Section 7.5: Percentage Yield
Tutorial 1 Practice, page 338
1. (a) Given: m Ni = 23.5 g
Required: theoretical yield of nickel carbonyl, m Ni(CO)4
Solution:
Step 1. List the given value, the required value, and the corresponding molar masses.
Ni(s) + 4 CO(g) → Ni(CO)4(g)
m Ni(CO)4
23.5 g
58.69 g/mol
170.73 g/mol
Step 2. Convert mass of given substance to amount of given substance.
1 molNi
nNi = 23.5 g !
58.69 g
nNi = 0.400 41 mol [two extra digits carried]
Step 3. Convert amount of nickel to amount of nickel carbonyl.
1 molNi(CO)
4
nNi(CO) = 0.400 41 molNi !
4
1 molNi
nNi(CO) = 0.400 41 mol
4
Step 4. Convert amount of nickel carbonyl to mass of nickel carbonyl.
! 170.73 g $
mNi(CO) = (0.400 41 mol ) #
&
4
" 1 mol %
mNi(CO) = 68.362 g [two extra digits carried]
4
Statement: The theoretical yield of nickel carbonyl is 68.4 g.
(b) actual yield of nickel carbonyl = 61.0 g
Use the theoretical yield of nickel carbonyl from (a).
actual yield
percentage yield =
! 100 %
theoretical yield
61.0 g
=
! 100 %
68.362 g
percentage yield = 89.2 %
Therefore, if 61.0 g of nickel carbonyl is collected, the percentage yield is 89.2 %.
2. Given: mCH 4 = 1.61 g ; m H 2O = 2.00 g ; actual yield of hydrogen = 5.0 g
Required: theoretical yield, m H 2 ; percentage yield of hydrogen
Solution:
Step 1. List the given values, the required value, and the corresponding molar masses.
CH4(g) + H2O(g) → CO(g) + 3 H2(g)
mH2
1.61 g
2.00 g
16.05 g/mol
18.02 g/mol
2.02 g/mol
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7.5-1
Step 2. Convert mass of given substances to amount of given substances.
1 molCH
4
nCH = 1.61 g !
4
16.05 g
nCH = 0.100 31 mol [two extra digits carried]
4
nH O = 2.00 g !
2
1 molH O
2
18.02 g
nH O = 0.110 99 mol [two extra digits carried]
2
Step 3. Determine the limiting reagent.
1 molH O
2
nH O = 0.100 31 molCH !
2
4
1 molCH
4
nH O = 0.100 31 mol [two extra digits carried]
2
Since the amount of water present is greater than the amount required to react with the methane,
methane is the limiting reagent.
Step 4. Convert amount of methane to amount of hydrogen.
3 molH
2
nH = 0.100 31 molCH !
2
4
1 molCH
4
nH = 0.300 93 mol [two extra digits carried]
2
Step 5. Convert amount of hydrogen to mass of hydrogen.
! 2.02 g $
mH = (0.300 93 mol ) #
&
2
" 1 mol %
mH = 0.607 88 g
2
Step 6. Calculate the percentage yield.
actual yield
percentage yield =
! 100 %
theoretical yield
0.50 g
! 100 %
0.607 88 g
percentage yield = 82 %
Statement: The theoretical yield of hydrogen is 0.608 g and the percentage yield is 82 %.
=
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7.5-2
Section 7.5 Questions, page 339
1. The theoretical yield is a prediction of the quantity of product expected based on the
stoichiometry of the reaction. The actual yield is the quantity of product actually collected when
the reaction is complete.
2. The actual yield rarely equals the theoretical yield because a number of factors limit the
quantity of product collected. These include the nature of the reaction, experimental conditions,
the presence of impurities in the reactants, and competing side reactions.
3. Reducing the number of steps in an experimental procedure reduces the chances of accidental
loss of product. The reacting materials may be lost due to measurements, through spillage during
the transfer of solutions, by splattering during heating, during the isolation and purification of the
product, or due to other handling errors, making the actual yield less than the theoretical yield.
4. (a) Not drying the mixture enough makes the final product heavier than expected, resulting in
a higher percentage yield.
(b) Adding insufficient sodium hydrogen carbonate has no effect on the percentage yield because
the reaction stops when the sodium hydrogen carbonate is consumed by the reaction. The final
mass of sodium chloride will be smaller but the percentage yield remains the same.
(c) Using impure sodium hydrogen carbonate produces less than the predicted amount of sodium
chloride, resulting in a decreased percentage yield.
(d) Some of sodium chloride product is lost as the reaction mixture splatters. Since not all
sodium chloride produced is collected, the percentage yield is reduced.
5. The presence of dichloromethane suggests that not all of the methane present initially is
converted into chloromethane. The percentage yield of chloromethane is therefore reduced.
6. (a) Zn(s) + 2 AgNO3(aq) → 2 Ag(s) + Zn(NO3)2(aq)
(b) Given: mZn = 3.00 g
Required: theoretical yield of silver metal, mAg
Solution:
Step 1. List the given value, the required value, and the corresponding molar masses.
Zn(s) + 2 AgNO3(aq) → 2 Ag(s) + Zn(NO3)2(aq)
m Ag
3.00 g
65.41 g/mol
107.87 g/mol
Step 2. Convert mass of given substance to amount of given substance.
1 molZn
nZn = 3.00 g !
65.41 g
nZn = 0.458 65 mol [two extra digits carried]
Step 3. Convert amount of zinc to amount of silver.
2 molAg
nAg = 0.045 865 molZn !
1 molZn
nAg = 0.091 730 mol [two extra digits carried]
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7.5-3
Step 4. Convert amount of silver to mass of silver.
! 107.87 g $
mAg = (0.091 730 mol ) #
&
" 1 mol %
mAg = 9.8949 g [two extra digits carried]
Statement: The theoretical yield of silver metal is 9.9 g.
(c) actual yield of silver metal = 7.2 g
Use the theoretical yield of silver metal from (b).
actual yield
percentage yield =
! 100 %
theoretical yield
7.2 g
=
! 100 %
9.8949 g
percentage yield = 73 %
Therefore, if 7.2 g of silver metal is collected, the percentage yield is 73 %.
7. (a) Given: percentage yield of chlorine = 95 %; theoretical yield of chlorine = 142 g
Required: actual yield of chlorine
Solution:
actual yield
percentage yield =
! 100 %
theoretical yield
actual yield = theoretical yield !
percentage yield
100 %
95 %
100 %
actual yield = 134.9 g [two extra digits carried]
Statement: The actual yield of chlorine is 135 g.
(b) Given: mCl2 = 134.9 g
= 142 g !
Required: mass of sodium chloride, m NaCl
Solution:
Step 1. Balance the chemical equation for the reaction. List the given value, the required value,
and the corresponding molar masses.
2 NaCl(aq) + 2 H2O(l) → Cl2(g) + 2 NaOH(aq) + H2(g)
m NaCl
134.9 g
58.44 g/mol
70.90 g/mol
Step 2. Convert mass of given substance to amount of given substance.
1 molCl
2
nCl = 134.9 g !
2
70.90 g
nCl = 1.903 mol [two extra digits carried]
2
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7.5-4
Step 3. Convert amount of chlorine to amount of sodium chloride.
2 molNaCl
nNaCl = 1.903 molCl !
2
1 molCl
2
nNaCl = 3.806 mol
Step 4. Convert amount of sodium chloride to mass of sodium chloride.
! 58.44 g $
mNaCl = (3.806 mol ) #
&
" 1 mol %
mNaCl = 220 g
Statement: To produce 135 g chlorine, 220 g of sodium chloride is required.
8. Given: nC = 4.00 mol ; actual yield of methane = 28.0 g
Required: percentage yield of methane
Solution:
Step 1. List the given value, the required value, and the corresponding molar masses.
2 C(s) + 2 H2O(g) → CH4(g) + CO2(g)
mCH 4
4.00 mol
12.01 g/mol
16.05 g/mol
Step 2. Convert amount of carbon to amount of methane.
1 molCH
4
nCH = 4.00 molC !
4
2 molC
nCH = 2.00 mol
4
Step 3. Convert amount of methane to mass of methane.
! 16.05 g $
mCH = (2.00 mol ) #
&
4
" 1 mol %
mCH = 32.1 g
4
The theoretical yield of methane is 32.1 g.
Step 4. Calculate the percentage yield.
actual yield
percentage yield =
! 100 %
theoretical yield
28.0 g
=
! 100 %
32.1 g
percentage yield = 87.2 %
Statement: The percentage yield of the reaction is 87.2 %.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7.5-5
9. (a) Given: m NH 3 = 3.4 g
Required: theoretical yield of ammonium sulfate, m(NH 4 )2 SO4
Solution:
Step 1. List the given value, the required value, and the corresponding molar masses.
2 NH3(g) + H2SO4(aq) → (NH4)2SO4(aq)
m(NH 4 )2 SO4
3.4 g
17.04 g/mol
132.17 g/mol
Step 2. Convert mass of given substance to amount of given substance.
1 molNH
3
nNH = 3.4 g !
3
17.04 g
nNH = 0.1995 mol [two extra digits carried]
3
Step 3. Convert amount of ammonia to amount of ammonium sulfate.
1 mol(NH ) SO
4 2
4
n(NH ) SO = 0.1995 molNH !
4 2
4
3
2 molNH
3
n(NH
) SO 4
4 2
= 0.9975 mol [two extra digits carried]
Step 4. Convert amount of ammonium sulfate to mass of ammonium sulfate.
! 132.17 g $
m(NH ) SO = (0.099 75 mol ) #
&
4 2
4
" 1 mol %
m(NH
) SO 4
4 2
= 13.18 g (two extra digits carried)
Statement: The theoretical yield of ammonium sulfate is 13 g.
(b) actual yield of ammonium sulfate = 10.4 g
Use the theoretical yield of ammonium sulfate from (a).
actual yield
percentage yield =
! 100 %
theoretical yield
10.4 g
=
! 100 %
13.18 g
percentage yield = 79 %
Therefore, if 10.4 g of ammonium sulfate is collected, the percentage yield is 79 %.
10. Given: m Fe3O4 = 35.0 g ; actual yield of iron = 15.0 g
Required: percentage yield of iron
Solution:
Step 1. List the given value, the required value, and the corresponding molar masses.
Fe3O4(s) + 2 C(s) → 3 Fe(s) + 2 CO2(g)
m Fe
35.0 g
231.55 g/mol
55.85 g/mol
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7.5-6
Step 2. Convert mass of given substance to amount of given substance.
1 moln
Fe3O4
nFe O = 35.0 g !
3 4
231.55 g
nFe O = 0.151 16 mol [two extra digits carried]
3
4
Step 3. Convert amount of magnetite to amount of iron.
3 molFe
nFe = 0.151 16 molFe O !
3 4
1 molFe O
3
4
nFe = 0.453 48 mol [two extra digits carried]
Step 4. Convert amount of iron to mass of iron.
! 55.85 g $
mFe = (0.5348 mol ) #
&
" 1 mol %
mFe = 25.327 g [two extra digits carried]
The theoretical yield of iron is 25.3 g.
Step 5. Calculate the percentage yield.
actual yield
percentage yield =
! 100 %
theoretical yield
15.0 g
=
! 100 %
25.327 g
percentage yield = 59.2 %
Statement: The percentage yield of the reaction is 59.2 %.
11. (a) N2(g) + 3 H2(g) → 2 NH3(g)
(b) Given: m N 2 = 1.00 g = 1000 g ; percentage yield of ammonia = 24.5 %
Required: actual yield of ammonia
Solution:
Step 1. List the given value, the required value, and the corresponding molar masses.
N2(g) + 3 H2(g) → 2 NH3(g)
m NH 3
1000 g
28.02 g/mol
17.04 g/mol
Step 2. Convert mass of given substance to amount of given substance.
1 molN
2
nN = 1000 g !
2
28.02 g
nN = 35.689 mol [two extra digits carried]
2
Step 3. Convert amount of nitrogen to amount of ammonia.
2 molNH
3
nNH = 35.689 molN !
3
2
1 molN
2
nNH = 71.378 mol [two extra digits carried]
3
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7.5-7
Step 4. Convert amount of ammonia to mass of ammonia.
! 17.04 g $
mNH = (71.378 mol ) #
&
3
" 1 mol %
mNH = 1216.3 g [two extra digits carried]
3
This is the theoretical yield of ammonia.
Step 5. Calculate the actual yield.
actual yield
percentage yield =
! 100 %
theoretical yield
actual yield = theoretical yield !
percentage yield
100 %
24.5 %
100 %
actual yield = 297.99 g [two extra digits carried]
Statement: The actual yield of ammonia is 298 g.
(c) Given: amount of nitrogen present = 35.689 mol; actual yield of ammonia = 297.99 g
Required: amount of nitrogen unreacted
Solution:
Step 1. List the given value, the required value, and the corresponding molar masses.
N2(g) + 3 H2(g) → 2 NH3(g)
nN2
297.99 g
= 1216.3 g !
28.02 g/mol
17.04 g/mol
Step 2. Convert mass of given substance to amount of given substance.
1 molNH
3
nNH = 297.99 g !
3
17.04 g
nNH = 17.488 mol [two extra digits carried]
3
Step 3. Convert amount of ammonia reacted to amount of nitrogen reacted.
1 molN
2
nN = 17.488 molNH !
2
3
2 molNH
3
nN = 8.7440 mol [two extra digits carried]
2
Step 4. Calculate the amount of nitrogen remaining.
35.689 mol – 8.744 mol = 26.945 mol
Statement: The amount of nitrogen that remains unreacted is 26.9 mol.
(d) Use the amount of unreacted nitrogen in (c).
! 28.02 g $
mN = (26.945 mol ) #
&
2
" 1 mol %
mN = 755 g
2
The mass of nitrogen that remains unreacted is 755 g.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7.5-8
12. Given: percentage yield of ethanoic acid = 85 %; actual yield of ethanoic acid = 2.5 kg
Required: mass of carbon monoxide, mCO
Solution:
Step 1. Convert 2.5 kg to a mass in grams.
1000 g
2.5 kg !
= 2500 g
1 kg
Calculate the theoretical yield of ethanoic acid.
actual yield
percentage yield =
! 100 %
theoretical yield
actual yield
! 100 %
percentage yield
2500 g
=
! 100 %
85 %
theoretical yield = 2941 g [two extra digits carried]
Step 2. List the known value, the required value, and the corresponding molar masses.
CH3OH(l) + CO(g) → HC2H3O2(l)
mCO
2941 g
28.01 g/mol 60.06 g/mol
Step 3. Convert mass of ethanoic acid to amount of ethanoic acid.
1 molHC H O
2 3 2
nHC H O = 2941 g !
2 3 2
60.06 g
theoretical yield =
nHC H O = 48.97 mol [two extra digits carried]
2
3
2
Step 4. Convert amount of ethanoic acid to amount of carbon monoxide.
1 molCO
nCO = 48.97 molHC H O !
2 3 2
1 molHC H O
2
3
2
nCO = 48.97 mol [two extra digits carried]
Step 5. Convert amount of carbon monoxide to mass of carbon monoxide.
! 28.01 g $
mCO = (48.97 mol ) #
&
" 1 mol %
mCO
= 1400 g
= 1.4 kg
Statement: The mass of carbon monoxide required to prepare 2.5 kg of ethanoic acid is 1.4 kg.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7.5-9
13. Given: percentage yield of reaction = 75.2 %; actual yield of hydrogen sulfide = 50.0 g
Required: mass of sulfur, mS
Solution:
Step 1. Calculate the theoretical yield of hydrogen sulfide.
actual yield
percentage yield =
! 100 %
theoretical yield
theoretical yield =
=
actual yield
! 100 %
percentage yield
50.0 g
! 100 %
75.2 %
theoretical yield = 66.489 g [two extra digits carried]
Step 2. List the known value, the required value, and the corresponding molar masses.
CH4(g) + 4 S(s) → CS2(g) + 2 H2S(g)
mS
66.489 g
32.07 g/mol
34.09 g/mol
Step 3. Convert mass of hydrogen sulfide to amount of hydrogen sulfide.
1 molH S
2
nH S = 66.489 g !
2
34.09 g
nH S = 1.9504 mol [two extra digits carried]
2
Step 4. Convert amount of hydrogen sulfide to amount of sulfur.
4 molS
nS = 1.9504 molH S !
2
2 molH S
2
nS = 3.9008 mol [two extra digits carried]
Step 5. Convert amount of sulfur to mass of sulfur.
! 32.07 g $
mS = (3.9008 mol ) #
&
" 1 mol %
mS = 125 g
Statement: The mass of sulfur required to produce 50.0 g of hydrogen sulfide is 125 g.
14. (a) Given: m P = 25.0 g ; percentage yield of reaction = 90.0 %
Required: actual yield of tetraphosphorus decoxide
Solution:
Step 1. List the known value, the required value, and the corresponding molar masses.
P4(s) + 5 O2(g) → P4O10(s)
m P4O10
25.0 g
123.88 g/mol
283.88 g/mol
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7.5-10
Step 2. Convert mass of phosphorus to amount of phosphorus.
1 molP
4
nP = 25.0 g !
4
123.88 g
nP = 0.201 81 mol [two extra digits carried]
4
Step 3. Convert amount of phosphorus to amount of tetraphosphorus decoxide.
1 molP O
4 10
nP O = 0.201 81 molP !
4 10
4
1 molP
4
nP O = 0.201 81 mol [two extra digits carried]
4
10
Step 4. Convert amount of tetraphosphorus decoxide to mass of tetraphosphorus decoxide.
! 283.88 g $
mP O = (0.201 81 mol ) #
&
4 10
" 1 mol %
mP O = 57.290 g [two extra digits carried]
4
10
Step 5. Calculate the actual yield of tetraphosphorus decoxide.
actual yield
percentage yield =
! 100 %
theoretical yield
actual yield = theoretical yield !
percentage yield
100 %
90.0 %
100 %
actual yield = 51.561 g [two extra digits carried]
Statement: The mass of tetraphosphorus decoxide collected from the first step is 51.6 g.
(b) Given: m P4O10 = 51.561 g ; percentage yield of reaction = 95.2 %
Required: actual yield of phosphoric acid
Solution:
Step 1. List the known value, the required value, and the corresponding molar masses.
P4O10(s) + 6 H2O(l) → 4 H3PO4(aq)
m H 3PO4
51.561 g
= 57.290 g !
283.88 g/mol
98.00 g/mol
Step 2. Convert mass of tetraphosphorus decoxide to amount of tetraphosphorus decoxide.
1 molP O
4 10
nP O = 51.561 g !
4 10
283.88 g
nP O = 0.181 63 mol [two extra digits carried]
4
10
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7.5-11
Step 3. Convert amount of tetraphosphorus decoxide to amount of phosphoric acid.
4 molH PO
3
4
nH PO = 0.181 63 molP O !
3
4
4 10
1 molP O
4
10
nH PO = 0.726 52 mol [two extra digits carried]
3
4
Step 4. Convert amount of phosphoric acid to mass of phosphoric acid.
! 98.00 g $
mH PO = (0.726 52 mol ) #
&
3
4
" 1 mol %
mH PO = 71.199 g [two extra digits carried]
3
4
This is the theoretical yield of phosphoric acid.
Step 5. Calculate the actual yield of phosphoric acid.
actual yield
percentage yield =
! 100 %
theoretical yield
actual yield = theoretical yield !
= 71.199 g !
percentage yield
100 %
95.2 %
100 %
actual yield = 67.8 g
Statement: The mass of phosphoric acid produced from the second step is 67.8 g.
15. Answers may vary. Sample answer: Atom economy and percentage yield are both indicators
of the efficiency of a chemical reaction. Atom economy refers to the percentage of atoms in the
reactants that are found in the products. If a chemical reaction is highly efficient, the atom
economy of the reaction should be high. Percentage yield gives the ratio of the actual quantity of
product compared to the predicted theoretical quantity.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7.5-12