Question

What is the percentage composition of `NH_4I`?

Answer 1

Well, the molar mass of ammonium iodide is `144.94*g*mol^-1`...

Explanation:

And so `%I-=(126.9*g*mol^-1)/(144.94*g*mol^-1)xx100%=87.6%`

And `%N-=(14.01*g*mol^-1)/(144.94*g*mol^-1)xx100%=9.67%`

And `%H-=(4xx1.00794*g*mol^-1)/(144.94*g*mol^-1)xx100%=2.80%`

This is always the problem when you use iodides...their salts are so heavy that you have to add a large mass to get the required equivalence. Iodides have some good solubilities in solvents such as acetone...