Answers to problem set 1 |
1. Cu3AsS4 = 3Cu+ As+ 4S 3(63.5)+ (74.9)+ 4(32.06) = 393.64 Cu = (190.5/396.64)x100 = 48.39% As = (74.9)/396.64)x100 = 19.03% (enargite) S = (128.24/393.64)x100 = 32.6% 2. Ag = (65.4/107.9) = 0.606 0.606/0.2029 = 3 As = (15.2/74.9) = 0.2029 0.2029/0.2029 = 1 S = (19.4/32.06) = 0.6051 0.6051/0.2029 = 3 3Ag+As+3S = Ag3AsS3 Ag3AsS3 (proustite) 3. PbMoO4 = PbO+ MoO3 (207.2+16)+((95.94+(16)3)) = 367 PbO = (223.2/367)x100 = 60.8% MoO3 = (133.94/367)x100 = 39.2% wulfeniteufen ite 4. Co3As2O8.8H2O = 3CoO+ As2O5+ 8H2O 3(58.9+16)+ 2(74.9)+5(16)+ 8(18)) = 598.5 CoO = (224.7/598.5) = 37.5% As2O5 = (229.8/598.5) = 38.4% H2O = (144/598.5) = 24.1% erythrite 5. Na2O = (22.7/62) = 0.366 (0.366/0.366) = 1 B2O3 = (51/70) = 0.728 (0.728/0.366) = 2 H2O = (26.3/18) =1.46 (1.46/0.366) = 4 Na2O+2B2O3+4H2O = Na2B4O7.4H2O other possible formulas: Na2B4O6(OH)2.3H2O (kernite) Na2B4O5(OH)4.2H2O Na2B4O4(OH)6.H2O Na2B4O3(OH)8 6. a. +2 b. +3 c. +5 d. +4 e. -2 7. a. 0.617: 1: 1.76 b. 1.63: 1: 5.37 8. a. (3, 6, 2) b. (1 ,0, 0) c. (5, 10, 2) d. (3, 3, 1) e. (9, 12, 4) f. (8, 15, 22) 9. BeO = (19.8/25) = 0.792 (0.792/0.786) = 1 Al2O3 = (80.2/102) = 0.786 (0.786/0.786) = 1 BeO+Al2O3=BeAl2O4------(chrysoberyl) (9+16) + ((2x27) + (3x16)) = 127 S. G. = (4x127/((6.023x10+23)x(227x10-24)) = (508/136.7) = 3.72 10. a. 3a: 2b: 1/2c b. (2, 3, 12) 11. a. 12 g. 12 b. 8 h. 12 c. 6 i. 12 d. 24 j. 8 e. 24 k. 8 f. 12 l. 6 |