Theoretical yield
The theoretical yield means the amount of a product that would be made, calculated
from the stoichiometry of the equation, using the moles of the limiting
reagent (see previous page)
Example Calculate
the theoretical yield of iron(II) hydroxide formed when 50cm3
of 1M FeSO4 is added to 100cm3 of 2M NaOH [Fe=56,
O=16, H=1]
Equation for the reaction:
2NaOH + FeSO4 →
Fe(OH)2 + Na2SO4
2 moles of sodium hydroxide react with 1 mole of iron(II) sulfate
From the data given,, moles of iron(II) sulfate = 0.05 x 1 = 0.05 moles
This would require 0.05 x 2 = 0.1 moles of sodium hydroxide
Moles of sodium hydroxide = 0.1 x 2 = 0.2 moles - the sodium hydroxide
is in EXCESS and the iron(II) sulfate is the limiting reagent.
From the equation, 1 mole of iron(II) sulfate makes 1 mole iron(II)
hydroxide
Therefore, 0.05 moles iron(II) sulfate theoretically makes 0.05 moles
of iron(II) hydroxide
Mr iron(II) hydroxide = 56 + 2(16 +1) = 90
Therefore, theoretical yield of iron(II) hydroxide = 90 x 0.05 = 4.5
g
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Percentage yield
The fraction of the theoretical yield is calculated by using the actual experimentally
obtained amount of product and dividing it by the amount expected in theory.
The percentage is obtained by multiplying this fraction by 100.
For example, if the theoretical yield is 8 g, but after collection we find
that there is only 6 g, then the percentage yield is 6/8 x 100% = 75% yield.
It does not matter whether moles or mass is used for the calculation as long
as the two units are not used at the same time.
Example: In an experiment
to prepare a sample of aspirin (C9H8O4),
13.8 g of salicylic acid (C7H6O3) was
reacted with 15g of ethanoic anhydride (C4H6O3)
according to the equation:
C7H6O3
+ C4H6O3 →
C9H8O4 + C2H4O2
Find the limiting reagent and theoretical yield.
Calculate the percentage yield if 2.5g of aspirin was obtained. [C=12,
O=16, H=1]
From the equation we see that moles of salicylic acid = moles of ethanoic
anhydride
Mr salicylic acid = (7 x 12) + (6 x 1) + (3 x 16) = 138
From experimental data, moles of salicylic acid = 13.8/138 = 0.1 moles
Mr ethanoic anhydride = (4 x 12) + (6 x 1) + (3 x 16) =
102
From experimental data, moles of ethanoic anhydride = 15/102 = 0.147
moles
The ethanoic anhydride is in excess and the salicylic acid is the limiting
reagent.
From the equation, moles of salicylic acid = moles of aspirin.
Theoretical yield of aspirin therefore = 0.1 moles
Mr aspirin = (9 x 12) + (8 x 1) + (4 x 16) = 180
Theoretical yield of aspirin = 180 x 0.1 = 18g
Actual experimental yield of aspirin = 2.5g
Percentage yield = 2.5/18 x 100 = 13.9%
(3 sig figs)
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Worked examples
Q362-01 6.5g zinc metal reacts
with 50cm3 of 1M copper sulfate solution CuSO4. After
filtering and drying 3g of copper metal was obtained. Calculate the percentage
yield of copper. [Zn=65, Cu=63.5, S=32, O=16]
Answer
|
Equation for the reaction: Zn + CuSO4 →
ZnSO4 + Cu
From the equation 1 mole of Zn requires 1 mole of CuSO4
From the question data given:
6.5g of zinc = 6.5/65 moles = 0.1 moles
50cm3 of 1M copper sulfate = 0.05 x 1 = 0.05 moles
There is not enough copper sulfate to react wth all the zinc. The
zinc is in EXCESS and the copper sulfate is the LIMITING reagent.
0.05 moles of copper sulfate makes the same number of moles of copper
= 0.05 moles
0.05 moles copper = 0.05 x 63.5 = 3.175
g
Mass obtained from experiment = 3.0g
Percentage yield = 3.0/3.175 x 100% = 94.5%
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Q362-02 5g of solid calcium
hydroxide, Ca(OH)2, reacts with 10g of solid ammonium chloride NH4Cl
on heating to produce solid calcium chloride, steam and 1.14 dm3
of ammonia gas, NH3, was collected at STP. Calculate the percentage
yield of ammonia. [N=14, H=1, Ca=40, Cl=35.5, O=16] [molar volume of a gas at
STP = 22.7 dm3]
Answer
|
Equation for the reaction: Ca(OH)2 + 2NH4Cl → CaCl2
+ 2NH3 + 2H2O
From the equation:
1 mole of calcium hydroxide reacts completely with 2 moles of ammonium
chloride
From experimental data:
5g of solid calcium hydroxide, Ca(OH)2, = 5/[40 + 2(16+1)]
moles = 0.0676 moles
this would need 2 x 0.0676 moles of ammonium chloride, but...
10g of solid ammonium chloride NH4Cl = 10/[14 + 4 + 35.5]
moles = 0.187 moles
There are more than enough moles of ammonium chloride - it is in
EXCESS. The calcium hydroxide is the LIMITING REAGENT.
From the equation:
1 moles of calcium hydroxide makes 2 moles of ammonia gas
Therefore theoretical moles of ammonia = 0.0676 x 2 = 0.1352 moles
NH3
Experimentally obtained moles = 1.14/22.7 = 0.050 moles
Percentage yield = 0.050/0.1352 x 100 = 37.1
%
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Q362-03 When 14.8g of magnesium
nitrate, Mg(NO3)2, was heated in a dry tube magnesium
oxide, nitrogen dioxide gas NO2, and oxygen were produced. When the
nitrogen dioxide was cooled it liquified and 2.0 g were collected. Calculate
the percentage yield of nitrogen dioxide. [N=14, H=1, Mg=24, O=16]
Answer
|
Equation for the reaction: 2Mg(NO3)2 →
2MgO + 4NO2 + O2
From the equation:
2 moles of magnesium nitrate produces 4 moles of nitrogen dioxide
therefore 1 mole of magnesium nitrate produces 2 moles of nitrogen
dioxide
From experimental data:
Moles of Mg(NO3)2 = 14.8/Mr = 14.8/[24
+ 2(14 + {3 x 16})] = 14.8/148 = 0.1 moles
Moles of nitrogen dioxide (theoretical) = 2.0/Mr = 1.0/[14
+ (2 x 16)] = 2.0/46 = 0.0435 moles
From the equation:
1 mole of magnesium nitrate produces 2 moles of nitrogen dioxide
Therefore theoretical moles = 2 x moles of Mg(NO3)2
= 0.2 moles
Experimentally obtained moles = 0.0435 moles
Percentage yield = 0.0435/0.2 x 100% = 21.7%
(3 sig figs)
|
Q362-04 Solid manganese(IV)
oxide (4.0 g) MnO2, oxidised 50cm3 of 4M hydrochloric
acid on heating producing chlorine gas, water and manganese(II) chloride solution.
If 0.8 dm3 of chlorine were produced calculate the percentage yield
of chlorine. [Mn=55, H=1, , Cl=35.5, O=16]
Answer
|
Equation for the reaction: MnO2 + 4HCl →
MnCl2 + Cl2 + 2H2O
From the equation:
1 mole of MnO2 reacts completely with 4 moles of HCl
From experimental data:
Moles MnO2 = 4.0/Mr = 4.0/87 = 0.046 moles
This would react fully with 4 x 0.046 moles acid = 0.184 moles
Moles HCl = 0.05 x 4 = 0.2 moles
There are more than enough moles of HCl - it is in EXCESS. The MnO2
is the LIMITING REAGENT.
From the equation:
1 mole of MnO2 produces 1 mole of chlorine
Therefore theoretical moles of chlorine = moles of MnO2
= 0.046 moles
Experimentally obtained moles of chlorine = vol in litres/22.7 =
0.8/22.7 = 0.0352 moles
Percentage yield = 0.0352/0.046 = 76.6
% (3 sig figs)
|
Q362-05 In an experiment a
solution containing 3.31 g of lead(II) nitrate reacts with a solution containing
2.00 g of sodium chloride. 2.30 g of lead(II) chloride solid was collected after
filtering and drying. What is the percentage yield of lead II chloride? [N=14,
Pb=208, Cl=35.5, O=16]
Answer
|
Equation for the reaction: Pb(NO3)2 + 2NaCl → PbCl2 +
2NaNO3
From the equation:
1 mole of lead(II) nitrate reacts completely wth 2 moles of sodium
chloride
From experimental data:
Moles of lead(II) nitrate = 3.31/331 = 0.01 moles (this would need
0.02 moles of NaCl)
moles of sodium chloride = 2.00/58.5 = 0.0342 moles
There are more than enough moles of sodium chloride - it is in EXCESS.
The lead(II) nitrate is the LIMITING REAGENT.
From the equation:
1 moles of lead(II) nitrate produces 1 moles of lead(II) chloride
Therefore theoretical moles = 0.01 moles of lead(II) chloride = 2.79g
Experimentally obtained mass = 2.30g
Percentage yield = 2.30/2.79 x 100 = 82.4%
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Q362-06 1.00 g of CaCO3 reacts
with 15cm3 of 2M HNO3. 1.12 g of Ca(NO3)2,
was obtained after crystallisation. Calculate the percentage yield of Ca(NO3)2.
[Ca=40, H=1, N=14, O=16]
Answer
|
Equation for the reaction: CaCO3 + 2HNO3 →
Ca(NO3)2 + CO2 + H2O
From the equation:
1 mole of calcium carbonate reacts with 2 moles of nitric acid
From experimental data:
1.00g of calcium carbonate = 1/100 = 0.01 moles - this would be expected
to react with 0.02 moles of nitric acid
15cm3 of 2M HNO3 is equal to 0.015 x 2 moles
= 0.03 moles
There are more than enough moles of nitric acid - it is in EXCESS.
The calcium carbonate is the LIMITING REAGENT.
From the equation:
1 mole of calcium carbonate makes 1 mole of calcium nitrate
Therefore theoretical moles of calcium nitrate = 0.01 moles = 0.01
x 164 = 1.64g
Experimentally obtained mass = 1.12g
Percentage yield = 1.12/1.64 x 100 = 68.3%
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Q362-07 In an experiment a
solution containing 5.85 g of sodium chloride reacted with a solution containing
12.75 g of silver nitrate. 8.2 g of silver chloride was obtained after filtering
and drying. Calculate the percentage yield of silver chloride. [Ag=108, Cl=35.5,
N=14, O=16]
Answer
|
Equation for the reaction: NaCl + AgNO3 →
NaNO3 + AgCl
From the equation:
1 mole of sodium chloride reacts with 1 moles of silver nitrate
From experimental data:
Moles of sodium chloride 5.85/58.5 = 0.1 moles, this would react
with 0.1 moles silver nitrate
Moles of silver nitrate = 12.75/170 = 0.075 moles
There are more than enough moles of sodium chloride - it is in EXCESS.
The silver nitrate is the LIMITING REAGENT.
From the equation:
1 moles of silver nitrate makes 1 moles of silver chloride
Therefore theoretical moles = 0.075 moles of silver chloride = 0.075
x 143.5 = 10.8g
Experimentally obtained mass = 8.2g
Percentage yield = 8.2/10.8 x 100 = 75.9%
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Q362-08 300dm3
of nitrogen were reacted over a catalyst with 300dm3 of hydrogen
and 30dm3 of ammonia was obtained (all gases measured at the same
temperature and pressure). Calculate the percentage yield of ammonia. [N=14,
H=1]
Answer
|
Equation for the reaction: N2 + 3H2 →
2NH3
From the equation:
Moles of gas are directly proportional to the gas volume, therefore
1 mole of nitrogen reacts with 3 moles of hydrogen
From experimental data:
300dm3 of nitrogen represents 4 volumes and 300dm3
of hydrogen represents 4 volumes
300dm3 volume of nitrogen is expected to react with 3
x 300dm3 = 900 dm3 of hydrogen.
There are not enough moles hydrogen, it is the LIMITING REAGENT.
From the equation:
Moles of ammonia produced = moles of 2 x hydrogen/3
Therefore theoretical volume = 2 x 300/3 = 200dm3
Experimentally obtained volume = 30dm3
Percentage yield = 30/200 x 100= 15.0%
|
Q362-09 In the contact process
for manufacture of sulfuric acid, 24.0 g of sulfur(VI) oxide was obtained from
40 g of sulfur(IV) oxide and 100 g of oxygen. Calculate the percentage yield
of sulfur(VI) oxide. [S=32, O=16]
Answer
|
Equation for the reaction: O2 + 2SO2 ⇋
2SO3
From the equation:
1 mole of oxygen reacts wth 2 moles of sulfur(IV) oxide making 2
moles of sulfur(VI) oxide
From experimental data:
moles of sulfur(IV) oxide = mass/RMM = 40/64 = 0.625 moles, this
would react with 0.625/2 = 0.313 moles oxygen
But, moles of oxygen = 100/32 = 3.125
There are more than enough moles of oxygen - it is in EXCESS. The
sulfur (IV) oxide is the LIMITING REAGENT.
From the equation:
2 moles of sulfur(IV) oxide makes 2 moles of sulfur(VI) oxide
Therefore theoretical moles of sulfur(VI) oxide from 0.625 moles
of sulfur(IV) oxide = 0.625 moles
Experimentally obtained moles of sulfur(VI) oxide = 24/80 = 0.3
Percentage yield = 0.3/0.625 x 100 = 48.0%
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Q362-10 In an organic ester
synthesis, 5.0 g of ethyl ethanoate (CH3COOC2H5)
was obtained from 10.0 g of ethanol (C2H5OH) and 10g of
ethanoic acid (CH3COOH). Calculate the percentage yield of the ester.
[C=12, H=1, O=16]
Answer
|
Equation for the reaction: C2H5OH + CH3COOH
⇋
CH3COOC2H5 + H2O
From the equation:
1 mole of ethanol reacts with 1 mole of ethanoic acid producing 1
mole of ethyl ethanoate.
From experimental data:
Moles of ethanol = 10/46 = 0.217 moles, moles of ethanoic acid =
10/60 = 0.167 moles
There are more than enough moles of ethanol - it is in EXCESS. The
ethanoic acid is the LIMITING REAGENT.
From the equation:
moles of ethanoic acid = moles of ethyl ethanoate
Therefore theoretical moles = 0.167 moles
Experimentally obtained moles = 5.0/88 = 0.0568 moles
Percentage yield = 0.0568/0.167 x 100 = 34.0%
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Now test yourself
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