dtermine the empirical formula for a compound that has 9.09g carbon, 1.52g hydrogen, and 14.4g fluorine.?

Answer 1

moles (Carbon) = 9.09g/12.01 = 0.757 mol
moles (Hydrogen) = 1.52g/1.0079g.mol-1 = 1.51 mol
moles (Fluorine) = 14.4g/18.9984g.mol-1 = 0.758 mol

Divide all numbers by 0.757, giving the approximate ratio of 1:2:1 (C:H:F) as the empyrical formula.

Answer 2

Divide each of the grams given by the respective molecular mass of the atom. For example, divide 9.09g of Carbon by 12.

You now have 3 numbers that represent the molecular ratios in the empirical formula. If the 3 digits are whole numbers, great. If not, divide all 3 by the smalles of the s numbers and you will get the number of each atom in the EF