Mass Relationships in Reactions
5. Percentage Yield
Learning outcomes
- I can define percentage yield.
- I can calculate percentage yield from theoretical and actual yields.
- I can explain factors that reduce percentage yield.
- I can interpret percentage yield values in experiments.
- I can solve problems involving percentage yield.
Percentage Yield
In a chemical reaction, the amount of product actually obtained is often less than the maximum amount predicted by stoichiometry.
Percentage yield compares the amount of product actually obtained with the amount that should theoretically have been produced.
It tells us how successful a reaction or experimental process was at producing the desired product.
The equation is:
percentage yield = (actual yield / theoretical yield) × 100%
where:
- actual yield = amount of product actually obtained
- theoretical yield = maximum amount of product predicted by stoichiometry
- percentage yield = actual yield expressed as a percentage of the theoretical maximum
Understanding Percentage Yield
Suppose a reaction has a theoretical yield of:
20.0 g
but only:
16.0 g
of product is actually collected.
Percentage yield:
percentage yield = (16.0 / 20.0) × 100%
= 80.0%
This means the experiment produced:
80% of the maximum predicted amount
The remaining 20% does not necessarily represent one single type of loss. Several different factors may have reduced the amount of product collected.
The Three Types of Yield
It is important to distinguish between three related ideas.
Theoretical Yield
The maximum amount of product predicted by stoichiometry.
Example:
theoretical yield = 50 g
Actual Yield
The amount of product actually obtained during the experiment.
Example:
actual yield = 42 g
Percentage Yield
A comparison between the actual and theoretical yields.
percentage yield = (42 / 50) × 100%
= 84%
So:
theoretical yield → predicted
actual yield → measured
percentage yield → compares the two
Why Use a Percentage?
Suppose two experiments produce:
Experiment A:
- theoretical yield = 10 g
- actual yield = 8 g
Experiment B:
- theoretical yield = 100 g
- actual yield = 80 g
Experiment B produces much more product, but both reactions have:
80% yield
Percentage yield allows us to compare reactions performed on different scales.
Interpreting Percentage Yield
A percentage yield close to:
100%
means the amount collected was close to the theoretical maximum.
A lower percentage indicates that less of the expected product was successfully obtained.
For example:
| Actual Yield | Theoretical Yield | Percentage Yield |
|---|---|---|
| 10 g | 10 g | 100% |
| 9 g | 10 g | 90% |
| 8 g | 10 g | 80% |
| 5 g | 10 g | 50% |
| 2 g | 10 g | 20% |
A higher percentage yield generally means a greater proportion of the theoretically possible product was obtained.
Worked Example: Basic Percentage Yield
A reaction has:
theoretical yield = 25.0 g
actual yield = 21.0 g
Calculate the percentage yield.
Use:
percentage yield = (actual yield / theoretical yield) × 100%
Substitute:
percentage yield = (21.0 / 25.0) × 100%
= 84.0%
Answer
Percentage yield = 84.0%
Worked Example: Magnesium Oxide
Consider:
2Mg + O₂ → 2MgO
Suppose stoichiometry predicts:
20.15 g MgO
but an experiment produces:
18.0 g MgO
Calculate the percentage yield.
percentage yield = (18.0 / 20.15) × 100%
≈ 89.3%
Answer
Percentage yield ≈ 89.3%
This means approximately 89% of the theoretically possible magnesium oxide was obtained.
Calculating the Theoretical Yield First
Sometimes the theoretical yield is not provided.
You must calculate it using stoichiometry before calculating percentage yield.
The complete pathway becomes:
reactant amount
↓
moles of reactant
↓
moles of product
↓
theoretical yield
↓
compare with actual yield
↓
percentage yield
Worked Example: Calculate Theoretical Yield and Percentage Yield
Consider:
2Mg + O₂ → 2MgO
Suppose:
24.3 g Mg
reacts with excess oxygen.
The experiment produces:
35.0 g MgO
Calculate the percentage yield.
Use:
M(Mg) = 24.3 g/mol
M(MgO) = 40.3 g/mol
Convert Mg to moles
24.3 ÷ 24.3 = 1.00 mol Mg
Use the mole ratio
Mg : MgO = 2 : 2
Therefore:
1.00 mol Mg → 1.00 mol MgO
Calculate theoretical yield
1.00 × 40.3 = 40.3 g MgO
Therefore:
theoretical yield = 40.3 g
Calculate percentage yield
Actual yield:
35.0 g
Therefore:
percentage yield = (35.0 / 40.3) × 100%
≈ 86.8%
Answer
Percentage yield ≈ 86.8%
Percentage Yield with a Limiting Reactant
If amounts of two reactants are provided, you must first identify the limiting reactant.
The limiting reactant determines the theoretical yield.
The pathway is:
reactant quantities
↓
convert to moles
↓
identify limiting reactant
↓
calculate theoretical yield
↓
compare with actual yield
↓
calculate percentage yield
Worked Example: Hydrogen and Oxygen
Consider:
2H₂ + O₂ → 2H₂O
Suppose:
10 g H₂
react with:
64 g O₂
and the experiment produces:
63 g H₂O
Use:
M(H₂) = 2 g/mol
M(O₂) = 32 g/mol
M(H₂O) = 18 g/mol
Convert reactants to moles
H₂:
10 ÷ 2 = 5 mol H₂
O₂:
64 ÷ 32 = 2 mol O₂
The reaction requires:
2 mol H₂ : 1 mol O₂
Two moles O₂ require:
4 mol H₂
We have 5 mol H₂.
Therefore:
O₂ is the limiting reactant.
Calculate the Theoretical Yield
From:
2H₂ + O₂ → 2H₂O
2 mol O₂ → 4 mol H₂O
Mass:
4 × 18 = 72 g H₂O
Therefore:
theoretical yield = 72 g
Actual yield:
63 g
Percentage yield:
(63 / 72) × 100%
= 87.5%
Answer
Percentage yield = 87.5%
Why Percentage Yield Is Often Less Than 100%
Real chemical experiments are not perfectly efficient.
There are many reasons why the actual yield may be lower than the theoretical yield.
Incomplete Reactions
Some reactions do not proceed until all of the limiting reactant has been converted into product.
If some reactant remains unreacted, less product forms.
For example, stoichiometry might predict:
10.0 g product
but incomplete reaction might result in only:
8.5 g product
This lowers the percentage yield.
Reversible Reactions
Some reactions can proceed in both directions.
Instead of completely converting reactants into products, the reaction may reach equilibrium.
At equilibrium:
- reactants remain
- products are present
- forward and reverse reactions continue
Because not all reactants become products, the yield may be lower than the theoretical maximum.
Side Reactions
Reactants may sometimes undergo unwanted reactions.
Instead of producing only the desired product:
reactants → desired product
some reactants may form other substances:
reactants → unwanted products
This reduces the amount of desired product.
Product Lost During Transfer
Some product may remain:
- inside a beaker
- on a stirring rod
- inside a flask
- in a funnel
- on filter paper
Every transfer creates an opportunity to lose a small amount of material.
For example, pouring a solid from one container into another may leave some material behind.
Product Lost During Filtration
When collecting a precipitate:
- small particles may pass through the filter
- some product may remain dissolved
- some may stick to the glassware
- some may be spilled
This lowers the actual yield.
Product Lost During Purification
Chemical products often need to be purified.
Processes such as:
- filtration
- recrystallization
- washing
- extraction
- distillation
can cause some desired product to be lost.
A highly pure product may therefore have a lower recovered mass.
Gas Loss
If the desired product is a gas, some may escape from the apparatus.
For example:
CaCO₃ → CaO + CO₂
If CO₂ is being collected, leaks in the apparatus can reduce the amount measured.
Impure Reactants
Suppose a sample is labelled:
10.0 g
but contains only:
8.0 g of the actual reactant
If the theoretical yield calculation incorrectly assumes all 10.0 g are pure reactant, it will predict too much product.
This can make the calculated percentage yield appear unusually low.
Product Decomposition
Sometimes the desired product can decompose during:
- heating
- drying
- storage
- purification
If some product breaks down after it forms, the final measured amount will be lower.
Mechanical Losses
Some losses have nothing to do with the chemistry itself.
Examples include:
- spilling material
- losing crystals during transfer
- leaving product on equipment
- breaking or damaging a sample
- losing fine particles
These are sometimes called mechanical losses.
Can Percentage Yield Equal 100%?
Yes.
A percentage yield of:
100%
means:
actual yield = theoretical yield
For example:
Theoretical:
15.0 g
Actual:
15.0 g
Percentage:
(15.0 / 15.0) × 100% = 100%
This represents perfect agreement between the measured and theoretical quantities.
In real laboratory work, exactly 100% is possible but should still be interpreted in light of measurement uncertainty and experimental conditions.
Can Percentage Yield Be Greater Than 100%?
A calculated percentage yield can sometimes be greater than 100%.
For example:
Theoretical yield:
10.0 g
Measured actual yield:
10.8 g
Percentage yield:
(10.8 / 10.0) × 100%
= 108%
This does not normally mean the reaction somehow produced more pure desired product than was theoretically possible.
Instead, something should be investigated.
Why Might Percentage Yield Exceed 100%?
The Product Is Wet
Suppose a solid product contains water.
The balance measures:
product + water
The measured mass is therefore too high.
The Product Contains Impurities
Other substances may be mixed with the product.
The measured mass then includes:
desired product + impurities
Unreacted Reactants Remain
The sample may contain some reactant that was not removed.
The measured material is therefore not pure product.
Incomplete Drying
This is particularly common when precipitates or crystals are collected.
Water or solvent remaining on the product increases its apparent mass.
Measurement Error
An incorrect balance reading or another measurement problem can produce an inaccurate actual yield.
Calculation Error
The theoretical yield may have been calculated incorrectly.
Possible mistakes include:
- incorrect molar mass
- unbalanced equation
- incorrect mole ratio
- wrong limiting reactant
- arithmetic error
Interpreting a Yield Greater Than 100%
Suppose:
percentage yield = 112%
A useful scientific conclusion is not:
The reaction was 112% efficient.
Instead:
The measured product mass exceeds the theoretical maximum, suggesting contamination, incomplete drying, measurement error, or an error in the theoretical-yield calculation.
Percentage yield should be interpreted scientifically rather than accepted without question.
Comparing Percentage Yields
Consider three experiments:
Experiment A recovered the greatest proportion of its theoretical maximum.
However, percentage yield alone does not tell us everything about whether a process is desirable.
We may also need to consider:
- purity
- safety
- cost
- reaction time
- energy use
- waste production
- environmental impact
Worked Example: Calcium Oxide
Calcium carbonate decomposes:
CaCO₃ → CaO + CO₂
Suppose:
200 g CaCO₃
are heated.
The experiment produces:
95.2 g CaO
Use:
M(CaCO₃) = 100 g/mol
M(CaO) = 56 g/mol
Calculate moles CaCO₃
200 ÷ 100 = 2.00 mol
Calculate theoretical CaO
Ratio:
1 mol CaCO₃ : 1 mol CaO
Therefore:
2.00 mol CaO
Theoretical mass:
2.00 × 56 = 112 g
Calculate percentage yield
percentage yield = (95.2 / 112) × 100%
= 85.0%
Answer
Percentage yield = 85.0%
Worked Example: Potassium Chlorate
Consider:
2KClO₃ → 2KCl + 3O₂
Suppose:
24.5 g KClO₃
decomposes.
The experiment collects:
8.16 g O₂
Use:
M(KClO₃) = 122.5 g/mol
M(O₂) = 32.0 g/mol
Convert KClO₃ to moles
24.5 ÷ 122.5 = 0.200 mol
Calculate theoretical O₂
Ratio:
2 mol KClO₃ : 3 mol O₂
Therefore:
0.200 × (3/2)
= 0.300 mol O₂
Mass:
0.300 × 32.0 = 9.60 g
Theoretical yield:
9.60 g O₂
Calculate percentage yield
(8.16 / 9.60) × 100%
= 85.0%
Answer
Percentage yield = 85.0%
Worked Example: Iron Oxide
Consider:
4Fe + 3O₂ → 2Fe₂O₃
Suppose:
112 g Fe
react with excess oxygen.
The experiment produces:
136 g Fe₂O₃
Use:
M(Fe) = 56 g/mol
M(Fe₂O₃) = 160 g/mol
Convert Fe to moles
112 ÷ 56 = 2.00 mol Fe
Calculate theoretical product
Ratio:
4 mol Fe : 2 mol Fe₂O₃
Therefore:
2.00 mol Fe → 1.00 mol Fe₂O₃
Theoretical mass:
1.00 × 160 = 160 g
Calculate percentage yield
(136 / 160) × 100%
= 85%
Answer
Percentage yield = 85%
Rearranging the Percentage Yield Equation
The main equation is:
percentage yield = (actual yield / theoretical yield) × 100%
We can rearrange it to find other quantities.
Finding Actual Yield
If percentage yield and theoretical yield are known:
actual yield = (percentage yield / 100) × theoretical yield
For example:
Theoretical yield:
80 g
Percentage yield:
75%
Actual yield:
(75 / 100) × 80
= 60 g
Answer
Actual yield = 60 g
Finding Theoretical Yield
If actual yield and percentage yield are known:
theoretical yield = actual yield × 100 / percentage yield
For example:
Actual yield:
36 g
Percentage yield:
80%
Theoretical yield:
36 × 100 / 80
= 45 g
Answer
Theoretical yield = 45 g
A Useful Reasonableness Check
If the percentage yield is below 100%, then:
actual yield < theoretical yield
For example:
actual = 40 g
theoretical = 50 g
makes sense.
But:
actual = 50 g
theoretical = 40 g
gives more than 100%.
That does not automatically mean the arithmetic is wrong, but it tells us that the result requires investigation.
Percentage Yield and Experimental Technique
Imagine two students perform the same reaction.
Both have a theoretical yield of:
12.0 g
Student A collects:
10.8 g
Student B collects:
8.4 g
Student A:
(10.8 / 12.0) × 100% = 90%
Student B:
(8.4 / 12.0) × 100% = 70%
If all other conditions were comparable, Student A recovered a larger proportion of the theoretically possible product.
Possible differences could include:
- transfer technique
- filtration technique
- reaction completion
- product recovery
- drying
- accidental loss
Percentage yield can therefore help evaluate an experimental procedure.
Percentage Yield and Purity Are Different
A high percentage yield does not automatically mean the product is pure.
Imagine an experiment has:
theoretical yield = 10.0 g
The collected sample has a mass of:
9.8 g
This gives:
98% yield
But if the 9.8 g contains impurities, the actual amount of desired product is lower.
Therefore:
yield describes how much material was obtained relative to the theoretical amount.
purity describes how much of the collected material is actually the desired substance.
These are different ideas.
High Yield Does Not Always Mean a Better Process
Suppose Process A has:
95% yield
but requires:
- a toxic solvent
- very high temperatures
- large amounts of energy
- difficult waste disposal
Process B has:
88% yield
but:
- uses safer materials
- operates at lower temperature
- produces less waste
- uses less energy
Depending on the purpose, Process B might be preferable.
Chemists consider more than percentage yield alone.
Percentage Yield in Industry
Percentage yield is extremely important in industrial chemistry.
Suppose a factory has:
theoretical production = 10,000 kg
but actually produces:
8,500 kg
Percentage yield:
(8,500 / 10,000) × 100%
= 85%
A 15% difference on a laboratory scale might mean a few grams.
On an industrial scale, it could represent:
1,500 kg of expected product not obtained
This can have major economic consequences.
Improving Percentage Yield
Chemists may improve percentage yield by:
- optimizing temperature
- optimizing pressure
- choosing appropriate catalysts
- improving mixing
- allowing sufficient reaction time
- reducing side reactions
- improving product separation
- reducing transfer losses
- improving filtration
- improving purification methods
- recycling unreacted materials
The best method depends on the reaction.
Percentage Yield and Green Chemistry
Improving yield can reduce waste because more of the starting material becomes useful product.
This can:
- conserve raw materials
- reduce waste disposal
- reduce production costs
- reduce environmental impact
However, percentage yield should be considered alongside other measures such as:
- atom economy
- energy requirements
- solvent use
- toxicity
- renewability of raw materials
A reaction with a high yield is not automatically environmentally friendly.
A Complete Percentage Yield Problem
Consider:
2Al + 3Cl₂ → 2AlCl₃
Suppose:
27.0 g Al
react with excess chlorine.
The experiment produces:
113.5 g AlCl₃
Use:
M(Al) = 27.0 g/mol
M(AlCl₃) = 133.5 g/mol
Convert Al to moles
27.0 ÷ 27.0 = 1.00 mol Al
Use the mole ratio
Al : AlCl₃ = 2 : 2
Therefore:
1.00 mol AlCl₃
Calculate theoretical yield
1.00 × 133.5 = 133.5 g
Identify actual yield
actual yield = 113.5 g
Calculate percentage yield
percentage yield = (113.5 / 133.5) × 100%
≈ 85.0%
Answer
Percentage yield ≈ 85.0%
Multi-Step Example with a Limiting Reactant
Consider:
N₂ + 3H₂ → 2NH₃
Suppose:
56 g N₂
react with:
9 g H₂
and:
45.9 g NH₃
are actually obtained.
Use:
M(N₂) = 28 g/mol
M(H₂) = 2 g/mol
M(NH₃) = 17 g/mol
Convert reactants to moles
N₂:
56 ÷ 28 = 2.00 mol
H₂:
9 ÷ 2 = 4.50 mol
The reaction requires:
1 mol N₂ : 3 mol H₂
For 2 mol N₂ we would need:
6 mol H₂
Only:
4.50 mol H₂
are available.
Therefore:
H₂ is limiting.
Calculate theoretical yield
Ratio:
3 mol H₂ : 2 mol NH₃
Therefore:
4.50 × (2/3) = 3.00 mol NH₃
Mass:
3.00 × 17 = 51.0 g NH₃
Calculate percentage yield
(45.9 / 51.0) × 100%
= 90.0%
Answer
Percentage yield = 90.0%
Common Mistakes
Reversing the Formula
Incorrect:
theoretical / actual × 100
Correct:
actual / theoretical × 100
Remember:
what you got / what you could have got × 100
Forgetting to Multiply by 100
For example:
18 / 20 = 0.90
This is the decimal form.
Percentage yield:
0.90 × 100% = 90%
Confusing Actual and Theoretical Yield
Actual yield is measured.
Theoretical yield is calculated.
Using the Wrong Limiting Reactant
If two reactants are given, first determine which is limiting.
The theoretical yield must be based on the limiting reactant.
Comparing Different Units
Do not calculate:
25 g / 0.50 mol
Both yields must be expressed in compatible units.
Usually:
g / g
or:
mol / mol
Assuming Every Yield Above 100% Is Impossible Data
A measured percentage greater than 100% can occur, but it usually indicates a problem such as:
- wet product
- contamination
- incorrect calculation
- measurement error
It should be investigated rather than interpreted as extra chemical efficiency.
Assuming High Yield Means High Purity
A contaminated product may have a large measured mass and therefore an apparently high percentage yield.
Yield and purity are different measurements.
Rounding Too Early
Keep several digits through intermediate calculations.
Round the final percentage appropriately.
Key Terms
Percentage yield — The actual yield expressed as a percentage of the theoretical yield.
Theoretical yield — The maximum amount of product predicted by stoichiometry.
Actual yield — The amount of product actually obtained experimentally.
Limiting reactant — The reactant consumed first and therefore responsible for determining theoretical yield.
Excess reactant — A reactant present in more than the required amount.
Stoichiometry — The quantitative relationship between reactants and products.
Incomplete reaction — A reaction in which not all available limiting reactant forms the desired product.
Side reaction — An unwanted reaction that forms products other than the desired product.
Mechanical loss — Loss of material through physical handling, transfer, filtration, spilling, or similar processes.
Purity — The proportion of a sample consisting of the desired substance.
Contamination — The presence of unwanted substances in a sample.
Reaction efficiency — A general description of how effectively a process produces its intended result; percentage yield is one useful measure of this.
Product recovery — The process of collecting and isolating the desired product after a reaction.
Reversible reaction — A reaction capable of proceeding in both forward and reverse directions.
Key Takeaways
- Percentage yield compares the actual yield with the theoretical yield.
- The equation is:
percentage yield = (actual yield / theoretical yield) × 100%
- Theoretical yield is calculated using stoichiometry.
- Actual yield is measured experimentally.
- A yield of 100% means the actual and theoretical yields are equal.
- Percentage yields are often below 100%.
- Incomplete reactions can reduce yield.
- Side reactions can reduce yield.
- Product can be lost during transfer, filtration, purification, drying, and other procedures.
- Reversible reactions may prevent complete conversion into products.
- Impure reactants can affect yield calculations.
- A percentage yield greater than 100% usually indicates contamination, incomplete drying, measurement error, or a calculation problem.
- Percentage yield can be used to compare experiments performed at different scales.
- When two reactant amounts are given, the limiting reactant must be identified before calculating theoretical yield.
- Percentage yield and purity are not the same thing.
- A high percentage yield does not automatically mean a process is safe, inexpensive, sustainable, or environmentally friendly.
- Improving percentage yield can reduce waste and improve economic efficiency.
- Percentage yield is important in laboratory chemistry, pharmaceuticals, manufacturing, and industrial chemical processes.
The central relationship to remember is:
PERCENTAGE YIELD = ACTUAL ÷ THEORETICAL × 100%
Check Your Understanding
Basic Percentage Yield
1. Define percentage yield in your own words.
2. A reaction has a theoretical yield of 20.0 g and an actual yield of 16.0 g. Calculate the percentage yield.
3. The theoretical yield is 50.0 g and the actual yield is 42.5 g. Calculate the percentage yield.
4. A reaction should produce 80 g but produces only 60 g. Calculate the percentage yield.
5. A reaction has an actual yield of 18 g and a theoretical yield of 24 g. Calculate the percentage yield.
6. A reaction has an actual yield of 39.6 g and a theoretical yield of 44.0 g. Calculate the percentage yield.
7. Explain what a percentage yield of 75% means.
8. Explain what a percentage yield of 100% means.
Rearranging the Equation
9. A reaction has a theoretical yield of 60 g and a percentage yield of 80%. Calculate the actual yield.
10. A reaction has a theoretical yield of 250 g and a percentage yield of 92%. Calculate the actual yield.
11. A reaction produces 36 g at an 80% yield. Calculate the theoretical yield.
12. A reaction produces 72 g at a 90% yield. Calculate the theoretical yield.
Stoichiometry and Percentage Yield
Use:
2Mg + O₂ → 2MgO
with:
M(Mg) = 24.3 g/mol
M(MgO) = 40.3 g/mol
13. Calculate the theoretical yield of MgO from 24.3 g Mg.
14. If 36.27 g MgO are actually produced, calculate the percentage yield.
15. Calculate the theoretical yield from 48.6 g Mg.
16. If the reaction in Question 15 has an 85% yield, calculate the actual mass of MgO obtained.
Use:
CaCO₃ → CaO + CO₂
with:
M(CaCO₃) = 100 g/mol
M(CaO) = 56 g/mol
17. Calculate the theoretical yield of CaO from 250 g CaCO₃.
18. If 119 g CaO are obtained, calculate the percentage yield.
19. If the percentage yield were 92%, calculate the actual mass of CaO that would be obtained from 250 g CaCO₃.
Limiting Reactant and Yield
Use:
2H₂ + O₂ → 2H₂O
with:
M(H₂) = 2 g/mol
M(O₂) = 32 g/mol
M(H₂O) = 18 g/mol
20. If 10 g H₂ react with 64 g O₂, identify the limiting reactant.
21. Calculate the theoretical yield of H₂O.
22. If 61.2 g H₂O are obtained, calculate the percentage yield.
23. If the percentage yield were 75%, calculate the actual yield.
More Challenging Problems
Use:
4Fe + 3O₂ → 2Fe₂O₃
with:
M(Fe) = 56 g/mol
M(O₂) = 32 g/mol
M(Fe₂O₃) = 160 g/mol
24. If 112 g Fe react with excess oxygen, calculate the theoretical yield of Fe₂O₃.
25. If 140 g Fe₂O₃ are obtained, calculate the percentage yield.
26. If the reaction operates at 92% yield, calculate the actual amount of Fe₂O₃ produced.
Use:
N₂ + 3H₂ → 2NH₃
with:
M(N₂) = 28 g/mol
M(H₂) = 2 g/mol
M(NH₃) = 17 g/mol
27. If 56 g N₂ react with 9 g H₂, identify the limiting reactant.
28. Calculate the theoretical yield of NH₃.
29. If 43.35 g NH₃ are obtained, calculate the percentage yield.
30. If the reaction instead operated at 95% yield, calculate the actual mass of NH₃ produced.
Analysis and Application
31. Give four reasons why percentage yield may be below 100%.
32. Explain how product loss during filtration affects percentage yield.
33. Explain how incomplete reactions affect percentage yield.
34. Explain how side reactions can lower percentage yield.
35. A student calculates a percentage yield of 108%. Give three possible explanations.
36. Explain why incomplete drying can produce an apparent percentage yield above 100%.
37. Explain the difference between percentage yield and purity.
38. Two reactions have yields of 92% and 85%. Explain why the reaction with the 92% yield is not automatically the better industrial process.
39. Explain why percentage yield is economically important in large-scale chemical manufacturing.
40. Describe the complete procedure for calculating percentage yield when the masses of two reactants and the actual mass of product are provided.