Topic 2: Percentage Composition and Empirical Formula
Class: SS1
Specific Objectives:
By the end of the lesson, students should be able to:
- Define percentage composition.
- Calculate percentage composition of elements in compounds.
- Define empirical and molecular formulae.
- Calculate empirical formula from percentage or mass composition.
- Differentiate between empirical and molecular formulae.
Instructional Materials:
- Periodic table
- Whiteboard and marker
- Chart showing worked examples
- Calculator
- Sample compounds and their formulae
Lesson Content:
Step 1: Percentage Composition
- Definition: The percentage by mass of each element in a compound.
- Formula:
\text{Percentage composition of an element} = \left(\frac{\text{Mass of the element in 1 mole}}{\text{Molar mass of the compound}}\right) \times 100\%
Example:
Calculate the percentage composition of water (H₂O).
- Molar mass of H₂O = 2(1) + 16 = 18 g/mol
- % of H = (2/18) × 100 = 11.11%
- % of O = (16/18) × 100 = 88.89%
Step 2: Empirical Formula
- Definition: The simplest whole-number ratio of atoms of elements in a compound.
- It may or may not be the same as the molecular formula.
- E.g., Molecular formula of glucose = C₆H₁₂O₆
- Empirical formula = CH₂O
Step 3: Steps to Calculate Empirical Formula
- Convert percentages or masses to moles:
\text{Moles} = \frac{\text{Mass or percentage}}{\text{Atomic mass (Ar)}}
- Multiply to get whole numbers (if necessary).
- Write the empirical formula using the ratio obtained.
Step 4: Worked Example (Empirical Formula Calculation)
Example:
A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen. Find its empirical formula.
Step 1: Convert to moles:
- C: 40 ÷ 12 = 3.33
- H: 6.7 ÷ 1 = 6.7
- O: 53.3 ÷ 16 = 3.33
Step 2: Divide by smallest (3.33):
- C: 3.33 ÷ 3.33 = 1
- H: 6.7 ÷ 3.33 ≈ 2
- O: 3.33 ÷ 3.33 = 1
Empirical Formula = CH₂O
Step 5: Molecular Formula
- Definition: Actual number of atoms of each element in a molecule.
- Relation with empirical formula:
\text{Molecular formula} = n \times \text{Empirical formula}
n = \frac{\text{Molar mass of compound}}{\text{Empirical formula mass}}
Example:
If molar mass of compound = 180 g/mol, and empirical formula is CH₂O (mass = 30 g/mol):
n = \frac{180}{30} = 6
\Rightarrow \text{Molecular formula} = (CH₂O)₆ = C₆H₁₂O₆
Evaluation:
- Define percentage composition.
- Calculate the percentage of sodium in NaCl. (Na=23, Cl=35.5)
- What is the empirical formula of a compound that contains 80% C and 20% H?
- Differentiate between empirical and molecular formula.
- A compound contains 27.3% carbon and 72.7% oxygen. Find the empirical formula. (C=12, O=16)
Homework:
- Calculate the percentage composition of elements in NH₄NO₃.
- Determine the empirical formula of a compound with 52.2% C, 13.0% H, and 34.8% O.
- If the molar mass of the compound in (2) is 46 g/mol, find its molecular formula.
Conclusion:
Understanding percentage composition and empirical formula helps in determining the actual makeup of chemical substances and is essential in chemical analysis and synthesis.
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