Acompound is found to contain 13.65 % carbon and 86.35 % fluorine by mass. what is the empirical formula for this compound

Greetings!

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

C = 12

F = 19

You can simply change the percentages into full grams

C = 13.64g

F = 86.35g

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

C = 13.64 ÷ 12 = 1.13

F = 86.35 ÷ 19 = 4.544

Then divide each number of moles by the smallest value:

C = 1.13 ÷ 1.13 = 1

F = 4.544 ÷ 1.13 = 4.02 ≈ 4

So that means that there are 1 Carbon and 4 Fluorine. This means the empirical formula would be:

CF₄

Hope this helps!

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