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D2.7 · Measure actual and theoretical yield and evaluate error

Learn to measure actual and theoretical yield and evaluate error through clear examples and targeted practice.

Ontario Grade 11 Chemistry

Quantities in Chemical Reactions

Measuring product and evaluating differences

A reaction can produce a solid that is collected and weighed. The mass collected is evidence of how much product was recovered, but it may be less than the amount predicted from the starting materials. To understand the difference, first connect the visible product to the particles in the reaction. Then use a balanced equation and measured amounts to calculate and assess yield.

What you will learn

1. From a measured product to a particle model

A balance measures mass. In a yield investigation, the mass of product collected after a reaction is the actual yield. Actual yield is the amount of product obtained in the experiment. It is measured, not predicted.
Before calculating a yield, recall two ideas. Mass is measured in units such as grams. A balanced chemical equation shows the relative amounts of reactants and products. Its coefficients represent ratios of particles, or of amounts measured in moles. A mole is a counting unit used for very large numbers of particles.
At the particle level, reactant particles rearrange to form product particles. The balanced equation accounts for the atoms in that rearrangement. The equation lets us predict how much product could form from a known amount of reactant, if the reaction follows the equation and the product is fully obtained.
That predicted maximum amount is the theoretical yield. It is calculated from the balanced equation and the measured starting amount. If more than one reactant is present, the reactant that can make the least product sets the theoretical yield. This is called the limiting reactant.
actual yield = measured amount of product

2. Calculate and compare yields

To calculate theoretical yield, start with the balanced equation. Convert the known reactant mass to moles using its molar mass. Molar mass is the mass of one mole of a substance, in grams per mole. Use the equation coefficients to find the corresponding amount of product, then convert that amount to grams if the product is being weighed.
Keep units at each step. Units help show whether the calculation is using mass or amount of substance and help catch a conversion error. Use the molar masses and atomic masses given in the course reference material.
Percent yield compares the measured actual yield with the theoretical yield. It is actual yield divided by theoretical yield, multiplied by one hundred percent. A result below one hundred percent means less product was collected than predicted. A result above one hundred percent is a warning to check the measurement, sample purity, or calculation; it does not show that the reaction made more product than the theoretical maximum.
Report a result to a sensible number of significant digits. The answer should not appear more precise than the measurements used. In a lab, the balance’s display and the recorded starting mass affect the precision of the result.
percent yield=actual yieldtheoretical yield×100%\text{percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%

3. Evaluate error using evidence

Error means a difference between a measured result and the value expected from the model or calculation. Evaluating error means using the procedure and observations to suggest a reasonable cause and explain its effect. Do not claim a cause as fact unless the evidence supports it.
For example, some product may remain in the reaction container or be lost during transfer. That would make the collected actual yield smaller than the amount formed. Product that is not fully dry could add water mass and make the measured actual yield appear larger. A balance that is not zeroed correctly can also shift the measured mass.
Check whether the measured product is the intended substance and whether the mass recorded includes only that product. Review arithmetic, units, significant digits, and the balanced equation. If the actual yield is unexpectedly high, consider whether extra material or moisture was included. If it is low, consider product loss or a reaction that did not produce all the amount predicted. These are possible explanations, not proof of what happened.
Percent yield does not identify the source of error by itself. It summarizes the comparison. A useful evaluation links a specific possible issue to the direction of the difference and to evidence from the procedure.

4. A reliable calculation routine

Record the measured starting amount and the actual product amount with units. Write and balance the reaction before using its coefficients. Calculate theoretical yield from the measured reactant amount, then calculate percent yield using actual yield over theoretical yield.
Finally, compare the two yields. State whether the actual yield is below, close to, or above the theoretical yield. Identify a plausible source of error only if it fits the procedure, and explain whether it could raise or lower the measured product amount.
This routine keeps observation, prediction, and evaluation separate. The measured product is the evidence. The balanced equation supports the prediction. The comparison and error discussion explain what the result may mean.

Worked example

Finding theoretical yield and percent yield

A sample of calcium carbonate has a mass of 10.0 g. It is heated, and 5.10 g of calcium oxide is collected. Use the balanced reaction below to calculate the theoretical yield of calcium oxide and the percent yield. Then suggest one possible source of a yield below one hundred percent.
  1. Read the reaction ratio
    The equation is balanced. One mole of calcium carbonate can produce one mole of calcium oxide. The state symbols show that the calcium carbonate and calcium oxide are solids and the carbon dioxide is a gas.
    CaCO3(s)→CaO(s)+CO2(g)\mathrm{CaCO_3(s)} \rightarrow \mathrm{CaO(s)} + \mathrm{CO_2(g)}
  2. Convert reactant mass to predicted product mass
    Using course molar masses, calcium carbonate is 100.09 g/mol100.09\ \mathrm{g/mol} and calcium oxide is 56.08 g/mol56.08\ \mathrm{g/mol}. The one-to-one mole ratio means the product mass is found from the reactant mass and the ratio of molar masses.
    10.0 g CaCO3×1 mol CaCO3100.09 g CaCO3×1 mol CaO1 mol CaCO3×56.08 g CaO1 mol CaO=5.60 g CaO10.0\ \mathrm{g\ CaCO_3} \times \frac{1\ \mathrm{mol\ CaCO_3}}{100.09\ \mathrm{g\ CaCO_3}} \times \frac{1\ \mathrm{mol\ CaO}}{1\ \mathrm{mol\ CaCO_3}} \times \frac{56.08\ \mathrm{g\ CaO}}{1\ \mathrm{mol\ CaO}} = 5.60\ \mathrm{g\ CaO}
  3. Compare measured and predicted product
    The collected mass is the actual yield, 5.10 g5.10\ \mathrm{g}. Divide it by the theoretical yield and multiply by one hundred percent. The starting mass has three significant digits, so report the percent yield to three significant digits.
    5.10 g5.60 g×100%=91.1%\frac{5.10\ \mathrm{g}}{5.60\ \mathrm{g}} \times 100\% = 91.1\%
  4. Evaluate one possible error
    The actual yield is lower than the theoretical yield. One possible reason is that some calcium oxide remained in the container or was lost while it was transferred. That would reduce the mass collected. The result alone does not prove that this occurred.
Answer: The theoretical yield is 5.60 g CaO5.60\ \mathrm{g\ CaO}. The percent yield is 91.1%.
Check: The predicted amount is greater than the amount collected, and the percent yield is below one hundred percent. The comparison is consistent with product loss, but does not establish its cause.

Common mistakes and how to avoid them

Calling the calculated maximum amount the actual yield.
Correction: Actual yield is measured in the experiment. Theoretical yield is calculated from the balanced equation and starting amount.
Using an unbalanced equation or ignoring its coefficients.
Correction: Balance the equation first. Its coefficients provide the mole ratio used to predict product.
Dividing theoretical yield by actual yield for percent yield.
Correction: Divide actual yield by theoretical yield, then multiply by one hundred percent.
Treating a low percent yield as proof of one specific error.
Correction: A low result may fit several causes. Name a possible cause and explain its effect without presenting it as proven.

Lesson summary

Check your understanding

Question 1

A student weighs the product collected after a reaction. What does this value represent?
  1. Theoretical yield
  2. Actual yield
  3. Percent yield
  4. The limiting reactant
Show answer and explanation
Actual yield
The product amount obtained and measured in the experiment is the actual yield.

Question 2

A reaction has a theoretical yield of 8.00 g and an actual yield of 6.00 g. What is the percent yield?
  1. 75.0%
  2. 133%
  3. 2.00%
  4. 14.0%
Show answer and explanation
75.0%
Divide actual by theoretical yield and multiply by one hundred percent: 6.00 g/8.00 g×100%=75.0%6.00\ \mathrm{g} / 8.00\ \mathrm{g} \times 100\% = 75.0\%.

Question 3

Some product is left behind during transfer. What is the most likely effect on actual yield?
  1. It will be measured as lower because less product is collected.
  2. It will be measured as higher because the equation changes.
  3. It will make theoretical yield equal to actual yield.
  4. It will have no possible effect on the collected mass.
Show answer and explanation
It will be measured as lower because less product is collected.
Product left behind is not included in the collected mass, so the measured actual yield is lower.

Key terms

Actual yield
The amount of product obtained and measured in an experiment.
Theoretical yield
The calculated maximum amount of product predicted from the balanced equation and measured reactant amount.
Molar mass
The mass of one mole of a substance, usually expressed in grams per mole.
Percent yield
The actual yield divided by the theoretical yield, multiplied by one hundred percent.
Error
A difference between a measured result and the value expected from a model or calculation.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Chemistry (SCH3U), expectation D2.7. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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