Chemistry · Stoichiometry

Amount of Substance (Mole)

The amount of substance n counts particles in portions of moles. Calculated from mass m and molar mass M, n = m/M is the basic formula of every stoichiometric calculation.

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Formula

LaTeX: n = \frac{m}{M}
n in mol · m in g · M in g/mol

Variables & units – Amount of Substance (Mole)

SymbolMeaningUnit
nAmount of substancemol
mMass of the substance portiong
MMolar mass (from the periodic table)g/mol

Derivation & background – Amount of Substance (Mole)

Since the 2019 SI reform the mole has been defined via the Avogadro constant: 1 mol contains exactly 6.02214076×10²³ particles. The amount of substance translates weighable masses into countable particle portions and thus makes reaction equations calculable, because reactions proceed in particle ratios, not mass ratios.

Exam blueprint

Validity range

Applies to any pure portion of substance; M must match the particle type counted (atom, molecule or formula unit). For mixtures it applies only per component.

Derivation steps

The molar mass is defined as mass per amount of substance; solving for n gives the formula.

  1. 1Definition of molar mass: M = m/n links mass and particle portion.
  2. 2Rearranging for the amount of substance gives n = m/M.

Rearrangements

Mass

This is how you calculate the mass to weigh in from a desired amount of substance.

Molar mass

This is how you identify unknown substances from measured values.

Task variant

How many moles are 8.0 g of NaOH (M = 40.0 g/mol)?

n = m/M = 8.0 g / 40.0 g/mol = 0.20 mol.

What is the mass of 0.25 mol of calcium carbonate CaCO₃ (M = 100.1 g/mol)?

m = n·M = 0.25 mol · 100.1 g/mol ≈ 25.0 g.

Common mistakes

Inserting the mass in kg while M is in g/mol.

Match units: insert m in grams or convert M.

Using M for the wrong particle type, for example O instead of O₂.

Always determine M for the actual formula: M(O₂) = 32.00 g/mol.

Equating amount of substance and mass.

n counts particle portions in moles; the mass additionally depends on M.

Exam context

  • Almost every stoichiometry task starts with n = m/M: yields, limiting reactants, gas volumes, titration.

These mistakes cost points in real exams. The set drills them until they stick.

Formula cluster

Stoichiometry backbone

n = m/M, c = n/V and N = n·N_A form the triangle of chemical amount calculations.

Worked example

8.0 g of sodium hydroxide NaOH (M = 40.0 g/mol): n = m/M = 8.0/40.0 = 0.20 mol. Conversely: 0.5 mol of water (M = 18.0 g/mol) weighs m = n·M = 0.5·18.0 = 9.0 g.

Applications

Stoichiometry (reaction equations), preparing solutions, titration, gas calculations, pharmaceutical dosing

Quanta exam set

Curated exam set for "Amount of Substance (Mole)":

Question (front)

Which formula describes Amount of Substance (Mole)?

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Question (front)

How do you rearrange n = m/M for Mass?

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Question (front)

Which common mistake happens with Amount of Substance (Mole)?

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Scientific sources

Common notations & search queries

n=m/Mn = m/Mn m M FormelStoffmenge berechnenMol FormelStoffmenge Formel Chemiemol berechnenamount of substanceMasse molare Masse StoffmengeMol Dreieck

Related formulas

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Frequently asked questions about Amount of Substance (Mole)

How do you calculate the amount of substance from the mass?+

Divide the mass by the molar mass: n = m/M. Insert the mass m in grams and the molar mass M in g/mol from the periodic table, then the amount of substance comes out in moles. Example: 8.0 g of sodium hydroxide NaOH with M = 40.0 g/mol give n = 8.0/40.0 = 0.20 mol. For compounds you first calculate M as the sum of the atomic masses, for NaOH 22.99 + 16.00 + 1.008 ≈ 40.0 g/mol. Make sure mass and molar mass refer to the same particle type: if you want the amount of oxygen gas, use M(O₂) = 32.00 g/mol, not M(O) = 16.00 g/mol.

What exactly is a mole?+

A mole is a counting unit for particles, just as a dozen denotes twelve items. Since the 2019 SI reform, one mole contains exactly 6.02214076×10²³ particles, the Avogadro constant. The trick of this number: one mole of carbon-12 weighs almost exactly 12 g, one mole of water about 18 g. So the molar mass in g/mol has the same numerical value as the atomic or molecular mass in u. This lets you count particles with a balance: weigh out 58.44 g of table salt and you have exactly one mole of NaCl formula units. Chemical reactions proceed in particle ratios, which is why all of stoichiometry calculates in moles.

Why does chemistry calculate with amounts of substance instead of masses?+

Because reaction equations describe particle ratios, not mass ratios. In 2 H₂ + O₂ → 2 H₂O two hydrogen molecules react with one oxygen molecule, but by no means two grams with one gram: by mass it is 4 g of hydrogen per 32 g of oxygen. The amount of substance translates the unwieldy particle number into a weighable quantity and makes the coefficients of the equation directly usable: n(H₂)/n(O₂) = 2/1. The typical exam procedure is therefore: convert the given mass into an amount with n = m/M, apply the ratio of coefficients, and finally translate back into the required mass with m = n·M.

How are amount of substance, mass and particle number related?+

The amount of substance n is the central hub: m = n·M with the molar mass M leads to the mass, N = n·N_A with the Avogadro constant N_A = 6.022×10²³ mol⁻¹ leads to the particle number. Example water: 9.0 g correspond to n = 9.0/18.0 = 0.50 mol, which is N = 0.5·6.022×10²³ ≈ 3.0×10²³ molecules. To go from particle number to mass, always route via the amount of substance: first n = N/N_A, then m = n·M. For gases there is a third bridge: at standard conditions one mole of gas occupies about 22.4 L, so amounts can also be obtained from volumes.

Which mistakes happen most often with n = m/M?+

Three classics. First, mixing units: inserting the mass in kilograms while M is in g/mol puts you off by a factor of 1000; always convert m to grams. Second, the wrong particle type: for chlorine gas M(Cl₂) = 70.9 g/mol applies, not 35.45 g/mol; for salts the complete formula unit counts, including all indices, so 74.10 g/mol for Ca(OH)₂. Third, conceptually confusing amount and mass: 1 mol of lead and 1 mol of helium contain the same number of particles but weigh 207 g and 4 g respectively. A quick plausibility check helps: verify the result unit (mol) and estimate the order of magnitude in your head.

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Create a curated FSRS exam set for n = m/M: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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How do you calculate with Amount of Substance (Mole)?

Here is how to work through a typical Amount of Substance (Mole) (n = m/M) task step by step:

  1. 1

    Task

    How many moles are 8.0 g of NaOH (M = 40.0 g/mol)?

    Solution path

    n = m/M = 8.0 g / 40.0 g/mol = 0.20 mol.

  2. 2

    Task

    What is the mass of 0.25 mol of calcium carbonate CaCO₃ (M = 100.1 g/mol)?

    Solution path

    m = n·M = 0.25 mol · 100.1 g/mol ≈ 25.0 g.