Percent yield, theoretical yield, actual yield: Exercises

Level 1 — Recognition

Goal: point at the right quantity and plug into the formula. No multi-step chains yet.

A student calculates that a reaction should make $50.0$ g of aspirin, but the balance reads $42.0$ g of dry product. Which number is the actual yield, which is the theoretical yield, and what is the percent yield?

Recall Solution 1.1

Theoretical yield = the paper prediction = $50.0$ g (what the maths says could form).
Actual yield = the balance reading = $42.0$ g (what really came out).
Percent yield: $\frac{42.0}{50.0} \times 100% = 84.0%$ Same unit (g) top and bottom, so grams cancel and the answer is a pure percentage.

Exercise 1.2

A reaction ran at 75.0% yield and its theoretical yield was $120$ g. What actual mass of product came out?

Recall Solution 1.2

Rearrange the percent-yield formula for the actual yield: $actual = theoretical \times \frac{percent}{100} = 120 \times 0.750 = 90.0 g$ WHY: percent yield is the fraction of the dream you achieved, so multiply the dream by that fraction.

Level 2 — Application

Goal: run the full mass → mole → mole → mass chain once, then take a ratio.

Exercise 2.1

Decompose $10.0$ g of calcium carbonate: $CaCO_{3} \to CaO + CO_{2}$ . You collect $4.8$ g of $CaO$ . Find the percent yield. ( $M (CaCO_{3}) = 100.0$ , $M (CaO) = 56.0$ g/mol.)

Recall Solution 2.1

Step 1 — moles of reactant. Grams cannot enter the recipe; moles can. $n (CaCO_{3}) = \frac{10.0}{100.0} = 0.100 mol$ Step 2 — mole ratio. Coefficients are $1 : 1$ , so one $CaCO_{3}$ gives one $CaO$ : $n (CaO) = 0.100 mol$ Step 3 — back to grams (a balance reads grams): $theoretical = 0.100 \times 56.0 = 5.60 g$ Step 4 — percent yield: $\frac{4.8}{5.60} \times 100% = 85.7%$

Exercise 2.2

$2 H_{2} + O_{2} \to 2 H_{2} O$ . Burn $4.0$ g of $H_{2}$ in excess oxygen and collect $32.0$ g of water. Percent yield? ( $M (H_{2}) = 2.0$ , $M (H_{2} O) = 18.0$ g/mol.)

Recall Solution 2.2

Step 1: $n (H_{2}) = \frac{4.0}{2.0} = 2.0$ mol. Step 2 — ratio $2 : 2 = 1 : 1$ : so $2.0$ mol of $H_{2} O$ theoretically. Step 3: theoretical mass $= 2.0 \times 18.0 = 36.0$ g. Step 4: $\frac{32.0}{36.0} \times 100% = 88.9%$ . Oxygen is stated as excess, so it is not the bottleneck — we ignore it for the ceiling.

Level 3 — Analysis

Goal: decide the limiting reagent yourself, then it caps the yield.

Exercise 3.1

$N_{2} + 3 H_{2} \to 2 NH_{3}$ . Mix $14$ g $N_{2}$ with $6.0$ g $H_{2}$ . Which reactant is limiting, and what is the theoretical yield of $NH_{3}$ in grams? ( $M (N_{2}) = 28.0$ , $M (H_{2}) = 2.0$ , $M (NH_{3}) = 17.0$ g/mol.)

Recall Solution 3.1

Step 1 — moles of each. $n (N_{2}) = \frac{14}{28.0} = 0.500 mol, n (H_{2}) = \frac{6.0}{2.0} = 3.00 mol$ Step 2 — product each could make alone (see the bar figure):

From $N_{2}$ : $0.500 \times \frac{2}{1} = 1.00$ mol $NH_{3}$ .
From $H_{2}$ : $3.00 \times \frac{2}{3} = 2.00$ mol $NH_{3}$ .

Figure — Percent yield, theoretical yield, actual yield

Step 3 — the smaller wins. $N_{2}$ makes only $1.00$ mol, so $N_{2}$ is limiting — it runs out first and caps the product. $H_{2}$ is left over. Step 4 — theoretical mass: $1.00 \times 17.0 = 17.0 g of NH_{3}$

Exercise 3.2

Continuing Ex 3.1: you actually obtain $13.6$ g of $NH_{3}$ . Find the percent yield, and how many grams of the excess reactant are left unreacted.

Recall Solution 3.2

Percent yield: $\frac{13.6}{17.0} \times 100% = 80.0%$ Leftover $H_{2}$ : the $0.500$ mol $N_{2}$ consumes $0.500 \times 3 = 1.50$ mol $H_{2}$ . Started with $3.00$ mol, so leftover $= 3.00 - 1.50 = 1.50$ mol. $mass left = 1.50 \times 2.0 = 3.0 g of H_{2}$ (The leftover is real mass on the balance but is not product — never count it in yield.)

Level 4 — Synthesis

Goal: run the chain backwards, or combine yield with limiting-reagent logic.

Exercise 4.1

A synthesis of ester ( $M = 88.0$ g/mol) is known to run at 65.0% yield. Your target is $22.0$ g of pure ester. How many moles of ester must the theoretical yield be, and hence how many moles of the limiting alcohol ( $1 : 1$ ratio) must you charge?

Recall Solution 4.1

Step 1 — theoretical mass (undo the yield loss): $theoretical = \frac{actual}{yield fraction} = \frac{22.0}{0.650} = 33.85 g$ WHY: you only keep $65%$ , so to keep $22.0$ g you must make more than $22.0$ g on paper. Step 2 — moles of ester needed: $n = \frac{33.85}{88.0} = 0.3846 mol$ Step 3 — alcohol needed. The ratio is $1 : 1$ , so charge at least $0.3846$ mol of the limiting alcohol. $n (alcohol) = 0.3846 mol \approx 0.385 mol$

Exercise 4.2

Two steps run in sequence. Step A converts reactant to intermediate at $80.0%$ yield; step B converts intermediate to final product at $75.0%$ yield. Starting from a step-A theoretical maximum of $1.00$ mol product, what is the overall percent yield and the final moles obtained?

Recall Solution 4.2

Key idea: each step keeps a fraction of what entered it, so fractions multiply. $overall fraction = 0.800 \times 0.750 = 0.600$ $overall percent yield = 60.0%$ Final moles $= 1.00 \times 0.600 = 0.600$ mol. WHY multiply, not average or add? Step B can only work on the $0.800$ mol that step A actually delivered; it then keeps $75%$ of that. Chaining fractions = multiplying them.

Level 5 — Mastery

Goal: full chain — balance, find limiting reagent, apply yield, work backwards, catch the >100% flag.

Exercise 5.1

Iron is smelted by $Fe_{2} O_{3} + 3 CO \to 2 Fe + 3 CO_{2}$ . You react $160$ g $Fe_{2} O_{3}$ with $84$ g $CO$ . The furnace yields $89.6$ g of iron. Find (a) the limiting reagent, (b) the theoretical yield of Fe, (c) the percent yield. ( $M (Fe_{2} O_{3}) = 160.0$ , $M (CO) = 28.0$ , $M (Fe) = 56.0$ g/mol.)

Recall Solution 5.1

Step 1 — moles. $n (Fe_{2} O_{3}) = \frac{160}{160.0} = 1.00 mol, n (CO) = \frac{84}{28.0} = 3.00 mol$ Step 2 — Fe each could make (see figure):

From $Fe_{2} O_{3}$ : $1.00 \times \frac{2}{1} = 2.00$ mol Fe.
From $CO$ : $3.00 \times \frac{2}{3} = 2.00$ mol Fe.

(a) Both give exactly $2.00$ mol — this is the stoichiometric (exact) mix, neither is in excess. We may call either one "limiting"; nothing is left over. (b) Theoretical yield $= 2.00 \times 56.0 = 112$ g of Fe. (c) Percent yield: $\frac{89.6}{112} \times 100% = 80.0%$

Exercise 5.2

Same reaction as Ex 5.1 (theoretical yield $112$ g Fe). A careless student weighs the iron before drying off adsorbed water and records $120$ g, then reports the percent yield. (a) What percent yield does the raw number give? (b) Why is it impossible, and what should they do?

Recall Solution 5.2

(a) Raw calculation: $\frac{120}{112} \times 100% = 107.1%$ (b) A percent yield above $100%$ is physically impossible — you cannot create iron atoms that were never in the ore (Conservation of Mass). The extra $8$ g is not product; it is trapped water/impurity inflating the balance reading. Fix: dry the sample to constant mass, then reweigh. Only then does the number fall below $100%$ and become a true yield.

Exercise 5.3 (capstone — work fully backwards)

A pharmaceutical route to a drug ( $M = 250.0$ g/mol) runs at $40.0%$ yield from a precursor ( $M = 200.0$ g/mol) in a $1 : 1$ mole ratio. Marketing demands $50.0$ g of pure drug. What mass of precursor must you start with?

Recall Solution 5.3

Walk the chain in reverse: product mass → theoretical mass → moles → precursor moles → precursor mass. Step 1 — theoretical drug mass (undo the $40%$ yield): $theoretical = \frac{50.0}{0.400} = 125.0 g$ Step 2 — moles of drug on paper: $n (drug) = \frac{125.0}{250.0} = 0.500 mol$ Step 3 — moles of precursor ( $1 : 1$ ratio, so equal): $n (precursor) = 0.500 mol$ Step 4 — mass of precursor: $0.500 \times 200.0 = 100.0 g of precursor$ So you must charge $100.0$ g of precursor to walk away with $50.0$ g of drug at $40%$ efficiency.

Recall One-line self-check before you leave

Overall two-step yields multiply ::: $0.80 \times 0.75 = 0.60 = 60%$ The limiting reagent is chosen by ::: the smallest moles-of-product it can make, not the smallest grams or moles of reactant A yield above 100% means ::: impurity/moisture inflating the mass, never real extra product To work backwards from a target, first ::: divide the target mass by the yield fraction to get the theoretical mass

Connections

Limiting Reagent — the ceiling used in every L3–L5 problem.
Mole Concept — the currency for all conversions here.
Molar Mass — every $\div M$ and $\times M$ step.
Balancing Chemical Equations — supplies each mole ratio.
Conservation of Mass — why >100% is impossible (Ex 5.2).
Empirical & Molecular Formula — same mass–mole reasoning reused.

Exercises — Percent yield, theoretical yield, actual yield