# How are the molar mass and molecular mass of any compound numerically the same?

This observation is really annoying me, and the internet isn't providing me with any solid answers. Either their definitions of molar mass completely differ, or they don't stick to their own definition while actually calculating the molar mass of a certain element or compound.

Now, according to Wikipedia,

The molar mass M is a physical property defined as the mass of a given substance (chemical element or chemical compound) divided by the amount of substance.

and it's unit is

g/mol

My understanding of this is that we're calculating the total mass of all the atoms(or molecules) present in 1 mole of that element(or compund). This means that we would have to take the atomic mass of that element and multiply it by the number of atoms present in 1 mole of that element, which is 6.022 × 10^23 atoms.

So for example, the relative molecular mass of water is 18.02 u, and it's almost the same as the molar mass of water, which is 18.015 g/mol.

So my question is, how is 18.02 u × (6.022 × 10^23) = 18.015 g/mol?

• Approximations? Conceptually, they should be identical - considering the same material, i.e. either the relative molecular mass and the relative molar mass, or both isotopic. Aug 19, 2018 at 9:46
• – user7951
Aug 19, 2018 at 9:51

So for example, the relative molecular mass of water is 18.02 u, and it's almost the same as the molar mass of water, which is 18.015 g/mol.

So my question is, how is 18.02 u × (6.022 × 10^23) = 18.015 g/mol?

That is not quite correct.

The relative molecular mass $M_\mathrm r$ of water is a dimensionless quantity, i.e. the unit for relative molecular mass is the unit one and not $\mathrm u$:

$$M_\mathrm r=18.015\,28$$

It's the average mass per molecule $m_\mathrm f$ that has the unit $\mathrm u$:

$$m_\mathrm f=18.015\,28\ \mathrm u$$

Note that

$$1\ \mathrm{u}=1.660\,538\,921(73)\times10^{-27}\ \mathrm{kg}$$

Also note that you are missing the unit of the Avogadro constant:

$$N_\mathrm A=6.022\,140\,857(74)\times10^{23}\ \mathrm{mol^{-1}}$$

Thus

\begin{align} M&=m_\mathrm f\cdot N_\mathrm A\\[6pt] &=18.015\,28\ \mathrm u\times N_\mathrm A\\[6pt] &=18.015\,28\times1.660\,538\,921\times10^{-27}\ \mathrm{kg}\times N_\mathrm A\\[6pt] &=18.015\,28\times1.660\,538\,921\times10^{-27}\ \mathrm{kg}\times6.022\,140\,857\times10^{23}\ \mathrm{mol^{-1}}\\[6pt] &=0.018\,015\,28\ \mathrm{kg\ mol^{-1}}\\[6pt] &=18.015\,28\ \mathrm{g\ mol^{-1}} \end{align}

• Note that this answer was valid until the redefinition of the SI base units came into force on 20 May 2019.
– user7951
Sep 7, 2019 at 20:59