Alexis Musso
CHEM 111H – 202
October 2 and 9, 2012
Experiment 18
Preparation and Analysis of Lead (II) Iodide:
Recovery, Recycling, and Reuse of the Metal
Abstract
This experiment’s objective was to study the chemistry of a metal by preparing a compound and determining its formula by gravimetric analysis. In part one, PbI
2 was prepared through a series of reactions starting with (PbCO
3
)
2
·Pb(OH)
2
and using acids and potassium iodide. The technique used was recrystallization in which a solid was dissolved using a hot solvent, and then the solution was cooled to effect precipitation. After filtering a volume of the solution, the collected solids of PbI
2
were dried then weighed to calculate percent yield. Then, in part two, purity of the PbI
2 produced was determined by comparing the percent lead obtained through gravimetric analysis with the theoretical percentage. Finally, the empirical formula of PbI
2
was determined. In total, the experiment was a success if the empirical formula determined for lead iodide corresponded to the theoretical empirical formula of PbI
2
.
Experimental
Stoichiometric Equations :

(PbCO
3
)
2
·Pb(OH)
2
+ 6HNO
3
→ 3Pb
2+
+ 6NO
3
-
+ 4H
2
O + 2CO
2
+ excess H
+

Pb(NO
3
)
2
+ 2OH
-
→ Pb(OH)
2
↓ + 2NO
3
-

NH
3
+ H
2
O → NH

Pb(OH)
4
-
+ OH
-

Pb(NO
3
2
+ 2HNO
3
→ Pb(NO
3
)
2
+ 2H
2
O
)
2
+ 2KI → PbI
2
+ 2KNO
3

PbI
2
(s) + HNO
3
(aq) → Pb
2+
(aq) + 2NO

Pb
2+
+ H
2
SO
4
→ PbSO
4
+ H
+
3
-
(aq) + H
+
(aq) + I
-
(aq)
Mathematical Formulas:

Theoretical yield PbI
2
= g (PbCO
3
)
2
·Pb(OH)
2
1 𝑚𝑜𝑙 (PbCO3)2·Pb(OH)2
(
775.62 𝑔 (PbCO3)2·Pb(OH)2
) (
3 𝑚𝑜𝑙𝑒𝑠 𝑃𝑏(𝑁𝑂
3
)
2
1 mole (PbCO3)2·Pb(OH)2
)(
1 𝑚𝑜𝑙𝑒 𝑃𝑏𝐼
2
1 𝑚𝑜𝑙 𝑃𝑏(𝑁𝑂
3
)
2
)(
461.00 𝑔 𝑃𝑏𝐼
2
1 𝑚𝑜𝑙 𝑃𝑏𝐼
2
)

Percent yield PbI
2 𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 𝑜𝑓 𝑃𝑏𝐼
2
(𝑔)
= theoretical yield of 𝑃𝑏𝐼
2
(g)
𝑥 100

Weight of Pb in PbSO
4
(g) = g PbSO
4
(
1 𝑚𝑜𝑙 PbSO4
303.26 𝑔 PbSO4
) (
1 𝑚𝑜𝑙 𝑃𝑏
1 mol PbSO4
)(
207.2 𝑔 𝑃𝑏
)
1 𝑚𝑜𝑙 𝑃𝑏

Percent Composition = 𝑡ℎ𝑒 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑡ℎ𝑒 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
x 100

Theoretical % lead in PbI
2 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑃𝑏
= 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑃𝑏𝐼
2
x 100

Percent error =
|𝑎𝑐𝑡𝑢𝑎𝑙−𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙|
x 100 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙

Moles of element in PbI
2 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡
= 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑃𝑏𝐼
2

Mole Ratio = 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 1 moles of element 2
>> use to find empirical formula Pb x
I y

Reagents:
Common Name Formula
Molar Mass
(g/mole)
Density
(g/cm 3 ) lead carbonate nitric acid ammonia solution
(PbCO
3
)
2
·Pb(OH)
2
HNO
3
NH
3
775.62
63.00
58.179
-
2-3 (vapor)
0.59 sulfuric acid potassium iodide
H
2
SO
4
KI
98.08
166.0
3.4
3.1
Water H
2
O 18 1.00
Source : http://www.chem.tamu.edu/class/majors/chem101h-lab/index.html, http://ull.chemistry.uakron.edu/erd/
Materials:
Part One:

1 small X stir bar

1 large X stir bar
 filter paper

1 Buchner funnel

1 suction flask

1 watch glass

1 Pasteur pipet + bulb

1 magnetic stirring hot plate








1 sand bath
1 support stand
Melting
Point
(°C)
>400
-42
-77
-64
680
0
2 magnetic stir retrievers
1 600 mL beaker
1 50 mL beaker
1 250 mL beaker
1 400 mL beaker
Analytical balance
Part Two:

1 small X stir bar
 filter paper
 waste beaker

1 tweezers

1 crucible + cover

1 triangle

1 Buchner funnel

1 suction flask

1 25 mL volumetric pipet

1 pipet pump









1 Pasteur pipet + bulb
1 ring stand
1 magnetic stirring hot plate
1 desiccator
1 Bunsen burner
1 100 mL beaker
1 100 mL volumetric flask
1 150 mL beaker
Analytical balance
Boiling
Point
(°C)
-
122
-
337
1330
100

Experimental Apparatus:
400 mL beaker
Support Stand
Magnetic Stirring Plate
Buchner funnel
Ring stand
Triangle Holder
Crucible
Bunsen Burner

Procedure
Part 1:
1.
200 mg of the lead compound was weighed into a 50 mL beaker
2.
5 mL 6 M HNO
3
was added and stirred to dissolve.
3.
6 M NH
3
was added in drops until a precipitate persisted.
4.
The precipitate was dissolved by slowing adding 1 M HNO
3
by drops.
5.
Next, the solution was transferred by rinsing to a 400 mL beaker with a large stir bar.
6.
600 mg KI was weighed into a 250 mL beaker and dissolved with 50 mL H
2
O. Then, this mixture was added to the 400 mL beaker.
7.
The total volume was brought to 150 mL and heated until the precipitate dissolved. Then, the beaker was transferred to a sand bath and the stir bar was removed.
8.
After the crystal formation slowed, the beaker was placed in a larger 600 mL ice bath.
9.
While the wash bottle was in the ice bath, the crystals were filtered.
10.
Then, the last of the crystals were washed with the cold water onto the filter paper. The air was sucked through until the filter paper was somewhat dry.
11.
The filter paper with crystals was transferred to a watch glass and stored.
12.
The waste was cleaned up with all the waste going into the waste jug.
Part 2:
1.
The crucible was heated to constant weight.
2.
In the meantime, the watch glass, filter paper, and PbI
2
were weighed.
3.
Then, 200 mg PbI
2
was transferred to a 100 mL beaker with stir bar. Extra PbI
2
was put in the waste jar.
4.
Next, the watch glass and filter paper were weighed. After, the filter paper was put in the plastic beaker on the desk.
5.
10 mL of 50% HNO
3
was added to the beaker of PbI
2
.
6.
The mixture was heated under the hood until PbI
2
was dissolved and no more red-brown gas was being emitted.
7.
Next, it was transferred to a 100 mL volumetric flask and diluted to the mark and mixed by inversion at least twenty times.
8.
25 mL of the solution was pipetted twice into a 150 mL beaker.
9.
The beaker was cooled thoroughly and 10 mL of concentration H
2
SO
4
was added.
10.
The beaker was then warmed until colorless or white fumes formed.
11.
Next, the beaker was cooled in an ice bath, filtered, and washed with a 0.5 M H
2
SO
4
.
12.
The filter paper and PbSO
4
were placed in the crucible and heated until the filter paper burned off.
13.
The steps of heating, cooling, and weighing were repeated until constant weight.
14.
The waste was cleaned up with all the waste going into the waste jug.

Results
Reduced Data Tables :
Chart 1: Preparation of Lead (II) Iodide
Weight of (PbCO
3
)
2
·Pb(OH)
2
(g) 0.2230
Weight of KI (g) 0.6036
Weight of watch glass, filter paper, and PbI
2
(g)
Used weight of PbI
2
product
Weight of watch glass and filter paper (g)
Total weight of PbI
2
product (g)
Theoretical yield of PbI
2
product (g)
Percentage yield
49.4345
0.2072
49.0716
0.3629
0.3976
91.27%
Chart 2: Gravimetric Analysis
Used sample weight of PbI
2
(g)
Weight of empty, heated crucible (g) 11.5290
Weight of crucible and PbSO
4
(g)
Weight of PbSO
4
(g)
Weight of lead in PbSO
4
(g)
% lead in sample of PbI
2
Theoretical % lead in PbI
2
Percent error
11.5711
Trial 1 Trial 2
0.2072
11.5286
11.5712
0.0426
0.0291
28.09%
44.95%
37.51%
Empirical Formula PbI
3
Note : In determining the empirical formula, the actual was PbI
2.56
which rounds up to PbI
3
verse the theoretical PbI
2
.
Example Calculations :

Total weight of PbI
2
= (watch glass + filter paper + PbI
2
) – ( watch glass + filter paper) o Known values: mass of watch glass + filter paper + PbI
2
= 49.4345 g, mass of watch glass + filter paper = 49.0716 g o Unknown variable: total weight of PbI
2
product o Total weight of PbI
2
product = (49.4345 g) – (49.0716 g) = 0.3629 g

Theoretical yield PbI
2
= g (PbCO
3
)
2
·Pb(OH)
2
1 𝑚𝑜𝑙 (PbCO3)2·Pb(OH)2
(
775.62 𝑔 (PbCO3)2·Pb(OH)2
) (
3 𝑚𝑜𝑙𝑒𝑠 𝑃𝑏(𝑁𝑂
3
)
2
1 mole (PbCO3)2·Pb(OH)2 o Known values: g PbCO
3
)(
1 𝑚𝑜𝑙𝑒 𝑃𝑏𝐼
2
1 𝑚𝑜𝑙 𝑃𝑏(𝑁𝑂
3
= 0.2230 g, molar mass (PbCO g/mol, mole ratio of Pb(NO
3
)
2
to (PbCO
3
)
2
·Pb(OH)
2
)
2
3
)
)(
2
461.00 𝑔 𝑃𝑏𝐼
2
1 𝑚𝑜𝑙 𝑃𝑏𝐼
2
·Pb(OH)
)
2
= 775.62
= 3:1, mole ratio of PbI
2
to
Pb(NO
3
)
2
= 1:1, molar mass PbI
2
= 461.00 g/mol o Unknown variable: theoretical yield of PbI
2 o 0.2230 g (PbCO
3
)
2
·Pb(OH)
2
1 𝑚𝑜𝑙 (PbCO3)2·Pb(OH)2
(
775.62 𝑔 (PbCO3)2·Pb(OH)2
) (
3 𝑚𝑜𝑙𝑒𝑠 𝑃𝑏(𝑁𝑂
3
)
2
1 mole (PbCO3)2·Pb(OH)2
)(
1 𝑚𝑜𝑙𝑒 𝑃𝑏𝐼
2
1 𝑚𝑜𝑙 𝑃𝑏(𝑁𝑂
3
)
2
)(
461.00 𝑔 𝑃𝑏𝐼
2
1 𝑚𝑜𝑙 𝑃𝑏𝐼
2
) =
0.3976 g PbI
2


Percent yield PbI
2 𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 𝑜𝑓 𝑃𝑏𝐼
2
(𝑔)
= theoretical yield of 𝑃𝑏𝐼
2
(g)
𝑥 100 o Known values: total yield of PbI
2
= 0.3629 g, theoretical yield of PbI
2
= 0.3976 g o Unknown variable: percent yield o Percent yield PbI
2
=
0.3629 𝑔
0.3976 g
𝑥 100 = 91.27% PbI
2
 grams PbSO
4
= (Weight of crucible + PbSO
4
) – (Weight of empty, heated crucible) o Known values: Weight of crucible and PbSO
4
= 11.5712 g, Weight of empty, heated crucible (g) = 11.5286 g o Unknown variable: weight of PbSO
4 o Weight of PbSO
4
(g) = 11.5712 g – 11.5286 g = 0.0426 g PbSO
4 o note : the second trials were used

Weight of Pb in PbSO
4
(g) = g PbSO
4
(
1 𝑚𝑜𝑙 PbSO4 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 PbSO4
) ( 𝑚𝑜𝑙𝑒 𝑟𝑎𝑡𝑖𝑜 𝑃𝑏
)(
PbSO4 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑃𝑏
)
1 𝑚𝑜𝑙 𝑃𝑏 o Known values: g PbSO
4
= 0.0426 g, molar mass PbSO
4
= 303.26 g/mol, mole ratio of Pb to PbSO4 = 1:1, molar mass Pb = 207.2 g/mol o Unknown variable: Weight of Pb in PbSO
4 o Weight Pb = 0.0426 g PbSO
4
(
1 𝑚𝑜𝑙 PbSO4
303.26 𝑔 PbSO4
) (
1 𝑚𝑜𝑙 𝑃𝑏
1 mol PbSO4
)(
207.2 𝑔 𝑃𝑏
) = 0.0291 g
1 𝑚𝑜𝑙 𝑃𝑏

Percent Composition = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
x 100 o Known values: mass Pb = 0.0582 g, mass PbI
2
= 0.2072 g o Unknown variable: percent composition o % Pb =
0.0582 g Pb
0.2072 𝑔 𝑃𝑏𝐼
2
x 100 = 28.09 % Pb

Theoretical % lead in PbI
2 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑃𝑏
= 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑃𝑏𝐼
2
x 100 o Known values: molar mass Pb = 207.2 g/mol, molar mass PbI
2
= 461.00 g/mol o Unknown variable: theoretical % Pb in PbI
2 o Theoretical % lead in PbI
2
=
207.2
g mol g
𝑃𝑏
461.00
mol
𝑃𝑏𝐼
2
x 100 = 44.95% Pb

Percent error =
|𝑎𝑐𝑡𝑢𝑎𝑙−𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙|
x 100 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 o Known values: actual % = 28.09%, theoretical % = 44.95% o Unknown variable: percent error o Percent error =
|28.09 −44.95|
x 100 = 37.51% error
44.95

Moles of element in PbI
2 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡
= 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑃𝑏𝐼
2 o Assuming 100 g of the compound, 28.09% Pb is 28.09 g Pb o Known values: mass of element = 28.09 g Pb, molar mass PbI
2 o Unknown variable: moles of element
= 461.00 g/mol o # Moles =
28.09 𝑔 𝑃𝑏 𝑔
461.00
𝑚𝑜𝑙
𝑃𝑏𝐼
2
= 0.0609 mol Pb


Mole ratio = 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 1 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 2 o Known values: moles of Pb = 0.0609 mol Pb, moles of I = 0.1560 mol I o Unknown variable: mole ratio o Mole ratio =
0.1560 mol I
0.0609 mol Pb
= 2.56 ≈ 3 o Therefore, the ratio of Pb to I is 1:3 , and the empirical formula is PbI
3
Discussion
In part one, a series of reactions were carried out to implement recrystallization of PbI
2
, in which the solid was dissolved using a hot solvent and the solution cooled to effect precipitation. First, 5 mL 6 M HNO
3
was added to (PbCO
3
)
2
·Pb(OH)
2
to separate out the Pb
2+ ions in the reaction: (PbCO
3
)
2
·Pb(OH)
2
+ 6HNO
3
→ 3Pb 2+ + 6NO
3
+ 4H
2
O + 2CO
2
+ excess H +
Then, 6 M NH
3
was added in drops until a precipitate persisted to transform Pb(NO
3
)
2
to
Pb(OH)
2
. Then, to get Pb(OH)
2 back in solution, 1 M HNO
3
was added by drops to slowly dissolve the precipitate. It was important to add just enough to re-dissolve Pb(OH)
2
so that excess acid was not produced. Finally, about 6 mg of KI was weighed out and added to the solution in order to produce PbI
2
in the reaction: Pb(NO
3
)
2
+ 2KI → PbI
2
+ 2KNO
3
. Because the limiting reagent was (PbCO
3
)
2
·Pb(OH)
2
, the grams were converted to moles using its molar mass. Then, because the mole ratio of Pb(NO
3
)
2
to (PbCO
3
)
2
·Pb(OH)
2
= 3:1 and the mole ratio of PbI
2
to Pb(NO
3
)
2
= 1:1, the number of moles of PbI
2 was obtained and converted to grams using its molar mass. x grams (PbCO
3
)
2
·Pb(OH)
2
𝟏 𝒎𝒐𝒍 (𝐏𝐛𝐂𝐎𝟑)𝟐·𝐏𝐛(𝐎𝐇)𝟐
(
𝟕𝟕𝟓.𝟔𝟐 𝒈 (𝐏𝐛𝐂𝐎𝟑)𝟐·𝐏𝐛(𝐎𝐇)𝟐
) (
𝟑 𝒎𝒐𝒍𝒆𝒔 𝑷𝒃(𝑵𝑶
𝟑
)
𝟐
𝟏 𝐦𝐨𝐥𝐞 (𝐏𝐛𝐂𝐎𝟑)𝟐·𝐏𝐛(𝐎𝐇)𝟐
)(
𝟏 𝒎𝒐𝒍𝒆 𝑷𝒃𝑰
𝟐
𝟏 𝒎𝒐𝒍 𝑷𝒃(𝑵𝑶
𝟑
)
𝟐
)(
𝟒𝟔𝟏.𝟎𝟎 𝒈 𝑷𝒃𝑰
𝟐
𝟏 𝒎𝒐𝒍 𝑷𝒃𝑰
𝟐
)
The solids collected from filtering were dried and weighed to calculate the yield of PbI
2
. The percent yield was then determined by comparing the actual yield to the theoretical yield of PbI
2
.
Next, in part two, about 200 mg of the lead (II) iodide from part one was measured out and
10 mL of nitric acid was added. The mixture was heated, transferred to a 100 mL flask, and diluted to the mark. Then, 50 mL of the solution was transferred to another beaker, 10 mL of concentrated H
2
SO
4
was added, and the beaker was warmed until white fumes formed. It was important to note that only 50 mL of the 100 mL of solution was used in the rest of the experiment because in the data manipulation the grams of lead calculated would need to be multiplied by two to get the entire amount. After the beaker was cooled, filtered, and washed, the filter paper and PbSO
4
were placed in the crucible and heated to burn off the filter paper. Once

the crucible was cooled, it could be measured. This allowed the amount of PbSO
4
to be calculated by difference of crucible plus PbSO
4
and the empty crucible weighed at the beginning.
The grams of PbSO
4
determined could be converted to moles using molar mass, and since the mole ratio of Pb to PbSO
4
is 1:1, this was the same number of moles of Pb, which was finally converted to grams. The grams of lead found was then multiplied by two to get the total lead in the experiment. This amount was further used to calculate percent lead in lead (II) iodide using percent composition: 𝒕𝒉𝒆 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒆𝒍𝒆𝒎𝒆𝒏𝒕
Percent Composition = 𝒕𝒉𝒆 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅
x 100
Finally, the purity of the lead (II) iodide produced was determined by comparing the percent lead obtained through gravimetric analysis with the theoretical percentage. Lastly, the empirical formula of lead (II) iodide was determined to be PbI
3
due to a percent error of 37.51%.
The experiment was unsuccessful due to some errors. In part one, one error that may have occurred is that some of the solid PbI
2
could have been lost while filtering if it got under the filter paper. In part two, an error occurred when the filter paper was disposed of in the waste jar before weighing a second time. However, the filter paper was retrieved and weighed but an inaccurate weight may have been obtained. Overall, because the solution was continually being transferred between beakers, flasks, and crucibles, continual loss of the product may have occurred adding up to a significant loss if it stuck to the pipettes and beakers from which it was being transferred or filtered each time.
Conclusion
The percent lead determined in the sample of lead (II) iodide was 28.09% while the theoretical percent was 44.95%, which gives a percent error of 37.51%. The empirical formula determined for lead (II) iodide was PbI
3
. Therefore, the experiment was fairly
unsuccessful because the theoretical empirical formula is PbI
2
.