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Chlorofluorocarbons, or CFCs, looked like ideal chemicals: unreactive, non-toxic, and good for refrigerators and spray cans. A young postdoctoral researcher asked a simple question about them: if nothing destroys them near the ground, where do they end up? Within two decades the answer had helped bring about a global treaty. This post covers Mario Molina's path, the reactions he and Sherwood Rowland proposed, two worked calculations, and what students can learn from him.
Who was Mario Molina?
The Nobel facts page records that Molina was born in Mexico City on 19 March 1943 and died there on 7 October 2020. He shared the 1995 Nobel Prize in Chemistry with Paul Crutzen and F. Sherwood Rowland "for their work in atmospheric chemistry, particularly concerning the formation and decomposition of ozone".
In his Nobel autobiography, Molina recalls turning a little-used bathroom at home into a laboratory, helped by an aunt who was a chemist. At 11 he was sent to a boarding school in Switzerland, because German was thought important for a future chemist. He studied chemical engineering at UNAM in Mexico, did research in Freiburg, then went to the University of California, Berkeley, in 1968 for a PhD in physical chemistry. He admits that his training was weak in mathematics and physics, and that quantum mechanics was "totally alien" to him then; he studied mathematics on his own in Paris and took physics and mathematics courses in his first year at Berkeley.
What question did he ask about CFCs?
In autumn 1973 he joined Rowland's group at the University of California, Irvine, as a postdoctoral fellow. Rowland offered a list of projects, and Molina picked one he knew little about: the environmental fate of CFCs. He writes that he searched systematically for processes that might destroy them in the lower atmosphere, and found none. They would eventually drift high enough to be broken apart by solar radiation. The question became what happens next.
What did the 1974 paper show?
Molina and Rowland published in *Nature* on 28 June 1974: "Stratospheric sink for chlorofluoromethanes: chlorine atom-catalysed destruction of ozone". The paper's main points:
CFCl₃ and CF₂Cl₂ had been detected throughout the lower atmosphere at about 6 and 10 parts per 10¹¹ by volume.
They are inert and could stay in the atmosphere for 40 to 150 years.
In the stratosphere, at 20 to 40 km, ultraviolet light in a window of about 175 to 220 nm splits off chlorine atoms.
Each chlorine atom then runs a catalytic chain:
Cl + O₃ → ClO + O₂
ClO + O → Cl + O₂
Add them and chlorine cancels: O₃ + O → 2O₂. The Cl atom comes out unchanged, ready to destroy more ozone. The paper notes that ClO reacts with O six times faster than NO₂ does at stratospheric temperatures, so the chlorine chain can be more efficient than the natural nitrogen oxide chain studied by Crutzen.
Worked example 1: why the light must be ultraviolet
The paper says CFCs are transparent to wavelengths longer than 290 nm, so they are not broken up in the lower atmosphere. Compare the photon energies. Use h = 6.626 × 10⁻³⁴ J s, c = 2.998 × 10⁸ m/s and Nₐ = 6.022 × 10²³ mol⁻¹.
For λ = 200 nm: E = hc/λ = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) ÷ (2.00 × 10⁻⁷) = 9.93 × 10⁻¹⁹ J per photon. Per mole of photons: 9.93 × 10⁻¹⁹ × 6.022 × 10²³ = 5.98 × 10⁵ J, about 598 kJ/mol.
For λ = 290 nm: E = 6.85 × 10⁻¹⁹ J per photon, about 413 kJ/mol.
Shorter wavelength means higher energy. Only the high-altitude, short-wavelength light that CFCs actually absorb can split them, which is why they survive until they reach the stratosphere.
Worked example 2: how thin is the ozone layer?
The 1995 press release says that if all the ozone in the atmosphere were compressed to surface pressure, it would form a layer only 3 mm thick. How many ozone molecules lie above each square centimetre?
Number density of a gas at 0 °C and 1 atm: n = P/(kT) = 101,325 ÷ (1.381 × 10⁻²³ × 273.15) = 2.69 × 10²⁵ m⁻³ = 2.69 × 10¹⁹ cm⁻³.
A column 0.3 cm tall and 1 cm² in area holds 0.3 cm³.
Molecules = 0.3 × 2.69 × 10¹⁹ ≈ 8.1 × 10¹⁸ per cm².
A huge number, yet a layer thinner than a coin, which is why the press release calls it an "Achilles heel".
Common mistakes:
Writing Cl + O₃ → Cl₂ + O₂ or leaving Cl out of the net equation. Cl is a catalyst and cancels.
Using wavelength in nm inside E = hc/λ without converting to metres.
Confusing stratospheric ozone, which shields us, with ground-level ozone, which is a pollutant. The press release discusses both.
What happened next?
The press release says the 1974 paper led to some restrictions in the late 1970s and early 1980s, but real urgency came in 1985 when Joseph Farman and colleagues reported large losses of ozone over Antarctica. The Montreal Protocol was signed in 1987. According to the 2023 UN-backed assessment, nearly 99% of banned ozone-depleting substances have been phased out, and the ozone layer is expected to recover to 1980 values by around 2066 over the Antarctic and 2040 for most of the world.
For school exams, ozone depletion appears in the Class 10 chapter on our environment. The JEE (Main) 2026 syllabus lists the environmental effects of freons, and the JEE (Advanced) 2026 syllabus has an environmental chemistry topic. The detector that found CFCs everywhere has its own story: read about James Lovelock.
What can a student learn from Molina?
Pick a problem you do not yet understand. He chose the CFC project because it meant learning a new field.
Fill gaps deliberately. He knew his maths and physics were weak and worked on them before and during his PhD.
Follow the chain to the end. "Nothing destroys CFCs here" was not the answer; it was the start of the question.
If catalytic cycles or photon energy calculations are where you get stuck, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can check each line.
Try this: find the energy per mole of photons at 250 nm (answer: about 479 kJ/mol).
