it's my first week of IB chem and I'm already confused. We learned how to find the number of atoms, but not specifically ions... is it the same? Similar? HELP


closed as off-topic by Mithoron, Todd Minehardt, R.M., andselisk, pentavalentcarbon Sep 9 '17 at 1:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Is it in an aqueous solution? $\endgroup$ – can-ned_food Sep 8 '17 at 23:11
  • 2
    $\begingroup$ @can-ned_food That's is completely irrelevant. $\endgroup$ – Zhe Sep 9 '17 at 0:00
  • $\begingroup$ You want to use the stoichiometry of the compound and Avogadro's number. $\endgroup$ – Zhe Sep 9 '17 at 0:01
  • 1
    $\begingroup$ @Zhe With all respect due, it is not irrelevant if it helps us understand the asker's confusion. It seems like there at least some here. This is someone pursuing an educator's curriculum, not a chemist's. $\endgroup$ – can-ned_food Sep 9 '17 at 0:18
  • $\begingroup$ @can-ned_food That specific issue doesn't help the OP answer their question, nor does it change the answer. It would be great if we didn't ask questions that may further confuse the issue by introducing variables that are not relevant to the problem at hand. $\endgroup$ – Zhe Sep 9 '17 at 13:16

$1$ $\ce{NaCl}$ contains $1$ $\ce{Na+}$ and $1$ $\ce{Cl-}$.

$1$ mol $\ce{NaCl}$ contains $1$ mol $\ce{Na+}$ and $1$ mol $\ce{Cl-}$.

$1$ mol ions = $6.02*10^{23}$ ions.

$1$ mol $\ce{NaCl}$ contains $6.02*10^{23}$ $\ce{Na+}$ and $6.02*10^{23}$ $\ce{Cl-}$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.