Monday, December 30th, 2019

  1. Yield of Rubidium Bromide was 2.39 grams
    1. Molar mass RbBr = 165.372
    2. Moles of Rubidium Bromide = 0.0144522651960 moles
    3. Theoretical yield of rubidium dichromate was 0.09022419596 moles or 34.91 grams
      1. Some of this was lost. Working on the assumption that I begal with 30 grams..... 
      2. Rubidium dichromate = 386.92 grams per mole
      3. 30 grams RbCr2O7 = 0.07753540 moles ⇒ ÷ 11 = 0.0070486727  moles
      4. 0.0070486727  moles of RbBr = 1.165​​​​​​​

Saturday, December 28th, 2019

Continuing Experiment to Convert Rubidium Dichromate to Rubidium Hydroxide

  1. The ammonium carbonate trick appears to have worked. Adding excess barium hydroxide precipitated out all of the dichromate/chromate. This was filtered off and kept as part of my barium salt collection. Ammonium carbonate was added to the filtrate which precipitated out the excess barium ion as barium carbonate. This was filtered off and kept to make another barium compound (barium bromide). 

Friday, December 27th, 2019

Begin Experiment to Convert Rubidium Dichromate to Rubidium Hydroxide

  1. The theoretical yield of the rubidium dichromate was 0.09022419596 moles of rubidium dichromate or 34.91 grams.
  2. Barium Hydroxide Octahydrate = 315.46 grams per mole
    1. 0.09022419596 moles of barium hydroxide octahydrate = 28.46 grams
    2. Solubility of barium hydroxide octahydrate = Soluble in water (72 mg/ml at 20°C), methanol, ethanol (slightly), and water (56 mg/ml at 15°C). Insoluble in acetone (practically) https://www.alfa.com/en/catalog/A12714/
  3. Barium Carbonate
    1. Solubility product of barium carbonate is 5.9 x 10-9. https://www.chm.uri.edu/weuler/chm112/refmater/KspTable.html
  4. Barium Sulfate
    1. The solubility product constant for barium sulfate is 1.1 x 10-10.
  5. ​​​​​​​Rubidium Hydroxide
    1. ​​​​​​​The Rubidium hydroxide, RbOH, is formed by dissolving the metal in water, or by the action of barium hydroxide on rubidium sulphate. It is a very deliquescent substance, and is volatilized by heat. Its melting-point is 301° C., and its density 3.203 at 11° C. The latent heat of fusion per mol. Is 1.614 Cal.; the heat of formation from the elements is 101.99 Cal. A monohydrate and a dihydrate have been described. http://rubidium.atomistry.com/rubidium_hydroxide.html
    2. Molar Mass = 102.475 https://en.wikipedia.org/wiki/Rubidium_hydroxide

Tuesday, December 24th, 2019

Hydriodic Acid 

  1. 15 grams of elemental iodide is combined with 30 mL of water in a 3 neck round bottom flask.
  2. Hydrogen sulfide is passed through the flask with good stirring and cooling until all of the iodine reacts. 
    1. Problem knowing when the reaction is complete
  3. ​​​​​​​Filtration of the mixture
    1. ​​​​​​​Filtration through glass wool was insufficient
  4. ​​​​​​​Addition of hypophosphorus acid to the mixture
    1. ​​​​​​​That actually worked very well
  5. ​​​​​​​Distillation of the mixture
    1. ​​​​​​​Sulfur was present in the distillate
  6. ​​​​​​​Conclusions:
    1. ​​​​​​​This method is plagued by elemental sulfur at every stage. Not a viable method for producing HI in the laboratory.

Saturday, December 21th, 2019

  1. Molar mass of uranyl acetate dihydrate = 424.15 grams per mole
    1. 2 grams of uranyl acetate dihydrate = 0.004715357 moles; 1.5 grams = 0.003536484 moles
  2. Molar mass of barium acetate monohydrate = 273.43 grams per mole
    1. 0.004715357 moles of barium acetate monohydrate = 1.2893200 grams
  3. Transferred the uranyl acetate dihydrate solution made on June 30, 2019 to a 250 mL beaker. Diluted with some distilled water to 50 mL and then heated with stirring to try to get the undissolved solid to dissolve.
  4. After some of the solid had dissolved the solution was decanted into another 250 mL beaker. Fresh water was added to the solid and stirring and heating were continued until it dissolved.
  5. Weighed out two 1.5 gram portions of barium acetate and dissolved each in 25 mL of distilled water.
  6. Each portion was combined with a separate uranyl acetate solution. One was done for practice and one was filmed for the video.
  7. Then concentrated aqueous ammonia was added to each solution (~3-5 mL) and an orange-yellow solid precipitated. 
  8. The solid was easily filtered off, pressed dry, and then put in a desiccator to fully dry over 3 A molecular sieves. 

Tuesday, December 17th, 2019

  1. Weight of crushed tellurium powder = 3 grams
    1. Atomic weight of tellurium = 127.6 grams per mole
    2. 3 grams of tellurium = 0.02351097178 moles of tellurium
  2. Ammonium iodide = 144.94 grams per mole
    1. 0.02351097178 moles of ammonium iodide = 3.407680 grams
  3. Sodium iodide = 149.89
    1. ​​​​​​​0​​​​​​​.02351097178 moles of sodium iodide = 3.524059560 grams
  4. Nickel(II) iodide = 312.502
    1. ​​​​​​​0​​​​​​​.02351097178 moles of nickel(II) iodide = 7.34722570 grams

Monday, December 16th, 2019

  1. Weight of empty evaporating dish = 56.70 grams
    1. Weight of evaporating dish + gold powder = 57.10 grams
    2. Weight of gold powder = 0.40 grams
    3. Atomic weight of gold = 196.966570 grams per mole
    4. 400 mg of gold = 0.00203080 moles of gold
  2. Molar mass of potassium bromide = 166.0028 grams per mole
    1. 0.00203080 moles of potassium bromide = 0.337 grams

The experiment appears to have failed. A dark solution was obtained but nothing crystallized from it. The next day a small amount of gold powder was found at the bottom of the flask but this could have been fine particles suspended in the liquid. Without processing the gold waste I cannot know for sure if any gold is missing. I will do this after the tellurium video is done. 

Sunday, December 15th, 2019

  1. Weight of empty evaporating dish = 56.70 grams
    1. Weight of evaporating dish + gold powder = 57.40 grams
    2. Weight of gold powder = 700 mg
    3. ​​​​​​​Atomic weight of gold = 196.966570 grams per mole
    4. 700 mg of gold = 0.00355390 moles of gold
  2. Molar mass of potassium bromide = 119.002 grams per mole
    1. 0.00355390 moles of potassium bromide = 0.4229 grams
  3. ​​​​​​​Added 5 mL of azeotropic HBr and ~2 mL of 48% nitric acid. Dissolved in an evaporating dish over boiling water.
  4. Added in the KBr and stirred.
  5. Allowed to evaporate over boiling water. 
  6. Reconstituted and evaporated several times to remove as much HBr as possible.
  7. The final product was dried out under vacuum. The product crystallized out as a dark purple solid which was red by transmitted light. .

Saturday, December 14th, 2019

1. Prepared a batch of hydrobromic acid as described in the Grimoire


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