Question

An element germanium crystallizes in a bcc type crystal structure with an edge of unit cell $288 pm$ and density of element is $7.2$ $gc{{m}^{-3}}$. Calculate the number of atoms present in $52 g$ of crystalline elements. Also calculate the atomic mass of the element.

Answer 1

The bcc type lattice stands for body centered cubic lattice. It has two effective numbers of atoms which are important for calculating the density or the molecular weight of the lattice. The number of molecules can be calculated after knowing the molecular weight of the lattice compound by the help of mole concept.

Complete step by step solution:
-Lattice is a 3-D arrangement of the particles which is repeated to give the complete structure of the compound formed. Its smallest portion is called a unit cell which has 3 edges and 3 angles between the edges.
-Coordination number is the number of the nearest neighbouring atoms associated to a particular atom present inside the lattice. Both the cations and the anions present in the crystal lattice have a particular coordination number which may or may not be the same.
-The effective number of the formula unit cells which is denoted by Z for bcc crystal lattice is 2. This is helpful in calculating the density of the lattice which is given by
$Density =$ $\dfrac{ZxM}{{{N}_{A}}x{{a}^{3}}}$
Where M=molecular weight of the compound and ${{N}_{A}}$ is avogadro’s number.
-We are given the value of density and also a. So we can easily find the molecular weight of the lattice. Putting the values in the above equation we get

& 7.2=\dfrac{2xM}{{{\left( 288x{{10}^{-8}} \right)}^{3}}x6.023x{{10}^{23}}} \\\ & \Rightarrow \text{M=}\dfrac{7.2x{{\left( 288x{{10}^{-8}} \right)}^{3}}x6.023x{{10}^{23}}}{2} \\\ & \Rightarrow M=\dfrac{7.2x2.39x{{10}^{-23}}x6.023x{{10}^{23}}}{2} \\\ & \Rightarrow \text{M=51}\text{.8}\approx \text{52} \\\ \end{aligned}$$ -Thus we get the value of molecular weight of the molecule to be 52 grams. Now we need to find the number of molecules in 52 grams of the lattice. We see that the molecular weight is equal to the weight of the germanium crystal given. It means that we are given 1 mole of the crystal and so the number of molecules will be equal to the avogadro’s number. **Therefore the number of atoms present in 52 g of crystalline element is ${{N}_{A}}=6.023x{{10}^{23}}$. Also the atomic mass of the element is 52 grams.** **Note:** We need to be correct in writing the SI units for all the quantities like the volume of the lattice and the weight of the molecule. They should be described in the units which resemble that of density only. Also the value of Z should be kept in mind for all types of lattices.