Example: NaCl. For details, see text p. 146.

Given an unknown, what do we need to know?

What the atoms are and where they are: = "SOLVE" the crystal structure.

What is the composition ?

Electron microprobe (recall: this week's lab) Na and Cl characteristic X-rays.

Using number of characteristic Na and Cl X-rays, determine that the sample is 39.4 sw % Na and 60.6 wt % Cl.

100g of the sample, 1.71 moles Na and 1.71 moles Cl (divide wt% by atomic wt)

Thus: Na:Cl = 1:1, thus NaCl.

Atomic weight NaCl is 58.443 g/mole

How dense is the sample ? Measurement of density: how?

g/mole / g/cm# = cm3/mole

cm3/mole x 10^24 = A3 / mole. For NaCl, 27.057 x 10^24 A3/mole

Avagadro's number: formula units/mole

fu/mole / A3/mole = fu/A3 (formula units per cubic angstrom). So how many formula units per unit cell ? Need to know how many A3 / unit cell (i.e, unit cell volume.

SINGLE CRYSTAL DIFFRACTION.

Note that row of diffraction spots is normal to lattice planes.

Note, by extension, the diffraction pattern follows from the orientation of the planes of atoms.

Note, the inverse relationship between interplanar spacing and distance between spots in single crystal diffraction pattern.

Note the angular relationship between planes of atoms can be extracted from the angles between rows of spots in the single crystal diffraction pattern.

THUS, * a,b,c and alpha, beta, and gamma can be determined from the pattern!!, thus get cell size and shape! Returning to NaCl: find a = 5.640 A, = b = c and all angles are 90 degrees.

Thus, volume of cell = 179.405 A3

.0223 fu / A3 x 179.406 A3 = 4.00075 fu /cell

Thus, each unit cell of the mineral contains (only) 4 Na and 4 Cl.

SPACE GROUPS

note the symmetry in the single crystal pattern