The moves are $1+2+3+5$ and $1+4+6$.

Show that Archimedes needs to use the bag twice. Let him first put into the bag the pieces weighing $1$, $2$, $3$, and $5$ kg, and then the pieces weighing $1$, $4$, and $6$ kg. In both cases, the bag does not break.

Let us prove that this could only happen if the $1$ kg piece was used both times. Indeed, if Archimedes, instead of the pieces weighing $1, \ldots, 6$ kg, used pieces weighing $w_1, \ldots, w_6$ kg, then these weights would satisfy the system of inequalities $w_1 + w_2 + w_3 + w_5 \le 11$, $w_1 + w_4 + w_6 \le 11$. Adding these inequalities, we get $w_1 + (w_1 + w_2 + \ldots + w_6) \le 22$. The sum in parentheses is the sum of six distinct natural numbers, so it is at least $1 + 2 + \ldots + 6 = 21$. Hence, $w_1 \le 22 - 21 = 1$. Therefore, $w_1 = 1$, that is, the $1$ kg piece is uniquely determined.

It remains to show that one move is not enough. If Archimedes puts only one piece in the bag, the bag will not break in any case, so no piece can be identified. If Archimedes puts more than one piece in the bag and the bag does not break, then if the $1$ kg piece is not in the bag, replacing any piece in the bag with it will not change the result; thus, Hiero cannot even tell whether the $1$ kg piece is in the bag. If the desired piece is in the bag, Hiero cannot tell which of the (at least two) pieces is the required one.

Note. After these two moves, the group of pieces weighing $2$, $3$, and $5$ kg, as well as the group weighing $4$ and $6$ kg, are also uniquely determined.