An Experiment to Determine the Empirical Formula of Lead Iodide

Chemicals:

• Distilled water
• Granulated lead
• 3M HNO3 (nitric acid)
• Potassium iodide, KI

Procedure:

1. About 0.130 and 0.150 g of granulated lead was weighed accurately on the electronic balance in a  250 cm3 beaker marked #1. and the mass of  lead was recorded to the nearest 0.001g
2. The process (1) was repeated with two other beakers marked #2 and #3.
3. Approximately 30 cm3 of 3 mol dm-3 nitric acid solution (HNO3) was placed in a 50 cm3 graduated cylinder.
4. 1/3 (10cm3) of 3 mol dm-3 nitric acid was poured onto each beakers containing lead.
5. Each beaker was covered with a watch glass.
6. The mixture was gently heated on an iron stand with cartridge burner in the fume hood until all of the lead has dissolved or until steam exits from under the watch glass
7. Approximately 20 cm3 of distilled water was added to each beaker.
8. The solution (beaker #1, 2 and 3) was heated again until it steams.
9. Each beaker was removed from heat.
10. Approximately 1.2 g of potassium iodide (KI) was weighed.
11. Potassium iodide was transferred to a clean 150 cm3 beaker.
12.  60 cm3 of distilled water was added to this beaker.
13. This beaker was heated until it steams.
14. The solution was cooled slightly.
15.  1/3 of this solution was poured into beaker #1,2 and 3.
16. Each beaker was heated and was stirred constantly until crystals were formed.
17. Each beaker was cooled in an ice bath for 5minutes.
18. 3 filter papers were weighed and were recorded to the nearest 0.001 g. Each filter was marked: filter “I”, “II” and “III”.
19. Each filter was placed on a long stemmed funnel.
20. The lead iodide in each beaker was filtered using corresponding filter papers.
21. The precipitate was washed twice with cold distilled water and transferred completely from the beaker to the filter paper. 
22. After the water was completely drained through the filter, the filter paper I, II and III were removed from the funnel and were dried by wrapping it in paper towel to absorb the water.
23. Each filter paper was air dried for one week at room temperature.
24. Each filter paper with lead iodide was weighed for each trial on the electronic balance to the nearest 0.001 g and the mass was recorded.

Data Collection and Processing

Raw Data:

Table 1: Mass of lead, filter paper and filter paper with lead iodide

Observations:

 Color of solution: Potassium iodide and nitric acid containing lead were all colorless.  When the potassium iodide was added in the beaker containing lead and nitric acid, the solution turned into bright yellow component.

State of solution: Potassium iodide and nitric acid containing lead were liquid.  When the potassium iodide was added in the beaker containing lead and nitric acid, an aqueous component is formed (lead iodide). Crystals were formed in the beaker.

Calculation of Empirical Formula Run 1 

Calculation of mass of lead iodide:

Mass of filter paper with lead iodide - Mass of filter paper= Mass of lead iodide

Mass of lead iodide = 1.129 ± 0.001 g  - 0.837± 0.001 g = 0.292 ± 0.002 g

Calculation of mass of iodide reacted:

Mass of lead iodide – Mass of used lead = Mass of iodide reacted

Mass of iodide reacted = 0.292 ± 0.002 g - 0.151 ± 0.001 g= 0.141 g ± 0.003 g

Calculation of amount of lead:

n= m/M

Mass of lead / molar mass of lead (Pb= 207.2 g mol–1) = mole of lead

mole of lead = 0.151 ± 0.001 g / 207.2 ± 0.1 g mol–1 = 7.29 x 10-4 moles

Uncertainty:

Balance: 0.001 / 0.151 x 100 = 0.66 %

Molar mass = 0.1 / 207.2 x 100 = 0.048 %

Total uncertainty = 0.66 % + 0.048 % = 0.71 %

Amount of lead = 7.29 x 10-4 ±0.71 % moles

Calculation of amount of Iodine:

n = m/M

Mass of iodine / molar mass of iodine (I= 126.9 g mol–1) = mole of iodine

mole of iodine = 0.141 ± 0.003 g / 126.9 ± 0.1 g mol–1= 1.11 x 10-3 moles

Uncertainty:

Balance: 0.003 / 0.141 x 100 = 2.1 %

Molar mass = 0.1 / 126.9 x 100 = 0.079%

Total uncertainty = 2.1 % + 0.079 % = 2.18 %

Amount of iodine =1.11 x 10-3 ± 2.18 %moles

Ratio: Moles of iodine: moles of lead

Ratio of Pb to I = moles of lead / moles of iodine

Ratio = (7.29 x10-4/7.29 x10-4)  : (7.29 x10-4 / 1.11 x10-3)= 1: 1.5

Uncertainty:

Moles of lead: 0.71 %

Moles of iodine = 2.18 %

Total uncertainty = 2.18 % + 0.71 % = 2.9 %

Ratio of lead to iodine= 1:1.5 ± 2.9 %

Similar calculations have been carried out for run 2 and 3.

Table 2: Final Result of Empirical Formula calculations

Conclusion:

The ratios of the mole of lead to Iodine were calculated and are 1: 1.6. We can assume that the compound is lead iodide PbI2 since the oxidation status of lead is either 2 or 4. Therefore, the theoretical ratio of iodine to lead is 1:2. Compared to the ratio in this experiment, the iodine in the theoretical ratio is greater by 0.4 than that of obtained calculation from our experiment.

Moreover, the percentage error of each calculation is below 5%, thus, the results are acceptable. If the percentage error is larger than 5%, the experiment is not valid.

Evaluation:

However, the result 1:1.6 obtained is slightly different from the theoretical possible ratio 1:2.  Hence, the component was not completely gathered. The product may have escaped when the watch glass was lifted, lost during filtration, or left on the wall of the beaker containing lead iodide.

Furthermore, the purity of lead will influence the ratio. The lead used might not be pure. If the lead was impure, other product apart from lead iodide would be formed.

To further improve the experiment, we must avoid lead iodide to escape, avoid lead iodide on the funnel, and filter all the lead iodide in the beaker. These would give greater ratio of iodine to lead.

Reference:

• http://en.wikipedia.org/wiki/Lead_iodide