Stoichiometry — Print · Grade 11 Science
Chapter: Stoichiometry & YieldStoichiometry
Learning goals
- I can read the mole ratio between any two substances directly from the coefficients of a balanced equation.
- I can carry out a mass-to-mass calculation by converting mass to moles, applying the mole ratio, and converting back to mass.
Readings for this unit
Before this unit
Balanced equations, mole conversion, and molar mass.
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Start with something to explain
Can a product weigh more than the reactant sample you started weighing?
Illustrative evidence case. A selected metal gains mass when oxygen is incorporated into its oxide.
Notice and wonder
Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.
Can a product weigh more than the reactant sample you started weighing?
Your first model
Draw a boundary and include the previously unweighed oxygen input.
Build the language from the idea
- stoichiometry
- The use of a balanced equation to calculate the amounts of reactants and products in a reaction.
- mole ratio
- A ratio between the amounts of any two substances in a reaction, taken directly from the coefficients of the balanced equation.
- mass-to-mass calculation
- Finding the mass of one substance from the mass of another by converting to moles, applying the mole ratio, and converting back to mass.
- given
- The substance whose amount you know at the start of a stoichiometry problem.
- unknown
- The substance whose amount you are solving for.
A balanced equation hides a superpower in plain sight. Once you know that 2H₂ + O₂ → 2H₂O, you know far more than which substances react — you know the exact numerical recipe that connects them. Stoichiometry is the branch of chemistry that cashes in that recipe, using the coefficients of a balanced equation to calculate precisely how much of one substance reacts with, or is produced from, a given amount of another. It is how a chemist answers the unit's essential question: how much will this reaction actually make?
The heart of stoichiometry is the mole ratio — a ratio between two substances read straight off the balanced coefficients. In 2H₂ + O₂ → 2H₂O, the coefficients 2, 1, and 2 tell you that 2 moles of hydrogen react with 1 mole of oxygen to make 2 moles of water. So the mole ratio of hydrogen to water is 2:2 (or 1:1), and of oxygen to water is 1:2. Crucially, these coefficients are ratios of moles, never of grams — a subtlety that trips up beginners, because 2 grams of hydrogen is not the same amount of stuff as 2 grams of water.
That warning points to the golden rule of the whole subject: coefficients count moles, not mass, so you can only apply a mole ratio to amounts measured in moles. This is exactly why the mole from the last lesson matters so much. If a problem hands you grams, you cannot leap straight to the answer — you must first translate those grams into moles, do your reasoning in the currency of moles, and then translate back. Skipping the conversion and multiplying masses by the coefficients is the single most common stoichiometry mistake.
The simplest stoichiometry problems are mole-to-mole: you already have an amount in moles and just multiply by the mole ratio. Take the ammonia synthesis N₂ + 3H₂ → 2NH₃. If you have 6.0 moles of hydrogen, how much ammonia can form? The mole ratio of NH₃ to H₂ is 2:3, so you multiply: 6.0 mol H₂ × (2 mol NH₃ ÷ 3 mol H₂) = 4.0 mol NH₃. Notice how the units of H₂ cancel, leaving moles of NH₃ — a sign you set the ratio up the right way.
The real workhorse is the mass-to-mass calculation, because in the lab you weigh things in grams, not moles. It follows one reliable four-step road map every time: start with the mass of your given substance, divide by its molar mass to get moles, multiply by the mole ratio to cross over to moles of your unknown, then multiply by the unknown's molar mass to reach its mass. In symbols: mass A → (÷ molar mass A) → moles A → (× mole ratio) → moles B → (× molar mass B) → mass B. It is the same path every problem, only the numbers change.
Walk it once with water. How many grams of water form from 4.04 g of hydrogen burning in excess oxygen (2H₂ + O₂ → 2H₂O)? Step one: 4.04 g H₂ ÷ 2.02 g/mol = 2.00 mol H₂. Step two: the mole ratio of H₂O to H₂ is 2:2, so 2.00 mol H₂ × (2 ÷ 2) = 2.00 mol H₂O. Step three: 2.00 mol H₂O × 18.02 g/mol = 36.04 g of water. This same road map lets engineers size a rocket's propellant, farmers calculate fertilizer output, and chemists scale any reaction with confidence — the balanced equation, run through the mole, predicts the yield exactly.
Model check: Coefficients give mole ratios, not mass ratios. It requires 6 mol H₂ and can produce 6 mol H₂O if reactants are sufficient.
Build your understanding
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Part 1 · Use a balanced equation as an amount map
Stoichiometry uses a balanced equation to relate amounts of reacting species. The coefficient ratio in N₂+3H₂→2NH₃ is 1:3:2 for entities or mole amounts, not grams. A selected half mole of nitrogen requires one and a half moles of hydrogen to form one mole of ammonia under complete reaction assumptions. An equation’s coefficient does not prove that one nitrogen molecule and three hydrogen molecules collide simultaneously; mechanism and stoichiometric balance are different descriptions.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
If 0.50 mol N₂ reacts fully, why is the required hydrogen 1.50 mol rather than 0.50 mol?
Visual reasoning — after your attempt
The balanced amount ratio is three moles of H₂ per mole of N₂, so multiply 0.50 by three.
Balanced coefficients compare moles of specified species for the stated reaction model. Begin with a correct equation; an unbalanced ratio makes later arithmetic meaningless. Convert the known quantity to moles, apply the coefficient ratio, then convert to the requested unit.
For N₂+3H₂→2NH₃, one mole N₂ corresponds to two moles NH₃ if the modeled reaction proceeds completely with enough hydrogen.
Pause and explain
What must happen before using a reaction mole ratio?
Check your reasoning after answering
The equation must be balanced and the reaction conditions and sufficient co-reactant assumption must be appropriate.
Part 2 · Show units through a conversion chain
A mass-to-mass calculation needs three links: mass of the starting species to its mole amount, the balanced mole ratio to the target, and the target molar mass back to mass. Keep units and species names on each factor. For 2H₂+O₂→2H₂O the hydrogen-to-water mole ratio is one-to-one, while their gram amounts differ greatly. The selected rounded molar masses simplify arithmetic; use one consistent reference set and round only at the final step.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Where does the oxygen mass in 90.0 g water come from if the starting hydrogen mass is only 10.0 g?
Visual reasoning — after your attempt
It comes from the oxygen reactant. Product mass includes atoms from both reactants, not only the species used to begin the calculation.
Mass-to-mass stoichiometry uses separate molar masses on either side of the mole ratio. A coefficient ratio is not a gram ratio. Keep units on each factor so the chain cancels grams of reactant, moles of reactant, and moles of product correctly.
28.0 g N₂ at 28.0 g/mol is one mole; the 2: 1 ratio gives two moles NH₃, or 34.0 g at 17.0 g/mol. Hydrogen supplies the extra mass.
Pause and explain
Why can product mass exceed the mass of one reactant?
Check your reasoning after answering
Another reactant contributes matter. Complete conservation compares all reacting inputs and outputs, not one selected starting mass.
Part 3 · State excess, completeness, and purity assumptions
A theoretical yield is the product amount predicted by a stated chemical model and available reacting quantities. An impure sample contains less target reactant than its total mass, an insufficient partner can limit conversion, and an equilibrium or competing pathway can reduce the modeled target formation. These are distinct limitations, not reasons to abandon conservation. State the reaction, purity, available amounts and completeness assumptions before presenting a number as an experimental outcome.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
What changes if only 80% of a 10.0 g starting sample is the specified reactant?
Visual reasoning — after your attempt
Only 8.00 g is available for that reactant’s stoichiometric conversion under the stated mass-fraction model.
A theoretical amount assumes the supplied limiting amount, enough other reactants, the stated composition, and complete conversion into the target product. Actual recovery can be lower or contaminated. Do not call a theoretical result a measured yield.
A 90% pure reactant requires adjusting the usable mass before mole conversion if impurities do not react in the modeled way. An excess label concerns amounts, not how concentrated a substance looks.
Pause and explain
What changes if the supplied reactant is not pure?
Check your reasoning after answering
The usable amount must be calculated from its composition. Using total sample mass as pure reactant can overestimate the theoretical product.
Part 4 · Check composition and plausible magnitude
Mass conservation is checked within a stated system boundary. The selected complete water reaction gives matching reactant and product mass totals under the supplied rounded molar masses. If a gas escapes an open container, the measured container contents can lose mass without violating conservation. Unreacted material also belongs in the final inventory. A product mass greater than one reactant’s starting mass is possible when atoms from a second reactant are incorporated; only a complete account can test closure.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Why can product mass exceed the starting mass of one reactant without violating conservation?
Visual reasoning — after your attempt
The product can also contain atoms supplied by other reactants. Compare the complete reacting inventory and any leftovers or transfers.
Verify both the conversion units and a mass account using all consumed reactants. A large or small answer may be reasonable depending on molar masses, but an amount beyond what the limiting reactant can support is inconsistent. Explain the result chemically as well as numerically.
Recalculate the product in reverse to recover consumed reactant amount. This checks arithmetic and ratio orientation but does not independently establish actual conversion or purity.
Pause and explain
What does a reverse calculation check and what does it not check?
Check your reasoning after answering
It checks ratio and arithmetic consistency. It does not prove the reaction occurred completely or the recovered product is pure.
Apply the whole unit · Predict ammonia and account for matter
Illustrative model: N₂+3H₂→2NH₃. Use 14.0 g pure N₂ with excess H₂, M(N₂)=28.0,M(H₂)=2.00,M(NH₃)=17.0 g/mol; assume complete target conversion.
Your task
Calculate theoretical ammonia mass and hydrogen consumed, show the conversion chain, and check total consumed mass. Explain why the result remains a prediction rather than a laboratory yield.
What your response should show
- Choose the appropriate chemical model or relationship and state its conditions.
- Show substitutions, units, particle or charge counts, and intermediate reasoning as relevant.
- Use supplied evidence to justify the conclusion; distinguish observation from inference.
- Identify a meaningful limitation, uncertainty, or next measurement.
Response guidance — after your attempt
n(N₂)=14.0/28.0=0.500 mol.
n(NH₃)=0.500×2=1.00 mol, giving 17.0 g.
n(H₂)=0.500×3=1.50 mol, giving 3.00 g.
Consumed mass=14.0+3.00=17.0 g, matching modeled product.
The theoretical product is 17.0 g under the complete-conversion and excess-hydrogen assumptions. Actual yield requires recovered mass and purity evidence. The mass increase relative to nitrogen alone is supplied by consumed hydrogen, not creation of matter. This is a calculation record, not an instruction to produce ammonia.
Scientific references and source-model foundations
Parent / teacher teaching notes
Use a units-labeled chain and require the stated reaction assumptions. Ask students to explain why 14 g nitrogen can lead to more than 14 g product. Ask the learner to explain each particle symbol, axis, unit and arrow before calculating. Distinguish hypothetical models and calculated curves from measured records. Public OpenSciEd excerpts here are selected foundation prompts, not complete high-school teacher editions or an endorsement. All required advanced explanations and practice are provided locally.
The key idea
The coefficients of a balanced equation give mole ratios (not mass ratios); a mass-to-mass problem follows one road map — mass A → moles A → (× mole ratio) → moles B → mass B.
Mole Ratio
mole ratio = (coefficient of wanted substance) : (coefficient of given substance)
taken directly from the coefficients of the balanced equation; it applies only to amounts in moles, never in grams.
Mass-to-Mass Road Map
mass A → (÷ M of A) → moles A → (× mole ratio) → moles B → (× M of B) → mass B
the four-step path for finding the mass of substance B produced from (or reacting with) a given mass of substance A.
Worked examples
Example 1
- Read the mole ratio from the coefficients: NH₃ to H₂ is 2 : 3.
- Multiply the given moles of H₂ by the ratio, arranged so H₂ cancels: 6.0 mol H₂ × (2 mol NH₃ ÷ 3 mol H₂).
- Compute: 6.0 × 2 ÷ 3 = 4.0 mol NH₃.
Example 2 · Worked application
- The H₂:H₂O coefficient ratio is 2:2.
- 3.0 mol H₂×(2 mol H₂O/2 mol H₂)=3.0 mol H₂O.
- The excess-reactant condition permits complete consumption in the ideal yield model.
Mole ratios come from coefficients, not molar masses.
Example 3
- Convert the given mass of H₂ to moles: 4.04 g ÷ 2.02 g/mol = 2.00 mol H₂.
- Apply the mole ratio of H₂O to H₂, which is 2 : 2 (or 1 : 1): 2.00 mol H₂ × (2 ÷ 2) = 2.00 mol H₂O.
- Convert moles of water to mass: 2.00 mol × 18.02 g/mol = 36.04 g H₂O.
Example 4 · Find and fix the mistake
- Test the claim against the stated conditions: Mole ratios come from coefficients, not molar masses.
- They mean equal mole quantities; different molar masses give different masses.
Mole ratios come from coefficients, not molar masses.
Example 5 · Reason from the visual model
- A theoretical yield is the product amount predicted by a stated chemical model and available reacting quantities. An impure sample contains less target reactant than its total mass, an insufficient partner can limit conversion, and an equilibrium or competing pathway can reduce the modeled target formation. These are distinct limitations, not reasons to abandon conservation. State the reaction, purity, available amounts and completeness assumptions before presenting a number as an experimental outcome.
- Only 8.00 g is available for that reactant’s stoichiometric conversion under the stated mass-fraction model.
Original assumption organizer, not a complete kinetic/equilibrium calculation or a general promise of quantitative conversion.
Avoid these mistakes
- Coefficients give mole ratios, not mass ratios.
Check your understanding
2H₂+O₂→2H₂O: what does 3 mol O₂ require?
Discuss the answer
It requires 6 mol H₂ and can produce 6 mol H₂O if reactants are sufficient.
Return to the phenomenon
Return to the initial case. Can a product weigh more than the reactant sample you started weighing? Revise your drawing or explanation using two named ideas and one stated limitation.
Compare explanations after your attempt
The product includes material from the metal and oxygen. Use balanced mole ratios and all system transfers rather than compare only one starting mass.
What should we investigate next?
Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.
Investigate, calculate and revise
Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.
Activity 1 · Return to the opening case
Use the opening phenomenon, unit definitions and explained Parts 1–4.
Your task
Return to the initial case. Can a product weigh more than the reactant sample you started weighing? Revise your drawing or explanation using two named ideas and one stated limitation.
Success criteria
- Use the stated observations and appropriate units.
- Connect a labeled model with the chemical mechanism.
- State assumptions and an evidence limitation.
- Explain a substantive revision or next test.
Hint
Keep species identity, material transfer and energy account distinct.
Response guidance — discuss after your attempt
The product includes material from the metal and oxygen. Use balanced mole ratios and all system transfers rather than compare only one starting mass.
Activity 2 · Transfer to a changed case
Cover the worked answer and use the explained models, relationships and notebook.
Your task
For 2Mg+O₂→2MgO, use Mg=24.3 and MgO=40.3 g/mol. Find theoretical MgO from 4.86 g pure Mg with excess oxygen. Explain the mass increase.
Success criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Hint
Name known quantities and conditions, show intermediate steps and check conservation.
Response guidance — discuss after your attempt
4.86/24.3=0.200 mol Mg. The 1:1 ratio gives 0.200 mol MgO and 0.200×40.3=8.06 g MgO. The extra mass comes from incorporated oxygen; the complete mass account includes it.
Words to know
- stoichiometry
- The use of a balanced equation to calculate the amounts of reactants and products in a reaction.
- mole ratio
- A ratio between the amounts of any two substances in a reaction, taken directly from the coefficients of the balanced equation.
- mass-to-mass calculation
- Finding the mass of one substance from the mass of another by converting to moles, applying the mole ratio, and converting back to mass.
- given
- The substance whose amount you know at the start of a stoichiometry problem.
- unknown
- The substance whose amount you are solving for.
Where the word 'stoichiometry' comes from
The tongue-twisting word stoichiometry was coined in 1792 by the German chemist Jeremias Benjamin Richter, who stitched together two Greek roots: stoicheion, meaning 'element,' and metron, meaning 'measure.' Literally, it means 'the measuring of elements,' which is exactly what you are doing when you use a balanced equation to calculate amounts. Richter was among the first to argue that chemistry obeys strict mathematical laws — that elements combine in fixed, calculable proportions rather than by rough guesswork. When you set up a mole-ratio calculation, you are practicing a more-than-two-hundred-year-old conviction that chemistry is, at its heart, a quantitative science you can work out with a pencil.
Let the units do the thinking
Experienced chemists rarely 'reason through' a stoichiometry problem in words; they let a technique called dimensional analysis, or the factor-label method, carry them. You write each conversion as a fraction arranged so the unit you want to cancel sits diagonally, top and bottom, and strikes itself out. Grams of A cancel to leave moles of A; moles of A cancel to leave moles of B; moles of B cancel to leave grams of B — the units literally guide your hand from the given to the unknown. If your units cancel down to the quantity you wanted, you almost certainly set the problem up correctly; if they cancel to something absurd like grams-squared per mole, you know to flip a factor before you ever touch the calculator.
For gases, you can skip the moles
When every substance in a reaction is a gas measured at the same temperature and pressure, stoichiometry offers a shortcut. Because equal volumes of gas contain equal numbers of molecules — Avogadro's insight again — the volume ratio equals the mole ratio. So for 2H₂ + O₂ → 2H₂O run entirely as gases, 2 liters of hydrogen react with exactly 1 liter of oxygen to make 2 liters of water vapor, the same 2 : 1 : 2 as the coefficients, with no conversion to moles required. The French chemist Joseph Gay-Lussac spotted this pattern of simple whole-number volume ratios in reacting gases around 1808, and it became one of the strongest early clues that matter is built from discrete particles.
Fueling a rocket by the mole
The Space Shuttle's main engines ran on the simplest reaction in this whole unit — hydrogen burning in oxygen, 2H₂ + O₂ → 2H₂O — but the stoichiometry hid a clever twist. The balanced equation calls for 2 moles of hydrogen per mole of oxygen, which works out to roughly an 8 : 1 ratio by mass, since oxygen is far heavier per mole. Yet engineers deliberately fed the engines a fuel-rich mixture closer to 6 : 1 by mass, with extra hydrogen. The leftover light hydrogen molecules shoot out of the nozzle faster than heavier exhaust would, squeezing out more thrust — a case where the 'perfect' stoichiometric ratio is knowingly broken to win better performance.
The laws that make the math possible
Stoichiometry only works because compounds have fixed recipes, an idea captured in two foundational laws. Around 1800 the French chemist Joseph Proust established the law of definite proportions: a given compound always contains its elements in the same fixed mass ratio — water is always about 8 parts oxygen to 1 part hydrogen by mass, whether it came from a river or a laboratory flask. Soon after, John Dalton's law of multiple proportions extended the pattern to elements that form more than one compound. Together these laws guarantee that a formula like H₂O means something rigid and reliable, which is the bedrock assumption every stoichiometry calculation quietly stands on.
In the real world
Life-support engineers use stoichiometry to size the lithium hydroxide canisters that scrub exhaled carbon dioxide from the air on submarines and spacecraft — too little and the crew suffocates, too much and mass is wasted.
Exit check
2H₂+O₂→2H₂O: what does 3 mol O₂ require?
Check your explanation after answering
It requires 6 mol H₂ and can produce 6 mol H₂O if reactants are sufficient.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Problem 1
- 3 : 1
- 3 : 2
- 1 : 1
- 2 : 3
Answer and explanation
3 : 2
The coefficients give H₂ : NH₃ = 3 : 2, read straight from the balanced equation.
Problem 2
Answer and explanation
4 mol
The mole ratio of H₂O to H₂ is 2 : 2 = 1 : 1, so 4.0 mol H₂ → 4.0 mol H₂O.
Problem 3
Answer and explanation
6 mol
9.0 mol H₂ × (2 mol NH₃ ÷ 3 mol H₂) = 6.0 mol NH₃.
Problem 4
Answer and explanation
44.01 g
The mole ratio of CO₂ to C is 1 : 1, so 1.00 mol C → 1.00 mol CO₂, and 1.00 mol × 44.01 g/mol = 44.01 g.
Problem 5
Answer and explanation
36.04 g
4.04 g ÷ 2.02 g/mol = 2.00 mol H₂; ratio H₂O:H₂ = 1:1 gives 2.00 mol H₂O; 2.00 mol × 18.02 g/mol = 36.04 g.
Problem 6
Hints
Mole ratios come from coefficients, not molar masses.
Answer and explanation
1.5×2/1=3.0 mol NH₃ theoretical yield.
Problem 7
Hints
State what the supplied observations or conditions establish, and separate that from any assumption.
Answer and explanation
Balanced coefficients relate particle counts/moles, so masses must be converted using each substance's molar mass.
Problem 8
Answer and explanation
Only 8.00 g is available for that reactant’s stoichiometric conversion under the stated mass-fraction model.
A theoretical yield is the product amount predicted by a stated chemical model and available reacting quantities. An impure sample contains less target reactant than its total mass, an insufficient partner can limit conversion, and an equilibrium or competing pathway can reduce the modeled target formation. These are distinct limitations, not reasons to abandon conservation. State the reaction, purity, available amounts and completeness assumptions before presenting a number as an experimental outcome.
Only 8.00 g is available for that reactant’s stoichiometric conversion under the stated mass-fraction model.
Response criteria
- Interpret labels, structures, axes, units and directions correctly.
- Explain the chemistry relationship or mechanism.
- Respect stated assumptions and distinguish observation from inference.
Problem 9
Answer and explanation
4.86/24.3=0.200 mol Mg. The 1:1 ratio gives 0.200 mol MgO and 0.200×40.3=8.06 g MgO. The extra mass comes from incorporated oxygen; the complete mass account includes it.
4.86/24.3=0.200 mol Mg. The 1:1 ratio gives 0.200 mol MgO and 0.200×40.3=8.06 g MgO. The extra mass comes from incorporated oxygen; the complete mass account includes it.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.