Determining the chemical formula of a polyiodide Salt

I've got a compound with formula $\ce{N(CH3)4.I(I2)_x}$. By mass I've got $\pu{0.138 g}$ of $\ce{I2}$ present out of $\pu{0.189 g}$ of polyiodide added. How do I find $x$?

This is the exact part of the question that I got stuck. It's more of the concept that I am not getting instead an equation that I can just use.

We begin with using $\pu{0.25 g}$ of tetramethylammonium. Then used $\pu{0.825 g}$ of iodine, and finally collected $\pu{0.8 g}$ of polyiodide salt.

Then we titrated $\pu{5 mL}$ of the polyiodide solution with $\pu{0.2 M}$ thiosulphate solution. Final titre used is $\pu{5.40 mL}$.

Calculated below:

• $\pu{1.09e-4 mol}$ of thiosulphate used to reach end point
• $\pu{5.43e-5 mol}$ of iodine present in the $\pu{5 mL}$ polyiodide solution
• $\pu{5.43e-4 mol}$ of iodine present in $\pu{50 mL}$ original sample
• $\pu{0.138 g}$ of $\ce{I2}$ present in original sample
• $\pu{0.189 g}$ is the full mass of original sample
• Thus giving 73 percent iodine by mass

So for the final ratio i got $1:2.7$ which can be rounded to 1:3. The formula thus is $\ce{NMe4.I(I2)_x}$, where I think $x=3$.

But I can't seem to make sense of this. I mean when $\ce{I(I2)_x} = (\ce{I2$x$}+1)$ could it mean $2x+1=3$, therefore making $x=1$?