Solution for each chapter and Solution manual For Chemistry 4e Canadian Olmsted

,Solution manual For Chemistry 4e Canadian
Olmsted
Hello all ,
We have all what you need with best price
Our email :

Our website :
testbanks-store.com

,Chemistry, 4e Instructor Solutions Manual Chapter 1



Chapter 1 Fundamental Concepts of Chemistry

Solutions to Problems in Chapter 1
1.1 You must commit to memory the correspondence between names and symbols of
various elements, but remembering them is simplified by the fact that most English
names and elemental symbols are related: (a) H; (b) He; (c) Hf; (d) N; (e) Ne; and (f)
Nb.

1.2 You must commit to memory the correspondence between names and symbols of
various elements, with special attention to those elements whose symbols come from
Latin words rather than their English names: (a) K; (b) Pt; (c) Pu; (d) Pb; (e) Pd; and
(f) P.

1.3 Associating an element’s name with its symbol requires memorization of both names
and symbols. The examples in this problem all begin with the letter “A” but their
names do not necessarily begin with “A” too: (a) arsenic; (b) argon; (c) aluminum; (d)
americium; (e) silver; (f) gold; (g) astatine; and (h) actinium.

1.4 Associating an element’s name with its symbol requires memorization of both names
and symbols. The examples in this problem all begin with the letter “B:” (a) bromine;
(b) beryllium; (c) boron; (d) berkelium; (e) barium; and (f) bismuth.

1.5 To convert a molecular picture into a molecular formula, count the atoms of each type
and consult (or recall) the colour scheme used for the elements. See Figure 1-3 of
your textbook for the colour scheme used in this and many other texts: (a) Br2; (b)
HCl; (c) C2H5I; and (d) C3H6O.

1.6 To convert a molecular picture into a molecular formula, count atoms of each type
and consult (or recall) the colour scheme used for the elements. See Figure 1-3 of
your textbook for the colour scheme used in this and many other texts: (a) PCl3; (b)
SF4; (c) N2O4; (d) C3H8O; and (e) H2S.

1.7 In writing a chemical formula, remember to use elemental symbols and subscripts for
the number of atoms: (a) CCl4; (b) H2O2; (c) P4O10; and (d) Fe2S3.

1.8 In writing a chemical formula, remember to use elemental symbols and subscripts for
the number of atoms: (a) C5H12; (b) SiF4; (c) N2O5; and (d) FeCl3.

1.9 Scientific notation expresses any number as a value between 1 and 10 times a power
of ten. Trailing zeros are retained only when they are significant: (a) 1.00000 × 105;
(b) 1.0 × 104; (c) 4.00 × 10-4; (d) 3 × 10-4; and (e) 2.753 × 102.

1.10 Scientific notation expresses any number as a value between 1 and 10 times a power
of ten. Trailing zeros are retained only when they are significant: (a) 1.75906 × 105;
(b) 6.05 × 10−5; (c) 2.5000 × 106; and (d) 2.500 × 109.

© 2021 John Wiley and Sons Canada, Ltd. 1-1

,Chemistry, 4e Instructor Solutions Manual Chapter 1


1.11 To do unit conversions, multiply by a ratio that cancels the unwanted unit(s). Refer
to Table 1-2 for the SI base units:

(a) 432 kg = 4.32 × 102 kg

 10−12 s  −10
(b) 624 ps   = 6.24 10 s
 1 ps 

 10−9g   10−3 kg  −9
(c) 1024 ng     = 1.024 10 kg
 1 ng   1 g 

 103 m 
 = 9.300 10 m
7
(d) 93 000 km 
 1 km 
 24 h   60 min  60 s 
 = exactly 8.64 10 s (assuming exactly 1 day)
4
(e) 1 day   
 1 day   1 h  1 min 
 2.54 cm  1 m  −3
(f) 0.0426 in    = 1.08 10 m
 1 in  100 cm 

1.12 To do unit conversions, multiply by a ratio that cancels the unwanted unit(s). Refer
to Table 1-2 for the SI base units:

 7 days   24 h   60 min  60 s 
 = 6.048 10 s (exactly 1 week)
5
(a) 1 week    
 1 week   1 day   1 h  1 min 
 1m  −3
(b)1.35 mm   = 1.35 10 m
 1000 mm 
 1000 m 
 = 1.5 10 m
3
(c) 15 km 
 1 km 
 1s  −6
(d) 4.567 μs  6  = 4.567 10 s
 10 μs 
 1 L  1 m  −6
(e) 6.45 mL    = 6.45 10 m
 1000 mL  1000 L 
(f) 47 kg = 4.7 10 kg
1




1.13 This is a unit-conversion problem involving summation and unusual units. First
convert all masses into kg, then put them into the same power of ten and add the
masses:

 3.168 grains   1g   10−3 kg 
mdiamonds = 5.0 10−1 carat   
−4
 = 1.0 10 kg
 1 carat   15.4 grains  1 g 

© 2021 John Wiley and Sons Canada, Ltd. 1-2

,Chemistry, 4e Instructor Solutions Manual Chapter 1



 10−3 kg  −3
mgold = 7.00 g   = 7.00 10 kg
 1 g 

mtotal = (7.00 10−3 kg) + (0.10 10−3 kg) = 7.10 10−3 kg

1.14 When values are added, they must all be in the same units and power of ten. The
mass of water must first be converted to grams:

 103 g 
m(water) = 2.000 kg   = 2.000 10 g
3

 1 kg 

m(sodium chloride) = 0.0065 × 103 g

m(sugar) = 0.047546 × 103 g

Sum and round: (2.000 + 0.0065 + 0.047546) × 103 g = 2.054 × 103 g

1.15 The question asks for the density of water expressed in SI units (kg/m3). Begin by
analyzing the given information. The mass of the container is given before and after
the water has been added. Thus, the mass of water can be obtained from the
difference between the masses of the filled and empty container:

m = 270.064 g – 93.054 g = 177.010 g H2O

 10−3 kg 
In SI units, m = 177.010 g   = 0.177010 kg
 1g 

Convert the units from inches to metres before computing the volume:

 1m 
r = 2.22 cm   = 0.0222 m
 100 cm 

 1m 
h = 11.43 cm   = 0.1143 m
 100 cm 

V = π r2h = (3.1416)(0.0222 m)2(0.1143 m) = 1.77 × 10-4 m3

Calculate the density by dividing the mass of water by the volume:

m 0.177010 kg
= = = 9.98 102 kg/m3 or 0.998 g/cm 3 (Round to three significant
V 1.77 10−4 m3
figures because the radius of the cylinder is known to only three significant figures.)

© 2021 John Wiley and Sons Canada, Ltd. 1-3

,Chemistry, 4e Instructor Solutions Manual Chapter 1


1.16 The question asks for the density of an organic liquid expressed in g/cm3. The mass
of the container is given before and after the organic liquid has been added. Thus, the
mass of liquid is the difference between the mass of the filled container and the empty
container:

m = 0.4827 g – 0.4763 g = 6.4 × 10−3 g

Divide the mass of liquid by the volume to obtain density:

m 6.4 10−3 g
= = = 0.80g cm3
V 8.00 10−3 cm3

(The result has two significant figures because the mass of the liquid has only two
significant figures.)

1.17 The question asks for a comparison of the masses of two objects of different
densities and shapes. For each object, m = V:

4
Au sphere:  = 19.3 g/cm3, V =   r 3 and r = diameter/2 = 1.00 cm
3

4(1.00 cm)3  19.3 g 
V= = 4.19 cm3 and mgold = 4.19 cm3  3 
= 80.9 g
3  1 cm 

Ag cube:  = 10.50 g/cm3 , V = lwh, and l = w = h = 2.00 cm

 10.50 g 
V = (2.00 cm)3 = 8.00 cm3 and msilver = 8.00 cm 3  3 
= 84.0 g
 1 cm 

The silver cube has more mass than the gold sphere.

1.18 The question asks for a comparison of the masses of two objects of different
densities and shapes. For each object, m = V:

Zn sphere:  = 7.14 g/cm3, V = (4/3)r3, and r = diameter/2 = 1.00 cm

V =(4/3)(1.00 cm)3 = 4.19 cm3 and m(zinc) = 4.19 cm3(7.14 g/cm3) = 29.9 g

Al cube:  = 2.70 g/cm3, V= l w h, and l=w=h=2.00 cm

V = (2.00 cm)3 = 8.00 cm3 and m(aluminum) =8.00 cm3 (2.70 g/cm3) = 21.6 g

The zinc sphere has more mass than the aluminum cube.

© 2021 John Wiley and Sons Canada, Ltd. 1-4

,Chemistry, 4e Instructor Solutions Manual Chapter 1


1.19 Volume and mass are related through density:

m  36.5 g 
V= = = 11.7 mL
  3.12 g/mL 

1.20 Mass and volume are related through density, m = V  :

 103 mL   0.70 g   10−3 kg 
m = 45.4 L     = 31.8 kg
 1 L   1 mL   1 g 

1.21 The molar mass of a naturally occurring element can be calculated by summing the
product of the fractional abundance of each isotope times its isotopic molar mass:

 0.337% 
 = 0.121 g/mol
36
Ar: 35.96755 g/mol 
 100% 

 0.063% 
 = 0.024 g/mol
38
Ar: 37.96272 g/mol 
 100% 

 99.600% 
 = 39.803 g/mol
40
Ar: 39.96238 g/mol 
 100% 

Elemental molar mass = 0.121 g/mol + 0.024 g/mol + 39.803 g/mol = 39.948 g/mol

1.22 The molar mass of a naturally occurring element can be calculated by summing the
product of the fractional abundance of each isotope times its isotopic molar mass:

 92.23% 
 = 25.80 g/mol
28
Si: 27.97693 g/mol 
 100% 
 3.10% 
 = 0.898 g/mol
28
Si: 28.97649 g/mol 
 100% 
 4.67% 
 =1.400 g/mol
28
Si: 29.97376 g/mol 
 100% 

Elemental molar mass = 25.80 g/mol + 0.898 g/mol + 1.400 g/mol = 28.10 g/mol

1.23 Mass–mole conversions require the use of masses in grams and molar masses in
grams per mole. To determine the number of moles, convert the mass into grams and
divide by the molar mass:




© 2021 John Wiley and Sons Canada, Ltd. 1-5

,Chemistry, 4e Instructor Solutions Manual Chapter 1


m
n=
M

 1 mol 
(a) n = 7.85 g   = 0.141 mol
 55.85 g 

 10−6 g  1 mol  −6
(b) n = 65.5 g    = 5.45  10 mol
 1 μg   12.01 g 

 10−3 g  1 mol  −4
(c) n = 4.68 mg    = 1.67  10 mol
 1 mg   28.09 g 

 106 g  1 mol 
 = 5.4110 mol
4
(d) n = 1.46 ton  
 1 ton   26.98 g 

1.24 Mass–mole conversions require the use of masses in grams and molar masses in
grams per mole. To determine the number of moles, convert the mass into grams and
divide by the molar mass:

(a) n = 3.67 kg (103 g/kg)(1 mol/47.87 g) = 76.7 mol

(b) n = 7.9 mg (10–3 g/mg)(1 mol/40.08 g) = 2.0 × 10–4 mol

(c) n = 1.56 g (1 mol/101.07 g) = 1.54 × 10–2 mol

(d) n = 9.63 pg (10–12 g/pg)(1 mol/98.9 g) = 9.7 × 10–14 mol

1.25 To calculate the number of atoms in a mass, convert the mass to grams, divide by
molar mass to obtain moles and multiply by Avogadro’s number to obtain the number
of atoms:

 10−3 g   1 mol   6.022 1023 atoms 
(a) # = 5.86 mg     = 3.92 10 atoms
20

 1 mg   9.012 g   1 mol 

 10−3 g   1 mol   6.022 1023 atoms 
(b) # = 5.86 mg     = 1.14 10 atoms
20

 1 mg   30.97 g   1 mol 

 10−3 g   1 mol   6.022 1023 atoms 
(c) # = 5.86 mg     = 3.87 10 atoms
19

 1 mg   91.22 g   1 mol 



© 2021 John Wiley and Sons Canada, Ltd. 1-6

,Chemistry, 4e Instructor Solutions Manual Chapter 1



 10−3 g   1 mol   6.022 1023 atoms 
(d) # = 5.86 mg     = 1.48 10 atoms
19

 1 mg  238.0 g  1 mol 

1.26 To calculate the number of atoms in a mass, convert the mass to grams, divide by
molar mass to obtain moles and multiply by Avogadro’s number to obtain the number
of atoms:

 103   1 mol   6.022 1023 atoms 
 = 3.68 10 atoms
25
(a) # = 2.44 kg   
 1 kg   39.92 g   1 mol 
 10   1 mol   6.022 10 atoms 
3 23

 =1.36 10 atoms
25
(b) # = 2.44 kg   
 1 kg   107.9 g   1 mol 
 103   1 mol   6.022 1023 atoms 
 =1.16 10 atoms
25
(c) # = 2.44 kg   
 1 kg   126.9 g  1 mol 
 10   1 mol   6.022 10 atoms 
3 23

 = 3.76 10 atoms
25
(d) # = 2.44 kg   
 1 kg   39.10 g   1 mol 

1.27 To calculate the molar mass of a compound, multiply each elemental molar mass by
the number of atoms in the formula and sum over the elements:

(a) CCl4: M = 12.01 g/mol C + 4(35.45 g/mol Cl) = 153.81 g/mol
(b) K2S: M = 2(39.10 g/mol K) + 32.07 g/mol S = 110.27 g/mol
(c) O3: M = 3(16.00 g/mol O) = 48.00 g/mol
(d) LiBr: M = 6.94 g/mol Li + 79.90 g/mol Br = 86.84 g/mol
(e) GaAs: M = 69.72 g/mol Ga + 74.92 g/mol As = 144.64 g/mol
(f) AgNO3: M = 107.87 g/mol Ag + 14.01 g/mol N + 3(16.00 g/mol O) = 169.88
g/mol

1.28 To calculate the molar mass of a compound, multiply each elemental molar mass by
the number of atoms in the formula and sum over the elements:

(a) (NH4)2CO3:

M = 2[(14.01 g/mol N) + 4(1.008 g/mol H)] + 12.01 g/mol C + 3(16.00 g/mol O) =
96.09 g/mol

(b) N2O: M = 2(14.01 g/mol N) + 16.00 g/mol O = 44.02 g/mol

(c) CaCO3: M = 40.08 g/mol Ca + 12.01 g/mol C + 3(16.00 g/mol O) = 100.09 g/mol

(d) NH3: M = 14.01 g/mol N + 3(1.008 g/mol H) = 17.03 g/mol


© 2021 John Wiley and Sons Canada, Ltd. 1-7

, Chemistry, 4e Instructor Solutions Manual Chapter 1


(e) Na2SO4:

M = 2(22.99 g/mol Na) + 32.07 g/mol S + 4(16.00 g/mol O) = 142.05 g/mol

(f) C4H10: M = 4(12.01 g/mol C) + 10(1.008 g/mol H) = 58.12 g/mol

1.29 Determine the molecular formula from the line drawing, taking into account the
“missing” carbon atoms at the ends and vertices of lines and the “missing” hydrogen
atoms attached to carbon atoms. To calculate the molar mass, multiply each elemental
molar mass by the number of atoms in the formula and sum over the elements:

(a) Tyrosine: 9 C atoms, 7 missing H atoms, 4 shown H atoms, 1 N atom, 3 O atoms.

C9H11NO3: M = 9(12.01 g/mol) + 11(1.008 g/mol) + 1(14.01 g/mol) + 3(16.00 g/mol)
= 181.19 g/mol

(b) Tryptophan: 11 C atoms, 8 missing H atoms, 4 shown H atoms, 2 N atoms, 2 O
atoms.

C11H12N2O2: M =11(12.01 g/mol) + 12(1.008 g/mol) + 2(14.01 g/mol) + 2(16.00
g/mol) = 204.23 g/mol

(c) Glutamic acid: 5 C atoms, 9 H atoms, 1 N atom, 4 O atoms.;

C5H9NO4: M = 5(12.01 g/mol) + 9(1.008 g/mol) + 1(14.01 g/mol) + 4(16.00 g/mol) =
147.13 g/mol

(d) Lysine: 6 C atoms, 14 shown H atoms, 2 N atoms, 2 O atoms.

C6H14N2O2: M = 6(12.01 g/mol) + 14(1.008 g/mol) + 2(14.01 g/mol) + 2(16.00
g/mol) = 146.19 g/mol

1.30 Determine the molecular formula from the line drawing, taking into account the
“missing” carbon atoms at the ends and vertices of lines and the “missing” hydrogen
atoms attached to carbon atoms. To calculate the molar mass, multiply each elemental
molar mass by the number of atoms in the formula and sum over the elements.

Biotin: 10 C atoms, 13 missing H atoms, 3 shown H atoms, 2 N atoms, 3 O atoms,1 S
atom.

C10H16N2O3S: M = 10(12.01 g/mol) + 16(1.008 g/mol) + 2(14.01 g/mol) + 3(16.00
g/mol) + 1(32.07 g/mol) = 244.32 g/mol


(b) Nicotinamide: 6 C atoms, 4 missing H atoms, 2 shown H atoms, 2 N atoms, 1 O
atom.

© 2021 John Wiley and Sons Canada, Ltd. 1-8