1. Empirical and Molecular Formulas
Chemical Compounds Unit
Notes #5

2. Writing a Formula
What if we wanted to be able to give a name/formula to an unknown compound, based on data?
Ex. A compound weighs g. It is 72.2% magnesium and 27.8% nitrogen by mass.
How could we determine its formula?

3. Empirical Formula
A compound’s empirical formula is the smallest whole number ratio of atoms in the compound.
The ratio is shown in the subscripts.
Ex) NO2 vs. N2O2
It does not necessarily tell us how many atoms are in the compound.

4. Some Empirical Formulas
Actual (Molecular) Formula HO
HO, H2O2, H3O3…
CH2O
CH2O, C2H4O2, C6H12O6 …
CH3O
CH3O, C2H6O2 …

5. Calculating the Empirical Formula
Ex. A compound weighs g. It is 72.2% magnesium and 27.8% nitrogen by mass.
Change the percents into masses.
Mg: 72.2%  72.2 g
N: 27.8%  27.8 g

6. Calculating the Empirical Formula
Find the moles of each element.
Mg: 72.2 g x 1 mole = 2.97 moles g
N: 27.8 g x 1 mole = 1.97 moles g

7. Calculating the Empirical Formula
Divide both numbers of moles by the smaller number.
We have 2.97 moles Mg and 1.97 moles N
Mg : 2.97 moles / 1.97 moles = 1.50
N: 1.97 moles / 1.97 moles = 1.00

8. Calculating the Empirical Formula
Multiply both quotients by the same number until they are both whole numbers.
Mg : 1.50 x 2 = 3.00
N: 1.00 x 2 = 2.00
The final answers become the subscripts in the empirical formula.
Mg3N2

9. Try it! A sample is 25.9% nitrogen and 74.1% oxygen. In grams:
In moles:
Divided:
Multiplied:
Final:
25.9 g N; g O
1.85 moles N; moles O
N: 1.00; O: 2.50
N: 1.00 x 2 = 2.00; O: 2.50 x 2 = 5.00
N2O5

10. Actual (Molecular) Formula
The molecular formula is the actual ratio of atoms in the compound.
Many molecular formulas can have the same empirical formula.
Empirical Formula
Actual (Molecular) Formula HO
HO, H2O2, H3O3…
CH2O
CH2O, C2H4O2, C6H12O6 …
CH3O
CH3O, C2H6O2 …

11. Calculating the Molecular Formula
A compound has the empirical formula CH2 and a molecular mass of 28.1 g/mol.

12. Calculating the Molecular Formula
Find the empirical formula
CH2
Find the molar mass of the empirical formula
1(12.01) + 2(1.01) = g/mol

13. Calculating the Molecular Formula
Divide the empirical mass by the molecular mass
28.1 g/mol = g/mol
Multiply each subscript by that quotient.
CH2 x 2.00 = C2H4

14. Try it!
A sample is 25.9% nitrogen and 74.1% oxygen, and has a molecular mass of 216 g/mol.
Empirical formula:
Empirical mass:
Molecular mass divided by empirical mass:
Subscripts multiplied:
N2O5
108 g/mol
216 / 108 = 2
(N2O5) * 2 = N2O5