1. Molecular Formula vs
Empirical Formula

2. Different compounds can have the same empirical formula but different molecular formulas.
Empirical Formula is a reduced form of Molecular formula
Molecular Formula may be the same as the empirical formula but is probably different.

3. An empirical formula is:
simplest whole-number ratio of atoms in a compound- Reduced form
A molecular formula is:
The actual formula of the molecule

4. Therefore….
CH2O is an empirical formula C6H12O6 is a molecular formula

5. Empirical Formulas
Step1: Change % to grams (g) - if you are given grams, skip this step Step 2: Convert masses to moles using molar mass Step 3: Divide all # of moles by the one that is the smallest Step 4: If dividing gave you .5, then multiply by 2 Step 5: If dividing gave you .3 or .7, then multiply by 3 Once you know the ratios, place them as subscripts in the formula

6. What is the important information from this problem? ( the given)
Empirical Practice #1: Find the empirical formula of a compound that is 33.38% Na, 22.65% S, and 44.9% O.
What is the important information from this problem? ( the given)

7. Empirical Practice #1: Find the empirical formula of a compound that is 33.38% Na, 22.65% S, and 44.9% O.
33.38gNa 22.65g S 44.9g O

8. What is the important information from this problem? ( the given)
Empirical Pracitice #2: A compound contains 3.26g of arsenic and 1.04g of oxygen. What is the empirical formula?
What is the important information from this problem? ( the given)

9. Empirical Pracitice #2: A compound contains 3. 26g of arsenic and 1
Empirical Pracitice #2: A compound contains 3.26g of arsenic and 1.04g of oxygen. What is the empirical formula?
3.26 g As 1.04 g O

10. Empirical Pracitice #2: A compound contains 3. 26g of arsenic and 1
Empirical Pracitice #2: A compound contains 3.26g of arsenic and 1.04g of oxygen. What is the empirical formula?
3.26 g As 1 mole mole g g O 1 mole mole = =1
=1.49 =
Step 4: If dividing gave you .5, then multiply by 2
Step 5: If dividing gave you .3 or .7, then multiply by 3

11. Empirical Pracitice #2: A compound contains 3. 26g of arsenic and 1
Empirical Pracitice #2: A compound contains 3.26g of arsenic and 1.04g of oxygen. What is the empirical formula?
3.26 g As 1 mole mole g g O 1 mole mole =
=1*2=2
=1.49*2=3 =
As2O3

12. What is the empirical formula of a compound that is 62. 10% C, 13
What is the empirical formula of a compound that is 62.10% C, 13.80% H & 24.10% N?

13. Molecular Formulas
1. To find the molecular formula you must: 2. Find the empirical formula if not given 3. Determine the molar mass of the empirical formula 4. MM molecular formula = X MM empirical formula 5. multiply each subscript in the empirical formula by “X”

14. What is the important information from this problem? ( the given)
Molecular Practice 4: The empirical formula of a compound is CH; the molecular molar mass is g/mol. What is its molecular formula?
What is the important information from this problem? ( the given)

15. Empirical: CH MMmolecular formula = 78.11 g/mole
Molecular Practice 4: The empirical formula of a compound is CH; the molecular molar mass is g/mol. What is its molecular formula?
Empirical: CH MMmolecular formula = g/mole

16. What is the important information from this problem? ( the given)
Molecular Practice #5:A compound has an empirical formula of CH3O and a molecular mass of 62 g/mol. What is its molecular formula?
What is the important information from this problem? ( the given)

17. Empirical: CH30 MMmolecular formula = 62.00 g/mole
Molecular Practice #5:A compound has an empirical formula of CH3O and a molecular mass of g/mol. What is its molecular formula?
Empirical: CH30 MMmolecular formula = g/mole

18. Molecular Practice #5:A compound has an empirical formula of CH3O and a molecular mass of g/mol. What is its molecular formula?
Empirical: CH30 MMmolecular formula = g/mole MMCH3O = g/mole =2
C2H6O2

19. A compound is 26.70% C, 2.2.0%O & 71.1% O. If its molecular molar mass is 90.00g/mole. What is its molecular formula?

20. Summary a.C6H8 b. N2O6 c. C6H12O6 d. BCl3
Determine the molecular and empirical formula for each:
a.C6H8
b. N2O6
c. C6H12O6
d. BCl3

21. Summary a.C6H8 molecular b. N2O6 molecular c. C6H12O6 molecular
Determine the molecular and empirical formula for each:
a.C6H molecular
b. N2O molecular
c. C6H12O6 molecular
d. BCl empirical

22. Summary a.C6H8 molecular -C3H4 (empirical)
Determine the molecular and empirical formula for each:
a.C6H molecular -C3H4 (empirical)
b. N2O molecular –NO3 (empirical)
c. C6H12O6 molecular- CH2O (empirical)
d. BCl empirical

23. Explain why some compounds do not have an empirical formula:?