Answer. In two operations.

Solution. Let the mass of one original piece be 1. If in the first operation there are $k$ pieces in each group, then after it each piece will have a mass of 1 or $k$; therefore, it is impossible to obtain eleven pieces of different masses in one operation.

We will show that the required result can be achieved in two operations. In the first operation, select 37 groups of 2 pieces each; after the operation, there will be 37 pieces with masses of 1 and 2. In the second operation, select 9 groups of 8 pieces each: in the $i$-th group $(1 \leqslant i \leqslant 9)$ there will be $i-1$ pieces of mass 2 and $9-i$ pieces of mass 1. Then two pieces of masses 1 and 2 will remain unused, and from the $i$-th group a piece of mass $9-i+2(i-1)=7+i$ will be obtained. Thus, 11 pieces with masses $1,2,8,9, \ldots, 16$ will be obtained, as required.

Remark. It can be shown that the method presented is the only possible one.
11.1. Initially, there are 111 pieces of clay of the same mass on the table. In one operation, you can choose several groups with the same number of pieces and in each group, combine all the clay into one piece. What is the minimum number of operations required to get exactly 11 pieces, any two of which have different masses?

(I. Bogdanov)