FUNDAMENTALS
A.1
(a) chemical;
(b) physical;
(c) physical
A.3
The temperature of the injured camper and the evaporation and
condensation of water are physical properties. The ignition of propane is a
chemical change.
A.5. (a) physical;
(b) chemical;
A.7
(a) intensive;
A.9
(a) 2×10−6 km = 2×103 m;
(c) chemical
(b) intensive;
(c) extensive;
(d) extensive
(b) 2×105 m;
(c) 2×107 nm = 2×104
m. Therefore, (b) is the largest
A.11
1.00 cup × (
A.13
d
1000 mL
1 pint
1 quart
0.946 L
)× (
)× (
)× (
) = 236 mL
1L
2 cups
2 pints
1 quart
m
V

  1 mL 
112.32 g


3 
 29.27 mL  23.45 mL   1 cm 
 19.3 g  cm 3
SM-3
A.15
m
m
, rearranging gives V 
V
d
 0.750 carat   200 mg   1 g 



3  
 3.51 g  cm   1 carat   1000 mg 
d
 0.0427 cm3
A.17 d 
m
, rearrange gives
V
V
m
d
20.0 g
 7.41 cm3
3
2.70 g .cm
Since the area is 1 centimeter, the thickness must be 7.41 cm.
V
A.19 The result should have 3 significant figures because the number 3.25
has the least significant figures. The results should be 0.989
A.21 (a) 4.82 nm × (
1000pm
) = 4.82 × 103 pm
1nm
1 min
1.83mL
1000mm 3
1cm3
(b) (
)× (
)× (
)× (
) = 30.5 mm3/s
3
min
60 s
1cm
1mL
(c) 1.88 ng × (
1g
1kg
)× ( 3 ) = 1.88 × 10−12 kg
9
10 ng
10 g
2.66g
106 cm 3
1kg
(d) (
)× (
)× (
) = 2.66 × 103 kg/m3
3
3
cm
1m
1000g
(e) (
A.23
0.044g
1L
1000mg
)× (
)× (
) = 0.044 mg/cm3
3
L
1000cm
1g
m
V
0.213 g




 1.100 cm  0.531 cm  0.212 cm 
 0.213 g 

3 
 0.1238 cm 
(a) d 
 1.72 g  cm 3
SM-4
This determination is more precise because the volume is not limited to
two significant figures as it is in part (b).
m
V
 41.003 g  39.753 g 


 20.37 mL  19.65 mL 
 1.250 g   1 mL 


3 
 0.72 mL   1 cm 
(b) d 
 1.7 g  cm 3
A.25
EK 
1 2
mv
2
1
 (4.2 kg)(14 km  h 1 ) 2
2
2
 1 h   1000 m 

 

 3600 s   1 km 
2
 32 kg  m 2  s 2
 32 J
A.27
m = 2.8 metric tons, vi  100 km  hr 1 , v f  50 km  hr 1
EK 
1 2
mv
2
EK (init ) 
 103 kg 
1
 2.8 metric tons  

2
 1 metric ton 
 100 km  1 hr  1 min.  
 1 hr  60 min.  60 sec.  




2
 4.32 kg  km 2  s 2  4.32 106 kg  m 2  s 2  4,320 kJ
EK (final)
 103 kg 
1
  2.8 metric tons  

2
 1 metric ton 
2
 50 km  1 hr  1 min.  
 1 hr  60 min.  60 sec.  




 0.27 kg  km 2  s 2  0.27 106 kg  m 2  s 2  270 kJ
EK (init)  EK (final)  (4,320  270) kJ = 4,050 kJ = 4.0 103 kJ (2 SF)
SM-5
This amount of energy could have been recovered, neglecting friction and
other losses, or used to drive the vehicle up a hill.
EP  mgh
g  9.81 ms 2
Setting potential energy equal to 4,050 kJ = 4.05 kg m2 s−2 and solving for
height gives
h
A.29
 4.05 106 kg  m2  s 2 

  147 m = 150 m (2 SF).
mg  (2800 kg)(9.81 ms2 ) 
Ep
EP  mgh
 1 kg 
 (40.0 g)(9.81 m  s 2 )(0.50 m) 

 1000 g 
 0.20 kg  m 2  s 2 for one raise of a fork.
For 30 raises, (30)(0.20 kg  m2  s 2 )  6.0 J.
A.31
We need to use the expansion given in Exercise A.22 to help solve this
problem. We also need to recognize that EP  egh for the small
difference in distance, h, can be represented by subtracting EP at distance
r between the proton and electron from EP at distance r + h.
Ep 
e2 h
= egh
4 0 r 2
So g 
A.33
e2 h  1 
e
when E p = egh.

2 
4 0 r  eh  4 0 r 2
The relationship between distance of separation and potential energy for
charged particles is given in section A.2, Equation 4.
E p  V (r ) 

q1 q2
(e)(e)

4 0 r
4 0 r
(1.602  1019 C) 2
4 (8.85419  10-12 C2  J 1  m 1 )(53  1012 m)
1 eV


 4.352  1018 J 
 27.17 eV=  27 eV (2 SF)
19 
 1.602  10 J 
SM-6
Considering the proton and electron beginning at rest and at infinite
separation sets the initial total energy to 0. Since the electron is not at rest
in a hydrogen atom, its total energy is represented by Equation 5:
E  EK  EP
We have only calculated the potential energy. The discrepancy between
the calculated value of the potential energy, −27.7 eV, and the measured
amount released, −13.6 eV, is the kinetic energy of the electron, 13.6 eV.
B.1
number of beryllium atoms 
mass of sample
mass of one atom

  1 kg 
0.210 g


26
1  
 1.50  10 kg  atom   1000 g 
 1.40  1022 atoms
B.3
(a) 5p, 6n, 5e; (b) 5p, 5n, 5e; (c) 15p, 16n, 15e; (d) 92p, 146n, 92e
B.5
(a)
B.7
194
Ir ; (b)
Element
22
Ne ; (c)
51
Symbol
V
Protons
Neutrons
Electrons
Mass number
Chlorine
36
Cl
17
19
17
36
Zinc
65
Zn
30
35
30
65
Calcium
40
Ca
20
20
20
40
Lanthanum
137
57
80
57
137
La
B.9 protons
B.11
(a) They all have the same mass. (b) They have differing numbers of
protons, neutrons, and electrons
SM-7
B.13
(a) 0.5519; (b) 0.4479; (c) 2.459 × 10-4; (d) 551.9 kg
B.15
(a) Scandium is a Group 3 metal.
(b) Strontium is a Group 2 metal.
(c) Sulfur is a Group 16 nonmetal. (d) Antimony is a Group 15 metalloid.
B.17
(a) Sr, metal;
(b) Xe, nonmetal;
(c) Si, metalloid
B.19
Fluorine, F, Z = 9, gas; Chlorine, Cl, Z = 17, gas; Bromine, Br, Z = 35,
liquid; Iodine, I, Z = 53, solid
B.21
(a) d block; (b) p block; (c) d block; (d) s block; (e) p block; (f) d
block
C.1
Container (a) holds a mixture (one is single compound and another is
single element); container (b) holds a single element.
C.3
The chemical formula of xanthophyll is C40H56O2.
C.5
(a) C3H7O2N;
C.7
(a) Cesium is a metal in Group 1; it will form Cs  ions.
(b) C2H7N
nonmetal in Group 17/VII and will form I  ions.
Group 16/VI nonmetal and will form Se 2 ions.
(b) Iodine is a
(c) Selenium is a
(d) Calcium is a Group
2 metal and will form Ca 2 ions.
C.9
(a)
10
Be2+ has 4 protons, 6 neutrons, and 2 electrons.
protons, 9 neutrons, and 10 electrons.
neutrons, and 36 electrons.
(d)
75
(c)
80
(a)
19
F ;
(b)
24
Mg 2 ;
(c)
128
17
O2− has 8
Br− has 35 protons, 45
As3− has 33 protons, 42 neutrons, and
36 electrons.
C.11
(b)
Te2 ;
SM-8
(d)
86
Rb 
C.13
(a) Aluminum forms Al3 ions; tellurium forms Te2 ions. Two
aluminum atoms produce a charge of 2  3  6. Three tellurium atoms
produce a charge of 3  2  6. The formula for aluminum telluride is
Al2 Te3 .
(b) Magnesium forms Mg 2 ions and oxygen forms O 2  ions.
A magnesium ion produces a charge of +2, which is required to balance
the charge on one O 2  ion. The formula for magnesium oxide is MgO.
(c) Sodium forms +1 ions; sulfur forms –2 ions. The formula for sodium
sulfide is Na 2S.
(d) Rubidium forms +1 ions and iodine forms –1 ions.
One iodide ion is required to balance the charge of one rubidium ion, so
the formula is RbI.
C.15
(a) HCl, molecular compound (in the gas phase); (b) S8, element
(molecular substance); (c) CoS, ionic compound; (d) Ar, element; (e)
CS2, molecular compound;
(f) SrBr2, ionic compound
C.17
(a) Group III; (b) aluminum, Al
C.19
The formula is Al2O3.
m
V
102 g




 2.5 cm  3.0 cm  4.0 cm 
d
 3.4 g  cm 3
D.1
(a) MnCl2. Mn forms 2+ ions and chlorine forms −1 ions.
(b) Ca3(PO4)2. Calcium forms +2 ions and the phosphate ion is PO43−.
(c) Al2(SO3)3. Aluminum forms +3 ions and the sulfite ion is SO32−.
(d) Mg3N2. Magnesium forms 2+ ions and the nitride ion is N 3 .
SM-9
D.3
(a) calcium phosphate;
vanadium(V) oxide;
D.5
(b) tin(IV) sulfide, stannic sulfide;
(c)
(d) copper(I) oxide, cuprous oxide
(a) TiO2 ; (b) SiCl4 ;
(c) CS2 ;
(d) SF4 ;
(e) Li 2S;
(f) SbF5 ;
(g) N2 O5 ; (h) IF7
D.7
(a) sulfur hexafluoride;
triiodide;
(b) dinitrogen pentoxide;
(d) xenon tetrafluoride;
(c) nitrogen
(e) arsenic tribromide;
(f)
chlorine dioxide
D.9
(a) hydrochloric acid;
acid;
(b) sulfuric acid;
(e) sulfurous acid;
D.11
(a) ZnF2
(b) Ba(NO3)2
D.13
(a) sodium sulfite;
(c) nitric acid;
(f) phosphoric acid
(c) AgI
(d) Li3N
(e) Cr2S3
(b) iron(III) oxide or ferric oxide;
oxide or ferrous oxide;
sulfate hexahydrate;
(d) acetic
(d) magnesium hydroxide;
(c) iron(II)
(e) nickel(II)
(f) phosphorus pentachloride;
(g) chromium(III) dihydrogen phosphate;
(h) diarsenic trioxide;
(i)
ruthenium(II) chloride
D.15 (a) CuCO3 Copper (II) carbonate; (b) K2SO3 potassium sulfite; (c) LiCl,
lithium chloride
D.17
(a) heptane;
(b) propane;
(c) pentane;
(d) butane
D.19
(a) cobalt(III) oxide monohydrate; Co2 O3  H 2 O ; (b) cobalt(II)
hydroxide; Co(OH) 2
D.21 E = Si; SiH4, silicon tetrahydride; Na4Si, sodium silicide
SM-10
D.23
(a) lithium aluminum hydride, ionic (with a molecular anion); (b)
sodium hydride, ionic
D.25
(a) selenic acid;
(b) sodium arsenate;
(c) calcium tellurite;
barium arsenate;
(e) antimonic acid;
(f) nickel(III) selenate
(d)
D.27 (a) alcohol; (b) carboxylic acid; (c) haloalkane
E.1
1.0 mole ×
6.022  1023Ag atom
144pm
1m
×
× 12
1 Ag atom 10 pm
1.0 mole Ag
= 8.67 × 1013 m = 8.67 × 1010 km
E.3
You will to add 18 Ca atoms to balance the 9 Br atoms on the left pan
because a Br atom is twice as massive as a Ca atom.
E.5
(a) moles of people 
6.0  109 people
6.022  1023 people  mol 1
 1.0  1014 mol
(b) time 
1 mol peas
 1.0  1014 s
14
1
1.0  10 mol  s
 1 h   1 day   1 yr 
6
(1.0  1014 s) 


  3.2  10 years
 3600 s   24 h   365 days 
E.7
(a) mass of average Li atom
 7.42 
 92.58 
24
23

 (9.988  10 g)  
 (1.165  10 g)
 100 
 100 
 1.153  1023 g  atom 1
molar mass  (1.153  1023 g  atom1 ) (6.022  1023 atoms  mol 1 )
 6.94 g  mol1
(b) mass of average Li atom
SM-11
 100  5.67 
 5.67 
24
23

 (9.988  10 g)  
 (1.165  10 g)
100
 100 


23
1
 1.1556  10 g  atom
molar mass  (1.1556  1023 g  atom 1 ) (6.022  1023 atoms  mol1 )
 6.96 g  mol1
E.9
(a) MgSO4  7 H2 O formula mass  246.48 g  mol1

  11 mol O atoms 
5.15 g
atoms of O  

1  
 246.48 g  mol   mol MgSO4  7 H 2 O 
(6.022  1023 atoms  mol1 )  1.38  1023


5.15 g
(b) formula units  
(6.022  1023 atoms  mol1 )
1 
246.48
g

mol


22
 1.26  10


5.15 g
(c) moles of H 2 O  7 
 0.146 mol
1 
 246.48 g  mol 
E.11
The percentage
10
B  100  percentage 11 B
 % 10 B 
molar mass  
 (mass
 100% 
10
 % 11 B 
11
B)  
 (mass B)
 100% 
 100%  % 11 B 

 (mass
100%


10
 % 11 B 
11
B)  
 (mass B)
 100% 
Rearranging gives
% 11 B 
100  molar mass  100  mass
mass 11 B  mass 10 B
10
B
100(10.81 g  mol1 )  100(10.013 g  mol 1 )

11.093 g  mol 1  10.013 g  mol 1
 73.8 %
%
10
B  26.2 %
SM-12
E.13
(a)
75 g
 0.65 mol In
114.82 g  mol 1
80 g
 0.63 mol Te
127.60 g  mol 1
75 g of indium contains more moles of atoms than 80 g of tellurium.
(b)
15.0 g
 0.484 mol P
30.97 g  mol1
15.0 g
 0.468 mol S
32.07 g  mol1
15.0 g of P has slightly more atoms than 15.0 g of S.
(c) Because the two samples have the same number of atoms, they will
have the same number of moles, which is given by
2.49  1022 atoms
 0.0413 mol
6.022  1023 atoms  mol1
E.15

  102.90 g Rh 
36 g Ga
(a) mRh = 
×
 = 53 g Rh
1
 69.72 g  mol Ga   1 mol Rh 

  102.90 g Rh 
36 g In
(b) mRh = 
 ×
 = 32 g Rh
1
 114.82 g  mol In   1 mol Rh 
E.17
(a) molar mass of Al2 O3  101.96 g  mol1
nAl2O3 
10.0 g
 0.0981 mol
101.96 g  mol1
N Al2O3  (0.0981 mol)(6.022  1023 atoms  mol 1 )  5.91  1022 molecules
(b) molar mass of HF  20.01 g  mol1
nHF 
25.92  103 g
 1.30  103 mol
1
20.01 g  mol
N HF  (1.30  103 mol)(6.022  1023 atoms  mol1 )
 7.83  1020 molecules
(c) molar mass of hydrogen peroxide = 34.02 g  mol1
SM-13
nH2O2 
1.55  103 g
 4.56  105 mol
1
34.02 g  mol
N H2O2  (4.56  105 mol)(6.022  1023 atoms  mol 1 )
 2.75  1019 molecules
(d) molar mass of glucose = 180.15 g  mol1
nglucose 
1250 g
 6.94 mol
180.15 g  mol1
Nglucose  (6.94 mol)(6.022  1023 atoms  mol1 )  4.18  1024 molecules
(e) molar mass of N atoms = 14.01 g  mol1
nN 
4.37 g
 0.312 mol
14.01 g  mol1
N N  (0.312 mol)(6.022  1023 atoms  mol1 )  1.88  1023 atoms
molar mass of N 2 molecules = 28.02 g  mol1
nN2 
4.37 g
 0.156 mol
28.02 g  mol1
N N2  (0.156 mol)(6.022  1023 atoms  mol1 )  9.39  1022 molecules
E.19
(a) molar mass of CuBr2 = 223.35 g  mol1

  1 mol Cu 2+ 
3.00 g CuBr2
2+
nCu2  
×
 = 0.0134 mol Cu
1
223.35
g

mol
CuBr
1
mol
CuBr
2  
2 

(b) molar mass of SO3 = 80.06 g  mol1


0.700 g SO3
3
nSO3 = 
 = 8.74 × 10 mol SO3
1
 80.06 g  mol SO3 
(c) molar mass of UF6 = 352.03 g  mol1
 1000 g   1 mol UF6   6 mol F 

nF- = 25.2 kg × 
× 
 = 430. mol F
× 
1
kg
352.03
g
UF
1
mol
UF

 
6  
6 
SM-14
(d) molar mass of Na2CO3 ·10H2O = 286.21 g·mol−1

 

2.00 g Na 2CO3 10H 2O
10 mol H 2O
nH2O = 
×

1
 286.21 g  mol Na 2CO3 10H 2O   1 mol Na 2CO3 10H 2O 
= 0.0699 mol H 2O
E.21
(a) number of formula units = (0.750 mol)( 6.022  1023 formula units
 mol1 )
 4.52  1023 formula units
(b) molar mass of Ag 2 SO 4  311.80 g  mol 1


 1000 mg 
2.39  1020 formula units
(311.80 g  mol1 ) 


23
1 
 1g 
 6.022  10 formula units  mol 
 124 mg
(c) molar mass of NaHCO 2  68.01 g  mol 1


3.429 g
(6.022  1023 formula units  mol1 )

1 
 68.01 g  mol 
 3.036  1022 formula units
E.23
(a) molar mass of H 2 O  18.02 g  mol1


18.02 g  mol1
 2.992  1023 g  molecule1

23
1 
6.022

10
molecules

mol




1000 g
(b) N H2O  
(6.022  1023 molecules  mol 1 )
1 
 18.02 g  mol 
 3.34  1025 molecules
E.25
(a) molar mass of CuCl2·4H2O = 206.53 g.mol−1


8.61 g CuCl2  4H 2O
n= 
 = 0.0417 mol CuCl2  4H2O
1
 206.53 g  mol CuCl2  4H 2O 
SM-15
(b) Because there are 2 mol Cl− per mole of compound, the number of
moles will be twice the amount in part (a), 0.0834 mol Cl−
(c) There are 4 mole of H2O for each mole compound. Therefore,
number of water molecules = (0.0417 × 4) mole H2O × (6.022 v 1023
molecules · mol−1) = 1.00 × 1023 H2O molecules
(d) fraction of mass due to O =
4(16.00 g  mol1 )
= 0.3099
206.53 g  mol1
E.27 (a)
 10 mol H 2 O
  18.01 g H 2 O   1 mol hydrated cpd 



  100
 1 mol hydrated cpd   1 mol H 2 O   264.166 g hydrated cpd 
 68.2% H 2 O
Therefore, 6.82 kg out of 10 kg was water.

  10 kg hydrated cpd  1 kg H 2 O 
?$
$72.00




L H 2 O  10 kg hydrated cpd  6.82 kg H2 O  1 L H2 O 
 $10.60 per L H 2 O
(b) Since the anhydrous compound costs $80.00 for 10 kg, or $8.00/kg,
and only 10−6.82 = 3.18 kg is NaHCO3 , a fair price would have been
$8.00
 3.18 kg = $25.44 .
kg
E.29
Zr atoms/mole = (90.0000 g
90


1atom
Zr) 

22
 1.4929 10 g 
= 6.028535 × 1023 atoms = 1 mole
Since there are more atoms per mole, each mole will weigh more and all
of the molar masses will be proportionately higher by
 6.028535 1023  6.02214 1023

100  = 0.10619%

23
6.02214 10


SM-16
(a) 12C  12.0000…. g·mol−1 × 1.0010619 = 12.013 g·mol−1
(b) 196.97Au  196.97…. g·mol−1 × 1.0010619 = 197.18 g·mol−1
E.31
Solve by factor label (dimensional analysis).
23
 1.074 g   2.50 10 H atoms 
? mole H atoms = 28.0 cm3 NaBH 4 

3 
3.93 g
 1 cm  

1 mol




23
 6.022 10 H atoms 
 3.18 moles H atoms
E.33
For example, Br has m1 = 80.9163 u, m2 = 78.9183 u
81
mean mass (u)
80.5
80
79.5
79
78.5
0
20
40
60
percent m1
E.35 (a) (Brisotope1 + Brisotope1) = 158, Brisotope1 = 79;
(Brisotope1 + Brisotope2) = 160, Br(isotope2) = 81
Bromine-79 and bromine-81 occur in nature.
(b) Bromine-79 is more abundant.
SM-17
80
100
F.1
C10H16O
10 × 12.01 g = 120.1 g × (
100
) = 78.90%
152.2264
16 × 1.0079 g = 16.1264 g × (
1 × 16.00 g = 16.00 g
F.3
× (
100
) = 10.59%
152.2264
100
) = 10.51%
152.2264
C7 H15 NO3
 100 

  52.15%
 161.20 g 
 100 
15 1.0079 g = 15.1185 g  
  9.3787%
 161.20 g 
7 12.01 g = 84.07 g
114.01 g = 14.01 g
3 16.00 g = 48.00 g
 100 

  8.691%
 161.20 g 
 100 

  29.78%
 161.20 g 
161.20 g
F.5
(a) M2 O
88.8% M
100.00%
For 100 g of compound, 88.8 g is M, 11.2 g is O.
 100 g M 2O   16.00 g O   1 mol O 
? g M 2O
=



mole M 2 O  11.2 g O   1 mol O   1 mol M 2O 
= 143 g  mol1 M 2 O
Therefore, 143−16 = 127 g/mol are due to M. Since there are 2 moles of
M
per mole of M2 O , the molar mass of M = 63.4 g/mol. That molar mass
matches Cu. (b) copper(I) oxide
F.7
(a) For 100 g of compound,
moles of Na 
32.79 g
 1.426 mol
22.99 g  mol 1
moles of Al 
13.02 g
 0.4826 mol
26.98 g  mol1
SM-18
moles of F 
54.19 g
 2.852 mol
19.00 g  mol1
Dividing each number by 0.4826 gives a ratio of 1 Al : 2.95 Na : 5.91 F.
The formula is Na 3 AlF6 .
(b) For 100 g of compound,
moles of K 
31.91 g
 0.8161 mol
39.10 g  mol1
moles of Cl 
28.93 g
 0.8161 mol
35.45 g  mol1
mass of O is obtained by difference:
100 g  31.91 g  28.93 g
 2.448 mol
16.00 g  mol1
moles of O 
Dividing each number by 0.8161 gives a ratio of 1.00 K : 1 Cl : 3.00 O.
The formula is KClO3 .
(c) For 100 g of compound,
moles of N 
12.2 g
 0.871 mol
14.01 g  mol1
moles of H 
5.26 g
 5.22 mol
1.0079 g  mol 1
moles of P 
26.9 g
 0.869 mol
30.97 g  mol1
moles of O 
55.6 g
 3.475 mol
16.00 g  mol1
Dividing each number by 0.869 gives a ratio of 1.00 N : 6.01 H : 1.00 P :
4.00 O. The formula is NH6 PO4 or [NH4 ] [H2 PO4 ], ammonium
dihydrogen phosphate.
F.9
(a) moles of P 
moles of Cl 
4.14 g
 0.134 mol
30.97 g  mol1
27.8 g  4.14 g
 0.667 mol
35.45 g  mol1
SM-19
Dividing each number by 0.134 mol gives a ratio of 4.98 Cl : 1 P. The
formula is PCl5 .
(b) Name is Phosphorus pentachloride.
F.11
For 100 g of compound,
 67.49 g 
Moles of C = 
= 5.619 mol
1 
 12.01g  mol 


4.60 g
Moles of H = 
= 4.563 mol
1 
 1.008g  mol 
 12.45g

Moles of Cl = 
= 0.3512 mol
1 
35.45g

mol




9.84 g
Moles of N = 
= 0.7024 mol
1 
 14.01g  mol 


5.62 g
Moles of O = 
= 0.3512 mol
1 
 16.00 g  mol 
Dividing each number by 0.3512 mol gives a ratio of 16.00 C : 13.00 H :
1.00 Cl : 2.00N : 1.00 O. The formula is C16H13ClN2O.
F.13
For 100 g of the osmium carbonyl compound,
moles of C 
15.89 g
 1.323 mol
12.01 g  mol 1
moles of O 
21.18 g
 1.324 mol
16.00 g  mol1
moles of Os 
62.93 g
 0.3309 mol
190.2 g  mol1
Dividing each number by 0.3309 mol gives a ratio of 4.00 C : 4.00 O :
1.00 Os. (a) The empirical formula is OsC4 O4 .
(b) The formula mass
of OsC4 O4 is 302.24 g  mol1 . The molar mass is 907 g  mol1 which is 3
times the formula mass, so the molecular formula is Os3 C12 O12 .
SM-20
F.15
For 100 g of caffeine,
moles of C 
49.48 g
 4.12 mol
12.01 g  mol 1
moles of H 
5.19 g
 5.15 mol
1.0079 g  mol 1
moles of N 
28.85 g
 2.059 mol
14.01 g  mol 1
moles of O 
16.48 g
 1.03 mol
16.00 g  mol 1
Dividing each number by 1.03 mol gives a ratio of 4.00 C : 5.00 H : 2.00
N : 1.00 O. The formula is C4 H5 N2 O with a molar formula mass of
97.10 g  mol1 . Because the molecular molar mass is twice this value, the
actual formula will be C8 H10 N 4 O2 .
F.17
Glucose (C6 H12 O6 ) has a molar mass of 180.15 g  mol1 and will have
the following composition:
6(12.01 g  mol1 )
%C 
 40.00%
180.15 g  mol1
%H 
12(1.0079 g  mol 1 )
 6.71%
180.15 g  mol1
%O 
6(16.00 g  mol1 )
 53.29%
180.15 g  mol1
Sucrose (C12 H22 O11 ) has a molar mass of 342.29 g  mol1 and will have
the following composition:
12(12.01 g  mol1 )
%C 
 42.10%
342.29 g  mol1
%H 
22(1.0079 g  mol1 )
 6.48%
342.29 g  mol1
SM-21
11(16.00 g  mol1 )
%O 
 51.42%
342.29 g  mol1
While the %H values for glucose and sucrose are too close to allow us to
distinguish between them by this value alone, %C (40.00 versus 42.10%)
and %O (53.29 versus 51.42%) values are sufficient that, when taken
together, can give us a reasonable amount of confidence in distinguishing
between them.
F.19
Calculate the mass percent carbon for each fuel from its formula.
ethene, C2 H 4
propanol, C3H 7 OH
heptane, C7 H16
2(12.01)
100  85.63% C
2(12.01) + 4(1.0079)
3(12.01)
100  59.96% C
3(12.01) + 8(1.0079) + 16.00
7(12.01)
100  83.91% C
7(12.01) + 16(1.0079)
ethene (85.63%) > heptane (83.91%) > propanol (59.96%)
F.21
(a) empirical formula: C2H3Cl, molecular formula: C4H6Cl2
(b) empirical formula: CH4N, molecular formula: C2H8N2
F.23
This problem requires that we relate unknowns to each other appropriately
by writing a balanced chemical equation and using other information in
the problem.
x NaNO3  y Na 2SO4  ( x  2 y ) Na +  x NO3 +y SO 42
1.61 g Na +
 0.07003 mol Na +  x  2 y
1
+
22.99 g  mol Na
5.37 g total  1.61 g Na + =3.76 g =(62.01 g  mol1 ) x + (96.07 g  mol1 ) y
SM-22
Rearrange and substitute:
3.76 = 62.01(0.07003  2y )  96.07 y
0.06065  0.07003  2 y  1.548 y
0.009385  0.4516 y
y  0.02078 moles sulfate, x = 0.02847 moles nitrate
Therefore, the mass of sodium nitrate in the mixture was
 85.00 g NaNO3 
0.02847 mol 
  2.42 g NaNO3
1 mol


2.42 g NaNO3
100  45.1% NaNO3 .
5.37 g total
G.1
(a) density;
(b) the abilities of the components to adsorb;
(c) boiling
points
G.3
(a) heterogeneous, decanting;
by filtration and distillation;
(b) heterogeneous, dissolving followed
(c) homogeneous, distillation
G.5
mass of AgNO3  (0.179 mol  L1 )(0.5000 L)(169.88 g  mol 1 )  15.2 g
G.7
(a) molarity of Na 2 CO3 
2.111 g
 0.079 67
(105.99 g  mol1 )(0.2500 L)
M
Na 2 CO3
V
(b)
(2.15  103 mol Na  (1 mol Na 2 CO 2 )
 1.35  102 L or 13.5 mL
1

(0.079 67 mol  L Na 2 CO3 )(2 mol Na )
V
(4.98  103 mol CO32 )(1 mol Na 2 CO 2 )
(0.079 67 mol  L1 Na 2 CO3 )(1 mol CO32 )
= 6.25  102 L or 62.5 mL
(50.0  103 g Na 2 CO3 )
(c) V 
(105.99 g  mol1 )(0.079 67 mol  L1 Na 2 CO3 )
 5.92  103 L or 5.92 mL
SM-23
G.9
(a) Weigh 1.6 g (0.010 mol, molar mass of KMnO 4  158.04 g  mol 1 )
into a 1.0-L volumetric flask and add water to give a total volume of 1.0 L.
Smaller (or larger) volumes could also be prepared by using a
proportionally smaller (or larger) mass of KMnO4 .
(b) Starting with 0.050 mol  L1 KMnO 4 , add four volumes of water to
one volume of starting solution because the concentration desired is onefifth of the starting solution. This relation can be derived from the
expression
Vi  molarityi  Vf  molarity f
where i represents the initial solution and f the final solution. But
Vf  Vi  Vd , where Vd represents the volume of solvent that must be
added to dilute the initial solution. Rearranging the first equation gives
Vi molarityf

Vf molarityi
Vi
molarityf

.
Vi  Vd molarityi
So if the ratio of final molarity to initial molarity is 1 : 5, we can write
Vi
1

Vi  Vd 5
5Vi  Vi  Vd
4Vi  Vd .
For example, to prepare 50 mL of solution, you would add 40 mL of water
to 10 mL of 0.050 mol  L1 KMnO 4 .
G.11 (a) V (0.778 mol  L1 )  (0.1500 L)(0.0234 mol  L1 )
V  4.51  103 L or 4.51 mL
(b) The concentration desired is one-fifth of the starting NaOH solution,
so the stockroom attendant will need to add four volumes of water to one
volume of the 2.5 mol  L1 solution. To prepare 60.0 mL of solution,
SM-24
divide 60.0 by 5; so 12.0 mL of 2.5 mol  L1 NaOH solution are added to
48.0 mL of water. (See the solution to G.9. )
G.13 (a) mass of CuSO 4  (0.20 mol  L1 )(0.250 L)(159.60 g  mol 1 )
 8.0 g
(b) mass of CuSO 4  5 H 2 O  (0.20 mol  L1 )(0.250 L)(249.68 g  mol1 )
 12 g
G.15
(a) Total moles of K+:
 0.500g KCl   0.500g K 2S   2 mol K  

+

1  
1  
 74.55g  mol   110.26g  mol  1mol K 2S 
 0.500g K 3PO4   3mol K  


1  
 212.27 g  mol  1mol K 3PO4 
= 6.71 × 10−3 mol + 9.07 × 10−3 mol + 7.07 × 10−3 mol = 2.29 × 10-−2
mol
 2.29 10 2 mol K 
Molarity of K+ = 
0.500L


−2
 = 4.58 × 10 M

(b) molarity of S2− :
 0.500g K 2S   1molS2   1

−3


 = 9.07 × 10 M
1  
110.26g

mol
1mol
K
S
0.500
L



2  
G.17 (a) mass of K 2SO 4  (0.125 mol  L1 )(1.00 L)(174.26 g  mol 1 )  21.8 g
(b) mass of NaF  (0.015 mol  L1 )(0.375 L)(41.99 g  mol1 )  0.24 g
(c)
G.19
mass of C12 H 22 O11  (0.35 mol  L1 )(0.500 L)(342.29 g  mol 1 )
 60. g
The total molar concentration of Cl− :
 0.50 g NaCl   0.30 g KCl    1


+
×

 = 0.13 M Cl
1  
1   
58.44
g

mol
74.55
g

mol
0.100
L

 
 

SM-25
G.21 We can show that fewer than 1 molecule of X would be left after only 70
doublings.
10 mL 
0.10 mol 6.02214  1023 molecules

 6.0  1020 molecules are
1000 mL
1 mol
in the first 10 mL aliquot of the solution. In order to find the number of
times the volume must be doubled to get to one molecule, we can solve for
the number of times this amount of molecules must be cut in half until it
equals 1.
n
1
6.0  1020 molecules     1 molecule
 2
1
log(6.0  1020 )  n log    log(1)
 2
20.8  n(0.301)  0
20.8  0.301n
n  69
The other 21 additional doublings involve solutions with no X remaining.
There can be no health benefits if there are no molecules of the active
substance, X, left in the solution.
G.23 Find the volume of concentrated HCl solution that is equivalent to 10.0 L
of the dilute HCl solution with respect to the number of moles of solute
present in each volume.
? mL con. HCl(aq)
 0.7436 mol HCl   36.46 g HCl 
=10.0 L dilute HCl(aq) 


 1 L dilute HCl(aq)   1 mol HCl 
 100 g con. HCl(aq)   1 cm3 



37.50 g HCl

 1.205 g 
 600. mL con. HCl
Therefore, 600. mL of concentrated HCl(aq) must be diluted up to a final
volume of 10.0 L by adding water in order to form 0.7436 M HCl(aq).
SM-26
H.1
(a) You cannot add a different compound or element to the chemical
equation that is not produced (or involved) in the chemical reaction. In
this case, O atom is not produced by the defined reaction.
(b) 2 Cu + SO2

2 CuO + S

H.3
2 SiH4 + 4 H2O
H.5
(a) NaBH4(s) + 2 H2O(l)  NaBO2(aq) + 4 H2(g)
(b) LiN3(s) + H2O(l) 
2 SiO2 + 8 H2
LiOH(aq) + HN3(aq)
(c) 2 NaCl(aq) + SO3(g) + H2O(l)  Na2SO4(aq) + 2 HCl(aq)
(d) 4 Fe2 P(s)  18 S(s)  P4S10 (s)  8 FeS(s)
H.7
(a) Ca(s) + 2 H2O(l)  H2(g) + Ca(OH)2(aq)
(b) Na 2 O(s)  H 2 O(l)  2 NaOH(aq)
(c) 3 Mg(s) + N2(g)  Mg3N2(s)
(d) 4NH3(g) + 7 O2(g)  6 H2O(g) + 4 NO2(g)
H.9
(I) 3 Fe2 O3 (s)  CO(g)  2 Fe3 O4 (s)  CO2 (g)
(II) Fe3 O4 (s)  4 CO(g)  3 Fe(s)  4 CO2 (g)
H.11 (I) N2 (g)  O2 (g)  2 NO(g)
(II) 2 NO(g)  O2 (g)  2 NO2 (g)
H.13
4 HF(aq)  SiO2 (s)  SiF4 (aq)  2 H 2 O(l)
H.15
C7 H16 (l)  11 O2 (g)  7 CO2 (g)  8 H 2O(g)
H.17
4 C10 H15 N(s)  55 O2 (g)  40 CO2 (g)  30 H 2 O(l)  2 N 2 (g)
SM-27
H.19 (I) H2S(g)  2 NaOH(s)  Na 2S(aq)  2 H2 O(l)
(II) 4 H2S(g)  Na 2S(alc)  Na 2S5 (alc)  4 H 2 (g)
(III)
2 Na 2S5 (alc)  9 O2 (g)  10 H 2O(l)  2 Na 2S2O3  5 H 2O(s)  6 SO2 (g)
H.21 (a) We can find the empirical formulas from the percent compositions.
First oxide:
P 43.64 g  30.97 g/mol = 1.409 mol
O 56.36 g  16.00 g/mol = 3.523 mol
3.523 mol
 2.500
1:2.5 or 2:5
P2 O5
1.409 mol
Second oxide:
P 56.34 g  30.97 g/mol = 1.819 mol
O 43.66 g  16.00 g/mol = 2.729 mol
2.729 mol
 1.500
1:1.5 or 2:3 P2 O 3
1.819 mol
These empirical formulas could be named diphosphorus pentoxide and
diphosphorus trioxide. The names according to the Stock system are
given below with the formulas of the actual compounds.
(b) P4O10 (phosphorus (V) oxide), P4O6 (phosphorus (III) oxide).
Since the molar masses of the empirical formulas are 142 g/mol and
110 g/mol, respectively, and these masses are both half as big as the molar
masses of the compounds, both molecular formulas are twice the empirical
formulas.
(c) P4(s) + 3 O2(g) → P4O6(s); P4(s) + 5 O2(g) → P4O10(s)
I.1
The picture would show a precipitate, CaSO4(s), at the bottom of the flask
composed of red and pink particles in a 1:1 ratio. Sodium (black) and
chloride (blue) ions, NaCl(aq), would remain throughout the solution.
I.3
You should use the reagent to only react with one of the ions and form
SM-28
precipitate, so that two ions can be separated.
(a) Diluted H2SO4 solution will be used as reagent.
Pb2+(aq) + SO42−(aq)  PbSO4(s)
(b) H2S solution will also be used as reagent.
Mg2+(aq) + S2− (aq)  MgS(s)
I.5
(a) CH3 OH, nonelectrolyte;
(b) BaCl2, strong electrolyte;
(c) KF,
strong electrolyte
I.7
(a) soluble;
(b) slightly soluble;
I.9
(a) Na  (aq) and I (aq); (b) Ag  (aq) and CO32 (aq), Ag 2 CO3 is insoluble.
The very small amount that
(c) insoluble;
(d) insoluble
does go into solution will be present as
Ag  and CO32 ions. (c) NH 4  (aq) and PO 43 (aq); (d)
Fe 2 (aq) and SO 4 2 (aq)
I.11
(a) Fe(OH)3 , precipitate ; (b) Ag 2 CO3 , precipitate forms ; (c) No
precipitate will form because all possible products are soluble in water.
I.13
(a) net ionic equation: Fe2 (aq)  S2 (aq)  FeS(s) ;
spectator ions: Na  , Cl
(b) net ionic equation: Pb 2 (aq)  2 I  (aq)  PbI2 (s) ;
spectator ions: K  , NO3
(c) net ionic equation: Ca 2 (aq)  SO 4 2 (aq)  CaSO 4 (s) ;
spectator ions: NO3 , K 
(d) net ionic equation: Pb 2 (aq)  CrO4 2 (aq)  PbCrO4 (s) ;
spectator ions: Na  , NO3
SM-29
(e) net ionic equation: Hg 2 2 (aq)  SO 4 2 (aq)  Hg 2SO 4 (s) ;
spectator ions: K  , NO3
I.15
(a) overall equation: (NH 4 ) 2CrO 4 (aq)  BaCl2 (aq)
 BaCrO 4 (s)  2 NH 4Cl(aq)
complete ionic equation:
2 NH 4  (aq)  CrO4 2 (aq)  Ba 2 (aq)  2 Cl (aq)
 BaCrO4 (s)  2 NH 4  (aq)  2 Cl (aq)
net ionic equation: Ba 2 (aq)  CrO 4 2 (aq)  BaCrO 4 (s) ;
spectator ions: NH 4  , Cl
(b) CuSO4 (aq)  Na 2S(aq)  CuS(s)  Na 2SO4 (aq)
complete ionic equation:
Cu 2 (aq)  SO4 2 (aq)  2 Na  (aq)  S2 (aq)
 CuS(s)  2 Na  (aq)  SO4 2 (aq)
net ionic equation: Cu 2 (aq)  S2 (aq)  CuS(s) ;
spectator ions: Na  , SO 4 2
(c) 3 FeCl2 (aq)  2 (NH 4 )3PO 4 (aq)
 Fe3 (PO 4 ) 2 (s)  6 NH 4Cl(aq)
complete ionic equation:
3 Fe2 (aq)  6 Cl (aq)  6 NH 4  (aq)  2 PO43 (aq)
 Fe3 (PO4 )2 (s)  6 NH 4  (aq)  6 Cl (aq)
net ionic equation: 3 Fe2 (aq)  2 PO43 (aq)  Fe3 (PO4 ) 2 (s) ;
spectator ions: Cl , NH 4 
(d) K 2 C2 O4 (aq)  Ca(NO3 )2 (aq)  CaC2 O4 (s)  2 KNO3 (aq)
complete ionic equation:
SM-30
2 K  (aq)  C2 O4 2 (aq)  Ca 2 (aq)  2 NO3 (aq)
 CaC2 O4 (s)  2 K  (aq)  2 NO3 (aq)
net ionic equation: Ca 2 (aq)  C2 O 4 2 (aq)  CaC2 O4 (s) ;
spectator ions: K  , NO3
(e) NiSO4 (aq)  Ba(NO3 )2 (aq)  Ni(NO3 )2 (aq)  BaSO4 (s)
complete ionic equation:
Ni 2 (aq)  SO4 2 (aq)  Ba 2 (aq)  2 NO3 (aq)
 Ni 2 (aq)  2 NO3 (aq)  BaSO4 (s)
net ionic equation: Ba 2 (aq)  SO 4 2 (aq)  BaSO 4 (s) ;
spectator ions: Ni 2 , NO3
I.17
(a) Ba(CH3 CO2 )2 (aq)  Li 2 CO3 (aq)  BaCO3 (s)  2 LiCH3CO2 (aq)
Ba 2 (aq)  2 CH3 CO 2  (aq)  2 Li  (aq)  CO32 (aq) 
BaCO3 (s)  2 Li  (aq)  2CH 3 CO 2  (aq)
net ionic equation: Ba 2 (aq)  CO32 (aq)  BaCO3 (s)
(b) 2 NH4 Cl(aq)  Hg 2 (NO3 )2 (aq)  2 NH4 NO3 (aq)  Hg 2 Cl2 (s)
2 NH 4  (aq)  2 Cl (aq)  Hg 2 2  (aq)  2 NO3 (aq) 
Hg 2 Cl2 (s)  2 NH 4  (aq)  2 NO3 (aq)
net ionic equation: Hg 2 2 (aq)  2 Cl  (aq)  Hg 2 Cl 2 (s)
(c) Cu(NO3 )2 (aq) + Ba(OH)2 (aq)  Cu(OH)2 (s)  Ba(NO3 )2 (aq)
Cu 2 (aq)  2 NO3 (aq)  Ba 2 (aq)  2 OH  (aq) 
Cu(OH) 2 (s)  Ba 2 (aq)  2 NO3 (aq)
net ionic equation: Cu 2 (aq)  2 OH  (aq)  Cu(OH) 2 (s)
I.19
(a) AgNO3 and Na 2 CrO4
(b) CaCl2 and Na 2 CO3
SM-31
(c) Cd(ClO4 )2 and (NH 4 )2S
I.21
(a) 2 Ag  (aq)  SO 4 2 (aq)  Ag 2SO 4 (s)
(b) Hg 2 (aq)  S2 (aq)  HgS(s)
(c) 3 Ca 2 (aq)  2 PO 43 (aq)  Ca 3 (PO 4 ) 2 (s)
(d) AgNO3 and Na 2SO 4 ; Na  , NO3 ;
Hg(CH3 CO 2 ) 2 and Li 2 S; Li  , CH3 CO 2 
CaCl2 and K 3 PO 4 ; K  , Cl 
I.23
white ppt. = AgCl(s), Ag+; no ppt. with H2SO4, no Ca2+; black ppt. =
ZnS, Zn2+
I.25
(a) 2 NaOH(aq) + Cu(NO3 )2 (aq)  Cu(OH)2 (s) + 2 NaNO3 (aq)
complete ionic equation:
2 Na  (aq)  2 OH  (aq) + Cu 2+ (aq) + 2 NO3 (aq)
 Cu(OH) (s) + 2 Na  (aq) + 2 NO (aq)
2
net ionic equation:
3
Cu 2+ (aq)  2 OH  (aq)  Cu(OH) 2 (s)
 40.0 mL  0.100 M 
(b) M Na + = 
 = 0.0800 M
50.0 mL


I.27
(a) Find the number of moles of potassium chromate per liter of solution.
? mol K 2 CrO4  3.50 g K 2 CrO4   1 mol K 2 CrO4   1000 mL 




L solution
 75.0 mL solution   194.20 g K 2 CrO4   1 L 
 0.240 M K 2 CrO4 (aq)
(b) Find the number of grams of potassium in the amount of potassium
chromate that was dissolved.
SM-32
 1 mol K 2 CrO4  2 mol K + 
? g K +  3.50 g K 2 CrO 4 


 194.20 g K 2 CrO4  1 mol K 2 CrO4 
 39.0983 g K + 


+
 1 mol K

+
 1.41 g K
(c) MgCrO4
J.1
(a) base;
(b) acid;
(c) base;
(d) acid;
(e) base
J.3
(a) overall equation: HF(aq)  NaOH(aq)  NaF(aq)  H2 O(l)
complete ionic equation:
HF(aq)  Na  (aq)  OH  (aq)  Na  (aq)  F (aq)  H 2 O(l)
net ionic equation:
HF(aq)  OH  (aq)  F (aq)  H 2 O(l)
(b) overall equation: (CH3 )3 N(aq)  HNO3 (aq)  (CH3 )3 NHNO3 (aq)
complete ionic equation:
(CH3 )3 N(aq)  H3O (aq)  NO3 (aq)
 (CH3 )3 NH  (aq)  NO3 (aq)  H 2O(l)
net ionic equation:
(CH3 )3 N(aq)  H 3O  (aq)  (CH 3 )3 NH  (aq)  H 2O(l)
(c) overall equation: LiOH(aq)  HI(aq)  LiI(aq)  H2 O(l)
complete ionic equation:
Li  (aq)  OH  (aq)  H3 O  (aq)  I  (aq)  Li  (aq)  I  (aq)  2 H 2 O(l)
net ionic equation: OH  (aq)  H3O  (aq)  2 H 2O(l)
J.5
(a) HBr(aq)  KOH(aq)  KBr(aq)  H2 O(l)
(b) Zn(OH)2 (aq)  2 HNO2 (aq)  Zn(NO2 )2 (aq)  2 H 2 O(l)
(c) Ca(OH)2 (aq)  2 HCN(aq)  Ca(CN)2 (aq)  2 H2 O(l)
(d) 3 KOH(aq)  H3 PO4 (aq)  K3 PO4 (aq)  3 H2 O(l)
SM-33
J.7
(a) acid: H 3 O  (aq); base: CH3 NH 2 (aq) ; (b) acid: HCl(aq);
base: C2 H5 NH2 (aq) ;
(c) acid: HI(aq) ; base: CaO(s)
J.9
Since X turns litmus red and conducts electricity poorly, it is a weak acid.
We can find the empirical formula from the percent composition.
C 26.68 g  12.01 g/mol = 2.221 mol
H 2.239 g  1.0079 g/mol =2.221 mol
O 71.081 g  16.00 g/mol = 4.443 mol
So the subscripts are 1:1:2 on the empirical formula.
(a) CHO2
(b) Since the molar mass of the empirical formula is 45.0 g∙mol−1 while
the molar mass of X is 90.0 g∙mol−1, the molecular formula is twice the
empirical formula or C2H2O4.
(c) The weak acid whose formula matches the one given in part (b) is
oxalic acid.
(COOH) 2(aq) + 2 NaOH(aq) → Na2C2O4(aq) + 2 H2O(l)
net ionic equation: (COOH) 2(aq) + 2 OH− → C2O42− (aq) + 2 H2O(l)
J.11
(a) C6H5O− (aq) + H2O(l) → C6H5OH (aq) + OH− (aq)
(b) ClO− (aq) + H2O(l) → HClO(aq) + OH− (aq)
(c) C5H5NH+ (aq) + H2O(l) → C5H5N(aq) + H3O+(aq)
(d) NH4+ (aq) + H2O(l) → NH3(aq) + H3O+ (aq)
J.13
(a) AsO43− (aq) + H2O(l) → HAsO42− (aq) + OH− (aq)
HAsO42− (aq) + H2O(l) → H2AsO4− (aq) + OH− (aq)
H2AsO4− (aq) + H2O(l) → H3AsO4(aq) + OH− (aq)
In each equation, H2O is the acid.
SM-34
(b)
 1 mol Na 3 AsO 4 
? mol Na +  35.0 g Na 3 AsO 4 

 207.89 g Na 3 AsO 4 
 3 mol Na + 


 1 mol Na 3 AsO 4 
 0.505 mol Na +
K.1
(a) 2 NO2 (g)  O3 (g)  N 2 O5 (g)  O2 (g)
(b) S8 (s)  16 Na(s)  8 Na 2S(s)
(c) 2 Cr 2 (aq)  Sn 4 (aq)  2 Cr 3 (aq)  Sn 2 (aq)
(d) 2 As(s)  3 Cl2 (g)  2 AsCl3 (l)
K.3
(a) Mg(s)  Cu 2 (aq)  Mg 2 (aq)  Cu(s)
(b) Fe2 (aq)  Ce4 (aq)  Fe3 (aq)  Ce3 (aq)
(c) H2 (g)  Cl2 (g)  2 HCl(g)
(d) 4 Fe(s)  3 O2 (g)  2 Fe2 O3 (s)
K.5
(a) +4;
(b) +4;
(c) −2;
(d) +5;
(e) +1;
(f) 0
K.7
(a) +2;
(b) +2;
(c) +6;
(d) +4;
(e) +1
K.9
(a) Methanol CH3 OH (aq) is oxidized to formic acid (the carbon atom
goes from an oxidation number of +2 to +4). The O 2 (g) is reduced to O 2 
present in water.
(b) Mo is reduced from +5 to +4, while some sulfur
(that which ends up as S(s)) is oxidized from –2 to 0. The sulfur present in
MoS2 (s) remains in the –2 oxidation state.
(c) Tl is both oxidized and reduced. The product Tl(s) is a reduction of
Tl (from +1 to 0) while the Tl3 is produced via an oxidation of Tl . A
SM-35
reaction in which a single substance is both oxidized and reduced is
known as a disproportionation reaction.
K.11 (a) Cl2 will be reduced more easily and is therefore a stronger oxidizing
agent than Cl  .
(b) N 2 O5 will be a stronger oxidizing agent because it will be readily
reduced. N 5 will accept e  more readily than will N  .
K.13 (a) oxidizing agent: H  in HCl(aq); reducing agent: Zn(s)
(b) oxidizing agent: SO 2 (g) ; reducing agent: H 2 S(g)
(c) oxidizing agent: B2 O3 (s); reducing agent: Mg(s)
K.15 (a) ClO4-  ClO2, Cl goes from +7 to +4; need a reducing agent
(b) SO42-  SO2, S goes from +6 to +4; need a reducing agent
K.17
CO2(g) + 4 H2(g) 
CH4(g)
+ 2 H2O(l)
Oxidation-reduction reaction;
CO2 is oxidizing reagent; H2 is reducing reagent
K.19 (a) oxidizing agent: WO3 (s) ; reducing agent: H 2 (g)
(b) oxidizing agent: HCl reducing agent: Mg(s)
(c) oxidizing agent: SnO2 (s); reducing agent: C(s)
(d) oxidizing agent: N2 O4 (g) ; reducing agent: N2 H4 (g)
K.21 (a) 3 N2H4(l)  4 NH3(g) + N2(g); (b) −2 in N2H4; −3 in NH3, 0 in N2;
(c) N2H4 is both oxidizing and reducing agent;
(d) Factor label (dimensional analysis) can be used to find the volume of
nitrogen.
SM-36
 1000 cm3   1.004 g   1 mol 
? L N 2 (g)  1.0 L N 2 H 4 (l) 


3 
 1 L   1 cm   32.0 g 
 1 mol N 2   28.0 g   24 L 




 3 mol N 2 H 4   1 mol   28 g 
 2.5 102 L N 2 (g)
K.23 (a) 2 Cr2+ + Cu2+  2 Cr3+ + Cu(s); (b) 2 e− transferred;
(c) Use factor label (dimensional analysis):
? mol NO3  50.5 g Cr(NO3 ) 2   1 mol Cr(NO3 ) 2 



L solution  250.0 mL solution   176.0 g Cr(NO3 ) 2 
 2 mol NO3   1000 mL 



 1 mol Cr(NO3 ) 2   1 L 
 2.27 M NO 
3
? mol SO 24  60.0 g CuSO 4   1 mol CuSO 4 



L solution  250.0 mL solution   159.62 g CuSO 4 
 1 mol SO 24   1000 mL 



 1 mol CuSO 4   1 L 
 1.50 M SO 2
4
K.25
(a) SiCl4(l) + 2 H2(g)  Si(s) + 4 HCl(g); silicon tetrachloride; Si4+
(b) SnO2(s) + C(s)  Sn(l) + CO2; tin(IV) oxide; Sn4+
(c) V2O5(s) + 5 Ca(l)  2 V(s) + 5 CaO(s); vanadium(V) oxide; V5+
(d) B2O3(s) + 3 Mg(s)  2 B(s) + 3 MgO(s); boron oxide; B3+
L.1
2 Na 2S2 O3 (aq)  AgBr(s)  NaBr(aq)  Na 3 [Ag(S2 O3 ) 2 ](aq)
(a) moles of Na 2S2 O3 needed to dissolve 1.0 mg AgBr
 1 g AgBr   1 mol AgBr   2 mol Na 2S2 O3 
 1.0 mg AgBr 



 1000 mg AgBr   187.78 g AgBr   1 mol AgBr 
 1.1  105 mol Na 2S2 O3
SM-37
(b) mass of AgBr to produce 0.033 mol Na 3 [Ag(S2 O3 )2 ]

  187.78 g AgBr 
1 mol AgBr
 0.033 mol Na 3 [Ag(S2 O3 ) 2 ] 


 1 mol Na 3 [Ag(S2 O3 ) 2 ]   1 mol AgBr 
 6.2 g AgBr
L.3
6 NH4 ClO4 (s)  10 Al(s)  5 Al2 O3 (s)  3 N 2 (g)  6 HCl(g)  9 H 2 O(g)
 1 mol NH 4 ClO4
(a) (1.325 kg NH 4 ClO4 ) 
 117.49 g NH 4 ClO4
  1000 g 


  1 kg 
 10 mol Al   26.98 g Al 


  507.1 g Al
 6 mol NH 4 ClO4   1 mol Al 
 1000 g Al   1 mol Al 
(b) 3500 kg Al 


 1 kg Al   26.98 g Al 
 5 mol Al 2 O3   101.96 g Al 2 O3 



 10 mol Al   1 mol Al 2 O3 
 6.613  106 g Al 2 O3 or 6.613  103 kg Al 2 O3
L.5
2 C57 H110 O6 (s)  163 O2 (g)  114 CO2 (g)  110 H 2 O(l)
 1 mol fat   110 mol H 2 O   18.02 g H 2 O 
(454 g fat) 



(a)
 891.44 g fat   2 mol fat   1 mol H 2 O 
 505 g H 2 O
 1 mol fat   163 mol O 2   32.00 g O 2 
(454 g fat) 



(b)
 891.44 g   2 mol fat   1 mol O 2 
 1.33  103 g O 2
L.7
2 C8 H18 (l)  25 O2 (g)  16 CO2 (g)  18 H 2 O(l)
d  0.79 g  mL1 , density of gasoline
 1000 mL   0.79 g gas   1 mol gas   18 mol H 2 O 
(3.785 L gas) 




 1 L   1 mL   114.22 g gas   2 mol C8 H18 
 18.02 g H 2 O 
3

  4.246  10 g H 2 O or 4.2 kg H 2 O
 1 mol H 2 O 
SM-38
L.9
(a) HCl  NaOH  NaCl  H 2 O
 0.234 mol HCl   1 mol NaOH 
17.40 mL 

  0.00407 mol
1000
mL
1
mol
HCl



concentration of NaOH 
0.00407 mol
 0.271 mol  L1
15.00 ×103 L
(b) (0.271 mol  L1 )(0.01500 L) (40.00 g  mol1 )  0.163 g NaOH
L.11
(a) Ba(OH)2 (aq)  2 HNO3 (aq)  Ba(NO3 ) 2 (aq)  2 H 2 O(l)
 9.670 g Ba(OH) 2 
? mol HNO3
 11.56 mL Ba(OH) 2 (aq)  

L HNO3 (aq)
 250. mL Ba(OH) 2 (aq) 
 1 mol Ba(OH) 2  2 mol HNO3 



171.36 g Ba(OH) 2  1 mol Ba(OH) 2 


 1L 
(25.0 mL HNO3 (aq)) 

 1000 mL 
 0.209 mol  L1
(b) mass of HNO3 in solution:
 63.02 g HNO3 
 0.209 mol 
  0.329 g

 (25.0 mL) 
 1000 mL 
 1 mol HNO3 
L.13
HX(aq)  NaOH(aq)  NaX(aq)  H2 O(l)
 0.750 mol NaOH 
(68.8 mL) 
  0.0516 mol NaOH
 1000 mL NaOH 
3.25 g HX corresponds to 0.0516 mol NaOH used
3.25 g
 63.0 g  mol1 = molar mass of acid
0.0516 mol
L.15
NaI(aq)
+ CuNO3(aq)  CuI(s) + NaNO3(aq)
 15.75 g CuI   1mol CuNO3  
1

M CuNO3 = 

 = 1.65 M
-1  
 190.45 g  mol   1mol CuI   0.0500 L 
SM-39
L.17
(a) Na 2 CO3 (aq)  2 HCl(aq)  2 NaCl(aq)  H2 CO3 (aq)
(b) First find the concentration of the diluted acid.
 0.832 g Na 2 CO3   1 mol Na 2 CO3 
? M HCl(aq) dilute = 


 0.100 L base solution   105.99 g Na 2 CO3 
 0.025 L base solution 


 0.031 25 L acid solution 
 2 mol HCl 

  0.126 M HCl(aq) dilute
 1 mol Na 2 CO3 
The original HCl solution is 100 times more concentrated than the solution
used for titration (diluted 10.00 mL to 1000 mL), so the original
concentration of the HCl solution is 12.6 mol  L1 .
L.19
I3  SnCl x  ( y  x)Cl  3 I  SnCl y
The information given can be used to find the molar mass of the reactant
in order to identify it.
 0.120 mol I3 
3

25.00 mL 
  3.00 10 mol I3
 1000 mL 
 19.0 g tin chloride 
30.00 mL 
  0.570 g tin chloride
1000 mL


If the reaction is 1:1 then the # moles of I3 is the same as the number of
moles of SnCl x . In that case, the molar mass of the tin chloride reactant is
0.570 g
=190. g  mol1 . This molar mass matches that of SnCl2 ,
3
3.00 10 mol
189.61 g  mol1 . Tin (II) chloride also has the correct mass percent tin.
118.71 g  mol1 Sn
100  62.6%
189.61 g  mol1 SnCl 2
Since the product compound is oxidized relative to the reactant, we can
expect it to be Sn(IV). The net ionic equation for the reaction is
SM-40
I3  Sn 2+  3 I  Sn 4+ . Another way to write a balanced reaction would
be I3  SnCl2 (aq)  2 Cl  3 I  SnCl 4 (aq) .
L.21
(a) S2O32− is both oxidized and reduced.
(b) Find the number of grams of thiosulfate ion in 10.1 mL of solution.
 1.45 g HSO (aq)  55.0 g HSO  
3
3


? g S2 O32  10.1 mL HSO3 (aq) 


 1 mL HSO (aq)  100 g HSO (aq) 
3
3



 1 mol HSO3  1 mol S2O32   112.0 g S2O32 

 
 
2 
 81.0 g HSO3  1 mol HSO3   1 mol S2 O3 
 11.1 g S O 2present initially
2
L.23
3
XCl4  2 NH3  XCl2 (NH3 )2  Cl2
The reactant and product that contain X are in a 1:1 ratio, so 3.571 g of the
reactant is equivalent to 3.180 g of the product. The molar mass of the
reactant is x + 4(35.453 g/mol) while that of the product is x + 2(35.453
g/mol)+2(14.01 g/mol) + 6(1.0079 g/mol). Therefore, we can set up the
following proportion in order to solve for x in g/mol:
x  141.8 3.571

x  104.97 3.180
1.1230 x  x  23.923
x  194.6 g  mol1 , or Pt
L.25
The number of moles of product is
2.27 g  208.23 g  mol1  0.0109 moles BaCl 2 . An equivalent number
of moles is represented by 3.25 g of BaBrx, so its molar mass is
3.25 g  0.0109 moles = 298 g  mol 1 BaBrx . Since 137.33 g is
attributable to Ba, 161 g must be Br. Each Br has a mass of 80.4 g/mol,
so there must be 2 moles of Br for each mole of Ba in the reactant.
x  2, BaBr2  Cl2  BaCl2  Br2
SM-41
L.27
First equation must be balanced as following:
Fe + Br2  FeBr2
3 FeBr2 + Br2  Fe3Br8
Fe3Br8 + 4 Na2CO3  8 NaBr + 4 CO2 + Fe3O4
The masses of Fe are needed to produce 2.5 t of NaBr:
 1000 kg   1000 g  1 mol NaBr   1 mol Fe3Br8 
2.50 t × 




 1 t   1 kg  102.9 g NaBr   8 mol NaBr 
 3 mol FeBr2  1 mol Fe   55.84 g Fe 
5
×


 = 5.09 × 10 g Fe
 1 mol Fe3Br8   1 mol FeBr2   1 mol Fe 
= 509 kg Fe
L.29
(a) M1V1 = M2V2 is used for calculation of dilution problems:
 1.00 L × 0.50 M 
V1 = 
 = 0.031 L = 31 mL
16.0 M


Pipette 31 mL of 16 M HNO3 into a 1.00 L volumetric flask which
contains about 800 mL of H2O. Dilute to the mark with H2O. Shake the
flask to mix the solution thoroughly.
 100 mL × 0.50 M 
2
(b) VNaOH = 
 = 2.5 × 10 mL
0.20 M


L.31
Total mass of tin oxide is: 28.35g – 26.45 g = 1.90 g
 1 mol Sn 
2
(a) 1.50 g Sn × 
 = 1.264 × 10 mol Sn
118.71
g
Sn


 1 mol O 
moles of O: (1.90  1.50)g × 
 = 0.025 mol O
 16.00 g O 
mole ratio of Sn:O is 1:2
empirical formula: SnO2
(b) tin(IV) oxide
SM-42
M.1
CaCO3 (s)  CaO(s)  CO2 (g)
theoretical yield:
 1 mol CaCO3   1 mol CO2   44.01 g CO2 
(42.73 g CaCO3 ) 



 100.09 g CaCO3   1 mol CaCO3   1 mol CO2 
 18.79 g CO2
actual yield:
17.5 g
 100%  93.1% yield
18.79 g
M.3
C x H y Cl z  ( x 
y
z
)O 2  x CO 2  y H 2 O+ Cl 2
2
2
1.52 g
2.224 g
 4.21103 mol Arochlor yields
1
360.88 g  mol
44.0 g  mol 1
 5.055 102 mol CO2
2.53 g
0.2530 g
 7.01103 mol Arochlor yields
1
360.88 g  mol
18.01 g  mol1
 1.405 102 mol H 2O
Therefore, x 
5.055  102
1.405  102

12.0
y

 2.00
and
4.21  103
7.01  103
12.011x  1.0079 y  35.453 z  360.88
12.011(12.0)  1.0079(2.00)  35.453 z  360.88
35.453z  214.7
z  6.06
Since the number or Cl atoms per Arochlor 1254 molecule must be a
whole number, the number of chlorine atoms is 6.
M.5
(a) P4 (s)  3 O2 (g)  P4 O6 (s)
P4 O6 (s)  2 O2 (g)  P4 O10 (s)
In the first reaction, 5.77 g P4 uses
 1 mol P4   3 mol O2   32.00 g O2 
(5.77 g P4 ) 


  4.47 g O2 (g)
 123.88 g P4   1 mol P4   1 mol O2 
SM-43
excess O2  5.77 g  4.47 g O2  1.30 g O2
In the second reaction, 5.77 g P4 uses

  1 mol P4 O6   2 mol O2   32.00 g O2 
5.77 g P4





1
 123.88 g  mol P4   1 mol P4   1 mol P4 O6   1 mol O2 
 2.98 g O2
limiting reagent: O2

  1 mol P4 O10   283.88 g P4 O10 
1.30 g O2
(b) 

  5.77 g P4 O10

1
2
mol
O
1
mol
P
O
32.00
g

mol
O
2 
4 10


2 

  1 mol P4 O6   219.88 g P4 O6 
1.30 g O2




(c)  32.00 g  mol1 O2   2 mol O2  1 mol P4 O6 
 4.47 g P4 O6 used
In the first reaction, 5.77 g P4 produces
 5.77 g P4
  219.88 g P4 O6  1 mol P4 O6 

  10.2 g P4 O6

1  
 123.88 g  mol   1 mol P4 O6   1 mol P4 
excess reagent: 10.2 g  4.47 g  5.7 g P4 O6
M.7
(a) Cu2+(aq) + 2 OH−(aq)  Cu(OH)2(s)
 1 mol NaOH 
(b) 2.00 g NaOH × 
 = 0.0500 mol NaOH
 40.0 g NaOH 
(0.0800 L)(0.500 M) = 0.0400 mol Cu(NO3)2
Moles of NaOH required to react with 0.04 mol Cu(NO3)2:
 2 mol NaOH 
0.0400 mol Cu(NO3 ) 2 × 
 = 0.0800 mol NaOH
 1 mol Cu(NO3 )2 
0.0800 mol > 0.0500 mol; therefore, NaOH is limiting reagent.
The mass of Cu(OH)2(s):
 1 mol Cu(OH)2   97.57 g Cu(OH) 2 
0.0500 mol NaOH × 
 = 2.44 g
×
2
mol
NaOH
1
mol
Cu(OH)

 
2 
SM-44
M.9
 1 mol CO2   1 mol C 
(0.682 g CO2 ) 

  0.0155 mol C
 44.01 g CO2   1 mol CO2 
(0.0155 mol C) (12.01 g  mol 1 C)  0.186 g C
 1 mol H 2 O   2 mol H 
(0.174 g H 2 O) 

  0.0193 mol H
 18.02 g H 2 O   1 mol H 2 O 
(0.0193 mol H) (1.0079 g  mol1 H)  0.0195 g H
 1 mol N 2   2 mol N 
(0.110 g N 2 ) 

  0.007 85 mol N
 28.02 g N 2   1 mol N 2 
(0.007 85 mol N) (14.01 g  mol1 N)  0.110 g N
mass of O  0.376 g  (0.186 g  0.0193 g  0.110 g)  0.061 g O
0.061 g O
 0.0038 mol O
16.00 g O
Dividing each amount by 0.0038 gives C : H : N : O ratios = 4.1 : 5.1 : 2.1
: 1. The empirical formula is C4 H5 N2 O.
The molecular mass of caffeine is 194 g  mol1 . Its empirical mass is
97.10 g  mol1 .
molecular formula  2  empirical formula  C8 H10 N 4 O2
2 C8 H10 N4 O2 (s)  19 O2 (g)  16 CO2 (g)  10 H2 O(l)  4 N2 (g)
M.11 Calculate the mass percentage composition; doing so eases the comparison
of data from multiple analyses.
 1 mol CO2   1 mol C   12.01 g C 
2.20 g CO2 × 
×
×
 = 0.600 g C
 44.01 g CO2   1 mol CO2   1 mol C 
0.600 g C
%C =
× 100% = 44.5% C
1.35 g unknown
SM-45
 1 mol H 2O   2 mol H   1.0079 g H 
0.901 g H 2O × 
×
×
 = 0.101 g H
 18.02 g H 2O   1 mol H 2O   1 mol H 
0.101 g H
%H =
× 100% = 7.48% H
1.35 g unknown
 1 mol N 2   2 mol N   14.01 g N 
0.130 g N 2 × 
×
×
 = 0.130 g N
 28.02 g N 2   1 mol N 2   1 mol N 
%N =
0.130 g N
× 100% = 26.0% N
0.500 g unknown
Oxygen must be present in the compound because the percentages of C, H,
and N only account for 78.0% of its composition. Combustion analysis
does not generate data directly for oxygen; we calculate it by difference:
%O = 100 − 78.0 = 22% O
To find the empirical formula, assume a sample size of 100 g and find the
mole ratios.
 1 mol C 
44.5 g C × 
 = 3.71 mol ×
 12.01 g C 
 1 mol H 
7.48 g H × 
 = 7.42 mol ×
 1.0079 g H 
 1 mol N 
26.0 g N × 

 14.01 g N 
 1 mol O 
22.0 g O × 

 16.00 g O 
1



 = 2.69 × 3 = 8.07
 1.38 mol 
1



 = 5.38 × 3 = 16.1
 1.38 mol 
1


= 1.86 mol × 
 = 1.35 × 3 = 4.05
 1.38 mol 
1


= 1.38 mol × 
 = 1.00 × 3 = 3.00
 1.38 mol 
The empirical formula is C8H16N4O3.
M.13 3 Ca(NO3 )2 (aq)  2 H3 PO4 (aq)  Ca 3 (PO4 )2 (s)  6 HNO3 (aq)
(a) The solid is calcium phosphate, Ca 3 (PO4 )2 .
 1 mol Ca(NO3 ) 2   2 mol H3 PO 4 
(b) (206 g Ca(NO3 ) 2 ) 


 164.10 g Ca(NO3 ) 2   3 mol Ca(NO3 ) 2 
 97.99 g H3 PO4 

  82.01 g H3 PO 4
 1 mol H3 PO4 
Therefore, Ca(NO3 )2 is the limiting reagent.
SM-46
 1 mol Ca(NO3 ) 2   1 mol Ca 3 (PO 4 ) 2 
(206 g Ca(NO3 ) 2 ) 


 164.10 g Ca(NO3 ) 2   3 mol Ca(NO3 ) 2 
 310.18 g Ca 3 (PO 4 ) 2 

  130 g Ca 3 (PO 4 ) 2
 1 mol Ca 3 (PO 4 ) 2 
M.15 If the 2-naphthol (144.16 g  mol1 ) were pure, it would give the following
combustion analysis:
%C 
10(12.01 g  mol 1 )
 100%  83.31% C
(144.16 g  mol 1 naphthol)
%H 
8(1.0079 g  mol 1 )
 100%  5.59% H
(144.16 g  mol 1 naphthol)
The observed percentages are low as is expected for a sample
contaminated with a substance that contains no C or H. Because the
sample does not contain C or H, the percent purity can be easily obtained
by
%purity (based on C) 
% found
77.48% mixture

 100%
% theoretical 83.31% pure naphthol
 93.00%
%purity (based on H) 
% found
5.20% mixture

 100%
% theoretical
5.59% pure naphthol
 93.0%
M.17 (a) C x H y O z  ( x 
y z
 ) O2  x CO2  y H 2O
2 2
2.492 g CO2
 5.664 102 mol CO2 = mol C
1
44.0 g  mol
0.6495 g H 2O
 3.608 102 mol H 2O = 7.216 102 mol H
1
18.01 g  mol
 12.011 g 
(5.664  102 mol C) 
  0.6803 g C
 mol 
 1.0079 g 
(7.216  102 mol H) 
  0.07273 g H
 mol 
SM-47
1.000 g compound  (0.6803 g + 0.07273 g) = 0.2470 g
0.2470 g O 16.00 g  mol1  1.544 102 mol O
5.664  102
 3.67
1.544  102
7.216  102
 4.67
1.544  102
The mole ratio of C:H:O is 3.67:4.67:1 or 11:14:3, so the empirical
formula is C11H14O3.
(b) The molar mass of the empirical formula is 194 g/mol, which is half of
388.46 g/mol. Therefore, the molecular formula of the compound is
C22H28O6.
M.19 Determine the number of moles of each element present in the compound
and then find their ratios to get the subscripts for the empirical formula.
0.055 g Cl
 1.55 103 mol Cl
1
35.453 g  mol
0.0682 g CO2
 1.55 103 mol CO2 = mol C
1
44.0 g  mol
0.0140 g H 2O
 7.78 104 mol H 2O = 1.56 103 mol H
1
18.01 g  mol
0.100 g compound
12.0 g C
1.0079 g H 

  0.055 g Cl +1.55 103 mol 
 1.55 10 3 mol 

mol
mol


 0.0247 g O
 1 mol 
3

  1.55 10 mol O
 16.00 g 
The mole ratio is 1:1:1:1, so the empirical formula is CHOCl.
As 0.100 g of the compound contains 1.55×10−3 moles of each element, its
molar mass is
0.100 g
= 64.52 g  mol1 . The molar mass of the
3
1.55×10 mol
empirical formula is 64.47 g∙mol−1, so the molecular formula is CHOCl.
SM-48
M.21 HA + XOH

H2O + XA
 2.45 g HA 
Moles of HA: 
= 0.0106 mol
1 
 231 g  mol 
 1.50 g XOH 
Moles of XOH: 
= 0.0120 mol
1 
125
g

mol


Molar mass of XA = (231−1.0) + (125−17) = 338 g∙mol−1
 2.91 g XA 
Mole of XA produced: 
= 8.61 × 10−3 mol
1 
338
g

mol


Theoretical yield: (based on the limiting reagent, HA)
 1 mol XA 
0.0106 mol HA × 
 = 0.0106 mol XA
 1 mole HA 
 8.61 × 103 mol XA 
Percentage yield: 
 × 100% = 81.2%
0.0106
mole
XA


SM-49