Worked examples — Percent yield, theoretical yield, actual yield

Before anything, one promise: we never compare grams to grams through coefficients. Coefficients count particles, and the only unit for counting particles is the mole. Every example obeys that. If a word here is new, Molar Mass (the grams-per-mole conversion) and Balancing Chemical Equations (where coefficients come from) are your prerequisites.

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The scenario matrix

Every problem this topic can hand you falls into exactly one of these case classes. The table lists them and points to the example that nails each one.

Cell	Case class	What makes it tricky	Example
A	Single reactant, excess of the other	Nothing hidden — clean chain	Ex 1
B	Two reactants, one limiting	Must test both, pick smaller product	Ex 2
C	Back-solve: given % yield, find reactant needed	Rearrange the formula	Ex 3
D	Reported yield > 100%	Impossible — diagnose impurity	Ex 4
E	Degenerate / zero input	Limiting reagent = 0 ⇒ theoretical = 0	Ex 5
F	Real-world word problem	Extract numbers from prose, watch purity	Ex 6
G	Exam twist: two products, asked for one	Ratio to the right product only	Ex 7
H	Limiting = exact stoichiometric match	Boundary: both give same product	Ex 8

Molar masses used throughout (g/mol): H = 1.0, C = 12.0, N = 14.0, O = 16.0, Na = 23.0, Cl = 35.5, Ca = 40.0. So $M({\text{CH}}_4)=16.0$, $M({\text{CO}}_2)=44.0$, $M({\text{NH}}_3)=17.0$, $M({\text{N}}_2)=28.0$, $M({\text{H}}_2)=2.0$, $M({\text{H}}_2\text{O})=18.0$, $M({\text{CaCO}}_3)=100.0$, $M(\text{CaO})=56.0$, $M(\text{NaCl})=58.5$, $M({\text{AgNO}}_3)=170.0$, $M(\text{AgCl})=143.5$.

The bar chart above is the single mental picture for every two-reactant case: draw one bar per reactant showing how much product it could make; the shorter bar wins and its height is the theoretical yield. The taller bar is just leftovers.

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Cell A — single reactant, other in excess

Forecast: Excess oxygen means methane alone decides everything. Guess: will percent yield be near, above, or below 100%? Jot a number.

1. Moles of ${\text{CH}}_4$: $n=\frac{6.4}{16.0}=0.40\text{mol}$. Why this step? Grams cannot enter a coefficient ratio. We must switch to the mole currency first.
2. Mole ratio ${\text{CH}}_4:{\text{CO}}_2=1:1$, so theoretical ${\text{CO}}_2=0.40\text{mol}$. Why this step? Each carbon atom in becomes one carbon atom out — conservation of atoms, expressed by the coefficients.
3. Back to grams: $0.40\times 44.0=17.6\text{g}$ theoretical. Why this step? The balance in the lab reads grams, so the theoretical must be in grams to be comparable.
4. Percent yield: $\frac{15.4}{17.6}\times 100=87.5\%$. Why this step? Yield = (what you got) ÷ (best possible), same unit (g), ×100.

Verify: Units: $\text{g}/\text{g}$ cancels → pure number, good. Sanity: $15.4<17.6$ so the answer is below 100%, as any real reaction must be. ✓

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Cell B — two reactants, one limiting

Forecast: Which runs out first — nitrogen or hydrogen? Guess before computing; the recipe needs 3 H₂ per N₂.

1. Moles: $n({\text{N}}_2)=\frac{14.0}{28.0}=0.50\text{mol}$; $n({\text{H}}_2)=\frac{6.0}{2.0}=3.0\text{mol}$. Why this step? Both reactants must be in moles to feed the counting recipe.
2. Product each could make (the two bars):
   • From ${\text{N}}_2$: $0.50\times \frac{2}{1}=1.0\text{mol}{\text{NH}}_3$.
   • From ${\text{H}}_2$: $3.0\times \frac{2}{3}=2.0\text{mol}{\text{NH}}_3$. Why this step? You can only make as much product as your scarcest ingredient permits. Test both, never assume.
3. Limiting reagent = ${\text{N}}_2$ since $1.0<2.0$. Theoretical $=1.0\times 17.0=17.0\text{g}$. Why this step? The shorter bar sets the ceiling; leftover ${\text{H}}_2$ just sits there. See Limiting Reagent.
4. Percent yield: $\frac{12.0}{17.0}\times 100=70.6\%$.

Verify: Leftover check: ${\text{N}}_2$ used = 0.50 mol needs $0.50\times 3=1.5\text{mol}{\text{H}}_2$; we had 3.0 mol, so 1.5 mol H₂ left over — consistent with H₂ being the excess. ✓ And $12.0<17.0$ ⇒ below 100%. ✓

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Cell C — back-solve from a known percent yield

Forecast: Since you lose 25%, must the theoretical be more or fewer moles than what 27.0 g represents? Guess the direction.

1. Theoretical mass: $\frac{\text{actual}}{\text{yield fraction}}=\frac{27.0}{0.750}=36.0\text{g}$. Why this step? Rearrange $\text{actual}=\text{theoretical}\times \text{fraction}$ to $\text{theoretical}=\text{actual}÷\text{fraction}$.
2. Convert to moles: $n=\frac{36.0}{90.0}=0.400\text{mol}$. Why this step? Reactant amounts are planned in moles via the Mole Concept, so the target must be in moles too.

Verify: Push it forward: $0.400\text{mol}\times 90.0=36.0\text{g}$ theoretical; $36.0\times 0.750=27.0\text{g}$ actual — matches the demand exactly. ✓ Direction check: theoretical (36.0 g) > actual (27.0 g), correct because we over-plan to cover the loss. ✓

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Cell D — a reported yield above 100%

Forecast: Theoretical CaO from 10.0 g CaCO₃ — will 6.5 g be below or above it? Guess.

1. Moles of ${\text{CaCO}}_3$: $n=\frac{10.0}{100.0}=0.100\text{mol}$. Why this step? Same rule — moles before ratio.
2. Ratio ${\text{CaCO}}_3:\text{CaO}=1:1$, theoretical $\text{CaO}=0.100\times 56.0=5.6\text{g}$. Why this step? One formula unit decomposes into exactly one CaO.
3. Apparent percent yield: $\frac{6.5}{5.6}\times 100=116.1\%$. Why this step? Blindly applying the formula gives an impossible number — that is the diagnostic signal.
4. Diagnose: $>100\%$ means you cannot have created atoms. The residue still contains undecomposed ${\text{CaCO}}_3$ (heating was incomplete), inflating the mass. Reheat to constant mass, then reweigh. Why this step? Conservation of Mass forbids >100%; excess mass is always impurity/unreacted material.

Verify: $6.5>5.6$, so the ratio exceeds 1 and ×100 exceeds 100 — arithmetically forced. The physics says impossible, so the data (not the formula) is at fault. ✓

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Cell E — degenerate / zero input

Forecast: Lots of hydrogen — surely some ammonia? Guess a number first.

1. Moles: $n({\text{H}}_2)=\frac{10.0}{2.0}=5.0\text{mol}$; $n({\text{N}}_2)=\frac{0}{28.0}=0\text{mol}$. Why this step? Zero grams of a reactant is a legal input; it becomes zero moles.
2. Product each could make:
   • From ${\text{H}}_2$: $5.0\times \frac{2}{3}=3.33\text{mol}$.
   • From ${\text{N}}_2$: $0\times \frac{2}{1}=0\text{mol}$. Why this step? Test both, exactly as always — a zero is not a special case, it just wins the "smallest" contest.
3. Limiting reagent = ${\text{N}}_2$ (the smaller, 0 < 3.33). Theoretical $=0\times 17.0=0\text{g}$. Why this step? With none of an essential reactant, the ceiling is zero. No amount of H₂ helps.
4. Percent yield: actual $0\text{g}$ over theoretical $0\text{g}$ is undefined ($0/0$). The correct statement is "no reaction — theoretical yield is zero," not a percentage. Why this step? Dividing by zero is not a number; recognise the degenerate case in words.

Verify: Sanity: a reaction needing an absent reactant produces nothing; theoretical $=0$. Percent yield formula requires theoretical $>0$ to be defined — flagging this is the whole point of the cell. ✓

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Cell F — real-world word problem

Forecast: Excess NaCl ⇒ silver nitrate is limiting. Guess whether the wet-chemistry loss will land the yield around 90%, 80%, or 70%.

1. Extract & convert: limiting reactant is ${\text{AgNO}}_3$ (NaCl is excess). $n({\text{AgNO}}_3)=\frac{8.50}{170.0}=0.0500\text{mol}$. Why this step? The word "excess" tells you which reactant to base the calculation on — read the prose carefully.
2. Ratio ${\text{AgNO}}_3:\text{AgCl}=1:1$, theoretical $\text{AgCl}=0.0500\times 143.5=7.175\text{g}$. Why this step? Each silver ion ends up in one AgCl unit.
3. Percent yield: $\frac{6.60}{7.175}\times 100=91.99\%$. Why this step? Actual (filtered, dried) ÷ theoretical ×100. Losses come from AgCl staying in solution or on the filter paper.

Verify: $6.60<7.175$ ⇒ below 100% ✓. Rounded: $91.99\%\approx 92.0\%$, a believable gravimetric yield. Units $\text{g}/\text{g}$ cancel ✓.

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Cell G — exam twist: two products, asked for one

Forecast: The water coefficient is 2, not 1 — will you make more moles of water than of methane? Guess.

1. Moles of ${\text{CH}}_4$: $n=\frac{4.0}{16.0}=0.25\text{mol}$. Why this step? Moles first, always.
2. Ratio ${\text{CH}}_4:{\text{H}}_2\text{O}=1:2$, theoretical ${\text{H}}_2\text{O}=0.25\times \frac{2}{1}=0.50\text{mol}$. Why this step? The twist: you must read the coefficient of the requested product (water = 2), not reuse the CO₂ ratio. Using 1:1 here is the classic exam trap.
3. Back to grams: $0.50\times 18.0=9.0\text{g}$ theoretical water. Why this step? Convert with water's molar mass, not methane's.
4. Percent yield: $\frac{6.3}{9.0}\times 100=70.0\%$.

Verify: Atom count: 4 H in one CH₄ → 4 H in two H₂O, so mole of water = 2× mole of methane; $0.50=2\times 0.25$ ✓. $6.3<9.0$ ⇒ below 100% ✓.

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Cell H — boundary case: exact stoichiometric match

Forecast: With the perfect recipe ratio, is there a limiting reagent at all? Guess yes/no.

1. Moles: $n({\text{N}}_2)=\frac{28.0}{28.0}=1.0\text{mol}$; $n({\text{H}}_2)=\frac{6.0}{2.0}=3.0\text{mol}$. Why this step? Moles first.
2. Product each could make:
   • From ${\text{N}}_2$: $1.0\times \frac{2}{1}=2.0\text{mol}$.
   • From ${\text{H}}_2$: $3.0\times \frac{2}{3}=2.0\text{mol}$. Why this step? Both bars are the same height — this is the knife-edge where neither is in excess.
3. Verdict: stoichiometrically balanced; both run out together. Theoretical ${\text{NH}}_3=2.0\times 17.0=34.0\text{g}$, and nothing is left over. Why this step? At exact match there is no "leftover" reactant — the boundary between Cell B's two possibilities.

Verify: Ratio check: $n({\text{H}}_2):n({\text{N}}_2)=3.0:1.0=3:1$, exactly the coefficient ratio, so neither is limiting alone. Both predictions equal 2.0 mol ✓.

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Recall Which cell am I in? (quick decision)

One reactant given / others "excess"? ::: Cell A — go straight down the chain. Two amounts given, no "excess" stated? ::: Cell B or H — test both reactants. Given a % yield, asked for reactant/mass? ::: Cell C — rearrange, divide by the fraction. Computed yield came out above 100%? ::: Cell D — impurity/incomplete; the data is wrong, not the formula. A reactant is zero or absent? ::: Cell E — theoretical is 0, percent yield undefined. Asked about a product whose coefficient isn't 1? ::: Cell G — use that product's coefficient and molar mass.

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Connections

• Limiting Reagent — Cells B, E, H live or die on identifying it.
• Mole Concept — the currency that makes every ratio legal.
• Balancing Chemical Equations — supplies the coefficients used in every ratio step.
• Molar Mass — the ÷M and ×M ends of the chain.
• Conservation of Mass — the law that forbids Cell D's >100%.
• Empirical & Molecular Formula — reuses the same mass–mole reasoning.