Abstract. Harborth [Elemente der Mathematik, Vol. 33 (5), 116-118, 1978] proved that every set of 10 points in the plane, no three on a line, contains an empty convex pentagon. From this it follows that the number of disjoint empty convex pentagons in any set of n points in the plane is least ⌊ n 10 ⌋. In this paper we prove that every set of 19 points in the plane, no three on a line, contains two disjoint empty convex pentagons. We also show that any set of 2m + 9 points in the plane, where m is a positive integer, can be subdivided into three disjoint convex regions, two of which contains m points each, and another contains a set of 9 points containing an empty convex pentagon. Combining these two results, we obtain non-trivial lower bounds on the number of disjoint empty convex pentagons in planar points sets. We show that the number of disjoint empty convex pentagons in any set of n points in the plane, no three on a line, is at least ⌊ 5n 47⌋. This bound has been further improved to 3n−1 28for infinitely many n.