Stoichiometry
Master stoichiometric calculations: mole ratios, mass-to-mass conversions, limiting reagents, theoretical yield, and percent yield.
Chemistry reference tablesWhat Is Stoichiometry?
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. The word comes from the Greek stoicheion (element) and metron (measure) — literally, measuring elements.
A balanced chemical equation tells you exactly how much of each substance is involved. The coefficients in front of each formula give you the ratio of moles — and from moles, you can calculate masses, volumes, and number of particles.
Every stoichiometry problem follows the same core pattern: use the balanced equation to build a mole ratio (also called a stoichiometric factor), then use that ratio to convert between substances.
Go deeperWhy this matters: your car computes stoichiometry in real time
Gasoline burns cleanly at an air-to-fuel mass ratio of about 14.7 to 1, the stoichiometric ratio for complete combustion. Your car’s oxygen sensor measures leftover O2 in the exhaust dozens of times per second, and the engine computer nudges the fuel injection toward that ratio continuously: too rich wastes fuel and emits CO; too lean overheats and produces nitrogen oxides. Rocket engineers run the same calculation with far less forgiveness, because every kilogram of mismatched propellant is dead weight lifted at enormous cost. Stoichiometry is not exam bookkeeping; it is running in every engine you have ever ridden behind.
Mole Ratios from Balanced Equations
The coefficients in a balanced equation directly give you the mole ratios between any two substances in the reaction. Consider the synthesis of ammonia:
3 H2 + N2 → 2 NH3
From this equation, you can derive several stoichiometric factors:
- 3 mol H2 : 1 mol N2
- 3 mol H2 : 2 mol NH3
- 1 mol N2 : 2 mol NH3
Each ratio can be flipped depending on what you're solving for. If you know the moles of one substance, multiply by the appropriate mole ratio to find the moles of another.
Example: How many moles of I2 react with 0.429 mol Al?
2 Al + 3 I2 → 2 AlI3
The mole ratio is 3 mol I2 per 2 mol Al:
0.429 mol Al × (3 mol I2 / 2 mol Al) = 0.644 mol I2
Go deeperCommon mistake: reading coefficients as grams
The 3 : 1 in 3 H2 + N2 → 2 NH3 counts molecules (and therefore moles), never grams. Three grams of hydrogen do not match one gram of nitrogen; the true mass ratio is 3 mol H2 (6.05 g) to 1 mol N2 (28.02 g), roughly 1 g of hydrogen per 4.6 g of nitrogen. Reading coefficients as masses inverts reality because molecules weigh wildly different amounts.
The discipline that prevents it: never apply a mole ratio to anything that is not already in moles. Grams must buy their ticket (division by molar mass) before boarding the ratio.
Mass-to-Mass Conversions
In the lab, you measure masses — not moles. A mass-to-mass stoichiometry problem adds two conversion steps around the mole ratio:
- Mass → Moles: Divide the given mass by the molar mass of that substance
- Moles → Moles: Apply the mole ratio from the balanced equation
- Moles → Mass: Multiply by the molar mass of the target substance
The pattern is always: grams → moles → mole ratio → moles → grams.
Example: What mass of NaOH is needed to produce 16 g of Mg(OH)2?
MgCl2 + 2NaOH → Mg(OH)2 + 2NaCl
Step 1: 16 g Mg(OH)2 ÷ 58.32 g/mol = 0.274 mol Mg(OH)2
Step 2: 0.274 mol Mg(OH)2 × (2 mol NaOH / 1 mol Mg(OH)2) = 0.549 mol NaOH
Step 3: 0.549 mol NaOH × 40.00 g/mol = 22.0 g NaOH
Go deeperWhy every route detours through moles
Students often ask why there is no direct gram-to-gram shortcut. The reason: a gram is not a count. One gram of H2 holds about fourteen times as many molecules as one gram of N2, so no fixed gram ratio can describe a reaction whose recipe is written in molecules. The mole is the common currency conversion: like exchanging pesos to dollars to yen, you convert mass into the universal counting unit, trade at the reaction’s fixed molecular exchange rate, and convert back out to the units you need. Once you see the mole as currency, the grams → moles → moles → grams pattern stops being a procedure to memorize and becomes the only route that could work.
Limiting Reagent
When two or more reactants are mixed in amounts that don't match the exact mole ratio from the balanced equation, one will run out first. The reactant that is completely consumed is the limiting reagent — it limits how much product can form. The other reactant is present in excess.
To identify the limiting reagent:
- Convert each reactant's mass to moles
- Divide each by its coefficient in the balanced equation
- The reactant with the smallest value is the limiting reagent
Alternative method: Calculate how much product each reactant could produce independently. The reactant that yields the lesser amount of product is limiting.
Example: Which is limiting when 3 mol H2 and 2 mol Cl2 react?
H2 + Cl2 → 2 HCl
The stoichiometric ratio is 1:1. You have 3:2. Since Cl2 has fewer moles relative to its coefficient (2/1 = 2) compared to H2 (3/1 = 3), Cl2 is the limiting reagent. It will be completely consumed, producing 4 mol HCl, with 1 mol H2 left over.
Go deeperWhy this matters: industry picks its limiting reagent on purpose
In a manufacturing plant, the limiting reagent is a business decision. If one reactant is expensive and another is cheap, engineers deliberately supply the cheap one in excess so that virtually every molecule of the costly one converts to product; the leftover cheap reagent is recycled or discarded. The same logic runs your grill: the propane is metered, and the air is unlimited excess.
So the skill you are practicing has two industrial faces: finding which reagent limits (analysis), and choosing which reagent should limit (design).
Go deeperTry it: hydrogen and oxygen, who runs out?
You mix 4 mol H2 with 3 mol O2 and ignite: 2 H2 + O2 → 2 H2O. Which reactant limits, how much water forms, and what remains? Work it, then check below.
Answer: Divide each amount by its coefficient: H2 gives 4/2 = 2; O2 gives 3/1 = 3. The smaller quotient wins: H2 is limiting. All 4 mol H2 react, producing 4 mol H2O and consuming 4/2 = 2 mol of O2, so 3 − 2 = 1 mol O2 remains unreacted. Note the trap the quotient method avoids: there are more moles of H2 than O2 in the flask, yet H2 is still the one that runs out.
Theoretical Yield and Percent Yield
The theoretical yield is the maximum amount of product that could be formed from a given amount of limiting reagent, calculated using stoichiometry. It assumes the reaction goes to completion with no losses.
In practice, you almost always get less product than the theoretical yield. Reasons include:
- Side reactions that produce unwanted products
- Incomplete reactions that don't go to 100% completion
- Mechanical losses during collection, transfer, or purification
The actual yield is the amount you actually obtain in the lab. The percent yield compares the two:
Percent yield = (actual yield / theoretical yield) × 100%
Both yields must be in the same units (grams, moles, etc.) for this formula to work.
Example: Reacting 1.274 g CuSO4 with excess Zn gives 0.392 g Cu. The theoretical yield of Cu is 0.508 g. Percent yield = (0.392 / 0.508) × 100% = 77.2%
Go deeperWhy this matters: percent yield is why some drugs cost so much
Complex pharmaceuticals are built in long sequences of reactions, and yields multiply. A ten-step synthesis averaging a respectable 80% per step delivers only 0.8010 ≈ 11% of the theoretical amount overall: nearly ninety percent of the starting material is lost along the way. That multiplication is why process chemists celebrate raising a single step from 80% to 95%, why routes with fewer steps beat elegant routes with many, and why a molecule’s price can reflect its synthetic distance more than its ingredients. Percent yield is the exchange rate between chemistry on paper and chemistry you can sell.
Calculating Excess Reagent Remaining
Once you identify the limiting reagent, you can calculate exactly how much of the excess reagent remains unreacted:
- Use the moles of the limiting reagent to calculate how many moles of the excess reagent are consumed (via the mole ratio)
- Subtract the consumed amount from the original amount of the excess reagent
Example: If 3 mol H2 and 2 mol Cl2 react (1:1 ratio), Cl2 is limiting. It consumes 2 mol H2. Remaining H2 = 3 − 2 = 1 mol H2 unreacted.
To convert this to grams, multiply by the molar mass: 1 mol H2 × 2.016 g/mol = 2.016 g H2 remaining.
Gas Stoichiometry at STP
At standard temperature and pressure (STP: 0 °C and 1 atm), one mole of any ideal gas occupies 22.4 L. This is called the standard molar volume.
This fact gives you an additional conversion factor for stoichiometry problems involving gases at STP:
1 mol gas = 22.4 L at STP
You can now convert between volume and moles without needing the ideal gas law (PV = nRT) — as long as the gas is at STP.
Example: What volume of O2 at STP is needed to react with 2.7 L of propane (C3H8) at the same conditions?
C3H8 + 5O2 → 3CO2 + 4H2O
When gases are at the same temperature and pressure, their volume ratios equal their mole ratios (Avogadro's law). Since the coefficient ratio is 1:5, you need 5 × 2.7 L = 13.5 L of O2.
For gases not at STP, use the ideal gas law: n = PV/RT to convert between volume and moles, then proceed with the normal stoichiometric mole ratio.
Go deeperCommon mistake: using 22.4 L/mol everywhere
The number 22.4 L/mol carries three conditions, and forgetting any of them breaks the answer. It applies only to gases (never liquids or solids: 1 mol of liquid water is 18 mL, not 22.4 L), only at STP (0 °C and 1 atm), and only to gases behaving ideally (a safe assumption at this level). At room temperature, 25 °C, one mole of gas occupies about 24.5 L; the ideal gas law handles those cases when you meet it in the gases topic.
Before reaching for 22.4, run the checklist: gas? STP stated? If either answer is no, the shortcut is not available.
Multi-Step Stoichiometry Problems
Many real stoichiometry problems combine several of the skills above into a single calculation. A typical multi-step problem might ask you to:
- Start with a mass of one reactant
- Identify the limiting reagent
- Calculate the theoretical yield in grams
- Determine the percent yield from an actual yield
- Find how much excess reagent remains
The key to solving these is to work one step at a time. Don't try to do everything in one equation. Follow the chain:
Given quantity → moles → mole ratio → moles → target quantity
At each step, check your units and make sure you're using the right mole ratio from the balanced equation. Dimensional analysis is your best tool for keeping track of conversions.
Stoichiometry Problem Workflow and Common Mistakes
Every stoichiometry problem follows the same core workflow:
- Write and balance the equation. No calculation is valid without a balanced equation.
- Convert the given quantity to moles (using molar mass, molarity, or molar volume at STP).
- Use the mole ratio from the balanced equation to find moles of the target substance.
- Convert moles of target to the requested unit (grams, liters, particles).
- For limiting-reagent problems: calculate moles of product from each reactant separately. The reactant that gives the smaller amount of product is the limiting reagent.
Common mistakes: using an unbalanced equation, skipping the grams-to-moles conversion and plugging grams directly into the mole ratio, confusing limiting and excess reagents, and forgetting that percent yield = (actual/theoretical) × 100%. A quick check: if your percent yield exceeds 100%, re-examine your theoretical yield calculation.
Key Equations
Learning Objectives
After studying this topic, you should be able to:
- Use mole ratios from a balanced equation to relate amounts of reactants and products
- Perform mass-to-mass stoichiometric calculations
- Identify the limiting reagent in a reaction
- Calculate the theoretical yield of a reaction
- Calculate percent yield from actual and theoretical yields
- Calculate the amount of excess reagent remaining after a reaction
- Perform stoichiometric calculations involving the volume of a gas at STP
How-To Procedure
How to Solve a Limiting Reagent Problem
- Write and balance the chemical equation if it is not already balanced.
- Convert the given mass of each reactant to moles using its molar mass.
- Divide each reactant's moles by its coefficient in the balanced equation.
- The reactant with the smallest result from step 3 is the limiting reagent.
- Use the moles of the limiting reagent and the mole ratio to calculate the moles of product (this is the theoretical yield in moles).
- Convert moles of product to grams using the product's molar mass to get the theoretical yield in grams.
- If an actual yield is given, calculate percent yield: (actual yield / theoretical yield) x 100%.
Worked Example
Calculating Moles of Product from a Given Reactant
Calculate the moles of each product made in the reaction below, starting with 2.4 mol NaOH.
NaOH + CuSO4 → Cu(OH)2 + Na2SO4
- Balance the equation: 2NaOH + CuSO4 → Cu(OH)2 + Na2SO4
- Use mole ratio for Cu(OH)2: 2.4 mol NaOH × (1 mol Cu(OH)2 / 2 mol NaOH) = 1.2 mol Cu(OH)2
- Use mole ratio for Na2SO4: 2.4 mol NaOH × (1 mol Na2SO4 / 2 mol NaOH) = 1.2 mol Na2SO4
1.2 mol Cu(OH)2 and 1.2 mol Na2SO4
Test Your Understanding
In a reaction, a student calculates a percent yield of 112%. Is this possible? What does it indicate about the experiment?
Self-Study Questions
What is stoichiometry?
What information does a balanced chemical equation provide?
What is a mole ratio and where does it come from?
Hint: Look at the coefficients in the balanced equation.
What is a limiting reagent and how do you identify it?
What is theoretical yield?
What is percent yield and why is it usually less than 100%?
What is an excess reagent?
How do you convert from grams of one substance to grams of another using stoichiometry?
What role does the balanced equation play in every stoichiometry calculation?
What is the molar volume of a gas at STP and how is it used?
Content Sources
Concept sections draw on the sources listed below. OpenStax material is available under the CC BY 4.0 license.