[Yusuf]

The question is determine the formula of a compound of iodine and copper formed when 0.6527g of copper reacted with 2.5380 grams of iodine.


I know what to do pretty much but their is something that confuses me, what do I do about Iodine being a diatomic in a natural state?


would I do

0.6527g of copper / 63.54g = .0102 mols


2.5380g of Iodine / 126.9g = .2 mols     

Cu + I2

or should I do

2.5380 * 2  (diatomic) of iodine /126.9g = .4 mols

and that would equal CuI4


or would I do

2.5380 / 126.9 * 2 ?

what would I do?
.

[Dan]

The molecular mass of I₂ is 254. If you use that, it will tell you how many mols of I₂ there are.

If you use 127, the atomic mass of I, it will tell you the number of mols of I atoms.

Also,

2.5380g of Iodine / 126.9g = .2 mols

You are out by a factor of 10.

[Vidya]

would I do

0.6527g of copper / 63.54g = .0102 mols


2.5380g of Iodine / 126.9g = .2 mols     

Cu + I2

or should I do

2.5380 * 2  (diatomic) of iodine /126.9g = .4 mols

and that would equal CuI4


or would I do

2.5380 / 126.9 * 2 ?

what would I do?
.

Try to understand this -
2.5380 grams of I2 X 1mole of I2 /molar mass of I2   X 2moles of I /moles of I2 = this will give you moles of I atoms
take out simplest ratio of moles of I and moles of Cu to take out the empirical formula.



Edit: Quote code corrected. Dan