Mole Conversion Calculator

The mole stopped being defined by carbon in 2019

For most of the twentieth century the mole was pegged to a mass of carbon — however many atoms happened to sit in exactly twelve grams of carbon-12. That definition made Avogadro's constant an experimental quantity: you had to go and measure it, and the accepted value shifted slightly as the measurements improved.

Since 20 May 2019 the relationship runs the other way. The mole is defined as exactly 6.02214076 × 10²³ elementary entities, full stop. Avogadro's constant is now an exact integer by definition and carbon has nothing to do with it.

The practical effect is almost nothing and the conceptual effect is considerable. A mole of carbon-12 now weighs 11.9999999958 grams rather than exactly 12 — a discrepancy nine decimal places down, which no laboratory balance will ever see, but which means the molar mass constant is an experimentally determined quantity rather than exactly 1 g/mol. If a modern reference states the molar mass constant with an uncertainty and an older one states it as exact, neither is in error; they are quoting different definitions.

One junction, four roads into it

A mole conversion is always a question about the same central quantity — the amount of substance, in moles — approached from one of four directions.

• From a mass, using the molar mass of the substance.
• From a number of particles, using Avogadro's constant.
• From a gas volume, using the molar volume at a stated temperature and pressure.
• From a solution, using concentration and volume together.

Nothing else is going on. Every "how many molecules are in X grams" problem is two of these steps placed end to end, with the amount in moles as the mandatory stopping point in the middle. Attempting to jump from grams straight to particles in one operation is the commonest way to scramble a calculation, because the two conversion factors then arrive in the same line and either can be inverted without it looking wrong.

Twenty-five grams of carbon dioxide, on paper

Take 25.0 g of CO₂ and find the number of molecules, the number of oxygen atoms, and the volume it would occupy as a gas.

Step one — molar mass. Carbon at 12.011 plus two oxygens at 15.999 gives 44.009 g/mol.

Step two — amount.

n = 25.0 g ÷ 44.009 g/mol = 0.568 mol

Step three — molecules. Multiply by Avogadro's constant:

0.568066 × 6.02214076 × 10²³ = 3.42 × 10²³ molecules

Step four — oxygen atoms. Each molecule contains two, so double it: 6.84 × 10²³.

Step five — volume as a gas. At 273.15 K and 100 kPa, one mole of an ideal gas occupies 22.711 L:

0.568066 mol × 22.711 L/mol = 12.90 L

Notice that the unrounded 0.568066 was carried into every later step and the rounding to three significant figures happened only at each reported answer. Rounding the amount to 0.568 before multiplying by Avogadro's constant changes the fourth digit of the molecule count, which is exactly the kind of drift that makes a correct method produce a mismatched number.

Formula units, molecules and atoms are three different counts

More marks are lost here than anywhere else in the topic, and the distinction is not subtle once it is stated.

A mole of a substance is a mole of whatever the formula describes, and the formula does not always describe a molecule.

• A mole of oxygen gas is a mole of O₂ molecules and therefore two moles of oxygen atoms.
• A mole of sodium chloride is a mole of formula units, which dissolves to give one mole of sodium ions and one mole of chloride ions — two moles of ions in total.
• A mole of aluminum sulfate, Al₂(SO₄)₃, releases five moles of ions per formula unit: two aluminum and three sulfate.
• A mole of copper sulfate pentahydrate contains five moles of water, which is mass that counts towards the molar mass and water that appears when the solid is heated.

Whenever a question asks for "particles", read back through it to see which particle it means. The arithmetic is trivial and the reading comprehension is where the error lives.

Why a gas volume answer may not match the book

The molar volume figure used above, 22.711 L/mol, applies at 100 kPa. A great many textbooks quote 22.414 L/mol instead, which is the value at 101.325 kPa — one standard atmosphere. IUPAC changed the recommended standard pressure to 100 kPa in 1982, and the older figure has stayed in circulation ever since.

The gap is about 1.3%, which for the carbon dioxide above is the difference between 12.90 L and 12.73 L. Neither is wrong. If a mark scheme expects one and you used the other, the fix is to state the conditions alongside the answer, which is good practice regardless.

A separate caution: both figures assume ideal behaviour. Real carbon dioxide is measurably non-ideal even at ambient conditions, so a gas-volume answer to four significant figures is false precision no matter which constant you used. And 24.0 L/mol, which appears in some school syllabuses, is the molar volume at 25 °C rather than at 0 °C — a third convention that catches people out because it is quoted without its temperature.

Concentration is the same junction from a different direction

For a solution, amount comes from concentration multiplied by volume, with the volume in cubic decimetres — a litre and a cubic decimetre being the same thing. A 0.150 mol/dm³ solution measured out as 25.0 cm³ contains 0.150 × 0.0250 = 3.75 × 10⁻³ mol. From there the road back to grams or particles is the one already described.

One property of concentration is worth carrying: molarity is defined per volume of solution, and volume changes with temperature. A solution made up at 20 °C is very slightly less concentrated at 40 °C, without a single molecule having gone anywhere. Where that matters — in precise physical chemistry rather than in ordinary bench work — molality, defined per kilogram of solvent, is used instead because mass does not expand.

What the amount in moles is actually for

The reason every stoichiometry problem funnels through moles is that a balanced chemical equation is a statement about numbers of particles and about nothing else. The coefficients say two molecules of this react with one of that; they say nothing about grams, and the masses on either side of an equation are never in the ratio of the coefficients.

So the amount in moles is the only currency in which a reaction can be reasoned about. Convert into it, do the chemistry with the coefficients, convert back out into whatever the question wanted — mass to weigh, volume to measure, or concentration to make up. A conversion error at the front of that sequence produces an answer that is internally consistent, correctly rounded, and wrong, which is why it is worth checking the molar mass before anything else.