[pizza1512]

I did a experiment at school but am not quite sure how to work out the amount of copper iodide produced from my results.  Please Help Me!! When copper(II) ions and an excess of iodide ions react, an iodide of copper is formed.

This is what I did:
Accurately weighed 0.51g of hydrated copper(II) sulphate into a 100cm³ beaker. Then dissolved the crystals in 10cm³ of distilled water.

Then dissolved approximately 1.3g of potassium iodide in a similar volume of distilled water in a second beaker. Mix the two solutions with stirring.

When there was no further change in colour, I added sodium metabisulphite solution drop-wise (in a fume cupboard) until there was no further changed in colour.

Filter off the copper iodide using a weighed filter paper. Washed the residue with 2 cm³ of sodium metabisulphite solution and then with about 5cm³ if distilled water. Dried the product by heating to constant mass.

I ended up with 1.37g of CuI₂ on the filter paper.  How do I work out the formula of copper iodide then?

Thanks

[enahs]

http://www.chemguide.co.uk/inorganic/transition/copper.html

Scroll down a lot (or just read everything on that page!).