30g Sample amount of Nitrogen atoms in Aluminum nitrate

30 gram sample of aluminum nitrate Al(NO3)_3 how many nitrogen atoms? mass of al nitrate is 212.995 right, and N mass in this is total of 42.07g. N=.19718 of the compound. now what? Divide 30g by mass of aluminum nitrate ? gives you moles.... help im lost

As far as you've got it's correct. 42.07/212.995 x 100%= 19.718 % by weight of nitrogen.
19.718 % x 30 grams=5.915 grams of nitrogen in the sample.
To express that in moles of nitrogen 5.915/14(molecular weight of N)=.423 moles.
To get the number of atoms of nitrogen multiply the number of moles by avogadrios number (6.02252 x 10^23. The result is 2.54 x 10^23 atoms.