Empirical & Molecular Formula – Study Notes

Core Definitions – Empirical vs. Molecular Formula

• Empirical Formula (EF): simplest whole-number ratio of atoms in a compound.

• Molecular Formula (MF): actual number of atoms in a molecule.

• Relationship: MF is an integer multiple of EF.

Percentage Composition

• Definition: percent by mass of each element in a compound.

• General expression: $\%[element] = \dfrac{\text{mass of element in 1 mol}}{\text{molar mass of compound}} \times 100$

• Carbohydrate sample (total $2.225\,\text{g}$: $0.890\,\text{g C}$, $0.148\,\text{g H}$, remainder O).

Calculating an Empirical Formula (General Procedure)

1. List element symbols.

2. If percentages are given, assume $100\,\text{g}$ of compound; convert % \rightarrow grams.

3. Convert grams to moles for each element (use $n = \dfrac{m}{M}$).

4. Divide each mole value by the smallest mole value to obtain ratios.

5. Adjust to whole numbers (round reasonably or multiply by a common factor).

Worked Example (Hydrocarbon 75 % C / 25 % H)

• Assume $100\,\text{g}$ \rightarrow $75.0\,\text{g C}$, 25.0\,\text{g H}}.

• Moles: $n<em>C = \dfrac{75.0}{12.01} \approx 6.24$; $n</em>H = \dfrac{25.0}{1.008} \approx 24.8$.

• Ratio: $\dfrac{6.24}{6.24} : \dfrac{24.8}{6.24} \approx 1 : 4$ \rightarrow EF = $\text{CH}_4$ (methane).

Calculating a Molecular Formula

• Requirements: empirical formula and experimental molar mass $M_r$.

• Steps:

  1. Compute empirical mass (EM).

  2. Evaluate $n = \dfrac{M_r}{\text{EM}}$ (should be integer).

  3. Multiply all subscripts in EF by $n$ to obtain MF.

• Example – Hydrazine

  • Composition: $87.4\,\%\,\text{N}$, $12.6\,\%\,\text{H}$.

  • EF determined (through standard steps) as $\text{NH}_2$.

  • EM = $14.01 + 2(1.008) = 16.026$.

  • Given $M_r = 32.06$ \rightarrow $n = \dfrac{32.06}{16.026} \approx 2$.

  • MF = $\text{N}<em>2\text{H}</em>4$.

• Practice questions include:

  • Carbohydrate sample $m = 0.8361\,\text{g},<br>M_r = 88$.

  • Nicotine: mass % $74.1\,\text{C}$, $8.7\,\text{H}$, 17.2\,\text{N};\n M_r \approx 160.

  • Phosphorus oxide: $43.6\,\%\,\text{P}$, 56.4\,\%\,\text{O};\n 0.006\,\text{mol} weighs $1.703\,\text{g}$.