A method that meets Bennett's now improved-upon lower bound is as follows. First let the masses be L < M < H
Choose a coin A and weight it against each of seven other coins - placing each of the weighted coins in one of three piles l (for lighter), s (for same) and h for heavier than coin A.
If you are lucky and A was M, then the piles will be 223, or 313, or 322. Then you know the unweighed coin is L, M, or H, respectively (and know A is M) so have done the job in 7 weighings total.
If A was L then the piles lsh will have number of coins 016 or 025 (depending on whether the final coin was L or not). And you know the final unweighed coin is L, or either M or H, respectively (and know A is L). So now choose a coin B from the 6 in the heavier pile (consisting M and H coins). Weight it against each of four remaining coins in that pile.
If B is M, then you will get sh = 22, or sh = 13. So the unweighed coin of the six is H or M respectively (and know B is M). This would complete the sorting in a total of 11 weighings.
If B is H, you will get ls = 22, or 31, confirming the final coin is M or H respectively (and B is H). Also for 11 weighings total.
By symmetry, if A is H you can follow an effectively identical process to that above and still get the sorting done in 11 goes.