How many ways are there to put 5 balls in 2 boxes if the balls are distinguishable but the boxes are not?
Since the boxes are indistinguishable, there are 3 possibilities for arrangements of the number of balls in each box.

Case 1: 5 balls in one box, 0 in the other box. We must choose 5 balls to go in one box, which can be done in $\binom{5}{5} = 1$ way.

Case 2: 4 balls in one box, 1 in the other box. We must choose 4 balls to go in one box, which can be done in  $\binom{5}{4} = 5$ ways.

Case 3: 3 balls in one box, 2 in the other box. We must choose 3 balls to go in one box, which can be done in $\binom{5}{3} = 10$ ways.

This gives us a total of $1 + 5 + 10 = \boxed{16}$ arrangements.