1. Empirical Formulas and Molecular Formulas

2. Empirical Formulas
shows the smallest whole-number ratio of the atoms in the compound
CH only means it is a 1:1 ratio between carbon and hydrogen, NOT the actual number of atoms in the compound
could be C2H2 (ecetylene), or C6H6 (benzene), or C8H8 (styrene)

3. Determining Empirical Formula of a Compound
What is the empirical formula of a compound that’s mass is 25.9% N and 74.1% O?
% is not really a unit of measurement we can use in chemistry at times
if there was 100g of this compound then 25.9g would be nitrogen and 74.1g oxygen
change to moles in order to convert the % to a #
25.9g N x 1 mol N = 1.85 mol N
14.0 g N
74.1g O x 1 mol O = 4.63 mol O
16.0 g O

4. What is the whole number ratio?
a formula is not just the ratio of atoms, it is also the ratio of moles
this leaves a ratio of N1.85O4.63 which is not a a possible formula/whole number
therefore, divide each by the lowest quantity (in this case, 1.85)
now is N1O2.5
again, still not a whole number, but if you were to double it, it would be N2O5
the empirical formula is N2O5

5. Example #2
Calculate the empirical formula of a compound composed of % C, % H, and %N.
Assume 100 g so…
38.67 g C x mol C = mole C g C
16.22 g H x mol H = mole H g H
45.11 g N x mol N = mole N g N

6. If we divide all of these by the smallest
3.220 mole C
16.09 mole H
3.219 mole N
C1H5N1 is the empirical formula
C3.22H16.09N3.219
If we divide all of these by the smallest
number (3.22), it will give us whole numbers and therefore the empirical formula

7. Example #3 (notice this is already in grams and not a percentage)
A compound is found to contain 16g of C and 4g of H. What is the empirical formula?
16 g C x mol C = 1.33 mole C g C
4 g H x mol H = 4 mole H g H
1.33 is too far off from 1
divide both by 1.33
1.33/1.33 = 1
4/1.33 = 3
empirical formula is CH3

8. Comparing Empirical and Molecular Formulas
Comparing Empirical and Molecular Formulas
Compound
Molecular Formula
Empirical Formula
Water
H2O
Hydrogen Peroxide
H2O2 HO
Glucose
C6H12O6
CH2O
Methane
CH4
Ethane
C2H6
CH3
Octane
C8H18
C4H9

10. Molecular Formulas
a molecular formula is the same as, or a multiple of, the empirical formula
it is based on the actual number of atoms of each type in the compound, NOT just the ratio
the following all have same empirical formula (CH2O at a ratio) but different molecular formulas
C2H4O2 is the molecular formula of ethanoic acid
CH2O is the molecular formula of methanol
C6H12O6 is the molecular formula of glucose

11. Determine the molecular formula from the empirical formula and the experimental molar mass
divide the experimental molar mass by the mass of one mole of the empirical formula
this results in the multiplier to convert to the molecular formula (this sounds confusing but it really isn’t)
ex: calculate the molecular formula of a compound whose molar mass is found to be 60.0g and has an empirical formula of CH4N
experimental molar mass of ? g
empirical molar mass of CH4N g
2 (CH4N) = C2H8N2
= 2

12. need to calculate the empirical formula first assume 100 g so…
Determining the empirical formula and then molecular formula of a compound
A compound is known to be composed of % Cl, 24.27% C and 4.07% H. Its molar mass is known to be g. What is its molecular formula?
need to calculate the empirical formula first
assume 100 g so…
71.65 g Cl x mol Cl = 2.02 mole Cl g Cl
24.27 g C x mol C = 2.02 mole C g C
4.07 g H x mol H = 4.07 mole H g H

13. Cl2.02C2.02H4.07 divide by lowest (2.02 mol )
ClCH2 is the empirical formula
if ClCH2 was the molecular formula, this would give a mass of 48.5 g
however, recall the problem asked for the molecular formula of a compound with a molar mass of g which is twice that of 48.5 g
therefore, Cl2C2H4 is the molecular formula