1. Let’s get started in an easy fashion. Review the physical properties of calcite (CaCO3) in your lab 3 data table or in the properties of minerals handout used for identification. Complete the following: 1pt
2. Use your data above to try out the MIK. Proceed through the steps using your physical property data above.
a) Does your data lead you to a unique solution? Explain why or why not. 1pt
b) What other important data were necessary to get you to calcite? 1pt
3. Try using known physical properties for other minerals to give you practice using the MIK. Consider chalcopyrite (CuFeS2), fluorite (CaF2) and augite (Mg-Fe-Al pyroxene).
a) Describe any problems you may have had at arriving at a given mineral: 1pt
4. Below are physical properties for a mineral that you have not worked with in lab. Use the MIK to identify the mineral.
Streak: dark brown
Specific gravity: 5.0 – 5.2
a) What is the mineral? (correct response has 11 letters) 2pts
b) What mineral group does it belong too (native elements, sulfides, etc.) 1pt
c) Click on the mineral name hot link in the MIK table. Observe the images of the 1pt mineral. Describe the crystal habit.
d) Scroll down to the “About” section. What is the mineral named after? 1pt
e) Scroll down to the Physical Properties section. Does this mineral have fracture or cleavage? Describe the kind of fracture or cleavage. 1pt
f) Scroll down to the “Chemical Properties” section. What is the IMA chemical formula? (omit any valence states) 1pt
5. In entry level geology you estimate the relative density or specific gravity of a mineral (low, average, high). However, you can calculate the theoretical density (g/cm3)if you know the unit cell volume and its atomic contents.
a) Generally, what is a mineral’s unit cell? 1pt
As an example of a theoretical density calculation, halite (NaCl) has a cubic unit cell with edge length = 5.6402A (unit = Angstrom (A) = 10-10 meter) or 5.6402 x 10-8 cm
Volume of halite’s unit cell = edge length x edge length x edge length
= 5.6402A x 5.6402A x 5.6402A or 5.6402A3
= 179.4252A3 or 1.7943 x 10-22 cm3
Mass = total mass of the atoms per unit cell
First we need to know the number of Na and Cl atoms per unit cell. [note: We did an example of this in lecture for galena (PbS).] Here is a model of the unit cell for halite:
So for each unit cell of halite there are 4 Na atoms and 4 Cl atoms. The ratio and be reduced to 1:1 which is consistent with the halilte’s chemical formula. The total number of chemical formulas in the unit cell is referred to as a Z number. For halite, the Z number is 4.
Next we need to determine the total mass of the 4 Na and 4 Cl atoms. To do this we need to know the atomic mass.
For Na = 22.9898 g/mole
For Cl = 35.453 g/mole
As you recall from your chemistry course, 1 mole = 6.02 x 1023 atoms = Avogadro’s Number, so
Mass of 4 atoms of Na = (22.9898/6.02 x 1023) x 4 = 1.5276 x 10-22 g
Mass of 4 atoms of Cl = (35.453/6.02 x 1023) x 4 = 2.3557 x 10-22g
Total mass of halite unit cell = 3.8833 x 10-22 g
Then theoretical density of halite = total mass of 3.8833 x 10-22 g = 2.1642 g/cm3
volume of 1.7943 x 10-22 cm3
b) Returning to www.mindat.org from 4f above, scroll down to the crystallography section of the mineral you identified in 4a. Calculate the theoretical density (g/cm3) from the reported cubic unit cell parameter (a), unit cell volume (V) and Z number =8. Clearly, show your calculations. 6pts
c) How does your calculated density compare with the reported value under the “Physical Properties” section? 1pt
6. Another mineral associated with the one you identified as the following physical properties:
Luster: non-metallic, vitreous or resinous
a) Use the MIK to name this mineral: 2pts
b) Do the two minerals you identified using the MIK belong to the same group? Explain why or why not. 1pt
c) What chemical metal element do these two minerals have in common? 1pt
Visit the following link: http://www.franklinborough.org/local-info/history/ and review.
d) What is significant about these two minerals from a commercial standpoint? 1pt
e) The mineral you identified in 6a has a very special optical property that contributes to the fascination of the locality as one of the most famous mineral collecting sites in the world. What is this optical property? 1pt
Equivalent Na atoms:
12 x ¼ =3
1 x 1 = 1
3+1 = 4
Equivalent Cl atoms
8 x 1/8 = 1
6 x ½ = 3
3 + 1 =4