Percent yield, theoretical yield, actual yield

WHAT are these three quantities?

HOW to derive theoretical yield from first principles

There is no magic formula to memorize — theoretical yield is just careful bookkeeping of atoms. Follow the chain:

$\text{mass reactant} \xrightarrow{\div M} \text{mol reactant} \xrightarrow{\times \frac{\text{coeff. product}}{\text{coeff. reactant}}} \text{mol product} \xrightarrow{\times M} \text{mass product}$

The limiting reagent is the reactant that runs out first, so it caps how much product forms. To find it: compute how much product each reactant could make on its own; the smallest answer wins and sets the theoretical yield.

Worked Example 1 — clean single-reactant case

Burn 8.0 g of methane completely: $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$. Oxygen is in excess. You collect 19.0 g of $\text{CO}_2$. Find percent yield.

Step 1 — moles of $\text{CH}_4$: $n = \dfrac{8.0}{16.0} = 0.50$ mol. Why this step? Grams can't be compared to the recipe; moles can.

Step 2 — mole ratio to $\text{CO}_2$: ratio is $1:1$, so $0.50$ mol $\text{CO}_2$ theoretically. Why this step? One carbon in, one carbon out — conservation of atoms.

Step 3 — theoretical mass: $0.50 \times 44.0 = 22.0$ g. Why this step? A balance reads grams, so we convert back to grams to compare with the lab result.

Step 4 — percent yield: $\dfrac{19.0}{22.0}\times100 = 86.4\%$. Why this step? Actual ÷ theoretical, same units (g), ×100.

Worked Example 2 — limiting reagent decides the theoretical yield

$\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$. React 28 g $\text{N}_2$ with 9.0 g $\text{H}_2$. Actual $\text{NH}_3$ obtained = 25 g. Find percent yield.

Step 1 — moles: $n(\text{N}_2)=\frac{28}{28}=1.0$ mol; $n(\text{H}_2)=\frac{9.0}{2.0}=4.5$ mol.

Step 2 — product each could make:

• From $\text{N}_2$: $1.0 \times \frac{2}{1} = 2.0$ mol $\text{NH}_3$.
• From $\text{H}_2$: $4.5 \times \frac{2}{3} = 3.0$ mol $\text{NH}_3$.

Why this step? Whichever gives less product is the true bottleneck — you can't make more product than your scarcest ingredient allows.

Step 3 — limiting reagent = $\text{N}_2$ (2.0 < 3.0). Theoretical = 2.0 mol → $2.0 \times 17.0 = 34.0$ g.

Step 4 — percent yield: $\dfrac{25}{34.0}\times100 = 73.5\%$.

Worked Example 3 — working backwards from percent yield

A reaction has a known 90.0% yield and you need 18.0 g of product ($M = 60$ g/mol). How many moles of product must the theoretical yield be?

Step 1: theoretical mass $= \dfrac{\text{actual}}{\text{yield fraction}} = \dfrac{18.0}{0.900} = 20.0$ g. Why this step? Rearranging the percent-yield formula: theoretical = actual ÷ (%/100).

Step 2: $n = \dfrac{20.0}{60} = 0.333$ mol needed on paper. Why this step? Now you scale up your reactants to hit this target despite the 10% loss.

Common Mistakes (Steel-manned)

Active Recall Flashcards

#flashcards/chemistry

Connections

• Limiting Reagent — sets the ceiling on theoretical yield.
• Mole Concept — the mole is the currency for every stoichiometric conversion.
• Balancing Chemical Equations — coefficients give the mole ratio.
• Molar Mass — converts between grams and moles both directions.
• Conservation of Mass — the reason atoms in = atoms out.
• Empirical & Molecular Formula — mass–mole reasoning reused.

Concept Map

Hinglish (regional understanding)