The formula
How to calculate stoichiometry
Stoichiometry converts between masses of reactants and products using the balanced equation. The route is always the same: mass to moles, apply the mole ratio, moles back to mass.
The defaults describe the thermal decomposition of calcium carbonate to calcium oxide and carbon dioxide — a 1:1 ratio, so 25 g of limestone yields about 14 g of quicklime and the rest leaves as gas.
The inputs, one by one:
- Mass of reactant (g)
- Molar mass of reactant (g/mol) — calcium carbonate is 100.09 g/mol
- Coefficient of the reactant
- Coefficient of the product
- Molar mass of the product (g/mol) — calcium oxide is 56.08 g/mol
No submit button: type and the answer moves. Your inputs end up in the link, so the page can be shared already filled in.
The calculation runs on exactly the numbers currently in the fields above, recomputed in full each time — there is no dependency on the order values are entered in, so adjusting one input to test a scenario and then changing it back leaves the result exactly where it started.
Why stoichiometry matters
This kind of calculation comes up in coursework, in a laboratory or field setting, and in professional practice, and the arithmetic is identical in every case — only the numbers being fed into it, and what is riding on getting them right, actually change.
Beyond a single check, the same calculation is worth rerunning whenever a measured input changes — a new reading, a corrected value, an updated assumption — since the result here always reflects exactly what is currently in the fields above rather than a value calculated once and then left stale.
A formula like this one is rarely the last step in a piece of work — the figure it produces usually feeds into a further calculation, a comparison against a published value, or a write-up that needs to state both the result and how confident it is. Getting this step right the first time, rather than propagating a small arithmetic slip through several more steps, is the main practical reason to check a manual calculation against a tool like this one before building on top of it.
In practice, a formula like this one is most often reached for at the exact moment a manual calculation needs checking against a deadline — a lab report due, a problem set to submit — which is precisely the situation where a small arithmetic slip is easiest to miss and most costly to leave uncorrected. Running the same inputs through an independent calculator catches that class of error reliably.
Worked example
Take the figures the calculator starts with:
- Mass of reactant: 25 g
- Molar mass of reactant: 100.09 g/mol
- Coefficient of the reactant: 1
- Coefficient of the product: 1
- Molar mass of the product: 56.08 g/mol
That gives:
- Mass of product: 14.0074 g
- Moles of reactant: 0.2498 mol
- Moles of product: 0.2498 mol
The figures above are the calculator's own default values, shown purely so the working is visible rather than hidden — the same steps apply exactly to your own numbers, entered in the fields at the top of this page.
Reading the result
Check the mass balance. The product mass should never exceed the reactant mass unless something else is being added to the reaction, which the ratio does not know about.
Where this goes wrong. Using an unbalanced equation. The coefficients are the whole calculation, and an equation that is not balanced gives a confidently wrong answer with no warning.
A result that is wrong by an exact factor of ten, a hundred or a similar round number is almost always a units error rather than a mistake in the formula itself — checking each input against the unit stated beside it is the fastest way to track it down.
The ratio of coefficients in the balanced equation. In 2H₂ + O₂ → 2H₂O the ratio of hydrogen to water is 2:2, or 1:1, while hydrogen to oxygen is 2:1.
No, this is the theoretical yield assuming complete reaction and no losses. Multiply by the expected percent yield to estimate what you would actually recover.
The headline figure is mass of product. With 25 g mass of reactant, 100.09 g/mol molar mass of reactant and 1 coefficient of the reactant, that comes to 14.0074 g. Change any field and the figure moves with it.
Generally, no more than the least precise input justifies — a result reported to six decimal places from inputs measured to two significant figures is implying a precision the calculation does not actually have. The calculator shows full precision so you can round appropriately for your own use.
Yes — the equation shown in the formula section above is the standard form used in textbooks and reference material for this calculation, not a simplified or approximate version.
Yes, in the sense that it applies the correct standard formula and returns an accurate result for the inputs given — but check your own course or publication's requirements for how results should be rounded, presented and referenced, since those conventions vary and are not something a calculator can know on your behalf.