How many molecules are there in 17.53 grams of $\ce{NH4ClO4}$?

I get to 1 mol = 118 then I get stuck.


closed as off-topic by Mithoron, Gaurang Tandon, airhuff, Tyberius, aventurin Jun 15 '18 at 5:21

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.

  • 1
    $\begingroup$ LOL - Technically none. There is no such thing as a "molecule" in a salt. There is not a unique to assign a particular $\ce{NH4^+}$ group to a particular $\ce{ClO4^-}$ group in the 3D crystal matrix $\endgroup$ – MaxW Jun 8 '18 at 1:48
  • $\begingroup$ How about not technically, trying to help teach my daughter and it's been a long time $\endgroup$ – The C Jun 8 '18 at 1:52
  • 1
    $\begingroup$ The site rules state that we're not supposed to do "homework" problems, but I'm a sucker for someone trying to help a daughter... $\endgroup$ – MaxW Jun 8 '18 at 2:29
  • $\begingroup$ The question you need to ask with problems like this is molecules of what? If ammonium perchlorate were a molecular compound this would be simple: the answer would be the moles of the compound. But it is an ionic salt with two components. Unless the question is specific about what it wants you to count the question is ambiguous. The answer could be the umber of moles of ammonium perchlorate or it could be twice that (counting the ammonium and perchlorate ions separately). $\endgroup$ – matt_black Jun 8 '18 at 9:57

Ammonium perchlorate has a molecular weight of 117.485 grams/mole. Since the mass was given to 4 significant figures you should carry at least that number in the molecular weight. I'd use all 6 and round the final result to 4 significant figures.

Moles Ammonium perchlorate = $\dfrac{17.53}{117.485}$

To six significant figures Avogadro's constant is $6.02214\times10^{23}$

$$\begin{align} \text{Number of molecules} &= \dfrac{17.53}{117.485}\times 6.02214\times10^{23}\\ &= 8.9856674639315657317955483678768\times10^{22}\\ &= 8.986\times10^{22}\text{(rounding to 4 sign figures)}\\ \end{align}$$

Note that $\dfrac{17.53}{117.5}\times 6.022\times10^{23} = 8.984\times10^{22}$ hence the reason to carry 2 extra significant figures in the intermediate calculations. Significant figures are a crude way to doing error propagation. Not exact, but such use keeps you from stupidly assuming that all the figures in $8.9856674639315657317955483678768\times10^{22}$ are significant.


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