What is the empirical formula of a compound with the following percent composition: 18% C, 2.5% H, 63.5% I, and 16.0% O?

The empirical formula is C₃H₅ IO₂

Explanation:

Assume 100 g of the compound is present. This changes the percents to grams:  

C = 18 g, H = 2.5 g , I = 63. 5 g , O = 16 g

Convert the masses to moles:

C = 18/ 12 = 1.5

H = 2.5/1 = 2.5

I = 63.5/ 127 =0.5

O = 16/16 = 1

Divide by the lowest, seeking the smallest whole-number ratio:

C = 1.5 / 0.5 = 3

H = 2.5/0.5=5

I = 0.5/0.5= 1

O = 1/0.5= 2

Now we can write the Empirical formula as,

C₃H₅ IO₂

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.