Why is weight of 1 mole of substance equal to atomic/molecular mass in grams?

According to me, it happens because mole has been defined in such a way. It is defined as the numbers of particles in $\pu{12 g}$ of $\ce{^{12}C}$. If it were $\pu{24 g}$ instead of $\pu{12 g}$, then the weight of 1 mole of substance would equal 2 times the atomic/molecular mass in grams.

Please correct me if I am wrong somewhere (or maybe everywhere).

  • $\begingroup$ Yes, you are basically correct. $\endgroup$
    – user467
    Commented Jun 6, 2014 at 19:57

5 Answers 5


Why is weight of 1 mole of substance equal to atomic/molecular mass in grams?

According to me, it happens because mole has been defined in such a way.

Yes! That is correct.

It is defined as the numbers of particles in 12 g of C12. If it were 24 g, instead of 12 g, then the weight of 1 mole of substance would equal 2 times the atomic/molecular mass in grams.

Also correct, assuming that the definition of unified atomic mass units (amu) remained the same. @Martin's answer is correct, but we can also arrive at the same conclusion using a simple dimensional analysis approach.

First we need the definition of an amu:

$$\pu{1 atom}~ \ce{^12C} = \pu{12 amu}$$

Now take the real definition of a mole:

$$\pu{1 mol}~ \ce{^12C} = \pu{12 g}$$

Now, divide the first equation by the second:

$$\frac{\pu{1 atom}~ \ce{^12C}}{\pu{1 mol}~ \ce{^12C}} = \frac{\pu{12 amu}}{\pu{12 g}}$$

Cross-multiply and reduce:

$$1~\frac{\pu{g}}{\pu{mol}~ \ce{^12C}} = 1~\frac{\pu{amu}}{\pu{atom}~ \ce{^12C}}$$

What this tells us is that the ratio of g/mol to amu/atom is exactly one - and we made sure it would work out that way by carefully choosing how we defined moles and amus.

Since the masses of all elements are defined in terms of amu, which are ultimately based on the mass relative to carbon-12, this ratio holds for all any atom or molecule.

Let's take it further and put in your hypothetical doubling of the number of grams of carbon-12 per mole:

$$\frac{\pu{1 atom}~ \ce{^12C}}{\pu{1 mol}~ \ce{^12C}} = \frac{\pu{12 amu}}{\pu{24 g}}$$

Now when we cross-multiply and reduce, we get:

$$2~\frac{\pu{g}}{\pu{mol}~ \ce{^12C}} = 1~\frac{\pu{amu}}{\pu{atom}~ \ce{^12C}}$$

Since we kept the definition of an amu the same, but changed that of the mole, what we see is that 1 mole of carbon-12 using these units would have to weigh 24 g (the molar mass), even though its atomic mass would still be 12.

That's a lot harder to keep track of mentally, and so it is a good thing for us that the definitions of amus and moles were chosen so carefully!

  • $\begingroup$ Avogadro's constant is the inverse of 1 amu...How is that related to mole concept? $\endgroup$ Commented May 30, 2014 at 3:40
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    $\begingroup$ The definitions of moles and a.m.u. in combination, yes. $\endgroup$
    – thomij
    Commented May 30, 2014 at 3:44
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    $\begingroup$ The use of the atomic mass unit is deprecated, it should be used the unified atomic mass unit. What is the difference? Amu was based on $\ce{^{16}O}$. $\endgroup$ Commented May 30, 2014 at 4:28
  • $\begingroup$ @Martin you are right about that - typically today, a.m.u. refers to the unified atomic mass unit. I'll edit my answer to reflect that. $\endgroup$
    – thomij
    Commented May 30, 2014 at 11:03
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    $\begingroup$ Actually $\mathrm{u}$ refers to the unified atomic mass, just to avoid confusion. $\endgroup$ Commented May 30, 2014 at 11:27

In short terms:

Your assumption is correct. If the definition of mole would have been based on 24 grams of carbon-12, all molecular weights would double. This is not the case and I highly doubt, that this definition will ever change.

In long terms:

This question is very definition specific. One should not be confused with the several types of definitions. All these definitions are based on the carbon-12 isotope. Hence the atomic mass (constant) has a specific value:

One twelfth of the mass of a carbon-12 atom in its nuclear and electronic ground state, $m_u = 1.660 5402 10~\times~10^{−27}~\text{kg}$. It is equal to the unified atomic mass unit.

The unified atomic mass unit is therefore a derived non-SI unit:

Non-SI unit of mass (equal to the atomic mass constant), defined as one twelfth of the mass of a carbon-12 atom in its ground state and used to express masses of atomic particles, $u\approx 1.660 5402 10~\times~10^{−27}~\text{kg}$.

The atomic weight (i.e. relative atomic mass) is a ratio (=number) and therefore has no unit:

The ratio of the average mass of the atom to the unified atomic mass unit.

Also based on carbon-12 is the definition of mole, which you stated correctly:

SI base unit for the amount of substance (symbol: mol). The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12. When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.

Derived from this is the Avogadro constant:

Fundamental physical constant (symbols: $L$, $N_A$) representing the molar number of entities: $L = 6.022 141 79 30 \times 10^{23}~\text{mol}^{−1}$.

Therfore one can derive several other commonly used properties, such as molecular mass and molar mass. In a biannual publication "Atomic weights of the elements" are published by IUPAC. In Pure Appl. Chem., 2013, Vol. 85, No. 5, pp. 1047-1078 (or here) it says:

The atomic mass, $m_\text{a}$, of an unbound neutral atom of carbon-12, $m_\text{a}(\ce{{}^{12}C})$, in its nuclear and electronic ground states is $12~u$ exactly, where $u$ is the unified atomic mass unit. The atomic weight (also called the relative atomic mass) of isotope $^i\text{E}$ of element $\text{E}$, symbol $A_\text{r}(^i\text{E})$, in material $\text{P}$ is

$A_\text{r}(^i\text{E}) > =\frac{m_\text{a}(^i\text{E})_\text{P}}{\frac{1}{12}m_\text{a}(\ce{{}^{12}C})}=\frac{m_\text{a}(^i\text{E})_\text{P}}{u}$

Thus, the atomic mass of $\ce{{}^{12}C}$ is $12~u$, and the atomic weight of $\ce{{}^{12}C}$ is $12~$ exactly. All other atomic weight values are ratios to the $\ce{{}^{12}C}$ standard value and thus are dimensionless numbers. The atomic weight of element $\text{E}$, $A_\text{r}(\text{E})$, in a material $\text{P}$ is determined from the relation

$A_\text{r}(\text{E})_\text{P}=\sum\left[x(^i\text{E})_\text{P}\times A_\text{r}(^i\text{E})\right]$

where $x(^i\text{E})_\text{P}$ is the mole fraction of isotope $^i\text{E}$ in material $\text{P}$ (also called the isotopic abundance).

Therefore the standard atomic mass of carbon is (given in an interval)

$m_{\text{a}}(\ce{C})=A_\text{r}(\text{C})_\text{graphite}\times u = [12.0096, 12.0116]u$, with


However, the atomic mass of a single molecule is always an integer multiple of $u$.

The molar mass of carbon can then be defined as \begin{aligned} M(\ce{C})&=m_{\text{a}}(\ce{C})\times L\\ &=[12.0096, 12.0116]\times10^{-3}~\text{kg/ mol}\\ &\approx 12.01~\text{g/ mol}~\text{(4 s.f.)} \end{aligned}

The whole relation becomes a little bit clearer when looking at bromine. There are two important Isotopes of bromine: $\ce{^{79}Br}$ and $\ce{^{81}Br}$ (ref). So in elemental bromine ($\ce{Br2}$) the molecules may have three different masses $m_{\text{a}}(\ce{^{79}Br2})=158u$, $m_{\text{a}}(\ce{^{81}Br2})=162u$ and $m_{\text{a}}(\ce{^{79}Br^{81}Br})=160u$.

The standard atomic mass of bromine is $m_{\text{a}}(\ce{Br})= [79.901, 79.907]u$. Therefore $M(\ce{Br})=[79.901, 79.907]\times 10^{-3}~\text{kg/ mol} \approx 79.90~\text{g/ mol}~\text{(4 s.f.)}$ (based on the reference value).

When calculating with molar masses $M$ one always has to keep in mind, that the used standard values are based on a (global) statistic.

Related definitions:

amount of substance, $n$, Also contains definition of: number of moles

Base quantity in the system of quantities upon which SI is based. It is the number of elementary entities divided by the Avogadro constant. Since it is proportional to the number of entities, the proportionality constant being the reciprocal Avogadro constant and the same for all substances, it has to be treated almost identically with the number of entities. Thus the counted elementary entities must always be specified. The words 'of substance' may be replaced by the specification of the entity, for example: amount of chlorine atoms, $n_\ce{Cl}$, amount of chlorine molecules, $n_{\ce{Cl2}}$. No specification of the entity might lead to ambiguities [amount of sulfur could stand for $n_\ce{S}$, $n_{\ce{S8}}$, etc.], but in many cases the implied entity is assumed to be known: for molecular compounds it is usually the molecule [e.g. amount of benzene usually means $n_{\ce{C6H6}}$], for ionic compounds the simplest formula unit [e.g. amount of sodium chloride usually means $n_{\ce{NaCl}}$] and for metals the atom [e.g. amount of silver usually stands for $n_{\ce{Ag}}$]. In some derived quantities the words 'of substance' are also omitted, e.g. amount concentration, amount fraction. Thus in many cases the name of the base quantity is shortened to amount and to avoid possible confusion with the general meaning of the word the attribute chemical is added. The chemical amount is hence the alternative name for amount of substance. In the field of clinical chemistry the words 'of substance' should not be omitted and abbreviations such as substance concentration (for amount of substance concentration) and substance fraction are in use. The quantity had no name prior to 1969 and was simply referred to as the number of moles.

relative molecular mass, $M_r$

Ratio of the mass of a molecule to the unified atomic mass unit. Sometimes called the molecular weight or relative molar mass.

relative molar mass Molar mass divided by $1~\text{g/ mol} (the latter is sometimes called the standard molar mass).


Like "1 dozen", "1 mole" refers to an exact number. Molar mass refers to the mass of a very specific number of molecules/atoms: 1 mole. The molar mass is the mass divided by the amount (number of individual entities such as atoms or molecules) of substance measured in moles.

Now 1 mole is equal of course to Avogadro's constant $\pu{N_A}$. To make matters clear $\pu{N_A = 6.022 \times 10^{23}/mol}$ but it is usually ok to speak of Avogadro's number as equal to the number of particles in 1 mole. It is therefore fair in practice to make the following substitution when performing computations:

$$\pu{1 mole = N_A = 6.022 \times 10^{23}}$$

withthe understanding that $\pu{N_A}$ always refers to the number of entities in 1 mole.

Now lets say we take 1 mole of carbon-12 atoms and determine its mass. It's $\pu{m = 12.0 g}$, which means the molar mass is

$$\pu{M = 12.0 g/1 mole = 12.0 g/mol}$$

But what if we want to know the mass of 1 atom of $\ce{^{12}C}$? Ok, substitute "1 mole" with Avogadro's number. That's just

$$\pu{M_{molecule} = 12.0 g/N_A = 12.0 g/6.022 \times 10^{23} = 1.66\times 10^{-27} kg = 1.66 yg}$$

But $\pu{1.66\times 10^{-27} kg}$ is a really tiny and cumbersome number, and the "yoctogram" is very rarely used. However we can define an atomic mass unit:

$$\pu{1 amu = 1 g/N_A (exactly)} $$


$$\pu{M_{molecule} = 12.0 amu}$$

This is much tidier! But that's the same as before, except that we changed "g/mol" with "amu", so the units are practically exchangeable.

However they are not always interchangeable (sometimes you should be careful): the mass of an atom or molecule refers rather obviously to a single atom or molecule, whereas the mass of a mole is a sum over all species in that mole, which may contain for instance a mix of isotopes. Of course, if you refer to an "average mass" then amu and g/mol are interchangeable.

[3]: Actually, chemical engineers distinguish between lb-mol and g-mol and so forth, but I'm a plain chemist, so I stick to chemist's (or SI) convention.


Take the magnitude of formula mass and append "grams" to obtain the mass of one gram-mole. If you have a formula weight magnitude and append "pounds," it is one pound-mole. One gram-mole contains one Avogadro number of formula units.

Like a "dozen eggs," a mole is simply a bigger basket. It allows you to take microscopic descriptions to real world scales. Moles are scaling factors for thermodynamics, as in molar volumes of gases.

  • $\begingroup$ I do understand what mole is, but why does 1 mole of an atom/molecule weigh the same as the atomic/molecular mass in grams? $\endgroup$ Commented Apr 26, 2014 at 5:31
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    $\begingroup$ That is how it is defined through Avogadro's number as the scale factor for a gram-mole. If the kilogram is redefined as a single crystal silicon ball, Avogadro's constant will be a defined value rather than a measured one. $\endgroup$
    – Uncle Al
    Commented Apr 27, 2014 at 18:02

It is about the definitions, but I suspect you are overcomplicating in thinking it isn't obvious.

A mole is an Avogadro number (6.022 x 1023) of things (the things could be anything but being a big number it is only a useful metric for atoms and molecules). Given this definition, if a molecule weighs the equivalent of a carbon atom then a mole of it will be 12g. A mole of hydrogen gas weighs about (about because it might have some deuterium) 2g as hydrogen gas consists of H2 molecules.

  • $\begingroup$ But why is the molecular weight of hydrogen 2 and it's molar weight also 2g? $\endgroup$ Commented Apr 27, 2014 at 14:42
  • $\begingroup$ One number is the other divided by avogadro's number. Changing the number of molecules or atoms doesn't change the ratio of their atomic or molecular weights. $\endgroup$
    – matt_black
    Commented Apr 27, 2014 at 15:38
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    $\begingroup$ The Avogadro number is based and therefore the result of the exact definition given by the OP. goldbook.iupac.org/M03980.html That is also the reason, why the molecular weights deviate from natural numbers. $\endgroup$ Commented Apr 27, 2014 at 16:14
  • $\begingroup$ 10 years later..."Given this definition, if a molecule weighs the equivalent of a carbon atom then a mole of it will be 12g" Why does the definition of a mole equaling to Avogadro's number necessarily mean the above? I think, after reading through Wikipedia and several answers on this topic, the question remains, WHY is the mole defined as the same numeric value in grams as the atomic mass is in daltons? $\endgroup$
    – suse
    Commented May 3 at 4:55

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