2.8 Empirical and molecular formula from % composition
🎯 Learning outcomes
After the completion of the chapter, the students should be able to:
- Find the empirical and molecular formula from the percentage composition.
Core Concepts
Empirical Formula: The simplest whole-number ratio of atoms of each element present in a compound.
Molecular Formula: The actual number of atoms of each element present in one molecule of the compound.
Relationship: Molecular Formula = n × (Empirical Formula), where n = Molecular Weight / Empirical Formula Weight
Vapour Density Relationship: For volatile organic compounds, the molecular weight can be determined via its Vapour Density (VD): Molecular Weight = 2 × Vapour Density
Worked Examples (Formulas from % Composition)
Example 1: Investigation of a Plant Extract from Khaptad
A plant sample was collected from the pristine alpine regions of Khaptad, Far-Western Nepal. Solvent extraction of the plant contents revealed the presence of a suspected organic acid. Elemental analysis showed that the compound consists of 40% Carbon, 6.66% Hydrogen, and 53.44% Oxygen. Nitrogen, sulphur, and halogens were absent. If the molecular weight of this acid is determined to be 60, calculate its empirical and molecular formulas.
Step 1: Determine the Empirical Formula using an Element Ratio Table
| Element | % by Mass | Atomic Mass | Relative Moles | Simple Ratio | Simplest Whole Ratio |
|---|---|---|---|---|---|
| Carbon (C) | 40.00% | 12 | 40.00 / 12 = 3.33 | 3.33 / 3.33 = 1 | 1 |
| Hydrogen (H) | 6.66% | 1 | 6.66 / 1 = 6.66 | 6.66 / 3.33 = 2 | 2 |
| Oxygen (O) | 53.44% | 16 | 53.44 / 16 = 3.34 | 3.34 / 3.33 = 1.003 | 1 |
Empirical Formula = CH₂O
Step 2: Determine the Molecular Formula
1. Empirical Formula Weight = 12 + (2 × 1) + 16 = 30 amu
2. Given Molecular Weight = 60
3. Find n: n = Molecular Weight / Empirical Formula Weight = 60 / 30 = 2
4. Molecular Formula = 2 × (CH₂O) = C₂H₄O₂ (Acetic Acid / Ethanoic Acid)
1. Empirical Formula Weight = 12 + (2 × 1) + 16 = 30 amu
2. Given Molecular Weight = 60
3. Find n: n = Molecular Weight / Empirical Formula Weight = 60 / 30 = 2
4. Molecular Formula = 2 × (CH₂O) = C₂H₄O₂ (Acetic Acid / Ethanoic Acid)
Example 2: Organic Compound via Vapour Density
An organic compound containing carbon, hydrogen, and oxygen gives the following mass composition: C = 40.687%, H = 5.085%, and O = 54.228%. The vapour density of the compound is measured to be 59. Calculate its empirical and molecular formulas.
Step 1: Determine the Empirical Formula
| Element | % by Mass | Atomic Mass | Relative Moles | Ratio | Simplest Whole Ratio |
|---|---|---|---|---|---|
| C | 40.687% | 12 | 40.687 / 12 = 3.390 | 3.390 / 3.389 = 1 | 2 (1 × 2) |
| H | 5.085% | 1 | 5.085 / 1 = 5.085 | 5.085 / 3.389 = 1.5 | 3 (1.5 × 2) |
| O | 54.228% | 16 | 54.228 / 16 = 3.389 | 3.389 / 3.389 = 1 | 2 (1 × 2) |
Note: Because the structural ratio for H is 1.5, we multiply all numbers by 2 to achieve integers.
Empirical Formula = C₂H₃O₂
Step 2: Determine the Molecular Formula
1. Molecular Weight = 2 × Vapour Density = 2 × 59 = 118 g/mol
2. Empirical Formula Weight = (12 × 2) + (1 × 3) + (16 × 2) = 24 + 3 + 32 = 59 amu
3. Find n: n = 118 / 59 = 2
4. Molecular Formula = 2 × (C₂H₃O₂) = C₄H₆O₄
1. Molecular Weight = 2 × Vapour Density = 2 × 59 = 118 g/mol
2. Empirical Formula Weight = (12 × 2) + (1 × 3) + (16 × 2) = 24 + 3 + 32 = 59 amu
3. Find n: n = 118 / 59 = 2
4. Molecular Formula = 2 × (C₂H₃O₂) = C₄H₆O₄
Practice Questions
- Write the relationship between the empirical formula and molecular formula of a compound.
- A plant was collected from Khaptad, far western Nepal. From the solvent extraction of the plant contents, an acid was suspected. From elemental analysis, the compound was found to consist of 40% carbon, 6.66% hydrogen and 53.44% oxygen. Element detection shows the absence of nitrogen, sulphur, and halogens. The molecular weight of the acid was found to be 60. Determine the empirical formula and molecular formula of the acid.
- An organic compound containing carbon, hydrogen and oxygen gave the following composition: C = 40.687%, H = 5.085% and O = 54.228%. The vapour density of the compound is 59. Calculate the empirical formula and molecular formula of the compound.
- Glucose contains 40% C, 6.67% H and 53.3% O by mass. Calculate its empirical formula.
🧪 Test Your Understanding
Try this quick quiz to check how well you’ve grasped empirical and molecular formula calculations.
📋 Empirical & Molecular Formula Quiz
