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!

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3

idk what is it? is this a question that you know the answer or u dont