What is the empirical formula of a compound that is 18.8% li, 16.3% c , and 64.9% o

Greetings!

To find the empirical formula you need the relative atomic mass of each element!

Li = 6.9

C = 12

O = 16

You can simply change the percentages into full grams

Li = 18.8g

C = 16.3g

O = 64.9

Then you use this to find the Number of moles = amount in grams / atomic mass

Li = 18.8 ÷ 6.9 = 2.7246

C = 16.3 ÷ 12 = 1.3583

O = 64.9 ÷ 16 = 4.0562

Then divide each number of moles by the smallest value:

Li = 2.7246 ÷ 1.3583 = 2.0

C = 1.3583 ÷ 1.3583 = 1

O = 4.0562 ÷ 1.3583 = 2.9 ≈ 3

So that means that there are 2 Li, 1 C, and 3 O

Empirical formula would be:

Li₂CO₃

Hope this helps!

im pretty sure its b

a corroding iron, if that's one of your choices