/L
192
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
viCi vfCf
vi
vC
f f
Ci
vi
= vi
= 25.0 mL
vi
0.100 m
ol/L
= 100.0 mL 0.400 m
ol/L
= 25.0 mL
or
vi
100.0 mL 0.100 mol/L
0.400 mol/L
The required initial volume of the stock solution is 25.0 mL.
(b) Procedure
1. Wear eye protection and a lab apron.
2. Calculate the volume of a 0.400 mol/L stock solution (25.0 mL, as shown above) that will be required to prepare
100.0 mL of a 0.100 mol/L solution by dilution.
3. Add 40 to 50 mL of pure water to a 100-mL volumetric flask.
4. Use a 25-mL volumetric pipet to transfer 25.00 mL of the stock solution into the 100-mL volumetric flask.
5. Add pure water until the final volume of 100.0 mL is reached.
6. Stopper the flask and mix the solution thoroughly.
19. Analysis
Solution A shows acidic properties, turning blue litmus red and conducting current; so according to our current knowledge it must be the sulfuric acid.
Solution B shows neutral molecular properties, not changing litmus and not conducting current; so according to our
current knowledge it must be the glucose.
Solution C shows basic properties, turning red litmus blue and conducting current; so so according to our current
knowledge it must be the calcium hydroxide.
Solution D shows neutral ionic properties, not changing litmus but conducting current; so according to our current
knowledge it must be the potassium chloride.
Making Connections
20. (a) Oil and water will not dissolve in each other because water is polar and oils are not. Essentially, water molecules
attract each other so strongly with hydrogen bonding that oil molecules cannot intermix with them. Depending
on the density of the oil, it will either float on top of the water in a thin layer, or sink to the bottom in discrete
globs.
(b) Oil on the surface of the water prevents the exchange of oxygen and carbon dioxide between the air and the water.
It also clogs the fine filaments of organisms, such as fishes’ gills and invertebrates’ tentacles, as well as coating
the feathers and fur of marine birds and mammals, thus destroying their insulation and waterproofing. Together,
these problems result in the death of much marine life, as well as making the area unsightly.
(c) Oil is cleaned up with steam, solvents, absorbent materials, and physical collectors (skimmers and scrapers).
Sometimes it is also treated to make it sink to the bottom, but this does not necessarily solve the problem.
(d) Risks of transportation by large oil tanker: oil leakage; introduction of alien species in ballast water; control is in
the hands of relatively few extremely large companies.
Benefits of transportation by large oil tanker: relatively inexpensive; relatively safe (if ships are well built and well
handled); the system is already established, including fleets of ships, ports, loading and off-loading facilities.
Alternatives: no transoceanic transportation of oil; transportation in smaller ships; air transportation.
Stopping tanker oil transport would mean a major change in our civilization, which largely runs on fairly cheap
energy from fossil fuels. But continuing this shipping of oil inevitably means environmental damage will occur. A
risk–benefit analysis should focus on minimizing the risks, in this case, for example, by requiring that tankers be
double-hulled for increased safety, or that tankers not be allowed to travel near sensitive areas such as the Galapagos
Islands.
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Chapter 6 The Nature and Properties of Solutions
193
21. Solutions collected might include examples such as:
• rubbing alcohol (isopropanol): 70% (V/V) CH3CHOHCH3(l) in water (a liquid solute in a liquid solvent); not
to be taken internally; toxic by ingestion;
•
decongestant (oxymetazoline hydrochloride): 0.05% in water (a solid solute in a liquid solvent); limit two uses
per nostril per day; not to be used by children; causes increase in nasal congestion when use is discontinued;
•
bleach (sodium hypochlorite): 5.25% NaOCl(s) in water (a solid solute in a liquid solvent); corrosive; toxic fumes;
reacts with acids and bases to produce toxic products;
•
window cleaner (household ammonia): NH3(g) in water (a gaseous solute in a liquid solvent); irritant to eyes, nose
and throat; concentrated fumes (not in the product mentioned) are toxic and may be fatal;
•
metal ornaments (pewter): 4-7% Sb, 1-2%Cu in tin, Sn(s) (solid solutes in a solid solvent); no safety precautions;
•
air: about 20% oxygen in nitrogen (gasesous solute in a gaseous solvent); no safety precautions.
22. Research should turn up examples such as
• glucose or dextrose (5%); to provide nourishment to a patient by placing sugar directly in the bloodstream for use
by body cells;
•
multivitamin infusion (1 vial vitamin concentrate in 1000 mL of IV fluid); to eliminate risk of vitamin deficiency
in patients receiving nutrition primarily through IV feeding;
•
antibacterial agent such as gatifloxacin (200 mg (100 ml) or 400 mg (200 ml) of gatifloxacin (2 mg/ml) in 5%
dextrose); to treat bacterial infections;
•
heart drug, esmolol HCl (10 mg/mL infusion by adding one 2500 mg ampule to a 250 mL container of a compatible intravenous solution); to control heartbeat in patients recovering from surgery;
•
lidocaine (2% (20 mg/mL)); a local anaesthetic, particularly recommended for use while setting broken bones.
23. A typical chart would show examples as follows:
(a) Animal waste washed by heavy rain into a stream that feeds a river, from which water is extracted for a town or city’s
water supply.
(b) Human waste migrates underground from an outdoor toilet or an underground septic tank to a well drilled downhill
from the waste.
(c) Industrial waste is released into a river system from which water is extracted for domestic purposes farther down river.
(d) Municipal waste leaches into the ground and into the aquifer from which well water is extracted for domestic use.
(e) Natural waste from nondomestic plants and animals is washed by flooding into a river from which water is removed
for municipal use.
source
metal refining
burning sulfurous
coal
leaking
transformers
pulp-mills
pollutant
sulfur dioxide
acid rain
PCBs
dioxins
airborne
airborne
ground leaching
rivers
respiratory
problems
debilitates
plants
(uncertain)
nausea, headache,
skin irritation, death
route
potential
health
effect
Note: Student research is required for this answer. Other controversial chemicals include alcohol, asbestos, benzene,
CFCs, chlorine, cigarette smoke, DDT, lead, mercury, nitrites, nitrogen oxides, ozone, phosphates, radon, and vinyl
chloride. See Controversial Chemicals: A Citizen's Guide by P. Kruus and I.M. Valeriote (ISBN: 0-9198868-22-3).
24. See a summary of Guidelines for Canadian Drinking Water Quality, published by Health Canada, for acceptable water
organism and impurity levels.
(a) One potentially dangerous contaminant is cryptosporidium, a microorganism and parasite. There is no legal
requirement to test for cryptosporidium, and thus there is no MAC for this contaminant. Many town water
supplies are not tested for cryptosporidium.
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Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
(b) In April/May 2001, there was an outbreak of illnesses in North Battleford, Saskatchewan, that were caused by
cryptosporidium.
(c) The symptoms are flulike with severe stomach cramps, nausea, and diarrhea. This outbreak resulted in at least
three deaths from the cryptosporidium in the drinking water. Only people with immune deficiencies died or were
seriously threatened by the contaminant. Cryptosporidium effects are not generally as severe as E. coli effects,
such as those experienced in Walkerton, Ontario, in 2000, where seven people died.
(d) A water advisory and then a boil-water order was issued. It was discovered that the drinking-water filtration
system malfunctioned after it was shut down for semi-annual maintenance. Provincial authorities considered
legislation to require testing for cryptosporidium. The counterargument was that the testing is imprecise and that
the cryptosporidium is in such low concentrations that its presence may not even be detected by the testing. The
water management design, with the drinking water intake pipe downstream from the used-water treatment plant,
was also a concern.
Exploring
25. The report will be in the form of a sales brochure containing information such as the pressure needed to extract pure
water through a reverse osmosis membrane from sea water (theoretically about 2500 kPa minimum, empirically about
60 000 kPa – 90 000 kPa in practice). There are many commercially operating plants in Israel and the United States
using sea water, and one in Yuma, Arizona using saline river water. Reverse osmosis used to be relatively expensive
but is rapidly decreasing in cost with more widespread use. A plant near Tampa Bay in Florida, scheduled for completion in 2002, will treat 45 million U.S. gallons (170 million litres) of sea water each day and produce enough fresh
pure water to supply about 10% of the needs of that area.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
26. Chlorine treatment is used to kill harmful organisms (protozoa, bacteria, and viruses) in drinking water. For example,
one Redi Chlor tablet will make one gallon of water suitable for drinking. One system, Pristine, by a Canadian
company, Advanced Chemicals, mixes solution A (inactivated chlorine dioxide) with solution B (5% phosphoric acid)
to produce activated chlorine dioxide, which is then mixed with the water to purify it. Information on ingredients and
effects can also be obtained from a package label.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
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Chapter 6 The Nature and Properties of Solutions
195
CHAPTER 7
SOLUBILITY AND REACTIONS
Reflect on your Learning
(Page 312)
1. Most of the liquids that we encounter in the home are solutions. This could be tested by cooling, evaporating, or
smelling — anything that would encourage the components of a solution to separate.
2. Dissolving rate is usually speeded up by stirring, warming the solution, or finely dividing (grinding) the solute. If
excess solid remains after prolonged stirring, probably no more solute will dissolve.
3. Evidence for a reaction might be colour change, formation of a precipitate, formation of a gas, or a noticeable temperature change. The dissolving of both salt and sugar involves the solid separating into particles too small to see. This
is a physical change. Like most physical changes, it is reversible: the water can be removed, leaving the solute in its
original state.
Try This Activity: Measuring the Dissolving Process
(Page 313)
(a) The slightly cloudy mixture of table salt in water indicates that some substance(s) did not dissolve. Because we know
that sodium chloride is soluble, table salt must be a mixture containing at least one low-solubility solid. The solution
of pickling salt was completely clear.
(b) The ingredients are salt, calcium silicate, potassium iodide, sodium thiosulfate (for Sifto brand). Calcium silicate is
likely the cause of the cloudiness because silicates (such as sand) are not soluble in water.
(c) About 5 to 6 teaspoons of pickling salt appeared to dissolve.
(d) The final volume is a little more than 100 mL, based on the approximate markings on the side of the Erlenmeyer flask.
(e) The volume would likely be around 100 mL but not 120 mL, because solids generally dissolve in water without
increasing the volume. This can be tested by measuring 20.0 mL of sodium chloride in a graduated cylinder and
adding this to 100.0 mL of water measured in another graduated cylinder.
Note: The answer is about 106 mL.
7.1 SOLUBILITY
PRACTICE
(Page 316)
Understanding Concepts
1. (a) About 20 g of K2SO4(s) will dissolve in 100 mL of water at 70°C.
(b) KNO3(s) and KCl(s) have equal solubilities at about 22°C.
(c) To calculate molar concentration we would need to know the volume of the solution. The graph only gives the
volume of water used.
(d) The only substance shown for which 100 g will dissolve in 100 mL of water is KNO3(s), at about 56°C.
2. (a) NaCl would precipitate first at temperatures above 31°C.
(b) KCl would precipitate first at temperatures below 31°C.
Applying Inquiry Skills
3. The graph created from Investigation 7.1.1 will normally differ somewhat from the one in Figure 2. The difference is
primarily experimental error, due to a less than precise design for the Investigation.
Try This Activity: Gas Solubility
(Page 317)
Some suggested answers are given.
(a) The lit match is extinguished. This suggests the presence of carbon dioxide, which does not support combustion.
(b) (Observations will vary depending on the relative humidity.) The air near the outside of the cold glass cools. As the
temperature of the air decreases, so does the solubility of water vapour in the air. Some water vapour condenses to
liquid water. A qualitative test for water is to use cobalt(II) chloride paper.
Copyright © 2002 Nelson Thomson Learning
Chapter 7 Solubility and Reactions
197
(c) The small gas bubbles on the inside wall of the glass are likely due to air coming out of solution as the temperature
of the water increases to room temperature. As the temperature increases in the cold water glass, the volume of gas
also increases. In the hot water, the temperature is decreasing so the bubbles would stay small or decrease in volume.
PRACTICE
(Page 318)
Understanding Concepts
4. A room temperature can of pop is more likely to spray: as gases are less soluble in warm liquids than in cold, the gases
will tend to come out of solution, so the pressure in the can is higher.
Reflecting
5. (a) Red blood cells contain hemoglobin molecules, that bind chemically to oxygen molecules, making the oxygen
appear more soluble.
(b) If blood held the same amount of oxygen as water, there would be only enough oxygen in our blood to keep our
cells alive for a few seconds, i.e., our lives would be different in the event of such a change by becoming dramatically shorter. To compensate, we would require completely different circulatory systems, for example, one that
supplied a much greater volume of blood to our organs.
PRACTICE
(Page 319)
Understanding Concepts
6. Temperature must always be stated when reporting a solubility.
7.
Solubility
Gases
Solids
Temperature
8. (a) CO2(g) H2O(l) → H2CO3(aq)
(b) H2CO3(aq) Ca(OH)2(aq) → CaCO3(s) 2 H2O(l)
(c) The second reaction can only occur if carbon dioxide is present for the first reaction, so the test is diagnostic.
(d) If the calcium hydroxide solution were too dilute, the calcium carbonate that forms might be below its solubility,
and stay in solution rather than forming a precipitate. Thus, the test would not work.
(e)
Solubility Curve for Calcium Hydroxide
Solubility (g/100 mL)
0.20
0.15
0.10
0.05
0
20
40
60
80
100
Temperature (°C)
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Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
(f) From the solubility curve drawn in (e), the solubility of Ca(OH)2(s) at 22°C is 0.17 g/100 mL, or 0.17 g/0.100 L.
v Ca(OH) 1.0 L
2
c Ca(OH)
2
0.17 g/100 mL 0.17 g/0.100 L
m Ca(OH)
2
m Ca(OH)
2
0.17 g
1.0 L
0.100 L
1.7 g
The minimum mass of calcium hydroxide required to make up 1 L of saturated solution will be about 1.7 g. The actual
mass one would use should be much more than this — say, 15 to 20 g, to ensure a large excess of solute, which in
turn ensures saturation.
Note: In this particular instance the volume of the solute added is very small compared to the volume of the
solvent; so the difference between solvent volume and solution volume may be considered negligible.
(g) The solubility of calcium hydroxide is anomalous — unlike most solids it decreases in solubility with increasing
temperature. The generalization we use is still valid and useful for most soluble solids. It just needs to be understood that generalizations are exactly that — statements that generally (not invariably) describe events correctly.
Applying Inquiry Skills
9. Prediction
(a) The solubility of potassium chlorate should increase with an increase in temperature, according to the generalization for the solubility of solids, and also according to Figure 2 on page 316 of the text.
Analysis
(b)
Solubility Curve for Potassium Chlorate
60
Solubility (g/100 mL)
50
40
30
20
10
0
20
40
60
80
100
Temperature (°C)
(c) According to the Evidence, the solubility of potassium chlorate increases with increasing temperature, for solution temperatures from 0°C to 100°C.
Making Connections
10. Gases such as oxygen are less soluble in warm water, so active fish that require high oxygen levels may not thrive in
water that is warmed by an outside source, such as a power plant. In addition, power plants may alter the ecology of
a body of water by preventing it from freezing over in the winter. This would attract waterfowl that would feed on
aquatic organisms normally under the ice at this time.
PRACTICE
(Page 325)
Understanding Concepts
11. (a) The high/low solubility cutoff point is 0.10 mol/L at SATP.
(b) The 0.10 mol/L cutoff is useful because most laboratory solutions are within an order of magnitude of this
concentration, and because most ionic solids are either markedly more or less soluble than this value.
(c) The table need only be used to determine the solubility of an ionic solid if the solid does not contain a group I
ion, an ammonium ion, or a nitrate ion — because such compounds are always highly soluble. To determine the
solubility of any other ionic solid, find the anion in its column heading, and look down the column to find the row
Copyright © 2002 Nelson Thomson Learning
Chapter 7 Solubility and Reactions
199
(d)
12. (a)
(b)
(c)
(d)
(e)
13. (a)
(b)
(c)
containing the cation. The row where the cation is found will identify the compound as having low or high solubility.
In chemistry class, the word soluble usually refers to compounds that the table classes as “high solubility.” The
word insoluble usually refers to the compounds that the table classes as “low solubility,” provided their solubility
is so low that no effect is noticed from dissolving.
NaOH(s)
high solubility
MnCl2(s)
high solubility
Al(OH)3(s)
low solubility
Ca3(PO4)2(s)
low solubility
CuSO45H2O(s) high solubility
Large crystals take more time to “grow,” so the sugar probably crystallized slowly from a saturated solution.
Solubility is not affected by crystal size, so the specialty sugar would have the same solubility as regular white
sugar.
The dissolving rate is slower for large crystals.
Applying Inquiry Skills
14. Experimental Design
Small quantities of a solid may be dissolved sequentially in a measured sample of a solute until no more will dissolve,
to determine the solubility; or a saturated solution of known volume can be evaporated to dryness, and the solubility
calculated from the measured mass of solid solute remaining.
Making Connections
15. Pollutants in natural water can enter the water cycle through runoff from agricultural areas or landfills, or industrial
tailings ponds. Some pollutants are toxic and/or noxious at extremely low concentrations, so they may be dangerous
even if they have very low solubility.
PRACTICE
(Page 326)
Understanding Concepts
16.
Solubility Curves
Solubility
NH4HCO3
NaHCO3
Temperature
Making Connections
17. In the Solvay process, the scientific knowledge of solubility effects made possible a new technique for producing a
commercial chemical much more efficiently — a classic example of science leading technology.
SECTION 7.1 QUESTIONS
(Page 326)
Understanding Concepts
1. (a) The presence of the solid is evidence that sodium bromide is precipitating, probably because the solvent is evaporating (or maybe because of a drop in temperature). The solution is therefore at the limit of its concentration for
this temperature.
(b) The concentration of sodium bromide in the remaining solution is at the maximum possible value. Such a solution is said to be saturated.
200
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
(c) The mixture could be converted to a single phase solution by adding more solvent, or by warming. Either process
would dissolve the solid present.
2. Solids generally become more soluble as temperature increases, while gases generally become less soluble as temperature increases.
Applying Inquiry Skills
3. Prediction
(a) Barium sulfate should become more soluble as temperature increases, according to the generalization about the
solubility of ionic solids.
Experimental Design
(b) The independent variable is the solution temperature. The dependent variable is the mass of barium sulfate. The
most important controlled variable is the volume of saturated solution taken each time.
Analysis
(c)
Solubility of Barium Sulfate
Mass of BaSO4(s) (mg/100 mL)
0.50
0.40
0.30
0.20
0.10
0
10
20
30
40
50
Temperature (°C)
(d) According to the evidence from this investigation, barium sulfate does become more soluble as temperature
increases.
Evaluation
(e) The experiment could be improved by just making and using a single saturated solution, and taking samples from
it at different temperatures.
(f) The prediction was verified, according to the evidence gathered.
(g) The solubility generalization is supported by the results of this experiment. One test of one compound is certainly
not enough evidence to justify stating such a generalization. Many repeated tests of many compounds would be
required for scientific acceptance and confidence.
Making Connections
4. If a non-aqueous solvent is used, it is probably because the dirt and grease on clothing dissolves better in such
solvents. Also, some clothing fabrics are damaged by water, but are not affected by non-aqueous solvents.
5. (a) Cold, fast-flowing streams will have a higher concentration of dissolved oxygen, because more air will be mixed
with turbulent water, and because gases dissolve better in colder water.
(b) Trout probably require more oxygen than carp.
(c) Thermal pollution would probably affect trout seriously, because warming their water will reduce its oxygen
concentration.
7.2 HARD WATER TREATMENT
PRACTICE
(Page 329)
Understanding Concepts
1. (a) Scale in kettles and soap scum are both evidence of hard water.
(b) Ground water is hard in areas where soluble calcium and magnesium minerals exist in the ground.
Copyright © 2002 Nelson Thomson Learning
Chapter 7 Solubility and Reactions
201
(c) The mixture could be converted to a single phase solution by adding more solvent, or by warming. Either process
would dissolve the solid present.
2. Solids generally become more soluble as temperature increases, while gases generally become less soluble as temperature increases.
Applying Inquiry Skills
3. Prediction
(a) Barium sulfate should become more soluble as temperature increases, according to the generalization about the
solubility of ionic solids.
Experimental Design
(b) The independent variable is the solution temperature. The dependent variable is the mass of barium sulfate. The
most important controlled variable is the volume of saturated solution taken each time.
Analysis
(c)
Solubility of Barium Sulfate
Mass of BaSO4(s) (mg/100 mL)
0.50
0.40
0.30
0.20
0.10
0
10
20
30
40
50
Temperature (°C)
(d) According to the evidence from this investigation, barium sulfate does become more soluble as temperature
increases.
Evaluation
(e) The experiment could be improved by just making and using a single saturated solution, and taking samples from
it at different temperatures.
(f) The prediction was verified, according to the evidence gathered.
(g) The solubility generalization is supported by the results of this experiment. One test of one compound is certainly
not enough evidence to justify stating such a generalization. Many repeated tests of many compounds would be
required for scientific acceptance and confidence.
Making Connections
4. If a non-aqueous solvent is used, it is probably because the dirt and grease on clothing dissolves better in such
solvents. Also, some clothing fabrics are damaged by water, but are not affected by non-aqueous solvents.
5. (a) Cold, fast-flowing streams will have a higher concentration of dissolved oxygen, because more air will be mixed
with turbulent water, and because gases dissolve better in colder water.
(b) Trout probably require more oxygen than carp.
(c) Thermal pollution would probably affect trout seriously, because warming their water will reduce its oxygen
concentration.
7.2 HARD WATER TREATMENT
PRACTICE
(Page 329)
Understanding Concepts
1. (a) Scale in kettles and soap scum are both evidence of hard water.
(b) Ground water is hard in areas where soluble calcium and magnesium minerals exist in the ground.
Copyright © 2002 Nelson Thomson Learning
Chapter 7 Solubility and Reactions
201
2. (a) Na2CO3(s) and Ca(OH)2(s)
(b) The added carbonate ions cause calcium and magnesium ions in the hard water to precipitate, because CaCO3
and MgCO3 have very low solubility.
3. (a) C CaCO
3
7.1 10-5 mol/L
MCaCO 100.09 g/mol
3
0.000071 mol
100.09 g
c CaCO 3
L
mol
c CaCO 0.0071 g/L 7.1 mg/L
3
The concentration of calcium carbonate in the treated water will be equal to the solubility; 7.1 mg/L.
(b) 7.1 mg/L is equal to 7.1 ppm.
(c) The treated water is classified as soft, according to Table 1.
4. A home water softener unit has to be regenerated because the resin in it has a finite capacity to attract hard water ions,
to replace sodium ions. The sodium ions go into the bath and washing machine, but no precipitate (bathtub ring)
2+ and Mg2+ ions can attach to the resin, the regeneration cycle replaces the hard water ions
forms. When no more Ca(aq)
(aq)
with fresh sodium ions. The hard water ions are flushed down the drain.
Making Connections
5. On a large scale, the amount of resin required for softening would be extremely expensive, and the volume of concentrated brine needed for regeneration would be enormous, and thus a problem to make and store. The quantity of
sodium and chloride ions in the water might pose a health problem for some people.
6. Detergents clean well in harder water, producing no “soap scum,” which was a huge benefit. However, they increased
some pollution problems. Early detergents were not very biodegradable, and phosphates from detergents created (and
sometimes still do create) eutrophication problems in lakes and rivers.
7. Home-softened water (from ion-exchange softeners) is often not routed to toilets (for economy), and sometimes not
to a kitchen tap that provides drinking water (for possible health reasons). The incidence of heart disease is statistically slightly lower in areas with hard water than in areas with naturally soft water. People with heart concerns, particularly those on low-sodium diets, may not wish to drink home-softened water (which is often high in sodium).
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
SECTION 7.2 QUESTIONS
(Page 330)
Understanding Concepts
1. Water is naturally hard in areas where ground water comes in contact with minerals (such as limestone) that contain
slightly soluble calcium and/or magnesium compounds.
2+ and Mg2+ are the ions mostly responsible for water hardness.
2. Ca(aq)
(aq)
3. (a) Laundry “scum” and kettle scale are the noticeable effects of hard water.
(b) One serious effect of hard water that is not readily noticeable is the formation of scale in pipes. If it is allowed to
build up unchecked, the scale can significantly reduce the flow of water through the pipe, and come close to
blocking it completely.
4. Na2CO3(aq) Ca(HCO3)2(aq) → 2 NaHCO3(aq) CaCO3(s)
5. (a) Calcium ions are more readily attracted than sodium ions to negatively charged sulfonate sites on a resin molecule. This is mostly because a calcium ion has a charge of 2+, twice as much as the charge of a sodium ion.
(b) During regeneration, there are very many more sodium ions present (at maximum concentration) than calcium
ions, so the chance that sodium ions will collide with and attach to the sulfonate sites becomes very much greater,
overcoming the fact that calcium ions are attracted more strongly than sodium ions.
Applying Inquiry Skills
6. Question
(a) Is water softened by passing through a pipe to which a magnet is attached?
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Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
Experimental Design
Hard water is run through a pipe to a tap, and samples are tested for hardness, with and without a magnet being
attached to the pipe.
(b) Evaluation
The original design is obviously useless in light of the new understanding of the claim made for the magnet; it is
not testing the claimed result.
New Experimental Design
Hard water is run through identical piping that has and has not magnets attached. After several months, the piping is
cut and the ease of removing the scaling from the inside of the pipe is determined.
Making Connections
7. high sodium ion
concentration
WATER
SOFTENING
cleaner laundry;
no scale formation
REGENERATION
high sodium
ion concentration
high calcium ion and
saltwater discharge to
the environment
7.3 REACTIONS IN SOLUTION
PRACTICE
(Page 332)
Understanding Concepts
1. (a) silver sulfide: low solubility
(b) magnesium nitrate: high solubility
(c) zinc carbonate: low solubility
2. (a) Precipitate forms: SrSO4(s)
(b) No precipitate forms
(c) Precipitate forms: CuSO3(s)
PRACTICE
(Page 335)
Understanding Concepts
3. Sr(NO3)2(aq) Na2CO3(aq) → SrCO3(s) 2 NaNO3(aq)
2+ 2 NO–
+
2–
+
–
Sr(aq)
3(aq) 2 Na(aq) CO3(aq) → SrCO3(s) 2 Na(aq) 2 NO3(aq)
2+ CO2–
Sr(aq)
3(aq) → SrCO3(s)
4. (a) Compounds could be any four of:
copper(II) nitrate, copper(II) sulfate, copper(II) acetate, copper(II) chloride, copper(II) bromide,
or copper(II) iodide.
Choose copper(II) nitrate as the example:
(b) 2 Al(s) 3 Cu(NO3)2(aq) → 3 Cu(s) 2 Al(NO3)3(aq)
2+ 6 NO–
3+
–
(c) 2 Al(s) 3 Cu(aq)
3(aq) → 3 Cu(s) 2 Al(aq) 6 NO3(aq)
2+ → 3 Cu 2 Al3+
(d) 2 Al(s) 3 Cu(aq)
(s)
(aq)
5. Cl2(g) 2 Br–(aq) → Br2(l) 2 Cl–(aq)
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Chapter 7 Solubility and Reactions
203
2+ OOCCOO2– → CaOOCCOO
6. Ca(aq)
(s)
(aq)
or (if the oxalate ion formula is written in condensed form)
2+ C O2–
Ca(aq)
2 4(aq) → CaC2O4(s)
+
2+ 2 Ag
7. Pb(s) 2 Ag(aq)
→ Pb(aq)
(s)
Making Connections
3+ 3 OH–
8. (a) Fe(aq)
(aq) → Fe(OH)3(s)
(b) Filtration to remove the precipitate is the most likely process.
Explore an Issue
Debate: Producing Photographs
(Page 336)
(a) At first glance, it seems that digital cameras are more environmentally friendly than film cameras: they don’t require
film, film canisters, developing paper, or processing chemicals. However, this conclusion is based upon the assumption that only the final products affect the environment.
(b) To argue against the proposition you need to recognize that the whole story of a product must be considered, including
everything from the extracting of resources to manufacturing through disposal. For example, are there components of
the digital camera that during manufacture cause the emission of toxins into the environment, or, when the camera
must be disposed of, are there environmental concerns?
(c) Recognize that the resolution requires an environmental perspective only; you need not consider, for example,
economic, social, and technological arguments. In your investigation, you could look at subtopics such as raw materials, manufacturing, the developing process, the developing technology (and its environmental impact back to its
origins), and disposal.
(d) Consider the logic of the presentation and the quantity of evidence used to support the position taken. Were all stages
from pre-manufacturing through post-disposal considered?
SECTION 7.3 QUESTIONS
(Page 336)
Understanding Concepts
2+ SiO2–
1. Pb(aq)
3(aq) → PbSiO3(s)
2+ 2 PO3–
2. 3 Ca(aq)
4(aq) → Ca3(PO4)2(s)
+
2+ 2 Ag
3. Cu(s) 2 Ag(aq)
→ Cu(aq)
(s)
Making Connections
4. Student answers will vary widely, as they will be specific to the regulations controlling local hazardous waste
facilities.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
7.4 WASTE WATER TREATMENT
PRACTICE
(Page 340)
Understanding Concepts
1. Problems from the release of untreated sewage include: transmission of disease, lowering of oxygen levels, and rapid
growth of aquatic plant life.
2. A high BOD reading is an indication that bacteria are using up oxygen to decompose organic material in the water.
This is a problem for any oxygen-using life form in the water.
204
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
3.
Primary Treatment
Secondary Treatment
Tertiary Treatment
• screening
• flotation
• settling
• filtration
• aeration
• precipitation
• chlorination
• reverse osmosis
• distillation
• precipitation
• other...
4. Material that flows into a rural septic system is generally removed only by bacterial decomposition. Ideally, only water
and biodegradable substances should be flushed. Homeowners should be careful not to overwhelm the decomposition
process by flushing large quantities of any pollutants.
Reflecting
5. What goes down our drains is likely to end up in the ground or surface water. Being careful about waste disposal at
home and at school can make a cumulative positive difference in the quality of our environment.
SECTION 7.4 QUESTIONS
(Page 340)
Understanding Concepts
1.
deliver to user
ammoniation
postchlorination
fluoridation
primary treatment:
screening, flotation,
settling, and filtration
softening
aeration
disinfection
secondary treatment:
aeration and chlorination
filtration
coagulation,
flocculation, and
sedimentation
tertiary treatment (discretional):
e.g., reverse osmosis,
steam distillation.
chemical precipitation
collection
ground or
surface
water source
Drinking Water Treatment
Waste Water Treatment
Applying Inquiry Skills
2. (a) Hypotheses
(1) The fish kill may be due to a lack of oxygen caused by sewage discharge of organic matter into the river.
or
(2) The fish kill may be due to discharge of toxic material or a disease-causing organism into the river.
(b) Prediction (1)
The fish kill is caused by a high BOD, due to excess discharge of organic matter in sewage upstream.
Copyright © 2002 Nelson Thomson Learning
Chapter 7 Solubility and Reactions
205
Experimental Design (1)
The water is tested for oxygen and for organic matter, every 500 m upstream from the fish-kill area to the industrial
town.
Prediction (2)
The fish kill is caused by toxic chemicals or disease organisms in the sewage discharge upstream.
Experimental Design (2)
The water is tested for chemicals and for organisms causing common diseases in fish, every 500 m upstream from the
fish-kill area to the industrial town.
Note: This is an example of a correlational study (see Appendix A1, pages 608 – 9). Technically, correlational
studies are inductive investigations without a hypothesis and a prediction. Note above that when forced, a hypothesis
and a prediction are similar, indicating that the investigation is inductive and that neither a hypothesis nor a prediction should be used.
Making Connections
3. Garbage disposal units decrease the amount of solid bagged waste from a household, which cuts costs and extends the
usefulness of landfills. However, the increase of organic matter in the sewage places more demand on the local waste
water treatment system. Most of this food waste could be diverted to a composting system.
4. Answers will depend on the regulations controlling local hazardous waste facilities. Unless the local area has a tertiary
treatment facility, there will automatically be an argument for improvement. The only logical long-term human goal
is to eventually have all waste water returned to the cycle in a form that puts no stress on the environment.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
7.5 QUALITATIVE CHEMICAL ANALYSIS
PRACTICE
(Page 342)
Understanding Concepts
1. (a) colourless
(b) blue
(c) yellow-brown
(d) orange
(e) colourless
(f) green
2. (a) yellow-red
(b) blue
(c) yellow
(d) violet
(e) colourless
3. (a) yellow-red
(b) light blue-grey
(c) bright red
(d) green
Applying Inquiry Skills
4. (a) Analysis
According to Table 2, Colours of Flames:
solution A contains K+ ions,
solution B contains Cu2+ ions,
solution C contains Na+ ions,
solution D contains Ca2+ ions, and
solution E contains Li+ ions and/or Sr2+ ions.
(b) Evaluation
The design of this experiment is too limited. It only identifies those positive ions (cations) in the solutions that
206
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
happen to produce coloured flames. Other possible cations, not to mention the anions in each of these solutions,
cannot be identified this way. As well, solution E may contain either or both of two cations, since they both
produce the same result.
PRACTICE
(Page 346)
Understanding Concepts
5. Qualitative analysis determines what is in a sample, and quantitative analysis determines how much is present.
6. A diagnostic test statement always includes procedure, evidence, and analysis steps. For example: If a gas is bubbled
through limewater, and a white precipitate forms, then the gas is likely to contain carbon dioxide.
7. Precipitates could be formed with the listed ions by adding:
(a) OH– , CO32– , PO43– , or SO32– aqueous ions.
(b) SO42– , CO32– , PO43– , or SO32– aqueous ions.
(c) Ag+ , Pb2+ , Tl+ , Hg22+ , or Cu+ aqueous ions.
(d) Ag+ , Pb2+ , Ca2+ , Ba2+ , Sr2+ , or Ra2+ aqueous ions.
Applying Inquiry Skills
8. (a) To precipitate carbonate ions from a sample, without at the same time precipitating sulfide ions, add a compound
that supplies any Group II cation, and an anion that is always soluble in combination (e.g., calcium nitrate, barium
nitrate, or magnesium nitrate).
2+ CO2– → CaCO
(b) One example: Ca(aq)
3(aq)
3(s)
9. Experimental Design
Calcium nitrate solution is added to the test solution sample. If a precipitate forms it is filtered, and silver nitrate solution is added to the filtrate (or to the original sample, if no precipitate formed in the initial test).
Making Connections
10. There are innumerable examples of qualitative analysis in society. Common examples include environmental tests for
lead (or other heavy metals) in water supply systems; continuous monitoring by household detectors (carbon
monoxide or natural gas in houses, propane in trailers); simple swimming pool or aquarium water-testing kits
(including quantitative analysis); or even home pregnancy tests. There are also hundreds of industrial and commercial
examples of qualitative analysis. There are many career opportunities as an analyst.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
SECTION 7.5 QUESTIONS
(Page 346)
Understanding Concepts
1. The precipitation reactions are:
2+ 2 Cl–
(a) Pb(aq)
(aq) → PbCl2(s)
2+ S2– → ZnS
(b) Zn(aq)
(aq)
(s)
+
(d) Ag(aq)
C2H3O–2(aq) → AgC2H3O2(s)
2+ + 2 PO3(e) 3 Ba(aq)
4(aq) → Ba3(PO4)2(s)
2+ 2 OH–
(f) Ca(aq)
(aq) → Ca(OH)2(s)
2. (a)
(b)
(c)
(d)
(e)
3. (a)
yellow-brown
colourless
blue
green
colourless
Calcium (yellow-red flame test) can be distinguished from the other ions by a flame test, but lithium and strontium both give bright red flame tests, and so cannot be distinguished from each other in this way.
Copyright © 2002 Nelson Thomson Learning
Chapter 7 Solubility and Reactions
207
(b) Adding sulfate (or carbonate, phosphate, or sulfite) ions to the two unidentified test solutions (aqueous sodium
sulfate, for instance) would precipitate the strontium ions, but not the lithium ions.
Applying Inquiry Skills
4. (a) Experimental Design
The solution is flame tested.
(b) Evaluation
The experimental design is seriously flawed, because both ions produce coloured flames. If the initial flame test
is bright red, there is no way to know whether potassium ions are present, because the violet colour will be hidden
by the strontium ion colour.
Alternative answer:
(a) Experimental Design
Sodium sulfate is added to the solution. Any precipitate is filtered and then the filtrate or original solution is flame
tested.
(b) Evaluation
The experimental design is not valid. Although you will know with certainty whether strontium ions are present
or not, based upon whether a precipitate forms or not, you will not be able to determine whether potassium ions
are present, due to the masking of the potassium flame colour by the sodium flame colour. If cobalt-blue glass is
employed in the materials and the procedure, then the design would be valid. (The cobalt-blue glass filters the
yellow sodium colour from the flame and allows one to determine whether potassium is present or not.)
Alternately, the cation for the sodium solution must be chosen to have a colourless flame test, e.g., hydrogen.
(Unfortunately, a list of colourless flame-test ions is not provided in the text and would have to be researched.)
5. Experimental Design
Sodium chloride solution is added to the sample. If a precipitate forms, it is filtered. Sodium hydroxide solution is
added to the filtrate (or sample, if no precipitate forms). If a precipitate forms, it is filtered. Sodium carbonate (or
sulfate) solution is added to the filtrate (or sample, if no precipitate forms).
Note: A precipitate in the initial step (when you use any soluble halide ion compound) indicates that thallium ions
are present. A precipitate in the second step (when you use any soluble hydroxide compound) indicates that calcium
ions are present. A precipitate in the third step (when you use any soluble sulfate, carbonate, phosphate, or sulfite
compound) indicates that barium ions are present. The sequence of steps is very important: sodium carbonate cannot
be added first, for example.
6. (a) The Experimental Design is satisfactory. The solution colour test can confirm the presence of copper(II) ions, but
not calcium ions; and any red in the flame test will confirm the presence of calcium ions.
(b) Alternative Experimental Design
Sodium sulfate solution is added to the sample solution. If a precipitate forms, it is filtered. Sodium carbonate
solution is added to the filtrate (or sample, if no precipitate forms).
Note: A precipitate in the initial step (when you use any soluble sulfate compound) indicates that calcium ions
are present. A precipitate in the second step (when you use any soluble sulfide, carbonate, phosphate, or sulfite
compound) indicates that copper(II) ions are present.
7. Any carbonated beverage is a home solution with a gaseous solute. A diagnostic test is: If a gas is bubbled through
limewater, and a white precipitate forms, then the gas contains carbon dioxide.
Another example would be household ammonia, or any of several spray window cleaners containing ammonia. The
diagnostic test would be the characteristic odour of ammonia.
8. Sodium carbonate is a typical home solution with a solid solute; e.g., as a water softener (washing soda) in laundry
detergents. Diagnostic tests for this example would be flame testing for sodium ions, and precipitation using calcium
chloride solution, for carbonate ions.
9. Experimental Design
Oxalic acid is added to sample solutions of nitrate or chloride compounds of as many metal cations as can be found
in the school laboratory supplies. Any precipitate formation is recorded.
10. Experimental Design
A solution of the product is first flame tested; then a solution of calcium chloride is added.
Note: A yellow flame test indicates sodium; a precipitate indicates carbonate ions.
208
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
Making Connections
11. Forensic chemists analyze tissue for many things. A typical test would be for the presence and amount of arsenic.
Quantity must be measured precisely to determine if a substance is present in a natural amount, or in an amount much
greater, which could perhaps indicate foul play.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
7.6 QUANTITATIVE ANALYSIS
PRACTICE
(Page 353)
Understanding Concepts
1. 2 NH3(g) H2SO4(aq) → (NH4)2SO4(aq)
24.4 mL
50.0 mL
2.20 mol/L
C
2.20 mol
0.0537 mol
nNH 0.0244 L
3
1L
1
nH SO
0.0537 mol 0.0268 mol
2
4
2
nH SO
0.0268 mol
2
4
CH SO
2
4
CH SO
4
2
0.0268 mol
0.0500 L
0.537 mol/L
or
CH SO
2
CH SO
2
2.20 mol NH
1 mol H2SO4
1
0.0244 L
NH3 3 1L
NH3
2
mol N
H3
0.0500 L
0.537 mol/L
4
4
The concentration of sulfuric acid at this stage is 0.537 mol/L.
2. 3 Ca(OH)2(aq) Al2(SO4)3(aq) → 3 CaSO4(s) 2 Al(OH)3(s)
v
25.0 mL
0.0250 mol/L
0.125 mol/L
nAl (SO )
2
4 3
nCa(OH)
2
vCa(OH)
2
vCa(OH)
2
vCa(OH)
2
vCa(OH)
2
or
0.125 mol
25.0 mL
3.13 mmol
1L
3
3.13 mmol 9.38 mmol
1
1L
9.38 mmol
0.0250 mol
375 mL
OH)2
3
mol Ca(
1 L Ca (OH)2
0.125 mol Al2(SO
4)3
25.0 mL
Al2
(SO4)3 SO4)3
1
mol Al2(
375 mL
The volume of calcium hydroxide solution reacted is 375 mL.
3. (a) 2 FeCl3(aq) 3 Na2CO3(aq) → Fe2(CO3)3(s) 6 NaCl(aq)
75.0 mL
v
Copyright © 2002 Nelson Thomson Learning
Chapter 7 Solubility and Reactions
209
0.200 mol/L
nFeCl
3
nFeCl
3
nNa CO
2
3
nNa CO
3
2
vNa CO
2
3
vNa CO
3
2
or
0.250 mol/L
0.200 mol
75.0 mL
1L
15.0 mmol
3
15.0 mmol 2
22.5 mmol
1L
22.5 mmol
0.250 mol
90.0 mL
eCl3
0.200 m
ol F
3
mol Na2
CO3
1 L Na2CO3
vNa CO 75.0 mL
FeCl
3 2
3
2m
ol FeCl3
0.250 m
ol Na2
CO3
1
FeCl3
L
vNa CO 90.0 mL
2
3
The minimum volume of sodium carbonate solution required for complete reaction is 90.0 mL.
(b) A reasonable volume of sodium carbonate solution would be at least 100 mL, thus providing about a 10%
excess to ensure a complete reaction.
Applying Inquiry Skills
4. (a) Prediction
2 NaOH(aq) ZnCl2(aq) → Zn(OH)2(s) 2 NaCl(aq)
20.0 mL
m
2.50 mol/L
nNaOH
nNaOH
nZn(OH)
2
nZn(OH)
2
mZn(OH)
2
mZn(OH)
2
99.40 g/mol
2.50 mol
20.0 mL
1L
50.0 mmol
1
50.0 mmol 2
25.0 mmol
99.40 g
25.0 mmol
1
mol
2.49 103 mg
mZn(OH) = 2.49 g
2
or
mol Zn(OH)
99.40 g Zn(OH)2
2.50 mol N
aOH 1 mZn(OH) 20.0 mL
NaOH
2 2
1L
N
aOH
2m
ol N
a(OH)
1
mol Zn(OH)2
mZn(OH) 2.49 103 mg
2
mZn(OH) = 2.49 g
2
According to the stoichiometric method, the mass of zinc hydroxide produced is predicted to be 2.49 g.
Note: The calculated answer of 2.485 g can be rounded to 2.48 g or 2.49, depending on the rounding rule used
in the classroom.
(b) Analysis
mZn(OH) 3.30 g 0.91 g 2.39 g
2
According to the evidence, the mass of zinc hydroxide that is actually produced is 2.39 g.
(c) Evaluation
difference 2.39 g 2.48 g 0.10 g
0.10 g
% difference 100% 3.8%
2.48 g
Note: The unrounded value of 2.485 g was used in the calculation. If the rounded value is used, the difference
is 0.09 g and the percentage difference is 3.6%.
210
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
The prediction was 3.8% higher than the value obtained, and so is judged to be verified by the experimental results.
A prediction within 5% is considered acceptably accurate (95% accurate) for this kind of lab work, with any difference probably just due to normal experimental error.
(d) The stoichiometric concept is supported by the results of this investigation, and judged to be acceptable because
the prediction was verified. There is good confidence in this judgment, and no need is seen to modify the concept.
SECTION 7.6 QUESTIONS
(Page 355)
Applying Inquiry Skills
1. Precipitating all the lead(II) ions will require adding an excess of a solution containing an anion (in this case, sulfate)
that forms a low-solubility compound with lead(II). Making a sodium sulfate solution is a logical choice, since sodium
sulfate is soluble, inexpensive, and easy to obtain.
First, the mass of sodium sulfate required must be calculated.
Pb(NO3)2(aq) Na2SO4(aq) → PbSO4(s) 2 NaNO3(aq)
2.0 L
m
0.34 mol/L
142.04 g/mol
n Pb(NO )
3 2
n Na SO
2
4
m Na SO
4
m Na SO
4
m Na SO
4
m Na SO
4
2
2
= 97 g
or
2
2
0.34 mol
= 2.0 L
0.68 mol
1L
1
= 0.68 mol 0.68 mol
1
142.04 g
= 0.68 mol 1
mol
0.34 m
ol Pb
(NO )
1m
ol N
a2SO4
142.04 g Na2SO4
= 3 2 1L
Pb
(NO3)2
1m
ol Pb(N
O3)2
1
mol N
a2SO4
= 97 g
A minimum mass of 97 g of sodium sulfate must be used.
To ensure a complete precipitation of lead(II) ions, using an excess of sodium sulfate will be necessary. Commonly
this amount should be about 10% more than the minimum required — say, 105 g of sodium sulfate, in this case.
The simplest process would be to obtain 105 g of sodium sulfate and dissolve it to make, say, 1.0 L of reacting solution. It is possible that not all of the solute will dissolve, depending on the water temperature. This problem could be
solved by increasing the solvent volume.
2. The experimental design is judged to be inadequate, because the mass of solid measured includes not only the precipitate but also the excess reactant and the second (soluble) product. The latter two chemicals crystallize out of the solution when the water is boiled away. The mass of solid remaining will be of more than one substance, with no way to
calculate amounts from the value. The precipitate should have been separated by filtration and then dried.
If we take the reaction in Question 3 as an example, precipitating and then crystallizing would result in a mixture of
the solids, aluminum nitrate (the excess reactant), aluminum sulfide (the precipitate), and sodium nitrate (the second
(but soluble) product).
3. (a) Prediction
3 Na2S(aq) 2 Al(NO3)3(aq) → Al2S3(s) 6 NaNO3(aq)
20.0 mL
m
0.210 mol/L
n Na S
2
n Na S
2
150.14 g/mol
0.210 mol
20.0 mL
L
4.20 mmol
Copyright © 2002 Nelson Thomson Learning
Chapter 7 Solubility and Reactions
211
n Al S
2 3
n Al S
2 3
m Al S
2 3
m Al S
2 3
or
1
4.20 mmol 3
1.40 mmol
150.14 g
1.40 mmol
1
mol
210 mg = 0.210 g
m Al S
0.210 mol N
a2S
1
mol A
l2S3
150.14 g Al2S3
20.0 mL
Na
2S 1L
N
a2S
3m
ol N
a2S
m
ol A
l 2S3
m Al S
210 mg = 0.210 g
2 3
2 3
According to the stoichiometric method, the mass of aluminum sulfide produced is predicted to be 0.210 g.
(b) Analysis
m Al S 1.17 g 0.97 g 0.20 g
2 3
According to the evidence, the mass of aluminum sulfide that is actually produced is 0.20 g.
(c) Evaluation
The design of the experiment is judged to be adequate, with no obvious flaws. It allowed the question to be
answered easily with simple materials, concepts, and procedures.
difference 0.20 g 0.210 g 0.01 g
0.01 g
% difference 100% 5%
0.210 g
The prediction was 5% higher than the value obtained, and is judged to be verified by the experimental results. A
prediction within 5% is considered acceptably accurate (95% + ...) for this kind of lab work, with any difference probably just due to normal experimental error.
The stoichiometric concept is the authority for this investigation. It is supported by the results of this investigation,
and judged to be acceptable because the prediction was verified. There is good confidence in this judgment, and there
is no need to modify the concept.
4. (a) Analysis
v AgNO = 100 mL = 0.100 L
3
m AgNO = 6.74 g 1.27 g 5.47 g
3
M AgNO = 169.88 g/mol
3
1 mol
nAgNO = 5.47 g 3
169.88 g
0.0322 mol
0.0322 mol
CAgNO = 3
0.100 L
CAgNO = 0.322 mol/L
3
The molar concentration of silver nitrate in solution is 0.322 mol/L.
Making Connections
5. The most common way to check the concentration of antifreeze is to measure its density, since density and concentration are proportional. Battery acid is measured the same way (to determine battery charge). The higher the concentration of solute, the denser the solution will be, and the higher in the solution a buoyant object will float. A calibrated
device used to measure liquid density in this way is called a hydrometer. Common density units for solutions are g/mL
and kg/L, but concentrations of automotive fluids are usually expressed in terms of what they are intended to do:
lowest working temperature (for antifreeze), or charge condition (for battery acid). It is also common to state solution
densities as a ratio with the density of pure water — giving a numerical value greater or less than 1 — called the
“specific gravity” of the solution.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
212
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
CHAPTER 7
REVIEW
(Page 358)
Understanding Concepts
1. Combinations (a), (b), (c), (d), and (f) will react, but only (a), (c), (d), and (f) will form precipitates. The reaction equations are:
2+ + 2 OH– → Cu(OH)
(a) Cu(aq)
2(s)
(aq)
+ + H– → H O
(b) H(aq)
2 (l)
(aq)
2+ + 2 PO3–
(c) 3 Ca(aq)
4(aq) → Ca3(PO4)2(s)
+ + Cl– → AgCl
(d) Ag(aq)
(s)
(aq)
+ Cl– → CuCl
(f) Cu(aq)
(s)
(aq)
2+ + CO2– → ZnCO
2. (a) Zn(aq)
3(s)
3(aq)
2+ + CO2– → PbCO
(b) Pb(aq)
3(s)
3(aq)
3+ + 3 CO2– → Fe (CO )
(c) 2 Fe(aq)
2
3 3(s)
3(aq)
2+ + CO2– → CuCO
(d) Cu(aq)
3(aq)
3(s)
+ + CO2– → Ag CO
(e) Ag(aq)
3(aq)
2
3(s)
2+ + CO2– → NiCO
(f) Ni(aq)
3(aq)
3(s)
(g) The choice of sodium carbonate is good because carbonate ions form low soluble compounds with most metallic
ions and the compound is soluble, common, and inexpensive.
3+ + 3 OH– → Al(OH)
3. (a) Al(aq)
(aq)
3(s)
and
2+ + SO2– → CaSO
Ca(aq)
4(s)
4(aq)
or
3+ + 3 SO2– + 3 Ca2+ + 6 OH– → 2 Al(OH)
2 Al(aq)
(aq)
(aq)
3(s) + 3 CaSO4(s)
4(aq)
Note: The effective precipitate for clarifying the water is the flocculent precipitate, Al(OH)3(s).
2+ + 2 PO3– → Ca (PO )
(b) 3 Ca(aq)
3
4 2(s)
4(aq)
2+ + 2 OH– → Mg(OH)
(c) Mg(aq)
(aq)
2(s)
3+ + 3 OH– → Fe(OH)
(d) Fe(aq)
(aq)
3(s)
2+
4. Cu(aq)
5. Ions of alkali metals, as well as hydrogen, ammonium, and nitrate ions, form compounds with high solubility.
+ could indicate any anion on the solubility chart
6. A violet flame indicates potassium ions. A precipitate with Hg(aq)
except sulfate, nitrate, or acetate. The compound in solution might be KCl, KBr, K2S, K2SO4, KOH, K3PO4, or ...
7. In aqueous solution:
(a) Cu+ is green, Cu2+ is blue.
(b) Fe2+ is pale green, Fe3+ is yellow-brown.
(c) CrO42– is yellow, Cr2O72– is orange.
2+ and Mg2+
8. Ca(aq)
(aq)
9. (a) Na2CO3(aq) CuSO4(aq) → Na2SO4(aq) CuCO3(s)
v
4.54 L
1.25 mol/L
nCuSO
214
4
0.0875 mol/L
0.0875 mol
4.54 L
0.397 mol
1L
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
nNa CO
2
or
3
vNa CO
2
3
vNa CO
2
3
vNa CO
2
3
vNa CO
3
2
1
0.397 mol 0.397 mol
1
1L
0.397 mol 1.25 mol
0.318 L
0.0875 mol C
uSO4
1
mol Na
1 L Na2CO3
2CO3
4.54L
CuSO
4 1
L C
uSO4
1m
ol Cu
SO4
1.25 m
ol Na
2CO3
0.318 L
The minimum volume of sodium carbonate solution required is 0.318 L.
(b) A suitable volume would be about 350 mL. (Assume an excess of 10%.)
10. The mass of zinc reacted (24.89 g 21.62 g) 3.27 g
Zn(s) 2 HCl(aq) → ZnCl2(aq) H2(g)
3.27 g
350 mL
350 mL
65.38 g/mol
C
1 mol
3.27 g 65.38 g
nZn
0.0500 mol
nZn
nZnCl
2
CZnCl
2
CZnCl
2
or
1
0.0500 mol 0.0500 mol
1
0.0500 mol
0.350 L
0.143 mol/L
1
mol Z
n
1 mol ZnCl
1
CZnCl 3.27 g Zn 2 2
65.38 g Z
n
1
mol Zn
0.350 L
CZnCl 0.143 mol/L
2
The molar concentration of zinc chloride solution is 0.143 mol/L.
Applying Inquiry Skills
11. The precipitated anion could be SO42–, CO32–, PO43–, or SO32–.
Since most sulfates are soluble, and most sulfites, carbonates, and phosphates are only slightly soluble, the original
solution could be tested with Zn(NO3)2(aq), Cu(NO3)2(aq), or Ni(NO3)2(aq), etc. If no precipitate forms, the anion must
be SO42–.
12. Experimental Design
The solution is tested with TlNO3(aq) (or Hg2(NO3)2(aq) or CuNO3(aq)) for the presence of halide ions. If a precipitate
forms it is filtered, and the filtrate (or original solution if no precipitate forms) is tested with Ca(NO3)2(aq) (or barium,
strontium, or radium nitrate) for the presence of sulfate ions.
13. Experimental Design
The solutions are tested with litmus to identify the acid and hydroxide (basic) compounds. The remaining solutions
are then tested for conductivity (to identify the ionic compound). To confirm that the final solution contains nitrogen
gas, it could be heated slightly. The formation of tiny bubbles would confirm the presence of a dissolved gas.
Note: The least confidence is for the nitrogen test. Nitrogen has one-half the solubility of oxygen gas in water —
0.00175 g/100 mL.
14. (a) Materials
• deep-seawater sample solution
•
1.00 mol/L Pb(NO3)2 (aq) stock solution
•
KI (aq) test solution
•
medicine dropper
•
50-mL pipet and bulb
Copyright © 2002 Nelson Thomson Learning
Chapter 7 Solubility and Reactions
215
•
250-mL beaker
•
400-mL beaker
•
filtration apparatus
•
filter paper
•
wash bottle of pure water
•
centigram balance
(b) Analysis
mass of PbCl2 (s) precipitate (4.58 g – 0.91 g) = 3.67 g
Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2 (s) 2 NaNO3(aq)
50.0 mL
C
n PbCl
2
n NaCl
3.67 g
278.10 g/mol
1 mol
= 3.67 g = 0.0132 mol
278.10 g
2
= 0.0132 mol = 0.0264 mol
1
0.0264 mol
C NaCl = 0.0500 L
C NaCl = 0.528 mol/L
or
1
mol Pb
Cl2
2 mol NaCl
1
C NaCl 3.67 g PbCl
2 278.10 g P
bCl2
1
mol P
bCl2
0.0500 L
C NaCl = 0.528 mol/L
According to the evidence and the stoichiometric concept, the concentration of sodium chloride in the seawater sample is
0.528 mol/L.
(c) Evaluation
The design of this experiment is judged to be adequate because it allowed the question to be answered easily, and
with confidence in the result. There are no apparent flaws and the equipment is simple and easy to use.
15. (a) Materials
• CuSO4(aq) sample solution
•
0.750 mol/L NaOH (aq) stock solution
•
medicine dropper
•
25-mL pipet and bulb
•
250-mL beaker
•
400-mL beaker
•
filtration apparatus
•
filter paper
•
wash bottle of pure water
•
centigram balance
(b) Analysis
mass of Cu(OH)2 (s) precipitate
(2.83 g – 0.88 g) = 1.95 g
CuSO4(aq) + 2 NaOH(aq) → Cu(OH)2 (s) 2 Na2SO4(aq)
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Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
25.0 mL
C
nCu(OH)
2
nCuSO
4
CCuSO
4
CCuSO
4
CCuSO
4
CCuSO
4
or
1.95 g
97.57 g/mol
1 mol
= 1.95 g = 0.0200 mol
97.57 g
1
= 0.0200 mol = 0.0200 mol
1
0.0200 mol
= 0.02500 L
= 0.799 mol/L
1
mol C
u(OH)2
1 mol CuSO4
1.95 g Cu(OH)
2 97.57 g C
u(OH)2
1
mol C
u(OH)2
= 0.799 mol/L
According to the evidence and the stoichiometric concept, the concentration of copper(II) sulfate in the sample is
0.799 mol/L.
(c) Evaluation
The design of this experiment is judged to be adequate because it allowed the question to be answered easily, and
with confidence in the result. There are no apparent flaws and the equipment is simple and easy to use.
Making Connections
16. For example, the hardness may be 250 ppm (250 mg/L). The soda-lime process could be used by adding washing soda,
Na2CO3(aq), and lime, Ca(OH)2(aq)/(s), to the water to precipitate the hard-water ions as carbonates; e.g., CaCO3(s).
Note: Answers will be specific to school/community location.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
17. Answers will vary, but might express concern about an old septic system at a cottage because of a danger of leakage,
which might release disease-causing organisms into ground water.
Note: Answers will be specific to school/community location. Student discussion should use concepts from this
chapter.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
18. (a) Recall that ppm = mg/L
O2(aq) (maximum) in 50 L is:
14.7 mg
50 L
735 mg at 0°C
L
and
8.7 mg
435 mg at 20°C
50 L
L
735 435 mg 300 mg
The difference in mass of oxygen that can be dissolved in 50 L of water at the two temperatures is 300 mg.
(b) Fish require O2 for respiration, so they might prefer 0°C water, in which O2 solubility is higher and obtaining
sufficient oxygen would be easier. On the other hand, the temperature of the surrounding water affects the activity
level of cold-blooded animals. The fish would not be able to move as fast at the colder temperature.
Copyright © 2002 Nelson Thomson Learning
Chapter 7 Solubility and Reactions
217
8.1 UNDERSTANDING ACIDS AND BASES
PRACTICE
(Page 367)
Understanding Concepts
1. Ionic compounds when melted will conduct a current (which suggests that they contain charged particles such as
ions), while molecular compounds in the liquid state will not (which suggests the absence of charged particles).
2. (a) acidic
(b) basic
(c) neutral
(d) acidic
(e) neutral
(f) basic
3. Acids are unlike other molecular compounds in that their aqueous solutions conduct a current, turn blue litmus red,
and neutralize bases.
4. According to the Arrhenius theory, litmus colour change in acidic solution is caused by H+(aq) ions, and litmus colour
change in basic solution is caused by OH–(aq) ions.
5. Empirically, an acid is a substance that, in solution, will turn blue litmus pink, conduct a current, and neutralize a base.
Theoretically, an acid is a substance that will release H+(aq) ions in solution.
+ OH–
6. (a) NaOH(s) → Na(aq)
(dissociation)
(aq)
+ C H O–
(b) HC2H3O2(l) → H(aq)
2 3 2(aq) (ionization)
+ HSO–
(c) H2SO4(l) → H(aq)
4(aq)
(ionization)
2+ 2 OH–
(d) Ca(OH)2(s) → Ca(aq)
(aq)
(dissociation)
Note: Acid–base concepts in this text do not address the ionization of polyprotic acids such as sulfuric acid. Single
ionization is assumed until later in the student’s course of study. (See the example with oxalic acid, H2C2O4(aq), shown
on page 364.)
Applying Inquiry Skills
7. (a) Analysis
Chemical 1 is the molecular compound, C12H22O11(s), (sucrose) because it dissolves well, to produce a nonconducting, neutral solution.
Chemical 2 is the ionic compound, KCl(s), because it dissolves well, to produce a conducting, neutral solution.
Chemical 4 is the ionic hydroxide, Ba(OH)2(s), because it dissolves well, to produce a conducting, basic solution.
Chemical 7 is the acidic molecular compound benzoic, HC7H5O2(s), because it dissolves somewhat, to produce a
slightly conducting acidic solution.
Chemicals 3, 5, and 6 are zinc, calcium phosphate, and paraffin wax. They are not individually identifiable from this
evidence. Because none of them dissolve, no observations of the solutions can be made.
(b) Evaluation
The experimental design could be improved by testing the conductivity of the pure substance (which would identify
the metal zinc), and by testing the melting points, which would distinguish ionic calcium phosphate (very high) from
molecular paraffin (fairly low).
Making Connections
8. Both acids and bases make good solutions for different types of batteries. Lead–acid and alkaline batteries are the
most common examples.
9. Acids (a), (c), (e), and (f) are used for energetic reactions and should be handled with care. Acid (b), carbonic acid, is
a weak acid that has low solubility, so it is not dangerous. In fact, it is an ingredient in most soft drinks. Acid (d),
acetic acid, is dangerous in pure form, but in vinegar is only about a 5% solution so is not very dangerous because the
solution is so dilute.
10. You can be electrocuted anywhere, if current has separate places to enter and to leave your body. If the pure water that
you are standing in is part of the circuit, it will be a very poor conductor, and you should be relatively safe. In fact,
220
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
dissolved substances (for example, from your boots or feet) would make the water in such a situation a relatively good
conductor. It is always best to assume that electricity is dangerous around water!
8.2 pH OF A SOLUTION
PRACTICE
(Page 371)
Understanding Concepts
1. Examples of products with specified pH values include skin creams, soils, shampoos, cat foods, wines, water (in rain,
wells, lakes, rivers, aquariums), and waste water outflows.
+ ] values are:
2. [H(aq)
(a) 1 10–11 mol/L
(b) 1 10–2 mol/L
(c) 1.0 10–4 mol/L
(d) 1.0 10–14 mol/L
3. pH values are:
(a) 3.0
(b) 5.0
(c) 7.00
(d) 10.00
+ ] values would have to change from 1 10–5 mol/L to 1 10–7 mol/L: a ratio of 100:1. The hydrogen ion
4. [H(aq)
concentration would have to decrease by a factor of 1/100.
5. Since log 1 100 exactly, by the definition of logarithms, the pH of a solution with a hydrogen ion concentration of
1 mol/L will be 0.0.
Note: Since the concentration is given to one significant digit, the pH is reported to one decimal place (one place
in the logarithm characteristic).
6. (a) vH+
100 L
CH+
1 10–3 mol/L
nH+
nH+
1 10–3 mol
100 L
1L
0.1 mol
The amount of hydrogen ion present in the wine is 0.1 mol.
(b) vH+
100 L
pH
8.00
CH+
1.0 10–8 mol/L
nH+
nH+
1.0 10–8 mol
100 L
1
L
1.0 10–6 mol 1.0 mmol
The amount of hydrogen ion present in the seawater is 1.0 mmol.
(c) vH+
100 L
CH+
10.0 mmol/L 0.0100 mol/L
0.0100 mol
nH+
100 L
1L
nH+
1.00 mol
The amount of hydrogen ion present in the stomach acid is 1.00 mol.
Reflecting
7. (a) No, setting a zero measured level would have to specify the precision of the zero measurement. A more precise
technology may turn an earlier measurement of zero into a non-zero value. Secondly, every chemical is toxic at
some concentration. There is no known chemical that needs to be reduced in concentration to zero (whatever that
means) in order to become non-toxic.
Copyright © 2002 Nelson Thomson Learning
Chapter 8 Acids and Bases
221
(b) Theoretically, a zero level would mean no entities (atoms, molecules, and/or ions) of the chemical present at all.
Empirically, there would be no way of confirming or refuting a theoretical level of zero. Although a theoretical
level of zero is possible, it is not probable, and, again, is impossible to test.
(c) A zero level is not measurable; a zero reading may always be taken to mean that the quantity of chemical is just
too low to be detected by the instrument. A measurable zero quantity is a contradiction in terms: it would require
a perfect measuring device and/or system.
(d) One possible answer: A limit should be set low enough so that there are no measurable effects from such a level,
even over a lengthy period of time.
Note: How long a time? Ahhh, now there’s a good point for discussion and opinion. ... The world becomes a
very much trickier place the moment students wrap their minds around the concept that there is no such thing as
“perfectly” safe.
PRACTICE
(Page 374)
Understanding Concepts
8. Solution pH is measured with pH paper, with indicators, and with pH meters.
9. (a) Pure water has a pH of 7.
(b) The hydrogen ion concentration of pure water is 10–7 mol/L.
10. (a) The calculation of values to complete spaces in Table 2 are, in order:
Oranges
+ ]
pH –log [H(aq)
–log [5.5 10–3 mol/L]
pH 2.26
Asparagus
+ ]
[H(aq)
10–pH
10–8.4
+ ]
[H(aq)
4 10–9 mol/L
Olives
+ ]
[H(aq)
10–pH
10–3.34
+ ]
[H(aq)
4.6 10–4 mol/L
Blackberries
pH
+ ]
5 –log [H(aq)
5 –log [4 3 10–4 mol/L]
pH
5 3.4
(b) Based on pH values only, oranges are the most acidic, and should have the most sour taste.
Making Connections
11. Answers should always present both sides of the risk/benefit concept. An example might be the use of acidic cleaners
to remove rust stains from plumbing fixtures. The aesthetic benefit of shiny plumbing fixtures carries with it the risk
of skin irritation or disfiguring marking of other surfaces, resulting from careless use of the acidic cleaner; and
disposal might be a problem if the cleaner is not greatly diluted.
SECTION 8.1–8.2 QUESTIONS
(Page 375)
Understanding Concepts
+ ] 1 10–3 mol/L for the fruit juice, and 1 10–12 mol/L for the household cleaner.
1. (a) [H(aq)
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Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
(b) The hydrogen ion concentration ratio, juice to cleaner, is
1 10–3 m
ol/L
1 109
,
–
1
2
1 10 m
ol/L
1
2. (a) pH
8.0
(b) pH
7.0
(c) pH
+ ]
–log [H(aq)
or a billion to one.
–log [2.5 10–6 mol/L]
pH
(d) pH
5.60
+ ]
–log [H(aq)
–log [1.3 10–4 mol/L]
pH
3.89
3. (a) Pickling vinegar is more acidic.
(b) Pickling vinegar has a higher hydrogen ion concentration.
+ ]
10–pH
(c) [H(aq)
10–2.4 mol/L
+ ]
[H(aq)
4 10–3 mol/L
+ ]
[H(aq)
10–pH
(standard vinegar)
10–2.2 mol/L
+ ]
[H(aq)
6 10–3 mol/L
(pickling vinegar)
4. (a) The hydrochloric acid has a lower pH because the molecules ionize to a much greater extent than the acetic acid.
>99%
+ Cl–
(b) HCl(aq) → H(aq)
(aq)
<50%
+ C H O–
HC2H3O2(aq) → H(aq)
2 3 2(aq)
(c) The HCl(aq) should be handled more carefully because it will react much faster, and so is more dangerous.
+ ] 10–10 mol/L
5. (a) [H(aq)
(pH paper)
Note: When a pH value has no digits in the characteristic at all, it represents a quantity that is known only to an
order of magnitude—that is, a quantity with no significant digits; where only the place value is known. This level of
certainty is fairly common when pH is measured with coarse-scale pH papers. Measured values with no significant
digits, or even those with only one significant digit, are rare. They seldom occur outside of systems like pH, that have
such a great range that logarithmic scales must be used to express them.
+ ]
(b) [H(aq)
10–pH
10–9.8 mol/L
+ ]
[H(aq)
+ ]
(c) [H(aq)
2 10–10 mol/L
(pH paper or pH meter)
10–pH
10–9.84 mol/L
+ ]
[H(aq)
+ ]
(d) [H(aq)
1.4 10–10 mol/L
(pH meter)
10–pH
10–9.836 mol/L
+ ]
[H(aq)
1.46 10–10 mol/L
Copyright © 2002 Nelson Thomson Learning
(very precise pH meter)
Chapter 8 Acids and Bases
223
+ ]
6. (a) [H(aq)
10–pH
10–5.4 mol/L
+ ]
[H(aq)
+ ]
(b) [H(aq)
4 10–6 mol/L
10–pH
10–5.72 mol/L
+ ]
[H(aq)
(c) pH
1.9 10–6 mol/L
+ ]
–log [H(aq)
–log [5 10–7 mol/L]
pH
(d) pH
6.3
+ ]
–log [H(aq)
–log [7.9 10–6 mol/L]
pH
5.10
Applying Inquiry Skills
7. Experimental Design
Each of six samples of solutions of different acids, of equal concentrations, will be tested for conductivity, to determine which are strong acids.
Note: Alternatively, pH paper, indicators, or a pH meter could also be used.
8. Prediction
Since a change in hydrogen ion concentration of 100 times (or 102) represents two pH units, it seems probable that
when 1.0 mL of vinegar is diluted 100 times, the pH (characteristic) will increase by 2.
Experimental Design
The pH of some 5% V/V household vinegar is measured to one place in the mantissa (i.e., one decimal place). A
1.0-mL sample of the household vinegar is diluted to 100 mL, and the pH of the diluted sample is measured.
Materials
• household vinegar (5% V/V acetic acid)
• wash bottle of pure water
• 10-mL graduated cylinder
• 250-mL graduated beaker
• stirring rod
• pH meter (or pH paper, precise to 0.1 unit )
Procedure
1. Measure and record the pH of some household vinegar.
2. Obtain 1.0 mL of the vinegar in the 10-mL graduated cylinder.
3. Use a wash bottle to rinse the vinegar sample into the beaker.
4. Add pure water to the beaker until the solution volume is 100 mL.
5. Measure and record the pH of the diluted vinegar.
Evidence
The pH of the initial vinegar sample is 2.4.
The pH of the diluted vinegar sample is 3.4.
Analysis
The volume to which 10 mL of vinegar must be diluted in order to increase its pH by two units is more than 1000 mL.
The pH value only changed by about 1, indicating that although the acid solution concentration was decreased 100
times, the hydrogen ion concentration seems to have decreased only about 1 101, or 10 times.
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Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
Evaluation
The experimental design is judged to be adequate because the experiment produces the evidence needed to answer the
(very simple) question with a high degree of certainty (even though the Prediction was found to be incorrect).
The procedure is also judged adequate because the steps are simple and clear.
The technological skills of the experimenter are judged adequate — no special skills are required, and the
procedure is very straightforward.
The results have a very high degree of certainty. None of the materials or procedure steps are likely to have any
unexpected uncertainty. The percentage difference cannot be calculated as such, but the answer differs from the
prediction by at least an order of magnitude.
The prediction is definitively falsified because the evidence is clearly different from the prediction.
The assumption (that the pH value varies proportionally with the negative of the logarithm of the concentration
of a weak acid) is not supported by the results of this experiment. It seems likely that there is a more complex relationship between weak acid concentration and hydrogen ion concentration than was assumed.
Note: This is a nice example of a case where a simple concept turns out to be more complex — offering an opportunity to have students learn by falsifying a prediction. Students’ only experience with dilution and pH to this point is
with the strong acid HCl(aq) (Investigation 8.2.1,) and from calculations involving strong acids. Students will normally
+ ] equation, or upon their experience from Investigation 8.2.1. Neither
form predictions based upon the pH –log[H(aq)
of these methods is useful when dealing with a weak acid, where the varying degree of ionization is a predominant
factor. Empirically (experimentally), the final volume (100 L or more) is required to change the vinegar concentration
by two pH units. A theoretical explanation comes in Chemistry 12.
9. The design is probably valid because toothpaste is diluted in your mouth as you brush, anyway — so the pH measured will be approximately correct. An alternative design might be to squeeze toothpaste directly onto pH paper strips.
Making Connections
10. Examples of possible Did You Know?:
Some plants are acid-loving and others are not. For example, evergreen trees are acid-loving: they grow best in soil
that has a pH of less than 7 and will actually change the pH of the soil in which they grow. If you remove an evergreen tree from a lawn and plant grass in that area, the grass probably won’t grow: grass is not acid-loving. The soil
will have to be neutralized by adding a base, such as lime, to the soil.
Flowers on some plants will change colour depending on the pH of the soil. Examples are rhododendrons and
azaleas, which are acid-loving but vary in colour depending upon the soil pH. The pH can be decreased into the acidic
range by adding citric fruit peels, pine needles, and peat moss to the soil.
Some vegetables, such as cabbage, cauliflower, broccoli, and turnips, like to grow in alkaline soil with a pH of
7.5-8.0. Horticultural lime is used by gardeners to “sweeten” the sour soil.
You can test a sample of your soil with pH paper or a pH meter, or take the soil sample to a nearby greenhouse.
11. An acid wash will clean the lettuce and also tend to remove pesticides and kill microorganisms on it. The likely acid
to use would be the one that is commonly edible: acetic acid. You can use a diluted vinegar wash in your sink or in a
spray bottle.
Reflecting
12. “Weak,” referring to acids, indicates the degree of ionization in water. It has nothing necessarily to do with concentration.
Copyright © 2002 Nelson Thomson Learning
Chapter 8 Acids and Bases
225
8.3 WORKING WITH SOLUTIONS
PRACTICE
(Page 376)
Applying Inquiry Skills
Name
1.
Use
Erlenmeyer flask
for temporarily containing reacting substances (specifically shaped
to allow mixing of contents by swirling) and also for approximate
measure of various volumes
graduated pipet
for very precise addition of various (smaller) volumes
graduated cylinder
for precise measurement of various (larger) volumes
volumetric flask and stopper
for very precise measurement (and mixing) of a single specific volume
buret
for very precise addition of various (small) volumes
graduated beaker
for temporarily containing reacting substances, and for
approximate measure of various (large) volumes
PRACTICE
(Page 377)
Making Connections
2. Typical answers might include information such as:
Teaching Chemistry requires a minimum four years of university training, with five or six years usually seen as preferable. Specifically, such a person would require a B.Sc. degree in physical sciences, plus background in mathematics,
as well as education course requirements. (Courses vary from province to province.) Teachers must be certified by the
provincial government. At publication date, a nationwide shortage of teachers is being felt in this subject area, making
future job prospects good.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
8.4 ACID–BASE THEORIES
8.4 ACID – BASE THEORIES
PRACTICE
(Page 379)
Understanding Concepts
1. Sour taste for acids is not an appropriate lab test, because many lab acids are dangerous: toxic and/or highly reactive.
It would be of practical use in a household kitchen.
2. (a) Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
fast (strong acid)
(b) Mg(s) + 2 HC2H3O2(aq) → Mg(C2H3O2)2(aq) + H2(g)
slow (weak acid)
(c) 2 HCl(aq) + CaCO3(s) → H2O + CO2(g) + CaCl2(aq)
fast (strong acid)
(d) 2 HC2H3O2(aq) + CaCO3(s) → Ca(C2H3O2)2(aq) + H2O(g) + CO2(g)
slow (weak acid)
3. The best properties to distinguish strong acids from weak acids would be the rates of reaction with active metals
and/or carbonates, the last two properties listed in Table 1. The other tests would distinguish acids from bases, but not
strong acids from weak acids.
4. Strong acids are molecular substances that theoretically ionize in aqueous solution to a very large extent (essentially
completely), to produce hydrogen ions. Weak acids are molecular substances that theoretically ionize in aqueous solution to a very small extent, to produce hydrogen ions.
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Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
5. (a) The concentration of hydrogen ions in the strong acid solution will be 100 ions per litre, and in the weak acid
solution will be 2 ions per litre.
(b) Strong acids react much faster and (in equal concentration) have a lower pH, because their solutions have more
hydrogen ions present.
Applying Inquiry Skills
6.
Prediction
Strong bases should react much faster and (in equal concentration) have a higher pH than weak bases, because
their solutions should have more hydroxide ions present.
Experimental Design
Solutions of several bases (independent variable) of equal concentrations (controlled variable) will be tested for
pH (dependent variable).
7. (a) Experimental Design
Solutions of several acids (independent variable) of equal concentrations (controlled variable) will be tested for
pH (dependent variable).
(b) Analysis
According to pH values, in order of decreasing strength, the acids are:
hydrochloric and nitric acid (equal in pH)
hydrofluoric acid
methanoic acid
ethanoic acid
hydrocyanic acid
8. Experimental Design
Equally concentrated solutions of the substances are tested for pH and for electrical conductivity.
Analysis of the Design
The reasoning behind this design is that, of the two acidic solutions, HCl(aq) will have a lower pH than HC2H3O2(aq),
because HCl(aq) is a strong acid, and HC2H3O2(aq) is a weak acid. Of the two neutral solutions, NaCl(aq) is ionic and
will conduct, and C12H22O11(aq) is molecular and will not conduct. The two bases can be distinguished by adding
Na2SO4(aq) solution to each, which will precipitate BaSO4(s) from the Ba(OH)2(aq) solution, but not from the KOH(aq)
solution.
Note: The Ba(OH)2(aq) solution will have a higher pH and conduct electricity better than the KOH(aq) solution,
because the dissociation of Ba(OH)2(aq) produces two hydroxide ions per formula unit, and the dissociation of KOH(aq)
produces only one hydroxide ion per formula unit (but the course of study in the text has not yet addressed this point).
Making Connections
9. Personal experience indicates that acids “eat away” materials quite slowly — even those “instant” lime and scale
removers advertised on television. The stronger acids are more dangerous. Entertainment media are unlikely to portray
acid reactivity accurately; exaggeration is their “selling” point.
10. Acid deposition (acid rain) is acidic “fallout” from the atmosphere, initially mostly produced by vehicle exhausts and
industrial pollution sources. The two predominant acids are nitric and sulfuric (both strong acids). Nitrous and
sulfurous acids (both weak acids) may also be present. All rainwater contains dissolved carbon dioxide (carbonic
acid), so normal unpolluted rainwater has a pH of about 5.6 — very slightly acidic. You might predict that strong acids
would affect the environment more, but living systems are so complex that this would almost certainly be an oversimplification.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
Reflecting
11. The maximum concentration of the hydrogen ions in a 0.01 mol/L acid solution is (with a few exceptions) 0.01 mol/L.
This maximum concentration will be attained by strong acids but not by weak acids. There is no way to predict what
+ concentration will be in a 0.01 mol/L unknown acid solution, with the concepts studied to this point. If the
the H(aq)
acid were identified as strong, or the % ionization value were provided, that information would aid the accuracy of
prediction.
Copyright © 2002 Nelson Thomson Learning
Chapter 8 Acids and Bases
227
PRACTICE
(Page 386)
Understanding Concepts
12. Early ideas about acids were that they were hydrogen compounds that ionized in water to produce hydrogen ions.
However, some acidic solutions were found to be solutions of compounds that contain no hydrogen atoms.
Furthermore, hydrogen ions are theoretically impossible structures.
13. In the original Arrhenius theory, acids are molecular substances that ionize in aqueous solution to produce hydrogen
ions, and bases are ionic substances that dissociate in aqueous solution to produce hydroxide ions. This theory predicts
and explains quite a few acidic and basic substances, but there are lots of exceptions — too many to leave the theory
unrevised.
14. We now assume acidic solution properties are due to the presence of hydrated protons, commonly called hydronium
+ .
ions, symbolized H3O(aq)
+ CN–
15. (a) HCN(aq) H2O(l) → H3O(aq)
(aq)
+ NO–
(b) HNO3(aq) H2O(l) → H3O(aq)
3(aq)
+ SO2–
(c) Na2SO4(aq) → 2 Na(aq)
4(aq)
H2O(l) → HSO–4(aq) 2+ 2 OH–
Sr(OH)2(aq) → Sr(aq)
(aq)
SO2–
4(aq)
(d)
16. (a)
(b)
(c)
(d)
followed by
OH–(aq)
weak acid
strong acid
weak base
strong base
PRACTICE
(Page 389)
Understanding Concepts
17. According to Brønsted-Lowry definitions, acids differ from bases in that acids lose (donate) protons in a protontransfer reaction, and bases gain (accept) protons in a proton-transfer reaction.
18. (Reactants are shown in bold type.)
(a) The acids are:
HF(aq) and HSO–3(aq)
The bases are:
F–(aq) and SO2–
3(aq)
(b) The acids are:
HC2H3O2(aq) and HCO–3(aq)
The bases are:
C2H3O–2(aq) and CO2–
3(aq)
(c) The acids are:
H3PO4(aq) and HOCl(aq)
The bases are:
H2PO–4(aq) and OCl–(aq)
19. (a) HSO–4(aq) HCO–3(aq) → SO2–
4(aq) H2CO3(aq)
(b) One acid–base pair is:
and the other pair is:
HSO–4(aq) and SO2–
4(aq)
H2CO3(aq) and HCO–3(aq)
20. Brønsted-Lowry theory removes restrictions to acid–base reactions in that the theory does not depend on the
characteristics of solutions, nor does it require that water be a solvent, or, in fact, that the reactants be dissolved.
228
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
SECTIONS 8.3–8.4 QUESTIONS
(Page 392)
Understanding Concepts
1. In strong acids a high percentage of molecules react with water to form hydronium ions. In weak acids a low
percentage of molecules react with water to form hydronium ions. It is this difference in hydronium ion concentration
that gives strong and weak acids their distinctive properties.
2. Soluble ionic hydroxides are strong bases, as indicated by their empirical properties in solution — they not only taste
bitter, feel slippery, turn red litmus blue, and neutralize acids like all bases, but they also have very high pH values
(eleven or above) and electrical conductivity, and react rapidly (compared to weak bases with equal concentration).
The latter high pH, conductivity, and reaction rate for strong bases are explained by theorizing that they exist (100%)
in water as hydroxide ions and electrically balancing cations (e.g., hydroxide ions and sodium ions).
3. Weak bases have the properties of strong bases, but to a lesser degree. In particular, compared to strong base solutions
with equal concentration, weak bases form aqueous solutions with pH values above seven (but not above, say, ten),
and have conductivity and reaction rates that are low. This is explained by the theory that weak bases are substances
that, when they dissolve, react only partially with water to form a low percentage of hydroxide ions in solution.
+ Br–
4. (a) HBr(g) H2O(l) → H3O(aq)
(aq)
+ OH–
(b) KOH(s) → K(aq)
(aq)
+ C H O–
(c) HC7H5O2(s) H2O(l) → H3O(aq)
7 5 2(aq)
+ S2– followed by S2– H O → HS– OH–
(d) Na2S(s) → 2 Na(aq)
(aq)
(aq)
2 (l)
(aq)
(aq)
5. (a) According to the Arrhenius concept, acids are substances that dissolve in water to produce hydrogen ions.
(b) According to the revised Arrhenius concept, acids are substances that react with water to produce hydronium
ions.
(c) According to the Brønsted-Lowry concept, acids are substances that donate protons in proton-transfer reactions.
6. (a) According to the Arrhenius concept, bases are substances that dissolve in water to produce hydroxide ions. Weak
bases are not explained by this concept.
(b) According to the revised Arrhenius concept, bases are substances that dissolve in water or react with water to
produce hydroxide ions. Weak bases react only slightly with water.
(c) According to the Brønsted-Lowry concept, bases are substances that accept protons in proton-transfer reactions.
Weak bases attract protons less strongly than strong bases.
7. According to the Brønsted-Lowry concept, an acid–base reaction involves the transfer of a proton from the acid to the
base.
8. (a) Both of the acids and bases in the reaction are listed. The reactant acid and base asked for in the question are given
in bold type.
The acids are
HCO–3(aq) and HS–(aq)
The bases are
2–
S2–
(aq) and CO3(aq)
(b) The acids are
H2CO3(aq) and H2O(l)
The bases are
OH–(aq) and HCO–3(aq)
2–
2–
9. (a)HSO–4(aq) PO3–
4(aq) → SO4(aq) HPO4(aq)
+ HPO2– → H O
–
(b)H3O(aq)
4(aq)
2 (aq) H2PO4(aq)
10. One acid–base pair is
and the other pair is
11. (a) One acid–base pair is
HCO–3(aq) and CO2–
3(aq),
H2CO3(aq) and HCO–3(aq)
+
H3O(aq)
and H2O(aq),
and the other pair is
H2SO3(aq) and HSO–3(aq)
(b) One acid–base pair is
H2O(aq) and OH–(aq),
and the other pair is
HSO–3(aq) and SO2–
3(aq)
Copyright © 2002 Nelson Thomson Learning
Chapter 8 Acids and Bases
229
Applying Inquiry Skills
12. Baking soda will react both with strong bases and with acids. It releases carbon dioxide gas upon reaction with acids.
Since it can react either as an acid or as a base, it is not simple to predict how it will react in any given situation.
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Making Connections
13. Uses of baking soda include:
baking — reacts with food acids to produce CO2(g) for leavening
brushing teeth — a mild non-abrasive non-toxic cleaner
acid spills — neutralizes them for cleanup
base spills — neutralizes them for cleanup
odour removal — reacts with acidic or basic odorous gases in refrigerators, kitchens, and carpets
firefighting — releases carbon dioxide, which smothers flames
cleaning — makes a solution for washing surfaces
8.5 ACID–BASE REACTIONS
PRACTICE
(Page 394)
Understanding Concepts
1. Acids react with active metals to produce hydrogen and an ionic compound; react with carbonate compounds to
produce carbon dioxide gas and water; and neutralize bases to produce water.
2. (a) 2 HBr(aq) Zn(s) → H2(g) ZnBr2(aq)
HBr(aq) NaOH(s) → H2O(l) NaBr(aq)
2 HBr(aq) Na2CO3(s) → H2CO3(aq) 2 NaBr(aq)
or 2 HBr(aq) Na2CO3(s) → H2O(l) CO2(g) 2 NaBr(aq)
(b) The first neutralization produces hydrogen, which is flammable and dangerous, and uses zinc, which is not
commonly available.
The second neutralization uses lye (a strong base), which is very corrosive and not easy to handle.
The third neutralization is practical. It uses washing soda, which is non-hazardous, inexpensive, and commonly
available; and produces no dangerous products.
3. (a) 3 H2C2O4(aq) 2 Al(s) → 3 H2(g) Al2(C2O4)3(s)
(b) H2C2O4(aq) CaCl2(aq) → 2 HCl(aq) CaC2O4(s)
+
or H2C2O4(aq) Ca2+
(aq) → 2 H (aq) CaC2O4(s)
(c) 3 H2C2O4(aq) FeCl3(aq) → 6 HCl(aq) Fe2(C2O4)3(s)
3+ → 6 H+ Fe (C O )
or 3 H2C2O4(aq) Fe(aq)
(aq)
2 2 4 3(s)
Iron(III) ions are removed from solution by reaction with oxalate, effectively preventing the body from using them.
Practice
(Page 399)
Understanding Concepts
4. Acids react with active metals to produce hydrogen and an ionic compound; react with carbonate compounds to
produce carbon dioxide gas and water; and neutralize bases to produce water.
5. The requirements are that the chemical reaction must be spontaneous, rapid, quantitative, and stoichiometric.
6. The two reactants in a titration are the sample, usually in an Erlenmeyer flask, and the titrant, usually in a buret.
7. A standard solution is one with a precisely known concentration.
230
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
8. Repetition in titration — as elsewhere in science — increases the reliability of the answer. Mistakes can be easily identified and omitted, and averaging several measured values always reduces the effect of normal experimental error.
Applying Inquiry Skills
9. 2 KOH(aq) H2SO4(aq) → K2SO4(aq) 2 H20(l)
9.44 mL
10.00 mL
0.0506 mol/L
C
0.0506 mol
nKOH 9.44 mL
0.478 mmol
1L
1
nH SO 0.478 mmol 0.239 mmol
2
4
2
0.239 mmol
CH SO 2
4
10.00 m
L
CH SO 0.0239 mol/L = 23.9 mmol/L
2
4
1 mol H2SO4
0.0506 mol KOH
1
or CH SO 9.44 mL
KOH
2
4
1LK
OH
2m
ol K
OH
10.00 mL
CH SO 0.0239 mol/L = 23.9 mmol/L
2
4
The concentration of sulfuric acid in the water is 23.9 mmol/L.
10. (a) Evidence
Table 4: Volume of 0.0161 mol/L Sodium Hydroxide Required to
Neutralize 10.00 mL of Diluted Oxalic Acid
Trial
1
2
3
4
Final buret reading (mL)
14.3
27.8
41.1
13.8
Initial buret reading (mL)
0.2
14.3
27.8
0.4
Volume of NaOH(aq) used (mL)
14.1
13.5
13.3
13.4
(b) Analysis
2 NaOH(aq) H2C2O4(aq) → K2C2O4(aq) 2 H20(l)
13.4 mL
10.00 mL
0.0161 mol/L
C
0.0161 mol
nNaOH 13.4 mL
0.216 mmol
1L
1
nH C O 0.216 mmol 0.108 mmol
2 2 4
2
0.108 m
mol
CH C O 2 2 4
10.00 m
L
CH C O 0.0108 mol/L
2 2 4
0.0161 mol NaOH
1 mol H2C2O4
1
or CH C O 13.4 mL
NaOH
2 2 4
1 L Na
OH
2m
ol N
aOH 10.00 mL
CH C O 0.0108 mol/L
2 2 4
The concentration of oxalic acid in the rust remover is 100 times the concentration of the diluted acid used in the
titration, or 1.08 mol/L.
(c) Evaluation
difference experimental value predicted value
1.08 mol/L 1.11 mol/L
difference 0.03 mol/L
% difference
difference
=  100%
predicted value
Copyright © 2002 Nelson Thomson Learning
Chapter 8 Acids and Bases
231
0.03 m
ol/L
= 100%
1.11 m
ol/L
= 3%
The % difference is 3%, which is quite acceptable for school lab work.
The prediction based on the manufacturer’s label is quite accurate from a scientific point of view: the
percentage difference is only 3%, which is quite acceptable.
Note: You might discuss with students that from a legal perspective there could be a problem. A commercial
label is a legally guaranteed minimum unless specifically stated otherwise, so if the solution really has a
concentration below 1.11 mol/L, the manufacturer of the rust remover may be in trouble.
11. (a) Evidence
Table 5: Titration of 10.00-mL Samples of HCl(aq)
with 0.974 mol/L Ba(OH)2(aq)
Trial
1
2
3
4
Final buret reading (mL)
15.6
29.3
43.0
14.8
Initial buret reading (mL)
0.6
15.6
29.3
1.2
Volume of Ba(OH)2(aq) added (mL)
15.0
13.7
13.7
13.6
Colour at endpoint
blue
green
green
green
(b) Analysis
Ba(OH)2(aq) 2 HCl(aq) → BaCl2(aq) 2 H20(l)
13.7 mL 10.00 mL
0.974 mol/L
C
0.974 mol
nBa(OH) 13.7 mL
13.3 mmol
2
21 L
nHCl
13.3 mmol 26.6 mmol
1
26.6 mmol
CHCl 10.00 m
L
CHCl 2.66 mol/L
2 mol HCl
0.974 mol Ba(OH
)
1
or CHCl 13.7 mL
Ba(OH)
2 2 1
mol Ba
(OH)2
10.00 mL
CHCl 2.66 mol/L
The concentration of hydrochloric acid is 2.66 mol/L.
12. 2 NaOH(aq) H2SO4(aq) → Na2SO4(aq) 2 H20(l)
11.48 mL
10.00 mL
0.484 mol/L
C
0.484 mol
11.48 mL
5.56 mmol
1L
1
nH SO 5.56 mmol 2.78 mmol
2
4
2.78 mmol 2
CH SO 2
4
10.00 m
L
CH SO 0.278 mol/L
2
4
1 mol H2SO4
1
0.484 m
ol NaOH
or CH SO 11.48 mL
NaOH
2
4
1 L N
aOH
2
mol N
aOH
10.00 mL
CH SO 0.278 mol/L
nNaOH
2
4
The concentration of sulfuric acid is 0.278 mol/L.
232
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
SECTION 8.5 QUESTIONS
(Page 401)
Understanding Concepts
1. Answers will vary, but may include the following typical acid reactions.
Reaction with active metals:
2 HCl(aq) Mg(s) → H2(g) MgCl2(aq)
Reaction with strong bases:
HNO3(aq) NaOH(s) → H2O(l) NaNO3(aq)
Reaction with carbonate compounds:
2 HCl(aq) K2CO3(s) → H2CO3(aq) 2 KCl(aq)
or 2 HCl(aq) K2CO3(s) → H2O(l) CO2(g) 2 KCl(aq)
2. Al(OH)3(s) 3 HCl(aq) → AlCl3(aq) 3 H2O(l)
0.912 g
v
78.01 g/mol 0.10 mol/L
1 mol
nAl(OH) 0.912 g 0.0117 mol
3
78.01 g
3
nHCl
0.0117 mol 0.0351 mol
1
1L
vHCl
0.0351 mol 0.10 mol
0.35 L
vHCl
1
mol A
l(OH)3
3m
ol H
Cl
1 L HCl
0.912 g Al(OH)
3 78.01 g A
l(OH)3
1
mol A
l(OH)3
0.10 m
ol H
Cl
or vHCl
0.35 L
vHCl
The volume of hydrochloric acid neutralized is 0.35L.
3. Ca(OH)2(s) H2SO4(aq) → CaSO4(s) 2 H2O(l)
1.0 106 g
v
74.10 g/mol
1.2 10–3 mol/L
nCa(OH)
nH SO
2
4
vH SO
2
4
vH SO
4
2
or vH SO
2
4
vH SO
4
2
2
1 mol
1.0 106 g 1.4 104 mol
74.10 g
1
4
1.4 10 mol 1.4 104 mol
1
1L
1.4 104 mol 1.2 10–3 mol
1.1 107 L
1
mol H
1 L H2SO4
a(OH)
1m
ol C
2SO4
1.0 106 g Ca(OH)
2 2 a(OH)2 1 m
74.10 g C
ol Ca(
OH)2
1.1 107 L
The volume of lake water sulfuric acid that can be neutralized is 1.1 107 L (or 1.1 104 m3).
Copyright © 2002 Nelson Thomson Learning
Chapter 8 Acids and Bases
233
Applying Inquiry Skills
4. Experimental Design (1)
A sample of sodium hydroxide solution is titrated with standard hydrochloric acid, and the concentration of sodium
hydroxide is calculated from the measured volume of hydrochloric acid using the stoichiometric method.
Materials (1)
• lab apron
• eye protection
• standard HCl(aq)
• NaOH(aq)
• bromothymol blue
• wash bottle of pure water
• 100-mL or 150-mL beake
• 250-mL beaker
• 50-mL buret
• 10-mL volumetric pipet
• pipet bulb
• ring stand
• buret clamp
• stirring rod
• small funnel
• 250-mL Erlenmeyer flask
• meniscus finder
Experimental Design (2)
A sample of sodium hydroxide solution is reacted with excess lead(II) nitrate solution, and the precipitate is filtered
and dried; and the concentration of sodium hydroxide is calculated from the measured mass of lead(II) hydroxide
precipitate using the stoichiometric method.
Materials (2)
• lab apron
• eye protection
• standard Pb(NO3)2(aq)
• NaOH(aq)
• wash bottle of pure water
• 250-mL beaker
• 400-mL beaker
• 10-mL volumetric pipet
• pipet bulb
• ring stand
• iron ring
• stirring rod
• small funnel
• filter paper
• centigram balance
234
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
5. Experimental Design
A sample of oxalic acid solution is titrated with standard sodium hydroxide solution, and the concentration of oxalic
acid is calculated from the measured volume of sodium hydroxide using the stoichiometric method.
Materials (1)
•
lab apron
•
eye protection
•
standard NaOH(aq) (known concentration)
•
H2C2O4(aq)
•
bromothymol blue
•
wash bottle of pure water
•
100-mL beaker
•
150-mL beaker
•
50-mL buret
•
10-mL volumetric pipet
•
pipet bulb
•
ring stand
•
buret clamp
•
tirring rod
•
small funnel
•
250-mL Erlenmeyer flask
•
meniscus finder
•
toxic waste disposal container
Procedure
1. Obtain about 50 mL of oxalic acid in a clean, dry 100-mL beaker.
2. Obtain about 70 mL of NaOH(aq) in a clean, dry, labelled 150-mL beaker.
3. Set up the buret with NaOH(aq) following the accepted procedure for rinsing and clearing the air bubble
(see Skills Handbook).
4. Pipet a 10.00-mL sample of oxalic acid into a clean Erlenmeyer flask.
(Caution: Oxalic acid is toxic. Use a pipet bulb.)
5. Add 1 or 2 drops of bromothymol blue indicator.
6. Record the initial buret reading to the nearest 0.1 mL.
7. Titrate the sample with NaOH(aq) until a single drop produces a permanent change from pale yellow to pale blue.
8. Record the final buret reading to the nearest 0.1 mL.
9. Dispose of the flask contents by rinsing into a toxic waste container.
10. Repeat steps 4 to 9 until three consistent results (to 0.1 mL) are obtained.
Copyright © 2002 Nelson Thomson Learning
Chapter 8 Acids and Bases
235
CHAPTER 8
SUMMARY
(Page 402)
Make a Summary
buret
hydrochloric acid • titrant
• standard solution
• known concentration (0.12 mol/L)
• strong acid (high conductivity;
fast reaction)
• 100% ionized (Arrhenius theory)
100%
+
–
• HCl(aq)→ H(aq) + Cl(aq)
stop cock
• > 99% reacted with water
(Brønsted-Lowry theory)
99%
+
–
• HCl(aq) → H2O(1) + H3O(aq) + Cl(aq)
Erlenmeyer flask
• very low pH
+
• pH = –log [H(aq)] = – log 0.12 = 0.92
gripewater • unknown concentration
indicator • endpoint provides empirical
• NaHCO3(s) dissociates
+
•
the acid-base reaction
•
HCL(aq) + NaHCO3(aq) → H2CO3(aq) + NaCl(aq)
–
+
+
–
or H3O(aq) + Cl(aq) + Na(aq) + HCO3(aq) →
or
+
–
H2CO3(aq) + Na(aq) + Cl(aq) (total ionic)
+
–
H3O(aq) + HCO3(s) → H2O(l) + H2CO3(aq)
acid
H+
base
base
–
• NaHCO3(s) → Na(aq) + HCO3(aq)
evidence for the end of
–
HCO3(aq)
–
HCO3(aq)
is a (weak) base (here)
is amphiprotic with
conjugate acid-base pairs of
–
HCO3(aq) – H2CO3(aq) (here) and
–
2–
HCO3(aq) – CO3(aq)
acid
proton transfer
conjugate pair
conjugate pair
Reflect on your Learning
Here is an example:
My thinking on what an acid and a base are has changed from an Arrhenius (ionization) concept to a Bronsted-Lowry
(proton-transfer) concept. I still see value in the Arrhenius concept and will continue to use it whenever it is suitable.
I now understand the difference between a strong and weak acid and a strong and weak base. Now I have to remember
to distinguish between strength and concentration. This takes mental effort
The fact that sometimes a chemical can act as an acid in one reaction and as a base in another reaction is interesting.
I now see how important empirical evidence is to the study of chemistry. The ultimate way of knowing in chemistry is
empirical, not theoretical as I had previously thought. Now I see where chemical knowledge really comes from. This is
what is exciting to me.
Titrations are neat. I especially like the sudden colour change at the endpoint, although I often overshoot the endpoint.
What I don’t understand is how to choose the right indicator from that long list at the end of the book, and how do indicators work anyway? Why do they change colour?
236
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
CHAPTER 8
REVIEW
(Page 403)
Understanding Concepts
1. (a) An acid is a substance that dissolves to form a conducting solution that turns blue litmus red, neutralizes bases,
reacts with active metals to form hydrogen gas, and reacts with carbonate compounds to form carbon dioxide gas.
(b) A base is a substance that dissolves to form a conducting solution that turns red litmus blue, and neutralizes acids.
2 (a) sodium hydroxide
base
(b) acetic acid
acid
(c) magnesium hydroxide
base
(d) hydrochloric acid
acid
(e) calcium hydroxide
base
(f) (aqueous) ammonia
base\
+ OH–
3 (a) basic:
NaOH(s) → Na(aq)
(aq)
(b) acidic:
+ C H O–
HC2H3O2(aq) → H(aq)
2 3 2(aq)
(c) basic:
2+ 2 OH–
Mg(OH)2(s) → Mg(aq)
(aq)
(d) acidic:
+ Cl–
HCl(aq) → H(aq)
(aq)
(e) basic:
2+ 2 OH–
Ca(OH)2(s) → Ca(aq)
(aq)
(f) basic:
+
NH3(aq) H2O(l) → NH4(aq)
OH–(aq)
+ ] is much greater in the hydrochloric acid solution.
4. (a) [H(aq)
(b) In hydrochloric acid, essentially all of the HCl molecules are dissociated into ions. In acetic acid, only a very
small percentage of HC2H3O2 molecules are dissociated.
(c) The same volume of NaOH(aq) would be required for each neutralization reaction.
5. In order of increasing pH, the solutions are:
HCl(aq), HC2H3O2(aq), NaCl(aq), NH3(aq), NaOH(aq)
6. (a) pH
+ ]
–log [H(aq)
–log [7.5 10–3 mol/L]
pH
(b) pH
2.12
+ ]
–log [H(aq)
–log [2.5 10–3 mol/L]
7. (a)
pH
2.60
+ ]
[H(aq)
10–pH
10–11.56 mol/L
+ ]
[H(aq)
+ ]
(b) [H(aq)
2.8 3 10–12 mol/L
10–pH
10–3.50 mol/L
+ ]
[H(aq)
3.2 10–4 mol/L
8. Electrical conductivity, pH, and rate of reaction with active metals can all be used to rank acids in terms of
strength.
9. (a) HNO2(aq) has a higher pH (is less acidic, more basic).
<50%
+ NO–
(b) HNO2(aq) H2O(l) → H3O(aq)
2(aq)
>99%
Copyright © 2002 Nelson Thomson Learning
Chapter 8 Acids and Bases
237
+ NO–
HNO3(aq) H2O(l) → H3O(aq)
3(aq)
As shown, nitric acid transfers protons to water completely, whereas in nitrous acid the transfer is much less than
50%, making the solution much less acidic.
10. [H+(aq)] is ten times as concentrated in pH 5 as in pH 6.
11.
Concept
oxygen
Main Idea
acids contain oxygen and
react with limestone
Limitations
doesn’t explain solutions of HCl(g), or why some oxides make
basic solutions
hydrogen
acids contain hydrogen and
react with active metals to
produce hydrogen gas
doesn’t explain acidic solutions of SO2(g), or basic solutions from
compounds containing hydrogen, e.g., NH3(g)
Arrhenius
acids dissolve in water to
produce H+(aq)
doesn’t explain nonaqueous solutions, or hydrogen
containing substances that can neutralize both acids and bases
12.
(a)
(b)
(c)
(d)
(e)
acid
base
conjugate acid
conjugate base
HSO–4(aq)
HSO–4(aq)
H2PO–4(aq)
HCO–3(aq)
HSO–3(aq)
HCO–3(aq)
HPO2–
4(aq)
H2BO–3(aq)
HS–(aq)
→H2CO3(aq)
SO2–
4(aq)
→H2PO–3(aq)
SO2–
4(aq)
→H3BO3(aq)
HPO2–
4(aq)
→H2S(aq)
CO2–
3(aq)
NH3(aq)
+
→NH4(aq)
SO2–
3(aq)
13. (a) NH3(aq)
ammonia
B-L acid or base
+
(b) NH4(aq)
ammonium ion
B-L acid
(c) NO–2(aq)
nitrite ion
B-L base
NO–3(aq)
nitrate ion
B-L base
(d)
2
PO3
4(aq) and HPO 4(aq)
14. (a) conjugate acid–base pairs are:
and: H2O(aq) and OH
(aq)
(b) PO3–
4(aq) is a strong base that can only react as a base with water, so it must form a basic solution.
–
15. (a) H2SO3(aq) + OH
(aq) → HSO3(aq) + H2O(l)
2–
(b) HSO
3(aq) + OH(aq) → SO3(aq) + H2O(l)
16. (a) NaHCO3(aq) + HCl(aq) → H2CO3(aq) + NaCl(aq)
+ → H CO
HCO–3(aq) + H3O(aq)
2
3(aq) + H2O(l)
(b) NaHCO3(aq) + NaOH(aq) → Na2CO3(aq) + H2O(l)
2–
HCO–3(aq) + OH
(aq) → CO3(aq) + H2O(l)
+
(c) HCO
3(aq) + H3O(aq) → H2CO3(aq) + H2O(l)
B
A
A
B
2–
HCO–3(aq) + OH
(aq) → CO3(aq) + H2O(l)
A
B
B
A
The conjugate acid–base pairs are:
+ and H O
H3O(aq)
and
H2CO3(aq) and HCO
2 (l)
3(aq)
H2O(l) and OH
(aq)
and
HCO–3(aq) and CO2–
3(aq)
(d) The hydrogen carbonate ion may act as either an acid or as a base; it is amphiprotic.
17. A chemical reaction suitable for titration must be spontaneous, fast, stoichiometric, and quantitative.
18. You must know the concentration of one reactant solution accurately to calculate precise results.
238
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
19. (a) titration: a laboratory procedure involving the carefully measured and controlled addition of a standard solution
from a buret into a measured volume of a sample solution (or the sample solution could be in the buret ...)
(b) titrant: the solution in the buret during a titration
(c) endpoint: the point in a titration at which a sharp change in a property occurs (e.g., a colour change)
20. The acid reacts with an active metal to produce hydrogen gas.
H2SO4(aq) + Zn(s) → ZnSO4(aq) + H2(g)
The acid is partially neutralized by a basic solution.
H2SO4(aq) + Na2CO3(aq) → Na2SO4(aq) + H2CO3(aq)
The acid is completely neutralized by a strong basic solution.
H2SO4(aq) + Zn(s) → ZnSO4(aq) + H2(g)
Applying Inquiry Skills
21. The equipment required for the precision given would be:
(a) a volumetric flask
(b) a 10-mL graduated cylinder or a 10-mL graduated pipet
(c) a 10-mL volumetric pipet
22. Experimental Design
The pH is measured for samples of an acid and of a base. The samples are diluted tenfold, and the pH measured again.
Materials
• HCl(aq)
• NaOH(aq)
• two 150-mL beakers
• 100-mL graduated cylinder
• pH meter
23. (a) Analysis
Solution 1 is NaHCO3(aq), because it is basic, but not with the highest pH, and conducts well.
Solution 2 is KNO3(aq), because it is neutral, and conducts well.
Solution 3 is H2SO3(aq), because it is acidic, but not with the lowest pH, and conducts poorly.
Solution 4 is HCl(aq), because it is acidic, with the lowest pH, and conducts well.
Solution 5 is NaOH(aq) , because it is basic, with the highest pH, and conducts well.
(b) Evaluation
The experimental design is judged adequate because the experiment produces evidence needed to answer the
question with a high degree of certainty.
24. (a) Evidence
Table 2: Titration of 10.00 mL of 0.120 mol/L Na2CO3(aq) with HCl(aq)
Trial
1
2
3
4
Final buret reading (mL)
17.9
35.0
22.9
40.1
Initial buret reading (mL)
0.3
17.9
5.9
22.9
Volume of HCl(aq) added
17.6
17.1
17.0
17.2
Colour at endpoint
red
orange
orange
orange
(b) Analysis
Na2CO3(aq) 2 HCl(aq) → H2CO3(aq) 2 NaCl(aq)
10.00 mL
0.120 mol/L
17.1 mL
C
0.120 mol
nNa CO 10.00 mL
1.20 mmol
2
3
1L
2
nHCl
1.20 mmol 2.40 mmol
1
Copyright © 2002 Nelson Thomson Learning
Chapter 8 Acids and Bases
239
CHCl
2.40 m
mol
17.1 m
L
CHCl
0.140 mol/L
or
CHCl
CHCl
0.140 mol/L
0.120 mol Na
2 mol HCl
1
2CO3
10.00 mL Na
2CO3 1 L Na
C
O
N
a
1
m
ol
C
O
17.1
mL
3
2
2
3
The concentration of the hydrochloric acid solution is 0.140 mol/L.
25. (a) Evidence
Table 3: Titration of 10.00 mL of H3PO4(aq) with 1.25 mol/L NaOH(aq)
Trial
1
2
3
4
Final buret reading (mL)
13.3
25.0
36.8
48.4
Initial buret reading (mL)
0.4
13.3
25.0
36.8
12.9
11.7
11.8
11.6
deep red
pale pink
pale pink
pale pink
Volume of NaOH(aq) added
Colour at endpoint
(b) Analysis
H3PO4(aq) 2 NaOH(aq) → H2CO3(aq) 2 NaCl(aq)
10.00 mL
11.7 mL
C
nNaOH
nH PO
3
4
nH PO
4
3
CH PO
3
4
CH PO
4
or CH PO
3
4
CH PO
4
3
3
1.25 mol/L
1.25 mol
11.7 mL
14.6 mmol
1L
1
14.6 mmol 2
7.31 mmol
7.31 mmol
10.00 m
L
0.731 mol/L
1 mol H3PO4
1.25 m
ol N
aOH
1
11.7 mL
NaOH
1 L N
aOH
2
mol N
aOH
10.00 mL
0.731 mol/L
The concentration of the original phosphoric acid solution in the rust remover is 7.31 mol/L — ten times the
concentration of the diluted titrated solution.
(c) Evaluation
The prediction, based on the manufacturer’s label, is verified by the results. The label seems to be quite accurate.
Making Connections
26. Personal experience indicates that acids “eat away” materials quite slowly — even the “instant” lime and scale
cleaners advertised on television. The student response to this question will normally indicate that the movie
industry generally greatly exaggerates (or even completely fabricates) the reactivity of acids for effect. Movies
featuring entertaining misconceptions about acids include:
The Plastic Man Comedy/Adventure Show
Contaminazione
The Navy vs. the Night Monsters
Alien, Aliens
27. Boric acid is a very weak acid, so that almost none of its molecules in solution react with water to produce hydronium ions.
240
Unit 3 Solutions and Solubility
Copyright © 2002 Nelson Thomson Learning
<50%
H3BO3(aq) H2O(l) → H3O+(aq) H2BO3–(aq)
For strong acids like HCl(aq), the dissolved molecules react with water almost totally, producing a high concentration of reactive hydronium ions.
>99%
HCl(aq) H2O(l) → H3O+(aq) Cl–(aq)
28. Experimental Design
The scale is removed by reaction with hydrochloric (shown here) or any other acid. A weaker acid reacts the same
way, but more slowly — and would be much safer to work with. The reasoning behind this design is presented below.
The hard-water scale (CaCO3(s) and/or MgCO3(s)) reacts with the acid to form soluble compounds, as shown in the
following reaction equations.
CaCO3(s) 2 HCl(aq) → CaCl2(aq) H2CO3(aq)
MgCO3(s) 2 HCl(aq) → MgCl2(aq) H2CO3(aq)
Materials
•
eye protection
•
rubber gloves
•
hydrochloric acid (commercial scale remover or 0.10 mol/L)
•
tap water
•
plastic or glass bowls/containers
•
plastic sponge
Procedure
1. Wear eye protection and rubber gloves.
2. Soak smaller scale-coated parts in the acid in a glass or plastic container.
3. Pour the acid solution into containers such as kettles that need to be descaled.
4. Wash larger scale-coated parts with the acid using a plastic sponge, being careful not to allow dripping onto other
surfaces.
5. Rinse all solutions thoroughly down the sink with plenty of water.
29. Students will normally find that pH adjustment of municipal water is straightforward: any cheap non-toxic acid or
base will do. Sodium hydroxide will increase pH while hydrochloric acid will decrease it, without adding any harmful
ions in the process (providing only very low concentrations are achieved).
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
30. Conifers have evolved to use nutrients that are most readily present in soils with lower pH values. Acid rain in general
lowers the pH in plants more than the normal level, resulting in varying degrees of stress for the plants.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
Exploring
31. Lewis acid–base theory focuses on electron pairs in bond formation, with an acid being the electron pair acceptor, and
a base the electron pair donor. This concept allows dealing with many reactions that do not involve Brønsted-Lowry
acids or bases.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
Copyright © 2002 Nelson Thomson Learning
Chapter 8 Acids and Bases
241
(b) The can is not crushed. It appears to cool slowly and when cooled contains about a quarter-can of water. The fact that
some water is sucked into the can agrees with the hypothesis. The fact that the can is not crushed as before shows that,
at best, the hypothesis is insufficient and, at worst, is on the wrong track.
[Ideal answer: Heat transfer within a gas is a slow process dependent mainly on convection. The hot air cools relatively slowly and therefore, the volume decreases slowly. As the volume decreases slowly, so does the pressure inside
the container. The higher atmospheric pressure on the outside slowly pushes some water inside the can. There is no
water vapour to quickly condense and suddenly change the volume of the gas. The (ideal) hypothesis is supported.]
(c) According to the original hypothesis, the ice water should produce the most significant crushing of the can and the
warm water should produce less of an effect than the original cold water. These predictions are based on the cooling
of the can — most with ice water, least with warm water.
[Ideal answer: Because the crushing is caused by the condensation of the water vapour, there will likely be little difference when using ice water or warm water because both of these are significantly cooler than 100°C.]
(d) For both the ice water and hot water from the tap, the can was crushed in much the same way as it was originally.
Observation of the cans did not show any obvious differences in the cans. The predictions appear to be falsified and
again the hypothesis appears to be unacceptable.
[Ideal answer: The evidence appears to agree with the prediction and the (ideal) hypothesis is supported.]
9.1 STATES OF MATTER
ACTIVITY 9.1.1 MOLECULAR MOTION
(Page 421)
Procedure
• Step 2: The shape and volume of the sample remain relatively constant. Individual spheres jiggle slightly while
remaining in roughly the same location.
• Step 3: Some individual spheres bounce away from the main group and return. The average volume stays relatively
constant. Individual spheres jiggle around more rapidly and randomly. The sample does not stay uniform. Spheres
move around with occasional spaces created, which disappear quickly.
• Step 4: The volume of the sample is now the volume of the square container. Individual spheres move in straight lines
until they collide with each other or the sides of the container. The spheres are all separated from each other and the
spacing between the spheres is many times the size of the spheres.
Analysis
(a) The plastic spheres are placed in a watch glass like a shallow bowl. Gravity forces the spheres to the bottom, or lowest,
point. This simulates the attractive forces among particles.
(b) The degree of order or organization, from greatest to least, is solid (Step 2), liquid (Step 3), and then gas (Step 4).
Disorder (lack of order or organization) is most apparent with the gas model.
Synthesis
(c) The model works best for gases and reasonably well for solids and liquids. The primary molecular motion of gas
molecules is translational and this is clearly shown. Vibrational motion is shown for solids and liquids. There is little
evidence of rotation but this may just be hard to see when the spheres are projected. The model works reasonably well
for a substance increasing in temperature and changing in phase from solid to liquid and then to gas.
A drawback of the model is the representation of relative molecular speeds among phases of matter at the same
temperature. To make the physical model work, the spheres have to move faster (i.e., at a higher temperature) to model
a gas rather than a liquid. Evidence indicates that this isn’t so for molecules of, say, a gas compared to a liquid for
substances at the same temperature.
(d) Diffusion could be shown by introducing a few different-coloured or -sized spheres into the mixture and watching the
movement of these spheres relative to the rest of the spheres present. Compressibility of gases could be shown by
adding a movable barrier (e.g., a piece of wood dowelling) and decreasing the size of the container available in which
the spheres can move.
254
Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
PRACTICE
(Page 422)
Understanding Concepts
1. In order of increasing strengths of forces among particles, the states of matter are gas, liquid, and solid. Maintaining
a specific shape and volume (as a solid does) requires particles to be held close together in fixed positions.
Maintaining a specific volume, but not shape (as a liquid does) requires particles to be held close together, but not in
fixed positions, so forces are less effective. Dispersing spontaneously (as a gas does) means that particle motion overcomes the effect of any attractive forces.
2. Solids have the highest degree of order, and gases the lowest. Order is directly related to strength of attraction between
particles.
3. (a) According to our model of gases, gases are compressible because they consist of individual particles separated
by large distances. Pressure can easily force the particles closer together.
(b) (Typical answer) Why are there spaces between gas particles? Because they have too much energy for the very
weak attractive forces between particles to be able bring them together.
(c) (Typical answer) Why do the particles have a lot of energy? Because they are moving very fast. Why do they
move very fast? Because they collide elastically, without losing energy. Why do they collide elastically?
Hmmm....
Making Connections
4. (a) Liquids are nearly incompressible, so sealed liquid systems (such as hydraulic lifts) transfer forces very well.
(b) A gas bubble in a brake line is compressible, preventing the total force from transferring to the brakes.
Reflecting
5. The most common student answer is that visual aids and analogies are the most effective models, particularly when a
physical model that can be manipulated is involved. Most students will agree that a variety of models is better than a
single example for understanding — sometimes because it makes the understanding more complete, and sometimes
because there is only one model that is comprehensible to the students.
SECTION 9.1 QUESTIONS
(Page 422)
Understanding Concepts
1. (a) Atomic combinations make metallic (metal with metal), ionic (metal ion with nonmetal ion), molecular
(nonmetal atom with nonmetal atom), or network covalent (bonded elements along the staircase line) substances.
(b) The bond types between particles in the substances are, in order: metallic, ionic, intermolecular (London dispersion, dipole–dipole, and hydrogen), and covalent bonds.
(c) A solid state results when attractive forces between particles of a substance are strong enough to hold the particle
in place, even for relatively high kinetic (motion) energy. Kinetic energy depends on temperature, while attractive force depends on the structure of the particles and the distance between them.
2. (a) Both are incompressible, because their particles are close together.
(b) Particles in liquids are free to change position, while those in solids are not. This is because the attractive forces
between particles in liquids are not strong enough to stop this translational particle motion, and in solids they are.
3. (a) Attraction between particles in gas phases is negligibly small.
(b) The prediminant motion for gas particles is translational (place-to-place) motion.
Copyright © 2002 Nelson Thomson Learning
Chapter 9 The Gas State
255
4.
State
solid
liquid
gas
Properties
Explanations
solids have definite shape
and volume
the attractive force(s) between particles is
sufficiently high to hold the particles in place
are virtually incompressible
the particles are too close together to be compressed
do not flow easily
the attractive force(s) between particles holds the
particles in place
liquids assume the shape
of the container but have a
definite volume
attractive forces have been overcome so the particles
can rotate and roll over each other and assume new
shapes but maintain the same volume
are virtually incompressible
the particles are still “touching each other” (repelling
each other, electrostatically)
flow readily
the particles can roll over each other
gases assume the shape
and volume of the container
the particles are independent of each other due to very
small attractive forces; the particles are only contained
by the container, not by intermolecular attractions
are highly compressible
the distance between the particles is large relative to
the size of the particle (20–30 times the diameter), so
there is empty space for compression
flow easily
there is little or no attractive force(s) among the
particles, so they are free to move independently in any
direction
Making Connections
5. Gas particles move very rapidly, so the flexible fabric air bag fills in an instant, before the motorist hits the steering
wheel of the car. Gases have no definite shape, so the flexible bag can mould to the motorist’s shape and spread the
restraining force evenly. Gases can move through small openings, so the gas from an air bag quickly diffuses through
small holes to deflate the bag after a collision, freeing the motorist.
9.2 GAS LAWS
Try This Activity: A Simulation of Gas Properties
(Page 423)
(a) The order of gas cylinders from most likely to least likely to explode is 1, 5, 3, 2, 4. The order is based on the
reasoning that smaller volumes and higher temperatures (for the same mass of gas) produce greater pressures. The
higher the pressure, the more likely the cylinder will be to explode.
– Cylinder 1 has the smallest volume and the highest temperature and, therefore, should be at the highest pressure.
– Cylinder 5 is also at the highest temperature but has a greater volume than 1. Cylinder 5 has twice the volume
of 3, but 5 has a temperature that is almost three times larger. Therefore, 5 likely has a higher pressure than 3.
– Cylinder 3 follows 5, as indicated above, but precedes 2 because 3 has a higher temperature and the same
volume as 2.
– Cylinder 2 is at the same temperature as 4 but 2 has a smaller volume. Therefore, the pressure in 2 should be
greater than in 4.
– Cylinder 4 has the greatest volume and the lowest temperature, and therefore, the least pressure.
Note: At this stage students would not know about the Kelvin temperature scale. Although the above
reasoning is correct, the comparisons should be done using Kelvin, not Celsius, temperatures. If Kelvin
temperatures are used, the order of 5 and 3 would likely be reversed.
(b) Variables that need to be considered are the amount of gas, and the volume, temperature, and maximum pressure
for the cylinder.
256
Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
PRACTICE
(Page 425)
Understanding Concepts
1. STP is 0°C and 101.325 kPa; SATP is 25°C and 100 kPa.
2. Table 3: Converting Pressure Units
Pressure
(kPa)
Pressure
(atm)
Pressure
(mm Hg)
(a)
0.50
0.0049
3.8
(b)
96.5
0.952
724
(c)
110
1.09
825
(d)
253
2.50
1.90 103
1 atm
(a) 0.50 kPa
0.0049 atm
101.325 kPa
760 mm Hg
3.8 mm Hg
0.50 kPa
101.325 kPa
1 atm
(b) 96.5 kPa
0.952 atm
101.325 kPa
760 mm Hg
96.5 kPa
724 mm Hg
101.325 kPa
101.325 kPa
(c) 825 mm Hg 110 kPa
760 mm H
g
1 atm
825 mm Hg 1.09 atm
760 mm Hg
101.325 kPa
(d) 2.50 atm 253 kPa
1
atm
760 mm Hg
2.50 atm 1.90 103 mm Hg (= 1.90 m Hg)
1
atm
3. Using only one pressure unit facilitates communication, understanding, and comparison of pressures.
Making Connections
4. When the bulb is released, the pressure inside the tube is lower than that outside. The outside air pressure pushes down
on the surface of the surrounding liquid, forcing the liquid into the tube.
PRACTICE
(Page 428)
Understanding Concepts
5. Atmospheric pressure is the force per unit area exerted by air on the surfaces of all objects.
6. p1 = 101 kPa
v1 = 0.650 L
p2 = ?
v2 = 0.250 L
Copyright © 2002 Nelson Thomson Learning
Chapter 9 The Gas State
257
p1v1
= p2v2
p2
pv
= 11
v2
pair
or
pair
pair
0.650 L
= 101 kPa 0.250 L
= 263 kPa
0.650 L
= 101 kPa 0.250 L
= 263 kPa
The final pressure of the air must be 263 kPa.
7. p1 = 98.0 kPa
v1 = 35.0 L
p2 = 25.0 kPa
v2 = ?
p1v1
= p2v2
v2
p v1
= 1
p2
vHe
or
vHe
vHe
35.0 L
= 98.0 kPa
25.0 kPa
= 137 L
98.0 kPa
= 35.0 L 25.0 kPa
= 137 L
The final volume of helium will be 137 L.
8. p1 = 3.0 atm
v1 = 110 mL
p2 = 2.0 atm
v2 = ?
p1v1
= p2v2
v2
p v1
= 1
p2
110 mL
= 3.0 atm 2 .0 atm
vballoon = 1.6 102 mL = 0.16 L
or
3 .0 atm
vballoon = 110 mL 2. 0 atm
vballoon = 1.6 102 mL = 0.16 L
The final volume of the balloon will be 0.16 L.
258
Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
9. p1 = 98 kPa
v1 = 32 kL
p2 = 150 kPa
v2 = ?
p1v1
= p2v2
v2
pv
= 11
p2
vair
or
vair
vair
98 kPa 32 kL
= 150 kPa
= 21 kL
98 kPa
= 32 kL 150 kPa
= 21 kL
The final volume of air in the diving bell will be 21 kL.
10. The movement of air “masses” within the atmosphere, and thinning with altitude cause local air pressure to depend
on your location and time.
Making Connections
11.
Evangelista Torricelli (1608–1647), Blaise Pascal (1623–1662), Guillaume Amontons (1663–1705)
What follows is a brief summary of development:
Evangelista Torricelli (1608–1647)
At Galileo’s urging, Torricelli first investigated the principle of air pressure scientifically and invented the
mercury barometer (a tube of mercury inverted in a dish of mercury) in 1643. The basic question was why water could
only be raised about 10 m by a vacuum pump. Torricelli’s suggestion was that the weight of air pushed the water up,
rather than the pump’s “vacuum” pulling it up. (This was also the first vacuum known to science.)
Blaise Pascal (1623–1662)
Pascal continued investigating and verified Torricelli’s hypothesis that air had definite weight that would necessarily decrease with increasing altitude. He carried a barometer up a mountain, Puy de Dôme. Pascal also made a
barometer with wine (less dense than water) that had a height of over 13 m. (The wine experiment was done in the
same year that Torricelli died. The SI unit of pressure, the pascal, Pa, is named after Blaise Pascal.)
Guillaume Amontons (1663–1705)
Amontons invented the first hygrometer (for measuring moisture in the air) in 1687 (the same year that Isaac
Newton published his laws of motion and universal gravitation), and the first barometer that did not use mercury —
so that it could be used on board ships at sea. He also invented an air thermometer that improved on Galileo’s version
by keeping the volume constant and measuring pressure (directly) and temperature (indirectly). Amontons, like
Edison, was deaf.
Note: The SIRs computer program has some excellent simulations of barometers and manometers.
PRACTICE
(Page 432)
Understanding Concepts
12. Absolute zero is about –273°C, or 0 K
13. T (K) = T (°C) 273
(a) T (K) = 0 273 273 K
(b) T (K) = 100 + 272 = 373 K
(c) T (K) = –30 273 243 K
(d) T (K) = 25 273 298 K
Copyright © 2002 Nelson Thomson Learning
Chapter 9 The Gas State
259
14. T (°C) = T (K) 273
(a) T (°C) = 0 273 –273°C
(b) T (°C) = 100 273 –173°C
(c) T (°C) = 300 273 27°C
(d) T (°C) = 373 273 100°C
15. Scientists have come very close to achieving absolute zero. One source quotes, for example, < 2 10–8 K by 1988.
According to the quantum mechanics theory, molecules at absolute zero do not have zero kinetic energy, but rather a
minimum possible amount called the “zero–point” energy. In theory, it seems that molecular motion would not cease
altogether.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
PRACTICE
(Page 434)
Understanding Concepts
16. v1 = 12.7 mL
T1 = 22°C = 295 K
v2 = ?
T2 = –11°C = 262 K
v1
v2
= T1
T2
v2
vbutane
or
vbutane
vbutane
T v1
= 2
T1
262 K 12.7 mL
= 295 K
= 11.3 mL
262 K
= 12.7 mL 295 K
= 11.3 mL
The final volume of butane will be 11.3 mL.
17. v1 = 2.05 L
T1 = 5°C = 278 K
v2 = ?
T2 = 21°C = 294 K
v1
v2
= T1
T2
T v1
v2
= 2
T1
294 K
2.05 L
= 278 K
vair
= 2.17 L
or
260
Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
vair
vair
294 K
= 2.05 L 278 K
= 2.17 L
The final volume of the air initially in the bottle will be 2.17 L. The volume of air that will leave the bottle due to the
volume increase will therefore be (2.17 2.05) L, or 0.12 L.
18. v1 = 1.5 L
T1 = 22°C = 295 K
v2 = ?
T2 = 100°C = 373 K
v1
v2
= T1
T2
v2
vair
or
vair
vair
T v1
= 2
T1
373 K
1.5 L
= 295 K
= 1.9 L
373 K
= 1.5 L 295 K
= 1.9 L
The final volume of the 1.5 L of air initially in the pan will be 1.9 L. The increase in the volume of air will be
(1.9 1.5) L, or 0.4 L.
(The 1.5 L is a description of the (nominal) pan size, not a measured volume. The above calculation assumed an initial
volume of 1.5 L of air in this pan.)
0.40 L
100% 26%
1.500 L
The percentage increase of the volume of air will be 26%.
(The same answer is obtained by early rounding or not. Rounding on paper, but no rounding in the calculator, is
preferred.)
19. v1 = 1.00 L
T1 = 20°C = 293 K
v2 = ?
T2 = 80°C = 353 K
v1
v2
= T1
T2
v2
T v1
= 2
T1
353 K 1.00 L
= 293 K
vair
= 1.20 L
vair
353 K
= 1.00 L 293 K
= 1.20 L
or
vair
Copyright © 2002 Nelson Thomson Learning
Chapter 9 The Gas State
261
The final volume of each litre of air will be 1.20 L.
Applying Inquiry Skills
20. The design of this experiment is judged inadequate. At typical room temperatures, a change of one degree only
changes the volume by about 1/300th. It seems very unlikely that this would change the total volume enough to
measure with a graduated cylinder. As well, the water will evaporate to some extent into the trapped air space, and the
cylinder itself will expand and contract with temperature changes. Likely the biggest problem would be keeping the
pressure of gas in the cylinder constant.
To work effectively, the gas would need to be trapped by a liquid with little tendency to evaporate (such as mercury)
in a long thin tube (so small changes show up as visible motion) in an apparatus that kept the gas at constant pressure.
Note:Guillaume Amontons (1663–1705) invented just such a device, significantly improving an earlier design
constructed by Galileo, which had suffered from all the drawbacks mentioned above.
21. (a) v1 = assume 1.000 L
T1 = –60°C = 213 K
v2 = ?
T2 = 540°C = 813 K
v1
v2
= T1
T2
T v1
v2
= 2
T1
813 K 1.000 L
= 213 K
vN
= 3.82 L
2
or
vN
2
vN
2
813 K
= 1.000 L 213 K
= 3.82 L
The final volume of each litre of nitrogen will be 3.82 L.
3.82 L
The ratio of volume increase is , or 3.82:1.00.
1.00 L
(b) The gas expansion in jet engines is also used to spin the turbine that powers the compressor that compresses the
gas–fuel mix before ignition.
PRACTICE
(Page 435)
Understanding Concepts
22.
p1 = 97 kPa
T1 = 22°C = 295 K
p2 = 350 kPa (minimum; any higher pressure breaks the tank)
T2 = ?
p1
T1
p2
= T2
T2
T p2
= 1
p1
295 K 350 kPa
97 k
Pa
= 1064 K = 791°C (maximum)
= Tair
or
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Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
Tair
350 kPa
= 295 K 97 kPa
Tair
= 1064 K = 791°C (maximum)
The maximum final air temperature would have been 791°C.
23. Kinetic molecular theory postulates that pressure increases with temperature (for a fixed quantity and volume of a gas)
because higher temperature means faster-moving molecules, which must then collide more frequently with container
walls, and so with more force per unit area.
Applying Inquiry Skills
24. (a) p1 = 125 kPa
T1 = 300 K
p2 = ?
T2 = 400 K
p1
T1
p2
p2
= T2
T2p1
= T1
400 K 125 kPa
= 300 K
pNe
= 167 kPa
pNe
400 K
= 125 kPa 300 K
= 167 kPa
or
pNe
The final pressure of neon gas is predicted to be 167 kPa.
(b) The difference is: 162 kPa 167 kPa 5 kPa
5
kPa
The % difference is: 100% 3%
167 kP
a
Making Connections
25. (a) p1 = 310 kPa
T1 = 21°C = 294 K
p2 = ?
T2 = 38°C = 311 K
p1
p2
= T1
T2
T2p1
p2
= T1
pair
311 K 310 kPa
= 294 K
= 328 kPa
or
pair
311 K
= 310 kPa 294 K
pair
= 328 kPa
Copyright © 2002 Nelson Thomson Learning
Chapter 9 The Gas State
263
The final total air pressure in the tire will be 328 kPa.
(b) The final gauge pressure reading will be (328 100) kPa = 228 kPa.
(c) If the tire pressure is set when the tire is hot, the pressure will fall as the tire cools, and will be too low at ambient
temperatures. This can cause tire failure and/or premature wear.
PRACTICE
(Page 438)
Understanding Concepts
26. p1 = 101 kPa
v1 = 50.0 mL
p2 = ?
v2 = 12.5 mL
p1v1
= p2v2
p v1
= 1
v2
p2
101 kPa 50.0 mL
12.5 mL
p2
= pgas
= 404 kPa
pgas
50.0 mL
= 101 kPa 12.5 mL
= 404 kPa
or
pgas
The final pressure of the gas in the syringe will be 404 kPa.
27. v1 = 0.10 L
T1 = 25°C = 298 K
v2 = ?
T2 = 190°C = 463 K
v1
v2
= T1
T2
T v1
v2 = 2
T1
vCO
2
or
vCO
2
vCO
2
463 K 0.10 L
= 298 K
= 0.16 L
463 K
= 0.10 L 298 K
= 0.16 L
The final volume of the carbon dioxide will be 0.16 L.
28. p1 = 150 kPa
T1 = 35°C = 308 K
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Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
p2 = 250 kPa
T2 = ?
p1
T1
p2
= T2
T2
T p2
= 1
p1
308 K 250 kPa
= 150 k
Pa
Tbutane
= 513 K = 240°C
Tbutane
308 K 250 kPa
= 150 k
Pa
Tbutane
= 513 K = 240°C
or
The temperature of butane for which the valve will open is 240°C.
29. p1 = 100 kPa
T1 = 20°C = 293 K
v1 = 5.00 L
p2 = 90 kPa
T2 = 35°C = 308 K
v2 = ?
p v1
1
T1
v2
p v2
= 2
T2
T2p1v1
= T1p2
308 K
100 kPa
5.00 L
294 K
90 kPa
= 5.8 L
=
vballoon
or
vballoon
vballoon
308 K
100 kPa
= 5.00 L 294 K
90 kPa
= 5.8 L
The final total volume of the balloon will be 5.8 L.
30. p1 = 800 kPa
T1 = 30°C = 303 K
v1 = 1.0 L
p2 = 100 kPa
T2 = 25°C = 298 K
v2 = ?
p v1
1
T1
pv
= 22
T2
Copyright © 2002 Nelson Thomson Learning
Chapter 9 The Gas State
265
v2
T2p1v1
= T1p2
298 K
800 kPa
1.0 L
303 K
100 kPa
= 7.9 L
=
vHe
or
vHe
vHe
298 K
800 kPa
= 1.0 L 303 K
100 kPa
= 7.9 L
The volume occupied by the helium gas at SATP will be 7.9 L.
31. All the previous calculations are done assuming that all gases behave similarly, and that the relationships used predict
behaviour for any gas.
32. p1 = 6.5 atm
T1 = 10°C = 283 K
v1 = 2.0 mL
p2 = 0.95 atm
T2 = 24°C = 297 K
v2 = ?
p v1
1
T1
p v2
= 2
T2
v2
T2p1v1
= T1p2
297 K
6.5 atm 2.0 mL
283 K
0.95 atm
= 14 mL
= vbubble
or
vbubble
vbubble
297 K
6. 5 atm
= 2.0 mL 283 K
0 .9 5 atm
= 14 mL
The final volume of the bubble will be 14 mL.
33. The most important assumption in all previous calculations is that in every case the amount of gas remains constant.
Making Connections
p
34. (a) As the temperature increases, the pressure must increase: they are directly related, as expressed by = k.
T
(b) Pressure and volume are inversely related, as expressed by pv = k, so if the pressure drops to 1/9 the previous
value, the volume should increase to 9 times its previous value.
Reflecting
35. The appropriate law is chosen by examining which variables change, and which remain constant. The combined gas
law may always be used: unchanged variables simply cancel.
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Unit 4 Gases and Atmospheric Chemistry
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SECTION 9.2 QUESTIONS
(Page 439)
Understanding Concepts
1. Table 6: Using Pressure Units
Pressure
(kPa)
Pressure
(atm)
Pressure
(mm Hg)
(a)
88.7
0.875
665
(b)
25.0
0.247
188
(c)
112
1.11
842
101.325 kPa
(a) 0.875 atm 88.7 kPa
1
atm
760 mm Hg
0.875 atm 665 mm Hg
1
atm
1 atm
0.247 atm
(b) 25.0 kPa
101.325 kPa
760 mm Hg
25.0 kPa
188 mm Hg
101.325 kPa
101.325 kPa
(c) 842 mm Hg 112 kPa
760 mm H
g
1 atm
842 mm Hg 1.11 atm
760 mm Hg
2. Table 7: Celsius and Kelvin
t (°C)
T (K)
(a)
25
298
(b)
–35
238
(c)
39
312
(d)
–65
208
3. (a) inverse
(b) direct
(c) direct
(d) amount
4. (a) p1 = 225 kPa
v1 = 27 L
p2 = 98 kPa
v2 = ?
p1v1 = p2v2
v2
p v1
= 1
p2
Copyright © 2002 Nelson Thomson Learning
Chapter 9 The Gas State
267
225 kPa 27 L
= 98 kPa
= 62 L
vair
or
225 kPa
= 27 L 98 kPa
= 62 L
vair
vair
The final volume of the escaped air will be 62 L.
(b) The new volume is 62/27 = 2.3 times larger than before. The pressure has dropped by the inverse factor, and is
0.44 times its previous value. So, while the pressure has dropped, the volume has increased.
5. p1 = 1.00 atm
T1 = 40.0°C = 313 K
v1 = 500 mL
p2 = 35.0 atm
T2 = ?
v2 = 23.0 mL
p v1
p v2
1
= 2
T1
T2
T2
T1p2v2
= p1v1
313 K 35.0 atm 23.0 mL
= Tfuel = 504 K = 231°C
or
Tfuel
35 .0 atm
23.0 mL
= 313 K 1.00 atm
500 mL
Tfuel = 504 K = 231°C
The final temperature of fuel gases in the cylinder will be 231°C.
Applying Inquiry Skills
6. (a) Hypothesis
The pressure of a fixed volume of a gas is directly proportional to its absolute temperature. The reasoning is that
increasing the temperature increases the average kinetic energy of the molecules which, in turn, increases the
pressure exerted by collisions within the vessel.
(b) Experimental Design
The metal sphere is immersed in a bath of ice water at the freezing point, and is left there until the gas temperature has become equal to the temperature of the water bath. The water bath is slowly raised in temperature, and
the gas pressure is measured at regular temperature intervals.
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Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
(c) Analysis
Converting temperatures in degrees Celsius to kelvin gives:
Table 8: Evidence
Temperature
Pressure
Temperature
(°C)
(kPa)
(K)
0
100
273
20
106
293
40
115
303
60
123
333
80
129
353
100
135
373
Graphing pressure as a dependent variable against temperature as an independent variable gives:
Pressure of Heated Air
150
Pressure (kPa)
140
130
120
110
100
260
280
300
320
340
360
380
Temperature (K)
A graph of the evidence clearly shows that the pressure–temperature relation is very close to a straight line. This
means that, according to the evidence gathered in this investigation, pressure is directly proportional to the absolute
(kelvin) temperature for the gas sample.
(d) Evaluation
The evidence is judged to be adequate. The design and materials are simple and straightforward, and provide
evidence that enables the question to be answered easily and without ambiguity.
The hypothesis is supported by evidence gathered in the investigation and, therefore, verified.
Making Connections
7. (a) p1 = 305 kPa
T1 = 130°C = 403 K
v1 = 1.0 L
p2 = 101 kPa
T2 = 93°C = 366 K
v2 = ?
p v1
1
T1
v2
p v2
= 2
T2
T2p1v1
= T1p2
Copyright © 2002 Nelson Thomson Learning
Chapter 9 The Gas State
269
=
366 K
305 kPa
1.0 L
403 K
101 kPa
vbubble
= 2.7 L
vbubble
305 kPa
366 K
= 1.0 L 101 kPa
403 K
= 2.7 L
or
vbubble
The final total volume of the water vapour bubble will be 2.7 L.
(b) The narrow shaft opening is necessary to restrict flow, so that pressure has the time to build up enough to spray the
water high above the surface.
9.3 COMPRESSED GASES
PRACTICE
(Page 442)
Understanding Concepts
1. (a) p1 = 300 kPa
v1 = 6.0 L
p2 = 100 kPa
v2 = ?
p1v1
v2
vair
or
vair
vair
= p2v2
p v1
= 1
p2
300 kPa 6.0 L
= 100 kPa
= 18 L
300 kPa
= 6.0 L 100 kPa
= 18 L
The air volume at surface conditions will be 18 L.
(b) If you try to hold your breath while rising from a deep dive using scuba gear, the decrease in outside pressure can
result in serious damage to the alveoli in your lungs, caused by the pressure differential inside and outside your
chest.
(c) Temperature is assumed to be constant, as is the amount of gas.
2. The contents of aerosol cans are already under high pressure. If heating raises the temperature, the pressure will
increase proportionally. If the pressure becomes too high, the can may rupture abruptly (explode), which could injure
someone nearby.
Making Connections
3. Typical answers might include careers such as:
Trucks have braking systems that use compressed air to exert very high forces; drivers must be certified to operate
vehicles using such systems.
Welders are trained to use a variety of pressurized gases: acetylene and oxygen that are mixed to form the fuel; inert
gases such as argon that keep air away from a weld so that it does not oxidize.
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Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
Anesthesiologists use a variety of anaesthetic gases in their work, and require training in the precise delivery and
monitoring of such gases.
4. It is very important that pressure regulators not be switched among tanks of various gases, particularly if one of them
is oxygen. A pressure regulator for oxygen tanks, used on another tank, such as propane or hydrogen (or vice versa),
could cause oxygen to mix with a combustible substance at very high pressure, which would almost certainly cause
an explosion. Oxygen regulators are colour coded, and must have a warning attached. Propane tanks are reverse
threaded so that an oxygen regulator cannot be attached to a propane tank (or vice versa), even if you try.
5. The Donald Duck voice made while exhaling helium is due to the speed of sound in helium, which is much higher
than in air. Helium atoms move much faster than air molecules at the same temperature, so the resonant sounds within
the vocal tract become distorted. The danger in this practice is not from (inert) helium; it is that inhaling helium deeply
and repeatedly can cause oxygen deprivation. To make Duck imitations safe, always make the helium inhalation
“shallow,” and take several deep breaths of air immediately afterward.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
SECTION 9.3 QUESTIONS
Making Connections
(Page 443)
1. Automobile tires are the most common commercial product containing compressed air. Others include scuba diving
tanks and portable marine horns. Many commercial products contain other compressed gases. Examples include
propane and butane fuel containers, inert (often nitrogen or argon) propellants in aerosol cans, and natural gas burners
in barbecues, fireplaces, and furnaces.
2. The safety hazard of compressed gases is primarily the danger of rupture (explosion) of the container, caused by overheating or failure of some part, resulting in injury to people nearby.
3. There may be dangers inherent in the substance itself; it may be toxic, flammable, or corrosive, for instance.
9.4 THE IDEAL GAS LAW
PRACTICE
(Page 445)
Understanding Concepts
1. Volume of a gas in an air (pneumatic) shock absorber can be reduced by decreasing the amount, by lowering the
temperature, or by raising the pressure.
2. An ideal gas would have zero attractive force between particles, and zero particle size. Real gases will be closest to
this condition for small entities at very low pressures and very high temperatures, because then their particles will be
farthest apart and moving fastest, minimizing the effect of intermolecular forces. The lower the intermolecular attraction — as in helium atoms, for example — the closer the behaviour to “ideal”.
=?
3.
nCH
4
p
= 210 kPa
T
= 35.0°C = 308.0 K
v
= 500 mL = 0.500 L
R
= 8.31 kPaL/(molK)
pv
nCH4
= nRT
pv
= RT
Copyright © 2002 Nelson Thomson Learning
Chapter 9 The Gas State
271
210 kPa
0.500 L
308.0 K
mol
K
= 8.31 L
kPa nCH4
= 0.0410 mol = 41.0 mmol
The amount of methane gas present in the sample is 41.0 mmol.
4.
m
= 50 kg
p
= 150 kPa
T
= 125°C = 398 K
vO = ?
2
M
= 32.00 g/mol
R
= 8.31 kPaL/(molK)
1 mol
nO = 50 kg 2
32.00 g
nO = 1.56 kmol
2
pv = nRT
nRT
vo = 2
p
8.31 L
kPa
1.56 kmol
398 K
mol K
150 kPa
=
vo
2
= 34 kL or 34 m3
The volume of oxygen gas is 34 kL or 34 m3.
5. (a) mC H
3 8
=?
p
= 96.7 kPa
T
= 22°C = 295 K
v
= 1.00 L (assumed)
M
= 44.11 g/mol
R
= 8.31 kPaL/(molK)
pv
nC H
3 8
= nRT
pv
= RT
96.7 kPa
1.00 L
L
Pa = 8.31 k
K
295 molK
nC H
3 8
mC H
3 8
mC H
3 8
= 0.0394 mol
44.11 g
= 0.0394 mol 1m
ol
= 1.74 g
The mass of 1.00 L is 1.74 g, so the density of propane gas is 1.74 g/L at the conditions specified.
(b) Since propane is denser than air, leaking propane will flow downward and collect in low areas, causing a severe explosion hazard. Propane-powered vehicles are not normally permitted in inside or underground parking lots, for this
reason.
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Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
6.
n
pv
Since M = , and n = , by substituting we arrive at:
m
RT
pv
M = for any gas that behaves approximately like an ideal gas.
m RT
Applying Inquiry Skills
7. (a)
Analysis
Mgas = ?
m = 1.25 g
p = 100 kPa
T = 0°C = 273 K
v = 1.00 L
R= 8.31 kPaL/(molK)
pv = nRT
pv
ngas = RT
100 kPa
1.00 L
kPa 273 = 8.31 L
K
molK
ngas = 0.0441 mol
1.25 g
Mgas = 0.0441 mol
Mgas = 28.4 g/mol
The molar mass of the gas sample is 28.4 g/mol.
Making Connections
pv
8. From the expression pv = nRT, rearranging gives n = RT , which may be interpreted as follows:
If pressure and volume of gas remain constant — as is the case in a hot-air balloon that is a specific size and always
open to the atmosphere — then the amount of gas contained is inversely proportional to the temperature. Warm air,
therefore, will have less gas per unit volume, or lower density, than the same air when cool. This is what makes hotair balloons buoyant.
9. (a)
densityCO
= ?
2
M
VSATP
densityCO
2
densityCO
2
= 44.01 g/mol
= 24.8 mol/L
44.01 g
1
mol
= 1
mol
24.8 L
= 1.78 g/L
The density of carbon dioxide gas is 1.78 g/L at SATP.
(b) Since carbon dioxide is much denser than air, more will be found close to the floor.
(c)
mCO= ?
p
= 98 kPa
T = 20°C = 293 K
v
= 1.00 L (assumed)
M = 28.01 g/mol
Copyright © 2002 Nelson Thomson Learning
Chapter 9 The Gas State
273
R = 8.31 kPaL/(molK)
pv = nRT
pv
nCO = RT
98 kPa
1.00 L
8.3
1
k
L
P
a
= 293 K
mo lK
nCO = 0.040 mol
28.01 g
mCO = 0.040 mol 1
mol
mCO = 1.1 g
The mass of 1.00 L is 1.1 g; thus, the density of carbon monoxide gas is 1.1 g/L at the stated conditions.
(d) Carbon monoxide is less dense than air, so a detector should be placed close to the ceiling.
(e) The temperature of the gases in question may be very high, which will reduce their densities noticeably.
Reflecting
10. The word “ideal” refers in most cases to a mental construct — a concept that may be defined but is not measurable
— reflecting the limit of thought as applied to something. Most scientific formulas, like v = d/t from physics, reflect
an ideal situation (any real speed calculated this way is an approximate average value). Concepts like “pure” when
applied to actual chemical substances are similarly “ideal.”
SECTION 9.4 QUESTIONS
(Page 448)
Understanding Concepts
1. Any real gas has small interparticle attractive forces. These forces will pull the particles together in a condensed state
when the particle motion is slow enough (temperature is low enough).
2. Pressure of a gas sample is determined by its amount (directly), its temperature (directly), and its volume (inversely).
3.
n = 30 mol
pair = ?
T = 40°C = 313 K
v = 50 L
R = 8.31 kPaL/(molK)
pv
pair
= nRT
nRT
= v
8.31 kPaL
30 mol 313 K
mol K
=
50 L
pair
= 1.6 103 kPa or 1.6 MPa
The pressure of air in the cylinder is 1.6 MPa.
4.
m = 10.5 g
p = 85.0 kPa
TNH
3
=?
v = 30.0 L
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Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
M = 17.04 g/mol
R = 8.31 kPaL/(molK)
nNH
1 mol
= 10.5 g 17.04 g
nNH
= 0.616 mol
3
3
pv = nRT
pv
TNH = 3
nR
30.0 L
85.0 kPa
= 8.31 L
kPa 0.616 mol mol K
TNH
3
= 498 K = 225 °C
The temperature of the ammonia in the container is 225 °C.
5.
n = 1.00 mol
100
p = 100 kPa (exactly) = atm (exactly)
101.325
T = 298 K (exactly)
v = 24.8 L
Rgas = ?
pv
Rgas
= nRT
pv
= nT
100
atm 24.8 L
101.325
= 1.00 mol 298 K
Rgas
= 0.0821 atmL/(molK)
The value of the gas constant to three significant digits, calculated from the evidence given, is 0.0821 atmL/(molK).
6. (a) m
= 1.49 g
p
= 117 kPa
T
= 42.0°C = 315.0 K
v
= 981 mL = 0.981 L
Mgas
R
=?
= 8.31 kPaL/(molK)
pv = nRT
pv
ngas = RT
117 kPa
0.981 L
= 8.31 L
kPa 315.0 K
molK
ngas = 0.0438 mol
1.49 g
Mgas = 0.0438 mol
Copyright © 2002 Nelson Thomson Learning
Chapter 9 The Gas State
275
Mgas
= 34.0 g/mol
The molar mass of the gas sample is 34.0 g/mol.
(b) If the compound has the formula XH3, where H represents hydrogen, then the atomic molar mass of “X” must be
(34.0 3(1.01) ) g/mol, which gives 31.0 g/mol. The atom must be phosphorus, P, which is 31.0 g/mol if rounded
to 3 digits, and the gas is therefore PH3(g).
9.5 AIR QUALITY
PRACTICE
(Page 453)
Understanding Concepts
1. The percentage composition by volume of water vapour in air varies widely.
2. The nitrogen in the atmosphere is converted to soluble compounds (useful to plants) by lightning and by nitrogenfixing bacteria. During decay of the plants (or the animals that ate them) other bacteria convert the nitrogen back to
elemental form and release it to the atmosphere.
3. Use of the internal combustion engine is the most significant human activity contributing nitrogen oxides to the atmosphere.
4. Nitrogen dioxide may be decomposed by ultraviolet (UV) light to produce nitrogen oxide and oxygen.
UV light
NO2(g) → NO(g) + O(g)
The atomic oxygen reacts with molecular oxygen to produce ground-level ozone which is toxic to humans.
O(g) + O2(g) → O3(g)
5. (a) Find the total amount of gases in 240 kL of air at SATP:
nair = ?
p
= 100 kPa
T = 25°C = 298 K
v
= 240 kL
R = 8.31 kPaL/(molK)
pv
= nRT
pv
= nair
RT
100 kPa
240 kL
= 8.31 L
kPa 298 K
molK
nair
= 9.69 kmol
Then, using percentage composition values from Table 1:
nN2 = 9.69 kmol 0.7808
= 7.57 kmol
nO2
= 9.69 kmol 0.2095
= 2.03 kmol
nAr
276
= 9.69 kmol 0.00934
Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
Thank you for your attention to this matter. We are prepared to make an oral presentation to Council at any point
in the future. If you see this type of report as being useful, our chemistry class would be willing to continue this work
year after year.
Sincerely,
Dr. Lantz’s Chemistry 11 class
cc. Principal C. Bisset
SECTION 9.5 QUESTIONS
(Page 454)
Understanding Concepts
1. The gases nitrogen, oxygen, argon, and carbon dioxide are the most abundant gases in air — ignoring water vapour.
2. Argon, neon, and krypton are the most abundant noble gases in air.
3. Water molecules are lighter than either oxygen or nitrogen molecules, so moist air should be less dense than dry air.
4. The three main air pollutants are noxious gases, such as sulfur dioxide; vapours from VOCs (volatile organic
compounds); and heavy metals and carcinogens (primarily from coal combustion).
5. Ozone is not a primary air pollutant because it is produced secondarily from the reaction of nitrogen oxides and VOCs.
6. Catalytic converters reduce the emission of nitrogen oxide, carbon monoxide, and unburned hydrocarbon from car engines.
7. The most common strategies to reduce NOx(g) and VOC emissions involve reduced use of automobiles (car pools,
public transport), and in general, any energy-saving strategy that reduces the burning of fossil fuels.
CHAPTER 9 SUMMARY
(Page 455)
MAKE A SUMMARY
1.
Empirical and Theoretical Properties of Solids, Liquids and Gases
State
Solid
Liquid
Gas
simple model
(a) empirical
properties
definite shape and
volume; virtually
incompressible; does
not flow easily
(b) possible
forces present
(at SATP)
(c) type of
motion of the
particles
(d) degree of order
Assumes the shape
and volume of the
container; highly
compressible; flows
easily
network covalent,
ionic, intermetallic,
and London plus
(maybe) dipole and
hydrogen bonding
vibrational
(mostly)
assumes the shape of
the container but has
a definite volume;
virtually
incompressible; flows
easily
intermetallic (e.g.,
Hg(l)), and London
plus (maybe) dipoledipole and hydrogen
bonding
vibrational, rotational,
and translational
least disordered
somewhat disordered
most disordered
Copyright © 2002 Nelson Thomson Learning
van der Waal (dipoledipole and London)
translational
(mostly)
Chapter 9 The Gas State
279
2.
Relationships Between Volume and Pressure, Temperature, and Amount
Variable
Pressure
Temperature
Amount
Graph
inverse
relationship:
double is half
(v vs. p)
direct relationship
above zero
(v vs. T)
direct relationship
(v vs. n)
Equation
v = k1/p and
p1v1 = p2v2
v = k2T and
v1/T1 = v2/T2
v = k3n and
v1/n1 = v2/n2
Variables
controlled
temperature and
amount
pressure and
amount
pressure and
temperature
3. The ideal gas law
pv = nRT
p is the gas pressure in, for example, kilopascals (kPa)
v is the gas volume in, for example, litres (L)
n is the amount in, for example, moles (mol)
R is the gas constant in, for example, kPa·L/(mol·K)
T is the kelvin (absolute) temperature in kelvins (K)
CHAPTER 9
REVIEW
(Page 456)
Understanding Concepts
1. At SATP chlorine is a gas, bromine is a liquid, and iodine is a solid, indicating that larger halogen molecules have
stronger intermolecular forces.
2. (a) Water and ice are both colourless and odourless, have a specific volume, and are negligibly compressible.
(b) Ice has a specific shape; water does not.
(c) Gases (like water vapour) are highly compressible because there are very large spaces between their molecules.
(d) In ice, H2O molecules are very ordered, with specific location and orientation; the only significant molecular
motion is internal vibration. In liquid water, the molecules are free to rotate, and to slowly change location relative to each other; so the molecules are more disordered. In water vapour, rapid molecular change of location is
a much greater effect; so the molecules are very much more disordered.
(e) In ice, H2O molecules are held in a specific location and orientation by hydrogen bonds between molecules. In
liquid water, the kinetic energy of the molecules is higher, overcoming the hydrogen bonding to the extent that
the molecules are free to rotate and to change location relative to each other. In water vapour, the kinetic energy
of the molecules has overcome the hydrogen bonds and the molecules are widely separated from each other, so
the characteristics of this state are based on molecular motion.
(f) The two major components of the atmosphere are nitrogen (about 78%) and oxygen (about 21%). Two minor
components of the atmosphere are argon (about 1%) and water vapour (variable, usually around 1–3%).
(g) Water is a very important component of Earth’s atmosphere because it falls as rain, which is essential for plant
and animal life, and the phase changes of water control the planet’s weather.
3. (a) The volume of a gas sample decreases proportionally with its pressure, assuming the amount and temperature
remain constant.
(b) The volume of a gas sample increases proportionally with its absolute temperature, assuming the amount and
pressure remain constant.
(c) The volume of a gas sample is directly proportional to the product of its amount and its absolute temperature, and
inversely proportional to its pressure.
1
2
4. (a) v 2/3 v, a final volume 2/3 of the original volume.
3
1
2
2
(b) v 4 v, a final volume 4 times the original volume.
1
1
280
Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
1
2
(c) v v, a final volume equal to the original volume.
2
1
5. A law is a descriptive generalization based on evidence; Charles’s Law is a statement of how the temperature of a gas
sample is observed to affect its volume at constant pressure.
A theory is a possible explanation based on reasoning; Charles’s Law may be explained by postulating that gases
expand when warmed because their molecules are moving faster, and thus will exert the same average pressure on
container walls only if there are fewer collisions — that is, if they are farther apart.
101.325 kPa
6. (a) 745 mm Hg 99.3 kPa
760 mm H
g
1 kPa
(b) 150 Pa
= 0.150 kPa
1000 Pa
101.325 kPa
(c) 2.50 atm 253 kPa
1
atm
7. (a) 0°C = (0 + 273) = 273 K
(b) 21°C = (21 + 273) = 294 K
(c) 37°C = (37 + 273) = 310 K
(d) –273°C = (–273 + 273) = 0 K
p1 = 100 kPa
8.
v1 = 28.8 L
p2 = 350 kPa
vH
2
= ?
p1v1 = p2v2
vH
2
vH
2
or
vH
2
vH
2
p v1
= 1
p2
100 kPa 28.8 L
= 350 kP
a
= 8.23 L
100 kPa
= 28.8 L 350 kPa
= 8.23 L
The final volume of hydrogen gas will be 8.23 L, assuming amount and temperature are constant.
1
9. (a) Since the pressure and volume are inversely proportional, the new volume will be: 300 mL 150 mL.
2
(b)
p1 = 125 kPa
T1 = 7°C = 280 K
pCO
2
= ?
T2 = 30°C = 303 K
p1
p2
= T1
T2
pCO
2
T p1
= 2
T1
303 K 125 kPa
= 280 K
pCO
2
= 135 kPa
or
Copyright © 2002 Nelson Thomson Learning
Chapter 9 The Gas State
281
pCO
2
pCO
2
303 K
= 125 kPa 280 K
= 135 kPa
The final pressure of carbon dioxide will be 135 kPa, assuming amount and volume remain constant.
(c) A can of carbonated pop sometimes overflows when opened because bubbles form so rapidly from the saturated
solution, that they can push liquid and foam out of the opening. This is due to the sudden drop in pressure above
the liquid. The volume containing the gas above the liquid changes from small to huge (the size of the atmosphere) when the container is opened.
10.
p1 = 96.7 kPa
T1 = 19.5°C = 292.5 K
p2 = 195 kPa
Tgas = ?
p1
p2
= T1
T2
Tp
Tgas = 12
p1
292.5 K 195 kPa
= Tgas = 590 K = 317°C
or
195 kPa
Tgas = 292.5 K 96.7 kPa
Tgas = 590 K = 317°C
The final gas temperature when the container breaks will be 317°C.
11.
p1
= 600 kPa
T1
= 150°C = 423 K
v1
= 10.0 kL
psteam
= ?
T2
= 110°C = 383 K
v2
= 18.0 kL
p v1
1
T1
p v2
= 2
T2
psteam
T2p1v1
= T1v2
383 K
600 kPa 10.0 kL
423 K
18.0 kL
= 302 kPa
=
psteam
or
psteam
psteam
10.0 kL
383 K
= 600 kPa 18.0 kL
423 K
= 302 kPa
The final pressure of the turbine steam will be 302 kPa.
282
Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
12. (a)
p1
= 103 kPa
v1
= 3000 m3
p2
= 97 kPa
vair
= ?
p1v1
= p2v2
vair
p v1
= 1
p2
103 kPa
3000 m3
97 kPa
vair
= 3.2 103 m3
vair
103 k
Pa
= 3000 m3 97 kPa
or
vair = 3.2 103 m3
The air volume increases by (3.2 3.000) 103 m3 = 2 102 m3.
(b) p1 = 103 kPa
v1 = 100 mL
p2 = 97 kPa
vCH = ?
4
p1v1
= p2v2
vCH
pv
= 11
p2
4
103 kPa 100 mL
vCH
4
vCH
4
or
= 0.11 L
103 kPa
= 100 mL 97 kPa
vCH
= 1.1 102 mL = 0.11 L
4
The final volume of methane gas in the vegetation must be 0.11 L.
(c) If an advancing storm is accompanied by a low-pressure system, then air in caves will increase in volume and the
excess will escape (detectable as wind), and gas bubbles in sunken vegetation might expand enough for it to
become buoyant — thus indicating the oncoming storm.
13.
m = 4.54 kg
p = 96.5 kPa
T = 12°C = 285 K
vC H = ?
3 8
M = 44.11 g/mol
R = 8.31 kPaL/(molK)
nC H
3 8
1 mol
= 4.54 kg 44.11 g
Copyright © 2002 Nelson Thomson Learning
Chapter 9 The Gas State
283
nC H
3 8
= 0.103 kmol
pv
= nRT
vC H
nRT
= p
3 8
8.31 L
kPa
0.103 kmol
285 K
mol K
=
96.5 kPa
vC H
3 8
= 2.53 kL
The volume of propane gas is 2.53 kL, or 2.53 m3.
14.
mHe = ?
p
= 102.7 kPa
T
= 18°C = 291 K
v
= 7.5 5000 3.8 104 L = 38 kL
M
= 4.00 g/mol
R
= 8.31 kPaL/(molK)
pv
= nRT
pv
nHe = RT
=
102.7 kPa
38 kL
8.31 kL
Pa 291 K
mol
K
nHe = 1.6 kmol
4.00 g
1
mol
mHe = 1.6 kmol
mHe = 6.4 kg
The mass of helium gas required is 6.4 kg.
15.
nair
=?
(total amount of the gases present)
p
= 99.5 kPa
T
= 21°C = 294 K
v
= 2.95 m 3.50 m 2.45 m 25.3 m3 = 25.3 kL
R = 8.31 kPaL/(molK)
pv = nRT
pv
nair = RT
99.5 kPa
25.3 kL
8.3
1
L
k
P
a
= 294 K
molK
nair = 1.03 kmol
The amount of air present in the room is 1.03 kmol (of various gases).
Applying Inquiry Skills
16. Hypothesis
(a) According to Boyle’s Law, the volume of gas in the syringe should be inversely proportional to the pressure applied,
and the product of the pressure and volume should be a constant value.
284
Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
Analysis
(b) For each trial the pv product is 5.0 103 kPaL , thus verifying the relationship pv = k (a constant value). This means
that p and v are inversely proportional.
According to the evidence, the pressure and volume of the gas in this investigation are in inverse proportion.
Evaluation
(c) Boyle’s law is supported by the evidence gathered in this investigation. (Even though Boyle’s law is not a hypothesis
any more, it is used under that heading in the format provided herein. Every scientific concept has uncertainty
associated with it; hypotheses have more uncertainty than other concepts.)
17. Experimental Design
A syringe is sealed at the tip and then placed inside a hot-water bath. The temperature of the syringe is gradually
decreased by cooling the water bath. Temperature is the manipulated variable, and gas volume in the syringe is the
responding variable. The results are graphed, and the theoretical temperature at zero volume is found by
extrapolation.
Making Connections
18. Two examples of natural phenomena involving gas laws are:
(i) tornadoes — which are generated when warm air rises and abruptly cools and shrinks, and
(ii) expansion in size or rupture of fish caught in deep water and brought to the surface, as the external pressure
exerted on gases in their bodies decreases.
Two examples of technological products involving gas laws are:
(i) aerosol (spray) cans, which use volume increase caused by pressure decrease as a propellant system, and
(ii) internal-combustion engines, which depend in part on energy from strongly heating gases, thus increasing the
pressure exerted on the piston that drives the crankshaft.
19. Propane tanks have expiry dates, after which they may not be refilled, to forestall the possible problem of corrosion
and leakage of an old tank. Old tanks must be disposed of according to strict regulations, because they almost always
contain some residual propane. This remaining propane may be mixed with air inside the old tank, possibly creating
a dangerous explosive device if the tank is exposed to flame or sparks.
20. Air bags are porous fabric bags that fill with gas extremely rapidly. They can cushion and/or prevent a person’s impact
on the interior of his/her own automobile in a crash, thus preventing or reducing injury and risk of death. Air bags
inflate so rapidly that they can cause injury in individuals who are sitting too close to them. This risk is particularly
serious for children, to the extent that some manufacturers have developed second-generation air bags that have
adjustable inflation rates. Some vehicles (especially pickup trucks) allow the driver to switch off the passenger air bag
circuit when transporting a child in a child safety seat.
21. (a) Some (of many) categories that can be used to classify the components of air are: essential (e.g., the oxygen we
need to breathe); inert (e.g., the argon and nitrogen comprising 4/5 of the atmosphere); important (e.g., the water
vapour content that affects weather); and hazardous (e.g., ground-level ozone, carbon monoxide, or nitrogen
oxide).
(b) An example of an atmospheric gas that may be discussed from several perspectives is carbon dioxide. Some
perspective statements might include:
Scientific — The atmospheric concentration of CO2(g) gas rose 7.4% from 1900 to 1970.
Technological — Power plants designed to burn natural gas produce much less carbon dioxide than those that burn
coal.
Economic — Power plants that produce less carbon dioxide are often much more expensive to run than those that
produce larger amounts.
Environmental — The increase in atmospheric concentration of carbon dioxide may be accelerating the rate of
warming of the atmosphere by enhancing the greenhouse effect, which may in turn be increasing the effects of climate
change on all living things.
Exploring
22. A typical answer might discuss argon (about 1% of the atmosphere), its uses (for example, in light bulbs to prevent
reaction or evaporation of the filament and in welding to prevent corrosion during the joining process), and its
commercial production (by fractional distillation of liquid air).
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
Copyright © 2002 Nelson Thomson Learning
Chapter 9 The Gas State
285
CHAPTER 10 GAS MIXTURES AND REACTIONS
Try This Activity: Producing a “Natural” Gas
(Page 459)
•
White cloudy mixture formed with many gas bubbles rising to the surface. Many droplets of liquid were spitting up
from the surface.
•
The lit match was extinguished.
•
When the glass was tipped to “pour” the gas over the lit match, the match was also extinguished.
(a) Nitrogen, oxygen, and small quantities of other gases are present in the air.
(b) Carbon dioxide is likely present after the reaction, because the gas does not support combustion (a characteristic of
carbon dioxide).
(c) If the gas is bubbled through limewater, and the limewater turns cloudy, then carbon dioxide is likely present.
(d) It suggests that the gas is more dense than air.
(e) Carbon dioxide is used as a fire extinguisher (and also as an effervescent).
(f) Carbon dioxide escapes from underground pockets of gas. Natural decomposition of limestone produces bubbles of
carbon dioxide that escape from the surface of lakes. (Other examples include volcanoes and some hot springs.)
10.1 MIXTURES OF GASES
PRACTICE
(Page 461)
Understanding Concepts
1. The pressure of CO2(g) is constant at 2 kPa.
2. (a) The total pressure is (593.3 + 157 + 11 + 0.5) mm Hg = 762 mm Hg, or the total pressure is (79.11 + 20.9 + 1.5
+ 0.07) kPa = 101.6 kPa.
101.325 kPa
(b) 762 mm Hg 102 kPa
760 mm Hg
The pressures of 760 mm Hg and 102 kPa (which is 101.6 kPa rounded to 3 significant digits) are equivalent.
3. The partial pressure of helium is (14.0 – 1.1) atm = 12.9 atm.
Reflecting
4. (a) Both alloys and the atmosphere are solutions — homogeneous mixtures.
(b) Alloy components are measured by mass, and the law of conservation of mass serves the same purpose for alloys
that Dalton’s law does for gas solutions.
PRACTICE
(Page 463)
Understanding Concepts
5. (a) Pressure exerted by a gas is the total force per unit area of the gas molecules striking the interior surface of the
container.
(b) Particles move in straight lines unless deviated by some outside force. Intermolecular forces are very small (negligible) for most gases, so the molecules travel in (essentially) straight lines.
6. (a) The pressure is directly proportional to the number of gas molecules present — so it should double.
(b) The pressure should still double — the kind of gas molecule is unimportant.
(c) Dalton’s law of partial pressures is based on the concept that the kind of molecule is unimportant — the pressure
depends only on the relative numbers of molecules.
7. If gases react, the number and kind of molecules, and therefore the pressure, will change because of the reaction.
Copyright © 2002 Nelson Thomson Learning
Chapter 10 Gas Mixtures and Reactions
287
PRACTICE
(Page 465)
Understanding Concepts
8. From Table 3, the vapour pressure of water at 23°C is 2.81 kPa, so that will be the partial pressure inside the container.
9. Use the Table 3 value for water vapour pressure at 20°C of 2.34 kPa. The partial pressure of nitrogen will be (98.1 –
2.34) kPa = 95.8 kPa.
10. (a) Calculate as before, using Table 3. The partial pressure of hydrogen will be (92.4 – 3.17) kPa = 89.2 kPa.
(b) p1 = 89.2 kPa
v1 = 275 mL
p2 = 100 kPa
v2 = ?
p1v1
v2
= p2v2
p v1
= 1
p2
89.2 kPa 275 mL
= v2
= 245 mL
or
v2
v2
89.2 kPa
= 275 mL 100 kPa
= 245 mL
The final volume of hydrogen would be 245 mL.
Applying Inquiry Skills
11. Experimental Design
Ammonia is collected by downward displacement of air in a fume hood, since it is much less dense than air.
Note:
In practice, ammonia liquifies easily under only moderate pressure because of hydrogen bonding — so
it is usually collected and transported as a liquid, called anhydrous (waterless) ammonia.
Making Connections
12. The principal gases above the liquid are carbon dioxide and water vapour. The total pressure is noticeably higher than
atmospheric, so must be significantly greater than 100 kPa.
SECTION 10.1 QUESTIONS
(Page 465)
Understanding Concepts
1. (a) The partial pressure of O2(g) is (385 – 240) kPa = 145 kPa.
(b) We assume that the total pressure is the sum of the partial pressures.
2. (a) The total pressure in is (79.3 + 21.3 + 0.040 + 0.67) kPa = 101.3 kPa, and the total pressure out is (75.9 + 15.5
+ 3.7 + 6.2) kPa = 101.3 kPa.
1 atm
(b) 101.3 kPa
= 1.000 atm
101.325 kPa
(c) Since your lungs are open to the atmosphere when breathing, inhaled and exhaled air must begin and end at the
same (ambient atmospheric) pressure. The change in proportions of the gases are biological evidence that animals
need oxygen to live — and chemical evidence that reactions have occurred to produce carbon dioxide and water
vapour.
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Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
3. Dalton’s law works well for any gases that behave similarly to the “ideal” gas — that have small molecules and low
intermolecular forces, and do not react.
4. Dalton’s law can be explained by two concepts: gas particles act independently; and pressure is caused by particle
collisions with the walls of the container.
5. Dalton’s law works perfectly only for “ideal” gases; but to three significant digits, for most common gases, it works
well.
Making Connections
6. Possibilities include the gradual absorption of emitted gases by physical, geological, and biological processes. For
example, carbon dioxide dissolves in water (oceans), and is used by plants to generate sugars through photosynthesis.
The carbon in the gas can be followed through biological processes until it is deposited as sediment and buried, eventually forming limestone.
Reflecting
7. Diagrams drawn by students should show the principle illustrated in Figure 2, in a similar fashion. The adding of
amounts must be shown as proportional to the adding of pressures.
Visual models are much easier for most people to understand than mathematical models.
10.2 REACTIONS OF GASES
PRACTICE
(Page 468)
Understanding Concepts
1. Avogadro’s theory was needed to relate reacting volumes to equations.
2. Avogadro used empirical observations of coefficient values from equations, and reacting volume ratios of gases.
3. C3H8(g)
+
5 O2(g) →
3 CO2(g) +
4 H2O(g)
5.00 L
v
Pressure and temperature conditions equal for all gases measured.
5
vO = 5.00 L 2
1
vO = 25.0 L
2
or
Note:
The text example on page 468 uses a mole ratio for conversion, but the volume ratio shown here is more
appropriate.
5 L O2
vO = 5.00 L
C
H8 3
2
1
LC
3H8
vO = 25.0 L
2
The volume of oxygen required is 25.0 L.
4. 2 NO(g)
+
2 H2(g) →
N2(g)
1.2 L
+
2 H2O(g)
v
Pressure and temperature conditions equal for all gases measured.
1
vN = 1.2 L 2
2
vN = 0.60 L
2
or
1 L N2
vN = 1.2 L
NO 2
2
L
NO
vN = 0.60 L
2
The volume of nitrogen produced is 0.60 L.
Copyright © 2002 Nelson Thomson Learning
Chapter 10 Gas Mixtures and Reactions
289
5. (a) 16 H2S(g)
+
→
8 SO2(g)
3 S8(s)
+
16 H2O(g)
248 kL
v
All gases measured at 250 kPa and 350°C.
8
vSO = 248 kL 2
16
vSO = 124 kL
2
or
8 L SO2
vSO = 248 kL
H
2S 2
16 LH
2S
vSO = 124 kL
2
The volume of sulfur dioxide needed is 124 kL.
(b) S8(s)
+
8 O2(g) →
8 SO2(g)
v
250 kL
All gases measured at 200 kPa and 450°C.
8
vO = 250 kL 2
8
vO = 250 kL
2
or
8 L O2
vO = 250 kL
SO
2 2
8L
S
O2
vO = 250 kL
2
The volume of oxygen required is 250 kL.
(c) 2 SO2(g)
+
O2(g)
→
2 SO3(g)
v
v
325 kL
All gases measured at the same temperature and pressure conditions.
2
vSO = 325 kL 2
2
vSO = 325 kL
2
or
2 L SO2
vSO = 325 kL
SO
3 2
2
L
SO3
vSO = 325 kL
2
The volume of sulfur dioxide required is 325 kL.
1
vO = 325 kL 2
2
vO = 163 kL
2
or
1 L O2
vO = 325 kL
SO
3 2
2L
S
O3
vO = 163 kL
2
The volume of oxygen required is 163 kL.
Reflecting
6. The law of combining volumes is similar to the law of definite proportions in that it shows combination in integer
ratios. It differs in that mass, and not volume, is compared; and also in that it does not apply to different substances
in reactions, but only to compound substances formed from the same elements.
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PRACTICE
(Page 471)
Understanding Concepts
7. vSO
= 50 mL
2
VSATP
= 24.8 L/mol
nSO
=?
2
nSO
2
nSO
2
50 m
L 1 mol
= 24.8 L
= 2.0 mmol
The amount of sulfur dioxide is 2.0 mmol.
8. (a) nNe
= 2.25 mol
VSTP
= 22.4 L/mol
vNe
=?
22.4 L
vNe = 2.25 mol 1
mol
vNe = 50.4 L
The volume of neon at STP is 50.4 L.
(b) p1
= 101.325 kPa
T1
= 0°C = 273 K
p2
= ?
T2
= 35°C = 308 K
p1
p2
= T1
T2
p T2
= 1
T1
p2
101.325 kPa 308 K
= pNe
= 114 kPa
pNe
308 K
= 101.325 kPa 273 K
= 114 kPa
or
pNe
The final pressure of the 50.4 L of warmed neon gas will be 114 kPa.
(c) The tube must be transparent to allow light from the neon to escape, and strong enough to hold the pressure of the gas
when heated.
9. mCO
= 0.13 g
2
VSATP
= 24.8 L/mol
MCO
= 44.01 g/mol
vCO
2
2
=?
nCO
2
nCO
2
1 mol
= 0.13 g 44.01 g
= 0.0030 mol
Copyright © 2002 Nelson Thomson Learning
Chapter 10 Gas Mixtures and Reactions
291
vCO
2
vCO
2
vCO
2
vCO
2
24.8 L
= 0.0030 mol 1
mol
= 0.073 L = 73 mL
or
24.8 L
1
mol
= 0.13 g 1
mol
44.01 g
= 0.073 L = 73 mL
The volume of carbon dioxide at SATP is 73 mL.
10. mC H = 50.0 g
8 18
VSTP
= 22.4 L/mol
MC H
= 114.26 g/mol
vC H
=?
8 18
8 18
1 mol
nC H = 50.0 g 8 18
114.26 g
nC H = 0.438 mol
8 18
22.4 L
vC H = 0.438 mol 8 18
1
mol
vC H = 9.80 L
8 18
or
22.4 L
1
mol
vC H = 50.0 g 8 18
1
mol
114.26 g
vC H = 9.80 L
8 18
The volume of octane vapour at STP would be 9.80 L.
11. m NO
2
= 1.00 Mg
VSATP
= 24.8 L/mol
M NO
= 46.01 g/mol
v NO
=?
2
2
n NO
1 mol
= 1.00 Mg 46.01 g
n NO
= 0.0217 Mmol = 21.7 kmol
v NO
24.8 L
= 21.7 kmol
1
mol
vNO
= 539 kL
v NO
24.8 L
1
mol
= 1.00 Mg 1
mol
46.01 g
v NO
= 0.539 ML = 539 kL
2
2
2
2
or
2
2
The volume of nitrogen dioxide at SATP would be 539 kL or 539 m3.
12. mH O = ?
2
v
= 1.00 L
VSATP = 24.8 L/mol
292
Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
M H O = 18.02 g/mol
2
1 mol
n H O = 1.00 L
2
24.8 L
n H O = 0.0403 mol
2
18.02 g
m H O = 0.0403 mol 2
1
mol
m H O = 0.727 g
2
or
1
mol
18.02 g
m H O = 1.00 L
2
24.8 L
1
mol
m H O = 0.727 g
2
The mass of water required would be 0.727 g.
Applying Inquiry Skills
13. (a) Prediction
The molar volume of oxygen at STP will be 22.4 L/mol, just as for any other gas, according to Avogadro’s theory.
(b) Analysis
mO = (46.84 – 45.79) g = 1.05 g
2
p1
= (95.2 – 2.64) kPa = 92.6 kPa
v1
= 848 mL
T1
= 22°C = 295 K
p2
= 101.325 kPa
vO
=?
T2
= 0°C = 273 K
2
MO = 32.00 g/mol
2
VSTP = ?
nO
1 mol
= 1.05 g 32.00 g
nO
= 0.0328 mol = 32.8 mmol
2
2
pv
p v2
11 = 2
T1
T2
vo
2
T2 p1v1
= T1p2
273 K 92.6 kPa 848 mL
295 K 101.325 kPa
= 717 mL
= vo
2
or
vo
2
vo
2
92.6 kP
a
273 K
= 848 mL 101.325 k
Pa
295 K
= 717 mL
Copyright © 2002 Nelson Thomson Learning
Chapter 10 Gas Mixtures and Reactions
293
The volume of oxygen at STP would be 717 mL. Since this is an amount of 32.8 mmol, the molar volume can now be
calculated as follows:
VSTP
717 m
L
= 32.8 mmol
VSTP
= 21.9 L/mol
According to the evidence from this investigation, the molar volume of oxygen at STP is 21.9 L/mol.
(c) Evaluation
The prediction is judged to be verified, because the value calculated from the evidence is in close agreement.
difference
= 22.4 – 21.9 L/mol = 0.5 L/mol
% difference
0.5 L
/mol
= 100% = 2%
22.4 L
/mol
(d) Evaluation
Avogadro’s theory is supported by the result of this investigation, because the result agrees well with the prediction made from this authority.
Making Connections
14. (a) MCO
= 44.10 g/mol
2
VSATP
densityCO
2
densityCO
2
densityCO
2
= 24.8 L/mol
=?
44.10 g/
mol
= 24.8 L/
mol
= 1.8 g/L (to 2 significant digits)
The density of carbon dioxide at SATP is 1.8 g/L.
(b) Carbon dioxide is denser than air, and so will stay near the base of a fire to displace oxygen to extinguish the fire.
(c) Carbon dioxide also works as an extinguisher because it will not support combustion.
SECTION 10.2 QUESTIONS
(Page 474)
Understanding Concepts
1. (a) Avogadro’s theory states that equal volumes of gases at equal temperature and pressure have equal numbers of
particles.
(b) Avogadro’s theory provides the concept of a molar volume for substances in the gas phase, allowing chemical
calculations to be made easily for reactions with components that are gaseous.
+
5 O2(g) →
4 NO(g)
+
6 H2O(g)
2. 4 NH3(g)
100 L
v
v
v
All gases measured at 800°C and 200 kPa.
5
vO = 100 L 2
4
vO = 125 L
2
or
5 L O2
vO = 100 L
NH3 2
4L
N
H3
vO = 125 L
2
The volume of oxygen required is 125 L.
294
Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
4
vNO = 100 L 4
vNO = 100 L
or
4 L NO
vNO = 100 L
NH3 4
L
NH3
vNO = 100 L
The volume of nitrogen oxide produced is 100 L.
6
vH O = 100 L 2
4
vH O = 150 L
2
or
6 L H2O
vH O = 100 L
NH3 2
4
L
NH3
vH O = 150 L
2
The volume of water vapour produced is 150 L.
(b) 2 NO(g) +
O2(g)
→
v
2 NO2(g)
750 L
All gases measured at 800°C and 200 kPa.
1
vO = 750 L 2
2
vO = 375 L
2
or
1 L O2
vO = 750 L
NO2 2
2L
N
O2
vO = 375 L
2
The volume of oxygen required is 375 L.
(c) 3 NO2(g) +
H2O(l) →
2 HNO3(aq)
100 L
+
NO(g)
v
All gases measured at the same temperature and pressure.
1
vNO= 100 L 3
vNO= 33.3 L
or
1 L NO
vNO= 100 L
NO2 3
L
NO2
vNO= 33.3 L
The volume of nitrogen monoxide produced is 33.3 L.
(d) No prediction can be made from the law of combining volumes, because the nitric acid and the ammonium nitrate
are in aqueous solution, not in gaseous form.
3. (a) nH
= 7.50 mol
2
vH
= ?
VSATP
= 24.8 L/mol
2
vH
2
vH
2
24.8 L
= 7.50 mol 1
mol
= 186 L
The volume occupied by the hydrogen is 186 L.
Copyright © 2002 Nelson Thomson Learning
Chapter 10 Gas Mixtures and Reactions
295
(b) p1
= 100 kPa
T1
= 25ºC = 298 K
v1
= 186 L
p2
= 1.2 kPa
T2
= –47ºC = 226 K
vH
= ?
p v1
1
T1
2
vH
2
p v2
= 2
T2
T2p1v1
= T1p2
226 K
100 kPa
186 L
= 298 K
1.2 kPa
= 1.8 × 104 L = 18 kL
vH
2
or
100 kPa
226 K
= 186 L 1.2 kPa
298 K
4
= 1.8 × 10 L = 18 kL
vH
2
vH
2
The final volume of hydrogen in the balloon will be 18 kL.
4.
vO
= 20% of 20.0 L = 4.0 L
2
VSTP
= 22.4 L/mol
nO
=?
2
1 mol
= 4.0 L
22.4 L
= 0.18 mol
nO
2
nO
2
The amount of oxygen present is 0.18 mol.
5.
m CO
= 1.00 t = 1.00 Mg
2
VSTP
M CO
v CO
2
= 22.4 L/mol
2
= 44.01 g/mol
=?
1 mol
n CO = 1.00 Mg 2
44.01 g
n CO = 0.0227 Mmol = 22.7 kmol
2
22.4 L
v CO = 22.7 kmol
2
1
mol
v CO = 509 kL
2
or
22.4 L
1
mol
v CO = 1.00 Mg 2
1
mol
44.01 g
v CO = 0.509 ML = 509 kL
2
The volume of carbon dioxide at STP would be 509 kL, or 509 m3.
6.
mO
=?
2
vO
2
296
= 1.9 kL
Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
VSATP
= 24.8 L/mol
MO
= 32.00 g/mol
2
1 mol
nO = 1.9 kL
2
24.8 L
nO = 0.077 kmol
2
32.00 g
mO = 0.077 kmol
2
1
mol
mO = 2.5 kg
2
or
32.00 g
1
mol
mO = 1.9 kL
2
24.8 L
1
mol
mO = 2.5 kg
2
The mass of oxygen consumed would be 2.5 kg.
10.3 THE OZONE LAYER
PRACTICE
(Page 479)
Understanding Concepts
1. Ozone intercepts mostly the highest-energy (shorter wavelength) UV radiation from the Sun. Some UV radiation is
absorbed by oxygen to become ozone, and some UV radiation is absorbed by ozone decomposing.
2. CFCs were developed as stable, non-toxic refrigerants, aerosol propellants, and foaming agents.
3. In the upper stratosphere, CFCs initiate reactions that increase the rate of decomposition of ozone.
4. An ozone “hole” is a (misleading) name for a region of very low ozone concentration; it is not a region where there
is no ozone.
5.
uv
CF3
+
Br(g)
→
CFBr2(g)
+
Br(g)
Br(g)
+
O3(g)
→
BrO(g)
+
O2(g)
BrO(g)
+
O(g)
→
Br(g)
+
O2(g)
6. Ozone depletion is less severe in the Arctic because it is not as cold as the Antarctic, and because there is more air
mixing due to prevailing winds.
7. Suntanning time should be decreased proportionally to a drop in the level of ozone in the stratosphere. Current
medical thinking is that there is no absolutely “safe” level of sunlight exposure; so the perceived benefits of outdoor
activities — and particularly of deliberate tanning — must be weighed against the increased risks of skin damage and
skin cancer.
SECTION 10.3 QUESTIONS
(Page 480)
Understanding Concepts
1. The Montreal Protocol is an agreement among nations to decrease the production and use of CFCs, to try to prevent
damage to stratospheric ozone.
2. Freeon-12 is a CFC refrigerant that is routinely recycled.
3. HFE can replace CFCs for many uses, and hydrocarbons can be used as refrigerants.
4. Canada’s Arctic Observatory and National Research Council contribute to research on effects of CFCs and the
development of alternative substances.
5. Stratospheric ozone helps us by filtering potentially harmful UV radiation; but at low levels of the atmosphere (in the
air we breathe) ozone is dangerous — a very reactive and toxic substance.
Copyright © 2002 Nelson Thomson Learning
Chapter 10 Gas Mixtures and Reactions
297
10.4 GAS STOICHIOMETRY
PRACTICE
(Page 483)
Understanding Concepts
1. 2 CH3OH(l) +
3 O2(g) →
15 g
2 CO2(g) +
4 H2O(g)
v
32.05 g/mol
nCH OH =
3
nCH OH =
3
22.4 L/mol (STP)
1 mol
15 g 32.05 g
0.47 mol
3
0.47 mol 2
0.70 mol
22.4 L
0.70 mol 1
mol
16 L
nO
=
nO
=
vO
=
vO
=
vO
1
mol C
H3OH
3
mol O2
22.4 L O
= 15 g CH
2
3OH 32.05 gC
H3OH 2 mol C
H3OH 1 mol O2
= 16 L
2
2
2
2
or
2
vO
2
The volume of oxygen needed is 16 L
2. 2 NaCl(l
→
2 Na(s) +
105 kg
nNa
=
nNa
=
nCl
2
=
nCl
2
vCl
2
vCl
2
vCl
2
vCl
2
or
=
22.99 g/mol
1 mol
105 kg 22.99 g
4.57 kmol
1
4.57 kmol 2
2.28 kmol
Cl2(g)
v
24.8 L/mol (SATP)
24.8 L
= 2.28 kmol
1
mol
= 56.6 kL
1
mol Cl2
24.8 L Cl2
a
1m
ol N
= 105 kgg Na
22.99 g Na
2
mol N
a
1
mol Cl2
= 56.6 kL
The volume of chlorine produced is 56.6 kL, or 56.6 m3.
3.
C3H8(g) +
5 O2(g) →
3 CO2(g) +
8 H2O(g)
m
v
44.11 g/mol
24.8 L/mol (SATP)
vO
2
nO
2
298
= 125 L 20% 25 L
1 mol
= 25 L
24.8 L
Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
nO
= 1.0 mol
2
1
= 1.0 mol 5
= 0.20 mol
44.11 g
= 0.20 mol 1m
ol
= 8.9 g
nC H
3 8
nC H
3 8
mC H
3 8
mC H
3 8
or
1
mol O2
1
mol C
44.11 g C3H8
3H8
= 25 L
O2 24.8 L
O2
5
mol O2
1m
ol C
3H8
= 8.9 g
mC H
3 8
mC H
3 8
The mass of propane that can be burned is 8.9 g.
Applying Inquiry Skills
4. (a) Prediction
→
2 H2O2(aq)
2 H2O(l) O2(g)
50.0 mL
v (lab p & T conditions)
0.88 mol/L
nH O
2 2
nH O
2 2
nO
2
nO
2
0.88 mol
= 50.0 mL
1
L
= 44 mmol
1
= 44 mmol 2
= 22 mmol
p = (94.6 – 2.49) kPa = 92.1 kPa (corrected for water vapour pressure)
T = 21°C = 294 K
vO = ?
2
R = 8.31 kPa L/(mol K)
pv
vO
2
= nRT
nRT
= p
8.31 k
Pa L
22 mmol
294 K
m
ol K
= 92.1 kPa
= 5.8 103 mL = 0.58 L
vO
2
or
vO
0.88 mol H
1
mol O2
8.31 kPa L O2
294 K
2O2
= 50.0 mL
H
2O2 1L
H
O
2
m
o
l
H
O
1
m
o
l
O
K
92
.
1
k
Pa
2 2
2 2
2
vO
= 5.8 103 mL = 0.58 L
2
2
According to the ideal gas law, the volume of oxygen at room pressure and temperature is predicted to be 0.58 L.
(b) Analysis
According to the evidence, the volume of oxygen gas produced at room conditions is 556 mL = 0.556 L.
(c) Evaluation
difference
= 0.556 – 0.58L = 0.02 L
Copyright © 2002 Nelson Thomson Learning
Chapter 10 Gas Mixtures and Reactions
299
0.02 L
% difference = 100% = 4%
0.58 L
The evidence agrees well with the predicted value, within 4%.
(d) Evaluation
The ideal gas law is judged to be verified by this investigation, since the evidence agrees well with the prediction.
Making Connections
5. (a) 2 H2(g) O2(g) → 2 H2O(g)
300 L
All gases measured at 40°C and 1.50 atm.
1
vO = 300 L 2
2
vO = 150 L
2
or
1 L O2
vO = 300 L
H2 2
2
L
H2
vO = 150 L
2
The volume of oxygen required is 150 L.
(b) A vehicle burning hydrogen fuel by using oxygen from the air (which contains nitrogen) could still produce NOx
pollutants if the combustion temperature were high enough.
SECTION 10.4 QUESTIONS
(Page 486)
Understanding Concepts
1. 2 Fe(s)
3 H2SO4(aq)
→
Fe2(SO4)3(aq)
10 g
3 H2(g)
v
55.85 g/mol
nFe
nFe
nH
2
22.4 L/mol (STP)
1 mol
= 10 g 55.85 g
= 0.18 mol
3
= 0.18 mol 2
nH = 0.27 mol
2
22.4 L
nH = 0.27 mol 2
1
mol
nH = 6.0 L
2
or
3
mol H
22.4 L H
e
1m
ol F
nH = 10 g Fe
2 2
2
55.85 g Fe
2
mol Fe
1
mol H2
nH = 6.0 L
2
The volume of hydrogen produced will be 6.0 L.
300
Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
2. CH4(g)
2 O2(g) →
+
2.00 ML, 0°C, 120 kPa
CO2(g) +
v (SATP)
vO
2
= 2.00 ML 1
vO
= 4.00 ML (at 0°C = 273 K, 120 kPa)
p1
= 120 kPa
T1
= 0ºC = 273 K
v1
= 2.00 ML
p2
= 100 kPa
T2
= 25ºC = 298 K
2
2
vO
2 H2O(g)
= ?
2
p v1
1
T1
vO
2
p v2
= 2
T2
T2p1v1
= T1p2
=
298 K
120 kPa 2.00 ML
273 K
100 kPa
vO
= 2.62 ML
vO
2 L O2
120 kPa
298 K
= 2.00 ML
CH4 1
L
CH4
100 kPa
273 K
= 2.62 ML
2
or
2
vO
2
The SATP volume of oxygen required is 2.62 ML, or 2.62 103 m3.
3. 2 NH3(g)
H2SO4(aq)
→
(NH4)2SO4(s)
75.0 kL, 10°C = 283 K, 110 kPa
m
R = 8.31 kPa L/(mol K)
132.16 g/mol
pv
nNH
3
= nRT
pv
= RT
110 kPa
75.0 kL
8.
31
k
P
a
L
= 283 K
K
1 mol nNH
3
n(NH ) SO
4
n(NH ) SO
4
4 2
4 2
m(NH ) SO
4
m(NH ) SO
4
m(NH ) SO
4
4 2
4 2
or
4 2
= 3.51 kmol
1
= 3.51 kmol 2
= 1.75 kmol
132.16 g
= 1.75 kmol
1
mol
= 232 kg
1 mol
NH
100 kPa
3 = 75.0 kL
NH3 8.31 kPa
1
L
NH3
283 K
Copyright © 2002 Nelson Thomson Learning
Chapter 10 Gas Mixtures and Reactions
301
(continued)
1
mol (NH4)2SO4
132.16 g (NH4)2SO4
m(NH ) SO
= 232 kg
4 2
4
The mass of ammonium sulfate that can be produced is 232 kg.
4. CH46H2O(s) →
CH4(g) 6 H2O(l)
1.0 kg
v (20°C = 293 K, 95 kPa)
124.17 g/mol
nCH 6H O
4
2
nCH 6H O
4
2
nCH
4
nCH
4
R = 8.31 kPa L/(mol K)
1 mol
= 1.0 kg 124.17 g
= 0.00805 kmol
1
= 0.00805 kmol 1
= 0.00805 kmol
pv = nRT
nRT
vCH = 4
p
vCH
4
or
vCH
4
8.31 k
Pa L
0.00805 kmol
293 K
m
ol = 95 kPa
= 0.21 kL
1 mol
CH
46H2O
= 1.0 kg CH
46H2O (continued)
vCH = 0.21 kL
1
mol CH4
1
mol CH
46H2O
8.31 kPa L CH
293 K
4 1
mol CH4 K
95 kPa
4
The volume of methane gas produced is 0.21 kL, or 0.21 m3.
Making Connections
5. One consumer reaction that produces and consumes gases is the burning of propane in an outdoor barbeque. The reaction is:
C3H8(g)
+
5 O2(g) →
3 CO2(g)
+
8 H2O(g)
One industrial reaction that consumes a gas is the production of ammonium sulfate fertilizer. The reaction is:
2 NH3(g)
H2SO4(aq)
→
(NH4)2SO4(s)
One laboratory reaction that produces and consumes gases is the burning of methane (natural gas) in a lab burner. The
reaction is:
CH4(g)
+
2 O2(g) →
CO2(g) +
2 H2O(g)
Reflecting
6. (a) All stoichiometry involves the numerical ratios of reactants and products, that is, the ratios of amounts in moles.
(b) In an industry using a chemical reaction, knowledge of stoichiometry is essential to determine how much of
reactant to purchase and use. For example, in the reaction
2 NH3(g) H2SO4(aq) → (NH4)2SO4(s)
used to make fertilizer, the reacting amounts of ammonia and sulfuric acid would have to be calculate beforehand.
(c) Adding “just the right amount” of baking powder to a baking mix is an example of consumer use of
stoichiometry.10.5 APPLICATIONS OF GASES
302
Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
CHAPTER 10 SUMMARY
MAKE A SUMMARY
(Page 491)
Evidence and Explanation
The law is explained by kinetic
molecular theory (e.g. each
molecule has the same kinetic
energy) and intermolecular force
theory (e.g. in a gas, there are
negligible intermolecular forces).
Empirical Description
Dalton’s
law of partial
pressures
The sum of individual
pressures equals total
pressure.
Quantitative work in the laboratory
indicates that individual pressures
add to the total measured
pressure.
Each gas acts
independently.
Different gas molecules exert the
same pressure (because they all
have the same average kinetic
energy at the same temperature).
Mixtures of gases are composed
of different molecules.
Molecules react in ratios that
create new molecules that follow
their bonding capacity.
This law is explained by atomic,
molecular, and bonding theory.
Copyright © 2002 Nelson Thomson Learning
mixtures
of
gases
Gases react in simple
whole-number ratios
by volume.
law of
combining
volumes
Applications and Careers
The law provides understanding
in careers such as commercial
diver, anesthetic technician,
meteorologist, and welder.
Natural phenomena explained by
this law include Earth’s atmosphere.
Technological applications
explained by this law include
gasoline vapour–air mixture in
automobile engines (before
reacting) and exhaust system gas
mixtures (after reaction).
Mixtures of gases may react
(according to the law of combining
volumes) or not (according to
Dalton’s law of partial pressures).
Natural gas, propane, and
acetylene react with an optimal
fuel-to-air mixture.
This law provides understanding
in careers such as automobile
designer and mechanic, chemical
engineer, welder, and barbecue
designer.
Chapter 10 Gas Mixtures and Reactions
305
CHAPTER 10 REVIEW
(Page 492)
Understanding Concepts
1. (a) The total pressure of a mixture of nonreacting gases is equal to the sum of the partial pressures of the individual
gases in the mixture.
(b) p total = p1 + p2 + p3 + …
2. Dalton’s law of partial pressures can be explained by these two concepts from the kinetic molecular theory:
(i) Gas pressure is caused by the collisions of particles (molecules, atoms, ions) with the walls of the container.
(ii) Gas molecules essentially act independently of each other.
Therefore, the total pressure (total of the collisions with the walls) is the sum of the individual pressures
(collisions of only one kind of particle) of each gas present.
Note: Point (ii) presupposes, of course, that students know/assume that all particles at the same
temperature have the same average kinetic energy.
3. p total = (230 + 13 + 7) kPa = 250 kPa
4. p O = {100 (exactly) – 3.17} kPa = 96.83 kPa
2
5. (a)When measured at the same temperature and pressure, volumes of gaseous reactants and products of chemical reactions are found to be (to three significant digits) in simple ratios of whole numbers.
(b) 4 C7H5(NO2)3(s) 21 O2(g) → 28 CO2(g) 6 N2(g) 10 H2O(g)
5.00 L
Pressure and temperature conditions equal for all gases measured.
28
vCO
= 5.00 L 2
21
vCO
= 6.67 L
2
or
vCO
2
vCO
2
28 L CO2
= 5.00 L
O2 21 L O
2
= 6.67 L
The volume of carbon dioxide produced is 6.67 L.
6
vN
= 5.00 L 2
21
vN
= 1.43 L
2
or
vN
2
vN
2
6 L N2
= 5.00 L
O2 21 L
O
2
= 1.43 L
The volume of nitrogen produced is 1.43 L.
10
vH O
= 5.00 L 2
21
vH O
= 2.38 L
2
or
vH O
2
10 L H2O
= 5.00 L
O2 21 L
O2
vH O
= 2.38 L
2
The volume of water vapour produced is 2.38 L.
306
Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
6. Avogadro’s idea is theoretical. It attempts to explain observed (empirical) gas behaviour by proposing that equal
volumes of gases at the same temperature and pressure contain equal numbers of molecules. This proposal is a product
of intellect, not observation, and so is a theory, not a law.
7. 2 NH4NO3(s) →
2 N2(g) 4 H2O(g) O2(g)
1.00 mol
vSATP
nN
2
nN
2
vN
2
vN
2
or
vN
2
vN
2
all gases:
2
= 1.00 mol 2
= 1.00 mol
24.8 L
= 1.00 mol 1
mol
= 24.8 L
vSATP
vSATP
24.8 L/mol (SATP)
2m
ol N
2
24.8 L N
= 1.00 mol NH
2
4NO3 2
mol N H4
NO3
1
mol N2
= 24.8 L
The volume of nitrogen produced is 24.8 L.
4
nH O = 1.00 mol 2
2
nH O = 2.00 mol
2
24.8 L
vH O = 2.00 mol 2
1
mol
vH O = 49.6 L
2
or
vH O
2
vH O
2
2m
ol H2O
24.8 L H2O
= 1.00 mol NH
4NO3 2m
ol NH4
NO3
1
mol H2O
= 49.6 L
The volume of water vapour produced is 49.6 L.
1
nO = 1.00 mol 2
2
nO = 0.500 mol
2
24.8 L
vO = 0.500 mol 2
1
mol
vO = 12.4 L
2
or
vO
2
vO
2
1
mol O
2
24.8 L O
= 1.00 mol NH
2
4NO3 2
mol NH
1
mol O2
4NO3
= 12.4 L
The volume of oxygen produced is 12.4 L.
The total volume of gases produced is (24.8 + 49.6 + 12.4) L = 76.8 L.
8. (a)
nAr = ?
p
= (100.0 + 0.400) kPa = 100.4 kPa
T
= 20°C = 293 K
v
= 125 mL = 0.125 L
R
= 8.31 kPa L/(mol K)
pv
= nRT
Copyright © 2002 Nelson Thomson Learning
Chapter 10 Gas Mixtures and Reactions
307
nAr
pv
= RT
100.4 kPa
0.125 L
8.31 k
Pa L
293 K
mol K
nAr
= 0.00515 mol = 5.15 mmol
The amount of argon gas in a bulb is 5.15 mmol.
=
(b) p1
= 100.4 kPa
T1
= 20°C = 293 K
p2
= ?
T2
= 200°C = 473 K
p1
p2
= T1
T2
T2p1
p2
= T1
473 K
100.4 kPa
= 293 K
= 162 kPa
pAr
or
473 K
= 100.4 kPa 293 K
= 162 kPa
pAr
pAr
(c)
The final argon pressure at the higher temperature will be 162 kPa.
nAr = ?
p
= 100.4 kPa
T
= 20°C = 293 K
v
= 0.915 L
R
= 8.31 kPa L/(mol K)
pv
nAr
= nRT
pv
= RT
100.4 kPa 0.915 L
293 K
mol K
Pa L
= 8.31 k
nAr
= 0.0377 mol (in each fluorescent tube)
nAr
=?
mAr
= 50 kg
MAr
= 39.95 g/mol
nAr
nAr
#tubes
308
1 mol
= 50.0 kg 39.95 g
= 1.25 kmol (in the steel tank)
1 tube
= 1.25 kmol
0.0377 mol
Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
= 33.2 ktubes (3.32 104 tubes)
#tubes
9. 2 ZnS(s)
3 O2(g) →
2 ZnO(s) 2 SO2(g)
1.00 t = 1.00 Mg
vSATP
97.44 g/mol
24.8 L/mol
1 mol
= 1.00 Mg 97.44 g
= 0.0103 Mmol = 10.3 kmol
2
= 10.3 kmol 2
= 10.3 kmol
24.8 L
= 10.3 kmol
1
mol
= 255 kL = 255 m3
nZnS
nZnS
nSO
2
nSO
2
vSO
2
vSO
2
or
vSO
2
vSO
2
2
mol SO2
24.8 L SO
1
mol Z
nS
= 1.00 Mg ZnS
2
97.44 g ZnS
2
mol ZnS
1
mol SO2
= 0.255 ML = 255 kL = 255 m3
The volume of sulfur dioxide produced will be 255 kL, or 255 m3.
10.
2 CaCO3(s)
2 SO2(g) O2(g)
→
2 CaSO4(s)
m
500 kL (STP)
100.09 g/mol
(a)
vCO
2
vCO
2
2 CO2(g)
vSTP
22.4 L/mol
2
= 500 kL 2
= 500 kL
or
vCO
2
vCO
2
2 L CO2
= 500 kL
SO
2 2L
SO2
= 500 kL
The volume of carbon dioxide gas produced at STP is 500 kL.
1 mol
(b)
nSO
= 500 kL
2
22.4 L
nSO
= 22.3 kmol
2
2
nCaCO
= 22.3 kmol 3
2
nCaCO
= 22.3 kmol
3
100.09 g
mCaCO
= 22.3 kmol
3
1
mol
mCaCO
= 2.23 × 103 kg = 2.23 Mg = 2.23 t
3
or
mCaCO
3
mCaCO
3
1 S
mol
O2
2
mol C
aCO
100.09 g CaCO
= 500 kL
SO
2 3 3
22.4 LS
O2
2
mol SO2
1m
ol C
aCO3
= 2.23 × 103 kg = 2.23 Mg = 2.23 t
The mass of calcium carbonate consumed is 2.23 t
11. SO3(g)
H2O(l)
→
1.00 t = 1.00 Mg
Copyright © 2002 Nelson Thomson Learning
H2SO4(aq)
v
Chapter 10 Gas Mixtures and Reactions
309
80.06 g/mol
0.12 mmol/L
1 mol
nSO = 1.00 Mg 3
80.06 g
nSO = 0.0125 Mmol
3
nH SO
2
4
nH SO
2
4
vH SO
2
4
vH SO
4
2
1
= 0.0125 Mmol 1
= 0.0125 Mmol
1L
= 0.0125 Mmol
0.12 mmol
= 1.0 × 108 L = 0.10 GL
or
vH SO
2
4
vH SO
4
2
1
mol S
O3
1
mol H
1 L H2SO4
2SO4
= 1.00 Mg SO
3 80.06 g SO3
1m
ol S
O3
0.12 m mol H
2SO4
= 1.0 × 108 L = 0.10 GL
The volume of sulfuric acid that could be formed is 0.10 GL.
12. 2 H2O(l)
→
2 H2(g) O2(g)
50.0 mL
v
18.02 g/mol
(a) vO
2
vO
2
Both gases measured at 23°C = 296 K and 103 kPa.
1
= 50.0 mL 2
= 25.0 mL
or
1 L O2
vO = 50.0 mL
H2 2
2L
H2
vO = 25.0 mL
2
(b)
The volume of oxygen gas produced is 25.0 mL.
p1
= 103 kPa
T1
= 23°C = 296 K
v1
= 50.0 mL
p2
= 101.325 kPa
T2
= 0°C = 273 K
vH
2
= ?
p v1
1
T1
vH
2
p v2
= 2
T2
T2p1v1
= T1p2
273 K 103 kPa 50.0 mL
296 K 101.325 kPa
= 46.9 mL
= vH
2
or
vH
2
310
273 K
103 kPa
= 50.0 mL 296 K
101.325 kPa
Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
vH
= 46.9 mL
2
The volume of the hydrogen at STP would be 46.9 mL.
(c)
Note: The mass of water may be calculated at this point in this question from the volume of either hydrogen or
oxygen, and if using hydrogen values, from the volume at either set of pressure and temperature conditions.
For best accuracy, and to avoid perpetuating possible errors in preceding steps, calculating from original values is usually preferable, where possible.
nH = ?
2
p
= 103 kPa
T
= 23°C = 296 K
v
= 50.0 mL
R
= 8.31 kPa L/(mol K)
pv
= nRT
pv
= RT
nH
2
103 kPa 50.0 mL
296 K
mol K
Pa L
= 8.31 k
nH
= 2.09 mmol
2
2
nH O = 2.09 mmol 2
2
nH O = 2.09 mmol
2
18.02 g
mH O = 2.09 mmol
2
1
mol
m
= 37.7 mg
H2O
or
mH O
2
1m
ol H2 K
2
mol H2O
18.02 g H2O
103 kPa
= 50.0 mL H2 296 K
8.31 kP
a 1 L
H2
2
mol H2
1m
ol H
2O
mH O = 37.7 mg
2
The mass of water decomposed is 37.7 mg.
13. According to Dalton’s law of partial pressures and Avogadro’s theory, the partial pressure of any gas in a mixture must
be proportional to its mole fraction. Since the mixture has 2.00 mol N2(g) and 3.00 mol H2(g), totalling 5.00 mol; and
the total pressure is 200 kPa;
3.00 m
ol
pH = 200 kPa 2
5.00 m
ol
pH = 120 kPa
2
pN = ptotal – pH = (200 – 120) kPa = 80 kPa
2
14. a)
b)
2
The partial pressure of nitrogen is 80 kPa; and the partial pressure of hydrogen is 120 kPa.
When temperature increases, the molar volume of a gas increases.
When pressure increases, the molar volume of a gas decreases.
Applying Inquiry Skills
15. (a) Prediction
The molar volume of propane at STP will be 22.4 L/mol, just as for any other gas, according to Avogadro’s theory.
(Assume propane behaves like the ideal gas.)
Copyright © 2002 Nelson Thomson Learning
Chapter 10 Gas Mixtures and Reactions
311
(b) Analysis
= (426.79 – 424.92) g = 1.87 g
= (98.23 – 2.49) kPa = 95.74 kPa
= 1065 mL
mC H
3 8
p1
v1
T1
= 21.0°C = 294 K
p2
= 101.325 kPa
vC H
=?
3 8
T2
= 0°C = 273 K
MC H
= 44.11 g/mol
3 8
VSTP
=?
nC H
3 8
nC H
3 8
p v1
1
T1
vC H
3 8
1 mol
= 1.87 g 44.11 g
= 0.0424 mol = 42.4 mmol
p v2
= 2
T2
T2 p1v1
= 273 K
95.74 kPa 1065 mL
294 K
101.325 kPa
= 934 mL
= vC H
3 8
or
vC H
3 8
vC H
3 8
95.74 kPa
273 K
= 1065 mL 101.325 kPa
= 934 mL
The volume of propane at STP would be 934 mL. Since this is an amount of 42.4 mmol, the molar volume can now
be calculated as follows:
VSTP
934 m
L
= 42.4 mmol
VSTP
= 22.0 L/mol
According to the evidence from this investigation, the molar volume of propane at STP is 22.0 L/mol.
(c) Evaluation
difference
= 22.4 – 22.0 L/mol = 0.4 L/mol
0.4 L
/mol
= 100% = 2%
22.4 L
/mol
The prediction is judged to be verified, because the value calculated from the evidence is in close agreement with it,
to within 2%.
% difference
(d) Evaluation
Avogadro’s theory is supported by the result of this investigation, because the result agrees well with the prediction
made from this authority.
16. Experimental Design
A cylinder of compressed gas is used to release a noble gas, which is collected at ambient conditions by water displacement. The amount is calculated from volume, temperature, and pressure measurements. The molar mass is calculated from the amount and the measured mass, and is used to identify the gas.
312
Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning
17. One natural and one technological use or source for each of the following gases:
(a) oxygen
from plant respiration
for welding
(b) methane
from plant decomposition
for fuel
(c) helium
from natural gas wells
for balloons
(d) air
Earth’s atmosphere
source of gases
(e) water vapour
from animal respiration
for humidifiers
(f) carbon dioxide
from combustion
fire extinguishers
Making Connections
18. (a) Freons were initially used as refrigerants, and soon became popular as aerosol propellants, and then later as the
foaming agent for making foam plastics, and as a non-stick, non-toxic solvent for use by the electronics industry.
(b) The production of Freons is banned in many countries, including Canada, because these compounds react in the
upper atmosphere to produce chlorine atoms — which, in turn, catalyze the decomposition of ozone. This
increases environmental damage from ultraviolet rays.
(c) mCF Cl
2
MCF Cl
2
VCF Cl
2
nCF Cl
2
nCF Cl
2
vCF Cl
2
vCF Cl
2
vCF Cl
2
vCF Cl
2
2
2
2
2
2
2
2
= 1.00 kg
= 120.91 g/mol
= 24.8 L/mol (SATP)
1 mol
= 1.00 kg 120.91 g
= 0.00827 kmol = 8.27 mol
24.8 L
= 8.27 mol 1
mol
= 205 L
or
2
2
24.8 L
1
mol
= 1.00 kg 1
mol
120.91 g
= 0.205 kL = 205 L
The volume of Freon at SATP would be 205 L.
19.
2 C2H5SH(g)
9 O2(g)
→
1.00 g
62.14 g/mol
nC H SH
2 5
nC H SH
2 5
nCO
2
nCO
2
vCO
2
vCO
2
vCO
2
vCO
2
or
4 CO2(g) 6 H2O(g)
v
v
2 SO2(g)
v
All gases: 24.8 L/mol (SATP)
1 mol
= 1.00 g 62.14 g
= 0.0161 mol = 16.1 mmol
4
= 16.1 mmol 2
= 32.2 mmol
24.8 L
= 32.2 mmol
1
mol
= 798 mL
1
mol C2
H5SH
4m
ol C
O2
24.8 L CO2
= 1.00 g C2
H5SH 62.14 g C2
H5SH 2 mol C2
H5SH
1
mol CO2
= 0.798 L = 798 mL
The volume of carbon dioxide produced is 798 mL.
6
nH O = 16.1 mmol 2
2
Copyright © 2002 Nelson Thomson Learning
Chapter 10 Gas Mixtures and Reactions
313
nH O
= 48.3 mmol
vH O
24.8 L
= 48.3 mmol
1
mol
vH O
= 1.20 103 mL = 1.20 L
2
or
2
2
1m
ol C2H
6m
ol H2O
24.8 L H2O
5SH
vH O= 1.00 g C2
H5SH 2
62.14 g C2H
2m
ol C2H
1
mol H2O
5SH
5SH
vH O
2
= 1.20 L
The volume of water vapour produced is 1.20 L.
nSO
2
nSO
2
vSO
2
vSO
2
or
2
= 16.1 mmol = 16.1 mmol 2
24.8 L
= 16.1 mmol
1
mol
= 399 mL
1
mol C2H
2
mol SO2
24.8 L SO
5SH
vSO = 1.00 g C2
H5SH 2
2
62.14 g C2H
2
mol C2H
1
mol S
O2
5SH
5SH
vSO
2
= 0.399 L = 399 mL
The volume of sulfur dioxide produced is 399 mL.
(b) Ethanethiol added to natural gas increases safety when the natural gas is used as a heating fuel, since any leak can be
detected quickly, before the area becomes an explosion hazard. However, the compound is hazardous to handle in
undiluted form before it is mixed with natural gas, and it produces the pollutant sulfur dioxide when burned.
20. A typical answer might include discussion of the production of nitrogen oxides, symbolized NOX, by heavy vehicle
traffic in metropolitan areas. NOX encourages smog formation, reacting with sunlight and oxygen to cause an increase
in ground-level ozone. Ozone, O3(g), is thought to be the primary agent causing smog damage to vegetation, and is a
toxic and irritant gas to humans, interfering with lung function.
UNIT 4 PERFORMANCE TASK
Report
The report will require extensive student research, the results of which cannot be simulated here. The report should cover
the main points:
• historical development of the technology;
•
scientific principles used in the technology, especially those relating to gases;
•
risks and benefits (to society and the environment) of use of the technology;
•
a summary of a related career.
UNIT 4 REVIEW
(Page 496)
Understanding Concepts
1. (a) The intermolecular forces in solid carbon dioxide must be very weak, because the molecules separate completely
at very low temperature.
314
Unit 4 Gases and Atmospheric Chemistry
Copyright © 2002 Nelson Thomson Learning