Answer to Question #146799 in General Chemistry for qwerty

Question

1. Above 882 oC, titanium has a BCC crystal structure, with a = 0.332 nm. Below this temperature, titanium has a HCP structure, with a = 0.2978 nm and c = 0.4735 nm. Determine the percent volume change when BCC titanium transforms to HCP titanium. Is this a contraction or expansion?
2. The element lithium has BCC packing with a body-centered cubic unit cell. The volume of the unit cell is 4.32 x 10−26 L. Calculate the density (g/cm3) of the element.
3. What is the volume of the unit cell? A metal crystallizes in a face-centered cubic lattice. The radius of the atom is 0.197 nm. The density of the element is 1.54 g/cm3.
4. An element crystallizes in a body centered cubic unit cell. The radius of this element is 2.2 x 10−10 m, and the density is 0.991 g/cm3. How many moles of atoms are within one unit cell?
5. Nickel has a FCC crystal structure, an atomic radius of 0.125 nm, and an atomic weight of 58.68 g/mol. Compute and compare its theoretical density with the experimental value of 8.90 g/cm3.

Answer 1

Q146799

1. Above 882 oC, titanium has a BCC crystal structure, with a = 0.332 nm. Below this temperature, titanium has a HCP structure, with a = 0.2978 nm and c = 0.4735 nm. Determine the percent volume change when BCC titanium transforms to HCP titanium. Is this a contraction or expansion?

Solution :

Step 1 : Volume of 1 unit of BCC

Volume per unit cell of BCC crystal is given by, V _BCC = a³ ;

We are given a = 0.332 nm. Substitute this in the formula, we have

V_BCC = (0.332 nm) ³ = 0.036594 nm³ ;

Step 2 : Volume of 1 unit of HCP

We are given a = 0.2978 nm and c = 0.4735 nm

Volume per unit cell of HCP is given by,

$V_{HCP} = \frac{3\sqrt{3}*a^2*c}{2}$

$V_{HCP} = \frac{3\sqrt{3}*0.2978^2*0.4735}{2}$

V_HCP = 0.1091 nm² ;  

Step 3 : To find volume per unit atom for both BCC and HCP ;

Each unit of BCC contains 2 atoms and each units of HCP contains 6 atoms.  

Volume per unit atom for BCC = 0.036594 nm³ /2= 0.01830 nm³ / atom.  

volume per unit atom for HCP = 0.1091 nm² /6= 0.018183 nm³ / atom ;  

Step 4 : To find the % change in volume due to allotropic transformation of Titanium  

% change in volume = $\frac{V_{HCP}-V_{BCC}}{V_{BCC}}*100;$

= $\frac{0.018183nm^3-0.01830nm^3}{0.01830nm^3}*100$ ;

= -0.639 % ;

Hence the % change in volume when BCC titanium transforms to HCP titanium = - 0.639 % .

The negative sign implies that there is a contraction when Titanium undergoes allotropic transformation from BCC to HCP.

2. The element lithium has BCC packing with a body-centered cubic unit cell. The volume of the unit cell is 4.32 x 10−26 L. Calculate the density (g/cm3) of the element.
Solution :

Each unit of BCC contains 2 atoms. So there would be 2 atoms of lithium in each unit cell.

Atomic mass of Li = 6.941g/mol ;

Step 1 : To find mass of single unit of BCC

Using Avagadro’s number we will find mass of 1 atom of Li.

Mass of 1 atom of Li = 6.941g Li /1mol Li * 1 mol Li /6.022 * 10²³ atoms of Li  

= 1.1526 * 10^-23 grams/ atom .  

Mass of single BCC unit of Li = 2 * mass of 1 atom of Li = 2 * 1.1526 * 10^-23 grams/ atom

= 2.3052 * 10^-23 grams.

Step 2 : Convert the given volume to cm³ .

In question, we are given the volume of single unit of BCC = 4.32 x 10−²⁶ L

Convert this to cm³ , using 1 L = 1000 cm³ ;

volume of single unit in cm³ = 4.32 x 10−²⁶ L * 1000 cm³ /1L

= 4.32 x 10−²³ cm³ ;

Step 3: Find density using mass and volume.  

$density = \frac{mass}{volume};$

$density = \frac{2.3052 * 10^{-23} grams}{ 4.32 x 10^{−23} cm^{3}}$

density = 0.5336 g/cm³ ;  

In question, we are given volume in 3 significant figure.

So our final answer must also be in 3 significant figure.

Hence the density of lithium is 0.534 g/cm³ ; 

3. What is the volume of the unit cell? A metal crystallizes in a face-centered cubic lattice. The radius of the atom is 0.197 nm. The density of the element is 1.54 g/cm3.

Solution:

For FCC unit cell,  

$$a=2R\sqrt{2}$$

$V_{FCC}=a^3 = (2R\sqrt{2} )^3$

$V_{FCC}=a=(2*0.197*\sqrt{2})^3 ;$

$V_{FCC}=0.173 nm^3;$

So the volume of the unit cell is 0.173 nm³ ;

4. An element crystallizes in a body centered cubic unit cell. The radius of this element is 2.2 x 10−10 m, and the density is 0.991 g/cm3. How many moles of atoms are within one unit cell?

Solution :

Each unit cell of Body centered cubic system contains 2 atoms.

and we know, 1 mol = 6.022 * 10²³ atoms.

So, moles of atoms in a single unit cell = 2 atoms * 1 mol /6.022 * 10²³ atoms.

= 3.32 * 10^-24 moles.

So there are 3.3 * 10^-24 moles of atoms in one unit cell of BCC.

5. Nickel has a FCC crystal structure, an atomic radius of 0.125 nm, and an atomic weight of 58.68 g/mol. Compute and compare its theoretical density with the experimental value of 8.90 g/cm3.

Solution :

Step 1: Find the volume of single unit of FCC.

For FCC crystal structure,

$V_{FCC}=a^3 = (2*R*\sqrt{2})^{3}$ ;

Convert 0.125 nm to cm;

Radius of atom in 'cm' = 0.125nm * 1m/10^-9 nm * 100cm/1m = 0.125 *10^-7 cm;

$V_{FCC}=a^3 = (2*0.125*10^{-7}*\sqrt{2})^{3};$

$V_{FCC} = (3.5355*10^{-8})^3;$

$V_{FCC}= 4.419 *10^{-23};$

Step 2 : To find mass of 1 unit of FCC.

Each unit of FCC contains 4 atoms.

Mass of 1 atom = 58.68 g/1mol * 1 mol/ 6.022 * 10²³ atoms = 9.744 * 10^-23 g/ atom.

So, mass of 1 unit of FCC = 4 * 9.744 * 10^-23 g/ atom = 3.898 * 10^-22 grams.

Step 3 : To find the theoretical density.  

$density=\frac{mass}{volume} = \frac{3.898*10^{-22}g}{4.419*10^{-23}cm^{3}}$

Theoretical density = 8.819 g/cm³ ;

which in 3 significant figure is 8.82 g/cm³ ;

The experimental density given in the question is 8.90 g/cm³ ;

The Theoretical density is less than the experimental density.