[owlpower]

2 Cu + I2 --> 2 CuI

Creating copper iodide from copper metal and iodine solid. NaI was used as a solvent (and I think also to supply I- ions) and a drop of acetic acid was added (to remove rust on Cu).

Retrived the CuI of 3.943mmol.
Used 1.97mmol of I2 and 3.915mmol of Cu.

If we calculate CuI yield based on I2, it would be 99.92%, which is greater than that for Cu, at 99.29%. Why is that so?
 
I am thinking that it is because of the addition of NaI, but I am not very sure as the 1.97mmol does not account for the NaI added.

[chenbeier]

There must be a calculation or weight error. If you retrieve 3,943 mmol CuI then the amount of copper has to be the same or more but not less 3,915 mmol. Other way around 3,915 mmol Cu can only get  the same number of CuI. The Iodine fits some how. 1,97 mmol I₂ get 3,94 mmol I^- approx. 3,943 mmol CuI.

The yield is oppsit by the way 100,72% or 100,08% Product/Educt.