Empirical formula and molecular formula of a compound (I)

Introduction

When chemists analyse an unknown compound in the lab, they usually don't get to "see" its molecular formula directly. Instead, they measure masses of elements present and work backwards. This gives the empirical formula — the simplest whole-number ratio of atoms in a compound, based on experimental data (Talbot, p.90). Today we'll learn how to calculate it, and how it connects to the true molecular formula.

Empirical vs Molecular Formula

• Empirical formula: simplest whole-number ratio of atoms of each element in a compound.
• Molecular formula: the actual number of atoms of each element in one molecule.

Sometimes these are identical, sometimes not. For example:

• Propane: empirical formula C₃H₈, molecular formula C₃H₈ (same)
• Propene: empirical formula CH₂, molecular formula C₃H₆ (different — molecular formula is a multiple of the empirical one)

As Talbot notes (p.90), for simple molecular compounds many different molecules can share the same empirical formula — CH₂ is the empirical formula for C₂H₄, C₃H₆, C₄H₁₀, and more. However, for ionic compounds (like NaCl or CaO) and giant molecular/covalent network compounds (like SiO₂), the empirical formula is the actual chemical formula — there's no separate "molecular formula" because these substances don't exist as discrete molecules.

Finding the Empirical Formula: Method

1. Write down the mass (or %) of each element given.
2. Convert each mass to moles using n = m / M.
3. Divide all mole values by the smallest one to get a ratio.
4. If needed, multiply through to get whole numbers.
5. Write the empirical formula using these whole-number subscripts.

Worked Example 1

Question: 0.035 g of nitrogen forms 0.115 g of an oxide of nitrogen. Find the empirical formula.

Answer:

Element:        N              O
Mass:           0.035 g        0.115 − 0.035 = 0.080 g
Moles:          0.035/14.01    0.080/16.00
              = 0.0025 mol   = 0.0050 mol
Ratio:          1              2

Empirical formula: NO₂ (check against Talbot's worked example, p.94 — note the ratio 1:2 gives NO₂, since mole values 0.0025 and 0.0050 simplify to 1:2)

Worked Example 2

Question: A compound is found to contain 40.