Empirical Formulas and

Molecular Formulas

Empirical Formula•Simplest ratio of atoms in a formula

•All ionic compounds exist as empirical formulas.

•Molecular compounds are not guaranteed to have the simplest ratio of atoms.

Molecular Formula•Some whole number multiple of the empirical formula.

Empirical Formula•C6H12O6 is the molecular formula of glucose.

•What is its empirical formula?

Problem Solving•A compound contains only carbon, hydrogen, and chlorine. A sample is known to contain 49.67%C, 1.39%H, and 48.92%Cl. The molecular weight of the compound is 289.90g/mol. What are the EF and MF of the compound?

Problem Solving•Make the assumption that you have 100g of the compound.

•How many grams of C would be in that particular sample? Of H? Of Cl?

Problem Solving•49.67gC x 1molC = ?

12.01gC•1.39gH x 1molH = ?

1.01gH•48.92gCl x 1molCl = ?

35.45gCl

Problem Solving•4.14molC•1.38molH•1.38molCl•Now, you will divide each of the moles by the smallest # of moles that you got.

Problem Solving•4.14molC = ? 1.38mol•1.38molH = ?1.38mol

•1.38molCl = ?1.38mol

Problem Solving•3C•1H•1Cl•This gives you your ratio of atoms in the EF.

Problem Solving•There are special rounding rules at this point.

•If the numbers are <.2, round down.

•If they are >.8, round up.

Problem Solving•If they are anywhere in between .2 and .8, you must multiply all of the numbers by a factor that will make them roundable.

Problem Solving•EF is C3HCl

•You will need to determine the molar mass of the EF (also known as the empirical weight, or EW).

Problem Solving•EW = 72.49g/mol•Since the MF is always some whole number multiple of the EF, the molar mass of the MF (the MW) will always be some multiple of the EW.

Problem Solving•Divide the MW by the EW to determine the multiple of the MF.

•MW = 289.90g/mol = ? EW 72.49g/mol

Problem Solving•MW = 4 EW•Thus, the MF is 4 times the EF.

•MF = C12H4Cl4

Problem Solving•Try another…•A compound consists 43.6%P and 56.4%O and has a molecular weight of 283.88g/mol. What are its EF and MF?