Lecture on Empirical Formula
Introduction (5 minutes)
Definition: The empirical formula of a compound represents the simplest whole-
number ratio of atoms of each element present in the compound. It does not
provide the actual number of atoms but the simplest ratio.
Importance: Understanding empirical formulas is essential in chemistry for
determining the composition of a substance, analyzing experimental data, and
relating molecular formulas to simpler ratios.
Basics of Empirical Formula Calculation (10 minutes)
Step 1: Identify the Mass of Each Element: Usually given in a problem or
determined through experiments like combustion analysis.
Step 2: Convert Mass to Moles: Use the atomic mass of each element to
convert the mass in grams to moles.
Step 3: Find the Simplest Ratio: Divide the moles of each element by the
smallest number of moles to get the simplest whole number ratio.
Conceptual Understanding (10 minutes)
Di erence between Molecular and Empirical Formulas: The molecular
formula shows the actual number of atoms of each element in a molecule,
whereas the empirical formula is the reduced form showing the simplest ratio.
Examples of Common Compounds: Water (H₂O), Hydrogen Peroxide (H₂O₂ and
HO), Benzene (C₆H₆ and CH).
Applications of Empirical Formulas (10 minutes)
Determining Molecular Formulas: If the molar mass of a compound is known,
you can determine the molecular formula using the empirical formula.
Chemical Analysis and Industries: Used in calculating compositions in
pharmaceuticals, material science, and chemical manufacturing.
Practical Examples and Calculation (15 minutes)
Example 1: Calculate the empirical formula for a compound containing 40%
carbon, 6.7% hydrogen, and 53.3% oxygen by mass.
Example 2: A compound consists of 27.29% potassium, 1.07% hydrogen,
14.82% carbon, and 56.82% oxygen. Determine the empirical formula.
Interactive Practice Questions (10 minutes)
1. Question 1: A compound is found to contain 52.14% carbon, 13.13% hydrogen,
and 34.73% oxygen by mass. Calculate its empirical formula.
2. Question 2: Find the empirical formula of a compound if 0.092 moles of iron
combine with 0.184 moles of oxygen.
Conclusion (5 minutes)
Recap of the importance of empirical formulas in understanding the basic
composition of chemical substances.
Encourage students to practice calculating empirical formulas with diverse
examples to master the concept.
Practice Solutions
Solution to Question 1:
1. Convert Percent to Grams (assume 100g total): C = 52.14g, H = 13.13g, O =
34.73g.
2. Convert Grams to Moles:
Moles of C = 52.14 g / 12.01 g/mol = 4.34 moles
Moles of H = 13.13 g / 1.008 g/mol = 13.03 moles
Moles of O = 34.73 g / 16.00 g/mol = 2.17 moles
3. Simplest Whole Number Ratio: Divide each by the smallest number of moles
(2.17):
C: 4.34 / 2.17 ≈ 2
H: 13.03 / 2.17 ≈ 6
O: 2.17 / 2.17 = 1
Empirical Formula: C₂H₆O
Solution to Question 2:
1. Given Moles: Fe = 0.092 moles, O = 0.184 moles.
2. Simplest Ratio: Divide each by the smallest number of moles (0.092):
Fe: 0.092 / 0.092 = 1
O: 0.184 / 0.092 = 2
Empirical Formula: FeO₂
