Percentage Composition
Examples: H₂O, CH₄, C₆H₁₂O₆, Fe₂O₃
What is Percentage Composition?
Percentage Composition (% Composition) is the mass percentage of each element in a chemical compound.
Also known as: Mass Percent (%)
Formula unit: Empirical Formula
Key Concepts:
- Mass Percent: (Element mass / Total mass) × 100
- Empirical Formula: Simplest ratio of atoms
- Molecular Formula: Actual number of atoms
- Molar Mass: Sum of all atomic masses
Formulas
Percentage Composition:
% Element = (Atomic mass × Count) / Molar mass × 100
% Element = (Atomic mass × Count) / Molar mass × 100
Total Molar Mass:
M = Σ(Atomic mass × Count)
M = Σ(Atomic mass × Count)
Individual Element Mass:
Mass(Element) = Atomic mass × Count
Mass(Element) = Atomic mass × Count
Sum of All Percentages:
Σ % = 100% (always)
Σ % = 100% (always)
Note: The sum of all percentages should always equal 100% (or very close, depending on rounding).
Examples
Water (H₂O)
H: 2.016 / 18.015 × 100 = 11.19%
O: 15.999 / 18.015 × 100 = 88.81%
H: 2.016 / 18.015 × 100 = 11.19%
O: 15.999 / 18.015 × 100 = 88.81%
Carbohydrate (C₆H₁₂O₆)
C: 72.064 / 180.156 × 100 = 40.00%
H: 12.096 / 180.156 × 100 = 6.71%
O: 95.996 / 180.156 × 100 = 53.29%
C: 72.064 / 180.156 × 100 = 40.00%
H: 12.096 / 180.156 × 100 = 6.71%
O: 95.996 / 180.156 × 100 = 53.29%
Sodium Chloride (NaCl)
Na: 22.990 / 58.443 × 100 = 39.34%
Cl: 35.453 / 58.443 × 100 = 60.66%
Na: 22.990 / 58.443 × 100 = 39.34%
Cl: 35.453 / 58.443 × 100 = 60.66%
Sulfuric Acid (H₂SO₄)
H: 2.016 / 98.079 × 100 = 2.06%
S: 32.06 / 98.079 × 100 = 32.69%
O: 63.996 / 98.079 × 100 = 65.25%
H: 2.016 / 98.079 × 100 = 2.06%
S: 32.06 / 98.079 × 100 = 32.69%
O: 63.996 / 98.079 × 100 = 65.25%
Iron Oxide (Fe₂O₃)
Fe: 111.694 / 159.687 × 100 = 69.94%
O: 47.997 / 159.687 × 100 = 30.06%
Fe: 111.694 / 159.687 × 100 = 69.94%
O: 47.997 / 159.687 × 100 = 30.06%
Methanol (CH₄O)
C: 12.011 / 32.042 × 100 = 37.48%
H: 4.032 / 32.042 × 100 = 12.58%
O: 15.999 / 32.042 × 100 = 49.94%
C: 12.011 / 32.042 × 100 = 37.48%
H: 4.032 / 32.042 × 100 = 12.58%
O: 15.999 / 32.042 × 100 = 49.94%
Technical Background
Difference: Molecular Formula vs. Empirical Formula
Two different compounds can have the same percentage composition but different molecular formulas:
- Formaldehyde: CH₂O – 40% C, 6.7% H, 53.3% O
- Glucose: C₆H₁₂O₆ – 40% C, 6.7% H, 53.3% O (identical!)
Both have the same empirical formula CH₂O and identical percentage composition!
Atomic Masses (IUPAC 2021):
| Element | Symbol | Atomic Mass | Element | Symbol | Atomic Mass | |
|---|---|---|---|---|---|---|
| Hydrogen | H | 1.008 | Carbon | C | 12.011 | |
| Nitrogen | N | 14.007 | Oxygen | O | 15.999 | |
| Phosphorus | P | 30.974 | Sulfur | S | 32.06 | |
| Chlorine | Cl | 35.453 | Sodium | Na | 22.990 | |
| Potassium | K | 39.098 | Calcium | Ca | 40.078 | |
| Iron | Fe | 55.845 | Copper | Cu | 63.546 |
Practical Applications:
- Quality Control: Verification of purity of chemical compounds
- Analytical Chemistry: Determination of unknown compositions through elemental analysis
- Material Development: Characterization of new materials and alloys
- Empirical Formula Determination: Finding molecular formulas from combustion data
- Pharmacy: Standardization and verification of pharmaceutical active ingredients
Calculation from Combustion Data:
When C, H, and O are known from combustion analysis, calculate the empirical formula from their mole ratios, then derive the percentage composition.
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