The best examples of percent yield calculations examples in chemistry

Before we talk formulas, let’s walk straight into the lab. The easiest way to learn this topic is to see several examples of percent yield calculations examples in chemistry that you can model your own work on.

Imagine you carry out a simple reaction in a beaker and measure how much product you actually collect. The balanced equation predicts a maximum (theoretical yield), but your actual yield is always less. Percent yield just compares those two:

\[\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%\]

That’s the backbone of every example of percent yield you’ll see below. The variety comes from how we get the theoretical yield.

Classic classroom examples of percent yield calculations examples in chemistry

Let’s start with the reactions you’re most likely to see in a first chemistry course. These are the best examples to build your intuition before moving to more complex syntheses.

Example 1: Magnesium burning in oxygen to form magnesium oxide

Reaction:

\\[2\,\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\,\text{MgO}(s)\]

Suppose a student burns 2.40 g of Mg in excess oxygen and collects 3.60 g of MgO.

Step 1 – Moles of Mg
Molar mass of Mg ≈ 24.3 g/mol.

\\[n_{\text{Mg}} = \frac{2.40\,\text{g}}{24.3\,\text{g/mol}} \approx 0.0988\,\text{mol}\]

Step 2 – Theoretical moles of MgO
From the equation, 2 mol Mg → 2 mol MgO, so the mole ratio is 1:1.

\\[n_{\text{MgO, theo}} = 0.0988\,\text{mol}\]

Step 3 – Theoretical mass of MgO
Molar mass of MgO ≈ 40.3 g/mol.

\\[m_{\text{MgO, theo}} = 0.0988\,\text{mol} \times 40.3\,\text{g/mol} \approx 3.98\,\text{g}\]

Step 4 – Percent yield
Actual yield = 3.60 g.

\\[\text{Percent yield} = \frac{3.60\,\text{g}}{3.98\,\text{g}} \times 100\% \approx 90.5\%\]

This is one of the most common examples of percent yield calculations examples in chemistry because it’s easy to set up in the lab and the numbers are straightforward.

Example 2: Copper(II) sulfate from copper and nitric acid

A slightly more realistic lab scenario: copper metal reacts with nitric acid to eventually form copper(II) sulfate after a sequence of steps. Assume the simplified overall stoichiometry is:

\\[\text{Cu}(s) + \text{H}_2\text{SO}_4(aq) + 2\,\text{HNO}_3(aq) \rightarrow \text{CuSO}_4(aq) + 2\,\text{NO}_2(g) + 2\,\text{H}_2\text{O}(l)\]

A student starts with 1.00 g of Cu and obtains 3.20 g of CuSO₄·5H₂O crystals after crystallization.

Step 1 – Moles of Cu
Molar mass of Cu ≈ 63.5 g/mol.

\\[n_{\text{Cu}} = \frac{1.00\,\text{g}}{63.5\,\text{g/mol}} \approx 0.0157\,\text{mol}\]

Step 2 – Theoretical moles of CuSO₄·5H₂O
Cu and CuSO₄ are 1:1 in the overall process, and each CuSO₄ forms one CuSO₄·5H₂O crystal.

\\[n_{\text{CuSO}_4\cdot 5\text{H}_2\text{O, theo}} = 0.0157\,\text{mol}\]

Step 3 – Theoretical mass of CuSO₄·5H₂O
Molar mass of CuSO₄·5H₂O ≈ 249.7 g/mol.

\\[m_{\text{theo}} = 0.0157\,\text{mol} \times 249.7\,\text{g/mol} \approx 3.92\,\text{g}\]

Step 4 – Percent yield
Actual yield = 3.20 g.

\\[\text{Percent yield} = \frac{3.20}{3.92} \times 100\% \approx 81.6\%\]

Here you see crystallization losses and side reactions lowering the yield — a pattern that shows up in many real examples in teaching labs.

Limiting-reactant based examples of percent yield calculations

Most real reactions don’t have a reactant in perfect excess. You need to identify the limiting reactant before you can even talk about theoretical yield. These examples of percent yield calculations examples in chemistry show that full workflow.

Example 3: Hydrogen gas from zinc and hydrochloric acid

Reaction:

\\[\text{Zn}(s) + 2\,\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)\]

Suppose you mix 5.00 g of Zn with 7.30 g of HCl and collect 0.150 g of H₂.

Step 1 – Moles of each reactant
Molar masses: Zn ≈ 65.4 g/mol, HCl ≈ 36.5 g/mol.

\\[n_{\text{Zn}} = \frac{5.00}{65.4} \approx 0.0764\,\text{mol}\] \\[n_{\text{HCl}} = \frac{7.30}{36.5} \approx 0.200\,\text{mol}\]

Step 2 – Find limiting reactant
Stoichiometry: 1 mol Zn reacts with 2 mol HCl.

Moles of HCl needed for 0.0764 mol Zn:

\\[0.0764\,\text{mol Zn} \times 2 = 0.1528\,\text{mol HCl}\]

You have 0.200 mol HCl, which is more than required, so Zn is limiting.

Step 3 – Theoretical moles of H₂
From the equation, 1 mol Zn → 1 mol H₂.

\\[n_{\text{H}_2,\,\text{theo}} = 0.0764\,\text{mol}\]

Step 4 – Theoretical mass of H₂
Molar mass of H₂ ≈ 2.02 g/mol.

\\[m_{\text{H}_2,\,\text{theo}} = 0.0764\,\text{mol} \times 2.02\,\text{g/mol} \approx 0.154\,\text{g}\]

Step 5 – Percent yield
Actual yield = 0.150 g.

\\[\text{Percent yield} = \frac{0.150}{0.154} \times 100\% \approx 97.4\%\]

This is a high-yield reaction, which is why it’s a favorite example of percent yield for gas-collection labs.

Example 4: Formation of water from hydrogen and oxygen

Reaction:

\\[2\,\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\,\text{H}_2\text{O}(l)\]

You react 4.00 g of H₂ with 32.0 g of O₂ and obtain 34.0 g of liquid water.

Step 1 – Moles of reactants
Molar masses: H₂ ≈ 2.02 g/mol, O₂ ≈ 32.0 g/mol.

\\[n_{\text{H}_2} = \frac{4.00}{2.02} \approx 1.98\,\text{mol}\]
\\[n_{\text{O}_2} = \frac{32.0}{32.0} = 1.00\,\text{mol}\]

Step 2 – Limiting reactant
Stoichiometry: 2 mol H₂ per 1 mol O₂.

To use all 1.00 mol O₂, you need 2.00 mol H₂. You only have 1.98 mol H₂, so H₂ is slightly limiting.

Step 3 – Theoretical moles of H₂O
From the equation, 2 mol H₂ → 2 mol H₂O, so 1:1 between H₂ and H₂O.

\\[n_{\text{H}_2\text{O, theo}} = 1.98\,\text{mol}\]

Step 4 – Theoretical mass of H₂O
Molar mass of H₂O ≈ 18.0 g/mol.

\\[m_{\text{theo}} = 1.98\,\text{mol} \times 18.0\,\text{g/mol} \approx 35.6\,\text{g}\]

Step 5 – Percent yield
Actual yield = 34.0 g.

\\[\text{Percent yield} = \frac{34.0}{35.6} \times 100\% \approx 95.5\%\]

This reaction is often used as one of the best examples of percent yield calculations examples in chemistry because the stoichiometry is simple and the concept of limiting reactant is crystal clear.

Synthesis-focused examples of percent yield calculations in chemistry

In real labs and industry, the stakes are higher: yield affects cost, safety, and environmental impact. Modern green chemistry and process optimization research (see resources from the U.S. Environmental Protection Agency) pay close attention to percent yield.

Example 5: Aspirin synthesis in an organic chemistry lab

Reaction (simplified):

\\[\text{Salicylic acid} + \text{Acetic anhydride} \rightarrow \text{Aspirin} + \text{Acetic acid}\]

Suppose a student uses 2.00 g of salicylic acid (molar mass ≈ 138.1 g/mol) and excess acetic anhydride, and isolates 2.90 g of aspirin (molar mass ≈ 180.2 g/mol).

Step 1 – Moles of salicylic acid

\\[n_{\text{SA}} = \frac{2.00}{138.1} \approx 0.0145\,\text{mol}\]

Step 2 – Theoretical moles and mass of aspirin
Stoichiometry is 1:1.

\\[n_{\text{aspirin, theo}} = 0.0145\,\text{mol}\] \\[m_{\text{theo}} = 0.0145\,\text{mol} \times 180.2\,\text{g/mol} \approx 2.61\,\text{g}\]

Step 3 – Percent yield

\\[\text{Percent yield} = \frac{2.90}{2.61} \times 100\% \approx 111\%\]

A yield over 100% is a red flag. This is one of the most instructive examples of percent yield calculations examples in chemistry because it forces you to think about purity and measurement error. The “extra” mass likely comes from trapped water, unreacted reagents, or impurities in the product.

This is exactly the kind of situation discussed in many undergraduate lab manuals and teaching resources from universities like MIT OpenCourseWare, which stress that percent yield is only as good as your experimental technique.

Example 6: Industrial ammonia synthesis (Haber process)

Reaction:

\\[\text{N}_2(g) + 3\,\text{H}_2(g) \rightleftharpoons 2\,\text{NH}_3(g)\]

In the Haber process, real plants rarely achieve 100% conversion in a single pass because the reaction is reversible and limited by equilibrium. However, percent yield is still calculated per pass.

Suppose a reactor feed contains 100. g of N₂ and excess H₂. At given conditions, the process engineers observe that 68.0 g of NH₃ are produced before recycling unreacted gases.

Step 1 – Moles of N₂
Molar mass of N₂ ≈ 28.0 g/mol.

\\[n_{\text{N}_2} = \frac{100.}{28.0} \approx 3.57\,\text{mol}\]

Step 2 – Theoretical moles and mass of NH₃
Stoichiometry: 1 mol N₂ → 2 mol NH₃.

\\[n_{\text{NH}_3,\,\text{theo}} = 2 \times 3.57 \approx 7.14\,\text{mol}\]

Molar mass of NH₃ ≈ 17.0 g/mol.

\\[m_{\text{theo}} = 7.14\,\text{mol} \times 17.0\,\text{g/mol} \approx 121\,\text{g}\]

Step 3 – Percent yield

\\[\text{Percent yield} = \frac{68.0}{121} \times 100\% \approx 56\%\]

This is a realistic single-pass yield. Modern plants use recycling and optimized conditions to improve overall efficiency, which is actively studied in chemical engineering programs (see resources from NIST on thermodynamics and equilibrium data).

This is one of the best examples of percent yield calculations examples in chemistry for connecting classroom stoichiometry to real-world industrial practice.

More real examples: decomposition and precipitation reactions

To round out the variety, let’s look at a decomposition reaction and a precipitation reaction. These often show up on exams as slightly more challenging examples of percent yield calculations.

Example 7: Decomposition of potassium chlorate

Reaction:

\\[2\,\text{KClO}_3(s) \rightarrow 2\,\text{KCl}(s) + 3\,\text{O}_2(g)\]

A sample of 12.3 g of KClO₃ is heated strongly, and 3.20 g of O₂ are collected.

Step 1 – Moles of KClO₃
Molar mass of KClO₃ ≈ 122.6 g/mol.

\\[n_{\text{KClO}_3} = \frac{12.3}{122.6} \approx 0.100\,\text{mol}\]

Step 2 – Theoretical moles of O₂
Stoichiometry: 2 mol KClO₃ → 3 mol O₂.

\\[n_{\text{O}_2,\,\text{theo}} = 0.100\,\text{mol KClO}_3 \times \frac{3\,\text{mol O}_2}{2\,\text{mol KClO}_3} = 0.150\,\text{mol O}_2\]

Step 3 – Theoretical mass of O₂
Molar mass of O₂ ≈ 32.0 g/mol.

\\[m_{\text{theo}} = 0.150\,\text{mol} \times 32.0\,\text{g/mol} = 4.80\,\text{g}\]

Step 4 – Percent yield

\\[\text{Percent yield} = \frac{3.20}{4.80} \times 100\% \approx 66.7\%\]

This example of percent yield highlights a common problem: gas losses during collection, incomplete decomposition, or leaks in the apparatus.

Example 8: Precipitation of barium sulfate

Reaction:

\\[\text{BaCl}_2(aq) + \text{Na}_2\text{SO}_4(aq) \rightarrow \text{BaSO}_4(s) + 2\,\text{NaCl}(aq)\]

Suppose you mix solutions containing 0.250 mol of BaCl₂ and 0.300 mol of Na₂SO₄, and after filtration and drying, you obtain 54.0 g of BaSO₄.

Step 1 – Limiting reactant
Stoichiometry is 1:1 between BaCl₂ and Na₂SO₄ to form BaSO₄.

You have 0.250 mol BaCl₂ and 0.300 mol Na₂SO₄, so BaCl₂ is limiting.

Step 2 – Theoretical moles and mass of BaSO₄
Moles of BaSO₄ formed = 0.250 mol.

Molar mass of BaSO₄ ≈ 233.4 g/mol.

\\[m_{\text{theo}} = 0.250\,\text{mol} \times 233.4\,\text{g/mol} \approx 58.4\,\text{g}\]

Step 3 – Percent yield

\\[\text{Percent yield} = \frac{54.0}{58.4} \times 100\% \approx 92.5\%\]

This precipitation reaction is one of the standard examples of percent yield calculations examples in chemistry because it brings in concepts like solubility, filtration, and drying — all of which affect the final yield.

Why percent yield matters in 2024–2025 chemistry

Percent yield isn’t just a classroom curiosity. In 2024–2025, it sits right at the intersection of green chemistry, cost control, and safety:

• In pharmaceutical manufacturing, low yields mean higher costs and more chemical waste. Regulatory bodies like the U.S. Food and Drug Administration expect tight control over process performance, and percent yield is a key metric.
• In materials science, researchers optimizing battery materials, catalysts, and polymers report percent yield in publications to compare synthetic routes.
• In environmental chemistry, maximizing yield reduces side products and waste, aligning with principles promoted by the EPA’s green chemistry program.

So when you practice with these examples of percent yield calculations examples in chemistry, you’re not just chasing a grade — you’re learning how chemists think about efficiency in the real world.

Quick pattern summary: how to tackle any percent yield problem

If you scan back through all eight real examples, you’ll notice the same pattern over and over:

• Start from mass or volume data for reactants.
• Convert to moles, then use the balanced equation to find the moles of product.
• That product amount becomes your theoretical yield (in grams).
• Compare your actual yield (measured in the lab) to the theoretical value.
• Use the percent yield formula to finish.

Once you see that structure, even the more complicated examples of percent yield calculations — multi-step syntheses, limiting reactant puzzles, or industrial processes — become manageable.

FAQ: common questions about percent yield and examples

Q1. Can you give another simple example of percent yield with moles instead of grams?
Yes. Suppose a reaction predicts 0.500 mol of product, but you only isolate 0.365 mol. The percent yield is:

\\[\text{Percent yield} = \frac{0.365}{0.500} \times 100\% = 73.0\%\]

This is the same idea as the other examples of percent yield calculations examples in chemistry, just staying in moles instead of converting to grams.

Q2. Why do some examples of percent yield calculations give values over 100%?
Any example of percent yield over 100% almost always signals a problem: wet product, contamination, unreacted starting material stuck to the product, or a balance that wasn’t zeroed correctly.

Q3. In real examples from research papers, do chemists always report percent yield?
In synthetic chemistry, yes, very often. Organic and inorganic synthesis papers usually list percent yield for each step. Other fields, like analytical chemistry or physical chemistry, may focus more on accuracy, precision, or signal intensity instead of yield.

Q4. How are industrial percent yield calculations different from classroom examples?
Industry often talks about conversion, selectivity, and overall yield across multiple steps. But the core math is the same as in the classroom examples of percent yield calculations: actual output divided by theoretical maximum, expressed as a percentage.

Q5. Where can I find more worked examples of percent yield calculations examples in chemistry?
Many university general chemistry sites and open courseware platforms publish worked problems. Look for resources from major universities or government-supported education sites, such as MIT OpenCourseWare or materials linked through USA.gov, to practice with more real examples and problem sets.

If you can follow the logic in these eight examples of percent yield calculations examples in chemistry, you’re in good shape. The numbers and chemicals will change, but the strategy — moles, stoichiometry, theoretical yield, percent yield — stays the same.