Exactly three of the four chlorine atoms of $\mathrm{CCl}{}_4$ are replaced by fluorine (Swarts reaction) to give a chlorofluorocarbon. Using the rule of 90, its Freon number is

The product is $\mathrm{CClF}{}_3$. The product is $\mathrm{CClF}{}_3$ with C = 1, H = 0, F = 3. The Freon number is that of digits (C-1)(H+1)(F) = 0,1,3, written 13. Check: 13 + 90 = 103 gives C1 H0 F3, so Cl = 4 - 3 = 1. So Freon-13.