# Gravimetric Analysis with empirical formula

• asz304
The Iodide ion is the ion that reacts with the silver ion in the solution to form AgI. In summary, to determine the empirical formula of the compound of cobalt and iodine, a 3.500 g sample was dissolved in water and treated with excess AgNO3 solution, producing 5.255g AgI. The mole ratios of AgI to Iodide were used to calculate the number of moles of Iodide in the sample, which was then used to find the grams of cobalt by difference. The Iodide ion was the ion that reacted with the silver ion to form AgI.

## Homework Statement

a 3.500 g sample of a compound of cobalt and iodine was dissolved in water and treated with excess AgNO3 solution producing 5.255g AgI. Determine the empirical formula.

A clearer view of the question:http://www.chem.mun.ca/courseinfo/c1050/Louise_Dawe/Basic%20Chemical%20Concepts%20I.pdf" [Broken]

question # 20.

## The Attempt at a Solution

I started by getting the mole of AgI and making a mole to mole ratio with CoxIy...and I'm not sure if what I did was right.
Thanks

Last edited by a moderator:
You do not yet know the mole ratios or atom ratios of the I to the Co. First use the 1:1 mole ratio for the reaction of AgNO3 to the Iodide ion. How many moles of Iodide were precipitated? That gives you the number of moles of I from the sample of the cobalt-iodine compound. Convert to grams; find the grams of cobalt by difference, since you know how much sample you used.

Do I need to find the moles of AgNO3 to find the 1:1 mole ratio? or get the moles of AgI from it's mass 5.255g? and to which Iodide are you referring to?

Thanks

AgNO3 was in excess to ensure all iodine has been precipitated as AgI (silver iodide). Exact amount of AgNO3 doesn't matter.