Ever wonder
moles and molecules and atoms
Avogadro's Number

(C) - Copyright, 2000 F.W. Boyle, Jr., Ph.D.

Have you ever wondered about all these things or at least have you ever been frustrated with homework problems that want you to calculate, manipulate and otherwise abuse your mind?

Then I hope this works for you!

Of course, the mole is a chemistry counting unit and is equal to Avogadro's Number (not that Avogadro guessed the number; we just honored his contributions to chemistry by naming the number after him.) Avogadro's Number (A.N. for short) has a magnitude (this means numerical value) of 6.022 x 1023. This is THE number of parts (usually atoms or molecules) that must be gathered together to have what we chemists call a mole of "stuff". Stuff is my way of saying "matter".

So if one could count A.N. of atoms of any element (Yep! The number stays the same regardless of the element, dealing with atoms or molecules dealing with compounds.) and could place those 6.022 x 1023 atoms in a pile, one would have a mole of that "stuff".

Here are some examples:

6.022 x 1023 atoms of silver (Ag - argentum, a latin term)


6.022 x 1023 atoms of sulfur (S)


6.022 x 1023 atoms of helium (He)


6.022 x 1023 atoms of gadolinium (Gd)


6.022 x 1023 atoms of uranium (U)


6.022 x 1023 atoms of silicon (Si)

Of course, you have read my paper on counting this many atoms and the time needed to do so. ;o) You recognize that no one could ever physically count to A.N. (Let alone count 6 million ballots by hand within a reasonable amount of time.)

I will come back to what chemists do to "count" out moles or substances but first lets look at some mathematical manipulations of moles, molecules, atoms, and A.N.

One of the simplest ways to deal with multiples and parts of moles is to use ratios. We all ready know one side of every ratio regardless of element and that is:

1 mole
6.022 x 1023 atoms

Now lets suppose (just because I want to) that we need to determine how many atoms there are in several different numbers of moles.

How many atoms are in 3 moles of Ag (silver)?

1 mole 3 moles
------------------ = ------------------
6.022 x 1023 atoms X atoms

Solving for X, we find it to be 18.066 x 1023 which converted to proper scientific notation is 1.8066 x 1024 atoms of Ag.

How many moles are in 0.37 mole of Cu (copper)?

1 mole 0.37 moles
------------------ = ------------------
6.022 x 1023 atoms X atoms

And X is found to equal 2.228 x 1023 atoms of Cu.

Note that only the names were changed to protect the innocent. Sorry, my fingers slipped. Note that the only change was to replace the mole value on the right side of the equality.

This same ratio can be used to solve for the number of moles that are equal to a given number of atoms. Here are a couple of examples.

How 3.011 x 10 22 atoms of Na (sodium) is equal to how many moles of Na?

1 mole Y moles
------------------ = ------------------
6.022 x 1023 atoms 3.011 x 1022 atoms

We solve for Y and find that 3.011 x 1022 atoms is equal to 0.05 moles or 5.0 x 10-2 moles of Na.

How many moles of Ne (neon) contain 3.6132 x 1023 atoms of Ne?

1 mole Y moles
------------------ = ------------------
6.022 x 1023 atoms 3.6132 x 1023 atoms

Completing the mathematics, one finds Y equal 6 moles.

Once again, the above ratio equality can be used for any calculation of moles for any element.

Now, you say what about those darn compounds. They have to be harder to deal with than this. Compounds have so many different elements in them. I reply, "No, the same ratios above works because A.N. is 6.022 x 1023 molecules when dealing with compounds. (I won't confuse you at this point by adding to the discussion the concept of formula units. They are really the same thing as a molecule only different. ;o)

How about some examples you exclaim!!!

The first question is: How many molecules are there in 5.35 moles of sucrose (table sugar, C12H22O11?

Of course, this must be a hard one, sucrose is such a "big" molecule. But the answer is, "No way! Hombre"

1 mole 5.35 moles
----------------------- = ------------------
6.022 x 1023 molecules X molecules

Multiplying and solving for X (just like we did for atoms), we find that 5.35 moles of sucrose (or any other substance) is equal to 32.218 x 1023 molecules of sucrose (or any other substance) which converts in scientific notation to 3.2218 x 1024 molecules.

Need I do more?? I think not!

A place where these calculations sometimes really get confusing is when the question, how many atoms of H at there in 5.35 moles of sucrose? "Whoa, you say! What are you trying to do to me?" How can a find the number of atoms of a substance when we are talking about molecules.

Just remember that all molecules are atoms stuck (chemists call it bonded) together in a semipermanent form. In sucrose as in all compounds, the subscript numbers following each atom tell us how many aomts of that particular element are in a single molecule of a particular substance.

So lets try figuring out how many atoms of each element are in a single molecule of sucrose. The formula is C12H22O11. We note that the subscript number are 12 for C (carbon), 22 for H (hydrogen) and 11 for O (oxygen). Translating to English, the formula tells us that there are 12 atoms of C, 22 atoms of H, and 11 atoms of O in each and every molecule of sucrose. And you probably ask, "So what?"

This means that once we know the number of molecules of sucrose in a given number of moles of sucrose, we can quickly calculate the number of each kind of atom.

From above, we find that 5.35 moles of sucrose contain 3.2218 x 1024 molecules of sucrose. We can set up another ratio like so.

For C:

12 atoms C W atoms C
------------------ = --------------------------
1 molecule sucrose 3.2218 x 1024 molecules

Solving for W, one finds that there are 38.6616 x 1024 pr 3.86616 x 1025 atoms of C (carbon) in 5.35 moles of sucrose. In order to calculate H atoms, one substitutes the value of 22 atoms H. For O, the substitution is 11 atoms O. Solve for W and you have the answer. The key is knowing the chemical formula and calculating the total number of molecules using A.N..

I hope this helps. Good luck on your studies.