Empirical formulas Knowledge-Based Study

Study notes

1. What is an Empirical Formula?

   In chemistry, chemical formulas convey the composition of substances. The empirical formula represents the simplest whole-number ratio of the different atoms present in a compound.

   In contrast, a molecular formula indicates the actual number of atoms of each element in a single molecule of the compound.

   • Example: Hydrogen peroxide has the molecular formula H2O2. Dividing both subscripts by their greatest common factor, 2, yields its empirical formula: HO.
   • Example: Glucose has the molecular formula C6H12O6. Dividing each subscript by 6 gives the empirical formula: CH2O.
   • Example: Water has the molecular formula H2O. Because the ratio 2:1 cannot be simplified further into whole numbers, its empirical formula is also H2O.

   Calculating Empirical Formulas

   Empirical formulas are determined experimentally from the masses or percentage compositions of the constituent elements. To determine the empirical formula, follow these fundamental steps:

   1. Assume a 100 g sample (if percentages are given): This allows you to directly convert each percentage to mass in grams (e.g., 40.0% carbon becomes 40.0 g of C).
   2. Convert masses to moles: Divide the mass of each element by its respective molar mass using the periodic table:
      $$n = \frac{m}{M}$$
   3. Determine the mole ratio: Divide all calculated mole values by the smallest mole value among them.
   4. Convert to whole numbers: If all resulting values are whole numbers (or within $\pm 0.05$ of a whole number), write the empirical formula directly. If fractions remain, multiply all values by an appropriate integer.

   Sample Problem

   A compound contains 72.4% iron (Fe) and 27.6% oxygen (O) by mass. Determine its empirical formula. (Molar masses: $\text{Fe} = 55.85\text{ g/mol}$, $\text{O} = 16.00\text{ g/mol}$)

   Step 1: Convert percentages to mass in 100 g

   • Mass of Fe = 72.4 g
   • Mass of O = 27.6 g

   Step 2: Convert mass to moles

   • Moles of Fe: $$n_{\text{Fe}} = \frac{72.4\text{ g}}{55.85\text{ g/mol}} \approx 1.296\text{ mol}$$
   • Moles of O: $$n_{\text{O}} = \frac{27.6\text{ g}}{16.00\text{ g/mol}} \approx 1.725\text{ mol}$$

   Step 3: Divide by the smallest mole value

   • Fe: $$\frac{1.296}{1.296} = 1.00$$
   • O: $$\frac{1.725}{1.296} \approx 1.33$$

   Step 4: Convert to simplest whole-number ratio

   Since 1.33 corresponds to $1\frac{1}{3} = \frac{4}{3}$, multiply all ratios by 3:

   • Fe: $1.00 \times 3 = 3$
   • O: $1.33 \times 3 = 4$

   The empirical formula of the compound is Fe3O4.

   Handling Fractional Mole Ratios

   When the mole ratio is not initially a whole number after dividing by the smallest value, do not round off arbitrarily. Instead, identify the decimal and multiply all coefficients by the appropriate integer:

   • Ending in ~0.50: Multiply all values by 2 (e.g., $1 : 1.5 ightarrow 2 : 3$)
   • Ending in ~0.33 or ~0.67: Multiply all values by 3 (e.g., $1 : 1.33 ightarrow 3 : 4$)
   • Ending in ~0.25 or ~0.75: Multiply all values by 4 (e.g., $1 : 1.25 ightarrow 4 : 5$)
   • Ending in ~0.20: Multiply all values by 5 (e.g., $1 : 1.20 ightarrow 5 : 6$)

   Connecting Empirical to Molecular Formula

   If the molar mass of the compound is known, the molecular formula can be found using the integer factor $n$:

   $$n = \frac{\text{Molar mass of compound}}{\text{Empirical formula mass}}$$$$\text{Molecular formula} = (\text{Empirical formula})_n$$