S1.4.4 Empirical and molecular formulas

Simplest and Actual Ratios: Empirical and Molecular Formulas

1. You’re a detective in a chemistry lab, analyzing a mysterious compound.
2. You know its elements and their relative amounts, but what does the compound look like at the molecular level? Is it a simple molecule, or does it consist of repeating units?

To uncover the full picture, you’ll need two essential tools: the empirical formula and the molecular formula.

The Empirical Formula: The Simplest Ratio

It doesn’t indicate the exact number of atoms in a molecule—just their relative proportions.

Why is the empirical formula useful?

• The empirical formula is especially valuable when analyzing experimental data, such as percentage composition.
• It serves as a stepping stone for determining the molecular formula, which provides more detailed information.

The Molecular Formula: The Actual Number of Atoms

It is either identical to the empirical formula or an integer multiple of it.

From Formulas to Percentages: Step-by-Step Calculations

It can be calculated using the formula:

$$\mathrm{\%}\text{Element}=\frac{\text{Mass of element in 1 mole of compound}}{\text{Molar mass of compound}}\times 100$$

From Percentages to Formulas: Step-by-Step Calculations

Now that you understand the difference between empirical and molecular formulas, let’s explore how to calculate them using experimental data.

1. Calculating the Empirical Formula from Percentage Composition

Suppose you’re given the percentage composition of a compound. Follow these steps:

1. Convert percentages to masses:
   Assume you have 100 g of the compound. This makes the percentages equivalent to masses in grams. For example, if a compound is 40% carbon, 6.7% hydrogen, and 53.3% oxygen, you can assume:
   • Carbon: 40 g
   • Hydrogen: 6.7 g
   • Oxygen: 53.3 g
2. Convert masses to moles:
   Use the formula $n=\frac{m}{M}$, where $n$ is the number of moles, $m$ is the mass in grams, and $M$ is the molar mass (relative atomic mass in g/mol). For the example above:
   • Carbon: $n=\frac{40}{12.01}\approx 3.33\mathrm{mol}$
   • Hydrogen: $n=\frac{6.7}{1.008}\approx 6.65\mathrm{mol}$
   • Oxygen: $n=\frac{53.3}{16.00}\approx 3.33\mathrm{mol}$
3. Find the simplest ratio:
   Divide all mole values by the smallest number of moles:
   • Carbon: $\frac{3.33}{3.33}=1$
   • Hydrogen: $\frac{6.65}{3.33}\approx 2$
   • Oxygen: $\frac{3.33}{3.33}=1$
4. The empirical formula is CH₂O.
5. Adjust for whole numbers (if necessary):
   If any ratios are not whole numbers, multiply all ratios by the smallest factor that converts them to integers. For example, a ratio like 1:1.5:1 would be multiplied by 2 to give 2:3:2.

2. Determining the Molecular Formula

Once you have the empirical formula, follow these steps to find the molecular formula:

1. Calculate the molar mass of the empirical formula:
   Add up the relative atomic masses of all atoms in the empirical formula. For CH₂O, the molar mass is:
   $$M_{\text{empirical}}=12.01+(2\times 1.008)+16.00=30.03\mathrm{g}/\mathrm{mol}$$
2. Divide the compound’s molar mass by the empirical formula’s molar mass:
   $$\text{Multiplier}=\frac{M_{\text{molecular}}}{M_{\text{empirical}}}$$
3. For a compound with a molar mass of 180.18 g/mol:
   $$\text{Multiplier}=\frac{180.18}{30.03}\approx 6$$
4. Multiply the empirical formula by this factor:
   $$\text{Molecular formula}=\text{Empirical formula}\times 6={\text{C}}_6{\text{H}}_{12}{\text{O}}_6$$

Reflection and Connections