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