1. School example: M&M’s example: Tylenol:
Given the ratio of students to teachers to administratives to maintenance people, can you say how many people are in the school?
M&M’s example:
Given the ratio of colours only, can you say how many candies make up the bag?
Tylenol:
Given the ratio of each type atom in the active ingredient, can you say how many of each type of atom there are in a molecule of it?

2. MOLECULAR FORMULA
C?H?O?

3. Molecular Formula Definition:
A formula that tells you the exact number
of atoms in one molecule of a compound.

4. Molecular Formula vs. Empirical Formula
Empirical formula – a formula that tells you the lowest ratio of atoms in a compound
Molecular formula – a formula that tells you the exact number of atoms in a molecule of a compound

5. Steps to determining the molecular formula
Determine the empirical formula
Determine the molar mass of the empirical formula
Determine the molar mass of the compound
Determine the ratio of the molar mass of the compound to the molar mass of the empirical formula
Multiply the value of the ratio by the subscripts in the empirical formula

6. EXAMPLE 1
What is the molecular formula of a compound with empirical formula CH3O and molar mass g/mol ?
Step 1 - Determine empirical formula
Given - CH3O

7. Step 2 - Determine the molar mass of the empirical formula
M CH3O = 1(12.01g/mol) + 3(1.01 g/mol) + 1(16.00 g/mol)
M CH3O = g/mol
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Step 3 - Determine the molar mass of the compound
Given – Mcompound = g/mol

8. Step 4 - Determine the ratio of the molar mass of the compound to the molar mass of the empirical formula
Mcompound = g/mol = 3
Mempirical formula g/mol

9. Therefore, the molecular formula of the compound is C3H9O3
Step 5 - Multiply the value of the ratio by the subscripts in the empirical formula
Molecular formula = 3(empirical formula)
= 3(CH3O)
= C3H9O3
Therefore, the molecular formula of the compound is C3H9O3

10. EXAMPLE 2
The percentage composition of a compound, as determined by a combustion analyzer, is 40.03% carbon, 6.67% hydrogen, and % oxygen. Using a mass spectrometer, the molar mass of the compound is found to be g/mol. What is the molecular formula of the compound?

11. Remember this? Percent to mass, Mass to mole, Divide by the smallest,
Multiply ‘til whole.

12. Step 1 – Empirical formula?
C = 40.03% mC = x 100g = g
100
H = 6.67% mH = x 100g = g
O = 53.30% mO = x 100g = g

13. nC = 40.03g C x 1mol C = 3.33 mol C
12.01 g C
nH = g H x 1 mol H = 6.60 mol H
1.01 g H
nO = 53.30g O x 1 mol O = 3.33 mol O
16.00 g O

14. Therefore, the empirical formula of the compound is CH2O
nC : nH : nO = : 6.60 : 3.33
nC : nH : nO = 1 : 1.98 : 1
nC : nH : nO = 1 : 2 : 1
Therefore, the empirical formula of the compound is CH2O

15. Step 2 - Determine the molar mass of the empirical formula
M CH2O = 1(12.01g/mol) + 2(1.01 g/mol) + 1(16.00 g/mol)
M CH2O = g/mol
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Step 3 - Determine the molar mass of the compound
Given – Mcompound = g/mol

16. Step 4 - Determine the ratio of the molar mass of the compound to the molar mass of the empirical formula
Mcompound = g/mol = 6
Mempirical formula g/mol

17. Therefore, the molecular formula of the compound is C6H12O6
Step 5 - Multiply the value of the ratio by the subscripts in the empirical formula
Molecular formula = 6(empirical formula)
= 6(CH2O)
= C6H12O6
Therefore, the molecular formula of the compound is C6H12O6