Moles from Mass Calculator n = m/M: Mass, Moles, Particles

The moles from mass calculator turns the grams on a balance into the “portions” of a chemical equation: n = m/M takes a mass m (g) and a molar mass M (g/mol) and returns the amount of substance n (mol); multiplying by the Avogadro constant NA then gives the number of particles N = n NA—how many moles 18 g of water is and how many molecules it contains, how many moles 2.922 g of salt is, and how many atoms 1 g of gold and 1 g of hydrogen each contain. The inverse also works: how many grams is 0.25 mol of salt, and what is the molar mass of a substance when 10 g is 0.2 mol.

Moles from Mass Calculator n = m/M: Mass, Moles, Particles

About this calculator

The moles from mass calculator turns the grams on a balance into the “portions” of a chemical equation: $n = m/M$ takes a mass $m$ (g) and a molar mass $M$ (g/mol) and returns the amount of substance $n$ (mol); multiplying by the Avogadro constant $N_A$ then gives the number of particles $N = n N_A$ —how many moles 18 g of water is and how many molecules it contains, how many moles 2.922 g of salt is, and how many atoms 1 g of gold and 1 g of hydrogen each contain. The inverse also works: how many grams is 0.25 mol of salt, and what is the molar mass of a substance when 10 g is 0.2 mol.

This is the first step of chemical stoichiometry. Reactions proceed by numbers of particles, but a balance only measures mass, and $M$ is the conversion rate between them: grams per mole. What a mole is, why recipes go by number rather than mass, and how the concentration measures convert are set out in full by the chemistry-cluster anchor Molarity Calculator; once $n$ is known, go there to prepare a solution, and to the Dilution Calculator to dilute one. This page covers only its three unique topics: how molar mass is computed from atomic weights, why $N_A$ has been an exact value since 2019, and how the 8 displayed decimals relate to your significant figures.

What it does not do: it does not look up molar masses (add the atomic weights from the chemical formula yourself; this page gives the method and common values); it does not convert kg ↔ g or mg ↔ g; it does not distinguish molecules, ions and atoms— $N$ counts the “elementary entities” represented by the formula; and it does not compute solution concentrations.

Formula and variables

n = \frac{m}{M},\qquad N = n\,N_A,\qquad N_A = 6.022\,140\,76 \times 10^{23}\ \text{mol}^{-1}

Inverse forms:

m = n\,M,\qquad M = \frac{m}{n}

Symbol	Meaning	Calculator field	Unit	Range and notes
$m$	sample mass	Mass m (g)	g	$m \ge 0$ ; default 18. Grams, not kilograms
$M$	molar mass: the mass of 1 mol of the substance	Molar mass M (g/mol)	g/mol	$M > 0$ ; default 18.015 (H₂O). Numerically equal to the relative molecular mass
$n$	amount of substance	Moles n (mol)	mol	output; can also be entered to solve for $m$ or $M$
$N$	number of elementary entities (molecules, atoms, ions or formula units)	Particles N	dimensionless	output; computed with $6.02214076 \times 10^{23}$ and displayed in exponent form
$N_A$	Avogadro constant	(built in)	mol⁻¹	exact value, SI 2019

Dimensions: $\text{g} \div \text{g/mol} = \text{mol}$ ; $\text{mol} \times \text{mol}^{-1} = 1$ . The unit g/mol of $M$ and the dimensionless relative molecular mass share the same numerical value; the reason is under Principles.

How to use this calculator

Open the panelthis page corresponds to “Math tools → Formulas → Amount of substance” (/#/mathtools/formula?calculator=calculator.chem-moles). On open, “Mass m” = 18 and “Molar mass M” = 18.015 are already filled in, “Moles n” carries a “calculate” marker and shows 0.99916736, and “Particles N” shows 6.017126e+23—18 g of water is slightly less than 1 mol (see Example 1).
Find the amount of substance and the particle countchange the mass or the molar mass and both outputs update together. Compute the molar mass by adding the standard atomic weights according to the chemical formula (a table of common values is under Principles).
Solve for massclear “Mass m”, enter $n$ and $M$ , and the mass field switches to “calculate” and shows $nM$ ; the particle count is given at the same time from the $n$ you entered.
Solve for molar massclear “Molar mass M”, enter $m$ and $n$ , and the field shows $m/n$ —useful for identifying an unknown or checking a formula.
Entering $n$ aloneclear $m$ and $M$ and enter only $n$ ; the particle count $= nN_A$ is correct, but the mass and molar mass fields show 0 (information is missing).
Do not fill in all threeif $m$ , $M$ and $n$ are all inputs, the engine computes $N$ from the $n$ you entered and ignores $m/M$ —and does not warn you if the two disagree. To recalculate $n$ , clear it. Similarly, with only $m$ filled in and $M$ cleared, both $M$ and $n$ display the same number as $m$ (the denominator falls back to 1); that is not a result.
Read the results $n$ , $m$ and $M$ keep at most 8 decimal places with trailing zeros removed; because $N \ge 10^{12}$ it is displayed in exponent form with 7 significant figures (for example 6.017126e+23, i.e. $6.017126 \times 10^{23}$ ).
Copy the formulathe copy button at the right of the panel title puts the LaTeX n = \frac{m}{M} on the clipboard.

Worked examples

All four examples were recomputed by the engine's chem-moles compute routine; the display convention is at most 8 decimal places with trailing zeros removed for $n$ , $m$ and $M$ , exponent form with 7 significant figures for $N$ , in units g, g/mol and mol.

Example 1: why 18 g of water is not exactly 1 mol

Default inputs $m = 18$ , $M = 18.015$ : $n = 18 / 18.015 = 0.99916736$ , $N = 0.99916736 \times 6.02214076 \times 10^{23} = 6.017126 \times 10^{23}$ . Interface shows: moles n 0.99916736, particles N 6.017126e+23.

Change the mass to 18.015: the interface shows n 1 and N 6.022141e+23—exactly 1 mol, and the particle count is $N_A$ . The 0.08% gap between 18 and 18.015 comes from the molar mass of H₂O being $2 \times 1.008 + 15.999 = 18.015$ , not a whole number: the atomic weights of hydrogen and oxygen are not integers (isotope mixtures and nuclear binding energy), so “18 g of water = 1 mol” is a textbook rounding.

Those $6.017 \times 10^{23}$ molecules fit in one large tablespoon. Counting them at a billion per second would take about 19 million years—which is why chemistry invented the mole, a “big dozen” that scales numbers down to something a person can read.

Figure 1: the two-step conversion of Example 1. On the left is what can be weighed, on the right what can be combined; M and N_A are the two conversion rates

Example 2: salt and a sugar cube—from solution preparation back to the balance

2.922 g of salt (the amount to weigh out in Example 2 of the Molarity Calculator), $M_\text{NaCl} = 22.990 + 35.45 = 58.44$ . $m = 2.922$ , $M = 58.44$ : the interface shows n 0.05 and N 3.011070e+22— $3.0 \times 10^{22}$ NaCl formula units, that is $3.0 \times 10^{22}$ Na⁺ plus the same number of Cl⁻. NaCl is an ionic crystal and has no “NaCl molecules”; $N$ counts formula units.

Solving for mass: how many grams is 0.25 mol of salt? Clear $m$ , enter $n = 0.25$ , $M = 58.44$ : $m = 14.61$ . Interface shows: mass m 14.61, N 1.505535e+23.

A sugar cube: 4 g of sucrose C₁₂H₂₂O₁₁, $M = 12 \times 12.011 + 22 \times 1.008 + 11 \times 15.999 = 342.297$ . $m = 4$ , $M = 342.297$ : the interface shows n 0.01168576 and N 7.037328e+21. For the same few grams, larger molecules mean fewer moles—this is why, in Example 3 of Molarity, “9 g of glucose is 0.05 mol while 9 g of salt is 0.154 mol”.

Example 3: the same 1 g, 200 times the atom count

1 g of gold, $M_\text{Au} = 196.96657$ : $m = 1$ , $M = 196.96657$ : the interface shows n 0.005077 and N 3.057443e+21.

1 g of hydrogen atoms, $M_\text{H} = 1.008$ : the interface shows n 0.99206349 and N 5.974346e+23. For the same 1 g, the number of hydrogen atoms is about 195 times the number of gold atoms, because a single gold atom is about 195 times heavier. $n \propto 1/M$ —with the mass fixed, the amount of substance is inversely proportional to the molar mass.

100 g of copper, $M_\text{Cu} = 63.546$ : the interface shows n 1.57366317 and N 9.476821e+23.

44 g of carbon dioxide, $M = 12.011 + 2 \times 15.999 = 44.009$ : $m = 44$ : the interface shows n 0.9997955 and N 6.020909e+23—the same “just short” of 1 mol as with 18 g of water. Only $m = 44.009$ gives exactly 1.

Example 4: solving for molar mass, grams and kilograms, and a display trap

Identifying an unknown: 10 g of a sample measures 0.2 mol. Clear $M$ , enter $m = 10$ , $n = 0.2$ : $M = 50$ . Interface shows: molar mass M 50, N 1.204428e+23. A value of 50 g/mol rules out most candidate formulas.

Mass from $n$ : how heavy is 2 mol of water? Clear $m$ , enter $n = 2$ , $M = 18.015$ : the interface shows m 36.03 and N 1.204428e+24.

The kilogram trap: entering 18 g as 0.018 (kg): $m = 0.018$ , $M = 18.015$ : the interface shows n 0.00099917 and N 6.017126e+20—1000 times too small. The field unit is g.

The display trap: in the default state ( $m = 18$ , $M = 18.015$ ), also enter 2 in “Moles n”; all three fields are then inputs and the particle count becomes 1.204428e+24—the engine computes $N$ from the 2 mol you entered and no longer looks at 18/18.015. It will not tell you that the three numbers contradict one another.

The inputs of Example 1 can be entered directly into the panel to reproduce it; change the mass to 18.015 and watch n become 1 and N become N_A; change the molar mass to 58.44 (salt) or 196.96657 (gold) and watch the number of moles in the same 18 g fall in inverse proportion to M.

Principle and derivation

How molar mass is computed

$M$ equals the sum of the standard atomic weights of all the atoms in the chemical formula, in g/mol. Standard atomic weights are maintained by the IUPAC Commission on Isotopic Abundances and Atomic Weights (CIAAW); common values:

Element	Conventional atomic weight	Element	Conventional atomic weight
H	1.008	Na	22.990
C	12.011	Cl	35.45
N	14.007	Cu	63.546
O	15.999	Au	196.966 570

So H₂O $= 2 \times 1.008 + 15.999 = 18.015$ , NaCl $= 22.990 + 35.45 = 58.44$ , CO₂ $= 12.011 + 2 \times 15.999 = 44.009$ , C₆H₁₂O₆ $= 6 \times 12.011 + 12 \times 1.008 + 6 \times 15.999 = 180.156$ . Atomic weights are not integers because a natural element is a mixture of isotopes (chlorine is about 76% ³⁵Cl and 24% ³⁷Cl, averaging 35.45) and because nuclear binding energy makes a nucleus slightly lighter than the sum of its nucleons. Heavy water D₂O has $M = 20.03$ : the same formula with a different isotope changes the molar mass.

Why g/mol and relative molecular mass share a number

The scale of relative atomic mass is “one twelfth of the mass of a ¹²C atom is 1”. Between 1971 and 2019 the mole was defined as “the number of atoms in 0.012 kg of ¹²C”, and together the two stipulations guaranteed that the mass of 1 mol of any substance in grams is numerically equal to its relative molecular mass. This is why the mole was given its size—not a natural constant but a human choice aligning the gram with relative mass. After 2019, when $N_A$ became exactly defined, the equality is exact only to the $10^{-9}$ level, which affects no practical calculation.

$N_A$ : from measured value to defined value

In 1811 Avogadro proposed that equal volumes of gases at the same temperature and pressure contain the same number of molecules, but he did not know what that number was. In 1909 Perrin measured $6.5$ – $7 \times 10^{23}$ by several independent methods including Brownian motion, named it “Avogadro's number”, and provided decisive evidence that molecules are real (the 1926 Nobel Prize in Physics). Over the following century $N_A$ was measured ever more precisely, finally reaching a relative uncertainty of $10^{-8}$ through X-ray density measurements of silicon-28 single-crystal spheres; the 9th edition of the SI, in force from 20 May 2019, simply defined it as the exact $6.022\,140\,76 \times 10^{23}\ \text{mol}^{-1}$ , so the mole no longer depends on the 12 g of carbon-12. That is the value built into the engine.

Why the two conversion steps are kept separate

$N = m N_A / M$ could be computed in one step, but chemistry almost never needs $N$ —equation coefficients are mole ratios, and solution preparation, titration and gas volumes (22.4 L/mol) all use $n$ . The purpose of $N$ is to let you see how large a mole is: a count of order $10^{23}$ is the direct expression of the fact that macroscopic matter is made of microscopic particles. The engine's convention for $N$ reflects this: it takes “the $n$ you entered, otherwise $m/M$ ”, so $n$ is the primary result and $N$ a secondary conversion.

Significant figures

The note after the examples already said it: the 8 displayed decimals are a software convention. A stoichiometric result is reported to the least number of significant figures among the inputs; the molar mass is taken one digit more precise than the mass. Since 2009 IUPAC has given atomic-weight intervals rather than single values for elements including H, Li, B, C, N, O, Si, S, Cl and Tl, because samples from different sources genuinely differ in isotopic composition; conventional values suffice for routine calculations, and only isotope analysis needs the intervals.

Assumptions

The sample is a pure substance: all of $m$ belongs to the substance named by the formula; mixtures must first be split by composition.
Molar mass matches the formula: $M$ is computed from the actual formula, including water of crystallisation (for example CuSO₄·5H₂O is 249.68).
Natural isotopic abundance: standard atomic weights assume terrestrial natural abundance; isotopically enriched samples are computed separately.
$N$ counts formula units: for ionic compounds the ion count must be multiplied by the number of ions in the formula.
SI units: g, g/mol, mol; no conversion is performed.
Interface conventions: $N$ prefers the $n$ you entered; with only $m$ filled in, $M$ and $n$ fall back to a denominator of 1 and display the same number as $m$ ; $N$ is shown in exponent form with 7 significant figures.

Scope and limitations

Can calculate: $n$ and $N$ from $m$ and $M$ ; $m$ from $n$ and $M$ ; $M$ from $m$ and $n$ ; any positive numbers.
Approximate only: the display precision exceeds any real weighing precision; conventional and interval atomic weights can differ in the 4th–5th significant figure.
Cannot calculate: the molar mass itself (add the atomic weights from the formula); the “average molar mass” of a mixture (a weighted average, by hand); gas volumes (use the ideal-gas equation); solution concentrations (use the molarity calculator).
Does not: convert units (kg, mg, μg); parse chemical formulas.
Numerics: $n$ to 8 decimal places with trailing zeros removed; $N$ in exponent form; no consistency check when $m$ , $M$ and $n$ are all filled in.

Common pitfalls

Entering the mass in kg: 0.018 gives 0.001 mol (Example 4). The field is in g.
Taking “18 g of water = 1 mol” as exact: 18/18.015 = 0.99917 (Example 1); textbooks round.
Missing a subscript in the molar mass: H₂O has two hydrogens (2 × 1.008), and Ca(OH)₂ has two O and two H.
Forgetting water of crystallisation: the $M$ of CuSO₄·5H₂O is 249.68, not 159.61.
Treating $N$ as the ion count: for 0.05 mol of NaCl, $N$ is the number of formula units, and the total number of ions is twice that (Example 2).
Filling all three fields: $N$ is computed from the $n$ you entered, with no warning when it contradicts $m/M$ (Example 4).
Reading 8 decimal places as 8 significant figures: the precision of the result is set by the balance and the atomic weights (see the note).
Giving relative molecular mass a unit: relative molecular mass is dimensionless; molar mass is the one in g/mol, and the two share a numerical value.

Typical use cases

Teaching

The size of the mole and the history of $N_A$ ; computing molar mass from atomic weights; significant figures and display precision; isotopes and non-integer atomic weights.

Laboratory weighing

Solving for the grams to weigh out from a target amount of substance (then going to the molarity calculator to prepare the solution); computing the moles added from a weighed mass for a reaction recipe.

Stoichiometry and yields

Turning reactant grams into mole ratios, identifying the limiting reagent; converting theoretical yields from moles back to grams.

Analysis and identification

Recovering a molar mass from a known mass and a measured amount of substance and comparing it with candidate formulas; computing isotopically labelled samples from their actual $M$ .

FAQ

Why does 18 g of water show 0.99916736 rather than 1?

The molar mass of water is 18.015 g/mol, not 18. Only 18.015 gives exactly 1 mol.

Where do I look up molar mass?

Add the standard atomic weights (IUPAC CIAAW) of each element in the formula; a table of common values is under Principles. Reagent bottle labels and databases also give it directly.

Why is the particle count displayed as 6.017126e+23?

Above $10^{12}$ the interface uses exponent form with 7 significant figures; that is $6.017126 \times 10^{23}$ .

Does N count molecules or atoms?

It counts the elementary entities represented by the formula: molecules for H₂O, atoms for Au, and Na⁺Cl⁻ formula units for NaCl. For the total number of atoms, multiply by the number of atoms in the formula (1 mol of H₂O contains 3 mol of atoms).

Did the 2019 redefinition of the mole change the results?

Not perceptibly. $N_A$ went from a measured to an exact value, differing from the old value by of order $10^{-9}$ .

Can I enter the mass in mg?

Not directly; convert to g first (1 mg = 0.001 g). A resulting amount in mol can also be converted to mmol (multiply by 1000).

What happens if all three fields are filled in?

The engine does not recalculate $n$ , and the particle count is given from the $n$ you entered. To let $n$ follow from $m/M$ , clear $n$ .

References and further reading

The chem-moles definition in the CalcX engine source src/data/formulasScience.ts (compute(known, solveFor) for n / m / M, with N computed as (n ?? m/M) × 6.02214076e23); every figure on this page was recomputed by that engine.
BIPM, The International System of Units (SI), 9th edition （访问日期：2026-09-09）—the 2019 edition; the current definition of the mole and the fixed value of $N_A$ .
NIST, CODATA 2018: Avogadro constant （访问日期：2026-09-09）— $N_A = 6.022\,140\,76 \times 10^{23}\ \text{mol}^{-1}$ (exact value).
IUPAC CIAAW, Standard Atomic Weights （访问日期：2026-09-09）—the source of the atomic-weight table on this page, including the note on interval atomic weights.
Wikipedia, Molar mass （访问日期：2026-09-09）—the relationship between molar mass and relative molecular mass, and how to compute it.
Wikipedia, Avogadro constant （访问日期：2026-09-09）—the history from Perrin's 1909 measurement to the 2019 defined value.
Wikipedia, Amount of substance （访问日期：2026-09-09）—amount of substance as a base quantity and the meaning of “elementary entities”.

Privacy

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Sources & review

Reviewed by: CalcX 编辑组
Updated: 2026-09-09

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