Grade by Grade
D. Quantities in Chemical Reactions · Quantities in Chemical Reactions

Theoretical, Actual and Percentage Yield

D2.6 D2.1 D2.5 D3.4About 70 minutes
Learning goals
  • I can tell the difference between theoretical yield and actual yield.
  • I can calculate percentage yield.
  • I can combine limiting-reagent and percentage-yield calculations.
  • I can work backwards from a desired actual yield to the mass of reactant needed.
  • I can explain why yields are usually less than 100% and what a yield over 100% means.

Words to know

theoretical yield
The maximum mass (or amount) of product that can form, calculated by stoichiometry from the limiting reagent.
actual yield
The mass (or amount) of product actually collected and measured in an experiment.
percentage yield
(actual yield ÷ theoretical yield) × 100%.
side reaction
A reaction that happens alongside the main reaction and uses up reactants to make unwanted products.
overall yield
The yield of a process with several steps: the product of the decimal yields of all the steps.

The recipe said 24 cookies

A cookie recipe promises 24 cookies. You make 21. Some dough stuck to the bowl, a few cookies were too close together and merged, and one fell on the floor. The recipe's 24 is a theoretical number; your 21 is the actual number. You got 21 ÷ 24 = 87.5% of what the recipe promised.

Chemists use the same idea every day. Stoichiometry predicts the most product a reaction can make. Real reactions almost always give less.

Three kinds of yield

  • Theoretical yield: the maximum amount of product, calculated from the limiting reagent with stoichiometry (lessons 9 and 10).
  • Actual yield: the amount you really collect and weigh.
  • Percentage yield:

percentage yield = (actual yield ÷ theoretical yield) × 100%

Both yields must be in the same units (usually grams).

Worked example

Example 1: Percentage yield

Limestone is heated to make lime: CaCO₃(s) → CaO(s) + CO₂(g). Heating 50.0 g of CaCO₃ produces 25.2 g of CaO. Find the percentage yield.

  1. Theoretical yield: n(CaCO₃) = 50.0 ÷ 100.09 = 0.49955 mol → n(CaO) = 0.49955 mol → m = 0.49955 × 56.08 = 28.01 g.
  2. Actual yield = 25.2 g (given).
  3. Percentage yield = 25.2 ÷ 28.01 × 100% = 89.95%.

Percentage yield ≈ 90.0% (89.95% rounded to three significant figures).

theoretical yield28.01 g CaO (calculated)actual yield25.2 g CaO (weighed)lost: 2.8 gpercentage yield = 25.2 ÷ 28.01 × 100% = 90.0%
The actual yield is compared with the theoretical yield. The difference is product that was lost or never formed.

Why yields are less than 100%

  • The reaction does not go to completion. Some reactions stop part-way, with reactants and products mixed together (you will study these equilibrium reactions in Grade 12 Chemistry, SCH4U).
  • Side reactions use some reactant to make unwanted products. For example, some magnesium reacts with nitrogen instead of oxygen.
  • Product is lost when it is transferred, filtered, washed or stuck to glassware.
  • Impure reactants: if a reactant is only 95% pure, you started with less than you thought.

What if a yield is over 100%?

A pure, dry product can never have a yield over 100%; that would mean creating matter. A yield over 100% means the product weighed too much: it was still wet, or it contained impurities (such as leftover reactant or another precipitate). The fix is to dry to constant mass and purify the product.

theoreticalyieldactualyieldreaction does notgo to completionside reactions makeother productsproduct lost intransfer or filteringLosses always make the actual yield smaller than the theoretical yield.
Three common reasons why the actual yield is smaller than the theoretical yield.
Worked example

Example 2: Limiting reagent plus yield

10.0 g of Al and 35.0 g of Cl₂ react: 2Al + 3Cl₂ → 2AlCl₃. A student collects 39.5 g of AlCl₃. Find the percentage yield.

  1. From the last lesson: Cl₂ is the limiting reagent, and the theoretical yield is 43.88 g AlCl₃.
  2. Percentage yield = 39.5 ÷ 43.88 × 100% = 90.02%.

Percentage yield = 90.0%. Always find the limiting reagent first; using the excess reagent would give a theoretical yield that is too large.

Worked example

Example 3: Working backwards (planning a synthesis)

Aspirin is made from salicylic acid: C₇H₆O₃ + C₄H₆O₃ → C₉H₈O₄ + C₂H₄O₂. A pharmacy lab wants 10.0 g of aspirin (C₉H₈O₄, M = 180.17 g/mol), and the process usually has a 75.0% yield. What mass of salicylic acid (M = 138.13 g/mol) should be used? (The other reactant is in excess.)

  1. Actual yield wanted = 10.0 g. Theoretical yield needed = actual ÷ (% yield ÷ 100) = 10.0 ÷ 0.750 = 13.33 g.
  2. n(aspirin) = 13.33 ÷ 180.17 = 0.07400 mol.
  3. Mole ratio 1 : 1, so n(salicylic acid) = 0.07400 mol.
  4. m = 0.07400 × 138.13 = 10.22 g.

Use 10.2 g of salicylic acid. Notice you need more reactant than a 100% yield would require.

Multi-step processes

Medicines and many other products are made in several steps. The overall yield is the product of the yields of the steps (as decimals). Even good yields shrink quickly: three steps of 80% each give only 0.80 × 0.80 × 0.80 = 0.512, a 51.2% overall yield. This is one reason chemists in industry work hard to improve every single step.

100%start80%after step 164%after step 251.2%after step 3Three steps, each with an 80% yield: 0.80 × 0.80 × 0.80 = 0.512 → 51.2% overall
In a multi-step synthesis, each step keeps only part of the product from the step before, so the overall yield falls quickly.
Just for fun

Teacher: "Your theoretical yield was 5.0 g, but you reported an actual yield of 7.5 g. Can you explain?"

Student: "I guess my product was very attached to the water." (A yield over 100% almost always means the product was still wet!)

Hands-on activity

Popcorn percentage yield

A grown-up should run the microwave or stove: popping corn and steam are very hot.

You need: plain popcorn kernels, a microwave-safe bowl with a vented lid (or a pot with a lid on the stove), a kitchen scale if you have one, and a plate.

  1. Count exactly 100 kernels. This is your "theoretical yield": if every kernel popped, you would get 100 popped pieces.
  2. Pop them (microwave about 2 to 3 minutes, stopping when popping slows to one pop every 2 seconds). Let them cool.
  3. Count the popped pieces (your actual yield) and the unpopped kernels.
  4. Calculate the percentage yield.
  5. List at least two reasons your yield was not 100%, using chemistry words ("incomplete reaction", "side reaction", "loss").
  6. Optional: repeat with a longer cooking time or a different brand. What happens to the yield? What happens to the number of burnt pieces?

What to look for: most kernels usually pop, but the yield varies a lot from batch to batch. Good reasons: some kernels did not get hot enough (incomplete reaction), some had cracked hulls and lost their steam without popping, and some burned (a "side reaction" that makes an unwanted product). Cooking longer may pop more kernels but also burn more, a trade-off chemists face too.

Your turn

1

The theoretical yield of a reaction is 12.0 g, and the actual yield is 10.2 g. Find the percentage yield.

2

2Mg(s) + O₂(g) → 2MgO(s). 2.43 g of Mg is burned and 3.85 g of MgO is collected. Find the percentage yield.

3

N₂ + 3H₂ → 2NH₃. In one pass through a reactor, 28.0 g of N₂ (with excess H₂) produces 5.11 g of NH₃. Find the percentage yield.

4

Give three reasons why an actual yield is usually less than the theoretical yield.

5

A student reports a percentage yield of 108% for a precipitate. Give the most likely explanation.

6

Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s). 6.54 g of zinc reacts with excess copper(II) sulfate. If the percentage yield is 92.0%, what mass of copper is collected?

More practice: warm-up

1

Write the formula for percentage yield.

2

Which yield is calculated, and which is measured?

3

From which reactant do you calculate the theoretical yield?

4

Theoretical yield 10.0 g, actual yield 8.0 g. Find the percentage yield.

5

Can a pure, dry product have a percentage yield above 100%?

More practice: on your own

1

2Na + Cl₂ → 2NaCl. 5.00 g of Na reacts with 5.00 g of Cl₂ and 7.50 g of NaCl is collected. Find the percentage yield.

2

A cement plant needs 1.00 t of lime, CaO, from CaCO₃ → CaO + CO₂. The process has a 92.0% yield. What mass of CaCO₃ is needed?

3

Fe₂O₃ + 3CO → 2Fe + 3CO₂. 500. g of Fe₂O₃ produces 325 g of iron. Find the percentage yield.

4

A medicine is made in three steps with yields of 90%, 85% and 70%. What is the overall yield?

5

In a precipitation lab, a student weighs the filter paper and precipitate right after filtering. Will the percentage yield be too high or too low? How should the procedure be improved?

Challenge

1

A company wants 500. g of aspirin. Its process has an 82.0% yield. What mass of salicylic acid is needed? (C₇H₆O₃ + C₄H₆O₃ → C₉H₈O₄ + C₂H₄O₂; M(C₇H₆O₃) = 138.13, M(C₉H₈O₄) = 180.17 g/mol)

2

In lesson 10, Example 3, 3.31 g of Pb(NO₃)₂ and 5.00 g of KI could form at most 4.61 g of PbI₂. A student collects 4.12 g of dry PbI₂. Find the percentage yield, and suggest one way the student could have lost product.

3

Process A makes a product with a 90% yield but uses a toxic solvent. Process B has a 70% yield but uses water as the solvent. Give two reasons a company might choose Process B.

Check your understanding

Quiz

1.

What is the actual yield?

2.

Theoretical yield 25.0 g, actual yield 20.0 g. What is the percentage yield?

3.

A reported yield is 115%. What is the most likely cause?

4.

Which reactant is used to calculate the theoretical yield?

5.

Two steps, each with a 50% yield. What is the overall yield?

6.

To get 9.0 g of product from a reaction with a 90% yield, the theoretical yield must be…

For parents and teachers

Big idea: The theoretical yield is what stoichiometry predicts from the limiting reagent; the actual yield is what is collected; percentage yield compares them. This completes D2.6 (limiting reagents and percentage yield) and prepares for the D2.7 lab in the next lesson.

How to help: The popcorn activity is fun and memorable (a grown-up should handle the hot steam). While eating it, ask why the yield was not 100%, and whether cooking longer would really help.

Common difficulties:

  • Calculating the theoretical yield from the excess reagent instead of the limiting reagent.
  • Dividing the wrong way (theoretical ÷ actual).
  • In backward problems, multiplying by the percentage yield instead of dividing (you need more reactant when yields are low).
  • Thinking a yield over 100% is "extra good"; it signals an error such as a wet product.