10
$\begingroup$

I don't understand the difference between "molecular mass" and "average atomic mass". They seem like the same thing to me. Is it that average atomic mass is just the weighted average of the "weights"/masses of the isotopes whereas the molecular mass is the average of "weights"/masses of the average atomic masses of each element in the molecule.

Also, what's the difference between molar mass and molecular mass? Is it just that molar mass is expressed in Daltons and molecular mass is expressed in g/mol?

The hardest part about chemistry is keeping track of which people are using which terms, and which terms are outdated. Is this one of these "oh, we use this term now," type of thing?

$\endgroup$
8
  • $\begingroup$ Could you perhaps explain what is confusing you. They seem to be two different concepts to me. $\endgroup$
    – bon
    Sep 29, 2015 at 8:55
  • $\begingroup$ Related: Quick and simple explanation of molar mass, molecular mass and atomic mass and Units of mass on the atomic scale $\endgroup$
    – user7951
    Sep 29, 2015 at 8:57
  • $\begingroup$ @bon... i edited my original question to be more clear. thanks! $\endgroup$
    – kara
    Sep 29, 2015 at 8:59
  • 2
    $\begingroup$ @AstronAUT technically, molar mass refers to the mass per mole (units g/mol), not the mass of one mole (units g) $\endgroup$ Sep 29, 2015 at 10:17
  • 3
    $\begingroup$ @AstronAUT no. your distance per hour is 10 km/h, but the distance you ran in 1 hour is 10 km. $\endgroup$ Sep 29, 2015 at 10:34

3 Answers 3

14
$\begingroup$

Atomic mass refers to the average mass of an atom. This has dimensions of mass, so you can express this in terms of daltons, grams, kilograms, pounds (if you really wanted to), or any other unit of mass. Anyway, as you said, this is an average of the masses of the isotopes, weighted by their relative abundance. For example, the atomic mass of $\ce{O}$ is $15.9994~\mathrm{u}$. $\mathrm{u}$ is short for unified atomic mass unit and 1 u is equivalent to $1.661 \times 10^{-24}~\mathrm{g}$. It is exactly the same as the dalton, but from what I've seen, the term dalton is used more when discussing polymers, biomolecules, or mass spectra.

Molecular mass refers to the average mass of a molecule. Again, this has dimensions of mass. It's just the sum of the atomic masses of the atoms in a molecule. For example, the molecular mass of $\ce{O2}$ is $2(15.9994~\mathrm{u}) = 31.9988~\mathrm{u}$. You don't need to calculate the relative isotopic abundance or anything for this because it's already accounted for in the atomic masses that you are using.

The term molar mass refers to the mass per mole of substance - the name kind of implies this. This substance can be anything - an element like $\ce{O}$, or a molecule like $\ce{O2}$. The molar mass has units of $\mathrm{g~mol^{-1}}$, but numerically it is equivalent to the two above. So the molar mass of $\ce{O}$ is $15.9994~\mathrm{g~mol^{-1}}$ and the molar mass of $\ce{O2}$ is $31.9988~\mathrm{g~mol^{-1}}$.

Sometimes you may come across the terms relative atomic mass ($A_\mathrm{r}$) or relative molecular mass ($M_\mathrm{r}$). These are defined as the ratio of the average mass of one particle (an atom or a molecule) to one-twelfth of the mass of a carbon-12 atom. By definition, the carbon-12 atom has a weight of exactly $12~\mathrm{u}$. This is probably clearer with an example. Let's talk about the relative atomic mass of hydrogen, which has an atomic mass of $1.008~\mathrm{u}$: $$A_\mathrm{r}(\ce{H}) = \frac{1.008~\mathrm{u}}{\frac{1}{12} \times 12~\mathrm{u}} = 1.008$$

Note that this is a ratio of masses and as such it is dimensionless (it has no units attached to it). But, by definition, the denominator is always equal to $1~\mathrm{u}$ so the relative atomic/molecular mass is always numerically equal to the atomic/molecular mass - the only difference is the lack of units. For example, the relative atomic mass of $\ce{O}$ is 15.9994. The relative molecular mass of $\ce{O2}$ is 31.9988.

So in the end, everything is numerically the same - if you use the appropriate units - $\mathrm{u}$ and $\mathrm{g~mol^{-1}}$. There's nothing stopping you from using units of $\mathrm{oz~mmol^{-1}}$, it just won't be numerically equivalent anymore. Which quantity you use (mass/molar mass/relative mass) depends on what you are trying to calculate - dimensional analysis of your equation comes in very handy here.


Summary:

  • Atomic/molecular mass: units of mass
  • Molar mass: units of mass per amount
  • Relative atomic/molecular mass: no units

A small (and unessential) note about the definition of the $\text{u}$. It's defined by the $\ce{^{12}C}$ atom, which is defined to have a mass of exactly $12 \text{ u}$. Now the mole is also defined by the $\ce{^{12}C}$ atom: $12 \text{ g}$ of $\ce{^{12}C}$ is defined to contain exactly $1 \text{ mol}$ of $\ce{^{12}C}$. And we know that one mole of $\ce{^{12}C}$ contains $6.022 \times 10^{23}$ atoms - we call this number the Avogadro constant. That means that $12 \text{ u}$ must be exactly equal to $(12 \text{ g})/(6.022 \times 10^{23})$, and therefore,

$$1 \text{ u} = \frac{1 \text{ g}}{6.022 \times 10^{23}} = 1.661 \times 10^{-24} \text{ g.}$$

$\endgroup$
0
4
$\begingroup$

The IUPAC Gold Book provides the ultimate reference on matters of chemical terminology.

relative atomic mass (atomic weight), $A_\mathrm{r}$
The ratio of the average mass of the atom to the unified atomic mass unit.

The relative atomic mass (average atomic mass as you put it) is the weighted average mass of all the isotopes of an element in a given sample, relative to the unified atomic mass unit, which is defined as one twelfth of the mass of a carbon-12 atom in its ground state.

relative molecular mass, $M_\mathrm{r}$
Ratio of the mass of a molecule to the unified atomic mass unit. Sometimes called the molecular weight or relative molar mass.

This is the sum total of the relative atomic masses of all the atoms in a molecule. For example, $\ce{H2O}$ has a relative molecular mass of $1.008 + 1.008 + 15.999 = 18.015$.

The Gold Book does not have an entry for 'molar mass' but it is a commonly used term.

The molar mass is the mass of a substance divided by its amount of substance (commonly called number of moles). It therefore has units of $\mathrm{mass~(amount~of~substance)^{-1}}$ and is commonly expressed in $\mathrm{g~mol^{-1}}$. The relative atomic or molecular mass is just the molar mass of that substance divided by $\mathrm{1~g~mol^{-1}}$ to yield a dimensionless quantity.

$\endgroup$
0
1
$\begingroup$

Let’s take oxygen ($\ce{O2}$) for example. Examples shall make it easier to understand.

We will use u, kg and g as the units of mass. The full form of u is the Unified atomic mass unit. Commonly people use amu (atomic mass unit) or Da (Dalton) as well. kg is kilogram and g is gram.

1 u = mass of one nucleon (proton/neutron; the constituents of the atomic nucleus). $\pu{1 u} = \pu{1.66 \times 10^{−27} kg}$.

Atomic Mass:
An oxygen molecule is constituted of two oxygen atoms. $\ce{O2}$ is basically $\ce{O=O}$ Atomic mass is the mass of one atom. The mass of one oxygen atom is $\pu{(15.9994 \pm 0.0004) u}$ or roughly $\pu{16 u}$.

Molecular Mass:
Mass of one molecule of oxygen i.e. of one $\ce{O2}$ molecule (the whole $\ce{O=O}$ entity). So the mass of one molecule of oxygen will be $2 \times \pu{16 u} = \pu{32 u}$.

Molar mass:
The mass of one mole of oxygen. 1 mole of oxygen = $\mathrm{6.022 \times 10^{23}}$ number of oxygen molecules.

Let us try to calculate and see how it goes.

1 molecule of $\ce{O2}$ weighs $\pu{32 u} = \pu{32 \times 1.66 \times 10^{−27}kg}$

One mole of oxygen $\mathrm{= 6.022 \times 10^{23}}$ of oxygen molecules So 1 mole of oxygen weighs $\pu{32 \times 1.66 \times 10^{−27} \times 6.022 \times 10^{23}kg} = \pu{0.031988864 kg} = \pu{31.988 g} = \text{ approximately } \pu{32 g}$.

1 mole of oxygen consists of a large number of molecules, hence we switched to a bigger unit (from u to g) for convenience. I hope you get the differences now.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.