IMAT Chemistry · Stoichiometry and Reactions

Reacting Masses, Limiting Reagent and Percentage Yield

A balanced equation is a ratio of moles: use it to find how much product a mass of reactant can make, which reactant runs out first, and what share of the maximum you actually got.

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Why this matters

This is the most asked page. Percentage yield appears in 2011, 2014, 2015 and the 2023 ministry paper, and the 2024 ministry paper asked twice about which reactant runs out and what is left over.

Concept 1 of 3: Reacting masses through the mole ratio

The coefficients in an equation count particles, not grams. Two moles of magnesium react with one mole of oxygen, though the masses are 48 g and 32 g. So every reacting-mass problem runs the same three steps: grams into moles, moles through the ratio, moles back into grams.

Definition

Step 1: convert the known mass to moles ( $n = m/M$ ).
Step 2: use the coefficients of the balanced equation to find the moles of the substance you want.
Step 3: convert those moles to a mass ( $m = nM$ ), or to a number of particles.
When a reactant is a solution, its moles come from $n = cV$ (see Solutions and Concentration).
For gases at the same temperature and pressure, volumes react in the same ratio as moles (see States of Matter and Gas Laws).

Mole ratio from the equation

\frac{n_B}{n_A} = \frac{b}{a}

$n_A, n_B$ moles of substances A and B
$a, b$ their coefficients in the balanced equation

Worked example

Limestone decomposes on heating:

\mathrm{CaCO_3} \rightarrow \mathrm{CaO} + \mathrm{CO_2}

. What masses of calcium oxide and carbon dioxide form from 25 g of calcium carbonate? (Ca = 40, C = 12, O = 16)

Practice this conceptself-check · 4 quick reps

Coefficients are a ratio of moles, not of grams

2\mathrm{Mg} + \mathrm{O_2} \rightarrow 2\mathrm{MgO}

, 2 g of magnesium does not react with 1 g of oxygen. Convert to moles before using the coefficients, and back to grams afterwards.

Concept 2 of 3: Limiting reagent and the reactant in excess

Making sandwiches with 10 slices of bread and 3 slices of cheese gives 3 sandwiches, not 5: the cheese runs out first. In a reaction, the reactant that runs out first is the limiting reagent. It alone decides how much product forms; some of the other reactant is left over.

Definition

The limiting reagent is used up completely and sets the amount of product.
The other reactant is in excess: some of it is left at the end.
To find the limiting reagent, divide the moles of each reactant by its coefficient. The smallest result is the limiting reagent.
Work out the product from the limiting reagent only.
Excess left = moles at the start minus moles used (found from the limiting reagent through the ratio).
Total mass is still conserved: mass of products plus mass of excess left equals the mass at the start.

Finding the limiting reagent

\text{limiting} = \text{smallest of } \frac{n_A}{a},\ \frac{n_B}{b}

$n_A, n_B$ moles of each reactant at the start
$a, b$ their coefficients in the balanced equation

Worked example

4.0 g of hydrogen is exploded with 16 g of oxygen:

2\mathrm{H_2} + \mathrm{O_2} \rightarrow 2\mathrm{H_2O}

. Which reactant is limiting? What mass of water forms, and what mass of which gas is left over? (H = 1, O = 16)

Practice this conceptself-check · 3 quick reps

The limiting reagent is not simply the smaller mass or the fewer moles

Compare moles divided by coefficients. 0.10 mol of magnesium is limiting against 0.075 mol of oxygen, because the equation needs twice as much magnesium as oxygen. Then calculate the product from the limiting reagent, never from the one in excess.

Concept 3 of 3: Percentage yield

The equation tells you the most product you could possibly get, the theoretical yield. Real experiments get less: some reactant does not react, side reactions make other products, and some product is lost while filtering, drying or transferring it. Percentage yield measures how close you came.

Definition

Theoretical yield: the mass (or moles) of product calculated from the limiting reagent with the balanced equation.
Actual yield: the mass of product really obtained.
Percentage yield = actual yield divided by theoretical yield, times 100. Compare product with the same product, in the same units.
Yield is below 100% because of incomplete or reversible reactions, side reactions and losses during separation.
A measured yield above 100% means the product is impure (for example, still wet).

Percentage yield

\%\,\text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100

actual yieldmass of product obtained, in g
theoretical yieldmaximum mass of product from the limiting reagent, in g

Worked example

50 g of calcium carbonate is heated and 21 g of calcium oxide is collected:

\mathrm{CaCO_3} \rightarrow \mathrm{CaO} + \mathrm{CO_2}

. What is the percentage yield? (Ca = 40, C = 12, O = 16)

Practice this conceptself-check · 4 quick reps

Compare product with product, not product with reactant

Percentage yield is actual product mass over the theoretical mass of the same product. Dividing the product mass by the mass of reactant used is a common wrong option, because the two substances have different molar masses.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Reacting masses through the mole ratio
Mole ratio from the equation
$\frac{n_B}{n_A} = \frac{b}{a}$
Limiting reagent and the reactant in excess
Finding the limiting reagent
$\text{limiting} = \text{smallest of } \frac{n_A}{a},\ \frac{n_B}{b}$
Percentage yield
Percentage yield
$\%\,\text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$

Watch out for (3)

Coefficients are a ratio of moles, not of grams
→ Reacting masses through the mole ratio
The limiting reagent is not simply the smaller mass or the fewer moles
→ Limiting reagent and the reactant in excess
Compare product with product, not product with reactant
→ Percentage yield

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