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

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.

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

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

• 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.
– user7951
Sep 10, 2019 at 19:08