The formula
How to calculate moles
The mole is the chemist's counting unit. It bridges the gap between masses you can weigh and the numbers of atoms and molecules that actually react, which always combine in whole-number ratios.
Since the 2019 SI redefinition, the mole is defined as exactly 6.02214076 × 10²³ elementary entities. The old definition, based on 12 grams of carbon-12, gave essentially the same number.
What to enter:
- Mass (g)
- Molar mass (g/mol) — water is 18.015 g/mol
The result updates on every keystroke. The URL updates too, which makes the filled-in version easy to bookmark or send to someone else.
Units matter more here than the arithmetic itself: the formula assumes a specific set of units for each input, stated next to the field, and converting into those units first is usually the difference between a correct result and one that is wrong by a clean power of ten.
Why moles 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.
It is useful for checking a manual calculation before submitting or acting on it, and equally useful for building intuition about a formula by adjusting one input at a time and watching how the result moves in response — a much faster way to understand a relationship than working through several versions of the algebra by hand.
This calculation sits in a long tradition of being done first by hand with tables and slide rules, then with a scientific calculator, and now with a page like this one — the underlying mathematics has not changed at any point in that history, only the speed and convenience of getting from the inputs to the answer. Understanding the formula itself, shown above, is still worth doing even when a tool computes it instantly, since it is what makes the result trustworthy rather than just fast.
Where a calculation like this one is part of a larger piece of work, it is generally worth running it with a round, easy-to-check set of numbers first — inputs of exactly 1, 10 or 100 — purely to confirm the formula is being applied correctly, before switching to the real measured values the actual result depends on.
Worked example
A concrete run-through, using the values already in the fields:
- Mass: 36 g
- Molar mass: 18.015 g/mol
That gives:
- Moles: 1.99833 mol
- Particles: 12.03425 ×10²³
- Volume if a gas at STP: 44.79067 litres
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
The gas volume assumes standard temperature and pressure — 0 °C and 1 atmosphere — where any ideal gas occupies 22.414 litres per mole. At 25 °C the molar volume is closer to 24.5 litres.
Where this goes wrong. Use the correct molar mass for the species in question. Oxygen gas is O₂ at 32 g/mol, not atomic oxygen at 16, and hydrated salts include the water of crystallisation in their formula mass.
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.
Exactly 6.02214076 × 10²³, the Avogadro constant. Since 2019 that is a defined value rather than a measured one.
Add the atomic masses of every atom in the formula, taken from the periodic table. Water is 2 × 1.008 + 15.999 = 18.015 g/mol.
It returns moles. With 36 g mass and 18.015 g/mol molar mass, that comes to 1.99833 mol. 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.