$A$ body-centred cubic $(BCC)$ element of density $10.3 \ g \ cm^{-3}$ has a cell edge of $314 \ pm$. Calculate the atomic mass of the element. Also, Lithium borohydride crystallizes in an orthorhombic system with four molecules per unit cell. The unit cell dimensions are $a = 6.8 \ \mathring{A}$, $b = 4.4 \ \mathring{A}$, and $c = 7.2 \ \mathring{A}$. If the molar mass is $21.76 \ g \ mol^{-1}$, calculate the density of the crystal.

(N/A) Part $1$: For a $BCC$ unit cell, the number of atoms per unit cell $Z = 2$. The formula for density is $d = \frac{Z \times M}{N_A \times a^3}$. Given $d = 10.3 \ g \ cm^{-3}$ and $a = 314 \ pm = 3.14 \times 10^{-8} \ cm$. Rearranging for molar mass $M$: $M = \frac{d \times N_A \times a^3}{Z} = \frac{10.3 \times 6.022 \times 10^{23} \times (3.14 \times 10^{-8})^3}{2} \approx 96.03 \ g \ mol^{-1}$.
Part $2$: For an orthorhombic system, the density is $d = \frac{Z \times M}{N_A \times V}$, where $V = a \times b \times c$. Given $Z = 4$, $M = 21.76 \ g \ mol^{-1}$, $a = 6.8 \times 10^{-8} \ cm$, $b = 4.4 \times 10^{-8} \ cm$, $c = 7.2 \times 10^{-8} \ cm$. $d = \frac{4 \times 21.76}{6.022 \times 10^{23} \times (6.8 \times 10^{-8} \times 4.4 \times 10^{-8} \times 7.2 \times 10^{-8})} = \frac{87.04}{6.022 \times 10^{23} \times 215.424 \times 10^{-24}} = \frac{87.04}{129.72} \approx 0.67 \ g \ cm^{-3}$.