A particular solid has a structure in which Ca atoms are located at cube corners

Question

A particular solid has a structure in which Ca atoms are located at cube corners, O atoms at the center of the cube faces, and Ti at the cube centers. For the purposed of this problem we will assume the ideal structure of this compound is cubic and the cube edge is 3.84 angstroms. a) what is the formula of this material (Show how you determined the formula) b)What is the theroretical density, in grams cm^-3, of the material (show your work). the formula is CaTiO4 but I don's get how to find the density because I do not have the volume???

Explanation / Answer

as Ca atoms are located at cube corners so no. of atom in the unit cell = (1/8)*8 =1 O atoms at the center of the cube faces so no. of atom in the unit cell= (1/2)*6 =3 and Ti at the cube centers so no. of atom in the unit cell = 1 so formula of this material wil be CaTiO3. density ? = ZM / a³N Z = no. of atoms in the unit cell = 5 M= molecular weight =135.94g/mol a=3.84 angstroms so V= a^3= 5.66*10^-23 cm^3 N =6.022x10²³ so ? = ZM / a³N on substitutiong values.. ?=19.93g/cm^3

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