Question

15. (5 pts) Determine the number of formula units contained in the unit cell of the hypothetical monoclinic mineral Trogladite (Zr2Al KSI3012) S.G. = 4.45; a = 10.38 A; b 5.55 A; c 5.86 Å; B = 142°

Answer 1

15. Formula unit is the number of Trogladite molecules present in one unit cell i.e. if the number is nZr₂AlKSi₃O₁₂ per unit cell then the number of formula units is "n".

Now, the given information is: Specific gravity = 4.45, which means if the density of water is 1 g/cm³, then the density ($\rho$) of Trogladite is 4.45 g/cm³. cell parameters are also given, a = 10.38 A^o; b= 5.55 A^o; c = 5.86 A^o; unique angle = 142^o.

The molecular weight of the solid, M = Atomic masses of all the elements multiplied by their number of atoms = {91.224 (Zr)*2 + 26.9815(Al)*1 + 39.0983(K)*1 + 28.0855(Si)*3 + 16(O)*12} g/mol = 524.7843 g/mol.

Thus the expression for the density of the mineral is:

$\rho =\frac{M\times Z}{N_{A}\times V}$

Where M is the molar mass, Z is the formula unit, N_A is the Avogadro's number = 6.023*10²³, V is the volume of the unit cell.

Now, Volume of the monoclinic unit cell of the mineral, $V=abcsin\beta= 10.38\times 5.55\times5.86 sin142^{o}$ (A^o)³

= 207.84 (A^o)³ = 207.84*10^-24 cm³

Thus, we can find out from the expression above, what is the formula unit.

$4.45=\frac{524.7843\times Z}{6.023\times 10^{23}\times 207.84\times 10^{-24}}$

$=>Z=1.06\sim 1$

$\therefore$Hence the number of formula unit is 1.