Chem 122
Name: Grace Cole
Section DM1
Name of Partner: Virginia Torres
Spring 2015
Instructor: Dr. Eric Cotton
Date of Expt: 3/12/15
Date of Submission: 3/19/15
Experiment #5:
SYNTHESIS OF Ni(en)3Cl2
Abstract: The percent yield of tris was determined by setting up a nickel complex reaction.
Ethylenediamine (en) solution reacted with NiCl2•6H2O (hydrate) to produce 0.7628 g of
Ni(en)3Cl2 (tris). Hydrate was found to be the limiting reactant, and the theoretical yield
was calculated using the balanced chemical equation provided by Hamilton et al.1 Using
the actual yield and the theoretical yield, the percent yield of tris was determined to be
111%.
Introduction: To synthesize Ni(en)3Cl2 (tris), ethylenediamine solution was added to
aqueous NiCl2•6H2O (hydrate). The subsequent reaction (provided by Hamilton et al.1) occurred
as follows:
NiCl2•6H2O(s) + 3H2NCH2CH2NH2(aq)  [Ni(H2NCH2CH2NH2)3]Cl2(s) + 6H2O(l)
Each nickel ion can bond to three en molecules to form the Ni(en)32+ complex ion.1 When the
complex ion bonds to chloride ions in the solution, the product tris precipitates. However, the
hydrate, ethylenediamine, and tris are all soluble in water, which poses a significant problem in
separating tris. Tris is less soluble in an acetone/water mixture, though, so acetone was added to
the reaction mixture to facilitate the precipitation of the product.
Using the measurements of the reactants and the above equation, the hydrate was found to be the
limiting reactant. The theoretical yield was determined using the mass of the hydrate and the
molar masses of the hydrate and of tris. After synthesizing the nickel complex, the percent yield
of the product was calculated using the percent yield equation (also provided by Hamilton et al.1)
below:
𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑌𝑖𝑒𝑙𝑑 =
𝐴𝑐𝑡𝑢𝑎𝑙 𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑂𝑏𝑡𝑎𝑖𝑛𝑒𝑑
× 100
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑
Experimental Section
Procedure by Hamilton et al.1 was followed.
Data & Observations:
Mass of NiCl2•6H2O:
(Obtained using the TARE feature)
Appearance of solid NiCl2•6H2O:
Appearance of NiCl2•6H2O solution:
Volume of en solution (en soln):
Concentration of en solution:
(from the label on the bottle)
Density of en solution (from step #4):
Appearance of en solution before
addition to the hydrate:
Appearance of reaction solution after
addition of en solution to hydrate:
Appearance of reaction mixture after
addition of acetone:
Color of product, NiCl2•6H2O,
after drying process:
mass of empty vial + product:
Mass of empty vial:
0.527g
Lime green, multi-sided sand form
Dark green
2.30 mL
25.0%
0.950 g/mL
clear
Dark purple
Neon purple
Light purple
23.1300 g
22.3672 g
Calculations & Results:
1) 0.527 g NiCl2•6H2O
2) 0.527 𝑔 ℎ𝑦𝑑 ×
1 𝑚𝑜𝑙 ℎ𝑦𝑑
237.686 𝑔
= 𝟎. 𝟎𝟎𝟐𝟐𝟐 𝒎𝒐𝒍 𝑵𝒊𝑪𝒍𝟐 ∙ 𝟔𝑯𝟐 𝑶
3) 2.30 mL EN soln
4) 2.30 𝑚𝐿 𝐸𝑁 𝑠𝑜𝑙𝑛 ×
5) 2.30 𝑚𝐿 𝐸𝑁 𝑠𝑜𝑙𝑛 ×
6) 2.30 𝑚𝐿 𝐸𝑁 𝑠𝑜𝑙𝑛 ×
0.950 𝑔
1 𝑚𝐿
= 𝟐. 𝟏𝟗 𝒈 𝑬𝑵 𝒔𝒐𝒍𝒏
0.950 𝑔 𝐸𝑁 𝑠𝑜𝑙𝑛
1 𝑚𝐿 𝐸𝑁 𝑠𝑜𝑙𝑛
0.950 𝑔 𝐸𝑁 𝑠𝑜𝑙𝑛
1 𝑚𝐿 𝐸𝑁 𝑠𝑜𝑙𝑛
×
25 𝐸𝑁
100 𝐸𝑁 𝑠𝑜𝑙𝑛
25 𝐸𝑁
× 100 𝐸𝑁 𝑠𝑜𝑙𝑛 ×
= 𝟎. 𝟓𝟒𝟔 𝒈 𝑬𝑵
1 𝑚𝑜𝑙 𝐸𝑁
60.104 𝑔
= 𝟎. 𝟎𝟎𝟗𝟎𝟗 𝒎𝒐𝒍 𝑬𝑵
7) 0.527 𝑔 ℎ𝑦𝑑 ×
1 𝑚𝑜𝑙 ℎ𝑦𝑑
237.686 𝑔 ℎ𝑦𝑑
8) 2.30 𝑚𝐿 𝑠𝑜𝑙𝑛 ×
0.950 𝑔 𝑠𝑜𝑙𝑛
1 𝑚𝐿 𝑠𝑜𝑙𝑛
×
×
1 𝑚𝑜𝑙 𝑡𝑟𝑖𝑠
1 𝑚𝑜𝑙 ℎ𝑦𝑑
25 𝐸𝑁
100 𝑠𝑜𝑙𝑛
= 𝟎. 𝟎𝟎𝟐𝟐𝟐 𝒎𝒐𝒍 𝒕𝒓𝒊𝒔
×
1 𝑚𝑜𝑙 𝐸𝑁
60.104 𝑔
×
1 𝑚𝑜𝑙 𝑡𝑟𝑖𝑠
3 𝑚𝑜𝑙 𝐸𝑁
= 𝟎. 𝟎𝟎𝟑𝟎𝟑 𝒎𝒐𝒍 𝒕𝒓𝒊𝒔
9) Hydrate is the limiting reactant
10) 0.00222 moles of tris
1 𝑚𝑜𝑙 ℎ𝑦𝑑
1 𝑚𝑜𝑙 𝑡𝑟𝑖𝑠
11) 0.527 𝑔 ℎ𝑦𝑑 × 237.686 𝑔 ℎ𝑦𝑑 × 1 𝑚𝑜𝑙 ℎ𝑦𝑑 ×
12)
309.902 𝑔 𝑡𝑟𝑖𝑠
1 𝑚𝑜𝑙 𝑡𝑟𝑖𝑠
= 𝟎. 𝟔𝟖𝟕 𝒈 𝒕𝒓𝒊𝒔
(0.7628 𝑔 × 100)
⁄0.687 𝑔 = 𝟏𝟏𝟏%
Please see attached for Summary of Calculated Results page.
Discussion of Results:
Percent yield of tris synthesis was determined by setting up a nickel complex reaction and
measuring the product tris. The table below contains the results of the above calculations and the
data from which the limiting reactant is determined.
Experimental Moles
Theoretical Yield
of Tris in Moles
Theoretical Yield
of Tris in Grams
Mass of product (actual
yield):
Percent Yield of Tris
Hydrate
0.00222
Ethylenediamine
0.00909
0.00222
0.00303
0.687 g
0.7628 g
111%
In order to synthesize tris, ethylenediamine solution and hydrate NiCl2•6H2O were
reacted together. The en solution was added to the hydrate after grinding and dissolving it in a
minimal amount of water. After the exothermic reaction was observed, acetone was added to the
reaction mixture to help the product precipitate. The reaction mixture was allowed to cool down
with the help of a slush bath. During this time, a suction filtration apparatus was set up,
consisting of a Buchner funnel set atop a filter flask, which was supported by a ring stand and
connected to the vacuum pump.
The cooled reaction mixture was then transferred to the Buchner funnel in the suction filtration
apparatus. By turning on the vacuum, the filtrate separated from the product and entered the
filter flask. To be certain that only tris was left in the Buchner funnel, the suction was broken.
The solid in the Buchner funnel was broken up and some cold acetone was added before the
suction was returned. Suction was left on for a while then turned off. Tris crystals were then
scraped off the Buchner funnel onto a watch glass, which was placed beneath an infrared lamp.
Finally, the mass of the purple product was determined by difference.
To determine the percent yield, the theoretical yield had to be calculated. The
experimental moles of the hydrate were found by its experimental mass and molar mass, and the
experimental moles of en were found by its experimental volume of solution, density of solution,
molar mass, and concentration. By using dimensional analysis on both reactants, hydrate was
determined to be the limiting reagent. Thus, the theoretical yield of tris in grams was calculated
to be 0.687 grams of tris. Using the equation provided by Hamilton et al.1, the percent yield was
calculated to be 111%, a figure too high to be accurate.
The high percent yield is likely due to inability to completely dry tris and separate it from
the filtrate. There were many steps taken to reduce the effects of this limitation - including
addition of acetone, a slush bath, and heat of an infrared lamp - but there must still have been
some extra mass clinging to the product.
Another area in which error could have been introduced was in the transfer of the tris crystals
from the Buchner funnel to the watch glass to the vial. Although a rubber policeman was used to
transfer crystals, there was still some residual substance left on containers. Of course, this error
would have produced a low percent yield and should have canceled out some of the erroneous
percent yield produced by the above limitation.
Conclusion: Nickel chloride hexahydrate and ethylenediamine were successfully reacted
together to form tris, a light purple crystal. Percent yield of the product tris was calculated to be
111% from the actual yield of 0.7628 g. The excessive yield was due to inability of fully isolate
tris.
References:
1. Hamilton, P. Yau, C. and Zaman, K., CHEM 122 Experiments in General Chemistry I
Laboratory, Academx Publishing services: Bel Air, MD, 2013; pp. 63-70
Answers to Post-Lab questions
1. The procedure calls for using between 0.500 to 0.650 g of the hydrate in the synthesis.
Would a student using 0.500 g of hydrate obtain a smaller percent yield of TRIS than a
student using 0.650 g? Explain your answer.
Since the percent yield of an experiment does not depend on the amount used at the start,
the difference between percent yields cannot be predicted. The change in the amount of
hydrate that a student uses would be reflected in both the theoretical yield and the actual
yield. Percent yield is a ratio that relates actual yield and theoretical yield, thus the
increase or decrease in starting amount will not affect the percent yield in a perceptible
manner. Even though the hydrate was the limiting reactant, the amount used at the start
will not affect the percent yield because the percent yield depends on the method not the
amount of reactants.
2. At the part when all of the limiting reactant has reacted, how many moles of the other
reactant remain unreacted? Show your calculations & explain your answer clearly.
When all the hydrate has reacted, there are 2.44 x 10-3 moles of en left.
Moles of en present in reaction mixture:
2.30 𝑚𝐿 𝑜𝑓 𝑒𝑛 𝑠𝑜𝑙𝑛 ×
0.950 𝑔 𝑒𝑛 𝑠𝑜𝑙𝑛 25 𝑝𝑢𝑟𝑒 𝑒𝑛
1 𝑚𝑜𝑙 𝑒𝑛
×
×
= 0.00909 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑒𝑛
1 𝑚𝐿 𝑜𝑓 𝑒𝑛 𝑠𝑜𝑙𝑛 100 𝑒𝑛 𝑠𝑜𝑙𝑛 60.104 𝑔 𝑒𝑛
Moles of en reacted:
0.687 𝑔 𝑡𝑟𝑖𝑠 ×
1 𝑚𝑜𝑙 𝑡𝑟𝑖𝑠
3 𝑚𝑜𝑙 𝑒𝑛
×
= 0.00665 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑒𝑛
309.902 𝑔 𝑡𝑟𝑖𝑠 1 𝑚𝑜𝑙𝑒 𝑡𝑟𝑖𝑠
Moles of en left over:
0.00909 𝑚𝑜𝑙 𝑒𝑛 (𝑝𝑟𝑒𝑠𝑒𝑛𝑡) − 0.006650 𝑚𝑜𝑙 𝑒𝑛(𝑟𝑒𝑎𝑐𝑡𝑒𝑑) = 𝟎. 𝟎𝟎𝟐𝟒𝟒 𝒎𝒐𝒍 𝒆𝒏 (𝑙𝑒𝑓𝑡 𝑜𝑣𝑒𝑟)
The difference of the moles of en present during the reaction and the moles of en reacted
is equal to the number of moles of en remaining unreacted in the reaction mixture.
3. Based on the number of moles of en you used, how many moles of hydrate would you
need for it to totally react? How would this information tell you which reactant is the
limiting reactant? Show your work and explain clearly.
In order for all 0.00909 experimental moles of ethylenediamine to react, there would
need to be 0.00303 moles of hydrate. There were only 0.00222 moles of hydrate in the
experiment, which is a smaller value than 0.00303 moles; thus, the hydrate was the
limiting reactant.
Moles of hydrate needed to react with 0.00909 moles of en:
1 𝑚𝑜𝑙 ℎ𝑦𝑑𝑟𝑎𝑡𝑒
0.00909 𝑚𝑜𝑙 𝑒𝑛 ×
= 0.00303 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 ℎ𝑦𝑑𝑟𝑎𝑡𝑒
3 𝑚𝑜𝑙 𝑒𝑛
4. The liquid that goes through the filter paper into the filter flask is called the “filtrate.”
What exactly is in your filtrate? List all the substances that you can think of that is in the
filtrate.
There is leftover en solution, soluble tris, acetone, and water in my filtrate.