Expert: George Maxwell - 10/15/2007

Question
Hi,
I recently did a lab in my chem 101 class, on tin iodide. There was a part which required me to find the empirical formula of the reaction between tin and iodine, I tried to figure it out, and I got SnI4. However, I really don't know how I got it, or even if it's correct. How do I find the empirical formula of tin iodine?

thank you,
noel s. dasta

Answer
Hi, and thanks for your question.

I'll outline the basics of Empirical formulae below, but to apply it to a specific example, I'll need the experimental information that came with the question: you can't calculate empirical formulae without experimental data. Hope this helps.

Emipirical formulae give the simplest ratio of atoms present in a compound. For example, a compound with the molecular formula C12H24 has the empirical formula CH2 - the ratio of C:H is 1:2.

You calculate empirical formulae from experimental data, usually from an analysis of the % by mass of each element present in a compound.

Firsty, you need to turn your % mass in to actual masses. The easiest way to do this is to imagine that your sample has a mass of 100g and just convert the  % in to grams. It doesn't actually matter what mass you use (because the ratio stays the same), it's just easier this way.

E.G if my compound is 72% Magnesium and 28% nitrogen, then I assume my sample compound contains 78g magnesium and 28g nitrogen.

I then work out how many moles of each element I have: moles = mass / rmm.

EG Mg: moles = mass (72g) / RMM (24g/mol, from periodic table) = 2.97 mol.

Do this for every element in your compound, and you have a basic molar ratio: e.g. if moles of Mg = 3 and moles of N = 2, then the empirical formula is Mg3N2.

If the molar ratios are awkward numbers (e.g. 1.43 and 3.45), just divide them both by the smallest number of moles. In this case, 1.43 / 1.43 gives you 1, and 3.45 / 1.43 gives you 2.5. You con now easily multiply up those values 0 1:2.5 is the same as 2:5.

It's really important to note that this last step is just convenient maths: you're not coming up with new information, you're just making the same ratio in to a more convenient number.

PS: Sn and I can combine in a number of ways, and one of the recognised compounds is Tin (IV) Iodide, or SnI4.