1.3.4 · D5Chemical Reactions & Stoichiometry
Question bank — Percent yield, theoretical yield, actual yield
Prerequisites worth having fresh: Limiting Reagent, Mole Concept, Balancing Chemical Equations, Molar Mass, and Conservation of Mass.
Look at the figure: the tall violet bar is the theoretical yield — the ceiling nailed in place by the limiting reagent. The shorter magenta bar is the actual yield you scoop off the balance. Percent yield is just "how tall is magenta compared with violet." A dashed line marks the roof; any bar that appears to poke through it (the faint orange sliver) is not extra product — it is impurity puffing up the mass.
True or false — justify
Adding more of the excess reactant will raise the theoretical yield.
False — theoretical yield is fixed by the limiting reagent alone; excess simply sits there as leftover and never becomes extra product.
A reaction that goes "100% to completion" always gives 100% percent yield.
False — completion means the limiting reagent is fully consumed, but product can still be lost to spills, filtration, or evaporation, so actual yield (hence percent yield) can be below 100%.
If you double both reactants (keeping their ratio), the percent yield stays the same.
True — theoretical yield doubles and, all else equal, actual yield roughly doubles too, so the ratio is unchanged; percent yield is a fraction, not an absolute amount.
Percent yield tells you how pure your product is.
False — it tells you how much of the maximum amount you recovered; purity is a separate question, and impurity actually inflates the measured mass (see the >100% trap).
Theoretical yield can be measured on a balance.
False — theoretical yield is a paper prediction from stoichiometry; only actual yield is weighed on a balance.
If two reactants are present in exactly their stoichiometric ratio, there is no limiting reagent.
Half-true — both run out at the same instant, so either can be used to compute theoretical yield; they give identical answers, so no separate "limiting" test is needed.
A higher percent yield always means a "better" reaction.
Not necessarily — a fast, cheap reaction at may beat a slow, expensive one at ; percent yield measures completeness of this run, not overall usefulness.
Percent yield can be computed using masses without ever converting to moles.
True if actual and theoretical are already both masses of the same product — the final ratio is unit-free. But the theoretical mass itself required a mole conversion first.
Spot the error
"I have 5 mol H₂ and 5 mol N₂ for . Equal moles, so neither is limiting."
Error — you compare product each could make, not raw moles. H₂ makes mol NH₃; N₂ makes mol; H₂ gives less, so H₂ limits.
"My percent yield came out to , so my reaction over-performed."
Error — atoms cannot be created (Conservation of Mass); a value over means the weighed product still held solvent, water, or unreacted material. Dry and reweigh.
"I divided grams of product by grams of reactant using the coefficients directly."
Error — coefficients count particles (moles), not grams. Convert to moles before the ratio and back to grams after.
"Percent yield = theoretical ÷ actual × 100."
Error — the ratio is actual on top: . Flipping it makes a good reaction look over .
"I used the excess reactant's moles to find theoretical yield because I had more of it."
Error — the reactant you have more of is exactly the one that does not cap the product; use the one that runs out first.
"Actual yield was 0.40 mol and I needed a yield, so theoretical is ."
Error — you divide by the fraction: theoretical . You need to aim higher than what you get, not lower.
"The reaction released gas that escaped, so my percent yield should be over ."
Error — escaping product lowers actual yield (you collected less than formed), pushing percent yield down, not up.
Why questions
Why must the two yields be in the same unit before taking the ratio?
So the units cancel and the ratio is a pure number; a "fraction of the maximum" is only meaningful when numerator and denominator measure the same thing.
Why do we insist on moles (not grams) when applying the coefficient ratio?
Balanced-equation coefficients count particles; grams of different substances weigh differently per particle, so only the mole is a fair currency for the recipe (Mole Concept).
Why does the smallest predicted product identify the limiting reagent?
Once the scarcest ingredient is used up the reaction stops, so it physically caps how much product can form — no other reactant can push past that ceiling.
Why can percent yield never legitimately exceed ?
Because theoretical yield already assumes perfect, complete conversion; nothing can beat perfection without extra mass sneaking in as impurity.
Why is theoretical yield unaffected by how sloppy your lab technique is?
It is a prediction of the ideal maximum from stoichiometry alone; technique affects only what you actually recover, i.e. the actual yield.
Why do chemists still bother computing theoretical yield if reality always falls short?
It is the benchmark — without the ideal maximum you cannot say whether a real yield of 25 g is great or terrible; percent yield needs that reference.
Why does purifying and re-drying a product sometimes lower your reported percent yield?
Drying removes trapped solvent/water that had been inflating the mass, revealing the true (smaller) amount of actual product.
Edge cases
What is the theoretical yield if the limiting reagent is present in zero amount?
Zero — no limiting reagent means no reaction can proceed for that product, so theoretical yield is 0 g and percent yield is undefined (division by zero).
If actual yield is measured as exactly the theoretical yield, what is percent yield and is it realistic?
— mathematically valid but physically rare; it usually flags either an idealized problem or an undetected impurity making up the difference.
A side reaction consumes some of your limiting reagent into a different product. What happens to percent yield of your target product?
It drops — less limiting reagent reaches the desired pathway, so actual yield of the target falls below the theoretical value (which assumed 100% went to that product).
Two different products form from one reaction. Can each have its own percent yield?
Yes — each product has its own theoretical yield (from the mole ratio) and its own measured actual yield, so each gets an independent percent yield.
The limiting reagent is 100% consumed but you recover no product at all. Percent yield?
— completion of the reagent does not guarantee recovery; total loss during workup gives actual yield 0, so .
If a reaction is reversible and reaches equilibrium before completion, what does that do to the ceiling?
It lowers the achievable actual yield below the theoretical maximum, since not all limiting reagent converts; percent yield reflects this gap even with perfect technique.
Connections
- Limiting Reagent — the source of most "spot the error" traps.
- Conservation of Mass — why a value over is impossible.
- Mole Concept — why grams-in-the-ratio is always wrong.
- Molar Mass — the gram↔mole bridge these questions lean on.
- Balancing Chemical Equations — coefficients that set every ratio.