The mass of one molecule of a compound is $2.19\times10^{-22}\ \mathrm g$. What is the molar mass of the compound?

My attempt: Using the formula moles=mass/molar mass, I found how to calculate molar mass using the formula molar mass=moles/mass. Therefore, I was under the impression that, surely, $6.02\times10^{23}$ divided by the mass of the compound would have resulted in the molar mass. However, the answer was wrong.

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    $\begingroup$ You know how many molecules of anything are in one mole of that particular anything, so start from there. $\endgroup$ – Todd Minehardt Aug 20 '15 at 12:29
  • $\begingroup$ @ToddMinehardt Well I know that I have to divide the amount of moles of this compound over its mass to get to the molar mass. I have tried to get the number of moles through two different but haven't been able to because the compound hasn't been specified , but came to the wrong result twice. $\endgroup$ – Tim Aug 20 '15 at 12:52
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    $\begingroup$ You might want to include the steps you took towards solving this problem in your post. There is a homework policy for this site which states (in part) that a good-faith effort has to be made by the OP in order to keep the question open. I answered your query based on the effort you detailed in the comment (above). $\endgroup$ – Todd Minehardt Aug 20 '15 at 13:57
  • $\begingroup$ Okay, I'll do that now. $\endgroup$ – Tim Aug 20 '15 at 14:30

To get the molar mass of your mystery compound, make use of the fact that there are $N_{\rm A}$ (Avogadro's number) of molecules of it in one mole of it.

To solve the problem, we multiply the molecular mass (given) by Avogadro's number (known) to get the mass of one mole of the mystery compound:

$${\rm molar \;mass} = N_{\rm A}\times 2.19\times 10^{-22}\,{\rm g} = (6.022\times 10^{23}\,{\rm mol}^{-1}) \times (2.19\times 10^{-22}\,{\rm g}) = 131.88\;{{\rm g}\over{\rm mol}}$$

  • $\begingroup$ Oh I see now. Instead of multiplying Avogadro's number by the mass of the compound, I was dividing the two. $\endgroup$ – Tim Aug 20 '15 at 13:14

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