Introduction
In the previous lesson you learned how to calculate an empirical formula from experimental mass or percentage composition data. Today we go one step further: using the molar mass of a compound, we'll convert an empirical formula into the true molecular formula — the formula that shows the actual number of atoms of each element in one molecule (Talbot, p. 90).
Why empirical formula isn't always enough
The empirical formula only shows the simplest whole-number ratio of atoms in a compound. Many different molecules can share the same empirical formula. For example, CH₂ is the empirical formula for C₂H₄, C₃H₆, C₄H₁₀ and several other compounds (Talbot, p. 90).
- For ionic compounds (e.g. NaCl, CaO) and giant covalent/molecular structures (e.g. SiO₂), the empirical formula is the actual chemical formula — there's no separate "molecular formula" to find.
- For simple molecular compounds, we need extra information — the molar mass — to pick the correct multiple of the empirical formula.
The method
- Determine the empirical formula (from mass/mole ratios).
- Calculate the molar mass of the empirical formula unit.
- Find the multiple, n, using:
n = (molar mass of molecular formula) / (molar mass of empirical formula)
- Multiply every subscript in the empirical formula by n to get the molecular formula.
Worked Example 1
Question: 0.035 g of nitrogen forms 0.115 g of an oxide of nitrogen. The molar mass of the compound is 92 g mol⁻¹. Find (i) the empirical formula and (ii) the molecular formula.
Answer:
- Mass of oxygen = 0.115 − 0.035 = 0.080 g
- Moles N = 0.035 / 14.01 = 0.0025 mol
- Moles O = 0.080 / 16.00 = 0.0050 mol
- Ratio N : O = 1 : 2 → empirical formula = NO₂
Now find n:
- Molar mass of NO₂ = 14.01 + (2 × 16.00) = 46.01 g mol⁻¹
- n = 92 / 46.01 ≈ 2
Molecular formula = N₂O₄ (Talbot, p. 94)
Worked Example 2
Question: A compound has empirical formula CH₂ and a molar mass of 42.08 g mol⁻¹. Find its molecular formula.
Answer:
- Molar mass of CH₂ = 12.01 + (2 × 1.01) = 14.03 g mol⁻¹
- n = 42.08 / 14.03 ≈ 3