How to find the mole ratio
Three steps, and the middle one is just reading.
- Balance the equation. The coefficients are the ratio, so they have to be right first. See balancing equations if the equation is not balanced yet.
- Read off the two coefficients, in the order the question asks. “The mole ratio of A to B” means A’s coefficient on top, B’s underneath.
- Simplify if you want to. 2 : 2 and 1 : 1 are the same ratio, and you multiply by the same number either way.
For 2 H₂ + O₂ → 2 H₂O, the ratio of H₂O to H₂ is 2 : 2, or 1 : 1. The ratio of H₂ to O₂ is 2 : 1. There is no arithmetic beyond reading the numbers: the coefficients are the ratio.
You will see this called a mole ratio, a molar ratio, a stoichiometric ratio or a mole-to-mole ratio. They all mean this. The tool above does the whole conversion for any pair in any of these reactions, and it is free to embed on a class site or LMS.
Mole ratios for 13 common reactions
Every pair in every reaction on this site, with the coefficients as they stand and in lowest terms.
| Reaction | Mole ratio | From the coefficients | In lowest terms | Multiply by |
|---|---|---|---|---|
| 2 H₂ + O₂ → 2 H₂O Formation of water | O₂ to H₂ | 1 : 2 | already lowest | × 1/2 |
| H₂O to H₂ | 2 : 2 | 1 : 1 | × 1 | |
| H₂O to O₂ | 2 : 1 | already lowest | × 2 | |
| N₂ + 3 H₂ → 2 NH₃ Haber process (ammonia) | H₂ to N₂ | 3 : 1 | already lowest | × 3 |
| NH₃ to N₂ | 2 : 1 | already lowest | × 2 | |
| NH₃ to H₂ | 2 : 3 | already lowest | × 2/3 | |
| 2 Na + Cl₂ → 2 NaCl Formation of table salt | Cl₂ to Na | 1 : 2 | already lowest | × 1/2 |
| NaCl to Na | 2 : 2 | 1 : 1 | × 1 | |
| NaCl to Cl₂ | 2 : 1 | already lowest | × 2 | |
| 2 Mg + O₂ → 2 MgO Burning magnesium | O₂ to Mg | 1 : 2 | already lowest | × 1/2 |
| MgO to Mg | 2 : 2 | 1 : 1 | × 1 | |
| MgO to O₂ | 2 : 1 | already lowest | × 2 | |
| 2 H₂O → 2 H₂ + O₂ Electrolysis of water | H₂ to H₂O | 2 : 2 | 1 : 1 | × 1 |
| O₂ to H₂O | 1 : 2 | already lowest | × 1/2 | |
| O₂ to H₂ | 1 : 2 | already lowest | × 1/2 | |
| CaCO₃ → CaO + CO₂ Decomposition of limestone | CaO to CaCO₃ | 1 : 1 | already lowest | × 1 |
| CO₂ to CaCO₃ | 1 : 1 | already lowest | × 1 | |
| CO₂ to CaO | 1 : 1 | already lowest | × 1 | |
| 2 H₂O₂ → 2 H₂O + O₂ Decomposition of hydrogen peroxide | H₂O to H₂O₂ | 2 : 2 | 1 : 1 | × 1 |
| O₂ to H₂O₂ | 1 : 2 | already lowest | × 1/2 | |
| O₂ to H₂O | 1 : 2 | already lowest | × 1/2 | |
| Zn + 2 HCl → ZnCl₂ + H₂ Zinc in hydrochloric acid | HCl to Zn | 2 : 1 | already lowest | × 2 |
| ZnCl₂ to Zn | 1 : 1 | already lowest | × 1 | |
| H₂ to Zn | 1 : 1 | already lowest | × 1 | |
| ZnCl₂ to HCl | 1 : 2 | already lowest | × 1/2 | |
| H₂ to HCl | 1 : 2 | already lowest | × 1/2 | |
| H₂ to ZnCl₂ | 1 : 1 | already lowest | × 1 | |
| Fe + CuSO₄ → FeSO₄ + Cu Iron in copper(II) sulfate | CuSO₄ to Fe | 1 : 1 | already lowest | × 1 |
| FeSO₄ to Fe | 1 : 1 | already lowest | × 1 | |
| Cu to Fe | 1 : 1 | already lowest | × 1 | |
| FeSO₄ to CuSO₄ | 1 : 1 | already lowest | × 1 | |
| Cu to CuSO₄ | 1 : 1 | already lowest | × 1 | |
| Cu to FeSO₄ | 1 : 1 | already lowest | × 1 | |
| AgNO₃ + NaCl → AgCl + NaNO₃ Silver nitrate + sodium chloride | NaCl to AgNO₃ | 1 : 1 | already lowest | × 1 |
| AgCl to AgNO₃ | 1 : 1 | already lowest | × 1 | |
| NaNO₃ to AgNO₃ | 1 : 1 | already lowest | × 1 | |
| AgCl to NaCl | 1 : 1 | already lowest | × 1 | |
| NaNO₃ to NaCl | 1 : 1 | already lowest | × 1 | |
| NaNO₃ to AgCl | 1 : 1 | already lowest | × 1 | |
| HCl + NaOH → NaCl + H₂O Acid–base neutralization | NaOH to HCl | 1 : 1 | already lowest | × 1 |
| NaCl to HCl | 1 : 1 | already lowest | × 1 | |
| H₂O to HCl | 1 : 1 | already lowest | × 1 | |
| NaCl to NaOH | 1 : 1 | already lowest | × 1 | |
| H₂O to NaOH | 1 : 1 | already lowest | × 1 | |
| H₂O to NaCl | 1 : 1 | already lowest | × 1 | |
| CH₄ + 2 O₂ → CO₂ + 2 H₂O Burning methane | O₂ to CH₄ | 2 : 1 | already lowest | × 2 |
| CO₂ to CH₄ | 1 : 1 | already lowest | × 1 | |
| H₂O to CH₄ | 2 : 1 | already lowest | × 2 | |
| CO₂ to O₂ | 1 : 2 | already lowest | × 1/2 | |
| H₂O to O₂ | 2 : 2 | 1 : 1 | × 1 | |
| H₂O to CO₂ | 2 : 1 | already lowest | × 2 | |
| C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O Burning propane | O₂ to C₃H₈ | 5 : 1 | already lowest | × 5 |
| CO₂ to C₃H₈ | 3 : 1 | already lowest | × 3 | |
| H₂O to C₃H₈ | 4 : 1 | already lowest | × 4 | |
| CO₂ to O₂ | 3 : 5 | already lowest | × 3/5 | |
| H₂O to O₂ | 4 : 5 | already lowest | × 4/5 | |
| H₂O to CO₂ | 4 : 3 | already lowest | × 4/3 |
Why 2 : 2 and 1 : 1 are both right
A ratio does not change when you divide both sides by the same number, so 2 : 2, 1 : 1 and 10 : 10 all say “the same amount of each”. Textbooks and answer keys usually print the lowest-terms version, while the balanced equation in front of you shows the raw coefficients. If your ratio looks different from the book’s, check whether one of you has simplified before assuming you are wrong. The number you multiply by is identical either way.
Using the ratio to hop between substances
A balanced equation is a recipe. A cookie recipe says “2 cups of flour make 1 dozen cookies”; 2 H₂ + O₂ → 2 H₂O says “2 moles of hydrogen and 1 mole of oxygen make 2 moles of water”. Stoichiometry is using those numbers to work out how much.
To hop from a known amount of one substance to an unknown amount of another, multiply by the matching ratio:
moles of want = moles of have × (coefficient of want ÷ coefficient of have)
Put what you want on top. Getting the ratio upside down is the single most common mistake in stoichiometry, and the giveaway is an answer that is out by exactly the ratio, or its square if you flip it twice.
Coefficients count particles, so these ratios are always in moles, never in grams.
Grams and moles: convert with molar mass
Lab measurements come in grams, but the ratio only speaks moles, so you translate with molar mass (grams per mole), summed from atomic masses on the periodic table. Going from grams to moles you divide by molar mass; going back from moles to grams you multiply. Building formulas and counting their atoms in the molecule builder is good practice for getting those molar masses right.
A worked example
How much water forms from 4.000 g of H₂ in 2 H₂ + O₂ → 2 H₂O?
- Grams → moles. The molar mass of H₂ is about 2.016 g/mol, so 4.000 g ÷ 2.016 g/mol = 1.984 mol H₂.
- Mole ratio. From the coefficients, H₂O : H₂ is 2 : 2 = 1, so 1.984 mol × (2 ÷ 2) = 1.984 mol H₂O.
- Moles → grams. The molar mass of H₂O is about 18.015 g/mol, so 1.984 mol × 18.015 g/mol ≈ 35.744 g H₂O.
Only step 2 crosses from one substance to another: that is the mole ratio doing its job. Steps 1 and 3 are pure unit conversion within a single substance.
You can check the answer without redoing it. Running the same three steps for oxygen gives 31.744 g of O₂ consumed, and 4.000 + 31.744 = 35.744 g, the mass of water produced. Mass is conserved in every balanced equation, so if the reactant masses do not add up to the product masses, something has gone wrong earlier.
The road map to remember
Every grams-to-grams problem follows the same path:
grams (given) → moles (given) → moles (find) → grams (find)
Divide by molar mass to enter the world of moles, multiply by the mole ratio to move between substances, then multiply by molar mass to leave it again. Try different reactions and amounts in the tool above and watch the same three steps repeat every time.