July 1st, 2019

1. Prepared 20 mL of a 1 molar copper sulfate solution in water to use as an analytical reagent in future work. 

Number of moles = (molarity)(liters of solution). Number of moles needed = (1)(20/1000) = 0.02 moles or 4.9937 grams copper sulfate pentahydrate. 

2. A 0.165 molar solution of sodium cobaltinitrite was prepared by dissolving 0.67 grams of sodium cobaltinitrite into 10 mL of water. 

3. A solution of sodium hypophosphite was prepared by dissolving 1.06 grams sodium hypophosphite in 10 mL water. 

Molar mass of sodium hypophosphite monohydrate is 105.99 grams per mole. (1)(10/1000) = 0.01 moles or 1.0599 grams needed to make a 1 molar solution.

4. Prepared a 1 molar solution of hydroxylamine hydrochloride by dissolving 0.6949 grams HONH2•HCl in 10 mL of water. 

Molar mass of hydroxylamine hydrochloride = 69.49 grams per mole. 0.01 moles or 0.6949 grams needed to make a 1 molar solution.

5. Prepared a 0.66 molar solution of lead acetate by dissolving 3.79 grams of lead acetate trihydrate in 15 mL of water. 

Molar mass of lead acetate trihydrate = 379.33 grams per mole. 0.01 moles = 3.7933 grams. 

6. Prepared a 0.66 molar solution of barium acetate by dissolving 3.50 grams of barium acetate dihydrate in 15 mL water. 

Molar mass of barium acetate dihydrate = 350.4896 grams per mole. 0.01 moles = 3.504896 grams.

7. Prepared 10 mL of a 1 molar solution of potassium arsenite in aqueous potassium hydroxide. 

Molar mass of arsenious oxide is 197.841 grams per mole. 0.01 moles = 1.97841 grams. 

July 2nd, 2019

1. Prepared a test batch of hydriodic acid by combining 60 grams of potassium iodide with 34.5 mL of 75% H3PO4 and distilling the mixture. I used a 400 mL liebig column as a condenser cooled by ice water and a 200 mL vigreux column to ensure that phosphoric acid did not contaminate the HI distillate. The heating mantle was set up via an AC voltage regulator and was set to approximately 85 volts. The receiver was also chilled in ice water. The entire apparatus was wrapped in aluminum foil to protect the HI from the sunlight. 

2. Prepared a stock solution of beryllium chloride by dissolving 150 mg of beryllium metal in 5 mL of room 6M HCl(aq). The beryllium reacted readily but not violently and took several hours to dissolve even with the acid being heated. The final volume of the solution was 15 mL of a solution which had a very light gray-ish tint.

Atomic weight of beryllium = 9.012 grams/mole, molar mass of BeCl2 = 79.9182 grams/mole. 
150 mg of beryllium = 0.0166447 moles Be ⇒ 0.0166447 moles BeCl2 = 1.330 grams BeCl2
0.0166447 moles / (15/1000) liters of solution = 1.109 molar solution.

July 3rd, 2019

1. Prepared more cobalt nitrate hexahydrate by dissolving almost all my remaining cobalt carbonate in a minimal amount of 12 M nitric acid and the reduced the volume by boiling to ~ 75 mL. The beaker was covered and placed in ice for an hour after which the crystals of cobalt nitrate hexahydrate were collected by filtration, allowed to come to room temperature while sitting on folded paper towels, and were pressed four times to remove as much water as possible. 10 mL of the leftover saturated solution captured during the filtration was added to the reagent collection while the solid crystals were put back into the regular inventory. 

2. I tested the batch of HI that I made the other day by reacting it with an excess of aluminum powder. I was wanting to make aluminum triiodide hydrate both to use to create a standard solution of aluminum ion for my reagent collection but also to use as a catalyst/reagent in its own right. The anhydrous salt can be made by direct combination of the elements but those reactions tend to involve a lot of vaporizing iodine and hence loss of a reagent that is a bit touchy if you buy too much too quickly. I was hoping that the HI and the hydrate would suffice. I did get some AlI3 product but not nearly as much as I should have. I tested the material left in the boiling flask after synthesizing the HI for the presence of iodide ion and the test was negative so I am sure all of the KI reacted. I do not know why but this time the use of the Vigreux column seems to have been detrimental rather than beneficial. Since I used up all the HI I had I am going to have to make another batch and this time I won't use the fractionating column and I will see if I get the much stronger product I am used to getting out of that synthesis. 

July 5th, 2019

1. Prepared a 0.479M solution of chromium(III) chloride by dissolving 1.52 grams of anhydrous chromium(III) chloride in 20 mL of distilled water.  

Molar mass of anhydrous chromium(III) chloride = 158.36 grams/mole. 1.52 grams = 0.009598 moles. (0.009598)(20/1000) = 0.479919 molar CrCl3

2. Prepared a batch of nitric acid but almost all of the azeotropic acid was lost when the round bottom flask I was using for a boiling flask cracked. I had set up to synthesize azeotropic nitric acid per the usual method (see Synthesis of Azeotropic Nitric Acid on the Home Chemists Grimoire page). I used a 1 liter round bottom flask for the boiling flask because I knew it wouldn't boil over. Once all of the dilute nitric acid distilled over the temperature very rapidly rose to ~120 oC and the azeotropic nitric acid begun to distill. I switched out the receiver and then I made the attempt to transfer the contents of the boiling flask into a smaller round bottom flask that would allow it to distill more easily. I set the extremely hot larger flask on a cork ring stand that was sitting in a spot on the bench that was wet. It wasn't puddles of water; just some condensation that came off the coolant reservoir. But apparently that is all that it took because suddenly to my horror I saw the cork ring stand begin to turn black and smoke profusely. I gently lifted up the flask, confirmed it was cracked, and then set it down again back on the ring stand. In my experience if you have a substantial amount of liquid in a cracked flask that flask will very likely finish breaking altogether when you tip it in your desperate bid to save a much product as you can. That means its contents will pour out all over the place and very likely splash me in the process. Since I am not partial to 3rd degree chemical burns I didn't even take the chance. Instead I got a shovel and the garden hose, stood as far back as I could, and then pushed the flask off the bench. The flask shattered on impact and even though I immediately began dousing it with water it created a massive cloud of nitric acid vapor. Thankfully the cloud dissipated without incident.

3. Prepared a 1.05M solution of nickel(II) chloride hexahydrate by dissolving 2.50 grams of  in 10 mL of distilled water.  

Molar mass of nickel(II) chloride hexahydrate = 237.69 grams/mole. 2.50 grams = 0.0105179  moles. (0.0105179)/(10/1000) = 1.05 molar. 

4. Prepared a 0.55M solution of bismuth(III) nitrate pentahydrate by dissolving 4.00 grams of Bi(NO3)2 in 5 mL of 12M HNO3 and 10 mL distilled water.  

Molar mass of bismuth(III) nitrate pentahydrate = 485.07 grams/mole. 4.00 grams = 0.008246 moles. (0.008246)/(15/1000) = 0.5497 molar.

5. Prepared a 0.91M solution of calcium(II) chloride hexahydrate by dissolving 2.00 grams of CaCl2•6H2O in 10 mL distilled water.

Molar mass of calcium(II) chloride hexahydrate = 219.068 grams/mole. 2 grams = 0.0091295 moles of  CaCl2•6H2O. (0.0091295)(10/1000) = 0.91295 molar.

6. Prepared a 0.52M solution of potassium ferricyanide trihydrate by dissolving 2.01 grams of K3[Fe(CN)6•3H2O in 10 mL distilled water.

Molar mass of potassium ferricyanide trihydrate = 383.28584 grams/mole. 2.01grams = 0.005244 moles K3[Fe(CN)6]. (0.005244)/(10/1000) = 0.5244 molar.

7. Prepared a 0.947M solution of potassium ferrocyanide trihydrate by dissolving 4.00 grams of K4[Fe(CN)6]·3H2O in 10 mL distilled water.

Molar mass of potassium ferrocyanide trihydrate = 422.388 grams/mole. 4.00 grams = 0.0094699 moles K4[Fe(CN)6]·3H2O. (0.0094699)/(10/1000) = 0.9469966 molar.

8. Prepared a 1.03M solution of potassium chromate by dissolving 4.00 grams of K2CrO4 in 20 mL distilled water.

Molar mass of potassium chromate = 194.19 grams/mole. 4.00 grams = 0.020598 moles K2CrO4. (0.020598)/(20/1000) = 1.02991915 molar.

July 6th, 2019

1. Prepared another batch of HI today using the recycled iodide. I added 5x the normal portion of 85% phosphoric acid and did not use a fractionating column. There was a lot more iodine vapor this time as can be seen in the image below and the product of the distillation was very dark in color. It also had particulates suspended in it. Once the crude HI was made it was transferred into a round bottom flask that was sealed with a glass stopper and the flask was put in a dark corner of the shed to cool and await further purification at a later time. 

2. Dissolved 2.05 grams of cadmium (0.008895926 molesin 8 mL of 12M nitric acid. Rinsed into a scintillation vial with water until the total volume of the solution was 16 mL. The final molarity was (0.008895926)/(16/1000) = 0.556 molar cadmium nitrate. 

3. Dissolved 2.02 grams of lanthanum oxide into 7 mL of 12M nitric acid which became a total volume of 14 mL after washings were added. 2.02 grams of La2O3 = 0.00619995 moles of La2O3 ⇒ 0.01239990 moles La(NO3)3. (0.01239990)/(14/1000) = 0.886 molar lanthanum nitrate.

4. Attempted to dissolve 1.00 grams of scandium oxide Sc2O3 by heating a mixture of the oxide and a solution composed of 5 mL 12M HNO3, 5 mL of distilled water, 5 mL of 16 M HNO3, and 30 mL of 11.5M HCl. 

July 8th, 2019

1. After allowing the scandium oxide to sit in the aqua regia mix overnight I heated the mix up to boiling to remove the acid that I had used in my attempt to dissolve the Sc2O3. After about an hour I noticed that the scandium oxide was beginning to dissolve! Another hour or so later it had completely dissolved in the acid mixture producing a yellow solution. I put this in the shed to sit overnight last night so I could continue reducing the volume of the solution and put it in a scintillation vial to add to the reagent collection. Today I found it crystal clear and still tinted slightly yellow. I am using gentle stirring and low heat to finish reducing the volume today. 

Scandium oxide Sc2O3 = 137.910 grams/mole. 1 gram Sc2O3 = 0.00725 moles = 0.0145 moles ScCl3 = 2.194 grams scandium chloride.
(0.0145 moles) / (13 mL / 1000 mL) = 1.115 molar ScCl3

2. Made up Bettendorff's Reagent (ref. Bettendorff's Test) by dissolving 2.25 grams of stannous chloride in 10 mL concentrated HCl and 10 mL distilled water. A few small pieces of granulated tin metal were added to the bottle to preserve the reagent and prevent it from aging quickly. 

Stannous chloride dihydrate = 225.63 grams/mole. 2.25 grams SnCl2•2H2O = 0.009972 moles. (0.009972) / (20/1000) = 0.49860 molar. 

3. 2.00 grams of sodium tungstate were dissolved in 10 mL of water to create a 0.606 molar sodium tungstate dihydrate solution.

Sodium tungstate molar mass = 329.85 grams/mole. 2.00 grams =  0.00606336 moles. (0.00606336) / (10/1000) = 0.6063 molar.

4. 1.50 grams of sodium molybdate were dissolved in 10 mL of water to create a 0.620 molar sodium molybdate dihydrate solution. 

Sodium molybdate dihydrate molar mass = 241.95 grams/mole. 1.50 grams = 0.0061996 moles. (0.0061996) / (10/1000) = 0.61996 molar

5. Dissolved 220 milligrams of cerium metal in boiling sulfuric acid. In the end some product was lost during the filtration step and it could not be accounted for. Thus the concentration of the final solution is not known. The final volume was 20 mL and at least 0.55 grams of cerium(IV) sulfate (out of a total possible 0.6348 grams) did make it through the filtration process. Taking the lowest likely amount of 0.55 grams that would give a molarity of 0.068 for the final cerium(IV) sulfate solution. 

6. Prepared a solution of potassium thiocyanate by dissolving 2 grams of KSCN (97.181 grams/mole) in 10 mL of deionized water to produce a 2.058 molar solution.

7. Prepared a solution of sodium thiosulfate by dissolving 2 grams of Na2S2O3 (158.11 grams/mole) in 10 mL of deionized water to produce a 1.265 molar solution. 

8. Dissolved 2.13 grams of gallium metal in nitric acid to produce a saturated solution of gallium nitrate in nitric acid. 

July 10th, 2019

1. Dissolved 1.52 grams of sodium selenite in 10 mL deionized water to produce a 0.644 molar solution. 

Sodium selenite molar mass = 263.023 grams/mole. 1.52 Na2SeO3 = 0.00644005 moles. (0.00644005) / (10/1000) = 0.6440050 molar.

2. Dissolved 2.70 grams of potassium tellurite in 15 mL of deionized water to produce a 0.662 molar solution. 

Potassium tellurite hydrate = 271.809 grams/mole. 2.70 grams = 0.0099334 moles. (0.0099334) / (15/1000) = 0.6622297 molar.

3. I concentrated the hydriodic acid that I made on July 6 by setting up the round bottom flask containing the crude HI as the boiling flask in a distillation train. The object was not to distill the HI but instead to boil all of the water out of it. According to NileRed this will produce a solution of 57% hydriodic acid if the boiling flask is removed from heat when the temperature at the distillation head reaches 125-127 oC. The final product was a dark solution with a volume of approximately 

July 12th, 2019

1. Dissolved 2.58 grams of magnesium sulfate heptahydrate in 15 mL ofdeionized water to produce a 0.698 molar solution (neglecting water of hydration). 

Magnesium sulfate heptahydrate molar mass = 246.47 grams/mole. 2.58 grams = 0.0104678 moles. (0.0104678) / (15/1000) = 0.69785 molar.

2. Dissolved 2.00 grams in 15 mL of deionized water to produce a 0.841 molar solution. 

Strontium chloride molar mass = 158.53 grams/mole. 2.00 grams = 0.0126159 moles. (0.0126159) / (15/1000) = 0.841060 molar.

July 14th, 2019

1. Dissolved 2.09 grams of tantalum metal into 8.95 grams of molten KOH. This produced a beige colored melt which took on dull, purplish tones as it cooled and absorbed water from the atmosphere. Once the melt was cooled it was put into a 1 liter beaker with enough deionized water to cover it with about 1" of water and this was left to sit overnight .

Atomic weight of tantalum = 180.94788 amu. 2.09 grams of Ta = 0.011550 moles. 


 

July 15th, 2019

1. The softened melt was extracted into water, crushed in order to pulverize it, and then stirred in deionized water for about an hour. 

2. The solids were filtered off and added to a beaker containing about 300 mL of deionized water. This was acidified with hydrochloric acid and this produced a yellow-green solution containing tantalic acid (the source of the color was unknown; it wasn't the tantalum so far as I know and I think it was from an impurity in the HCl) which on boiling precipitated a gelatinous mass of tantalum pentoxide. This was left overnight to see if any more tantalum pentoxide would precipitate. 

3. The roughly 700 mL of KTaO3 solution filtrate separated from the solids treated in step 2 was centrifuged at 4000 rpm for ~5 minutes in ~80 mL portions to remove the suspended particles of tantalum pentoxide. This produced a clear solution which still contained a touch of opalescence indicating that very tiny particles of Ta2O5•nH2O remained suspended in the liquid. All of the opalescence could be removed by filtering the solution through a 0.2 μm pore filter. ~20 mL of this well filtered solution was added to the collection of reagents as potassium tantalate in KOH.

4. Four ~10 mL aliquots of the centrifuged KTaO3 were separated into test tubes as pictured below. The tube on the far left contains ammonium tantalate precipitated by adding saturated aqueous ammonium chloride to the KTaO3 aliquot. The center left tube contains sodium tantalate made in the same way as the ammonium tantalate except sodium chloride was used. The center right tube contains potassium pertantalate made by adding excess 35% H2O2 to the KTaO3 aliquot followed by addition of absolute ethanol. The tube on the far right is a control containing a KTaO3 aliquot to which only absolute ethanol has been added. 

 

July 20th, 2019

1. Dissolved 5.57 grams of cesium chloride in 10 mL of deionized water to produce a 3.308 molar solution. 

Molar mass of cesium chloride = 168.36 grams/mole. 5.57 grams of cesium chloride = 0.0331 moles. (0.0331) / (10 / 1000) = 3.30838679 molar

July 21st, 2019

1. Weight of empty 100 mL beaker #1 = 52.45 grams. Weight of beaker #1 + Hg = 52.98 grams. Mass of Hg = 0.53 grams.

1A. The Hg was transferred to a clean 150  mL Erlenmeyer flask to prevent splashing or aerosols formed by the action of bubbles from escaping the reaction flask during the reaction.

1C. A stir bar was added to the flask and it was put on the hot plate. 10 mL of concentrated HNO3 was added and stirring was turned on.

1D. Once the mercury had dissolved it was transferred to a scintillation vial along with 10 mL of rinsings of the flask for a total volume of 20 mL of 0.132 molar aqueous mercury(II) nitrate. 

Atomic weight of mercury = 200.59 amu. 0.53 grams of Hg = 0.00264220 moles. (0.00264220) / (20 / 1000) = 0.132110 molar solution.

2. Weight of empty 100 mL beaker #2 = 50.94 grams. Weight of beaker #2 + Hg = 51.61 grams. Mass of Hg = 0.67 grams.

2A. I decided to synthesize mercury(I) acetate to be my representative mercurous salt in the reagent library. The Hg was left in the beaker to facilitate evaporation of as much of the nitric acid as possible.

2B. Approximately 3 mL of hot concentrated nitric acid was used to dissolve the mercury and the volume was then raised to approximately 20 mL. This was then evaporated with stirring down to about 10 mL. It was at this point I discovered an error: mercury(I) acetate is synthesized from mercury(I) nitrate and not mercury(II) nitrate. This was a stupid error and illustrates what can happen when you don't pre-plan everything as carefully as you should. 

2C In order to recover from this mistake I reacted the mercury(II) nitrate solution I had produced with  Bettendorff's Reagent. Careful addition of this reagent selectively precipitated out mercurous chloride (excess reagent will reduce mercurous chloride to elemental mercury). I obtained a good volume of a snow white precipitate of Hg2Cl2 which began as clumps that were progressively broken up into powder by the action of the stirring. I bottled this solution and put it on the shelf to allow the mercurous chloride (which had become quite finely grained by this point) to settle out overnight. Tomorrow I will decant the liquid and add deionized water only to the vial to produce a cleaner mercurous chloride solution. If necessary I may repeat this process a few times to get as much of the Sn ion out of solution as possible. 

0.67 grams of Hg = 0.003340 moles. Molar mass of Hg(NO3)2 = 324.60 grams/mole. 0.003340 moles of Hg(NO3)2 = 1.084 grams. 

July 22nd, 2019

1. I decanted the mercurous chloride, removing as much of the clear aqueous solution as possible with a pipette without disturbing the bed of solid Hg2Cl2. Once this was removed the bottle was refilled with deionized water, capped, and shaken thoroughly. 

2. Once the solid mercurous chloride had fully settled back out of the liquid the process was repeated: the clear solution was decanted with a pipette, the bottle was refilled with deionized water, capped, shaken thoroughly, and then put back on the shelf to allow the solid to settle back out of suspension. This was done a total of three times so that as much of the Sn4+ ion as possible. 

3. All of the liquid obtained from decanting the solution was collected and treated with sodium sulfide in order to precipitate the tin and mercury as their highly insoluble sulfides. These were filtered from the liquid and transferred to waste storage while the liquid was discarded. 

July 25th, 2019

1. Prepared 10 mL of 1M solution of alkaline potassium cyanide by dissolving 0.65 grams of KCN into aqueous KOH. Cyanide solutions must remain alkaline! 

Molar mass of potassium cyanide = 65.12 grams/mole. 0.01 moles of KCN = 0.6512 grams. (0.01) /(10/1000) = 1 molar.

2. Prepared sodium antimonate NaSbO3 by reacting antimony powder with 35% hydrogen peroxide in hot aqueous sodium hydroxide. This was a proof of concept run which served the secondary purpose of allowing me a chance to become familiar with antimonate and its quirks. 

3. A cube of zirconium metal weighing 5.71 grams which had resisted all attempts to dissolve it with acids was reacted with molten KOH and KNO3 in a porcelain crucible. This did nothing. Potassium chlorate was then added to the mixture and allowed to melt. This too had no effect. Finally several grams of sodium bisulfate was added to the crucible and the zirconium was allowed to stew in this fused mixture for about 10-15 minutes. Once this was complete the burner was turned off and the crucible was allowed to cool to room temperature. The cool crucible was put in a 1 liter beaker to which ~600 mL of water was added. The solidified melt was then allowed to dissolve revealing the zirconium cube looking a bit smaller but still largely intact! After soaking it in water overnight to remove as much of the crust off the metal as possible it was weighed the next morning and came out to 4.23 grams. Thus despite everything I threw at it only 1.48 grams of zirconium actually reacted! This makes it the most resistant metal that I have worked with thus far. 

July 29th, 2019

1. Attempting the germanium tetraiodide experiment (see the blog entry for today for an explanation of the rationale for this experiment) on a small scale to see if it is viable. 0.46 grams of germanium was placed into a 250 mL round bottom flask equipped with a magnetic stir bar. Also 3.2 grams of crystal iodine and 25 mL of chloroform were added to the flask before it was hooked up to a reflux condenser (a 400 mm  Liebig). 

Atomic weight of germanium = 72.64 amu. 0.46 grams of germanium = 0.00633 moles. Since the tetraiodide is to be synthesized 0.02533 moles of I atoms are required. This means that 0.012665 moles of I2 are required. Molar mass of diiodine = 253.8089 grams/mole. 0.012665 moles of I2 = 3.21454 grams. 

2. After a short while several more chunks of Ge metal (each weighing from 0.5 to ~1 gram) were added to the reaction flask when no visible results were seen. Granted it would be very difficult to see any kind of result as the iodine and the germanium tetraiodide would both be dissolved in the chloroform and the iodine would render it's color so dark as to make it impossible to detect the orangeish GeI4

3. After refluxing for over an hour the heat was turned off and the boiling flask was allowed to cool. While there were dark solid crystals clinging to the side of the boiling flask it was impossible to tell if these were I2 or GeI4 crystals (or a mixture of both). 

4. About 10 mL of concentrated HI was added to the flask and the reflux was continued with vigorous stirring. After approximately 1 hour the heat and stirring were removed and the flask was allowed to cool. 

5. The flask was sealed with a glass stopper once taken off the reflux column. A small amount of the solution was dribbled onto a hot ceramic plate. The chloroform evaporated to leave a crust of iodine crystals. These then evaporate away more slowly leaving virtually no visible residue. A slight orangish discoloration was observed on the ceramic plate but it was so miniscule as to not be worth noting. This suggests that this is not a viable synthesis method for GeI4 most likely due to the inability for the system to hit a high enough temperature to get the reaction going. 


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