Limiting Reactants & Percent Yield
The Reality of the Lab: When Ingredients Run Out.
This is the final chapter of Stoichiometry. You'll need the skills from the previous lesson.
← Go back to Stoichiometric Calculations.
The Problem: The Sandwich Shop Dilemma
Imagine you're making sandwiches. Your recipe is simple: 2 slices of bread + 1 slice of cheese = 1 sandwich. You look in your pantry and find you have 10 slices of bread but only 3 slices of cheese.
How many sandwiches can you make? You have enough bread for 5 sandwiches, but only enough cheese for 3. The cheese is your limiting reactant. It limits the total number of sandwiches you can produce. You'll make 3 sandwiches and have 4 slices of bread left over (the excess reactant).
Chemistry works the exact same way. In a real lab, you rarely have the perfect stoichiometric amounts of your reactants. One will always run out first, limiting how much product you can make.
Part 1: Identifying the Limiting Reactant
Remember our sandwich shop? To figure out the limit, we had to check both ingredients separately:
- How many sandwiches can we make with 10 slices of bread? 5 sandwiches.
- How many sandwiches can we make with 3 slices of cheese? 3 sandwiches.
We then compared the two possible outcomes and chose the smaller number. That told us our true limit was 3 sandwiches, and cheese was the limiting ingredient.
The Chemical Method: A Race to the Finish Line
We do the exact same thing in chemistry. We'll perform two separate calculations—a "race" for each reactant to see how much product it can make on its own. The reactant that produces the least amount of product is the one that runs out first. It's our limiting reactant.
Let's walk through an example for the reaction 2H₂ + O₂ → 2H₂O. Imagine we start with 4 grams of H₂ and 40 grams of O₂.
Race 1: How much water can 4g of H₂ make?
Using stoichiometry (grams → moles → moles → grams), we find that 4g of H₂ can produce about 35.7 grams of H₂O.
Calculation Breakdown:
- Grams to Moles (H₂): First, convert the starting mass of H₂ to moles using its molar mass (~2.016 g/mol).
4.0 g H₂ ÷ 2.016 g/mol = 1.984 mol H₂ - Mole Bridge (H₂ to H₂O): Use the mole ratio from the balanced equation (2H₂ → 2H₂O) to find the moles of H₂O produced.
1.984 mol H₂ × (2 mol H₂O / 2 mol H₂) = 1.984 mol H₂O - Moles to Grams (H₂O): Finally, convert the moles of H₂O to grams using its molar mass (~18.015 g/mol).
1.984 mol H₂O × 18.015 g/mol = 35.74 g H₂O
Race 2: How much water can 40g of O₂ make?
Using stoichiometry again for the other reactant, we find that 40g of O₂ can produce about 45.0 grams of H₂O.
Calculation Breakdown:
- Grams to Moles (O₂): Convert the starting mass of O₂ to moles using its molar mass (~31.998 g/mol).
40.0 g O₂ ÷ 31.998 g/mol = 1.250 mol O₂ - Mole Bridge (O₂ to H₂O): Use the mole ratio from the balanced equation (1O₂ → 2H₂O) to find the moles of H₂O produced.
1.250 mol O₂ × (2 mol H₂O / 1 mol O₂) = 2.500 mol H₂O - Moles to Grams (H₂O): Convert the moles of H₂O to grams using its molar mass (~18.015 g/mol).
2.500 mol H₂O × 18.015 g/mol = 45.04 g H₂O
The Verdict:
Since H₂ produces a smaller amount of water (35.7g is less than 45.0g), we know that H₂ is the limiting reactant. We will run out of hydrogen long before we run out of oxygen.
The smaller of the two results, 35.7g H₂O, is our Theoretical Yield. It's the absolute maximum amount of product we can possibly make.
Now, use the interactive calculator below to prove this for yourself! Change the input values and watch how the "limiting" and "excess" labels change based on which "race" produces the smaller result.
H₂
O₂
H₂O
What About the Leftovers? Calculating Excess Reactant
Once you know the limiting reactant, you can figure out exactly how much of the excess reactant is left over. The process is a simple "Start - Used = Left" calculation.
- Start with the Theoretical Yield: Use your true theoretical yield (the smaller amount calculated in Part 1) to work backwards.
- Calculate Amount Used: Perform a stoichiometry calculation to determine how many grams of the excess reactant were *actually needed* to produce the theoretical yield.
- Subtract to Find the Remainder: Subtract the amount of excess reactant used from the initial amount you started with.
Example: How much O₂ is left?
From our example, we know H₂ is limiting and produces 35.74g of H₂O. How much O₂ did we use to make that much water?
35.74g H₂O → moles H₂O → moles O₂ → grams O₂
This calculation shows that 31.73g of O₂ were used. Since we started with 40g of O₂, the amount left over is:
40.00 g (start) - 31.73 g (used) = 8.27 g O₂ left over
Part 2: Reality Check (Percent Yield)
Your theoretical yield is a perfect-world calculation. But in a real lab, things happen: some product might stick to the glassware, side reactions might occur, or the reaction may not go to completion. The amount you *actually* produce in the lab is called the Actual Yield.
Percent Yield is a measure of your experiment's efficiency. It's a simple ratio that tells you how close you got to the perfect result.
Why is Percent Yield Rarely 100%?
In practice, achieving a 100% yield is nearly impossible due to:
- Experimental Error: Spills, incomplete transfers of solids, or measurement inaccuracies.
- Side Reactions: Reactants may form unintended "side products" that aren't what you wanted to make.
- Equilibrium: Many reactions are reversible, meaning they don't fully proceed in one direction.
A "good" yield can vary dramatically. For a simple precipitation, a chemist might expect over 90%. For a complex, multi-step organic synthesis, a yield of 40% might be considered excellent!
Percent Yield Calculator
Congratulations! You've Mastered Stoichiometry!
You now have all the core skills for chemical arithmetic. The next logical step is to explore the world of solutions and concentrations.