Wednesday, January 29th, 2020

1. Tin(IV) Bromide Experiment

  1. Atomic Weight of Tin = 118.71 grams per mole
  2. Molar Mass of Br2 = 159.808 grams per mole
  3. Sn + 2Br2 → SnBr4
    1. ​​​​​​​5 grams of tin = 0.042119450 moles of tin
    2. Moles of Br2 needed = 0.08423890 
    3. Mass of Br2 needed = 13.4620 grams of Br2

Tuesday, January 28th, 2020

1. Selenocyanate Experiment

  1. Atomic Weight of Selenium = 78.96 grams per mole
  2. Potassium Selenocyanate
    1. K4Fe(CN)6 + 4Se = 4KCNSe + FeC2 + N2
    2. 1 gram of selenium = 0.012664640 moles of selenium
    3. Potassium ferrocyanide = 422.39 grams per mole
      1. For 1 gram selenium 1.33735 grams of potassium ferrocyanide are required
  3. ​​​​​​​Phosphorus Triselenide
    1. ​​​​​​​P2Se3
    2. ​​​​​​​​​​​​​​Phosphorus = 30.973762 grams per mole

Saturday, January 25th, 2020

1. Palladous Bromide Experiment

  1. Palladium is 106.42 grams per mole
    1. 0.78 grams = 0.00732944935162 moles of palladium
  2. ​​​​​​​Two molar equivalents of ammonium bromide are required
    1. ​​​​​​​0.00732944935162 x 2 = 0.01465889870 moles
    2. Ammonium bromide = 97.94 grams per mole
    3. 1.43569 grams of ammonium bromide

Wednesday, January 22th, 2020

1. I obtained 150 mg of chloroauric acid after losing all the rest of the gold during the barium aurate experiment. Begin the scandium aurichloride 3ScCl3•2AuCl3•21H2O experiment. 

  1. Chloroauric acid molar mass = 339.785 grams per mole
    1. 150 milligrams of chloroauric acid = 0.0004414556263519 moles of chloroauric acid
  2. Thus 0.000662183439527936 moles of scandium chloride are needed.
    1. ​​​​​​​Molar mass of scandium chloride = 151.31 grams per mole
    2. The required mass of scandium chloride is 0.100194 grams

Friday, January 11th, 2020

1. I spent the last few days finishing up the processing of the rubidium dichromate into three different rubidium salts: bromide, iodide, and sulfate. The final products appear to be contaminated with some sodium and perhaps some potassium. However, since all the salts except the sulfate were heated until they melted into a free flowing liquid ammonium ion is only a tiny fraction of the total contaminants. 

2. I attempted to synthesize barium aurate by reacting precipitated gold with barium peroxide. However there was no apparent reaction between the gold and the barium peroxide. However, a lavender colored salt is what is left in the crucible and I don't think that barium peroxide should decompose to a lavender colored compound. Could it be that some of the gold reacted and some didn't? But if that was so then why wasn't a "bright green mass" obtained? On adding a thick layer of barium peroxide over the previous melt and then reheating over the Meker burner on high until the mass was red hot yielded a greenish-white crust around the edges (presumably barium oxide) but in the center where the gold was the melt was a reddish purple that is much more vivid than the pale lavender color it had previously. 

Saturday, January 11th, 2020

1. Prepared ammonium sulfate by reacting sulfuric acid with concentrated ammonia. 

2. Synthesis of Hydriodic Acid (Uncle Fester's Secrets of Methamphetamine Manufacture page 160, paragraph 2)

  1. Reactions
    1. 2P + 3I2 → 2PI3
    2. PI3 + H2O → 3HI + H3PO3
  2. To then use the iodine, one first puts the mixture into a large sep funnel. Add a couple hundred ml of toluene and shake. The toluene will dissolve about 40 grams of iodine. Separate off this toluene solution.
    1. Solubility: the solubility of iodine, expressed in g/kg of solvent at 25 °C is: H2O, 0.34; benzene, 164.0; CCl4, 19.2; CHCl3, 49.7; ethyl acetate, 157; ethanol, 271.7; diethyl ether, 337.3; n‐hexane, 13.2; toluene, 1875. Soluble glacial acetic acid; relatively insol dichloromethane.
  3. In the example batch given in this section, we are using 300 ml of water and 30 grams red P.
    1. 5 grams of red P is suspended in 50 mL of water.
      1. Molar mass of red P = 30.973761998 grams per mole
      2. 5 grams of red P = 0.1614272753 moles of red P
      3. The required amount of iodine is (0.1614272753÷2)x3 = 0.24214091295‬ moles
      4. 0.24214091295‬ moles of iodine is 61.457518
        1. ​​​​​​​Atomic weight of I2 is 253.8089 grams per mole 
      5. ​​​​​​​Theoretically 61.45 grams of iodine is soluble in 32.7 mL of toluene at 25 °C
  4. Add the toluene to the 300 ml of water and 30 grams of red P.
  5. Then shake to react the iodine with the phosphorous. Keep the mixture cool.
  6. Then separate off the toluene from the water/Hi mixture.
  7. Wash it with a 300 ml portion of water which will be used in the next batch. This will remove HI from the toluene, and save it for the next batch. Finally, pour the toluene back into the sep funnel with the iodine and shake to get another load of iodine. One repeats this process until all the red P has been consumed. 

Friday, January 10th, 2020

Begin Ruthenium Experiment

  1. Atomic weight of Ruthenium is 101.07 grams per mole
    1. 1.07 gram of ruthenium is 0.0105867220 moles of ruthenium
  2. Reaction: Ru + 3KNO3 + 2KOH → K2RuO4 + 3KNO2 + H2O
    1. Potassium Hydroxide = 56.1056 grams per mole
      1. 0.0105867220 moles of potassium hydroxide is 0.59397439 grams of potassium hydroxide
    2. Potassium Nitrate = 101.1032 grams per mole
      1. ​​​​​​​​​​​​​​0.0105867220 moles of potassium nitrate is 1.070351479 grams of potassium nitrate
  3. ​​​​​​​​​​​​​​Potassium Ruthenocyanide K4Ru(CN)6
    1. ​​​​​​​The melt formed by fusion of ruthenium in potassium hydroxide and nitrate, and thus containing potassium ruthenate,
      1. ​​​​​​​Added in 1.11 grams potassium nitrate and 0.68 grams of potassium hydroxide. A large portion of the ruthenium did not react. This was isolated by decantation and put back into the crucible for another round of fusion with the KOH and KNO3.
    2. was dissolved in water and boiled with potassium cyanide.
    3. The orange colour was quickly bleached, the ruthenocyanide produced being subsequently isolated by crystallisation.
      1. ​​​​​​​The color turned dark yellow and stayed that way. There was a dark particulate suspended in the solution which was removed by centrifugation leaving a dark yellow solution.
      2. When a portion of the solution (before centrifugation so the dark particulate was still suspended in this portion) was reacted with a ferric chloride solution a dark particulate was also produced which was suspended in solution. This solution did not give a purple precipitate with ferric chloride. Instead a brownish-red precipitate formed. It was darker in color than normal ferric hydroxide which was produced by precipitating the ferric chloride solution with sodium hydroxide. The filtrate was a light yellow. 
Precipitate from Uncentrifuged Solution and Ferric Chloride

Filtrate from Ferric Chloride Precipitate (left) and the Centrifuged Unreacted Solution (right)

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