Worked example 1
For N₂ + 3 H₂ → 2 NH₃, how many moles of NH₃ form from 6.0 mol H₂ with excess N₂?
Try it first: Which two coefficients does this question connect?
0 of 3 steps revealed.
Mole ratios/Mass to mass stoichiometryCourse level: High School / HonorsCourse level: AP Chemistry
Mole Ratios and Mass-to-Mass Stoichiometry
What you'll be able to do: Use the coefficients of a balanced equation to convert between amounts of any two substances in a reaction.
Best after: Balancing Chemical Equations, The Mole and Molar Mass
Introduction
A balanced equation is a recipe written in moles. Its coefficients are the only legitimate bridge between one substance and another, and every mass-to-mass calculation in chemistry passes across that bridge. Once you can run the route grams to moles to moles to grams, most quantitative reaction problems become the same problem.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Extract a mole ratio from a balanced chemical equation.
- Convert moles of one substance to moles of another.
- Carry out a complete mass-to-mass stoichiometry calculation.
- Track units through each conversion and report an answer with correct significant figures.
Coefficients are mole ratios
For 2 H₂ + O₂ → 2 H₂O, the coefficients say that 2 mol H₂ reacts with 1 mol O₂ to make 2 mol H₂O. They do not say 2 g reacts with 1 g.
Coefficients count particles, and only moles count particles, so any ratio you take from an equation must be applied to moles.
2 H₂ + O₂ → 2 H₂O
Coefficients are mole ratios, never mass ratios.
The stoichiometry road map
Almost every problem in this lesson follows one path: mass of known → moles of known (divide by M) → moles of unknown (multiply by the mole ratio) → mass of unknown (multiply by M). Draw the road map before you calculate. Knowing which arrow you are on prevents the classic error of multiplying when you should divide.
Write the complete conversion before evaluating it. For 10.0 g Al and excess CuSO₄ in 2 Al + 3 CuSO₄ → Al₂(SO₄)₃ + 3 Cu:
10.0 g Al × (1 mol Al / 26.98 g Al) × (3 mol Cu / 2 mol Al) × (63.55 g Cu / 1 mol Cu) = 35.3 g Cu.
The first factor changes units within Al, the second changes the chemical entity, and the third changes units within Cu. Only the middle factor comes from the equation. Product mass can exceed the supplied aluminum mass because copper atoms come from CuSO₄; conservation concerns all reactants and products together.
munknown = (m_known / M_known) × (coeffunknown / coeff_known) x Munknown
Writing the ratio the right way up
Put the known substance on the bottom of the ratio fraction so its unit cancels. Converting 5.0 mol H₂ to water uses (2 mol H₂O / 2 mol H₂). If your units do not cancel, the fraction is upside down. This single check catches most wrong answers.
You can use the same equation backwards to plan a synthesis. If 10.0 g Cu is required, start with grams of Cu, divide by Cu's molar mass, multiply by 2 mol Al/3 mol Cu, then multiply by Al's molar mass. Reversing which substance is known reverses the relevant ratio, not the underlying equation. Check the intermediate entity after each arrow; simply inverting every numerical value without its unit is unsafe.
Write every conversion as a fraction and cross out units as you go.
Working with volumes and solutions
Only the entry and exit steps change when the question gives a solution instead of a solid. Moles from a solution come from n = c x V with V in litres, and moles of a gas at standard conditions come from the molar volume. The middle mole-ratio step is identical, which is why this road map keeps working in later units.
A single-reactant yield calculation assumes that all other required reactants are available in excess and that the stated reaction proceeds completely. If two finite reactant amounts are supplied, calculate which one limits before reporting a product amount. If an actual experimental yield is requested, a percent yield or measured recovery may also be needed. Stoichiometry predicts the amount allowed by the balanced reaction, not the reaction rate or guaranteed laboratory recovery.
Significant figures and rounding
Coefficients are exact counting numbers and never limit significant figures, and molar masses are ordinarily supplied to enough digits that they do not limit the reported result. The measured quantity in the problem sets the precision.
Carry extra digits through intermediate steps and round only at the end, otherwise a three-step calculation can drift in the last digit.
Rounding at each intermediate step is the most common source of near-miss answers.
Multi-step sequences (Challenge extension)
When a product of one reaction becomes the reactant of the next, run the road map twice and chain the mole ratios: moles A → moles B (equation 1) → moles C (equation 2).
Do not convert back to grams in between unless the question asks for that intermediate mass. This section is optional enrichment beyond the core objectives.
Challenge material, safe to skip on a first pass.
Solutions enter the same mole-ratio route
For an aqueous reactant, multiply molarity (mol/L) by volume in litres to obtain moles before applying the balanced equation. The concentration of a solution is not itself a reaction ratio: 0.200 M means 0.200 mol per litre, while a coefficient compares moles of different substances.
If a gas product is reported as a volume, convert the resulting product moles using the stated temperature and pressure; 22.4 L/mol is only an approximation for an ideal gas at 0 °C and 1 atm.
Keep the units visible so you can see where each kind of ratio enters.
Key ideas
Definition
Mole ratio
The ratio of coefficients between two substances in a balanced equation, used as a conversion factor between their amounts in moles.
Rule
Moles are the only bridge
You cannot convert grams of one substance into grams of another in one step. Every route passes through moles.
Rule
Balance before you calculate
A mole ratio taken from an unbalanced equation is wrong, and the resulting answer is usually off by a small whole-number factor.
Assumption
The reaction goes to completion
Basic stoichiometry assumes complete reaction with the reactant of interest fully consumed. Limiting reactant and yield lessons relax this.
Rule
Coefficients are exact
Equation coefficients are exact counts. Measured inputs and the precision of supplied reference values determine final precision; molar masses are not exact constants.
Equation
Mole ratio conversion
nB = nA x (coeffB / coeffA)
- nA = moles of the known substance
- nB = moles of the substance you want
- coeff = coefficient from the balanced equation (exact)
- use when = moving between two substances in the same reaction
- limits = the equation must be balanced, and A must not be in short supply
Equation
Mass-to-mass road map
mB = (mA / MA) × (coeffB / coeffA) x MB
- mA = mass of the known substance, in g
- MA, MB = molar masses of the two substances, in g/mol
- mB = mass of the wanted substance, in g
- use when = the question gives grams and asks for grams
- limits = assumes complete reaction of A and a pure sample
Worked examples
Worked example 2
How many grams of CO₂ are produced when 25.0 g of C₃H₈ burns completely? C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
Try it first: Sketch the road map grams → moles → moles → grams before calculating.
0 of 4 steps revealed.
Worked example 3
What mass of Al is needed to produce 10.0 g of Cu? 2 Al + 3 CuSO₄ → Al₂(SO₄)₃ + 3 Cu
0 of 3 steps revealed.
Worked example 4
Excess NaHCO₃ reacts with 50.0 mL of 0.200 M HCl. What mass of CO₂ forms? HCl + NaHCO₃ → NaCl + H₂O + CO₂.
Try it first: Start with moles of HCl from M × L.
0 of 3 steps revealed.
Common mistakes
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Applying the coefficient ratio directly to grams. | Coefficients count particles. 2 H₂ + O₂ means 2 mol to 1 mol, which is 4 g to 32 g, not 2 g to 1 g. | Convert to moles before you use any coefficient. |
| Writing the mole ratio upside down. | The known substance's unit fails to cancel, and the answer is wrong by the square of the ratio in direction. | Check that the known unit cancels before evaluating the arithmetic. |
| Reading ratios from an unbalanced equation. | The coefficients are not the true reacting proportions, so every downstream number inherits the error. | Recount atoms on both sides before extracting any ratio. |
| Rounding the moles value before the final step. | Rounding twice in a three-step chain can shift the final digit and fail a tolerance check. | Carry at least one extra digit and round once at the end. |
| Using the molar mass of the known substance on the way out. | The exit step converts the unknown substance, so it needs the unknown's molar mass. | Label each molar mass with its substance as you write it down. |
Scroll horizontally if needed to read all columns.
After the lesson · Question bank
Independent practice
Work through new problems on your own. The lesson's quick checks and worked examples stay above if you need to review a step.
22 topic questions across 3 difficulty levels.
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What you should now know
Convert the known mass to moles, apply the mole ratio from the balanced equation, then convert back to the units the question asks for. Grams never convert directly to grams; the mole ratio is the only step that changes substance.
- Coefficients in a balanced equation are mole ratios, never mass ratios.
- The road map is mass → moles → mole ratio → moles → mass.
- Write ratios so the known unit cancels; if it does not, the fraction is inverted.
- Coefficients are exact; use sufficiently precise molar masses and round the measured result once at the end.
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Sources and further reading
- Chemistry 2e, Section 4.3: Reaction StoichiometryPaul Flowers, Klaus Theopold, Richard Langley, William R. Robinson · OpenStax, Rice University · Chapter 4.3View source
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The current online Chemistry 2e edition lists CC BY-NC-SA 4.0. The source version used for this lesson and its applicable rights remain unverified.
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Recorded attribution / license claim (applicability unverified): Chemistry 2e, OpenStax, Rice University, licensed CC BY 4.0.
Source rights and review status - General Chemistry bookshelf (LibreTexts)LibreTexts, University of California, DavisView source
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