1. Empirical and Molecular Formulas

2. Definitions
Empirical formula – the lowest whole-number ratio of the atoms of the elements in a compound.
Molecular formula – chemical formula that shows the actual number of atoms in a compound.

3. Which is it: empirical, molecular, or possibly both??NO
N2O4
C6H12O6
Ba(OH)2
C4H8

4. Empirical Formulas H2O2 H2O H2O CH2O CaCl2 CaCl2 HO C6H12O6
The simplest whole-number ratio of atoms in a compound.
H2O2 HO
Both are divisible by 2.
H2O
H2O
Already simplified
CH2O
C6H12O6
All are divisible by 6.
CaCl2
Correct ionic formulas are always empirical.
CaCl2

5. Finding an Empirical Formula
Grams  mole
Divide by smallest
Round ‘til whole
An unknown compound contains:
grams of C
grams of H
grams of O
What is the empirical formula of the compound?

6. Step 1: Change grams to moles
g C
g H
g O
1 mol O
g C
1 mol C
g H
1 mol H
g O
12.01 g C
1.01 g H
16.00g O
mol C
mol H
mol O

7. Step 2: Divide by smallest mole value Step 3: Round to the whole
mol C
mol H
mol O
0.0067
0.0067
0.0067
mol C
2 mol H
1 mol O
1 mol C
2 mol H
1 mol O

8. 1 mol C
2 mol H
1 mol O
CH2O
This is the empirical formula.

9. Let’s try another –
This time, using
percent composition!!

10. An unknown compound is known to have the following composition:
11.11% hydrogen
88.89% oxygen
What is the empirical formula of the compound?
In these problems we will choose to use a 100 gram sample.
So, in this sample we have grams H and grams O.

11. Step 1: Change grams to moles.
11.11 g H
1 mol H
1.01 g H
= mol H
88.89 g O
1 mol O
16.00 g O
= 5.56 mol O

12. Step 2: Divide by smallest mole value.
Step 3: Round to a whole number.
5.56 mol O
11.00 mol H
5.56
5.56 2 1

13. The ratio of moles of hydrogen
to moles of oxygen is 2:1
H2O

14. Going one step further… Finding molecular formulas from empirical formulas.
Remember: The empirical formula is the simplest whole number ratio; whereas, the molecular formula shows the actual number of each atom.
Step 1: Find molar mass of empirical formula.
Step 2: Divide the molar mass of the compound by the molar mass of the empirical formula.
Step 3: Multiply the empirical formula by that number.

15. Practice Problem
CH2O is the empirical formula. The molar mass of the actual compound is 180 g/mol. What is the molecular formula?

16. Step 1: Find molar mass of empirical formula.
(1 x 12.01) + (2 x 1.01) + (1 x 16.00) = g/mol
Step 2: Divide the molar mass of the compound by the molar mass of the empirical formula.
180 g / 30 g = 6
Step 3: Multiply the empirical formula by that number.
6 (CH2O)
= C6H12O6 The Molecular Formula

17. You try one… Empirical formula – CH4 The actual molecular mass is 64g.
Find the molar mass of empirical formula  C + 4(H) = 16g
Divide molecular mass by empirical mass  64g / 16g = 4
Multiply the empirical formula by that number  4 (CH4)
C4H16  molecular formula

18. One more thing to mention…
Hydrates vs. Anhydrates
Hydrates – salts with water molecules attached
BaCl2 • 2 H2O
Name the salt, then use a prefix + word “hydrate”—Barium Chloride Dihydrate
Anhydrates – the salt without its water
BaCl2

19. Finding the Formulas of Hydrates --Similar to Finding an Empirical Formula
1. Find the mass of the water by subtraction.
The mass of the hydrate and the anhydrate will be given.
2. Change g to mol
Mass of water to mol
Mass of salt to mol
3. Divide by the smallest mole value
4. Round to a whole # if necessary.

20. Finding the Formulas of Hydrates
BaCl2 • X H2O
Hydrate mass (mass with water) – g
Anhydrate mass (mass w/o water) g
Step 1: Find the mass of the water.
Mass H2O = g – g = g H2O
For the formula we want the ratio of moles of water to moles of anhydrate.

21. Step 2: Change grams to moles
Grams of Salt
20.80 g BaCl2
208 g BaCl2
1 mol
= 0.1 mol BaCl2
Grams of Water
3.60 g H2O
18 g H2O
1 mol
= 0.2 mol H2O

22. BaCl2 • 2 H2O 1 2 Barium chloride dihydrate
Step 3: Divide by the smallest mole value.
0.1 mol BaCl2
0.2 mol H2O
0.1
0.1 1 2
Step 4: Round to a whole # if necessary.
BaCl2 • 2 H2O
Barium chloride dihydrate

23. Calculating Percent Water in a Hydrate
Calculating the percent water in a hydrate is similar to calculating percent composition:
% H2O = mass of the water x 100
mass of the hydrate

24. Example Problem
In the lab, a five gram sample of hydrous copper (II) nitrate is heated. If 3.9 grams of the anhydrous salt remains, what is the percent water in the hydrate?
Mass of water: 5 g – 3.9 g = 1.1 g H2O
Mass of hydrate: 5 g
(1.1 g H2O / 5 g total) x 100
= 22 % H2O