Unit 3 Review, pages 356-363
Knowledge
1. (c)
2. (b)
3. (c)
4. (a)
5. (b)
6. (a)
7. (a)
8. (b)
9. (b)
10. (d)
11. (b)
12. (b)
13. (a)
14. (d)
15. (b)
16. (a)
17. False. Proper placement of a decimal point when prescribing a medication is vital for
accuracy of administration to the patient.
18. False. A mole may be used to count any entity.
19. False. Avogadro’s constant is the number of entities per mole of a substance.K/U]
20. False. A mole of one element has the same number of entities as a mole of another element.
21. False. The units used for reporting molar mass are grams/mole (g/mol).
22. False. 1 mol of H2 has 6.02 × 1023 molecules. Or 1 mol of H2 has 1.204 × 1024 atoms.
23. False. The metals combined to make an alloy are not always present in the same amounts.
24. False. The proportion of an element in a compound does not vary with the mass of the
sample of the compound.
25. True
26. False. The mass of sample required to produce a mass spectrum is usually 1 µg or larger.
27. True
28. True
29. False. Stoichiometry is the area of chemistry that deals with quantitative relationships among
reactants and products.
30. False. The reactant that is not entirely consumed in a chemical reaction is the excess reagent.
31. False. The actual yield is the amount of product actually produced.
32. The molar mass of a monatomic element is equivalent to the mass given on the periodic
table. For example, argon is a monatomic element or gas. As a gas, its molecules consist of one
atom. Its molar mass is therefore 39.95 u.
33. A mass spectrum is a compound’s “fingerprint” because the fragment pattern produced by
the mass spectrometer is often unique to the substance. The spectrum, along with the
compound’s molecular mass, allows the determination of the compound’s molar mass and,
therefore, its identity.
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-5
Understanding
34. Given: mass of acetic acid, mHC H O = 400 g
2
3
2
Required: amount of acetic acid, nHC H O
2
3
2
number of molecules of acetic acid, N HC H O
2
3
2
Solution:
Step 1. Determine the molar mass of acetic acid.
M HC H O = 4 M H + 2 M C + 2 M O
2
3
2
!
!
!
g $
g $
g $
= 4 # 1.01
+
2
12.01
+
2
16.00
#"
#"
mol &%
mol &%
mol &%
"
M HC H O = 60.06 g/mol [three extra digits carried]
2
3
2
Step 2. Determine the amount of acetic acid in 400 g.
mHC H O
2 3 2
nHC H O =
2 3 2
M HC H O
2
3
2
! 1 mol $
= 400 g #
&
" 60.06 g %
nHC H O
2
3
= 6.6666... mol
= 7 mol
2
Step 3. Determine the number of molecules in 400 g.
N HC H O = nHC H O ! N A
2
3
2
2
3
2
"
molecules %
= (7 mol ) $ 6.02 ! 1023
'
#
mol &
N HC H O = 4 ! 1024 molecules
2
3
2
Statement: There are 7 mol and 4 ! 1023 molecules of acetic acid in 400 g of acetic acid.
35. Given: number of silver atoms, N Ag = 8.0 ! 1022 atoms
Required: amount of silver, nAg
Solution: Determine the amount of silver.
N Ag
nAg =
NA
"
%
1 mol
= 8.0 ! 1022 atoms $
'
# 6.02 ! 1023 atoms &
nAg = 1.3 ! 10 –1 mol
Statement: There are 1.3 ! 10–1 mol of silver in 8.0 ! 1022 atoms.
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-6
36. Given: amount of iron(II) sulfide, nFeS = 0.500 mol
2
Required: amount of iron(III) oxide, nFe O ; mass of iron(III) oxide, mFe O
2
3
2
Solution: The chemical equation is balanced.
Step 1. List the given and required values.
4 FeS2(s)
+ 11 O2(g)
→ 2 Fe2O3(s)
0.500 mol
nFe O
2
3
+ 8 SO2(g)
3
Step 2: Convert amount of iron sulfide to amount of iron(III) oxide, using the mole ratio.
2 molFe O
2 3
nFe O = 0.500 mol FeS2S !
2 3
4 molFeS
2
nFe O = 0.125 mol
2
3
Step 3. Determine the molar mass of iron(III) oxide.
M Fe O = 2 M Fe + 3M O
2
3
!
!
g $
g $
= 2 # 55.85
+ 3# 16.00
&
mol %
mol &%
"
"
M Fe O = 159.7 g/mol [one extra digit carried]
2
3
Step 4. Convert amount of iron(III) oxide to mass of iron(III) oxide.
!
g $
mFe O = (0.250 mol ) # 159.7
&
2 3
"
mol %
mFe O = 39.9 g
2
3
Statement: The amount of iron(III) sulfide produced is 0.250 mol and the mass produced is
39.9 g.
37. (a) Given: number of gold mines, Ng = 50 = 5.0 × 10
mass of Canadian gold, mC = 1.0 × 105 kg
Required: average mass of gold produced per Canadian mine, GC
Solution: Determine the average gold production per mine.
" 1.0 ! 105 kg % " 1.0 ! 103 g %
GC = $
'$
'
# 5.0 ! 10 & # 1 kg &
= 0.02 ! 108 g
GC = 2.0 ! 106 g
Statement: The average gold production for each Canadian mine was 2.0 × 106 g.
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-7
(b) Given: mass of Canadian gold, mC = 1.0 × 105 kg
mass of worldwide gold, mI = 2.3 × 106 kg
Required: percentage of world’s gold that is Canadian, % C
Solution: Determine the Canada’s god production as a percent.
m
% C = C ! 100 %
mI
=
1.0 ! 105 kg
2.3 ! 106 kg
! 100 %
= 0.4347 ! 10 –1 ! 100 % [two extra digits carried]
= 0.04347 ! 100 %
% C = 4.3 %
Statement: Canada produced 4.3 % of the worldwide mass of gold in 2008.
(c) Given: mass of Canadian gold, mC = 1.0 × 105 kg
selling price of gold, s = $43/g
Required: value of Canadian gold, VC
Solution: Determine the dollar value of Canadian gold.
" 1.0 ! 103 g % " $43 %
5
VC = 1.0 ! 10 kg $
'$
'
# 1 kg & # 1 g &
= $43 ! 108 dollars
VC = $4.3 ! 109 dollars
Statement: The value of the Canadian gold was $4.3 × 109 in 2008.
38. Given: molar mass of nitrogen atom, MN = 14.01 g/mol
Avogadro’s constant, NA = 6.02 × 1023 atoms/mol
Required: mass of nitrogen atom, mN
Solution:
M
mN = N
NA
$
! 14.01 g $ !
1 mo l
=#
#
&
&
23
" 1 mol % " 6.02 ' 10 atoms %
mN = 2.33 ' 10 –23 g/mol
Statement: The mass of one atom of nitrogen is 2.33 × 10–23 g/atom.
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-8
39. Given: calcium hydroxide, Ca(OH)2(s)
Required: molar mass of calcium hydroxide, M Ca(OH)
2
Solution:
Step 1. Look up molar masses of elements.
MCa = 40.08 g/mol; MO = 16.00 g/mol; MH = 1.01 g/mol
Step 2. Add molar masses of elements.
M Ca(OH) = M Ca + 2 M O + 2 M H
2
= 40.08
!
!
g
g $
g $
+ 2 # 16.00
+ 2 # 1.01
&
mol
mol %
mol &%
"
"
M Ca(OH) = 74.10 g/mol
2
Statement: The molar mass of calcium hydroxide is 74.10 g/mol.
40. Given: amount of titanium, nTi = 0.473 mol
Required: mass of titanium, mTi
Solution:
! 47.87 g $
mTi = #
& 0.473 mol
" 1 mol %
(
)
mTi = 22.6 g
Statement: The mass of titanium in 0.475 mol is 22.6 g.
41. Given: mass of zinc, mZn = 4.20 g
Required: amount of zinc atoms, nZn
Solution:
! 1 mol $
nZn = 4.20 g #
&
" 65.39 g %
nZn = 0.0642 mol
Statement: The amount of zinc in 4.20 g of zinc is 0.0642 mol.
43. The molar mass of a substance is the mass of 1 mol of that substance. The atomic mass of a
substance is the weighted average of the masses of all the naturally occurring isotopes of that
element, and is the mass of a single atom. The molar mass and the atomic mass of a monoatomic
element have the same numerical value, but different units. The unit for molar mass is g/mol; the
unit for atomic mass
is u.
44. Given: mass of aluminum, mAl = 10.00 g
Required: number of aluminum atoms, NAl
Solution:
Step 1. Look up the molar mass of aluminum.
MAl = 26.98 g/mol
Step 2. Calculate the amount of aluminum in 10.00 g.
! 1 mol $
nAl = (10.00 g) #
&
" 26.98 g %
nAl = 0.3706 mol
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-9
Step 3. Calculate the number of atoms in the sample.
" 6.022 ! 1023 atoms %
N Al = (0.3706 mol ) $
'
#
&
1 mol
N Al = 2.23 ! 1023 atoms
Statement: There are 2.23 × 1023 atoms of aluminum in 10.00 g.
45. (a) Since the sample has the same mass as one mole of the substance, the sample must
contain one mole of phosphoric acid molecules. One mole of a substance always contains a
number of molecules equal to Avogadro's constant. The sample therefore contains
6.02 × 1023 phosphoric acid molecules
(b) Given: number of phosphoric acid molecules, N H PO = 6.02 × 1023
2
4
Required: total number of atoms, N atoms
Solution:
Step 1. Determine the number of atoms in one molecule of phosphoric acid.
" 3 H + 1 P + 4 O = 8 atoms
H3PO4 !!
Step 2. Determine the number of atoms in the sample.
N atoms = 8 ! N A
= 8(6.02 ! 1023 atoms)
N atoms = 4.82 ! 1024 atoms
Statement: There are 4.82 × 1024 atoms in 1 mol of phosphoric acid.
46. Given: mass of mercury, mHg = 0.50 g
Required: number of atoms of mercury, NHg
Solution:
Step 1. Look up the molar mass of mercury.
MHg = 200.59 g/mol
Step 2. Calculate the amount of mercury in 0.50 g.
! 1 mol $
nHg = 0.50 g #
&
" 200.59 g %
nHg = 0.002 492 [two extra digits carried]
Step 3. Determine the number of atoms in 0.50 g of mercury.
" 6.02 ! 1023 atoms %
N Hg = (0.002 492 mol ) $
'
#
&
1 mol
N Hg = 1.5 ! 1021 atoms
Statement: There are 1.5 × 1021 atoms in 0.50 g of mercury.
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-10
47. Given: mass of nitric acid, mHNO = 6.0 mg
3
Required: number of molecules of nitric acid, N HNO
3
Solution:
Step 1. Determine the molar mass of nitric acid.
M
= 1M H + 1M N + 3M O
HNO3
= 1.01
!
g
g
g $
+ 14.01
+ 3# 16.00
mol
mol
mol &%
"
g
[two extra digits carried]
HNO3
mol
Step 2. Determine the amount of nitric acid in 6.0 mg.
#
g &#
1 mol &
nHNO = (6.0 mg ) % 1.0 ! 10"3
(
%
(
3
mg ' $ 6.302 ! 10 g '
$
= 6.302 ' 10
M
nHNO = 9.9469 ! 10"4 mol [three extra digits carried]
3
Step 3. Determine the number of molecules of mercury.
# 6.02 ! 1023 molecules &
"4
N HNO = (9.9469 ! 10 mol ) %
(
3
$
'
1 mol
N HNO = 5.7 ! 1019 molecules
3
Statement: There are 5.7 × 1019 molecules of nitric acid in 6.0 mg.
48. Given: mass of compound, m = 100 g
percent of hydrogen, % H = 37 %
percent of oxygen, % O = 63 %
Required: mass of hydrogen, mH; mass of oxygen, mO
Solution:
mO = % O ! m
mH = % H ! m
= 0.37(100 g)
= 0.63(100 g)
mH = 37 g
mO = 63 g
Statement: In the 100 g sample, 37 g is hydrogen and 63 g is oxygen.
49. Given: mass of silver chloride, mAgCl = 8.66 g
total mass of potassium nitrate and potassium chloride, mreactants = 20.0 g
Required: percentage of potassium nitrate, % KNO3; percentage of potassium chloride, % KCl
Solution:
Step 1. Determine the molar mass of silver chloride.
M = 1M Ag + 1M Cl
AgCl
g
g
+ 35.45
mol
mol
g
= 143.32
[two extra digits carried]
mol
= 107.87
M
AgCl
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-11
Step 2. Determine the percentage of chlorine in silver chloride.
M Cl
% Cl =
! 100 %
M AgCl
=
35.45 g/mol
143.32 g/mo1
! 100 %
% Cl = 24.73 % [one extra digit carried]
Step 3. Determine the mass of chlorine in the product.
m = % Cl ! mAgCl
Cl
= 0.2473 ! 8.66 g
m = 2.1416 g [two extra digits carried]
Cl
Step 4. Determine the molar mass of potassium chloride.
M = 1M K + 1M Cl
KCl
g
g
+ 35.45
mol
mol
g
M = 74.55
[one extra digit carried]
KCl
mol
Step 5. Determine the percentage of chlorine in potassium chloride.
M Cl
% Cl =
! 100 %
M KCl
= 39.10
=
35.45 g/mol
74.55 g/mo1
! 100 %
% Cl = 47.55 % [one extra digit carried]
Step 6. Determine the mass of potassium chloride in the 20.0 sample.
Since the chlorine in the product can only have come from the potassium chloride, the mass of
chlorine in the silver chloride and the mass of the chlorine in the potassium chloride are equal,
2.1416 g (two extra digits carried).
Step 7. Determine the mass of potassium chloride present in the solution.
m
m = Cl
KCl
% Cl
2.1416 g
=
0.4755
m = 4.503 g [one extra digit carried]
KCl
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-12
Step 8. Determine the mass of the other compound, potassium nitrate, in the original solution.
m
= 20.0 g – m
AgNO3
m
KCl
= 20.0 g ! 4.503 g
= 15.496 g [two extra digits carried]
AgNO3
Step 9. Determine the percent compositions of potassium chloride and potassium nitrate.
m
% KCl = KCl ! 100 %
20.0 g
=
4.503 g
20.0 g
! 100 %
% KCl = 22.5 %
% KNO3 = 100 % – % KCl
= 100 % 22.5 %
% KNO3 = 77.5 %
Statement: The percentage of potassium nitrate in the original solution is 77.5 % and the
percentage of potassium chloride is 22.5 %.
50. Given: iron(III) oxide, Fe2O3
Required: percentage by mass of iron, % Fe; percentage by mass of oxygen, % O
Solution:
Step 1. Calculate molecular mass of the compound.
M Fe O = 159.7 u
2
3
Step 2. Calculate the percentage of each element.
! 111.7 u $
! 48.00 u $
% Fe = #
(100 %)
% O= #
(100 %)
&
" 159.7 u %
" 159.7 u &%
% Fe = 69.94 %
% O = 30.06 %
Step 3. Check that the percentages add up to 100%.
69.94 % + 30.06 = 100 %
Statement: The percentage of iron is 69.96 % and the percentage of oxygen is 30.06 %.
51. Given: nicotine, C10H14N2
Required: percentage by mass of nitrogen in nicotine, % N
Step 1. Calculate the molecular mass of nicotine.
M C H N = 162.26 u
10
14
2
Step 2. Determine the percentage by mass of nitrogen.
! 28.02 u $
%N= #
(100 %)
" 162.26 u &%
% N = 17.3 %
Statement: The percentage by mass of nitrogen in nicotine is 17.3 %.
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-13
52. Given: glucose, C6H12O6
Required: percentage of hydrogen in glucose, % H
Solution:
Step 1. Calculate the molecular mass of glucose.
M
= 180.18 u
C6 H12O6
Step 2. Determine the percentage by mass of hydrogen.
! 12.12 u $
%H= #
(100 %)
" 180.18 u &%
% H = 6.73 %
Statement: The percentage of hydrogen in glucose is 6.73 %.
53. Given: mass of alloy of tin, antimony, and lead, malloy = 500 g
percentage of tin, % Sn = 10 %
percentage of antimony, % Sb = 16 %
percentage of lead, % Pb = 74 %
Required: mass of tin, mSn; mass of antimony, mSb; mass of lead, mPb
Solution:
mSn = % Sn ! malloy
mSb = % Sb ! malloy
mPb = % Pb ! malloy
= 0.10 ! 500 g
mSn = 50 g
= 0.16 ! 500 g
mSb = 80 g
= 0.74 ! 500 g
mPb = 370 g
Statement: The mass of tin is 50 g, the mass of antimony is 80 g, and the mass of lead is 370 g.
54. Given: aluminum : sulfur = 1.00 : 1.78
Required: empirical formula of compound, AlxSy
Solution:
Al xS y = Al1S1.78
= Al1!2S1.78! 2
Al xS y = Al2S3
Statement: The empirical formula of aluminum–sulfur compound is Al2S3.
55. Given: percentage of carbon, % C = 80 %
percentage of hydrogen, % H = 20 %
molar mass of CxHy = 30.01 g/mol
Required: empirical formula of carbon–hydrogen compound; molecular formula of carbon–
hydrogen compound
Solution: A 100.0 g sample of this compound contains 80 g of carbon and 20 g of hydrogen.
Step 1. Determine the amount of each element in the compound.
! 1 mol $
! 1 mol $
nC = 80 g #
n
=
20
g
&
# 1.01 g &
H
" 12.01 g %
"
%
nC = 6.661 [two extra digits carried]
Copyright © 2011 Nelson Education Ltd.
nH = 19.80 [two extra digits carried]
Unit 3: Quantities in Chemical Reactions
U3-14
Step 2. Determine the simplest ratio.
nC 6.661 mol
=
nC 6.661 mol
nC
=1
nC
Step 3. Determine the empirical formula.
ratio of carbon : hydrogen atoms = 1 : 3
Statement: The empirical formula is CH3.
Step 4. Determine the molar mass of CH3.
M CH = 1M C + 3M H
nH 19.80 mol
=
nC 6.661 mol
nH
= 2.972 [two extra digits carried]
nC
3
!
!
g $
g $
= # 12.01
+ 3# 1.01
&
mol %
mol &%
"
"
g
[one extra digit carried]
CH 3
mol
Step 5. Solve for the mass multiple, x.
molar mass of compound
x=
molar mass of empirical formula
M
=
= 15.04
30.01 g/mol
15.04 g/mol
x = 2.001 [two extra digits carried]
Step 6. Determine the molecular formula.
The molar mass of the compound is twice the mass of the empirical formula.
C x H y = C1!2 H 3!2
C x H y = C2 H 6
Statement: The molecular formula of the compound is C2H6.
56. The empirical formula of a compound gives the number of atoms of each element in the
compound in the simplest ratio. The molecular formula gives the exact number of atoms of each
element in a molecule.
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-15
57. Given: actual yield of CS2 = 500 g; percentage yield = 86.0 %
Required: theoretical yield of CS2
Solution:
actual yield
percentage yield =
! 100 %
theoretical yield
theoretical yield =
=
actual yield
! 100 %
percentage yield
500 g
! 100 %
86.0 %
theoretical yield = 581 g
Statement: The theoretical yield of carbon disulfide is 581 g.
58. Write a balanced chemical equation for the reaction.
O2(g) + 2 H2(g) → 2 H2O(g)
Since oxygen is the limiting reagent, use the amount of oxygen to calculate the amount of water
that can be obtained. The mole ratio of oxygen to water is 1 mol O2 : 2 mol H2O
59. Given: nH 2 = 15 mol
Required: amount of ammonia, n NH 3
Solution:
Step 1. List the given value and the required value.
3 H2(g) + N2(g) → 2 NH3(g)
n NH 3
15 mol
Step 2. Convert amount of hydrogen to amount of ammonia.
2 molNH
3
nNH = 15 molH !
3
2
3 molH
2
nNH = 10 mol
3
Statement: 10 mol of ammonia can be produced from 15 mol hydrogen.
60. Given: nN 2 O5 = 1.75 mol
Required: amount of nitric oxide, nNO
Solution:
Step 1. List the given value and the required value.
2 N2O5(s) → 4 NO(g) + 3 O2(g)
nNO
1.75 mol
Step 2. Convert amount of dinitrogen pentoxide to amount of nitric oxide.
4 molNO
nNO = 1.75 molN O !
2 5
2 molN O
2
5
nNO = 3.50 mol
Statement: When 1.75 mol of dinitrogen pentoxide decomposes, 3.50 mol of nitric oxide will be
formed.
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-16
61. Given: mNa = 28.8 g
Required: mass of fluorine, mF2
Solution:
Step 1. List the given value, the required value, and the corresponding molar masses.
NaF(s) → 2 Na(s) + F2(g)
mF2
28.8 g
22.99 g/mol 38.00 g/mol
Step 2. Convert mass of sodium to amount of sodium.
1 molNa
nNa = 28.8 g !
22.99 g
nNa = 1.2527 mol [two extra digits carried]
Step 3. Convert amount of sodium to amount of fluorine.
1 molF
2
nF = 1.2527 molNa !
2
2 molNa
nF = 0.626 35 mol [two extra digits carried]
2
Step 4. Convert amount of fluorine to mass of fluorine.
! 38.00 g $
mF = (0.626 35 mol ) #
&
2
" 1 mol %
mF = 23.8 g
2
Statement: When sodium fluoride is decomposed, 23.8 g of fluorine will be produced with
28.8 g of sodium.
62. Four practical advantages of applying the concept of limiting reagents are: cost reduction,
decreasing environmental impact, environmental cleanup, and efficiency of reactant utilization.
63. Given: nNa = 1.8 mol ; nBr2 = 1.4 mol
Required: limiting reagent
Solution:
Step 1. List the given values.
2 Na(s) + Br2(l) → NaBr2(s)
1.8 mol 1.4 mol
Step 2. Use the amount of one reactant to find the stoichiometric amount of another.
1 molBr
2
nBr = 1.8 molNa !
2
2 molNa
nBr = 0.9 mol
2
Statement: Since the amount of bromine present is greater than the required amount, sodium is
the limiting reagent.
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-17
64. Given: mFe = 84.0 g ; mH 2 O = 50.0 g
Required: limiting reagent
Solution:
Step 1. List the given values and the corresponding molar masses.
3 Fe(s) + 4 H2O(g) → Fe2O4(s) + 4 H2(g)
84.0 g
50.0 g
55.85 g/mol 18.02 g/mol
Step 2. Convert mass of given substances to amount of given substances.
1 molFe
nFe = 84.0 g !
55.85 g
nFe = 1.5040 mol [two extra digits carried]
nH O = 50.0 g !
2
1 mol
H 2O
18.02 g
nH O = 2.7747 mol [two extra digits carried]
2
Step 3. Use the amount of one reactant to find the stoichiometric amount of another.
4 molH O
2
nH O = 1.5040 molFe !
2
3 molFe
nH O = 2.0053 mol
2
Statement: Since the amount of water present is greater than the required amount, iron is the
limiting reagent.
65. Given: mC 2 H 4 O = 880 kg ; mH 2 O = 400 kg
Required: limiting reagent
Solution:
Step 1. List the given values and the corresponding molar masses.
C2H4O(g) + H2O(g) → HOCH2CH2OH(l)
800 kg
400 kg
44.06 g/mol 18.02 g/mol
Step 2. Convert mass of given substances to amount of given substances.
1000 g 1 molC2 H 4 O
nC H O = 800 kg !
!
2 4
44.06 g
1 kg
nC H O = 18 157 mol [two extra digits carried]
2
4
nH O = 400 kg !
2
1000 g
1 kg
!
1 mol
H 2O
18.02 g
nH O = 22 197 mol [two extra digits carried]
2
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-18
Step 3. Use the amount of one reactant to find the stoichiometric amount of another.
1 molH O
2
nH O = 18 157 molC H O !
2
2 4
1 molC H O
2
4
nH O = 18 157 mol
2
Statement: Since the amount of water present is greater than the required amount, ethylene
oxide is the limiting reagent.
66. Since the ratio of carbon monoxide to oxygen in the balanced equation is 2:1, 20 carbon
monoxide molecules require 10 oxygen molecules for the formation carbon dioxide. Since there
are more oxygen molecules than is required, oxygen is the excess reagent. When 10 oxygen
molecules are used up in the reaction with 20 carbon monoxide molecules, 20 oxygen molecules
remain.
Analysis and Application
67. Answers may vary. Sample answer: I would be sure to include the units of measure for each
numerical quantity in the medical information. Common units for a particular measure can vary
from country to country. In Canada and Europe, for example, blood glucose is measured in
millimoles per litre, whereas in the United States the unit for blood glucose is milligrams per
decilitre. If the data I send to the hospital do not include units, the hospital staff might
misinterpret the meaning of the numbers.
68. Answers may vary. Sample answer: I would advise my cousin not to take my father’s
medicine. It is possible that their conditions, while similar, are different enough to require
different medications. My cousin might actually ingest a medical formulation that could be a
health risk. In addition, medical formulations are often adjusted for age, body mass, and medical
history. Differences in these factors between my father and my cousin could cause my father’s
medication to be significantly different than my cousin’s.
69. (a) Adding an inaccurate quantity of a chemical treatment could harm the fish in the tank if
the added quantity is larger than the correct amount and chemicals in the product become toxic to
fish at high levels. Similarly, adding a quantity that is too little might result in insufficient
remediation of a toxic condition in the tank, which could also harm the fish.
(b) The tank owner cannot assume that three drops is still the proper amount to add. The new
product might have the same chemicals, but in different concentrations. The new product might
also have different chemicals. The tank owner should look at the product label to find the proper
quantity to add.
70. Answers may vary. Sample answers:
(a) The grandmother could be consuming processed foods that are made with large amounts of
salt.
(b) She should read the labels on processed foods that she consumes to check their salt content.
She should avoid eating those with high salt content. She should also eat fresh foods as much as
possible because no salt is added to them before sale.
71. Chemists can use the composition of a compound, which gives the proportion of the masses
of the elements in the compound, and the molar mass of the compound to determine the chemical
formula of a compound. The mass of each element is first converted into amount in moles. The
ratio of the amounts gives the subscripts in the empirical formula. Using the molar mass of the
compound, the mass multiple can be solved to give the molecular formula of the compound.
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-19
72. (a) The spectrum can be used to identify the compound.
(b) The molecular mass of the compound is given by the peak with the largest mass in the
spectrum, which is 44 u. The heights of the peaks also quantitatively define the ratios in which
different molecular fragments form. Knowledge of the fragmentation pattern and the molecular
mass may lead to a determination of the compound’s molecular formula, which is a quantitative
statement of the ratio in which the atoms combine.
73. Answers may vary. Sample answer: A procedure for estimating the number of peanuts in a
bag of peanuts in the shell is given.
Step 1. Count a designated number, or group, of peanuts with shells, such as 10 per group.
Step 2. Determine and record the mass of the peanuts while still in the shell.
Step 3. Open the shells. Count and record the number of individual nuts.
Step 4. Determine and record the mass of the entire bag.
Step 5. Follow these calculations steps:
mass of bag
= equivalent number of groups
average mass of one group
Estimate the number of individual nuts in the bag.
equivalent number of peanuts = number of whole groups ! number of peanuts per group
This procedure uses mass to estimate the values in the substance in question. It may be used to
estimate the number of iodine molecules, I2, in a designated sample mass of iodine. The number
of molecules can then be multiplied by 2 to determine the number of individual atoms present in
the sample.
74. Given: amount of calcium chloride requested, nCaCl2 = 5 mol;
mass of calcium chloride supplied, mCaCl2 = 5 kg = 5.0 × 103 g
Required: amount of calcium chloride supplied, nCaCl2 = 5 mol
Solution:
Step 1. Calculate the molar mass of calcium chloride.
M CaCl = 1M Ca + 2 M Cl
2
= 40.08
2
!
g
g $
+ 2 # 35.45
mol
mol &%
"
g
2
mol
Step 2. Convert amount of chemical supplied to mass.
M CaCl = 110.98
!
g $
mCaCl = 5 mol # 110.98
&
2
"
mol %
mCaCl = 554.9 g
2
Statement: The requested amount of calcium chloride has a mass of 554.9 g. The stockroom
sent 5 kg, which is too much de-icing salt.
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-20
75. To convert from mass to moles use the following procedure:
• Determine the molar mass of the substance.
• Divide the mass of the substance by its molar mass.
amount of substance = mass of substance ÷ molar mass of substance
To convert from moles to mass use the following procedure:
• Determine the molar mass of the substance.
• Multiply the amount of the substance by its molar mass.
mass of substance = amount of substance × molar mass of substance
A summary of the calculation steps are shown in the following flow chart.
76. (a) The ratio of carbon to hydrogen atoms is lowest in CH, so this formula is the empirical
formula.
(b) (i) Given: CH (ii) C2H2 (iii) C6H6
Required: percentage of carbon, % C; percentage of hydrogen, % H in CH
Solution: Determine the percentage of each element in each formula using molecular mass.
Step 1. Calculate the molecular mass of the empirical formula of the compound.
MCH = 13.02 u
Step 2. Calculate the percentage of each element.
% C=
MC
M CH
12.01 u
13.02 u
% C = 92.24 % [one extra digit carried]
=
%H=
MH
M CH
1.01 u
13.02 u
% C = 7.76 % [one extra digit carried]
=
Step 3. Check that the percentages add up to 100 %.
92.24 % + 7.76 % = 100 %
Statement: The percentage of carbon in the empirical formula, CH, is 92.2 % and the percentage
of hydrogen is 7.76 % in CH.
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-21
(b) (ii) Given: C2H2 (iii) C6H6
Required: percentage of carbon, % C; percentage of hydrogen, % H in C2H2
Solution: Determine the percentage of each element in each formula using molecular mass.
Step 1. Calculate the molecular mass of the empirical formula of the compound.
MCH = 13.02 u
Step 2. Calculate the percentage of each element.
% C=
MC
%H=
2
MC H
2
2
24.02 u
=
26.04 u
% C = 92.24 % [one extra digit carried]
MH
2
MC H
2
2
2.02 u
=
26.04 u
% H = 7.76 % [one extra digit carried]
Step 3. Check that the percentages add up to 100 %.
92.24 % + 7.76 % = 100 %
Statement: The percentage of carbon in the empirical formula, C2H2, is 92.2 % and the
percentage of hydrogen is 7.76 % in C2H2.
(b) (iii) Given: C6H6
Required: percentage of carbon, % C; percentage of hydrogen, % H in C6H6
Solution: Determine the percentage of each element in each formula using molecular mass.
Step 2. Calculate the percentage of each element.
% C=
MC
MC H
6
%H=
6
6
72.06 u
78.12 u
% C = 92.24 % [one extra digit carried]
=
MH
6
MC H
6
6
6.06 u
78.12 u
% H = 7.76 % [one extra digit carried]
=
Step 3. Check that the percentages add up to 100 %.
92.24 % + 7.76 % = 100 %
Statement: The percentage of carbon in the empirical formula, C6H6, is 92.2 % and the
percentage of hydrogen is 7.76 % in C2H2.
For each compound, the percentage compositions for carbon and hydrogen are the same.
(c) Required: molar masses of CH, C2H2, and C6H6
Solution: from part (b)
M CH = 13.02 g/mol
M C H = 26.04 g/mol
2
2
M C H = 78.12 g/mol
6
6
Statement: Compounds (i) and (ii) each have a molar mass that is a multiple of the molar mass
of the empirical formula. The multiple is indicated by the subscripts.
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-22
(d) Since the percentage composition for each formula is the same, the diagram that best
represents 92.2 % carbon and 7.76 % hydrogen is diagram A.
77. Given: molar mass of compound CxHyOz, M Cx H y O z = 180.1 g/mol
percentage composition of carbon, % C = 54.5 %
percentage composition of hydrogen, % H = 9.2 %
percentage composition of oxygen, % O = 36.3 %
Required: molecular formula for compound, CxHyOz
Solution: Let there be a 100.0 g sample of this compound.
Step 1. Determine the mass of each element in a 100 g sample of the compound.
mC = % C ! 100 g
mO = % O ! 100 g
mH = % H ! 100 g
= 0.545 ! 100 g
mC = 54.5 g
= 0.363 ! 100 g
= 0.092 ! 100 g
mO = 36.3 g
mH = 9.2 g
Step 2. Determine the amount of each element in the compound.
! 1 mol $
nC = 54.5 g #
&
" 12.01 g %
nC = 4.5379 mol [one extra digit carried]
! 1 mol $
nH = 9.2 g #
&
" 1.01 g %
nH = 9.1089 mol [one extra digit carried]
! 1 mol $
nO = 36.3 g #
&
" 16.00 g %
nO = 2.2688 mol [one extra digit carried]
Step 3. Determine the simplest ratio.
nC 4.5379 mol
nH 9.1089 mol
=
=
nO
2.2688 mol
nO 2.2688 mol
nC
nH
= 4.014
= 2.000
nO
nO
Step 4. Determine the empirical formula.
carbon : hydrogen atoms : oxygen atoms = 2 : 4: 1
The empirical formula of the compound is C2H4O.
Copyright © 2011 Nelson Education Ltd.
nO
nO
nO
nO
=
2.2688 mol
2.2688 mol
=1
Unit 3: Quantities in Chemical Reactions
U3-23
Step 5. Determine the molar mass of C2H4O.
MC H O = 2 MC + 4 M H + MO
2
4
!
!
g $
g $
g
= 2 # 12.01
+ 4 # 1.01
+ 16.00
&
&
mol %
mol %
mol
"
"
g
2 4
mol
Step 6. Solve for the mass multiple, x.
molar mass of compound
x=
molar mass of empirical formula
M C H O = 44.06
=
180.1 g/mol
44.06 g/mol
x = 4.088
Step 7. Determine the molecular formula.
The molar mass of the compound is four times the mass of the empirical formula.
C x H y O z = C2!4 H 4!4O1!4
C x H y O z = C8 H16O 4
Statement: The molecular formula of the compound is C8H16O4.
78. Given: density of magnesium ribbon, d = 1.3 g/m
length of ribbon, l = 0.25 m
Required: number of magnesium atoms in ribbon, NMg
Solution:
Step 1. Determine the mass of ribbon.
mMg = d ! l
"
g%
= $ 1.3
(0.25 m)
m '&
#
mMg = 0.325 g [one extra digit carried]
Step 2. Determine the amount of magnesium, in moles.
! 1 mol $
nMg = (0.325 g) #
&
" 24.31 g %
nMg = 0.013 37 mol [two extra digits carried]
Step 3. Determine the number of magnesium atoms.
N Mg
" 6.02 ! 1023 atoms %
= (0.013 37 mol ) $
'
#
&
1 mol
N Mg = 8.05 ! 1021 atoms
Statement: The number of magnesium atoms in a 0.25 m ribbon is 8.05 × 1021 atoms.
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-24
79. Given: iron, Fe(s); iron(II) chloride, FeCl2(s); iron(III) bromide, FeBr3(s)
Required: percentage composition of iron in each compound
Analysis: % of element =
mass of element
mass of compound
! 100 %
Solution:
Step 1. Calculate the molecular mass of each substance.
M Fe = 55.85 u ; M FeCl = 126.75 u ; M FeBr = 295.55 u
2
3
Step 2. Determine the percentage by mass of iron in each compound.
In 1 mol of FeCl2,
In 1 mol of FeBr3,
55.85 u
55.85 u
% of Fe =
! 100 %
% of Fe =
! 100 %
126.75 u
295.55 u
% of Fe = 44.06 %
% of Fe = 18.90 %
Statement: The percentage composition of iron in iron(II) chloride is 44.06 %. The percentage
composition of iron is iron(III) bromide is 18.90 %. The percentage composition of iron is
greater in iron(II) chloride than iron(III) bromide.
By determining the molecular masses of the compounds and then calculating the quotient of the
molecular mass of iron to the molecular mass of the compound, I was able to determine the
percentage composition of iron in each compound.
80. The percentage of oxygen in carbon monoxide is 57.1 % compared to 72.7 % in carbon
dioxide. Thus, the percentage of oxygen increases when carbon monoxide is converted to carbon
dioxide.
81. (a) The difference in mass will not affect the percentage composition.
(b) The law of definite proportions states that a compound always contains the same proportion
of elements by mass. Therefore, the percentage composition must be the same, for the same
compound, no matter what the size of the sample.
82. The molecular formula provides the exact chemical composition of the compound to be
produced and is what should be provided to the manufacturer. Two compounds may have the
same empirical formula and very different molecular formulas, so the empirical formula does not
provide sufficient information. For example, an empirical formula of CH2O could represent
molecular formulas of CH2O, C2H4O2, or C3H6O3, and so on.
83. Answers may vary. Sample answer:
Step 1. Determine the composition of gastric acid.
Step 2. Analyze the antacid tablet to determine the mass of neutralizing substance (CaCO3) in
each tablet.
Step 3. Use the balanced equation for the neutralization reaction to determine the mole ratio of
acid to calcium carbonate.
Step 4. Use the mole ratio to determine the amount of acid that can be neutralized by the amount
of calcium carbonate in one tablet.
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-25
84. Answers may vary. Sample answer: Stoichiometric amounts are those amounts of reactants
and products that are in the same ratio as given in the balanced chemical equation. Assuming that
the reaction goes to completion and all the reactants are used up, the theoretical yield of product
is produced. Actual yield is the amount of product that is actually collected in the reaction
process. Three factors that may have an impact on the actual yield are:
There may be impurities in the reactants.
Other reactions may consume some of the reactants.
Some of the reactants or products may be lost.
85.
86. At the completion of the reaction, the limiting reagent is completely consumed. If copper was
the limiting reagent, there would be no copper wire remaining when the reaction stopped. If
unused copper wire remained, then it was not the limiting reagent.
87. I would first write the balanced chemical equation for the reaction. I would then use the mass
of the limiting reagent and its molar mass to calculate the amount of limiting reagent present.
Next I would use the mole ratio to determine the amount of the other reagent that is consumed.
Knowing the amount consumed, I could then use the molar mass of the excess reagent to predict
the mass that will be consumed. To find the mass of the excess reagent that would remain, I
would subtract the mass consumed from the designated starting mass of the excess reagent.
88. There are 6 molecules of A, 10 molecules of B, and 6 molecules of C in the diagram.
From the ratio 1A : 2B : 1C, 6 molecules of A require stoichiometric quantities of 12 molecules
of B and 6 molecules of C for complete reaction. Since there are only 10 molecules of B, B will
be the limiting reagent.
89. The percentage yield is calculated by comparing the actual mass of the product with the
theoretical mass or by comparing the actual amount of the product with the theoretical amount.
The ratio is expressed as a percentage using the following formula.
percentage yield =
actual yield
theoretical yield
! 100 %
90. The use of a recipe to prepare a meal or baked good is similar to the conducting a chemical
reaction and yielding a product. The ingredients (reactants) are listed along with their quantities
(amounts of reactants), and order of addition. Specific instructions for mixing, warming, or
stirring are often given in both baking and chemical reactions. The yield is often estimated for
both. The expected amount in a recipe is similar to the theoretical percentage yield in a chemical
reaction. Both amounts will be affected by similar variables such as incorrect measurements,
inadequate mixing, and using ingredients (reactants) that are less than the optimum quantity.
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-26
91. The calculation would require multiple steps. The theoretical yield and the actual yield are in
different units. One quantity has to be changed so that both quantities have same unit. An
addition step of converting the unit of one quantity to the unit of the other quantity is required.
For example, first convert the mass of the theoretical yield to a mass in milligrams.
1000 mg
14.79 g = 14.79 g !
1g
= 14 790 mg
Then, calculate the percentage yield.
actual yield
percentage yield =
! 100 %
theoretical yield
=
6733 mg
14 790 mg
! 100 %
percentage yield = 45.52 %
Percentage yield is not reported using any units. It is reported in percent.
Evaluation
92. Answers may vary. Sample answers:
(a) Forensic scientists must be able to conduct or use the results of chemical analyses in
evaluating the meaning of evidence collected at crime scenes or in similar contexts. People in
this career must understand concepts such as empirical and molecular formulas and how
analytical techniques such as mass spectrometry work because they have to identify the chemical
substances to evaluate the meaning of the evidence.
(b) Medical assistants and technicians are often responsible for preparing and/or administering
medications. They must fully understand the importance of accuracy because mistakes in
dosages can cause serious adverse health consequences for patients. In addition, medical
assistants and technicians must understand the meaning and importance of units of measure
provided with dosage instruction and possibly calculate dosages for people with different body
masses.
93. (a) The traditional Inuit diet is low in sodium, and Inuit have historically had a low incidence
of high blood pressure. Now that they have begun to ingest more salt, Inuit are suffering more
and more from high blood pressure. This is strong evidence that salt intake is causing the greater
incidence of high blood pressure in Inuit.
(b) Inuit have begun to eat more chemically preserved and processed foods, which often contain
high levels of salt. In addition, Inuit have acquired a taste for salt, much like the rest of the
Canadian population, and are adding salt to the meals they prepare.
(c) Answers may vary. Sample answer: I believe Inuit, like all Canadians, have the right to make
choices in their lives about what they eat. The government can develop and publish information
about the dangers of excessive salt consumption and make it publicly available. They could even
recommend that the information be incorporated into school lessons. However, the government
does not have the right to require Inuit or any other Canadian to eat less salt.
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-27
94. (a) The store manager made a few errors in the calculations with regard to significant digits.
In step 1, 5 is a counting number and has an infinite number of significant digits. The answer to
this step should be 0.38 g based on the 2 significant digits of the measured quantity 1.9 g.
In step 2, each of the measured quantities has 4 significant digits so the answer should also have
the 4 significant digits. In step 3, the number of significant digits for the rounded answer is
determined by the least number of significant digits in the quantities used for calculation, which
is 2 in this case. The correct steps are as follows:
1.9 g
average mass of one BB =
= 0.38 g
5
mass of BBs in container = 239.3 g ! 115.8 g = 123.5 g
123.5 g
number of BBs in container =
g
0.38
BB
= 325 BBs [one extra digit carried]
number of BBs in container = 330 BBs or 3.3 ! 102 BBS
(b) The store manager should not regard the answer as an exact measure of the number of BBs.
The answer is based on data with some degree of uncertainty due to errors of measurement.
Thus, the answer should not be treated as a perfect measure of the number of ball bearings.
(c) The store manager could have counted out a greater number of BBs with which to find the
average mass of one BB. This would yield more significant digits in the average mass, which
determines the number of significant digits in succeeding steps. Alternatively, the manager could
have used a scale that gave a readout with another significant digit.
(d) The number of significant digits in an answer calculated from measured quantities represents
the degree of certainty we have in the answer. A greater number of significant digits means a
more reliable answer.
95. The chemist’s proposed procedure will not yield useful results. The chemist is planning on
using the mass of the water absorbed to find the mass of hydrogen in the sample tested and,
similarly, the mass of carbon dioxide absorbed to find the mass of carbon in the sample. The
mass of the oxygen in the sample could then be determined by the difference. However, the
water scrubber and carbon dioxide scrubber could absorb water and carbon dioxide from the air,
giving falsely high readings. Furthermore, the sample also contains chlorine. The chlorine could
possibly be absorbed in either the water scrubber or the carbon dioxide scrubber, thus
introducing significant error into the measurement of the masses of water and carbon dioxide
absorbed. The chlorine could also escape the apparatus in some form. The “missing” mass could
not be attributed specifically to oxygen alone, so the amount of oxygen in the sample could not
be reliably determined.
96. Answers may vary. Sample answers:
(a) The plan is likely to succeed. Sodium hydroxide does react with carbon dioxide. Just as
astronauts in orbit use lithium hydroxide to absorb carbon dioxide that accumulates in their space
cabins, sodium hydroxide could be used to absorb carbon dioxide in a combustion apparatus.
(b) I believe the plan will work. Carbon dioxide gas and sodium hydroxide solid will react to
product sodium bicarbonate solid and liquid water. Together, sodium bicarbonate mixed with
water forms an effective cleaning agent.
The balanced chemical equation is
CO2(g) + 2 NaOH(s) → Na2CO3(s) + H2O(l)
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-28
Reflect on Your Learning
97. Answers may vary. Students’ answers should include an example of an idea related to the
relative quantities of ingredients or chemicals used for a process at home or in the workplace,
such as the preparation of foods, the manufacturing of drugs or other chemical products.
98. Answers may vary. Students’ answers should be unexpected findings resulting from their
understanding of the quantitative relationships in chemical reactions or from their research on
topics suggested in the career links and web links. They could, for example, relate to the concept
that equal amounts are not the same as equal masses.
99. Answers may vary. Students’ answers will likely list topics in this unit that are related to
their everyday activities, such as the limiting reagents in baking and how it affects the product,
and more examples of percentage yields from various processes such as the petroleum oil
production.
100. Answers may vary. Students will likely mention that they are worried about health problems
related to the intake of too much salt, likely through processed foods. They should pose solutions
or suggestions as a result of their research.
Research
101. Answers may vary. In general, branched-chain hydrocarbons burn better in automobile
engines. 2,2,4-trimethyl pentane (iso-octane) is a very smooth-burning hydrocarbon that is used
as a standard for rating the burning properties of gasoline blends, hence the term “octane
number.” Students’ answers should include drawings of molecular structures.
102. Answers may vary. A spectrophotometer consists of a spectrometer for producing light of
any selected colour (wavelength) and a photometer for measuring the intensity of light. For
example, a chemical reaction may result in colour development. The colour development is
linked to the concentration of a substance in a solution. In a visible light spectrophotometer, the
concentration can be measured by determining the extent of absorption of light at the appropriate
wavelength. The percentage of light absorbed by the sample can be mathematically related to the
concentration of the coloured compound. Students’ answers should include a schematic diagram
of their findings.
103. Answers may vary. Students’ reports should mention that many low-salt and salt-substitute
products contain potassium chloride. This compound has health consequences of its own for
certain individuals. Reports should include data about the toxicity and health risks of potassium
chloride. Students’ reports should also note that replacing table salt with salt substitutes can be a
successful low-sodium strategy for some people. There are also salt-substitute products that are
based on formulations without potassium chloride.
Copyright © 2011 Nelson Education Ltd.
Unit 3: Quantities in Chemical Reactions
U3-29