1. Chapter 11: More on the Mole
Table of Contents
Chapter 11: More on the Mole
11.4: Empirical and Molecular Formulas
11.5: Hydrates

2. (moles element) x (mass of element)
11.4: Empirical and Molecular Formulas
Percent composition: the percent by mass of each element in a compound
(moles element) x (mass of element)
Mass of compound
% by mass= x100
The sum of the mass percents = 100%

3. 11.4: Percent composition
What is the percent composition of H2O? H:
(2mol H) x
(1.01g/mol H)
x 100 =
11.2% H
18.02 g H2O O:
(1mol O) x
(16.00g/mol O)
x 100 =
88.80% O
18.02 g H2O
The percent composition of H2O is 11.2% H and 88.80% O.

4. What is the percent composition of NaHCO3?
(1mol Na) x
(22.99g/mol Na)
x 100 =
27.37% Na
84.01 g NaHCO3 H:
(1mol H) x
(1.008g/mol H)
x 100 =
1.2% H
84.01 g NaHCO3 C:
(1mol C) x
(12.01g/mol C)
x 100 =
14.3% C
84.01 g NaHCO3 O:
(3mol O) x
(16.00g/mol O)
x 100 =
57.14% O
84.01 g NaHCO3

5. 11.4: Empirical and Molecular Formulas
Why? This process can be used to determine the composition of an unknown compound
Empirical Formula: the formula with the simplest whole number ratios

6. 11.4: Empirical and Molecular Formulas
Finding Empirical Formulas
You have an substance made of % Sulfur and 59.95% Oxygen. What is its empirical formula?
1) Assume percentages are masses
S = g
O = g

7. 11.4: Empirical and Molecular Formulas
You have an substance made of % Sulfur and 59.95% Oxygen. What is its empirical formula?
2) Convert grams to moles → ratio
S = g
O = g
Ratio S : O = : 3.747
1 mol S =
1.249 mol S
32.07 g S
1 mol O =
3.747 mol O
16.00 g O

8. 11.4: Empirical and Molecular Formulas
You have an substance made of % Sulfur and 59.95% Oxygen. What is its empirical formula?
3) Simplify ratio: divide each by smallest moles
S = mol S
O = mol O
Empirical formula: SO3
= 1 mol S
1.25
= 3 mol O
1.25

9. 11.4: Empirical and Molecular Formulas
EXAMPLE 1: Determine the empirical formula for a compound with % C, 8.16% H, and 43.20% O.
1) Assume percentages are masses
C = g
H = 8.16 g
O = g

10. 11.4: Empirical and Molecular Formulas
EXAMPLE 1: Determine the empirical formula for a compound with % C, 8.16% H, and 43.20% O.
2) Convert grams to moles → ratio
C = g
H = 8.16 g
O = g
1 mol C =
4.050 mol C
12.01 g C
1 mol H =
8.10 mol H
1.008 g H
1 mol O =
2.700 mol O
16.00 g O

11. 11.4: Empirical and Molecular Formulas
EXAMPLE 1: Determine the empirical formula for a compound with % C, 8.16% H, and 43.20% O.
3) Simplify ratio: divide each by smallest moles
C = mol C
H = 8.10 mol H
O = mol O
= 1.5 mol C
2.700
= 3 mol H
2.700
= 1 mol O
2.700

12. 11.4: Empirical and Molecular Formulas
EXAMPLE 1: Determine the empirical formula for a compound with % C, 8.16% H, and 43.20% O.
3) Simplify ratio: whole numbers
C = 1.5 mol
H = 3 mol
O = 1 mol
Empirical formula: C3H6O2
x 2 =
3 mol C
x 2 =
6 mol H
x 2 =
2 mol O

13. 11.4: Empirical and Molecular Formulas
Practice:
Ex 2) A blue solid has 36.84% nitrogen and 63.16% Oxygen. What is its empirical formula?
Ex 3) Aspirin contains 60.00% carbon, 4.44% hydrogen, and 35.56% oxygen. What is its empirical formula?
N2O3
C9H8O4

14. 11.4: Empirical and Molecular Formulas
Sometimes substances with different properties have the same percent composition and empirical formula. WHAT?
Molecular Formula: the actual number of atoms in a molecule/formula unit of a substance

15. mass of empirical formula
11.4: Empirical and Molecular Formulas
Finding Molecular Formulas
The molar mass of acetylene is g/mol. The mass of its empirical formula (CH) is g/mol
Divide:
So… the molecular formula is 2 times the empirical formula
correct molar mass
mass of empirical formula
26.04 g/mol
13.02 g/mol = = 2

16. 11.4: Empirical and Molecular Formulas
Finding Molecular Formulas
The molar mass of acetylene is g/mol. The mass of its empirical formula (CH) is g/mol
2) Multiply:
CH x 2 = C2H2

17. mass of empirical formula
11.4: Empirical and Molecular Formulas
Finding Molecular Formulas
EXAMPLE 4: Succinic acid has a molar mass of g/mol and an empirical formula of C2H3O2
Divide:
So… the molecular formula is 2 times the empirical formula
correct molar mass
mass of empirical formula
118.1 g/mol
59.04 g/mol = = 2

18. 11.4: Empirical and Molecular Formulas
Finding Molecular Formulas
EXAMPLE 4: Succinic acid has a molar mass of g/mol and an empirical formula of C2H3O2
2) Multiply:
C2H3O2 x 2 = C4H6O4

19. 11.4: Empirical and Molecular Formulas
Practice:
Ex 5) A photographic developing fluid has a chemical composition of 65.45% C, 5.45% H, and 29.09% O. Its molar mass is g/mol. What is its molecular formula? (Find its empirical formula first)
Ex 6) A colorless liquid is composed of 46.68% nitrogen and 53.32% oxygen has a molar mass of g/mol. What is the molecular formula?
C6H6O6
N2O2