Okay, this isn't a complete solution as this does seem intricate but maybe this is a good starting point.
Note there are $5! = 120$ possible ways the weights can be determined. Each weighing will have $3$ possible outcomes. The weight balances, the left is heavier, and the right is heavier. If you can figure out a way to weigh things so that each outcome covers the same number of possibilities, the first weigh will reduce your point distributions to $40$ the second weighing to $14$ the third to $5$, the fourth to $2$ and the fifth to $1$.
That would be ideal.
But it can't be done.
So I evaluated how the distributions are accounted for by different combinations of weigh.
Example: If I weigh one weight (call it A) and another (call it B) then A < B accounts for 60 cases. B > A for 60 cases. But B = A will never occur. This is not a very desirable distribution.
If I weigh one weight (A) against two weights (BC) then, A > B account for A = 4; B= 1;C=2 OR A=4; B=2; C=1 OR A=5;B=1;C=2 OR A=5;B=2; C= 1; etc. basically A > BC has 6 possible solutions. A = BC has 8 possible solutions and A < BC has 106 solutions. This is a TERRIBLE distribution.
I won't bore you but if I weigh two against two I get $AB < CD$ has $48$ solutions. $AB = CD$ has $24$ solutions and $AB > CD$ has $48$ solutions. This is the best distribution.
Here's where I gave up and took a leap of faith. I figured that the second weighing between $AC$ and $BE$ would be the best second weighing for even distributions. It intuitively seem to redistribute the light pair among the heavy pair and to get the 5th unknown in early.
But I dunno.