How to Calculate Empirical and Molecular Formulas in Chemistry

1. What is an Empirical Formula?

• Definition: The simplest whole-number ratio of elements in a compound.
• Example: The empirical formula of hydrogen peroxide (H₂O₂) is HO.

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2. What is a Molecular Formula?

• Definition: The actual number of atoms of each element in a molecule.
• Difference between molecular and empirical formulas.
• Example: Hydrogen peroxide has a molecular formula of H₂O₂, which is a multiple of its empirical formula, HO.

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3. Steps to Determine Empirical Formulas

• Step 1: Convert mass percentages to grams.
• Step 2: Convert grams to moles for each element using atomic masses.
• Step 3: Divide each mole value by the smallest mole value.
• Step 4: If necessary, multiply ratios to get whole numbers.
• Example: Explanation of the “assume 100 g sample” approach when working with percentages.

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4. Example Calculation: Empirical Formula of Lead Chloride

• Given: 0.6884 g of lead (Pb) and 0.2356 g of chlorine (Cl).
• Steps:
  • Convert grams to moles: $$0.6884\text{g Pb}\times \frac{1\text{mol}}{207.2\text{g}}=0.003322\text{mol Pb}$$$$0.2356\text{g Cl}\times \frac{1\text{mol}}{35.45\text{g}}=0.006629\text{mol Cl}$$
  • Divide by the smallest number of moles: $$\frac{0.003322}{0.003322} = 1 \, \text{(Pb)}$$
    $$\frac{0.006629}{0.003322}\approx 2\text{(Cl)}$$
  • Result: Empirical formula is $\text{PbCl}_2$.

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5. Steps to Determine Molecular Formulas

• Step 1: Calculate the empirical formula mass (EFM).
• Step 2: Divide the molar mass of the compound by the empirical formula mass to find the multiplier.
• Step 3: Multiply the subscripts in the empirical formula by the multiplier.
• Example: Emphasis on how molecular formulas are whole-number multiples of empirical formulas.

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6. Example Calculation: Molecular Formula of a Gasoline Additive

• Given Data:
  • Percent composition: 71.65% chlorine (Cl), 24.27% carbon (C), 4.07% hydrogen (H).
  • Molar mass: 98.96 g/mol.
• Steps:
  • Convert Percentages to Grams (Assuming 100 g sample):
    • 71.65 g Cl, 24.27 g C, 4.07 g H.
  • Convert Grams to Moles: $$71.65\text{g Cl}\times \frac{1\text{mol}}{35.45\text{g}}=2.02\text{mol Cl}$$$$24.27\text{g C}\times \frac{1\text{mol}}{12.01\text{g}}=2.02\text{mol C}$$$$4.07\text{g H}\times \frac{1\text{mol}}{1.01\text{g}}=4.03\text{mol H}$$
  • Determine Empirical Formula:
    • Simplest ratio is $\text{Cl}_1 \text{C}_1 \text{H}_2$, so the empirical formula is $\text{CClH}_2$.
  • Calculate Empirical Formula Mass: $$\text{<br>}(35.45\text{g/mol})+(12.01\text{g/mol})+(2\times 1.01\text{g/mol})=49.48\text{g/mol}$$
  • Find the Multiplier for Molecular Formula: $$\frac{98.96\text{g/mol}}{49.48\text{g/mol}}\approx 2$$
  • Determine Molecular Formula:
    • Multiply each subscript in $\text{CClH}_2$ by 2, giving $\text{C}_2 \text{Cl}_2 \text{H}_4$.
  • Result: Molecular formula is $\text{C}_2 \text{Cl}_2 \text{H}_4$.