Percent yield, theoretical yield, actual yield: Visual walkthrough

We will follow one reaction the whole way through so the pictures connect: $N_{2} + 3 H_{2} \to 2 NH_{3}$ This is the making of ammonia. Read it as a recipe: take 1 box of nitrogen molecules, 3 boxes of hydrogen molecules, get 2 boxes of ammonia molecules.

Step 1 — Atoms are counted, not weighed

WHAT. Before any formula, look at what a chemical equation is actually saying. The number in front of a formula — the coefficient — counts whole molecules, like "2 eggs, 3 cups". It never counts grams.

WHY. A balance in the lab reads grams, but nature does its bookkeeping in counts of particles. One nitrogen molecule is heavier than one hydrogen molecule, so "equal grams" does not mean "equal numbers". If we compare reactants by weight we compare the wrong thing. We must compare by how many.

PICTURE. Look at the figure: the same recipe drawn as boxes of molecules. One $N_{2}$ box plus three $H_{2}$ boxes rearrange into two $NH_{3}$ boxes. Count the little balls: nitrogen atoms in $=$ nitrogen atoms out, hydrogen in $=$ hydrogen out. Nothing is created or destroyed — that is Conservation of Mass shown as a picture.

Figure — Percent yield, theoretical yield, actual yield

Step 2 — The mole: turning "count" into something we can weigh

WHAT. We cannot count individual molecules — there are far too many. So we bundle them into a standard pack called the mole. One mole is a fixed huge number of molecules (Avogadro's number). Think of it as "one dozen", just enormously bigger.

WHY. We need a bridge between the world of counting (where the recipe lives) and the world of weighing (where the balance lives). The mole is that bridge: a mole of any substance is a countable amount that also has a definite mass. This is the Mole Concept.

PICTURE. The figure shows two side-by-side scales. On the left, a jar labelled "1 mole of $N_{2}$ " tips the scale to $28$ g. On the right, "1 mole of $H_{2}$ " tips it to only $2$ g. Same count of molecules, very different weights — that gap is exactly why we can't compare grams directly.

Step 3 — Weigh in, convert to moles

WHAT. We start our real problem. Put $28$ g of $N_{2}$ and $9.0$ g of $H_{2}$ into the reactor. Convert each mass to moles using $n = m / M$ .

$n (N_{2}) = \frac{28}{28} = 1.0 mol n (H_{2}) = \frac{9.0}{2.0} = 4.5 mol$

$28$ and $9.0$ — masses we chose to weigh out, in grams.
$28$ and $2.0$ (the denominators) — molar masses of $N_{2}$ and $H_{2}$ .
$1.0$ and $4.5$ — the counts we can now feed into the recipe.

WHY. Step 1 warned us: the recipe speaks in counts. So the very first move is to translate our grams into moles. Only now are the two reactants speaking the same language.

PICTURE. The figure shows two funnels. Grams pour in the top, pass through a filter labelled " $\div M$ ", and moles drip out the bottom. Notice $9.0$ g of the light hydrogen produces more moles ( $4.5$ ) than $28$ g of the heavy nitrogen ( $1.0$ ) — light molecules pack more count per gram.

Step 4 — Ask each reactant: "how much product could YOU make?"

WHAT. Use the recipe's mole ratio on each reactant separately, pretending it alone controls the reaction.

From $N_{2}$ : the recipe says $1 N_{2} \to 2 NH_{3}$ , ratio $\frac{2}{1}$ . $1.0 \times \frac{2}{1} = 2.0 mol NH_{3}$
From $H_{2}$ : the recipe says $3 H_{2} \to 2 NH_{3}$ , ratio $\frac{2}{3}$ . $4.5 \times \frac{2}{3} = 3.0 mol NH_{3}$

In each product the fraction is $\frac{coefficient of product}{coefficient of reactant}$ — that is exactly the counting recipe from Step 1, used as a multiplier.

WHY. Why compute both? Because we don't yet know which ingredient runs out first. We ask each one "if you were in charge, how many ammonia packs?" and let the answers reveal the bottleneck.

PICTURE. Two parallel arrows, each carrying its reactant's moles through a gear labelled with the ratio, landing on a target that reads how much $NH_{3}$ it demands: $2.0$ from nitrogen, $3.0$ from hydrogen.

Step 5 — The smaller answer wins: the limiting reagent

WHAT. Compare the two answers: $2.0$ mol (from $N_{2}$ ) versus $3.0$ mol (from $H_{2}$ ). The smaller one, $2.0$ mol, is what actually forms. Nitrogen is the limiting reagent.

WHY. You can only build as many finished packs as your scarcest ingredient allows. Hydrogen wants to make $3.0$ mol, but it can't — it needs partner nitrogen, and nitrogen quits after $2.0$ mol. So nitrogen sets the ceiling. This is the Limiting Reagent rule: least product wins.

PICTURE. A see-saw. On one side " $N_{2}$ demands $2.0$ ", on the other " $H_{2}$ demands $3.0$ ". The see-saw tips to the lower number, and a ceiling line is drawn at $2.0$ mol — everything above it (the leftover hydrogen) is greyed out as "excess, just sits there".

Step 6 — Moles back to grams: the theoretical yield

WHAT. We found $2.0$ mol of $NH_{3}$ can form. Convert this count back to a weight, because the balance reads grams.

$m = n \times M = 2.0 \times 17.0 = 34.0 g$

$2.0$ — moles of product the limiting reagent allows (from Step 5).
$17.0$ — molar mass of $NH_{3}$ (weight of one pack).
$34.0$ g — the theoretical yield: the perfect, no-losses maximum.

WHY. We entered the mole-world to use the recipe; now we must leave it, because reality is measured on a scale in grams. The round trip is: grams $\to$ moles $\to$ (recipe) $\to$ moles $\to$ grams.

PICTURE. The full conversion chain drawn as one pipeline: mass in, $\div M$ to moles, $\times$ ratio to product moles, $\times M$ back to product mass. The $34.0$ g glows at the end as the "dream maximum".

Step 7 — Reality vs. the dream: percent yield

WHAT. In the lab you actually scoop out and weigh $25$ g of ammonia — the actual yield. Compare it to the $34.0$ g dream:

$Percent yield = \frac{actual}{theoretical} \times 100% = \frac{25}{34.0} \times 100% = 73.5%$

$25$ — actual yield, measured on the balance.
$34.0$ — theoretical yield, from Step 6.
Both in grams, so the units cancel and leave a pure fraction; $\times 100$ makes it a percent.

WHY. A yield is "how much of the best possible did we get". Best possible is the denominator; what we got is the numerator. Same unit top and bottom is essential — otherwise the ratio is meaningless.

PICTURE. A vertical bar shows the full $34.0$ g "theoretical" height in cyan; a shorter amber bar fills it up to $25$ g. The filled fraction, $73.5%$ , is written across it. The empty top slice is labelled "lost: stuck to walls, unreacted, escaped".

Step 8 — The degenerate & edge cases

WHAT & WHY. Never leave a scenario undrawn. Three boundary situations:

Actual $=$ theoretical (100%). The amber bar fills the whole cyan bar. Perfect, lossless — almost never happens in reality, but it is the ceiling.
Actual $>$ theoretical ( $> 100%$ ). The amber bar overshoots the top. This is impossible for pure product (you can't create atoms — Conservation of Mass). It flags impurity: trapped water or solvent adding fake mass. Fix: dry and reweigh.
Actual $= 0$ (0%). No product recovered. Stoichiometrically this means the limiting reagent produced zero moles of product: either none of it reacted (wrong conditions — no catalyst, too cold, no activation), or every particle formed was lost/decomposed before weighing. The recipe still holds — the mole ratio would give a positive ceiling — but the reacted amount of limiting reagent was zero, so the realised product is zero. The formula stays valid: $0/34.0 = 0%$ .

PICTURE. Three bars side by side — exactly full (green, 100%), overflowing (amber warning, "impurity!"), and empty (0%) — so you can recognise each at a glance.

Recall Reveal: what does >100% really mean?

A yield above 100% ::: is physically impossible for pure product; it signals extra mass from impurity, moisture, or unreacted leftovers — purify and reweigh.

The one-picture summary

Here is the entire walkthrough in a single diagram: grams of each reactant flow up into moles, each mole-stream is tested against the recipe, the smaller demand sets the ceiling (limiting reagent), that ceiling becomes theoretical grams, and the measured actual grams is divided into it for the final percent.

Recall Feynman retelling — the whole story in plain words

You've got a pile of nitrogen and a pile of hydrogen and you want to make ammonia. First, you can't compare the piles by weight, because nitrogen molecules are heavy and hydrogen ones are light — so you count them instead, in packs called moles. You weigh each pile and divide by its pack-weight to get the number of packs. Now you use the recipe ("1 nitrogen box + 3 hydrogen boxes → 2 ammonia boxes") to ask each pile, "how many ammonia boxes could you make?" Whichever pile answers with the smaller number is the boss — it runs out first and caps everything. That cap, converted back to grams, is your dream amount: the theoretical yield. Then you actually run the reaction, scrape the ammonia off the walls, weigh it — that's the actual yield. Divide actual by dream, times 100, and you've got your score. If your score comes out over 100, you didn't do magic — your product is dirty, so dry it and weigh again. And if it comes out zero, nothing reacted (or nothing survived to the balance).

Connections

Limiting Reagent — the see-saw of Step 5.
Mole Concept — the pack that bridges count and weight.
Balancing Chemical Equations — supplies the ratio gears of Step 4.
Molar Mass — the $\div M$ and $\times M$ conversions.
Conservation of Mass — why $> 100%$ is impossible.
Empirical & Molecular Formula — same mass–mole reasoning reused.
Hinglish version →