How To Calculate Mmol

Dealing with huge numbers can be annoying. The same is true for really small numbers. It can be a lot of zeroes to write out! To deal with this issue, scientists came up with a way to standardize talking about how many atoms or molecules are reacting in a given situation.

The mole (abbreviation mol) is the SI unit for the amount of a substance. It is defined as about 6.022 x 10²³ atoms or particles or things. Saying mol is definitely easier than having to say 6.022 x 10²³.

If someone says there is 1 mol of umbrellas outside, that means there are 6.022 x 10²³ umbrellas. If someone says there is 1 mol of pieces of dust, that means there are 6.022 x 10²³ pieces of dust. It doesn't matter the size or shape of the object. A mol is a mol. It's just a number.

In the same way that a mol is a number, so is an mmol. The "m" in front of mol stands for "milli." Thus, mmol would be said as "millimole." An mmol is a thousandth of a mol:

\(1\enspace mmol\enspace =\enspace 0.001\enspace mol\)

Or:

\(1000\enspace mmol\enspace =\enspace 1\enspace mol\)

If you have 0.33 grams of sodium chloride (NaCl), how many mols is that? How many mmols?

To find the molar mass of sodium chloride, you first go to the periodic table. Underneath the element symbol, you will find the molar mass. For sodium, it is 22.99 g/mol while for chlorine it 35.45 g/mol. Next, you add these two numbers together to get 58.44 g/mol of NaCl.

This means that sodium chloride has a molar mass (mass per mol) of 58.44 g/mol. Now, you can use that for your calculation. But how? How can you find the number of mols of sodium chloride in 0.33 g?

Remember your dimensional analysis. You should recognize that the starting amount has the unit "g" and that the molar mass has the unit "g" as well. The leftover unit is "mol," which is exactly what you want in this case. So how would you put it all together?

\(0.33g\enspace NaCl\Biggl(\dfrac{1\enspace mol\enspace NaCl}{58.44g\enspace NaCl}\Biggr)\enspace =\enspace 5.6\times10^{-3}mol\enspace NaCl\)

You know you set it up correctly because the units cancel. The "g" on the top and the "g" on the bottom cancel to leave you with mol. Then, after doing the math, you know that in 0.33g of NaCl there are 0.0056 mol of NaCl.

Now, how about the mmol of NaCl in 0.33 g of NaCl? You can use the conversion factor shown above to help you out here:

\(0.33g\enspace NaCl\Biggl(\dfrac{1\enspace mol\enspace NaCl}{58.44g\enspace NaCl}\Biggr)\Biggl(\dfrac{1000mmol}{1\enspace mol}\Biggr) =\enspace 5.6mmol\enspace NaCl\)

Since you know that there are 1,000 mmol in 1 mol, you can use that as your conversion factor. Applying that, you see that there are 5.6 mmol of NaCl in 0.33g of NaCl. Saying 5.6 mmol is much easier than 0.0056 mol. That's why using mmol can be much more convenient.

Practice some more of that dimensional analysis with a slightly trickier question: How many mmol of NiSO₄ are there in 10 mL of a 0.1 M NiSO₄ solution?

Step 1: Figure out the units hidden inside "M." Remember M means molar which is mol/L.

Step 2: Figure out how many mL are there in 1 L. There are 1,000 mL in 1 L.

Step 3: Set up the conversion so that you end up with mmol as the only unit that has not canceled out!

Here is how that would look:

\(\Biggl(\dfrac{0.1\enspace mol\enspace NiSO_4}{1L}\Biggr)\Biggl(\dfrac{1000mmol}{1\enspace mol}\Biggr)\Biggl(\dfrac{1L}{1000mL}\Biggr)10mL =\enspace 10mmol\enspace NiSO_4\)

This means that in 10 mL of a 1 M NiSO₄ solution there are 10 mmol of NiSO₄.

When trying to calculate mmol of a substance always remember to set the problem so that you can cross out all the units until you are only left with the unit you want. That way you know you did the problem correctly.