How to calculate the number of atoms of each element in Mg(NO3)2? [closed]








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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.




[$\ce{Mg(NO3)2}$](https://en.wikipedia.org/wiki/Magnesium_nitrate) 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](https://i.stack.imgur.com/8s46D.png)](https://i.stack.imgur.com/8s46D.png)


**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](https://en.wikipedia.org/wiki/Avogadro_constant).
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.


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