ANAND CLASSES Study Material and Notes to learn Empirical Formula and Molecular Formula with detailed explanations, solved examples, worksheets, and key differences. Perfect for JEE, NEET, and CBSE Class 11 Chemistry preparation.
🔬 Empirical Formula
📌 Definition:
The empirical formula of a chemical compound represents the simplest whole number ratio of atoms of each element present in the compound.
🔍 Key Characteristics:
- It does not show the actual number of atoms, just their simplest ratio.
- Often derived from percentage composition or mass ratio.
- It is not unique – different compounds may have the same empirical formula.
📊 Example 1: Glucose (C₆H₁₂O₆)
Let’s reduce this molecular formula:
- C : H : O = 6 : 12 : 6
- Simplify by dividing each number by 6:
→ C : H : O = 1 : 2 : 1
✅ Empirical Formula = CH₂O
📊 Example 2: Hydrogen Peroxide (H₂O₂)
- H : O = 2 : 2 → Simplified = 1 : 1
✅ Empirical Formula = HO
🧠 How to Determine the Empirical Formula from Percentage Composition:
💡 Steps:
- Assume 100 g of the compound → Convert percentages to grams.
- Convert grams to moles using molar mass.
- Divide all mole values by the smallest mole to get the simplest ratio.
- Round off to the nearest whole number (if needed).
- Write the empirical formula using those values.
🔍 Example:
A compound contains:
- 40% C
- 6.7% H
- 53.3% O
Step 1: Convert to grams (assume 100 g)
- C = 40 g, H = 6.7 g, O = 53.3 g
Step 2: Convert to moles $$\text{Moles of C} = \frac{40}{12} = 3.33 $$
$$ \text{Moles of H} = \frac{6.7}{1} = 6.7$$
$$ \text{Moles of O} = \frac{53.3}{16} = 3.33$$
Step 3: Divide all by the smallest value (3.33):
- C = 1, H = 2, O = 1
✅ Empirical Formula = CH₂O
🌟 Molecular Formula
📌 Definition:
The molecular formula of a compound shows the actual number of atoms of each element present in one molecule of the compound.
🔍 Key Characteristics:
- It is an exact multiple of the empirical formula.
- It represents the actual composition of a molecule.
- Requires knowledge of molar mass.
💡 Formula:
$$\text{Molecular Formula} = (\text{Empirical Formula}) \times n$$
Where, $$n = \frac{\text{Molar Mass of Compound}}{\text{Empirical Formula Mass}}$$
📊 Example using CH₂O as Empirical Formula:
Let’s say molar mass of the compound is 180 g/mol
- Empirical formula mass of CH₂O = 12 + 2 + 16 = 30 g/mol
- Now,
$$n = \frac{180}{30} = 6$$
- Multiply empirical formula by 6:
$$\text{Molecular Formula} = (CH₂O) \times 6 = C₆H₁₂O₆$$
✅ Final Answer: C₆H₁₂O₆
⚖️ Difference Between Empirical and Molecular Formula
| Feature | Empirical Formula | Molecular Formula |
|---|---|---|
| Shows | Simplest ratio | Actual number |
| Example (Glucose) | CH₂O | C₆H₁₂O₆ |
| Derived from | % composition | Empirical formula + Molar mass |
| Unique for each compound? | ❌ No | ✅ Yes |
Great! Below are solved examples on Empirical and Molecular Formulas in a step-by-step format with tables, perfect for Class 11, NEET, and JEE. These will help you master the concept through real application.
🧪 Solved Examples on Empirical and Molecular Formula
✅ Example 1:
A compound contains 40% Carbon, 6.7% Hydrogen, and 53.3% Oxygen by mass. The molar mass is 180 g/mol. Find the empirical and molecular formula.
📊 Step-by-Step Table:
| Element | % Composition | Mass (g) (Assume 100 g) | Atomic Mass (g/mol) | Moles = Mass ÷ Atomic Mass | Simplest Ratio | Rounded Ratio |
|---|---|---|---|---|---|---|
| C | 40.0% | 40.0 g | 12 | 40 ÷ 12 = 3.33 | 3.33 ÷ 3.33 = 1.00 | 1 |
| H | 6.7% | 6.7 g | 1 | 6.7 ÷ 1 = 6.7 | 6.7 ÷ 3.33 ≈ 2.01 | 2 |
| O | 53.3% | 53.3 g | 16 | 53.3 ÷ 16 = 3.33 | 3.33 ÷ 3.33 = 1.00 | 1 |
✅ Empirical Formula = CH₂O
📘 Empirical Formula Mass = 12 + (2 × 1) + 16 = 30 g/mol
🔄 Calculate n:
$$n = \frac{\text{Molar Mass}}{\text{Empirical Mass}} = \frac{180}{30} = 6$$
🧪 Molecular Formula = (CH₂O) × 6 = C₆H₁₂O₆
✅ Example 2:
A compound has 85.7% Carbon and 14.3% Hydrogen by mass. Its molar mass is 42 g/mol. Find the empirical and molecular formula.
📊 Step-by-Step Table:
| Element | % Composition | Mass (g) | Atomic Mass | Moles = Mass ÷ Atomic Mass | Simplest Ratio | Rounded Ratio |
|---|---|---|---|---|---|---|
| C | 85.7% | 85.7 g | 12 | 85.7 ÷ 12 = 7.14 | 7.14 ÷ 7.14 = 1 | 1 |
| H | 14.3% | 14.3 g | 1 | 14.3 ÷ 1 = 14.3 | 14.3 ÷ 7.14 = 2 | 2 |
✅ Empirical Formula = CH₂
📘 Empirical Formula Mass = 12 + (2 × 1) = 14 g/mol
🔄 Calculate n:
$$n = \frac{42}{14} = 3$$
🧪 Molecular Formula = (CH₂) × 3 = C₃H₆
✅ Example 3:
A compound contains 29.1% Na, 40.5% S, and 30.4% O. Find the empirical formula.
📊 Step-by-Step Table:
| Element | % Composition | Mass (g) | Atomic Mass | Moles = Mass ÷ Atomic Mass | Simplest Ratio | Rounded Ratio |
|---|---|---|---|---|---|---|
| Na | 29.1% | 29.1 g | 23 | 29.1 ÷ 23 ≈ 1.26 | 1.26 ÷ 1.26 = 1 | 1 |
| S | 40.5% | 40.5 g | 32 | 40.5 ÷ 32 ≈ 1.27 | 1.27 ÷ 1.26 ≈ 1.01 | 1 |
| O | 30.4% | 30.4 g | 16 | 30.4 ÷ 16 ≈ 1.90 | 1.90 ÷ 1.26 ≈ 1.51 | ≈1.5 → 3/2 |
🚨 Multiply all by 2 to remove fraction:
| Element | Adjusted Ratio |
|---|---|
| Na | 1 × 2 = 2 |
| S | 1 × 2 = 2 |
| O | 1.5 × 2 = 3 |
✅ Empirical Formula = Na₂S₂O₃
(This is Sodium Thiosulfate)
✅ Practice Question (with Solution)
Q. A compound contains 24.27% carbon, 4.07% hydrogen, and 71.65% chlorine. The molar mass is 98.96 g/mol. Find the empirical and molecular formula.
✍️ Step-by-step Solution:
1. Convert % to grams (assume 100 g):
- C = 24.27 g
- H = 4.07 g
- Cl = 71.65 g
2. Convert to moles:
- C = 24.27 ÷ 12 = 2.02 mol
- H = 4.07 ÷ 1 = 4.07 mol
- Cl = 71.65 ÷ 35.5 = 2.02 mol
3. Divide by smallest (2.02):
- C = 1, H = 2, Cl = 1
Empirical Formula = CH₂Cl
Empirical formula mass = 12 + 2 + 35.5 = 49.5 g/mol
4. Molar Mass = 98.96 g/mol $$n = \frac{98.96}{49.5} \approx 2$$
Molecular Formula = (CH₂Cl) × 2 = C₂H₄Cl₂
🧪 MCQs on Empirical and Molecular Formula
Q1. The empirical formula of a compound is CH₂. Its molecular mass is 56 g/mol. What is the molecular formula?
A. CH₂
B. C₂H₄
C. C₄H₈
D. C₃H₆
✅ Correct Answer: C. C₄H₈
Explanation:
Empirical formula mass = 12 + (2 × 1) = 14 $$n = \frac{56}{14} = 4$$ $$ \text{Molecular formula} = (CH₂) × 4 = C₄H₈$$
Q2. Which of the following is always the same for a compound?
A. Empirical formula
B. Molecular formula
C. Molecular mass
D. Percentage composition
✅ Correct Answer: D. Percentage composition
Explanation:
The percentage composition remains constant for a given compound, regardless of how its formula is written.
Q3. A compound contains 92.3% carbon and 7.7% hydrogen. What is its empirical formula?
A. CH
B. CH₂
C. C₂H₆
D. C₃H₈
✅ Correct Answer: A. CH
Explanation:
Moles of C = 92.3 ÷ 12 ≈ 7.69
Moles of H = 7.7 ÷ 1 = 7.7
Ratio ≈ C : H = 1 : 1
→ Empirical Formula = CH
Q4. The empirical formula of benzene is:
A. CH
B. C₂H₂
C. C₆H₆
D. C₃H₃
✅ Correct Answer: A. CH
Explanation:
Molecular formula of benzene = C₆H₆
Empirical formula = Simplest ratio = CH
Q5. Which of the following pairs has the same empirical formula?
A. C₂H₂ and C₆H₆
B. C₂H₆ and C₄H₁₀
C. H₂O and H₂O₂
D. C₆H₁₂O₆ and C₂H₄O₂
✅ Correct Answer: A. C₂H₂ and C₆H₆
Explanation:
- C₂H₂ = 2:2 = CH
- C₆H₆ = 6:6 = CH
→ Same empirical formula = CH
📝 Worksheet: Empirical and Molecular Formula
🧪 Section A: Multiple Choice Questions (MCQs)
Q1. A compound contains 36% carbon and 64% oxygen. What is its empirical formula?
A. CO
B. CO₂
C. C₂O
D. C₂O₃
Q2. The empirical formula of a compound is CH and its molar mass is 78 g/mol. The molecular formula is:
A. CH
B. C₆H₆
C. C₂H₂
D. C₃H₃
Q3. A compound has 26.7% C, 2.2% H, and 71.1% O. What is the empirical formula?
A. CHO
B. C₂H₂O₃
C. C₂H₄O₂
D. CH₂O
Q4. Which statement is true?
A. Molecular formula is always a multiple of the empirical formula.
B. Empirical formula is always the actual formula.
C. Empirical and molecular formulas are always the same.
D. None of these.
Q5. Which compound has the same empirical and molecular formula?
A. Glucose (C₆H₁₂O₆)
B. Benzene (C₆H₆)
C. Water (H₂O)
D. Ethylene (C₂H₄)
✍️ Section B: Short Answer Type
Q6. Define empirical formula and molecular formula with suitable examples.
Q7. A compound contains 54.5% carbon, 9.1% hydrogen, and 36.4% oxygen. Calculate the empirical formula.
Q8. Explain how you would determine the molecular formula if you are given the empirical formula and molar mass.
🧮 Section C: Numerical Problems
Q9. A compound contains 70% iron and 30% oxygen by mass. Calculate the empirical formula.
(Atomic masses: Fe = 56, O = 16)
Q10. A compound contains 85.7% carbon and 14.3% hydrogen. Its molar mass is 42 g/mol. Find the empirical and molecular formulas.
✅ Answer Key (for teachers)
| Q. No | Answer |
|---|---|
| Q1 | A. CO |
| Q2 | B. C₆H₆ |
| Q3 | D. CH₂O |
| Q4 | A. |
| Q5 | C. H₂O |
💡 “Do You Know?”
- Many ionic compounds like NaCl, K₂SO₄ do not have molecular formulas – only empirical.
- Some compounds can have same empirical but different molecular formulas:
e.g. CH → C₂H₂ (acetylene), C₆H₆ (benzene)
📘 Quick Revision Points
- Empirical Formula = Simplest ratio of atoms.
- Molecular Formula = Actual number of atoms.
- n = Molecular Mass / Empirical Formula Mass
- Ionic compounds are always written as empirical formulas.
