Calculate empirical formulae and molecular formulae, given appropriate data

Objective: Learn how to calculate empirical and molecular formulae from experimental data.

We’ll use mole as a “shopping basket” that always holds the same number of items – the Avogadro constant.

A mole is a unit that tells us how many atoms, ions or molecules are in a sample.

It is defined as the amount of substance that contains the same number of entities as there are atoms in 12 g of pure carbon‑12.

Avogadro constant: \$N_A = 6.022 \times 10^{23}\;\text{entities mol}^{-1}\$

Think of a mole as a supermarket basket that always holds exactly \$6.022 \times 10^{23}\$ items, no matter what the items are.

In chemistry we often measure mass, but reactions happen at the level of atoms and molecules.

The mole bridges the gap between the microscopic world (atoms) and the macroscopic world (grams).

With the mole, we can convert:

• Mass → Number of entities: \$n = \dfrac{m}{M}\$
• Number of entities → Mass: \$m = n \times M\$
• Number of entities → Moles: \$n = \dfrac{N}{N_A}\$

The empirical formula is the simplest whole‑number ratio of atoms in a compound.

Steps to find it:

1. Convert masses of each element to moles.
2. Divide all mole values by the smallest mole value.
3. Round to the nearest whole number (or a simple fraction).
4. Write the formula using the rounded numbers as subscripts.

Example:

Combustion of 1.00 g of a hydrocarbon gives 2.44 g CO₂ and 0.88 g H₂O.

• Carbon from CO₂: \$m_{\text{C}} = 2.44\,\text{g} \times \dfrac{12.01}{44.01} = 0.666\,\text{g}\$
• Hydrogen from H₂O: \$m_{\text{H}} = 0.88\,\text{g} \times \dfrac{2.02}{18.02} = 0.099\,\text{g}\$
• Moles: \$n{\text{C}} = 0.666/12.01 = 0.0554\,\text{mol}\$, \$n{\text{H}} = 0.099/1.008 = 0.0983\,\text{mol}\$
• Divide by smallest (\$0.0554\$): \$C:1\$, \$H:1.77 \approx 2\$
• Empirical formula: \$\boxed{\text{CH}_2}\$

The molecular formula tells the exact number of atoms in one molecule.

To find it:

1. Determine the empirical formula.
2. Find the empirical formula mass (EFM).
3. Divide the given molar mass (MM) by the EFM to get the ratio \$x\$.
4. Multiply each subscript in the empirical formula by \$x\$ to get the molecular formula.

Example:

• Empirical formula: \$\text{C}2\text{H}4\$ (EFM = \$2\times12.01 + 4\times1.008 = 28.05\$ g mol⁻¹).
• Given molar mass: \$56.10\$ g mol⁻¹.
• \$x = 56.10/28.05 = 2\$.
• Molecular formula: \$\text{C}4\text{H}8\$.

• Always check units: grams → moles → atoms. A common mistake is mixing up grams and moles.
• Round only at the end: Keep full precision during calculations to avoid rounding errors.
• Use the smallest mole value: This ensures the simplest whole‑number ratio.
• Remember Avogadro’s number: \$6.022\times10^{23}\$ is the key to converting between moles and atoms.
• Check your answer: Verify that the empirical formula mass times the ratio equals the given molar mass.

A sample of a compound weighing 3.00 g yields 4.00 g CO₂ and 2.00 g H₂O on complete combustion. The compound’s molar mass is 84.12 g mol⁻¹. Find the empirical and molecular formula.

Solution outline:

1. Convert CO₂ and H₂O masses to moles of C and H.
2. Determine the mole ratio → empirical formula.
3. Compute EFM and divide molar mass by EFM → ratio \$x\$.
4. Multiply empirical subscripts by \$x\$ → molecular formula.

Try it yourself before checking the answer in the next section!

Step 1: CO₂ → \$4.00\,\text{g} \times \dfrac{12.01}{44.01} = 1.09\,\text{g C}\$ → \$n_{\text{C}} = 1.09/12.01 = 0.0908\,\text{mol}\$

H₂O → \$2.00\,\text{g} \times \dfrac{2.02}{18.02} = 0.224\,\text{g H}\$ → \$n_{\text{H}} = 0.224/1.008 = 0.222\,\text{mol}\$

Step 2: Ratio \$C:H = 0.0908 : 0.222 \approx 1 : 2.44 \approx 1 : 2.5\$ → empirical formula \$\text{C}1\text{H}{2.5}\$ → round to \$\text{C}2\text{H}5\$.

Step 3: EFM = \$2\times12.01 + 5\times1.008 = 29.05\$ g mol⁻¹.

Step 4: \$x = 84.12/29.05 = 2.90 \approx 3\$.

Molecular formula: \$\text{C}6\text{H}{15}\$.