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   

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