Emirical formula

Obtaining the Mass (m)

• To derive the empirical formula from the masses of elements:
  • Start by determining the mass of each element in the compound.
  • Use mole ratios to convert these masses.
  • Conclude with the empirical formula of the compound.

Steps to Calculate Empirical Formula from Masses of a Compound

1. Calculate amount (in mol) of each element in the compound using the formula: $n = \frac{m}{M}$
   • where (m) is the mass of the element and (M) is the molar mass of the element.
2. Convert the obtained moles to a whole-number ratio by:
   • Dividing each mole value by the smallest number of moles calculated in the previous step.
3. Establish the simplest whole-number ratio of the atoms of each element present.
4. Write the empirical formula based on this ratio.

Worked Example 7.4.2: Determining the Empirical Formula from Percentage Composition

• Given percentages are: 52.2% Carbon (C), 13.0% Hydrogen (H), 34.8% Oxygen (O) (calculation: 100 - 52.2 - 13.0).

1. Assume 100 g sample:
   • m(C) = 52.2 g
   • m(H) = 13.0 g
   • m(O) = 34.8 g
2. Calculating moles:
   • $n(C) = \frac{52.2}{12.0} = 4.35 ext{ mol}$
   • $n(H) = \frac{13.0}{1.0} = 13.0 ext{ mol}$
   • $n(O) = \frac{34.8}{16.0} = 2.18 ext{ mol}$
3. Complete Simplification:
   • Ratios:
     • C: $\frac{4.35}{2.18} = 2.0$
     • H: $\frac{13.0}{2.18} = 6.0$
     • O: $\frac{2.18}{2.18} = 1.0$
4. Empirical formula: CHO -> C₂H₆O

Chemical Analysis of Warfarin

• Composition: Warfarin is composed of Carbon, Hydrogen, and Oxygen.
• Use of Elemental Microanalysis:
  • Burn the compound in air to analyze it. Combustion products are measured to find the composition:
  • Carbon dioxide (absorbed by NaOH), water (absorbed by Mg(CIO)).
• Calculate from elemental data:
  • $n(CO₂) = \frac{m}{M}$
  • $n(C) = n(CO₂)$
• From the water produced:
  • $n(H₂O) = \frac{m}{18.0}$
  • $n(H) = 2 \times n(H₂O)$

Determining the Mass of Carbon and Hydrogen in Organic Compound

1. From the mass of CO₂ produced:
   • $n(C) = n(CO₂)$ gives the total mol of carbon.
   • $m(C) = n(C) \times 12.0 ext{ g}$
2. For hydrogen from the water:
   • $n(H₂O) = \frac{m}{18.0}$
   • $n(H) = 2 \times n(H₂O)$
   • $m(H) = n(H) \times 1.0 ext{ g}$

Molecular Formula Versus Empirical Formula

• Molecular formula indicates exact numbers of atoms in a molecule whereas empirical formula shows simplest whole-number ratio.
• A molecular formula contains whole number multiples of empirical formulas determined using: $ext{Number of empirical units} = \frac{\text{Molar mass of compound}}{\text{Molar mass of empirical formula}}$
  • Example: For empirical formula CH,
  • Molar mass calculated (C=12, H=1) = 13 g/mol
  • Molar mass compound = 78 g/mol → 6 x CH → C₆H₁₂.

Percentage Composition

• Total mass proportions can be expressed as % composition of elements in compound using:
  \text{% by mass} = \frac{\text{mass of element in 1 mol}}{\text{molar mass of compound}} \times 100
• Example: For CO₂:
  • Molar mass CO₂ = 44 g/mol, Carbon mass = 12 g →
    \text{% of C} = \frac{12}{44} \times 100 = 27.3\%

Practical Examples in Molecular Science

• Understanding molecular formulas helps interpret chemical equations and substance behaviors.
• Thorough understanding of empirical and molecular formulas is primordial for various fields of chemistry, like pharmacology and materials science.