Partners: Sonya Pasia and Kristen Kobayashi
20 September 2011
Zinc Iodide (ZnI2) was an interesting binary compound to experiment with. In this experiment, weakly acidified water (25mL distilled water with 18 drops 5M acetic acid solution) was used as an aid to bring molecules of the zinc and iodide atoms together, by dissolving iodine molecules, so that bonding would transpire to produce a reaction. Deprived of water, the Zn and I2 molecules would not be capable of moving close enough to each other, and a reaction would not occur. Deprived of acid, the reaction of Zn + I2 would have resulted in 2HI(aq) rather than ZnI2 (s), and it wouldn’t have appeared to follow the Law of …show more content…
The percent composition of ZnI2 (s) was found by first adding the masses of the elements together to determine the g/mol of the compound, (64.41 g/mol Zn + 126.9 (2) g/mol I = 319.21 g/mol ZnI2.) and then dividing the mass of each component by the g/mol of the compound and multiplying by 100. (64.41 g/mol Zn/319.21 g/mol ZnI2*100=20.49% Zn) (253.8g/mol I2/319.21 g/mol ZnI2=79.32% I2) The large difference in the percentages is the reason for the excess Zinc metal that was noted towards the beginning of the experiment.
The empirical formula of the recorded data was determined by calculating the moles in amount of each element consumed by the compound. For Zn: 0.514 g (1mol/65.41g) = 0.0083 mol Zn and for I2: 1.985 g I2 (1 mol/126.9g)= 0.031 mol I2 were the results. By using the smallest number to divide each by, or 0.0083, the empirical formula was found to be ZnI4 by means of 0.0083molZn/0.0083=1, and 0.031 mol I2/0.0083=3.7, or approximately 4. This means that more Zn metal should have been consumed by the compound to create ZnI2. One possible reason for this not happening would be that the Zn metal shards were too big for the reaction