What is the empirical formula for a compound consisting of 72.2% magnesium and 27.8% nitrogen by mass?

Answer 1

Hi, Aaron,

Ah, these are always fun! OK, so you've been given an unknown substance which appears to consist of just two elements: magnesium (72.2% of the mass), and nitrogen (the remaining 27.8%). It's always wise to add those two values up to make sure they equal 100% ... and they do, so let's move on!

The "trick" for determining the empirical formula of an unknown is setting the weight of the unknown as a convenient number to work from -> this is typically 100 gm, and you'll see why in a moment. Now, let's assume you had a 100 gm sample of this mystery "Mg-N" unknown. Since you know the mass percentages, and you know (or you can look up) the atomic masses of those two elements, you can then determine their molar ratios in this unknown:

For a 100 gm sample, 72.2 gm must be Mg, and 27.8 gm must be N. If we then divide each of those mass amounts by their respective atomic masses, we obtain:

72.2 gm ÷ 24.30 gm/mole Mg = 2.97 mole Mg (in a 100 gm sample).

27.8 gm ÷ 14.01 gm/mole N = 1.97 mole N (in a 100 gm sample).

Since our choice of a 100 gm sample was arbitrary, all we know is that the empirical formula for this unknown compound must be Mg₃N₂. The "three parts magnesium to two parts nitrogen" empirical ratio will be constant no matter how big or small the sample is!

Good luck,

Chris