Introduction
Last lesson we learned how to calculate an empirical formula from experimental data such as masses or percentage composition. Today we go one step further: using the empirical formula and the molar mass of a compound to work out its actual molecular formula. This is a key skill for IB Paper 1 and 2 calculation questions, and it connects nicely to how chemists identify unknown compounds in real laboratories (Talbot, p.317).
Why empirical formula isn't always enough
The empirical formula only tells us the simplest whole-number ratio of atoms in a compound — it doesn't tell us how many atoms are actually in one molecule.
- CH₂ is the empirical formula for many different molecules: C₂H₄, C₃H₆, C₄H₁₀, and more (Talbot, p.90).
- Propene (C₃H₆) and propane (C₃H₈) are a good illustration — propene's empirical formula is CH₂, while propane's empirical formula is the same as its molecular formula, C₃H₈ (Talbot, p.90).
- For ionic compounds (like NaCl or CaO) and giant covalent/molecular structures (like SiO₂), there's no separate "molecule," so the empirical formula is the chemical formula (Talbot, p.90).
This is why, for molecular compounds, we need one more piece of information: the molar mass of the actual compound.
The method
n = molar mass of molecular formula / molar mass of empirical formula
Steps:
- Calculate the empirical formula from given data (as before).
- Work out the molar mass of that empirical formula.
- Divide the compound's actual molar mass by the empirical formula's molar mass to find n.
- Multiply every subscript in the empirical formula by n to get the molecular formula.
Worked Example 1
Question: 0.035 g of nitrogen reacts with oxygen to form 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 (based on Talbot, p.94):
- Mass of oxygen = 0.115 − 0.035 = 0.080 g
- Moles of N = 0.035 / 14.01 = 0.0025 mol
- Moles of O = 0.080 / 16.00 = 0.0050 mol
- Ratio N : O = 0.0025 : 0.0050 = 1 : 2
- Empirical formula = NO₂
Now find the molecular formula:
- Molar mass of empirical formula NO₂ = 14.01 + (2 × 16.00) = 46.01 g mol⁻¹
- n = 92 / 46.01 ≈ 2
- Molecular formula = (NO₂)₂ = N₂O₄
Worked Example 2
Question: A compound has the empirical formula CH₂O and a molar mass of 180 g mol⁻¹. Find its molecular formula.
Answer:
- Molar mass of CH₂O = 12.01 + (2 × 1.01) + 16.00 = 30.03 g mol⁻¹
- n = 180 / 30.03 ≈ 6
- Molecular formula = (CH₂O)₆ = C₆H₁₂O₆ (this is glucose!)
Worked Example 3
Question: A hydrocarbon has empirical formula C₂H₅ and molar mass 58.12 g mol⁻¹. Find its molecular formula.
Answer:
- Molar mass of C₂H₅ = (2 × 12.01) + (5 × 1.01) = 29.07 g mol⁻¹
- n = 58.12 / 29.07 ≈ 2
- Molecular formula = (C₂H₅)₂ = C₄H₁₀ (butane)
Key takeaways
- The molecular formula is always a whole-number multiple (n) of the empirical formula: n = molar mass (molecular) / molar mass (empirical).
- You need the compound's actual molar mass to go from empirical to molecular formula — empirical data alone isn't enough (Talbot, p.90).
- For ionic and giant covalent compounds, the empirical formula is the chemical formula — there's no separate molecular formula.
- Always double-check your value of n is a sensible whole number before multiplying the subscripts.
- This process (combustion analysis → empirical formula → molecular formula) is also the first step real chemists use to identify unknown organic compounds (Talbot, p.317).