# What is the approximate mass of one water molecule?

Oct 29, 2015

You can determine that by using the molar mass in $\text{g/mol}$ of the water molecule and converting so that you get the absolute mass in $\text{g}$.

The molar mass is $\text{MM"_("H"_2"O")" = 18.015 g/mol}$.

If you don't remember that, you can also look it up on Wikipedia, or refer to a periodic table and add up the molar masses of two hydrogen atoms and one oxygen atom.

$\text{MM"_("H"_2"O") = 2*"1.0079 g/mol H" + "15.999 g/mol O}$

$= \text{18.015 g/mol H"_2"O}$

Now we just need the relationship between the units $\text{mol}$ and $\text{number of things in a mol}$.

You may have been taught Avogadro's number, which is $6.0221413 \times {10}^{23} \text{things/mol}$, and you can use that for the stoichiometric conversion. You multiply by the reciprocal to cancel out the units because the goal for the final units is "absolute mass"/("molecule H"_2"O").

$\left(\text{18.015 g H"_2"O")/cancel("mol H"_2"O") xx cancel("1 mol H"_2"O")/(6.0221413xx10^(23) "molecules of H"_2"O}\right)$

$= \textcolor{b l u e}{2.9915 \times {10}^{- 23} \text{g/molecule H"_2"O}}$