Answer to Question #146795 in General Chemistry for qwerty

Question

1. Show that the atomic packing factor for FCC is 0.74.
2. Show that the atomic packing factor for BCC is 0.68.
3. Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol. Compute and compare its theoretical density with the experimental value, 7.87 g/cm3.
4. Rhodium has an atomic radius of 0.1345 nm and density of 12.41 g/cm3. Determine whether it has an FCC or BCC crystal structure.
5. The unit cell for tin has tetragonal symmetry, with a and b lattice parameters of 0.583 and 0.318 nm, respectively. If the density, atomic weight, and atomic radius are 7.30 g/cm3, 118.69 g/mol, and 0.151 nm, respectively, compute the atomic packing factor.

Answer 1

1. $APF$ is the fraction of solid sphere volume in unit cell.

$APF=$ $Total volume of sphere\over Total unit cell volume$ $=$ $V_s\over V_c$

$V_s=(4).$ $4\over3$ $\Pi R^3=$ $16\over 3$ $\Pi R^3$

$V_c=$ $16R^3$ $\sqrt{2}$

$APF=$ $V_s\over V_c$ $=$ ${{16\over3}\Pi R^3\over 16R^3\sqrt{2}}$ $=0.74$

2. $BCC=0.68$

In terms of $a\infin$ $R$

Atoms per unit cell

$a=$ $4\over \sqrt{3R}$

Coordination $=no.8$

$APF=0.68$

3. In finding the density of $Fe$

$\rho=$ $_nA_{Fe}\over V_cN_A$

For $BCC, n=2$ atoms /unit cell, and

$V_c=$ $($ $4R\over V3$ $)^3$

$\rho=$ $_nA_{Fe}\over {({4R\over V3})}^3N_A$

${(2atoms/unit)(55.85g/mol)}\over[(4)(0.124\times 10^{-7}/unit cell(6.023\times10^{23}atoms/mol)]$ $=7.90g/cm^3$

The actual value of $Fe=7.87g/cm^3$

4. $Rd=0.1345nm$

$\rho=12.41g/cm^3$

$A_w=102.91g/mol$

$\rho=$ $nA_w\over V_cN_A$

$V_{c_{BCC}}=a^3_{BCC}=$ ${({4R\over V3})}^3$ $n_{BCC}=2$

$V_{c_{BCC}}=a^3_{FCC}=(2R\sqrt{2})^3$ $n_{FCC}=4$

For FCC, $\rho=$ $4(102.91g/mol)\over{{({(4)\times (0.1345\times 10^{-7}cm^3\over \sqrt{3}})}}$ $\times 6.023\times 10^23atoms/mol$

$=11.41g/cm^3$

It is $FCC$

5. $a=0.583nm$

$b=0.318nm$

$\rho=7.30g/cm^3$

$A_w=118.69g/mol$

$A_R=0.151nm$

$n=$ $\rho V_cN_A\over A_{sn}$

$=(7.3g/cm^3)(5.83^2)(3.18\times 10^{-24})(6.023\times 10^{23}atoms/mol)$

$=4atoms/ unit cell$

$\therefore$ $APF=$ $V_s\over V_c$ $=$ $(4)({4\over 3}\Pi R^3)\over(a)^2(b)$

$=$ $4({4\over 3}\Pi(1.52\times 10^{-8}cm^3))\over(a5.83\times 10^{-8}cm^2)(3.18\times 10^{-8})$

$=0.544$