Stoichiometry Chemical Analyses and Formulas Stoichiometry Chemical analyses

Stoichiometry Chemical Analyses and Formulas

Stoichiometry • Chemical analyses of oxygen bearing minerals are given as weight percents of oxides. • We need to be able to recalculate oxide analyses to cations per given number of oxygens to derive a chemical formula.

Oxides of Lithophile Cations • • • Si. O 2 Al 2 O 3 Mg. O Ca. O Na 2 O Ti. O 2 Cr 2 O 3 Mn. O Fe 2 O 3 Fe. O K 2 O H 2 O (4+) (3+) (2+) (1+)

Example 1 Weight percents to formula • Oxide • Wt% Mol. Wt Oxide Moles Oxide Cation Oxygen • Si. O 2 • Mg. O • 59. 85 60. 086 40. 15 40. 312 100. 0 . 9960 1. 9920. 9960 2. 9980 • Mole ratios Mg : Si : O = 1 : 3 • Mg. Si. O 3

Example 2 Formula to weight percents • • Kyanite is Al 2 Si. O 5 Calculate the weight percents of the oxides Si. O 2 Al 2 O 3

Example 2 Formula to weight percents: Kyanite: Al 2 Si. O 5 • • • Oxide Moles PFU Si. O 2 1 Al 2 O 3 1 Formula weight Mol. Wt Oxide 60. 086 101. 963 Grams Oxide 60. 086 101. 963 162. 049 Wt% Percent 37. 08 62. 92

Example 3: Solid Solutions Weight percents to formula • Alkali Feldspars may exist with any composition between Na. Al. Si 3 O 8 and KAl. Si 3 O 8. • Formula has 8 oxygens: (Na, K)Al. Si 3 O 8 • The alkalis may substitute in any ratio, but total alkalis to Al is 1 to 1.

Example 3: Solid Solutions Weight percents to formula • Oxide • Wt% Mol. Wt Oxide • • 68. 20 60. 086 19. 29 101. 963 10. 20 61. 9796 2. 32 94. 204 100. 00 Si. O 2 Al 2 O 3 Na 2 O K 2 O Moles Oxide Cation Oxygen 1. 1350 2. 2701 0. 1892 0. 3784. 5676 0. 1646 0. 3291. 1646 0. 0246 0. 0493. 0246 3. 0269 x 2. 6430= 8. 000 Mole ratios Na 0. 87 K 0. 13 Al 1. 00 Si 3. 00 O 8 calculated as cations per 8 oxygens

Simple Solid Solutions • Na. Al. Si 3 O 8 - KAl. Si 3 O 8 Alkali Feldspars • • Mg. Si. O 3 - Fe. Si. O 3 Enstatite-Ferrosilite (pyroxene) Mg. Ca. Si 2 O 6 -Fe. Ca. Si 2 O 6 Diopside-Hedenbergite Mg 2 Si. O 4 - Fe 2 Si. O 4 Forsterite-Fayalite Mg 3 Al 2 Si 3 O 12 - Fe 3 Al 2 Si 3 O 12 Pyrope - Almandine

Example 4 Weight Percent Oxides from Formula • Given the formula En 70 Fs 30 for an orthopyroxene, calculate the weight percent oxides. • En = enstatite = Mg 2 Si 2 O 6 • Fs = ferrosilite = Fe 2 Si 2 O 6 • Formula is (Mg 0. 7 Fe 0. 3)2 Si 2 O 6 = (Mg 1. 4 Fe 0. 6)Si 2 O 6

Example 4 (Mg 1. 4 Fe 0. 6)Si 2 O 6 Weight Percent Oxides from Formula • • • Oxide Si. O 2 Mg. O Fe. O Moles PFU 2 1. 4 0. 6 • Formula weight Mol. Wt Oxide 60. 086 40. 312 71. 846 Grams Oxide 120. 172 56. 437 43. 108 Wt% Percent 54. 69 25. 69 19. 62 219. 717 100. 00

Example 5 Weight Percent Oxides from Formula • A pyroxene is a solid solution of 40% jadeite (Na. Al. Si 2 O 6) and 60% acmite (Na. Fe. Si 2 O 6). • Calculate the weight percent oxides • Formula is Na(Al 0. 4 Fe 0. 6)Si 2 O 6

Formula is Na(Al 0. 4 Fe 0. 6)Si 2 O 6 Calculate Weight Percent Oxides • • Oxide Moles PFU Si. O 2 2. 0 Al 2 O 3 0. 2 Fe 2 O 3 0. 3 Na 2 O 0. 5 Formula weight Mol. Wt Oxide 60. 086 101. 963 159. 692 61. 980 Grams Oxide 120. 172 20. 393 47. 908 30. 990 219. 463 Wt% Percent 54. 76 9. 29 21. 83 14. 12 100. 00

Coupled Substitutions • Plagioclase feldspar Na. Al. Si 3 O 8 - Ca. Al 2 Si 2 O 8 • Jadeite-diopside Na. Al. Si 2 O 6 - Ca. Mg. Si 2 O 6

Coupled Substitution 40% Anorthite; 60% Albite Calculate Weight percent Oxides • First write the formula • Anorthite is Ca. Al 2 Si 2 O 8 • Albite is Na. Al. Si 3 O 8 • An 40 Ab 60 is Ca. 4 Na. 6 Al 1. 4 Si 2. 6 O 8

An 40 Ab 60 is Ca. 4 Na. 6 Al 1. 4 Si 2. 6 O 8 • • Oxide Moles PFU Si. O 2 2. 6 Al 2 O 3 0. 7 Ca. O 0. 4 Na 2 O 0. 3 Formula weight Mol. Wt Oxide 60. 086 101. 963 55. 96 61. 980 Grams Oxide 156. 22 71. 37 22. 38 18. 59 268. 58 Wt% Percent 58. 17 26. 57 8. 33 6. 92 100. 00

Example Given Analysis Compute Mole percents of Jadeite and Diopside • Jadeite is Na. Al. Si 2 O 6 • Diopside is Ca. Mg. Si 2 O 6

Jadeite - Diopside • Oxide • • Si. O 2 Al 2 O 3 Mg. O Ca. O Na 2 O Wt% Mol. Wt Oxide 56. 64 60. 086 7. 21 101. 963 13. 30 40. 312 18. 46 55. 96 4. 38 94. 204 99. 99 Normalize to 8 oxygens Moles Oxide Cation Oxygen. 9426. 0707. 3299. 0246 • Na. 3 Ca. 7 Al. 3 Mg. 7 Si 2 O 6 = • 30% Jadeite 70% Diopside . 9426. 1414. 3299. 0493 1. 8852. 2121. 3299. 0246 2. 8278 x 2. 6430= 8. 000

Unit Cells and Mineral Density • • • Unit cell is basic repeat unit of structure. Parallel-piped box: a, b, c (Å), a, b, g (º) Å = 10 -8 cm Avogadro’s number (# atoms/mole)=6. 02 x 1023 If you know the contents of the box and the size of the box you can calculate the density

Cell Volume (1Å3=10 -24 cm 3) • • • Cubic Tetragonal Hex/Trigonal Orthorhombic Monoclinic Triclinic V = a 3 V = a 2 c sin 120º V = abc sin b • V = abc (1 + 2 cos a cos b cos g - cos 2 a - cos 2 b - cos 2 g)1/2

Density • Volume of Avogadro’s number of unit cells = AV • The number of formula units per unit cell = Z – Z is a small integer 1 to about 16. • Weight of Avogadro’s number of unit cells = Z* FW

Example Density Calculation • Calculate the density of ferberite (Fe. WO 4), which is monoclinic with • a = 4. 73; b = 5. 70; c = 4. 95; b = 90. 01; Z = 2. • Calculate the gram formula weight: – 1 Fe (55. 847) = – 1 W (183. 85) = – 4 O (15. 9995) – FW = 55. 847 183. 85 63. 998 303. 695 g

Example Density Calculation: Ferberite Fe. WO 4 • • • V = abc sin b = (4. 73)(5. 70)(4. 95)(sin 90. 01º) V = 133. 46 Å3 V = 1. 335 x 10 -22 cm 3 r = ZFw/AV = 2 (303. 70) / 6. 02 1023 *1. 335 10 -22 r = 7. 56 g/cm 3