Colours In Kool-Aid
Experiment 3. Colours in Kool-Aid
Abstract
Tetramethylammonium iodide was dissolved in a mixture of of Iodine and 95% ethanol while gently heated in a fume hood. Once cooled, crystals started to form in the beaker, which were then collected and washed with hexane through a series of processes involving vacuum filtration. Then a mixture of 5mh of the crystals and 5ml of 95% ethanol was created and tested on a slice of potato to analyse the iodine content. After, with sodium thiosulphate in the buret, the iodide solution was titrated until it turned orangey-yellow, then had 2ml of starch added to it and titrated again until it turned colourless.
Data and Results
Figure 1. Amount of Reagents | Measurement | Weight of Tetramethylammonium …show more content…
Yield of product. Amount formed of Tetramethylammonium triiodide (g) | 2.63g | Expected amount formed (g) | 1.77g | % Yield | 148.58% |
Figure 3. Potato Slice Experiment Time Lapse | Observation | 3 seconds | Potato turns yellow | 12 seconds | Potato turns dark blue-grey |
Figure 4. Titration Analysis Data and Results | Titration #1 | Titration #2 | Moles of thiosulphate added | 1.37x10-4 | 1.32x10-4 | Moles of Me4NI5 implied | 6.85x10-5 | 6.58x10-5 | Weight of Me4NI5 implied | 0.04854g | 0.04663g | Purity of Me4NI5 | 1.84% | 1.77% | Average purity | 1.80% | Standard Deviation | 0.05 | %RSD | 2.77% |
Calculations
Refer to page 4 and 5.
Discussion
Me4N+I- + 2 I2 Me4N+I5-
Tetramethylammonum Iodide added to 2 Iodine gives you tetramethylammonium pentaiodide.
The product was crystallised when the solid compound had hot solution dissolve it. Once the solution is set to cool, it can’t hold all the solute molecules any longer causing them to begin to leave the solution and form solid crystals. The chilled solution is then vacuum filtered to isolate the pure crystals by rinsing them with …show more content…
It helps indicate the percent of theoretical yield that was to be obtained from the final product of the experiment. The percent yield in my experiment was 148.58%. This could be due to a variety of reasons. The materials could have been inaccurately weighed. Also, the crystals that were produced may have not been thoroughly dried causing the extra weight of the moisture to produce a false high percent yield. There maybe too many impurities in the solute or what materials that I was working with might have been
Abstract
Tetramethylammonium iodide was dissolved in a mixture of of Iodine and 95% ethanol while gently heated in a fume hood. Once cooled, crystals started to form in the beaker, which were then collected and washed with hexane through a series of processes involving vacuum filtration. Then a mixture of 5mh of the crystals and 5ml of 95% ethanol was created and tested on a slice of potato to analyse the iodine content. After, with sodium thiosulphate in the buret, the iodide solution was titrated until it turned orangey-yellow, then had 2ml of starch added to it and titrated again until it turned colourless.
Data and Results
Figure 1. Amount of Reagents | Measurement | Weight of Tetramethylammonium …show more content…
Yield of product. Amount formed of Tetramethylammonium triiodide (g) | 2.63g | Expected amount formed (g) | 1.77g | % Yield | 148.58% |
Figure 3. Potato Slice Experiment Time Lapse | Observation | 3 seconds | Potato turns yellow | 12 seconds | Potato turns dark blue-grey |
Figure 4. Titration Analysis Data and Results | Titration #1 | Titration #2 | Moles of thiosulphate added | 1.37x10-4 | 1.32x10-4 | Moles of Me4NI5 implied | 6.85x10-5 | 6.58x10-5 | Weight of Me4NI5 implied | 0.04854g | 0.04663g | Purity of Me4NI5 | 1.84% | 1.77% | Average purity | 1.80% | Standard Deviation | 0.05 | %RSD | 2.77% |
Calculations
Refer to page 4 and 5.
Discussion
Me4N+I- + 2 I2 Me4N+I5-
Tetramethylammonum Iodide added to 2 Iodine gives you tetramethylammonium pentaiodide.
The product was crystallised when the solid compound had hot solution dissolve it. Once the solution is set to cool, it can’t hold all the solute molecules any longer causing them to begin to leave the solution and form solid crystals. The chilled solution is then vacuum filtered to isolate the pure crystals by rinsing them with …show more content…
It helps indicate the percent of theoretical yield that was to be obtained from the final product of the experiment. The percent yield in my experiment was 148.58%. This could be due to a variety of reasons. The materials could have been inaccurately weighed. Also, the crystals that were produced may have not been thoroughly dried causing the extra weight of the moisture to produce a false high percent yield. There maybe too many impurities in the solute or what materials that I was working with might have been