Question

A series of minerals consists of a crystal in which $F{e}^{2+}$ and $M{g}^{2+}$ ions may substitute for each other causing a substitutional impurity defect without changing the volume of the unit cell. In the mineral, the oxide ions exist as $fcc$ with $S{i}^{2+}$ occupying $\frac{1}{4}th$ of the octahedral voids and divalent ions occupying $\frac{1}{4}th$ of the tetrahedral voids. The density of forsterite (Magnesium silicate) is 3.21 $g/cc$ and that of fayalite (Ferrous silicate) is 4.34 $g/cc$. If the density of mineral is 3.88 g/cc, then which of the following statements is/are CORRECT.

• A. Forsterite = $M{g}_{2}Si{O}_4$
  Fayalite = $F{e}_{2}Si{O}_4$
• B. Mineral contains 40.71% Forsterite and 59.29% Fayalite
• C.
  Mineral contains 59.29% Forsterite and 40.71% Fayalite
• D. Forsterite = $MgS{i}_2{O}_2$
  Fayalite = $FeS{i}_2{O}_4$

Answer 1

Answer: (A), (B)

The correct options are
A

Forsterite = $M{g}_{2}Si{O}_4$
Fayalite = $F{e}_{2}Si{O}_4$

B

Mineral contains 40.71% Forsterite and 59.29% Fayalite

Accorrding to given data:
$O^{2-}=8\times \frac{1}{8}+6\times \frac{1}{2}=4$
$S{i}^{4+}=\frac{1}{4}\times \text{octahedral void}=\frac{1}{4}\times 4=1$
$M^{2+}=\frac{1}{4}\times \text{tetrahedral void}=\frac{1}{4}\times 8=2$

Therefore
Forsterite = $M{g}_{2}Si{O}_4$
Fayalite = $F{e}_{2}Si{O}_4$
Let forsterite make up x% and fayalite make up (100 - x)% of the mineral:
$\frac{x\times 3.21+(100-x)\times 4.34}{100}=3.88$
x = 40.71
The mineral contains of 40.71% forsterite and 59.29% fayalite.