Empirical Formula of Lead Iodide
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Published: Tue, 29 May 2018
If lead is dissolved in nitric acid and then allowed to react with potassium iodide, an insoluble lead iodide compound is formed. The mass of iodide in the lead iodide compound can be calculated by deducting the initial mass of lead from the mass of the precipitate. From the experimental data one can calculate then the ratio of moles of lead to moles of iodine in the compound, and thus determine the empirical formula.
Volume of nitric acid
Volume of potassium iodide solution
Volume of demineralized water
Mass of lead
Mass of the lead iodine compound
Method for Controlling Variables:
In this experiment, the mass of lead was accurately weighed with a calibrated electronic balance.
The volume of the nitric acid, dem. Water and potassium iodide solution was always measured using the same graduated cylinder.
The mass of the lead iodide compound was always determined on the same calibrated electronic balance.
- 3 pieces of 250 cm3 beakers (each one labeled #1, #2 and #3)
- 3 watch glasses
- 6 Iron stands
- 3 Cartridge burners
- Fume hood
- 3 pieces of Filter paper
- 3 pieces of Long-stemmed funnels
- Electronic Balance (300 g , readability ï‚± 0.001 g)
- 10 cm3 Graduated Cylinder (0.2 cm3, gradation ï‚± 0.1 cm3)
- 20 cm3 Graduated Cylinder (0.5 cm3, gradation ï‚± 0.1 cm3)
- 100 cm3 Graduated Cylinder (1 cm3, gradation ï‚± 0.5 cm3)
- 150 cm3 Erlenmeyer flask
- Ice bath
- 0.130 – 0.150 grams Granulated Lead
- 3 x 10 cm3 of 3 mol dm-3 Nitric Acid (HNO3)
- 1.2 g Potassium Iodide (KI)
- 120 cm3 Distilled Water
About 0.130 – 0.150 g granulated lead was weighed accurately on an electronic balance in a 250 cm3 beaker glass (labeled #1) and its mass was recorded.
The process was repeated for the two other beakers labeled # 2 and # 3.
3 x 10 cm3 of 3 mol dm-3 nitric acid, HNO3 solution was measured in a graduated cylinder and poured onto each sample of lead in the beakers #1, #2 and #3.
The beakers were covered with a watch glass, and heated gently on an iron stand with a cartridge burner in a fume hood until the steam exited from under the watch glass.
Reminder: Never boil it.
You may reheat it periodically if a gas evolution stops.
After all of the lead was reacted, approximately 20 cm3 of distilled water was added to each beaker.
The solution was heated until it steamed.
The beaker was removed and set aside for several minutes.
Approximately 1.2 g potassium iodide, KI was weighed onto weighing paper and transferred to a 150 cm3 Erlenmeyer flask.
Then 60 cm3 of distilled water was added to each beaker.
The solution was heated until it steamed and then cooled down slightly.
Then 20 cm3 of it was added slowly while stirred into beaker 1.
The mixture was set aside to cool down for 5 minutes while the procedure was repeated with beakers 2 and 3.
To be able to grow the crystals in size, the precipitate was heated gently with constant stirring for 5 – 10 minutes.
The mixture was cooled down using an ice bath.
A piece of filter paper was weighed to the nearest 0.001 gram on an electronic balance; its mass was recorded and was placed in a long-stemmed funnel.
Additional demineralized water was used to wash the precipitate and it was completely transferred from the beaker to the filter paper.
The filter paper was removed from the funnel and was placed on a labeled watch glass and was allowed to dry for one week.
The filtration was repeated for the solids of trials 2 and 3.
Once the filter paper and solids for each run are completely dry, the filter paper and solids were weighed for each trial on an electronic balance and the masses were recorded.
Table 1: Measurements of the Masses of Lead, Filter Paper and Lead Iodine Compound
Beaker / Filter Paper #
Mass of Lead
/ g ±0.001 g
Mass of Filter Paper
/ g ±0.001 g
Mass of Filter Paper plus Lead Iodide
/ g ±0.001 g
Observations: After dissolving the lead in nitric acid a clear colorless solution was obtained. When adding the potassium iodine solution a yellow crystalline precipitate formed.
Data Processing and Presentation:
Calculation of empirical formula trial 1:
MASS OF LEAD IODINE:
Mass of filter paper plus lead iodide – Mass of Filter paper = Mass of Lead Iodide
FILTER PAPER #1
1.383 g – 1.080 g = 0.303 g of Lead Iodide
Balance: 0.001/1.383 * 100 = 0.072 %
Balance: 0.001/1.080 * 100 = 0.093 %
Total uncertainty = 0.072 % + 0.093 % = 0.160 %
The mass of the lead iodine compound is 0.303 g ±0.16 %
MASS OF IODINE:
Mass of Lead iodide – Mass of Lead = Mass of Iodine
0.303 g – 0.153 g = 0.150 g of Iodine
Balance: 0.001/0.303 * 100 = 0.33 %
Uncertainty lead iodine compound = 0.16 %
Balance: 0.001/0.153 * 100 = 0.65 %
Total uncertainty = 0.33 % 0.16 % + 0.65 % = 0.98 % 0.81 %
The mass of the iodine compound is 0.150 g ±0.98% 0.81 %
Calculation of Moles of Lead (Pb):
n = m / M = 0.153 g / 207.19 g mol-1 = 0.000738 mol of Lead
Balance: 0.001/0.153 * 100 = 0.65 %
Uncertainty of molar mass : 0.01 / 207.19 * 100 = 0.0048 %
Total uncertainties = 0.65 % + 0.0048 % = 0.65 %
Moles of lead = 7.38 x 10-4 mol ±0.65 %
Calculation of Moles of Iodine (I):
n = m / M = 0.150 g / 126.90 g mol-1 = 0.00118 mol of Iodine
Balance: 0.001/0.150 * 100 = 0.67 %
Moles of lead = 1.18 x 10-3 mol ±0.67 %
uncertainty of iodine = 0.81 %
uncertainty of molar mass = 0.01 / 126.90 * 100 = 0.0079 %
Total uncertainties = 0.98 % + 0.0078 % = 0.99 %
Similar calculations have been carried out for trial # 2 and 3.
Ratio of Pb and I:
Ratio of Pb : I = 0.000711 mol : 0.00117 mol = 1.00 : 1.64 = 1 : 2
Table 2: Final result of the empirical formula calculation
Lead to iodine
7.38 x 10-4 ±0.65 %
1.18 x 10-3 ±0.67% 0.99 %
1 : 1.60
7.14 x 10-4 ±0.68%
1.21 x 10-3 ±0.65% 0.85 %
1 : 1.69
6.81 x 10-4 ±0.71%
1.11 x 10-3 ±0.71% 0.88 %
1 : 1.63
7.11 x 10-4 ±0.68%
1.17 x 10-3 ±0.68%
1 : 1.64
Total uncertainties: 0.68 % + 0.91 = 1.6 %
Theoretical Value – Experimental Value x 100 = Percentage Error
2.00 – 1.64 x 100 = 18 %
Conclusion and Evaluation:
In this experiment the empirical formula of a lead iodine compound was performed.
The average ratio of lead to iodine obtained was 1 : 1.64, and when rounded off would give a ratio of 1 : 2 such that its deviation from the true value is 0.36 mol which primarily depends on the oxidation status of lead.
The reason for the deviation of the result to the theoretical value is might be the purity of the lead, a loss of lead compound, a spilling, or an error into how the measurements were taken.
Comparing the result of the percentage error to the result of the uncertainties of the measurements shows that there is a big difference into the percentage error which cannot be explained. Conventionally, both values should be close to each other. Therefore the must have be variability with the error in the measurements particularly the impurity of the lead or the loss of lead throughout filtration.
The purity of the lead has a big impact on the result of the experiment which again is caused by a loss of a lead compound during filtration.
An improvement that could be done is to have less spillage, measure the volume of chemicals more precisely.
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