CALCULATION INVOLVING UNIT CELL DIMENSIONS

From the unit cell dimensions, it is possible to calculate the volume of the unit cell. Knowing the density of the metal. We can calculate the mass of the atoms in the unit cell. The determination of the mass of a single atom gives an accurate determination of Avogadro constant.

Suppose edge of unit cell of a cubic crystal determined by X – Ray diffraction is a, d is density of the solid substance and M is the molar mass, then in case of cubic crystal

Volume of a unit cell = a3  

Mass of the unit cell = no. of atoms in the unit cell × mass of each atom = Z × m  

Here Z = no. of atoms present in one unit cell  

m = mass of a single atom  

Mass of an atom present in the unit cell = m/NA  

∴ Density d = mass of unit cell / volume of unit cell = Z.m / a3  

d = Z.M. / a3 × NA  

Note:Density of the unit cell is same as the density of the substance  

Illustration 28. An element having atomic mass 60 has face centred cubic unit cell. The edge length of the unit cell is 400 pm. Find out the density of the element?  

Solution:          Unit cell edge length = 400 pm  

                        = 400 × 10–10 cm  

                        Volume of unit cell = (400 × 10-10)3 = 64 × 10-24 cm3  

                        Mass of the unit cell = No. of atoms in the unit cell × mass of each atom  

                        No. of atoms in fcc unit cell = 8 × 1/8 + 6 × 1/2 = 4  

                        ∴ Mass of unit cell = 4 × 60 / 6.023 × 1023  

                        Density of unit cell = mass of unit cell / volume of unit cell = 4 × 60 / 6.023 × 1023× 64 × 10–24 = 6.2 g/cm3    

Illustration 29. An element has a body centred cubic (bcc) structure with a cell edge of 288 pm. The density of the element is 7.2 g/cm3. How many atoms are present in 208 g of the element?  

Solution:          Volume of unit cell = (288×10–10)3 cm3 = 2.39 × 10-23cm3  

                        Volume of 208 g of the element = mass / volume = 208 / 7.2 = 28.88cm3  

                        No of unit cells in this volume = 28.88 / 2.39 × 10–23 = 12.08 × 1023  

                        Since each bcc unit cell contains 2 atoms  

                        ∴ no of atom in 208 g = 2 × 12.08 × 1023 = 24.16 × 1023 atom    

Illustration 30. A compound formed by elements X & Y, Crystallizes in the cubic structure, where X is at the corners of the cube and Y is at the six face centers. What is the formula of the compound? If side length is 5A°, estimate the density of the solid assuming atomic weight of X and Y as 60 and 90 respectively.    

Solution:          From eight corner atoms one atoms (X) contributes to one unit cell.  

                        From six face centres, three atoms (Y) contributes to one unit cell.  

                        So, the formula of the compound is XY3.  

                        As we know that,  

                        p = n × Mm / NA × a3  =, here n = 1  

                        Molar mass of XY3  

                        Mm = 60 + 3 × 90 = 330 gm  

                        p = 1×330 / 6.023 ×1023 × (5×10–8)3 gm/cm3  

                        a = 5Å = 5 × 10–8 cm  

                        = 330 / 6.023 ×1023 × 125 × 10–24 gm/cm3   = 4.38 gm / cm3  

Illustration 31. Lithium borohydrides, LiBH4, crystallizes in an orthorhombic system with 

4 molecules per unit cell. The unit cell dimensions are: a = 6.81 Å, b = 4.43 Å and c = 7.17 Å. Calculate the density of the crystal. Take atomic mass of Li = 7, 

B = 11 and H = 1 a.m.u.  

Solution:          Molar mass of LiBH4 = 7 + 11 + 4 = 22 g mol-1  

                        Mass of the unit cell = 4 × 22 gmol–1 / 6.02 × 1023 mol–1 = 14.62 × 10–23 g  

                        Volume of the unit cell = a × b × c  

                       = (6.81 × 10-8 cm) (4.43 × 10-8 cm) (7.17 × 10-8 cm)  

                                              = 21.63 × 10-23 cm3  

                        ∴   Density of the unit cell = Mass / Volume = 14.62 × 10–23 g / 21.63 × 10–23cm3 = 0.676 gcm–3    

Illustration 32. A binary solid (A+B) has a rocksalt structure. If the edge length is 400 pm and the radius of cation is 75 pm, the radius of anion is  

                        (A) 100 pm                                           (B) 125 pm  

                        (C) 250 pm                                           (D) 325 pm    

Solution:          (B)    

Illustration 33. The vacant space in bcc lattice unit cell is about  

                        (A) 32%                                                (B) 10%  

                        (C) 23%                                                (D) 46%    

Solution:          (A)    

Illustration 34. A substance has density of 2 kg dm-3 & it crystallizes to fcc lattice with 

edge-length equal to 700pm, then the molar mass of the substance is  

                        (A) 74.50gm mol-1                                  (B) 103.30gm mol-1  

                        (C) 56.02gm mol-1                                  (D) 65.36gm mol-1    

Solution:          (B)p = n × Mm / NA × a3 

                          2 = 4 × Mm / 6.023 × 1023 × (7 × 10–8)3  

                        (since, effective number of atoms in unit      cell = 4)  

                        On solving we get Mm = 103.03 gm / mol              

Exercise 9.        

(i)   Iron has body centred cubic lattice structure. The edge length of the unit cell is found to be 286 pm. What is the radius of an iron atom?  

(ii) KF has rock – salt structure  

      Calculate the value of Avogadro’s number. Density of KF = 2.48 g/cm3  

      And distance between K+ & F- in KF = 268 pm    

Exercise 10.  

An element crystallises as F.C.C. with density as 5.20 g/cc and edge length of the side of unit cell as 300 pm. Calculate mass of that element which contains 3.01 ´ 1024 atoms.

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