Standard level
The term yield means "that which is produced" in a chemical reaction.
There are many reasons why the actual yield is lower than the amount calculated using the stoichiometry of the equation. There may be other reactions involving the same reactants that compete with the expected reaction and make different products - this is often the case in organic chemistry. There may be product lost during the purification procedure. Some of the product may get decomposed in the course of extraction. The actual reaction may not go to completion, instead an equilibrium is established.
Syllabus ref: R2.1.4Reactivity 2.1.4 - The percentage yield is calculated from the ratio of experimental yield to theoretical yield.
- Solve problems involving reacting quantities, limiting and excess reactants, theoretical, experimental and percentage yields.
Guidance
Tools and links
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
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)
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% |
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 % |
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% |
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% |
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% |
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% |
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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