Chemistry 121
Mines, Fall 2015
Answer Key, Problem Set 3 – Complete Version (with explanations)
1. 2.72; 2. MP*; 3. NT1; 4. NT2; 5. NT3; 6. NT4; 7. 3.31; 8. 3.32; 9. 3.34; 10. 3.36; 11. 3.39; 12. 3.40;
13. 3.42; 14. 3.44; 15. 3.46; 16. 3.48; 17. 3.50; 18. 3.52; 19. 3.54; 20. NT5
-------------------------------1. 2.72.
Magnesium has three naturally occurring isotopes with the following masses and natural abundances:
Isotope
Mass(amu)
Abundance (%)
Mg-24
23.9850
78.99
Mg-25
24.9858
10.00
Mg-26
25.9826
11.01
Calculate the atomic mass of magnesium and sketch its mass spectrum.
Answer: 24.31 amu
㫪ork / Reasoning:
First of all, the atomic mass should be a bit above 24 amu because the vast majority of Mg have
a mass close to 24 amu, and the remaining isotopes have masses larger than 24 amu. To get
the precise value, multiply each isotopic mass by its fractional abundance and sum:
(weighted average)
atomic mass = 0.7899(23.9850 amu) + 0.1000(24.9858 amu) + 0.1101(25.9826 amu)
= 18.9457… + 2.49858 + 2.86068… = 24.3050…= 24.31 amu
(reasonable; a bit over 24
amu, as estimated)
Mass Spectrum (see Fig. 2.17):
2. MP (Problem in Mastering only; no answer key prepared for this problem)
3. NT1. Problem 2.74 in Tro.
The atomic mass of copper is 63.546 amu. Do any copper isotopes [atoms] have a mass of
63.546 amu? Explain.
Answer: No. The value of 63.546 amu is a weighted average of the two naturally occurring isotopes
on Earth (Cu-63 and Cu-65; See Example 2.5). So there are only two “kinds” of Cu atoms
and thus (only) two isotopic masses—one smaller than the average and one larger than it.
More specifically, all of the Cu-63 atoms have a mass less than 63.546 amu and all of the
Cu-65 atoms have a mass greater than it. But no Cu atoms have the precise mass of the
weighted average.
4. NT2.
The isotopic mass of Cu-63 is 62.9396 amu and its natural abundance (on Earth) is 69.17%. The atomic mass of
Cu (on Earth) is 63.546 amu. If 1 g = 6.022 x 1023 amu (this is a mass “conversion” relationship—it has nothing to
do with Cu per se), how many atoms would there be in a 1.000-gram (i.e., 6.022 x 1023-amu) sample of
(a) Cu-63 (i.e., a sample of copper that has only Cu-63 atoms in it. This would be very expensive to produce here on
Earth, but it could be done!)?
(b) a “regular” sample of Cu on planet Earth?
(c) Do the individual atoms in the two samples above have different masses? Be specific. What causes fewer atoms
to be in the second sample than the first?
PS3-1
Answer Key, Problem Set 3
Answers: (a) 9.568 x 1021 atoms; (b) 9.477 x 1021 atoms;
(c) Not most of them! In fact, 69.17% of the atoms in Sample 2 have identical masses to
all of the atoms in Sample 1. But 30.83% of the atoms in Sample 2 have a greater
mass. That is why there are fewer total atoms in the sample—if some of them weigh
more, then it will take fewer of them to add up to a certain mass (like 1.000 g)
Work / Reasoning for (a) & (b):
Each sample has a mass of 6.022 x 1023 amu. Each atom in (a) has a mass of 62.9396 amu.
Thus, the number of atoms in the first sample (to 4 SF) is just:
6.022 x 10 23 amu
 9.5679...x 10 21  9.568 x 10 21 atoms (all Cu-63 atoms)
62.9396 amu / atom
Since the (average) atomic mass of Cu is 63.546 amu, the number of atoms in the “regular”
sample of Cu (with both isotopes present in the ratio of their natural abundances) is just:
6.022 x 10 23 amu
 9.4765...x 10 21  9.477 x 10 21 atoms (some Cu-63 and some Cu-65)
63.546 amu / atom
5. NT3 Problem 3.24 in Tro. (i) Determine the number of each type of atom in each [one] formula [unit of each
substance]. (ADD part (ii): Determine the number of each type of ion in one formula unit of each substance.)
Literal “Expansion” of
formula (no charges)*
# of each type of atom
(in one FU)
# of each type of ion
(in one FU)
Ca; NO2; NO2
one Ca; two N; four O
one Ca2+; two NO2-
(b) CuSO4
Cu; S; O; O; O; O
one Cu; one S; four O
one Cu2+; one SO42-
(c) Al(NO3)3
Al; NO3; NO3; NO3
one Al; three N; nine O
one Al3+; three NO3-
(d) Mg(HCO3)2
Mg; HCO3; HCO3
one Mg; two H; two C; six O
(a) Ca(NO2)2
one Mg2+; two HCO3-
* This column was only added to try to clarify the literal meaning of subscripts in a formula; it is not
necessary for you to have this in your homework papers.
6. NT4 Problems 3.29 & 3.30 in Tro. Classify each compound as ionic or molecular (ADD part (ii);
For each ionic
compound write the (full) formula of the cation (i.e., including its charge as a superscript on the right. Be careful
here!)
Ionic or
Molecular?
3.29 (a) CO2
molecular
(b) NiCl2
ionic
(c) NaI
ionic
(d) PCl3
molecular
3.30 (a) CF2Cl2
molecular
(b) CCl4
molecular
(c) PtO2
ionic
(d) SO3
molecular
Reasoning
nonmetal (C)
listed first*
metal (Ni)
listed first
metal (Na)
listed first
nonmetal (P)
listed first*
nonmetal (C)
listed first*
nonmetal (C)
listed first*
metal (Pt)
listed first
nonmetal (S)
listed first*
Formula of
Cation
N/A
Ni2+
Na+
Reasoning
N/A
chloride is -1 and there are two of them  -2;
one Ni must have +2 for FU to be neutral
Na always forms a +1 ion in ionic compounds
(Type I; Gp 1A)
N/A
N/A
N/A
N/A
N/A
Pt4+
N/A
PS3-2
N/A
oxide is -2 and there are two of them  -4;
one Pt must have +4 for FU to be neutral
Answer Key, Problem Set 3
* And the nonmetal is not a N in NH4+
7. 3.31 & 8. 3.32.
Based on the molecular views, classify each substance as an atomic element, a molecular
element, an ionic compound, or a molecular compound.
3.31
2.32
3.31 (a) Molecular element; only one kind of atom in sample ( “element”), but atoms are
“grouped” in distinct pairs (called molecules) ( “molecular”)
(b) Molecular compound; four identical molecules are shown (“molecular”); each has more
than one kind of atom in it (“compound”)
(c) (Mon)Atomic element; only one kind of atom in sample (“element”), and no distinct
“groupings” (molecules) are apparent
3.32 (a) Ionic compound; two different kinds of “atom” in sample ( “compound”), but no distinct
“groupings” (molecules) are apparent. Rather, atoms are arranged in
an alternating “lattice” structure ( “ionic”)
(b) Molecular element; like 2.31(a) above
(c) Molecular compound; like 2.31(b) above
9. 3.34.
Write a formula for the ionic compound that forms between each pair of elements.
Elements*
(a) Ag and Cl
(b) Na and S
(c) Al and S
(d) K and Cl
Cation Formed
Ag+
(Type I, memorize)
Na+
(Type I, Gp 1A)
Al3+
(Type I, memorize)
K+
(Type I, Gp 1A)
Anion Formed1
Cl-
Formula of Ionic Compound Formed
AgCl
([Ag+, Cl-] is net neutral)
(Gp 6A is 2 away from
8A)
Na2S
([Na+, Na+, S2-] is net neutral)
as in (b)
Al2S3
([Al3+, Al3+, S2-, S2-, S2-] is net neutral)
KCl
([K+, Cl-] is net neutral)
(Gp 7A is 1 away from
8A)
S2-
Cl(Gp 7A is 1 away from
8A)
*In original problem, the element names were given, not the symbols. I wrote the symbols here to save space.
1
All of these anions are monatomic (because the compounds are binary). Thus the charges can be
determined from the position of the parent element on the periodic table. For each column “away from”
Gp 8A, you “get” a single negative charge.
PS3-3
Answer Key, Problem Set 3
10. 3.36.
Write a formula for the compound that forms between potassium and each polyatomic ion.
NOTE: K always forms a +1 ion when it is involved in ionic compounds (Type I, Gp 1A)
Formula
of Anion*
Formula of Ionic
Compound with K+
(a) carbonate
CO32-
K2CO3
(b) phosphate
PO43-
K3PO4
(H+ & PO43-)
= HPO42-
K2HPO4
C2H3O2-
KC2H3O2
(c) hydrogen phosphate
(d) acetate
Reasoning (basically, you need to figure out the
subscript of K using the neutrality principle)
one CO32- requires two K+’s
to be net neutral
one PO43- requires three K+’s
to be net neutral
one HPO32- requires two K+’s
to be net neutral
one C2H3O2- requires one K+
to be net neutral
* See Note and footnote to table in Problem #10 below about learning polyatomic ions.
11. 3.39 & 12. 3.40.
Formula of
Compound
3.39 (a) SnO
(b) Cr2S3
(c) RbI
(d) BaBr2
3.40 (a) BaS
Give each ionic compound an appropriate [its IUPAC] name.
Anion1 and Name
Cation2 and Name
Sn2+, tin(II) ion
O2-, oxide
S2-, sulfide
I-, iodide
(Snx + O2- is neutral  x = 2+)
Cr3+, chromium(III) ion
(Crx + Crx + S2- + S2- + S2- is neutral  x = 3+)
Rb+ (Type I, Gp 1A),
rubidium ion
Name of Compound
tin(II) oxide
chromium(III) sulfide
rubidium iodide
Br -, bromide
Ba2+ (Type I, Gp 2A),
barium ion
barium bromide
S2-, sulfide
Ba2+ (Type I, Gp 2A),
barium ion
barium sulfide
(b) FeCl3
Cl -, chloride
(c) PbI4
I-, iodide
(d) SrBr2
Br -, bromide
Fe3+, iron(III) ion
(Fex + Cl- + Cl- + Cl- is neutral  x = 3+)
Pb4+, lead(IV) ion
(Pb + I + I- + I- + I- is neutral  x = 4+)
x
-
Sr2+ (Type I, Gp 2A),
strontium ion
iron(III) chloride
lead(IV) iodide
strontium bromide
Reasoning: For ionic compounds, the name is just “cation name + anion name” (no prefixes
[except for those that are part of the name of certain anions]). But to get the full cation name
for those derived from Type II metals requires a) determination of the charge of the anion, and
2) application of the neutrality principle.
1 All of these anions are monatomic. Thus the charges can be determined from the position of the
parent element on the periodic table (See Problem 7). You do not actually need this step for the
compounds with cations of Type I metals, but I included it anyway (in smaller font) for
completeness.
2 If the metal is of Type I (Gp 1A & Gp 2A metals plus Al, Zn, and Ag [in my class]), you do not
include a Roman numeral in parentheses to indicate the charge on the cation. But any other
metal is of Type II (more than one cation is possible) and the Roman numeral in parentheses
indicating its charge must be included.
PS3-4
Answer Key, Problem Set 3
13. 3.42.
Name each ionic compound containing a polyatomic ion.
NOTE: You must familiarize/memorize the names and formulas (which includes the charge!) of the polyatomic
ions on the list I gave you in order to do questions such as this one on an exam.*
Formula of
Compound
Anion and Name
Cation and Name (as in #9 above)
(a) Ba(OH)2
OH-, hydroxide
Ba2+, barium ion
(c) NaBrO4
BrO4-, perbromate*
(d) Fe(OH)3
OH-, hydroxide
(e) CoSO4
SO42-, sulfate*
(f) KClO
barium hydroxide
[Type I, Gp 2A]
NH4+, ammonium ion (memorize!)
I-, iodide
(b) NH4I
Name of Compound
Na+, sodium ion
ammonium iodide
sodium perbromate
[Type I, Gp 1A]
Fe3+, iron(III) ion
iron(III) hydroxide
(Fex + OH- + OH- + OH- is neutral  x = 3+)
Co2+, cobalt(II) ion
cobalt(II) sulfate
(Cox + SO42- is neutral  x = 2+)
ClO-, hypochlorite*
K+, potassium ion
potassium hypochlorite
[Type I, Gp 1A]
* As noted in my PowerPoint, I would first learn the four “common” –ate ions (NO3-, SO42-, PO43-, and CO32-), and
then recognize/memorize that the halogen-containing “-ate” ions are analogous to NO3- (ClO3-, BrO3-, and IO3-).
Then learn that “-ites” have one fewer O than –ates, and that where there are four ions, the one with one more
than the -ate is the “per___ate” and the one with one fewer than the -ite is the hypo___ite.
14. 3.44.
Write a formula for each ionic compound.
Name of Compound
Formula of Cation
Formula of Anion
(a) copper(II) chloride
Cu2+
(b) copper(I) iodate
Cu+ *
IO3- (like nitrate)
(c) lead(II) chromate
Pb2+ *
CrO42- (memorize)
Ca2+ (Type I, Gp 2A)
(d) calcium fluoride
(e) potassium hydroxide
(f) iron(II) phosphate
K+
Cl-
*
(Type I, Gp 1A)
Fe2+ *
(Gp 7A)**
Formula of Compound***
CuCl2
[two “1-”’s + one “2+” = 0]
CuIO3
[one “1-” + one “1+” = 0]
PbCrO4
[one “2-” + one “2+” = 0]
F- (Gp 7A)**
CaF2
[two “1-”’s + one “2+” = 0]
OH- (memorize)
KOH
[one “1-” + one “1+” = 0]
Fe3(PO4)2
[two “3-”’s + three “2+”’s =
PO43- (memorize)
0]
* Charge given by the Roman numeral in parentheses.
** -ides with element stems are monatomic—get charge from column in Periodic Table
*** Use neutrality principle along with charges on the ions.
15. 3.46 Give the name from the formula or the formula from the name for each hydrated ionic compound [i.e. each
“hydrate”]
NOTE: Hydrates are the only substances whose formulas contain a coefficient. The coefficient
always goes after the “dot” and before “H2O” and it indicates the number of water
molecules per formula unit of ionic compound. Coefficients are usually reserved for usage
in balanced chemical equations (that represent chemical change).
(a) cobalt(II) phosphate octahydrate
Co2+, PO43-  Co3(PO4)2; “octahydrate”  “8 waters”; combine to get Co3(PO4)2 • 8H2O
(b) and (c) omitted from key at this time
PS3-5
Answer Key, Problem Set 3
(d) LiNO2 • H2O
Li+ is “lithium ion”; NO2- is “nitrite”  lithium nitrite; “1 water”  “monohydrate”;
combine to get: lithium nitrite monohydrate
16. 3.48.
Name each molecular compound.
NOTE: These are all binary compounds involving two nonmetals. That means they are molecular
compounds, and they are named with prefixes such as di, tri, penta, etc, and use the full element name
for the first element and “stem + -ide” format for the second. It is unfortunate that for every part in this
problem, the authors chose compounds that have the first element with no subscript. Make sure you
know how to deal with ones where that is not the case. You need to know the prefixes for 1-10 only.
(a) SO3
1 = mono, 3 = tri, but “mono” is omitted for 1st element  sulfur trioxide
NOTE: This is not “sulfite” ion because there is no charge in the formula! This is why it
is so important to show the charges on all ion formulas!
(b) SO2
1 = mono, 2 = di, but “mono” is omitted for 1st element  sulfur dioxide
(c) BrF5
1 = mono, 5 = penta, but “mono” is omitted for 1st element  bromine pentafluoride
NOTE: “fluorine” has the “U” before the “O”. Please do not write “flourine” or “flouride”!
(d) NO
1 = mono, but “mono” is omitted for 1st element  nitrogen monoxide
(e) XeO3
1 = mono, 3 = tri, but “mono” is omitted for 1st element  xenon trioxide
17. 3.50.
Write a formula for each molecular compound.
(a) boron tribromide
no subscript = 1, tri = 3
 BBr3
(b) dichlorine monoxide
di = 2, mono = 1
 Cl2O
(c) xenon tetrafluoride
no subscript = 1, tetra = 4  XeF4
(d) carbon tetrabromide
no subscript = 1, tetra = 4  CBr4
(e) diboron tetrachloride
di = 2, tetra = 4
18. 3.52.
 B2Cl4
Name each acid.
Strategy: 1) Determine the anion from which the acid is “derived” (i.e., remove all H+’s)
2) Name the anion
3) Name the acid based on the anion ending: -ide = hydro___ic acid
The stem of the
anion goes in the
-ate = ___ic acid
blank here.
-ide = ___ous acid
Name of Anion
Formula of Anion
Name of Acid
Formula of Acid
(from which acid is derived)
(and anion ending)
Cl-
chloride (-ide)
hydrochloric acid
(b) HClO2
ClO2-
chlorite (-ite)
chlorous acid
(c) H2SO4
SO42-
sulfate (-ate)
sulfuric acid*
(d) HNO2
NO2-
nitrite (-ite)
nitrous acid
(a) HCl
* By the “rules”, this should be “sulfic acid”, but for S (and P), the full element name is used instead of the
stem (somebody liked how that sounded better?). If you don’t remember that on an exam, I’ll let it slide.
PS3-6
Answer Key, Problem Set 3
19. 3.54.
Write formulas for each acid
Strategy: 1) Determine which “ending” is in the acid name (second column below)
2) Determine the anion ending from #1 above
3) Determine the name of the anion from the stem in the acid name and the anion
ending (#2 above)
4) Write the formula of the anion
5) Add as many H+’s as needed to obtain a neutral formula (unit)—use that as the
subscript for H in the acid formula.
Name of Acid
(a) phosphoric acid
(b) hydrocyanic acid
(c) chlorous acid
hydro__ic, ___ic,
or ___ous?
Anion
Ending
Name of
Anion
Formula of
Anion
Formula of
Acid
___ic acid
-ate
phosphate
PO43-
H3PO4
HCN
HClO2
hydro___ic
-ide
cyanide
CN-
___ous
-ite
chlorite
ClO2- *
* Because chlorate is like nitrate (3 O’s) and –ites have one fewer O than –ates.
20. NT5.
Write the name or formula for each of the following compounds:
(a) HC2H3O2
H first  acid; C2H3O2- is acetate ion; -ates become "-ic acids"
 acid name is acetic acid
(b) NH4NO2
nonmetal first, but not binary so check for NH4. It DOES have NH4 first, so it is
ionic, with cation = NH4+ ion (ammonium ion). Anion = NO2-, which is nitrite
 ammonium nitrite
(c) Co2S3
metal first  ionic compound; Co is Type II, so look at anion first. S2- 
charge on Co is 3+ (neutrality principle; total neg. charge = 3 x -2 = -6 so two
Co's must total +6)
 cobalt(III) sulfide (note: S2- is sulfide, not sulfate or sulfite!)
(d) ICl
binary with nonmetal (other than H) first  molecular compound (non-acid) 
use prefixes to indicate # of atoms of each in one molecule
 iodine monochloride (remember to omit mono for 1st element)
(e) Pb3(PO4)2
metal first  ionic compound; Pb is Type II, so look at anion first. PO43(phosphate)  charge on Pb is 2+ (neutrality principle; total neg. charge = 2 x -3
= -6 so three Pb's must total +6)
 lead(II) phosphate
(f) KIO3
metal first  ionic compound; K is Type I (1A), so K+. IO3 must be -1, IO3- is
iodate ion
 potassium iodate
(g) H2CO3
H first  acid; CO32- is carbonate ion; -ates become "-ic acids"
 acid name is carbonic acid
(h) Sr3N2
metal first  ionic compound; Sr is Type I (2A), so Sr2+. N3- is nitride ion
 strontium nitride
PS3-7
Answer Key, Problem Set 3
(i) Al2(SO3)3
metal first  ionic compound; Al is Type I (3A), Al3+; SO3 must be 2-;
(neutrality principle) SO32- is sulfite ion
 aluminum sulfite
(j) SnO2
metal first  ionic compound; Sn is Type II, so look at anion first. O2- (oxide)
 charge on Sn is 4+ (neutrality principle; total neg. charge = 2 x -2 = -4 so one
Sn must total +4)
 tin(IV) oxide
(k) HClO
H first  acid; ClO- is hypochlorite ion; -ites become "-ous acids"
 acid name is hypochlorous acid
(l) ammonium hydrogen phosphate
NH4+, HPO42-  (NH4)2HPO4 (neutrality principle)
(m) chromium(VI) sulfide
Cr6+, S2-  CrS3 (neutrality principle)
(n) silicon dioxide
SiO2 (prefixes tell you the subscripts)
(o) sodium sulfite
Na+, SO32-  Na2SO3 (neutrality principle)
(p) aluminum hydrogen sulfate
Al3+, HSO4-  Al(HSO4)3 (neutrality principle)
(q) nitrogen trichloride
NCl3 (prefixes tell you the subscripts)
(r) hydrobromic acid
hydro___ic acid  -ide anion  bromide  Br-  HBr
(s) bromous acid
ous acid  -ite anion  bromite  BrO2-  HBrO2
(t) perbromic acid
ic acid  -ate anion  perbromate  BrO4-  HBrO4
(u) potassium hydrogen sulfide K+, HS- (sulfide is S2- + H+ = HS- [hydrogen sulfide])  KHS (neutrality princ.)
PS3-8