Can I compare masses instead of moles to find the limiting reactant?

No. Mole ratios come from the coefficients of the balanced equation, and different substances have very different molar masses, so a larger mass can easily contain fewer moles. Always convert every given amount to moles first, then apply the ratio. Comparing grams directly is the most frequent source of wrong answers on this topic.

Does the limiting reactant change if I use different amounts?

Yes. Being limiting is not a property of the substance; it depends entirely on the quantities you mix relative to the balanced equation. The same chemical can be limiting in one trial and in excess in the next. That is why the identification step must be redone for every new set of starting amounts.

Why is percent yield almost never 100 percent?

Product gets lost at every practical step — some sticks to glassware during transfers, some stays dissolved in the filtrate, some is lost to side reactions, and some reactions reach equilibrium before all the limiting reactant is used. Impure starting materials also reduce the product recovered. A yield in the seventies or eighties is common for a routine laboratory synthesis.

Which reactant do I use to calculate theoretical yield if the problem does not say?

Always the limiting reactant. If you are unsure which one it is, calculate the product amount from each reactant separately and keep the smaller result. That smaller number is the theoretical yield, and the reactant that produced it is the limiting reactant.

CHEM-6.5

Limiting Reactants & Percent Yield

Learn how to spot the limiting reactant, calculate theoretical yield, and find percent yield — with a full worked mass-mass problem and the mistakes to avoid.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Limiting Reactants & Percent Yield, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Limiting Reactants & Percent Yield free

What this lesson covers

Recipes have a built-in catch: if you have ten slices of bread but only three slices of cheese, you are making three sandwiches, no matter how much bread is left on the counter. Chemical reactions behave exactly the same way. Real reactions almost never start with perfectly matched amounts of reactants, so one of them runs out first and shuts the reaction down.

In this lesson you will learn to identify which reactant is limiting and which is in excess, use the limiting reactant to calculate the theoretical yield of product, and compare that prediction to the amount actually collected in the lab using percent yield. Every step builds on the mole ratios and mass-mass stoichiometry you already know — the new skill is deciding which starting amount to trust before you convert.

Why One Reactant Runs Out First

A balanced equation gives a fixed stoichiometric ratio between reactants. For

2\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3

every 2 mol of aluminum demands exactly 3 mol of chlorine. If the amounts you mix do not match that 2:3 ratio, one reactant will be consumed completely while some of the other is left sitting in the flask.

The reactant that runs out first is the limiting reactant. It controls how much product can possibly form, because once it is gone the reaction stops. Anything left over belongs to the excess reactant.

The single most common misconception is that the reactant with the smaller mass, or the smaller number of moles, must be limiting. Neither is reliable. Consider 2.0 mol of aluminum with 2.5 mol of chlorine: chlorine has more moles, yet 2.0 mol Al requires 3.0 mol of

\text{Cl}_2

, so chlorine still runs out first. Coefficients matter as much as amounts, and mass is even less trustworthy because different substances have different molar masses.

A related misconception is that the limiting reactant is somehow "more reactive" or "stronger." It is not a chemical property at all — it is purely a bookkeeping result of the quantities you happened to combine. Change the amounts you weigh out and the same substance can switch from limiting to excess.

Because of this, every limiting reactant problem must start the same way: convert all given amounts to moles. Comparisons made in grams are meaningless.

Two Reliable Methods for Finding the Limiting Reactant

Once both reactants are in moles, there are two standard approaches. Both give the same answer; pick one and use it consistently.

The comparison method picks one reactant and asks how much of the other it would need. If the amount needed exceeds what you have, the other reactant is limiting.

The product method converts each reactant separately into moles of the same product. Whichever reactant produces the smaller amount of product is limiting, and that smaller number is already your theoretical yield in moles.

Step	Comparison method	Product method
1	Convert both reactants to moles	Convert both reactants to moles
2	Use the mole ratio to find how much of reactant B is required by all of reactant A	Use mole ratios to convert each reactant into moles of product
3	Compare required to available	Compare the two product amounts
4	Not enough available means B is limiting	The smaller product amount identifies the limiting reactant
Bonus	Tells you directly how much excess is left	Gives theoretical yield immediately

The product method is usually faster on problems that ask for yield, since the answer falls out of the comparison. The comparison method is handier when the question asks how much excess reactant remains.

Where students go wrong: flipping the mole ratio. Write the ratio as a fraction with the unit you want on top and the unit you have on the bottom, and check that the unwanted unit cancels. If you convert 0.635 mol

\text{Cl}_2

using

\frac{3\ \text{mol Cl}_2}{2\ \text{mol AlCl}_3}

, the units do not cancel and the answer is wrong.

Theoretical Yield and Leftover Excess

The theoretical yield is the maximum amount of product the reaction could make, calculated from the limiting reactant alone. Once you know which reactant is limiting, the rest of the problem is ordinary mass-mass stoichiometry:

\text{g limiting} \rightarrow \text{mol limiting} \rightarrow \text{mol product} \rightarrow \text{g product}

Never run this chain on the excess reactant. Doing so predicts more product than the reaction can physically deliver, and it is the error that most often turns an otherwise correct solution into a wrong one. A quick self-check: if you calculated two different theoretical yields, one from each reactant, the correct theoretical yield is always the smaller one.

Many problems also ask how much excess reactant is left over. That takes two more steps. First find how much of the excess reactant was actually consumed, starting from the moles of limiting reactant and using the reactant-to-reactant mole ratio. Then subtract:

\text{excess remaining} = \text{excess initially present} - \text{excess consumed}

Be careful to subtract in consistent units — either both in moles or both in grams, never a mix.

One useful conceptual check: mass is conserved. The total mass of everything you started with equals the mass of product formed plus the mass of leftover excess reactant. If your numbers do not roughly balance, something upstream went wrong. In the aluminum and chlorine example that follows, 15.0 g plus 45.0 g of reactants must equal the grams of

\text{AlCl}_3

plus the grams of leftover aluminum.

Percent Yield: Comparing the Lab to the Math

Theoretical yield is a prediction. The actual yield is what you actually weigh after the reaction, the filtration, the drying, and the transfers. These rarely match, and percent yield measures how close they came:

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

Both quantities must be in the same units — grams over grams or moles over moles. Since it is a ratio, the units cancel and the answer is a pure percentage.

Actual yield is normally lower than theoretical yield. Reasons include product lost when transferring between containers, product left dissolved in solvent or stuck to filter paper, side reactions that consume reactants in unintended ways, reactions that reach equilibrium before finishing, and impure starting materials.

A percent yield above 100 percent is not a triumph — it is a signal of experimental error. The usual cause is a product that was not fully dried, so trapped water or solvent is being weighed along with the product. Impurities in the collected solid do the same thing. Since the theoretical yield is the absolute maximum permitted by conservation of atoms, exceeding it means the measurement, not the chemistry, is off.

Students often mix up which number goes on top. Remember that actual is what you got and theoretical is the target, so actual sits over theoretical, and a normal reaction gives a number below 100 percent. If you compute 118 percent for a routine synthesis, check whether you inverted the fraction before blaming wet crystals.

Putting the Steps in Order

A dependable sequence handles nearly every problem in this topic.

Start by balancing the equation — an unbalanced equation gives wrong mole ratios and poisons everything downstream. Convert each given reactant mass to moles using molar mass. Use mole ratios to determine which reactant is limiting. Convert the limiting reactant to moles of the requested product, then to grams if the question asks for mass. That result is the theoretical yield. Finally, if an actual yield is given, divide and multiply by 100.

Question asks for	What you need
Limiting reactant	Moles of each reactant plus the balanced mole ratio
Theoretical yield	Limiting reactant only, run through the full stoichiometry chain
Excess remaining	Excess initial minus excess consumed
Percent yield	Actual divided by theoretical, times 100
Actual yield from a percent	Theoretical times the percent as a decimal

That last row shows a variation worth practicing: given a theoretical yield of 56.4 g and a known 85.0 percent yield, the expected actual yield is

56.4 \times 0.850 = 47.9

g. Chemists use this constantly when planning how much starting material to buy for a synthesis.

Two habits prevent most errors. Label every number with its substance, not just its unit — writing "0.635 mol

\text{Cl}_2

" rather than "0.635 mol" makes a flipped ratio obvious. And carry extra digits through the middle of the calculation, rounding only at the end to the correct number of significant figures based on the given data.

Key terms

Limiting reactant.: The reactant that is completely consumed first, which sets the maximum amount of product the reaction can form.
Excess reactant.: A reactant present in more than the amount required by the mole ratio; some of it remains unreacted when the reaction stops.
Theoretical yield.: The maximum mass or moles of product predicted by stoichiometry, calculated from the limiting reactant only.
Actual yield.: The mass of product actually recovered and measured in the laboratory after the reaction and purification.
Percent yield.: The ratio of actual yield to theoretical yield times 100 percent, a measure of a reaction's practical efficiency.
Stoichiometric ratio.: The fixed whole-number ratio of moles among reactants and products given by the coefficients of a balanced equation.
Mole ratio.: A conversion factor built from two coefficients in a balanced equation, used to convert moles of one substance to moles of another.

Worked example

Aluminum reacts with chlorine gas to form aluminum chloride:

2\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3

. A student combines 15.0 g of Al with 45.0 g of

\text{Cl}_2

and recovers 48.2 g of

\text{AlCl}_3

. Identify the limiting reactant, find the theoretical yield, determine the mass of excess reactant remaining, and calculate the percent yield.

Step 1 — Convert both reactants to moles. Molar mass of Al is 26.98 g/mol and of

\text{Cl}_2

is 70.90 g/mol.

n_{\text{Al}} = \frac{15.0}{26.98} = 0.556\ \text{mol Al}

n_{\text{Cl}_2} = \frac{45.0}{70.90} = 0.635\ \text{mol Cl}_2

Step 2 — Identify the limiting reactant with the comparison method. All 0.556 mol of Al would require

0.556\ \text{mol Al} \times \frac{3\ \text{mol Cl}_2}{2\ \text{mol Al}} = 0.834\ \text{mol Cl}_2

Only 0.635 mol of

\text{Cl}_2

is available, which is less than the 0.834 mol required, so chlorine is the limiting reactant and aluminum is in excess. Notice that chlorine has both the larger mass and the larger mole count, yet it still runs out first — the coefficients decide.

Step 3 — Theoretical yield, using the limiting reactant only.

0.635\ \text{mol Cl}_2 \times \frac{2\ \text{mol AlCl}_3}{3\ \text{mol Cl}_2} = 0.423\ \text{mol AlCl}_3

Molar mass of

\text{AlCl}_3

26.98 + 3(35.45) = 133.33

g/mol.

0.423\ \text{mol} \times 133.33\ \text{g/mol} = 56.4\ \text{g AlCl}_3

Step 4 — Excess aluminum remaining. Aluminum consumed:

0.635\ \text{mol Cl}_2 \times \frac{2\ \text{mol Al}}{3\ \text{mol Cl}_2} = 0.423\ \text{mol Al}

0.423 \times 26.98 = 11.4\ \text{g Al used}

15.0 - 11.4 = 3.6\ \text{g Al left over}

Mass check:

56.4 + 3.6 = 60.0

g, which equals the 15.0 g plus 45.0 g started with. Mass is conserved.

Step 5 — Percent yield.

\%\ \text{yield} = \frac{48.2}{56.4} \times 100\% = 85.5\%

A yield in the mid-eighties is typical for a synthesis with transfer and purification losses.

Practice questions

For the reaction

2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}

, a flask contains 4.0 mol of

\text{H}_2

and 3.0 mol of

\text{O}_2

. Which statement is correct?

Oxygen is limiting because it has fewer moles
Hydrogen is limiting, and 1.0 mol of $\text{O}_2$ remains unreacted
Oxygen is limiting, and 2.0 mol of $\text{H}_2$ remains unreacted
Neither is limiting because the amounts are close to each other

Answer: Hydrogen is limiting, and 1.0 mol of $\text{O}_2$ remains unreacted

The ratio required is 2 mol

\text{H}_2

per 1 mol

\text{O}_2

. The 4.0 mol of

\text{H}_2

needs only

4.0 \times \frac{1}{2} = 2.0

mol of

\text{O}_2

, and 3.0 mol is available, so oxygen is in excess and hydrogen runs out first. Oxygen consumed is 2.0 mol, leaving

3.0 - 2.0 = 1.0

mol unreacted. Choosing the reactant with the smaller mole count is the trap here — coefficients must be applied before comparing.

Nitrogen and hydrogen react according to

\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3

. A chemist starts with 28.0 g of

\text{N}_2

and 9.00 g of

\text{H}_2

and isolates 27.5 g of

\text{NH}_3

. Determine the limiting reactant, the theoretical yield of ammonia, and the percent yield.

Answer: Nitrogen is limiting; theoretical yield is 34.1 g $\text{NH}_3$ ; percent yield is 80.6 percent.

Moles:

28.0 / 28.02 = 0.999

mol

\text{N}_2

and

9.00 / 2.016 = 4.46

mol

\text{H}_2

. All the nitrogen would need

0.999 \times 3 = 3.00

mol of

\text{H}_2

, and 4.46 mol is available, so hydrogen is in excess and nitrogen is limiting. Ammonia produced:

0.999 \times \frac{2}{1} = 2.00

mol, and

2.00 \times 17.03 = 34.1

g. Percent yield is

\frac{27.5}{34.1} \times 100 = 80.6\%

. Note that hydrogen has over four times as many moles as nitrogen and is still not limiting.

A student calculates a theoretical yield of 12.6 g of copper sulfate crystals but weighs 13.4 g of product, reporting a yield of 106 percent. Explain why this result cannot reflect the actual chemistry, and give the most likely experimental cause.

Answer: Conservation of atoms makes the theoretical yield an absolute maximum, so the extra mass must come from something other than product — most likely water or solvent that was not fully dried off the crystals.

Theoretical yield is calculated from the atoms supplied by the limiting reactant, and atoms cannot be created during a reaction. Recovering more than the theoretical amount therefore means the balance is weighing something besides pure product. The classic cause is incomplete drying, since hydrated or damp crystals carry extra water mass; other possibilities are unreacted starting material or impurities mixed into the solid. The fix is to dry the sample to a constant mass, weighing repeatedly until the reading stops changing.

FAQ

Can I compare masses instead of moles to find the limiting reactant?: No. Mole ratios come from the coefficients of the balanced equation, and different substances have very different molar masses, so a larger mass can easily contain fewer moles. Always convert every given amount to moles first, then apply the ratio. Comparing grams directly is the most frequent source of wrong answers on this topic.
Does the limiting reactant change if I use different amounts?: Yes. Being limiting is not a property of the substance; it depends entirely on the quantities you mix relative to the balanced equation. The same chemical can be limiting in one trial and in excess in the next. That is why the identification step must be redone for every new set of starting amounts.
Why is percent yield almost never 100 percent?: Product gets lost at every practical step — some sticks to glassware during transfers, some stays dissolved in the filtrate, some is lost to side reactions, and some reactions reach equilibrium before all the limiting reactant is used. Impure starting materials also reduce the product recovered. A yield in the seventies or eighties is common for a routine laboratory synthesis.
Which reactant do I use to calculate theoretical yield if the problem does not say?: Always the limiting reactant. If you are unsure which one it is, calculate the product amount from each reactant separately and keep the smaller result. That smaller number is the theoretical yield, and the reactant that produced it is the limiting reactant.

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