8 - Molecular and Empirical Formula
Molecular Composition and Empirical Formula
Definitions
Mass Percent:
Calculated as the mass of an element in 1 mole of the compound divided by the molar mass of the compound, multiplied by 100%.
Example: For C₆H₁₂O₆, calculate mass percent for Carbon, Hydrogen, and Oxygen.
Calculating Mass Percent
Carbon:
Mass of Carbon = 6 x 12.01 g/mol = 72.06 g
Molar mass of C₆H₁₂O₆ = 180.18 g/mol
Percent Mass = (72.06 g / 180.18 g) x 100% = 40.00%
Hydrogen:
Mass of Hydrogen = 12 x 1.008 g/mol = 12.096 g
Percent Mass = (12.096 g / 180.18 g) x 100% = 6.71%
Oxygen:
Percent Mass = 100% - (40.00% + 6.71%) = 53.29%
Empirical vs. Molecular Formulas
Empirical Formula:
The simplest whole-number ratio of elements in a compound.
Molecular Formula:
The actual number of atoms of each element in a molecule. It may be the same as the empirical formula or a multiple of it.
Examples of Formulas
Example Compound: C₆H₁₂O₆
Empirical Formula = CH₂O (divided by 6)
Another Compound: C₆H₅O₂
Empirical Formula: C₃H₅O (divided by 2)
Solving for Empirical Formula
Scenario Calculation Example
Given:
Carbon = 37.51%,
Hydrogen = 4.20%,
Oxygen = 58.29% (assumed as 100g basis)
Convert to grams:
C = 37.51 g, H = 4.20 g, O = 58.29 g
Convert grams to moles:
C: 37.51 g / 12.01 g/mol = 3.12 mol
H: 4.20 g / 1.008 g/mol = 4.17 mol
O: 58.29 g / 16.00 g/mol = 3.64 mol
Calculate ratios (divide by smallest number, 3.12):
C = 1
H = 1.33
O = 1.17
Making whole numbers:
Multiply by 6: C₆H₈O₇ (empirical formula)
Finding Molecular Formula
Given Molecular Mass: 384.25 g/mol
Empirical Formula Mass: 192.12 g/mol
Ratio: 384.25 / 192.12 = 2
Molecular Formula = 2(C₆H₈O₇) = C₁₂H₁₆O₁₄
Tips for Multiplying Ratios to Whole Numbers
If decimal ends in 0.5, multiply by 2.
If decimal ends in .33, multiply by 3.
If decimal ends in .66, multiply by 3.
If decimal ends in 0.25, multiply by 4.
Example Problem with Isogenol
Given: 0.500g of Isogenol combusted to yield CO₂ and H₂O
Calculation involves determining grams of Carbon and Hydrogen from CO₂ and H₂O produced, then finding mass of Oxygen through subtraction from original mass.