Inorganic Chemistry
Experiment 3: Tin (IV) Iodide
Aim:
The primary objective of this procedure is to investigate the various properties of tin(IV) iodide.
Introduction:
Tin is a very important element as it is classified as a group 14 element. Examples of elements in this group include carbon, silicon and lead. It can be found in two different oxidation states, +2 and +4, almost identical to the element directly below it, which is lead (Pb). In terms of its properties, it is very similar to silicon. This is because they are on the metal/non-metal border. Another similarity that they share is that they are both excellent semiconductors. Tin is therefore an element very much of interest in the semiconductor industry.
Pre-Practical Questions:
…show more content…
It must be performed under dry conditions. This is due to the fact that in the presence of water (or aqueous solution), tin is hydrolysed. The following reaction occurs:
SnI4 + 4H2O ----> Sn(OH)4 + 4HI
The bonding in this molecule must clearly be covalent as 1,1,1-trichloroethane must be non-polar in order to act as a solvent. The rule ‘like dissolves like’ must be taken into account. The electronegativity difference between tin and iodine is 0.7, which means that it is polar covalent. The electronegativity difference between tin and fluorine is 2.02, making it an ionic molecule. As neither molecule is polar, tin(IV) tetrafluoride would be more soluble in the non-polar solvent.
Procedure:
Part 1:
…show more content…
This figure is slightly higher than its actual value, which should be approximately 81%. This was obtained as a result of finding the atomic mass of iodine. The atomic mass of the iodine was then multiplied by 4 as there were 4 atoms of iodine in tin(IV) iodide. This was then placed over the molecular mass of tin(IV) iodide and multiplied by 100 to be expressed as a percentage. This value could have been more accurate to its real value (i.e. approximately 81%) if the titration had been stopped as soon as the solution went yellow. At this point, the final colour of the solution is yellow due to the fact that some ICl2- (iodine dichloride) formed. This compound is formed from the reaction of iodine monochloride (ICl) with excess hydrochloric acid (HCl). In the preparation of the tin(IV) iodide, the crude % yield was found to be 61.02596306%. This yield could have been higher if the mixture in the round-bottomed flask was allowed to reflux for a longer time period. In addition to this, the crude yield could have been higher if the tinfoil was cut up into smaller pieces. This would have meant more surface area exposed to the mixture of iodine, acetic anhydride and glacial acetic acid. More surface area means more molecules have more space to react with the solution. This also means that the reaction would have occurred at a much
Then, an amount of KI (solid) about a size that would fit on a match head was dissolved in 0.05 of Potassium Iodate solution and about 1 mL of water and 1 mL of 1 M HCl were added, which exhibited a weak positive test for IO_3^- (aq). After the weak positive test, an amount of KI (solid) about a size that would fit on a match head was dissolved in about 1 mL of water and 1 mL of 1 M HCl, which exhibited a negative
Also, both their Ph level was 7 which mean that they are neutral and not acidic and reaction with iodine solution was exactly the same. Therefore, with all the experiments conducted and analyzed, icing sugar is the mystery substance. c) Q: Which properties, physical or chemical, were most useful in identifying the mystery powder? Explain your answer.
Firstly, an amount of 40.90 g of NaCl was weighed using electronic balance (Adventurer™, Ohaus) and later was placed in a 500 ml beaker. Then, 6.05 g of Tris base, followed by 10.00 g of CTAB and 3.70 g of EDTA were added into the beaker. After that, 400 ml of sterilized distilled water, sdH2O was poured into the beaker to dissolve the substances. Then, the solution was stirred using the magnetic stirrer until the solution become crystal clear for about 3 hours on a hotplate stirrer (Lab Tech® LMS-1003). After the solution become clear, it was cool down to room temperature. Later, the solution was poured into 500 ml sterilized bottle. The bottle then was fully wrapped with aluminium foil to avoid from light. Next, 1 mL of 2-mercaptoethanol-β-mercapto was added into fully covered bottle. Lastly, the volume of the solution in the bottle was added with sdH2O until it reaches 500 ml. The bottle was labelled accordingly and was stored on chemical working bench.
Iodine turns into a blue/black color when in the presence of starch, after using iodine if the blue/black color is absent then the starch has been used usually making a halo around the inoculum, resulting in a positive result. If it stays blue/black then the starch is still present meaning the organism cannot produce amylase causing a negative result. My color stayed blue/black and there was no evidence of a halo, meaning my organism is negative for producing amylase. (handout, amylase)
and brought to a boiling temperature then filtered out the final product is converted to
...for the original titration, shown in Table 5. This could be due to perhaps usage of the wrong indicator, or of not stopping the titration exactly when the color changed.
3. By increasing the pressure (only really significant in reactions involving gases). 4. By the use of a suitable catalyst. 5.
The procedure for this experiment can be found in Inorganic Chemistry Lab Manual prepared by Dr. Virgil Payne.
However, in order to measure the rates of reaction, sodium thiosulphate and starch are added. Sodium thiosulphate is added to react with a certain amount of iodine as it is made. Without the thiosulphate, the solution would turn blue/black immediately, due to the iodine and starch. The thiosulphate ions allow the rate of reaction to be determined by delaying the reaction so that it is practical to measure the time it takes for the iodine to react with the thiosulphate. After the all the thiosulphate has reacted with the iodine, the free iodine displays a dark blue/black colour with the starch. If t is the time for the blue/black colour to appear, then 1/t is a measure of the initial rate.
Thickett, Geoffrey. Chemistry 2: HSC course. N/A ed. Vol. 1. Milton: John Wiley & Sons Australia, 2006. 94-108. 1 vols. Print.
This is the first reaction in the Harcourt Essen experiment. The iodine is oxidised to produce I2 wh...
Figure 1: The hexacyanoferrate iron (III) ion: contains 6 ligands which are bound to the central metal therefore has a coordination number of 6 (n=6¬). (Hussain,2007) [13]
Analyze each fraction by spotting 10 times with capillary tubes on a TLC plate, which is exposed to iodine vapor for 15 minutes.
Results & Discussion:The actual, theoretical, and percent yield of sodium chloride (NaCl) was found to be 1.14g, .700g, and 61.4%,
In this experiment three different equations were used and they are the Stoichiometry of Titration Reaction, Converting mL to L, and Calculating the Molarity of NaOH and HCl (Lab Guide pg. 142 and 143).