Problem Set #4 Solutions

1. a) The coordination number of Si is 4. The coordination number of Al is 6. One expects the coordination number of Ca to be relatively large (6-8) because it is a large cation.

b) H₂O groups are present.

c) The tetrahedra share corners in pairs. Tetrahedra also share corners with octahedra. Octahedra share edges.

d) Since we have recognized bi-tetrahedral units in the structure, it must be a sorosilicate. Among the sorosilicates in our text, only one contains the right elements and possesses H₂O groups: lawsonite.

e) The chemical formula of Lawsonite is

CaAl₂Si₂O₇(OH)₂(H₂O) =

CaAl₂Si₂O₈(H₂O)(H₂O) =

CaAl₂Si₂O₈ + 2H₂O

Where the first term is the chemical formula of anorthite. The higher density of lawsonite is initially puzzling because H and O are relatively "light", that is they have a small atomic mass compared with the other atoms in the formula. The higher density can be understood in terms of coordination number. In anorthite , Al is in 4-fold coordination, whereas it is 6-fold coordination in lawsonite. Higher coordination numbers are generally associated with higher density.

2. a) Lattice points are shown as small open circles; b) the unit cell by dark lines. Notice that the unit cell is centered-rectangular.

c) Since all angles are unequal, the mineral is triclinic.

3.

a) Forsterite: Mg₂SiO₄, orthorhombic, a=4.75 Å, b=10.20 Å, c=5.98 Å, Z=4.

The volume of the unit cell is

V = abc = 289.73 ų

The mass of the unit cell is four times the mass of the formula

M = 4(2x24.305 g/mol + 28.0855 g/mol + 4x16 g/mol) = 562.782 g/mol

The density is the ratio of mass to volume

r = M/V = (562.782 g/mol)/( 289.73 ų)(1 mol/6.022x10²³)(10⁸ Å/1 cm)³ = 3.23 g/cm³

b) Pyrope: Mg₃Al₂Si₃O₁₂, cubic, a=11.46 Å, Z=8.

V = a³ = 1505.1 ų

M = 3225 g/mol

r = 3.56 g/cm³

c) Enstatite: MgSiO₃, orthorhombic, a=18.22 Å, b=8.81 Å, c=5.17 Å, Z=16.

V = abc = 829.88 ų

M = 803.112 g/mol

r =3.2 g/cm³

d) Beryl: Be₃Al₂Si₆O₁₈, hexagonal, a=9.23 Å, c=9.19 Å, Z=2.

The volume of a hexagonal cell is the height of the cell c multiplied by the area of the base. The base is a trapezoid with sides of length a and an angle g between them. The area of such a trapezoid is

a²sin(g)

Since g =60˚, we have for the volume of the unit cell

V=a²c√3/2=678.03 ų

M=1075.006 g/mol

r =2.63 g/cm³

Calculated densities agree with reported specific gravity to within 1 %. However, the calculated densities are systematically lower. This may be due to the presence of vacancies or impurities in the structure. Alternatively, it may indicate a small but systematic bias in x-ray diffraction determinations of the size of the unit cell.