I have trouble relating to the mole concept. For example in a question like the following:

How many moles of $\ce{K}$ are in 1 mole of $\ce{KOH}$?

How is it that in 1 mol of $\ce{KOH}$ there is 1 mol of $\ce{K}$ and x moles of $\ce{OH}$? How is it that 1 mol of $\ce{KOH}$ has 6.02×1023 atoms, and then $\ce{K}$ has 6.02×1023 atoms as well. In addition, $\ce{OH}$ will have 6.02×1023 atoms, but if we add these two it's greater than the original number of atoms in 1 mol of $\ce{KOH}$?

  • $\begingroup$ 1 Mol of KOH has 3*6.02*10^23 atoms, because there are three atoms in KOH. $\endgroup$ – Lighthart Feb 26 '15 at 5:37
  • $\begingroup$ Welcome to Chemistry.SE. Have a look at the tour to get familiar with this site. If necessary, mathematical expressions and equations can be formatted using $\LaTeX$. Please note that there is a homework policy here. $\endgroup$ – Klaus-Dieter Warzecha Feb 26 '15 at 5:44
  • $\begingroup$ hmm, ok so if I'm looking at say 1 mol of H2SO4, then the total atoms would be 3*6.02*10^23 as well? or would it be *7 instead of *3? | why is it that people say 1 mol = 6.02*10^23 if actually the KOH would contain 3 * as many total atoms in 1 mol? $\endgroup$ – user14652 Feb 26 '15 at 5:45

I had the same problem as well, until I sketched out the actual molecules.

When you have one mole of $\ce{KOH}$, you have one mole of the collective "molecule".

If you were to break it down into $\ce{K}$ and $\ce{OH}$, you would get 1 mole of $\ce{K}$, and 1 mole of $\ce{OH}$. This is because you need 1 mole of each to make 1 mole of $\ce{KOH}$.

It's kind of like if you have one hundred keyboards, you would get one hundred 'Enter' keys, one hundred 'Up arrow key' and so on...

  • $\begingroup$ ah ok, an interesting analogy; so when someone says i have 1 mol of KOH, it's not actually 6.02*10^23 total but 3*avg constant, and each atom is 1 mol = 6.02*10^23 combining towards the total? $\endgroup$ – user14652 Feb 26 '15 at 5:48
  • $\begingroup$ The concept of mole is just a number. When we say we have 1 mol of KOH, it means exactly that, there are 1 mole of KOH, not 1 mole of atoms. (100 keyboards, not 100 keys). $\endgroup$ – quidproquo Feb 26 '15 at 5:50
  • $\begingroup$ Ok, this is starting to make a bit more sense. 1 mol KOH = 1 mol K 1 mol OH (1 keyboard, 1 f key, 1 g key) 1 mol H2SO4 = 2 mol H 1 Mol SO4 (3 moles of ions) (1 custom keyboard = 2 space bars, 1 escape key) I could then take 2 mol of H and say there will be 2*(6.02*10^23) H atoms, etc. correct? If so, thank you very much! $\endgroup$ – user14652 Feb 26 '15 at 5:56
  • $\begingroup$ I think you've got it! For the H2SO4 example, you would have 2 mol of H, and 1 mol of the ion SO4, which you could count as 1 mol of S, and 4 mol of O. Just understand that the concept of mole is no different then the idea of "dozen". It is just a unit of measurement. $\endgroup$ – quidproquo Feb 26 '15 at 6:01

One mole of $\ce{KOH}$ means that there are Avogadro number of (6.023 × 10²³ molecules) $\ce{KOH}$ molecules and we need not consider the sum of number of constituent atoms. For example, 1 mole of argon contains 6.023 × 10²³ argon atoms. 1 mole of ozone contains 6.023 × 10²³ ozone molecules (and also consists of 3 moles of oxygen atoms).

  • $\begingroup$ Fantastic. This is all coming together now and I have a general idea of it all; enough to stop questioning it so much. $\endgroup$ – user14652 Feb 26 '15 at 20:28