[jayjay112]

A non-metal forms four fluorides. Three of them contain 13.01%, 30.98% and 51.15% F by weight respectively.  Identify the non-metal, and write the formula for the fourth fluoride.

(i) 19/13.01 x 86.99 = 127.04g so non-metal is Iodine.

(ii) I can get the formula for the second flouride and the 3rd
eg. 19/30.98 x 69.02 = 42.34g

127/42.34 = 3 so formula = IF3

I dont know the percentage for the fourth???
So I cant get the formula for the fourth fluride. The answer is IF5

Can anyone help please thanks.

[ARGOS++]

Dear jayjay112;

If 4.86% for fluoride content is really given, then you may think about that as formula all multiple of IF_1/3 will give the identical ratio or percentage, also for example:  I₅F_1.666.
So there is one lowest with no fractional numbers, even if it doesn’t exist in real.

Good Luck!
                    ARGOS⁺⁺

[jayjay112]

sorry no, 4.86 doesnt exist i just assumed thats what it was because i added up all % and took away from 100 and was
left with 4.86%.

How do i know what the fourth fluride is???

[ARGOS++]

Dear jayjay112;

Then please check n in the formula:  n * IF_1/3  =  I_nF_{(n * 1/3)} for no fraction as a result.

Good Luck!
                    ARGOS⁺⁺