Final answer:
The crystalline compound where atom A forms a ccp lattice and atom B occupies 2/3rd of the tetrahedral voids has the empirical formula A4B5, based on the consideration that there are twice as many tetrahedral voids as atoms in a ccp lattice and 2/3 of those are occupied by B atoms.
Step-by-step explanation:
To determine the formula of a crystalline compound where atom A occupies the face-centered cubic (ccp) lattice positions and atom B occupies 2/3rd of the tetrahedral voids, we first need to understand the fundamental aspects of ccp lattices and the tetrahedral holes within them. In a ccp lattice, there are 4 atoms per unit cell (considering there's one atom at each corner of the cube and they are shared by 8 other cubes, so ⅛ each, and six half atoms from the face centers).
Since each atom forms 8 tetrahedral voids in ccp, the number of tetrahedral voids is twice the number of atoms in the ccp lattice. With atom A forming the ccp lattice, we have 4 A atoms, leading to 8 tetrahedral voids. If atom B fills 2/3 of these tetrahedral voids, this means 2/3 * 8 = 16/3, which simplifies to approximately 5.33 voids are filled by B. Therefore, for every 4 atoms of A there are about 5.33 atoms of B. By approximation, this ratio is 4:5.33, which requires us to find the smallest whole number ratio, leading us to conclude the empirical formula of the compound is A4B5.