Popular in Course
Popular in Geology
This 13 page Class Notes was uploaded by Jessica Braun MD on Saturday September 12, 2015. The Class Notes belongs to GEOL 284 at West Virginia University taught by Helen Lang in Fall. Since its upload, it has received 19 views. For similar materials see /class/202706/geol-284-west-virginia-university in Geology at West Virginia University.
Reviews for Mineralogy
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 09/12/15
Dr Helen Lang De t of Geoloquot amp Geo39ra hr West Virginia University FALL 2008 GEOLOGY 284 MINERALOGY f V of r Vail l i if One mole of an element is the amount of that element whose weight in grams is equal to its atomic weight A mole of any element always contains the same number of atoms 6022 x 1023 atoms called Avogadro s number Elements combine in integral numbers of NM A mole of quartz which has the chemical formula SiO2 has 1 mole 6022x1023 atoms ofSi 28086 grams 2 moles 2 x 6022x1023 atoms of O 2 x 15999 qrams Molecular weight on quartz SiOZ us 60084 gramsm One mole of quartz contains 6022x1023 molecules of SiO2 1 l Remember minerals must have a welldefined chemical composition The composition range of a mineral is conveniently expressed by a general chemical formula eg quartz SiO2 feldspar CaNaK1FeAlSi4O8 olivine MgFe22SiO4 garnet CaMgFe2Mn3AlFe3 2Si3012 l L t If you have a chemical analysis of a particular mineral sample you can calculate a mineral formula from it or normalize it The method we will use is a little different than calculating a general empirical fuc J might have done in Chemistry because we want to end up with a certain number of oxygens or sulfurs or something Our method is slightly different than Perkin s method His method includes an unnecessary and confusing step K I n j T M 0 Fquot 70 7 la 0quot 53937quot 5951 x1331 1A E 3 ow L quot rill 1t M will T ll ll ll 5 a 39439 A 7 EJu A U x 7 L 2 1 Vii9 s L i i i K C u v k 1 Our goal Calculate a mineral formula from the mineral analysis we re given the feldspar analysis of Box 15 Feldspar formulas are written with 8 oxygens All feldspars must fit the general formula CaNaK1FeAlSi4O8 Ca Na and K fit in the same type of position site in the mineral structure that s why they re in parentheses similarly Fe Al and Si fit in the same type of position or site I will always tell you how many oxygens to use The first step is to figure out how many moles of each cation go with the 8 moles of oxygen The last ste is to write our result as a s ecific horizontal chemical formula Perkins Box 15 modified Th39s Is the H w d we mineral get this analysus FeldsparAnalysus Mol Wt cations anions Oxide Wt Moles of Oxide of oxide in oxide in oxide in mineral Oxide Cation gmole grams SiOz 6008 1 2 6590 AI203 10196 2 3 1945 F6203 15968 2 3 103 CaO 5608 1 1 061 M20 6196 2 1 712 K20 9420 2 1 620 total 10031 Mineral formula Moles of Cation Moles of Oxygen Moles of Cation per 8 Ox How do we know the Cation Each oxygen has a 2 charge Each other element has only a few common ions each with its own charge See inside front cover of your textbook Stable compounds are neutral zero charge The charge of the cation in each oxide is whatever is necessary to make the molecule neutral Example FeZO3 Neg charge 3x2 6 charge of 2Fe cations must be 6 each is Fe3 Perkins Box 15 modified Feldspar Analysis Oxide Si02 Al203 Fe203 CaO NaZO K20 total MOI Wt ofoxide in oxide in oxide in mineral Oxide 6008 10196 15968 5608 6196 942 cations anions Oxide Wt Moles of NN lNN l 6590 1945 103 061 712 620 10031 This is the total moles Cation 109 Si4 0191 AI3 0006 Fe3 0011 Ca2 0115 Na 0066 K oxygen we have We want 8 oxygens so we need a normalization or fudge factor to get to 8 oxygens Multiply Moles Cation x Factor to get this Moles of Ca on 1097 0382 0013 0011 0230 0132 Factor ox in formula 2oxygens Moles of 2194 0572 0019 0011 0115 0066 829772687 Perkins Box 15 modified Feldspar Analysus Oxide Mol Wt cations anions Oxide Wt Moles of Cation Moles of of oxide in oxide in oxide in mineral Oxide Cation Si02 6008 1 2 659 1097 Si4 1097 A203 10196 2 3 1945 0191 AI3 0382 Fe203 15968 2 3 103 0006 Fe3 0013 CaO 5608 1 1 061 0011 Ca2 0011 Map 6196 2 1 712 0115 Na 0230 K20 942 2 1 62 0066 K 0132 total 10031 Finally we can write the formula Moles of Oxyen 2194 0572 0019 0011 0115 0066 2977 CaOO3NaO62K035FeOO4A104Si29408 Moles of Cation per 8 Ox 2948 1025 0035 0029 0618 0354 5008 Always check to see that your formula fits the general formula for that mineralll Feldspar General Formula CaNaK1FeAlSi408 Our Calculated Specific Formula CauuoNa FeUU lA lU lSi O8 U04KUOO 434 CaNaK 100 FeAlSi 402 If it fits calculation is probably OK if not review calculation to find your error Feldspar Formula CaNaK1FeAlSi408 Correct Formula for our feldspar is this 03003N3062K035FeOO4A104Si294O8 Not this a that would mean 4 x 004 Fe 4 x 104 Aland 4 x 294 Si Keep three decimal places throughout the calculation you can round on at me eno Remember to multiply the Mole oxide by the ll ll i in the oxide to get Moles Cation Remember to multiply the Mole Oxide g i39 times the tjrl rr 151 Lil in the oxide to get the Moles Oxygen Perkins Box 15 modified Feldspar Analysrs Oxide Mol Wt cations anions Oxide Wt Moles of Cation of oxide in oxide in oxide in mineral Oxide SiOz 6008 1 2 659 1097 Si4 A203 10196 2 3 1945 0191 AI3 F6203 15968 2 3 103 0006 Fe3 CaO 5608 1 1 061 0011 Ca2 Na20 6196 2 1 712 0115 Na K20 942 2 1 62 0066 K total 10031 Iron is often reported as FeO not Fe203 how will that These columns change the can be omitted calculation Moles of Cation 1097 0382 0013 0011 0230 0132 Moles of Oxyen 2194 0572 0019 0011 0115 0066 2977 Moles of Cation per 8 Ox 2948 1025 0035 0029 0618 0354 5008
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'