What is the percentage composition of #NH_4I#?

1 Answer
Jan 1, 2018

Answer:

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...