Proforma 1.1(1) - Tin tetraiodide and complexes of tin PDF

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Tin tetraiodide and complexes of tin...

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Experiment 1.1 Tin tetraiodide and complexes of tin Name:

(a)

Date: 02/02/17

Summarise the aim and key results of the experiment. (Approximately 100-150 words). SnI4 was synthesised from refluxing tin and iodine in CHCl3 with the aim of making two unknown Sn complexes. Complex A was made by reacting SnI4 with PPH3 and complex B was made by reacting SnI4 with Pyridinium Chloride. The aim of the experiment was to determine the structures of the unknown Sn complexes and to analyse the iodine content in SnI4 by titrating against KI03. The yields of SnI4 (27.0%), complex A (3.75%) and complex B (65.4%)) was determined. The structure of both complexes where determined as: complex A – {Sn(I4)(PPh3)2} and complex B- {Sn (Cl)6}2- 2(C5H6N)+ both in octahedral coordination. Finally the % of iodine was calculated in SnI4 (34.4%), this is compared to the theoretical value of 40.5% meaning a 15.1% error in the results from impurities.

4 marks (b) What are the yield and melting point for SnI 4 and the yield for the complexes A and B? (Your answer should include the yield calculations). SnI4- 1.25 g (product obtained) I2 is the limiting reagent 3.75g/ (126.9*2) = 0.0147mol (I2) in a 2:1 ratio (0.0147*626.33)/2= 4.63g (expected yield) (1.25g/4.63g)*100 = 27.0% yield Complex A – 0.02g (product obtained) SnI4 is the limiting reagent, SnI4-0.29g used in 1:1 ratio (0.29g/626.33)=4.63*10^-4 mol (4.63*10^-4 mol*1150.9)=0.533g (expected yield)

(0.02g/0.53g)*100 = 3.75% yield Complex B-0.20g (product obtained) SnI4 is the limiting reagent, SnI4 used-0.39g in 1:1 ratio (0.39/626.3)= 6.23*10^-4 mol (6.23*10^-4 mol)*491.4 = 0.31g (expected yield) (0.20g/0.31g)= 65.4% yield

4 marks (b)

Using ChemDraw, draw the molecular structures, including any possible isomers, for complexes A and B.

complex B

Complex A ( trans)

Complex A (cis)

2 marks (d)

The experimentally determined elemental analysis (section 1.1.5) are: Complex A - Dark orange-brown crystals H 2.5, C 37.4, P 5.4, I 44.0%

Complex B - White crystals H 2.5, C 24.5, N 5.6, Cl 43.3 %.

Show that these experimental values are consistent with the molecular formulae you propose for these complexes. Complex A – Sn C36 H30 P2 I4 (mr 1150.9) H- (30*1)/1150.9)*100 = 2.6% C-((12*36)/1150.9)*100=37.5% P-((30.97*2)/1150.9)*100=5.4% I-(126.9*4)/1150.9)*100= 44.1% Complex B – Sn C10 H12 N2 Cl6 (mr 491.4) H-(12*1)/491.4)*100= 2.4% C-((10*12)/491.4)*100=24.4% N-((14*2)/491.4)*100= 5.7% Cl-((35.5*6)/491.4)*100=43.4%

Therefore both complex A and B are consistent with the proposed structures from these calculations.

4 marks (e)

Determine the redox equations for the iodine analysis you carried out (section 1.1.4). You should demonstrate that IO3-  2I-.

2I03- + 10I- + 12H+ --------- 6I2 +6H20 (equation 1) 3I03- +6I2 +18H+ ---------- 15I+ + 9H20 (equation 2) 5I03 + 10I- + 6I2 +30H+ -------- 6I2 +15I+ +15H20 (joint redox equation) divide by 5 I03- + 2I- + 6H+ -------- 3I+ + 3H20 (Final redox equation) Therefore I03- = 2I-

2 marks (f)

Taking into account that IO3-  2I-, calculate the percentage of iodine in your SnI 4 product. Calculate the theoretical value and compare this with your experimental value.

Experimental Amount of KI03 (mean titre) = 36.3 cm^3 5 gdm^3/ 214 = 0.0234 M (0.0234M*36.3Cm^3)/1000) = 8.48*10e-4 mol (KIO3) 8.48*10e-4mol*2 = 1.70*10e-3mol (I2) due to IO3- = 2I1.70*10e-4mol*(126.9*2) = 0.43g ( I2 in SnI4) Product gained (SnI4) =1.25g 0.43g/1.25g = 34.4% (amount of I2 in SnI4

Theoretical (126.9*2)/ (126.9+126.9+118.7) = 40.5% (40.5-34.4)40.5= 15.1% difference comparing experimental to theoretical most likely due to impurities from errors in the experiment

4 marks Note that your report must be computer generated and that your submission should include the following:  completed COSHH form...