I am running on the assumption you meant the elements Nitrogen and Iodine, as 'L' is not an element. The calculation applies the same regardless of your elements.
Converting atoms/molecules to grams requires the use of Avogadro's Number, 6.022 x1023, which is also know as a mole. Additionally, we will need the atomic weight of the molecules, which is obtained from the periodic chart, and shows us how many grams per mole each element is. The path to conversion will look like (# molecules) -> (# moles) -> (# grams)
Note: Atomic Weight for Nitrogen and Iodine are 14.007g/mol, and 126.900g/mol, respectively.
Next we need to figure out the total atomic weight of our compound, N2I6. This is done by multiplying the atomic weights by the number of molecules, and adding together:
Atomic Weight of N2I6 = 2(14.007) + 6(126.900) = 789.414 g/mol.
Now we multiple through our conversion rates until we get to our solution:
(8.2 x 1023 molecules) ÷ (6.022 x 1023 molecules/mole) x (789.414 g/mol) = [# grams]
This stoichiometry can get very confusing if you have trouble figuring out whether to multiply or divide your conversion value; there are several tricks to lodge it into memory, and my favorite is the use of conversion grids.
--
AJ
