Solution Manual For Introduction to General, Organic, and Biochemistry, 12th Edition

Name: Solution Manual For Introduction to General, Organic, and Biochemistry, 12th Edition
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Solution Manual For Introduction to General, Organic, and Biochemistry, 12th Edition condenses textbook information into an easy-to-follow study resource.

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1
Chapter 1: Matter, Energy, and Measurement
QC 1.1 Multiplication: (a) 4.69  105 (b) 2.8  10-15
Division: (a) 1.94  1018 (b) 1.37  105
QC 1.2 (a) o F = 9
5
o C + 32 = 9
5 64.0 o C + 32 = 147 o F
(b) o C = 5
9 ( o F - 32) = 5
9 (47 o F - 32) = 8.3o C
QC 1.3 8.55 mi 1.609 km
1 mi



 = 13.8 km
QC 1.4 332 m
s
1 km
1000 m





 1 mi
1.609 km




60 s
1 min





 60 min
1 hr





 = 743 mi/hr
QC 1.5
50. mLsol
hr
1.5 g antibiotic
1000. mL sol





 1000 mg antibiotic
1 g antibiotic





 1 hr
60 min





 = 1.3 mg antibiotic/min
QC 1.6 Mass of Ti = 17.3 mL 4.54 g Ti
1 mL



 = 78.5 g Ti
QC 1.7 d = m/V = 56.8 g
23.4 mL = 2.43 g/mL
QC 1.8 The specific gravity of a substance is the density of this substance divided by the
density of water, which is 1.000 g/mL.
Specific gravity = 1.016 = d
1.000 g/mL
d = 1.016 g/mL
1.2 A major reason for the increase of the average life expectancy of humans in the last 80
years has been great progresses in medical science. Diseases that were once fatal have
either been eradicated or cures developed. The causes and cures for many major diseases
are now better understood and treatments more effective.

Chapter 1: Matter, Energy, and Measurement
2
1.4 (a) Chemical change: burning gasoline is converted to carbon dioxide and water.
(b) Physical change: ice forming from liquid water is still H2O, just a different state of
matter.
(c) Physical change: boiling oil is still oil, just a different state of matter.
(d) Physical change: melting lead remains lead, just a different state of matter.
(e) Chemical change: elemental Fe has been converted to rust, Fe2O3.
(f) Chemical change: nitrogen and hydrogen converted ammonia, NH3, involves a change
in chemical composition.
(g) Chemical change: the chemical components in food are converted to energy, carbon
dioxide, and water thus changing chemical composition.
1.6 (a) 403,000 (b) 3,200 (c) 0.0000713 (d) 0.000000000555
1.8 (a) 2.10  1013 (b) 1.07  104 (c) 2.93  10-12 (d) 8.4  101 (e) 1.36  10-26
1.10 (a) 7.74  103 (b) 8.808  10-2 (c) 1.3022  102
1.12 3.25  10-10
1.14 (a) 3 (b) 2 (c) 1 (d) 4 (e) 5
1.16 (a) 2.5  104 (b) 4.1 (c) 15.5
1.18 (a) 10963.1 (b) 244 (c) 172.34
1.20 (a) 1 kg = 1000 g (b) 1 mg = 0.001 g
1.22 (a) 100 cm (b) 230 mL (c) 75 kg (d) 15 mL
(e) 50 mg (f) 100 mm (g) 8 g
1.24 Temperature conversions: o C = 5
9 ( o F - 32) and K = 273 + o C
(a) 5
9 (320 o F - 32) = 160 o C and 273 + 160 o C = 433 K
(b) 5
9 (212 o F - 32) = 100 o C and 273 + 100 o C = 373 K
(c) 5
9 (0 o F - 32) = -18 o C and 273 + (-18) o C = 255 K
(d) 5
9 (-250 o F - 32) = -157 o C and 273 + (-157) o C = 116 K

Chapter 1: Matter, Energy, and Measurement
3
1.26 Unit conversions:
(a) 42.6 kg 2.205 lb
1 kg





 = 93.9 lb (b) 1.62 lb 453.6 g
1 lb



 = 735 g
(c) 34 in 2.54 cm
1 in



 = 86 cm (d) 37.2 km 1 mi
1.609 km



 = 23.1 mi
(e) 2.73 gal 3.785 L
1 gal





 = 10.3 L (f) 62 g 1 oz
28.35 g





 = 2.2 oz
(g) 33.61 qt 1 L
1.057 qt





 = 31.80 L (h) 43.7 L 1 gal
3.785 L



 = 11.5 gal
(i) 1.1 mi 1.609 km
1 mi



 = 1.8 km (j) 34.9 mL 1 fl oz
29.57 mL



 = 1.18 fl oz
1.28 First, convert one of the bottle's units to the other.
9.5 fl. oz . 29.57 mL
1 fl. oz.





 1 cc
1 mL



 = 2.8  10 2 cc, which is less than 300. cc bottle
1.30 The key here is to first convert speed from mph to kph. Yes, you would reach Ottawa
within an hour.
Speed = 75 mi
1 hr
1.609 km
1 mi



 = 120 kph
Time to reach Ottawa = 80 km 1 hr
120 km



 = 0.7 hr (rounded to one significant figure)
1.32 Car efficiency = 25.00 mi
gal





 1.609 km
1 mi




1 gal
3.785 L





 = 10.63 km/L
1.34 186 lb 1 kg
2.205 lb





 2.0 mg drug
1 kg





 1 cc IV sol
10. mg drug





 1 mL IV sol
1 cc IV sol



 = 17 mL IV sol
1.36 36 lb child 1 kg
2.205 lb





 20. mg Velosef/day
1 kg





 = 3.3  10 2 mg/day
36 lb child 1 kg
2.205 lb





 20. mg Velosef /day
1 kg





 1 mL sol
208 mg Velosf





 = 1.6 mL sol
day

Chapter 1: Matter, Energy, and Measurement
4
1.38 750 mg lidocaine 150 mL IV sol
500. mg lidocaine





 1 min
5 mL IV sol





 = 5  101 min
The answer is rounded to one significant figure because both 1 min and 5 mL only contain
one significant figure.
1.40 Vol of pantoprazole IV = 40. mg PAN 1 mL IV sol
0.4 mg PAN





 = 1  10 2 mL PAN IV solution
Vol of MgSO 4 IV = 5 g MgSO 4
1 mL IV sol
0.02 g MgSO 4







 = 3  10 2 mL MgSO 4 IV solution
Combined volumes of IV solutions = 4  102 mL (rounded to one significant figure)
1.42 At low temperatures, most substances exist as solids.
1.44 d rock sample = mass/volume = 1.075 kg
334.5 mL





 1000 g
1 kg





 = 3.214 g/mL
1.46 Volume of Ti = 163 g Ti 1 mL Ti
4.54 g Ti





 = 35.9 mL Ti
1.48 Answer rounded to two significant figures:
Mass methanol = 280 mL methanol 0.791 g methanol
1 mL methanol



 = 2.2  10 2 g methanol
1.50 (a) The densities of O2 and CO2 can be calculated as follows:
d O2  10.00 g O 2
6702 mL O 2








1000 mL O 2
1 L O 2







 = 1.492 g/L
d CO2  10.00 g CO 2
5058 mL CO 2








1000 mL CO 2
1 L CO 2







 = 1.977 g/L
(b) Carbon dioxide (a gas that does not support combustion) being denser than oxygen will
sink to the level of the fire and temporarily displace the oxygen. By cutting off the
supply of oxygen (a fuel for the fire), the fire will be extinguished.

Chapter 1: Matter, Energy, and Measurement
5
1.52 Matter-energy can neither be created nor destroyed (law of conservation of energy), solar
panels work by converting light energy into electrical energy.
1.54 Assuming that the 180 lb man is measured to three significant figures:
135lbDrug dose = 135 lb man 445 mg drug
180 lb man = 334 mg drug
 
 
 
1.56 (a) Volume (b) Volume (c) Mass (d) Density
(e) Temperature (f) Velocity
1.58 Vroom = length x height x width = (5.3 m)(2.0 m)(4.2 m) = 45 m 3
m air = d air Vair = 1.25  10 -3 g
cm 3





 (45 m 3 ) 100 cm
1 m






3
1 kg
1000 g





 = 56 kg air
1.60 KE = 1/2(mv2)
m = 127 lb 453.6 g
1 lb





 = 5.76  10 4 g
v = 14.7 mi
hr





 1.609 km
1 mi





 1000 m
1 km





 100 cm
1 m




1 hr
60 min





 1 min
60 sec





 = 657 cm/s
KE = 1/2(5.76  10 4 g)(657 cm/s) 2 = 1.24  10 10 g cm 2 s -2 = 1.24  10 10 ergs
1.62 Convert dollars/L into dollars/gal, then compare:
Gasoline price Montreal = $1.22
L
3.785 L
gal





 = $4.62/gal
Gasoline is less expensive in Potsdam ($3.93/gal) than in Montreal ($4.62/gal).
1.64 75 mL
1 hr





 10. gtts
1 mL




1 hr
60 min





 = 13 gtts/min flow rate
1.66
dose = 14 lb cat 2.5 mg
1 lb cat





 = 35 mg per 12 hours
total drug to be delivered = 7 days 24 hr
day





 1 dose
12 hr





 35 mg
dose





 = 4.9  10 2 mg

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Chapter 1: Matter, Energy, and Measurement
6
1.68 51.4 lbs 1 kg
2.205 lbs





 1.3 mg drug
1 kg





 1 mL stock
15 mg drug





 = 2.02 mL of stock solution
1.70 Convert each quantity into a common unit (grams): (a) is the largest and (d) is the smallest.
(a) 41 g
3
(b) 3 10 mg 1 g
1000 mg = 3 g
 
 
 
 
6
(c) 8.2 10 g
 6
1 g
10 g
 = 8.2 g
 
 
 
 
-8
(d) 4.1310 10 kg 1000 g
1 kg
-5
= 4.1310 10 g
   
 
 
1.72 Travel time = 1490. mi 1.609 km
1 mi
1 hr
220. km
 
 
 
= 10.9 hr
 
 
 
1.74 1.00 mL of butter = 0.860 g and 1.00 mL of sand = 2.28 g
(a) mixture
3.14 g mixture
d = 1.57 g/mL
2.00 mL

(b) First calculate the volumes of sand and butter, which totals 1.60 mL
sandV 1.00 g sand 1.00 mL sand
2.28 g sand
butter
= 0.439 mL
V 1.00 g butter
 
 
 
 
 1.00 mL butter
0.860 g butter = 1.16 mL
 
 
 
 
mixture
2.00 g
d = = 1.25 g/mL
1.60 mL
1.76 60. mg meperidine 1 mL sol
75 mg meperidine





 = 0.8 mL solution injected
1.78 Aspirin tablets that contain 325 mg of aspirin contain more aspirin than 81 mg tablets. The
325 mg dose of aspirin is usually taken for pain, while the 81 mg daily dose of aspirin is
usually taken to help prevent heart attacks.
1.80 First, it is important to know exactly what the treatment involves and the mechanism of
action. Also, did the study compare the treatment group with a control group consisting of
ear infection patients that didn’t receive the treatment? Finally, did the scientist test the
treatment for safety?

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Chapter 1: Matter, Energy, and Measurement
7
1.82 The folk medicine will be extracted and separated into pure materials. Each pure material
will be tested for biological activity. The pure materials that are identified as biologically
active will have their chemical structures determined, and often, a chemical synthesis
developed for mass production.
1.84 flow rate = 1 L sol
12 hr





 1000 mL
1 L sol





 1 hr
60 min





 15 gtts
mL



 = 21 gtts/min
1.86 (a) Dose (in milligrams) of Tylenol calculated to two significant figures:
Dose = 42 lb child 1 kg child
2.205 lb child





 15 mg Tylenol
1 kg child





 = 2.9  10 2 mg Tylenol
(b) Dose in milliliters of Tylenol solution calculated to two significant figures:
Dose = 42 lb child 1 kg child
2.205 lb child





 15 mg Tyl
1 kg child





 5.0 mL sol
160 mg Tyl





 = 8.9 mL sol

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8
Chapter 2: Atoms
QC 2.1 (a) NaClO3 (b) AlF3
QC 2.2 (a) The mass number is 15 + 16 = 31.
(b) The mass number is 86 + 136 = 222.
QC 2.3 (a) The element has 15 protons, making it phosphorus (P); its symbol is15
31 P .
(b) The element has 86 protons, making it radon (Rn); its symbol is86
222 Rn .
QC 2.4 (a) The atomic number of mercury (Hg) is 80; that of lead (Pb) is 82.
(b) An atom of Hg has 80 protons; an atom of Pb has 82 protons.
(c) The mass number of this isotope of Hg is 80 + 120 = 200; the mass number for this
isotope of Pb is 82 + 120 = 202.
(d) The symbols of these isotopes are80
200 Hg and82
202 Pb .
QC 2.5 The atomic number of iodine (I) is 53. The number of neutrons in each isotope is 125 -
53 = 72 for iodine-125 and 131 - 53 = 78 for iodine-131. The symbols for these two
isotopes are53
125 I and53
131 I .
QC 2.6 The atomic weight is 6.941 amu, which is nearer to 7 amu than 6 amu. Therefore,
lithium-7 is the more abundant isotope. The relative abundances for these two isotopes
are 92.50 percent for lithium-7 and 7.50 percent for lithium-6.
QC 2.7 This element has 13 electrons and, therefore, 13 protons. The element with atomic
number 13 is Aluminum (Al).
2.2 (b), (c), (d), (f), (g), (h), and (k): True
(a) False: matter is divided into pure substances and mixtures.
(e) False: mixtures can be separated into their component pure substances.
(i) False: technetium, promethium, and all of the elements beyond uranium are artificial.
(j) False: H, O, C, N, Ca, and P are the six most important elements in the human body.
(l) False: the combining ratio is based on a ratio of atoms, not a ratio of masses.Al

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Chapter 2: Atoms
9
2.4 (a) Oxygen (b) Lead (c) Calcium (d) Sodium (e) Carbon (f) Titanium
(g) Sulfur (h) Iron (i) Hydrogen (j) Potassium (k) Silver (l) Gold
2.6 Given here is the element, its symbol, and its atomic number:
(a) Americium (Am, 95) (b) Berkelium (Bk, 97) (c) Californium (Cf, 98)
(d) Dubnium (Db, 105) (e) Europium (Eu, 63) (f) Francium (Fr, 87)
(g) Gallium (Ga, 31) (h) Germanium (Ge, 32) (i) Hafnium (Hf, 72)
(j) Hassium (Hs, 108) (k) Holmium (Ho, 67) (l) Lutetium (Lu, 71)
(m) Magnesium (Mg, 12) (n) Polonium (Po, 84) (o) Rhenium Re, 75)
(p) Ruthenium (Ru, 44) (q) Scandium (Sc, 21) (r) Strontium (Sr, 38)
(s) Ytterbium (Yb, 70), Terbium (Tb, 65), and Yttrium (Y, 39)
(t) Thulium (Tm, 69)
2.8 (a) K2O (b) Na3PO4 (c) LiNO3
2.10 (a) The law of conservation of mass states that matter can be neither created nor destroyed.
Dalton's theory explains this because if all matter is made up of indestructible atoms,
then any chemical reaction just changes the attachments between atoms and does not
destroy the atoms themselves.
(b) The law of constant composition states that any compound is always made up of
elements in the same proportion by mass. Dalton's theory explains this because
molecules consist of tightly bound groups of atoms, each of which has a particular
mass. Therefore, each element in a compound always constitutes a fixed proportion of
the total mass.
2.12 No. CO and CO2 are different compounds, and each obeys the law of constant composition
for that particular compound.
2.14 (b), (c), (e), (f), (g), (h), (k), (m), (o), (p), (q), and (s): True
(a) False: electrons and protons have equal, but opposite charges. Electrons are much
lighter in mass than protons.
(d) False: 1 amu = 1.6605  10-24 grams.
(i) False: opposite charges attract each other.
(j) False: the size of the atom includes the space occupied by the electrons. The nucleus is
a small fraction of the size of the atom.
(l) False: the mass number is the number of protons and neutrons.
(n) False: 1H has no neutrons, 2H has one neutron, and 3H has two neutrons.
(r) False: atomic weights are weighted averages of the known isotopes and their
abundances.
(t) False: density is mass/volume.
2.16 The statement is true. The element is determined by the number of protons (the atomic
number).

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Chapter 2: Atoms
10
2.18 (a) The element with 22 protons is titanium (Ti).
(b) The element with 76 protons is osmium (Os).
(c) The element with 34 protons is selenium (Se).
(d) The element with 94 protons is plutonium (Pu).
2.20 Each would still be the same element because the number of protons has not changed.
(a) The element is scandium (Sc), and its symbol is21
47 Sc .
(b) The element is titanium (Ti), and its symbol is22
50 Ti .
(c) The element is silver (Ag), and its symbol is47
109 Ag .
(d) The element is thorium (Th) and its symbol is90
248 Th .
(e) The element is argon (Ar) and its symbol is18
38 Ar .
2.22 Radon (Rn) has an atomic number of 86, so each isotope has 86 protons. The number of
neutrons is mass number - atomic number.
(a) Radon-210 has 210 - 86 = 124 neutrons
(b) Radon-218 has 218 - 86 = 132 neutrons
(c) Radon-222 has 222 - 86 = 136 neutrons
2.24 Two more neutrons: tin-120
Three more neutrons: tin-121
Six more neutrons: tin-124
2.26 (a) An ion is an atom or group of bonded atoms with an unequal number of protons and
electrons.
(b) Isotopes are atoms with the same number of protons in their nuclei but a different
number of neutrons.
2.28 Rounded to four significant figures, the calculated value is 12.01 amu. The value given in
the Periodic Table is 12.011 amu.98.90
100 ´ 12.000 amu
æ
èç
ö
ø÷ + 1.10
100 ´ 13.003 amu
æ
èç
ö
ø÷ = 12.01 amu
2.30 Carbon-11 has 6 protons, 6 electrons, and 5 neutrons.
2.32 Americium-241 (Am) has atomic number 95. This isotope has 95 protons, 95 electrons and
241 - 95 = 146 neutrons.

Loading page 11...

Chapter 2: Atoms
11
2.34 (a), (f), (g), (h), and (i): True
(b) False: main group elements go from groups 1A through 8A
(c) False: very roughly, the metalloids exist in a diagonal, starting in the lower right corner,
moving up to upper middle of the Periodic Table. Nonmetals exist above the
diagonal, metals below it.
(d) False: there are more metals than nonmetals
(e) False: horizontal rows are called periods.
2.36 (a) Groups 2A(2), 3B (3), 4B (4), 5B (5), 6B (6), 7B (7), 8B (8,9, and 10), 1B (11), and 2B
(12) contain only metals. Note that Group 1A contains one nonmetal, hydrogen.
(b) No group contains only metalloids.
(c) Only Groups 7A (17) and 8A (18) contain only nonmetals.
2.38 In the Periodic Table, elements of the same group should have similar properties: As, N,
and P; I and F; Ne and He; Mg, Ca, and Ba; K and Li.
2.40 (a) aluminum > silicon (b) arsenic > phosphorus
(c) gallium > germanium (d) gallium > aluminum
2.42 (a), (b), (c), (e), (g), (h), (j), (k), (l), (m), (n), (q), (r), and (s): True
(d) False: principal energy level 1 contains a maximum of two electrons, the principal
energy level 2 contains a maximum of eight electrons, the principal energy
level 3 contains a maximum of 18 electrons, and the principal energy level 4
contains a maximum of 32 electrons.
(f) False: a 2s electron is easier to remove than a 1s electron because it is further away
from the influence of the positively charged nucleus.
(i) False: The three 2p orbitals are arranged perpendicular to each other.
(o) False: paired electrons have spins in the opposite direction.
(p) False: each box represents an orbital and each orbital can accommodate two electrons.
When an orbital is completely filled, the electrons must be paired with spins in
the opposite direction.
(t) False: group 6A elements have two unpaired electrons.
2.44 The group number indicates the number of electrons in the valence shell of the element.
2.46 (a) Li(3): 1s22s1 (b) Ne(10): 1s22s22p6 (c) Be(4): 1s22s2
(d) C(6): 1s22s22p2 (e) Mg(12): 1s22s22p63s2
2.48 (a) He(2): 1s2 (b) Na(11): 1s22s22p63s1 (c) Cl(17): 1s22s22p63s23p5
(d) P(15): 1s22s22p63s23p3 (e) H(1): 1s1
2.50 In (a), (b), and (c); the outer-shell electron configurations are the same. The only
difference is the principal energy level of the valence shell being filled.

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Chapter 2: Atoms
12
2.52 The element X might be in Group 2A, all of which have two valence electrons. It might
also be helium.
2.54 (a), (b), (c), (f), and (h): True
(d) False: helium is a group 8A element and has two valence electrons.
(e) False: the period number has nothing to do with the number of elements.
(g) False: period 3 has eight elements.
2.56 (a), (b), (c), (d), and (f): True
(e) False: ionization energy decreases going from top to bottom within a column of the
Periodic Table.
2.58 (a) The additional electron in the valence shell exceeds the positive charge in the nucleus
causing the electrons in the valence shell to be held less tightly, thus increasing the
atomic radius. Also, the additional electron in the valence shell introduces new
repulsions causing the electron cloud to expand.
(b) With the loss of a valence electron, the positive charge in the nucleus exceeds the
collective negative charge of the electrons, causing the valence electrons to be held
more tightly, thus decreasing the atomic radius. Also, one less electron in the valence
shell reduces the electron-electron repulsions causing the electron cloud to contract.
2.60 Ionization energies generally increase going from left to right in a period of the Periodic
Table due to the increasing positive charge in the nucleus. As the nucleus becomes more
positively charge within a period, the valence electrons become more difficult to remove in
an ionization.
2.62 Sulfur and iron are essential components of proteins, and calcium is a major component of
bones and teeth.
2.64 Calcium is an essential element in human bones and teeth. Because strontium behaves
chemically much like calcium, strontium-90 gets into our bones and teeth and gives off
radioactivity for many years directly into our bodies.
2.66 Copper can be made harder by hammering it.
2.68 (a) Metals (b) Nonmetals (c) Metals
(d) Nonmetals (e) Metals (f) Metals
2.70 The atomic radius decreases going from left to right across a period in the Periodic Table.
Atomic radius increases going down a group (column) in the Periodic Table.
(a) Radium (Ra) (d) Neon (Ne)
(b) Beryllium (Be) (e) Fluorine (F)
(c) Lithium (Li) (f) Astatine (At)

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Chapter 2: Atoms
13
2.72 (a) Phosporous-32 has 15 protons, 15 electrons, and 32 - 15 = 17 neutrons.
(b) Molybdenum-98 has 42 protons, 42 electrons, and 98 - 42 = 56 neutrons.
(c) Calcium-44 has 20 protons, 20 electrons, and 44 - 20 = 24 neutrons.
(d) Hydrogen-3 has 1 proton, 1 electron, and 3 - 1 = 2 neutrons.
(e) Gadolinium-158 has 64 protons, 64 electrons, and 158 - 64 = 94 neutrons.
(f) Bismuth-212 has 83 protons, 83 electrons, and 212 - 83 = 129 neutrons.
2.74 For elements with atomic numbers less than that of iron (Fe), the number of neutrons is
close to the number of protons. Elements with atomic numbers greater than that of iron
have more neutrons than protons. Therefore, heavy elements typically have more neutrons
than protons.
2.76 Rounded to four significant figures, the atomic weight of naturally occurring boron is 10.81
amu. The value given in the Periodic Table is 10.811 amu.19.9
100 ´ 10.013 amu
æ
èç
ö
ø÷ + 80.1
100 ´ 11.009 amu
æ
èç
ö
ø÷ = 10.8 amu
2.78 It would take 6.0 x 1021 protons to equal the mass of a grain of salt.
2.80 The atomic number of this element is 54, which means that it is xenon (Xe). This isotope
of xenon has 54 protons, 54 electrons, and 131 - 54 = 77 neutrons.
2.82 (a) Ionization energy generally decreases down a column in the Periodic Table, so the
ionization energy of Tennessine should be less than that of At(85).
(b) Ionization energy generally increases from left to right across a row in the Periodic
Table, so ionization energy of Tennessine should be greater than Ra(88).
2.84 (a) The first electron is removed from the 2s orbital. The removal of each subsequent
electron requires more energy because, after the first electron is removed, each
subsequent electron is removed from a positive ion, which strongly attracts the
remaining electrons. The second ionization energy is especially large because the
electron is removed from the filled first principal energy level, meaning that it is
removed from an ion that has the same electron configuration as helium.
(b) Li+ ion has a smaller radius than Li because with the loss of a valence electron from
Li, the positive charge in the nucleus exceeds the collective negative charge of the
electrons, causing the valence electrons to be held more tightly, thus decreasing the
atomic radius. Also, one less electron in the valence shell of Li+ reduces the electron-
electron repulsions causing the electron cloud to contract.1.0 x 10-2 g NaCl
1.67 x 10-2 4 g/ proton = 6.0 x 1021 protons

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Chapter 2: Atoms
14
2.86 Increasing size: C < B < Al < Na
There are relatively small decreases in atomic radii going from left to right in the Periodic
Table due to the increasingly stronger pull on the electrons because of the additional
protons, therefore, the size of the atomic radii can be compared as C < B and Al < Na.
Going to a higher numbered period in the Periodic Table results in a much larger increase
in atomic radii, therefore, both Na and Al are larger than C and B.
2.88 Going down Group 3A, we have boron, aluminum, gallium, indium, and thallium. The
electron configuration of each 3A element tells us that the valence shell s orbitals are
completely filled and that the valence shell p orbital has one electron.
2.90 The Ca3+ ion is not found in chemical compounds because calcium only needs to lose two
electrons to achieve a stable noble gas electronic configuration. Losing three electrons
brings the electronic configuration away from a stable noble gas electronic configuration.
2.92 If the bromine is all used up, it is assumed that it is the limiting reagent. The reaction of
magnesium and bromine is as follows:M g + Br2 ® MgBr2
Mg reacted = 1.80 g Br2 1 mol Br2
159.8 g Br2
æ
è
ç
ç
ö
ø
÷
÷
1 mol Mg
1 mol Br2
æ
è
ç
ç
ö
ø
÷
÷
24.305 g Mg
1 mol Mg
æ
è
ç
ö
ø
÷ = 0.274 gMg remaining = 7.12 g (Mg initial mass) - 0.274 g (Mg reacted) = 6.85 g
2.94 Answers in bold. The mass numbers derived reflect the most abundant isotope.
Symbol Atomic
number
Atomic
weight
Mass
number
# of
protons
# of
neutrons
# of
electrons
H 1 1.0079 1 1 0 1
Li 3 6.941 7 3 4 3
Al 13 26.9815 27 13 14 13
Fe 26 55.845 58 26 32 26
Pt 78 195.084 195 78 117 78
Ca 20 40.078 37 20 17 20
S 16 32.066 32 16 16 16
2.960.4818 109 Ag atomic mass( ) + 0.5182(106.905 amu) = 107.868 amu109 Ag atomic mass = 52.47 amu
0.4818109 Ag atomic mass = 108.9 amu

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Chapter 2: Atoms
15
2.98 If 30Si has an isotopic abundance of 3.09%, then 28Si + 29Si = 96.91% of the mix.
Therefore, if 28Si = x and 29Si = y, then (x + y) = 0.9691 and y = (0.9691 – x ).(1) 27.977(x) amu + 28.086 (y) amu + 29.974 ´ (0.0309) amu = 28.086 amu
(2) replace y with (0.9691 – x) and then solve for x
(3) 27.977(x) amu + 28.086 (0.9691 – x) amu + 29.974 ´ (0.0309) amu = 28.086 amu
x = 0.9217
The isotopic abundance is: 28Si = 92.17%, 29Si = 4.74%, and 30Si = 3.09% (given)
2.100 (a) Ar (b) Al (c) Si (d) Cl (e) Mg (f) P (g) Ar (h) S

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Solution Manual For Introduction to General, Organic, and Biochemistry, 12th Edition

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