I'm working on chemistry and I'm not quite sure how to solve this question? I haven't worked with moles before and whenever I try to figure out how to do these types of question I only end up confusing myself more.

How many Mg atoms are in 5.7892 mol of $\ce{Mg(NO3)2}$?

How many N atoms does the sample contain?

How many O atoms does the sample contain?

I haven't any clue how to do this so a walkthrough on how to solve these types of problem would be nice.


1 Answer 1


$\ce{Mg(NO3)2}$ has (Figure 1):

  • One $\ce{Mg}$ atom
  • Two $\ce{NO3}$ groups, which are composed of:
    • One $\ce{N}$ atom
    • Three $\ce{O}$ atoms

Magnesium nitrate

Figure 1: we have one $\ce{Mg}$, two $\ce{N}$ ($2 \times 1$) and six $\ce{O}$ ($2 \times 3$) in $\ce{Mg(NO3)2}$.

Now all you need is to know that 1 mole equals $6.022\,140\,76\times 10^{23} \approx 6.022 \times 10^{23}$ particles. Just multiply:

How many Mg atoms are in 5.7892 mol of $\ce{Mg(NO3)2}$?

$$1 \times 5.7892 \text{ mol $\ce{Mg}$} = 1 \times 5.7892 \times 6.022 \times 10^{23} \text{ atoms $\ce{Mg}$} \\= 3.48625624 \times 10^{25} \text{ atoms $\ce{Mg}$}$$

And so on.

  • 1
    $\begingroup$ Your equation is not correct since there are different dimensions on both sides of the equal sign (on the left amount of substance in mol, on the right a dimensionless number). It would be better if you use the Avogadro constant instead of the Avogadro number four your calculation. $\endgroup$
    – user7951
    Commented Sep 10, 2019 at 19:08

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