What is the empirical formula for a compound that is 17.15 % carbon, 1.44 % hydrogen, and 81.41 % fluorine?

1 Answer
Jul 14, 2014

Thanks for the empirical formula question.

An empirical formula is the lowest whole-number ratio of moles (or atoms) in a chemical formula.

The first step in determining an empirical formula is to determine the relative number of moles in the compound. Since the percents are by mass already, we are going to use the percents as if they are grams and convert to moles. The denominators of each element are taken from the periodic table.

$$                                   [mole ratios](http://socratic.org/chemistry/stoichiometry/mole-ratios)


carbon = 17.15/12.01 = 1.428

Hydrogen = 1.44/1.008= 1.429

fluorine = 81.41/19.00 = 4.285

Now, it is obvious that we cannot have parts of atoms in our chemical formula, so the next step is to divide all the mole ratios by the SMALLEST MOLE RATIO. We do this to try to get the ratios down to the smallest whole numbers.

Carbon: 1.428/1.428 = 1

Hydrogen: 1.429/1.428 = 1

Fluorine: 4.285/1.428 = 3

We're in luck!! These are all whole numbers and we can get our empirical formula from these ratios.

The empirical formula is: CH ${F}_{3}$.

Hope this helps.