In the first part of the experiment, approximately 2 g of zinc and 2 g of iodine were weighed into two separate weighing boats. In the actual weighing, the accurate masses of the zinc and the iodine were 1 8885 g and 2.399 g, respectively. Next, a clean and dry 125 mL Erlenmeyer flask was weighed. When the flask was empty, it weighed 95.9610 g. Afterwards, when the initially weighed zinc and iodine were added, the flask and its contents weighed 100.120 g. Five mL of vinegar (an acid) was then added into the same flask for the desired reaction to take place. It was made sure that before adding the vinegar that the flask and its contents were placed at the fume hood, and ware directed away from anyone inside the laboratory to avoid any accident that may be caused by the chemical reaction. Inside the hood, the flask was swirled to create significant contact between the zinc and the iodine particles for them to react. The reaction time is approximately 10 minutes; when the solution color does not change, the reaction has finished. If the solid precipitate remains on the bottom, then one of the reactants is in excess.
In the second part of the experiment, the product formed by reacting zinc and iodine was separated from the whole mixture. This was done by first weighing a clean and dry 100 mL beaker, and then decanting the liquid out from the flask into the beaker. The mass of the beaker obtained was 61.56 g. A small amount of solids, however, were still left in the Erlenmeyer flask. Additional vinegar of around 1 mL volume was used to rinse the flask for three times. The liquid accumulated was decanted into the same beaker from the first decant. After rinsing, the solid still remaining inside the flask was considered as the excess reagent. This was then dried on a preheated hotplate at 225°C and then weighed once again to determine the mass of the remaining reactant.
In the last part of the experiment, the mass of the dry product was determined. The hotplate was also used in this part to dry the product. A good indicator that the product has been dried completely is that the cracking sound heard while heating eventually subsides. Lastly, the 100 mL beaker containing the product was determined to have a mass of 161.780 g. and the weight of the product was determined by difference.
Upon addition of vinegar to the zinc-iodine mixture, it was observed that after 2 minutes the mixture changed to dark brown, and 8 minutes thereafter the solution turned colorless. After the 2 minute mark, dissolution of the solid iodine happened in the aqueous vinegar that is why an observed darkening of the solution was observed. The dark brown color is accounted to the presence of iodide ions. Secondly, after the 8 minute mark, the solution turned colorless indicating that the iodide ions have reacted with zinc, to form the product. At all times of the reaction, excess gray solid was found at the bottom of the flask. This is due to the presence of solid zinc, the excess reactant.
In terms of the actual yield, it is first important to identify the reaction equation. In this reaction, the equation to be used is:
Zn + I2 → ZnI2.
In order to determine the limiting the reagent, the molar quantities of reagents has to be calculated. The iodine mass was 2.399 g. and molar mass 253.8 g/mol (diatomic), therefore 9.45·10-3 moles of iodine was be available for the reaction. For zinc, with mass 1.8885 g and molar mass 65.38 g/mol. 28.8·10-3 moles was available. According to the reaction, 1 mole of zinc reacts with 1 mole of iodine. For the studied experiment, zinc is predicted to be the excess reagent. Since the reaction equation is only one-is-to-one, 9.45·10-3 molecules of zinc iodide arc expected to form. Theoretically, at a mass of 319.18 g/mol. the yield of zinc iodide should be 3.016 g.
Percent yield on the other hand represents the ratio between the actual to theoretical yield, multiplied the 100 percent. Accordingly, the amount of excess zinc may be obtained by subtracting 9.45·10-3 to the 0.029 moles of zinc. The result is 0.019 moles of zinc, or 1.27 g of zinc.
The yield of ZnI2 is calculated basing on I2 quantity. The theoretical yield of ZnI2 is 0.00945 mol, which corresponds to 3.016 g. The weights of Zn and I2 used for the reaction are 1.8885 g and 2.399 g, respectively; the total weight is 4.284 g.
The practical weight of the product and the excess Zn is 4.218.
The yield is Y=mpractmtheor=4.2184.284∙100%=98.4% (mpract and mtheor stand for the practically found and theoretically calculated weights, respectively.
Discussion
The yield of the reaction was found 98.4%. The results, however, exhibit deviation from the theoretical quantities since there were some losses of the final product during the laboratory operations.
Whenever there is yield less than 100%, based on the reactants given, it is a typical situation. This may be due to ineffective contact between reactants, such as lack of stirring or swirling. It may also be due to losses while weighing, such that a higher mass is recorded whilst only a lesser amount is made to react. Lastly, it may also be due to a wrong decanting process when only a few, and not all product, is transferred into a separate flask to be dried and weighed.
The experiment obeys the conservation of mass: the weight of the reactants equals the weight of the product and the excess reagent (Table 1).
There is some difference in the actually obtained product and the theoretically calculated due to experimental error, such that the materials involved in the reaction does not lose its weight while undergoing a reaction, but only comes in another form such as another molecule.
Calculate the theoretical yield for the Zn/I2 reaction using the amounts of Zn and I2 that you actually measured out in Part 1.
Zn=1.8895g I2=2.399g, MZnI2=319.188gmol ;MZn=65.38gmol;
M( I2)=2·125.904gmol=253.808g/mol
Zn+I2→ZnI2
n(Zn) = 1.8895g65.38 g/mal=0.0289 mol n(I2) = 2.399g253.808 g/mal=0.00945 mol
Based on your work in Question 1, what is the excess reagent and what is the limiting reagent? Calculate how many grams of excess reagent remain.
Express reagent in Zn, limiting reagent is I2
ZnI2=319.188gmol x ∙0.0945 mol=3.0163 g
Zn + I2=4.2875g
2) – 1) = 1.2872 g
The quantity of Zn that remained unreacted is (0.0289 – 0.00945) mol·65.38 g/mol= 1.2716 g. The sum of the formed ZnI2 and the excess Zn is 3.0163 + 1.2716 = 4.2880 g
Based on your in-lab observations, identify the excess reagent. How did you make this identification? Does this result agree with your answer to Question 2?
The excess reagent is Zn, since it was left as the precipitate. The observations agree with the theoretical calculations.
What is the actual yield of product?
95.961+61.57=157.5319 g
161.750-157.5319=4.2181 g
This weight represents the weight of the remaining Zn and ZnI2.
Using the theoretical yield, calculate the % yield. What are possible sources of error that your group encountered that could have resulted in a lower % yield? If your % yield is greater than 100%, comment on why that might be.
Y=mpractmtheor=4.21814.284∙100%=98.4%
Because when we were putting the Zn we lost some.
Do you have conservation of mass in your experiment?
Yes, we have.
