Synthesis and Analysis of a Nickel Complex
Give an experimental section for this laboratory experiment. BOTH WEEKS
Synthesis of a Nickel complex was determined by adding ethylenediamine and ethanol. After obtaining close to 1.000g of nickel(II) sulfate hexahydrate, with your partner as instructed in step one, we quantitatively transferred it to a 250mL beaker with no more than 3mL of water. After the substance had dissolved we added the assigned amount of en to the substance slowly so it could attach to the nickel around the water, as instructed in step 2. Following step 3 in the lab manual we began to add ethanol slowly while stirring to precipitate out the nickel complex. Once the substance settled to the bottom we weighed a new container with the label, filter paper, and beaker, all included in step 4 of the lab manual. Following directions from steps five and six, we used the Buchner funnel to separate the precipitate from the liquid. Once the liquid was washed thoroughly with ethanol we let it dry and moved it to the previously weighed beaker. After this we set it in our lab drawer for it to dry over next week.
In this part of the experiment we ran the substance from week one through several tests to analyze our results. Using directions from step one and two, we weighed the Ni-en complex from last week. After this, we weighed about 0.30g of the Ni-en for our own calculations. Once I had 0.3g I quantitatively transferred it to a Erlenmeyer flask. Once the substance had dissolved in 50mL of DI, water green indicator was added and, HCl was titrated into the liquid until the substance turned a light yellow color, as indicated by step 3. Following steps 4 through five I added 6mL of NH4OH and heated the solution near boiling. While this occurred I add 30mL of 1% DMG while stirring. Using steps 6 and 7 in the lab manual, I pre-weighed a 100mL beaker with a filter paper inside. Once the solution had cooled I filtered the precipitate using the Buchner funnel and washed the Precipitate with ethanol. Once dry I put the precipitate in the 100mL and heated it in a drying oven. Once the substance heated for 20 minutes and had cooled I weighed the end results to make my calculations.
2. Completed Data tables Parts 1 through 4 (from pages 30, 32, 33 and 34). Be sure to include your lab partner’s name during the synthesis steps, and title each table appropriately.
Tables 1+2: Synthesis Calculations and Results of the Ni-en complex partner (s): Matthew Dovel
|Mass of NiSO4∙6H20||1.032g|
|Volume en added||5.4mL|
|Mass of filter paper, beaker, label||51.212g|
|Mass of filter paper, beaker, label, Ni-en||52.437g|
|Mass of your Ni-en complex||1.225g|
Table 3: Analysis and Calculations of en by HCl Titration
|Mass of Ni-en complex analyzed||0.306g|
|Concentration of HCl(aq)||0.129M|
|Initial buret reading||1.28mL|
|Final buret reading||32.49mL|
|Volume of HCl added||31.21mL|
|Moles HCl added||0.00403mol|
|Moles en in Ni-en complex analyzed||0.00201mol|
|Mass of en in Ni-en complex analyzed||0.121g|
Tables 4+ 5: Analysis and Calculations of Ni by DMG
|Mass of filter paper and beaker||50.690g (100mL beaker)|
|Mass of filter paper, beaker and Ni(DMG)2||50.921g|
|Mass of Ni(DMG)2||0.231g|
|Moles Ni2+ in Ni-en complex analyzed||0.000799mol|
|Mass Ni2+ in Ni-en complex analyzed||0.0469g|
|Moles SO42- in Ni-en complex analyzed||0.000799mol|
|Mass SO42- in Ni-en complex analyzed||0.0768g|
3. Completed Results tables Part 6 (from page 35).
Table 6: Analysis and Calculations of Hydrates in Ni-en complex
|Mass Ni2+ + mass Ni2+ + mass SO42-||0.245g|
|Mass H2O in Ni-en complex analyzed||0.061g|
|Moles H20 in Ni-en complex analyzed||0.0034mol|
4. Show the calculation to determine the moles of HCl that was added during the titration.
31.21mL HCl 1L / 1000mL= 0.00403mol HCl
5. Show the calculation to determine the moles of ethylenediamine, en, in your Ni-en complex you analyzed. Be sure to include any appropriate chemical reactions that were used in the calculation.
en + 2HCl H2en2+ + 2Cl–
0.00403mol HCl 1mol en/ 2mol HCl= 0.00201mol en
6. Show the calculation to determine the grams of ethylenediamine in your Ni-en complex you analyzed.
0.00156mol en 60.10g en/ 1mol en= 0.121g en
7. Starting with the mass of Ni(DMG)2 (that is the red precipitate), show the calculation to obtain the moles of Ni2+ that is present in your Ni-en complex you analyzed. Be sure to include any appropriate chemical reactions that were used in the calculation.
Ni2+ + 2DMG– Ni(DMG)2
0.231g Ni(DMG)2 1mol Ni(DMG)2 1mol Ni2+/ 288.94g Ni(DMG)2/ 1mol Ni(DMG)2= 0.000799mol Ni2+
8. Show the calculation to determine the grams of Ni2+ in your Ni-en complex you analyzed.
0.000799mol Ni2+ 58.69g Ni2+/ 1mol Ni2+= 0.0469g Ni2+
9. Show the calculation to determine the grams of sulfate, SO42–, in your Ni-en complex you analyzed.
1mol Ni2+= 1mol SO42-
0.000799mol Ni2+ = 0.000799mol SO42-
0.000799mol SO42- 96.06g SO42-/ 1mol SO42-= 0.0768g SO42-
10. Show the calculation to obtain sum of the en, Ni2+, and SO42– masses in your Ni-en complex.
0.0768g SO42- + 0.0469g Ni2+ + 0.121g en= 0.245g Ni2+, SO42-, en
11. Show the calculation to obtain the grams of water in your Ni-en complex you analyzed.
0.306g Ni2+, en, SO42-, H2O
-0.245g Ni2+, en, SO42-
12. Show the calculation to determine the moles of water in your Ni-en complex you analyzed.
0.061g H2O 1mol H2O/ 18.015g H2O= 0.0034mol H2O
13. Using the moles of each species in your Ni-en complex, show the calculations to determine the empirical formula for your Ni-en complex. Show how you determined ALL ratios and show the formula
0.0034mol H2O/ 0.000799mol Ni2+ = 4.25 approximately=4 H2O
0.000799mol Ni2+/ 0.000799mol Ni2+= 1 Ni2+
0.00201 en/ 0.000799mol Ni2+= 2.51 approximately= 3 en
NiSO4 6H2O+Yen [(Ni)x(en)y(H2O)z](SO4)x+6-ZH2O
Since ethylenediamine is a bidentate only three molecules are needed to complete each Nickel complex. Therefore, since there are 3 molecules of en then the Nickel complex is complete and the 4 hydrates are connected to the Sulfate.
NiSO4 6H2O + 3en [Nien3]SO4 4H2O +2H2O
14. Knowing your Ni-en complex’s empirical formula, write the overall balanced reaction that took place describing your synthesis.
NiSO4 6H2O + 3en [Nien3]SO4 4H2O +2H2O
15. Show the theoretical yield calculation for your Ni-en complex you synthesized.
(1.032g NiSO4 6H2O + 3en 1mol NiSO4 6H2O/ 262.9g NiSO4 6H2O + 3en) (1mol [Ni(en)3]SO4 4H2O/ 1mol NiSO4) 407.11g [Ni(en)3]SO4 4H2O/ 1mol [Ni(en)3]SO4 4H2O= 1.598g [Ni(en)3]SO4 4H2O
16. Show your percent yield calculation for your Ni-en complex you synthesized.
1.225g Ni-en/1.598g Ni-en 100= 76.66% yield
17. Answer the following questions. Show all work!
- Give the correct chemical formulas for the following:
a) potassium chromate b) sulfurous acid c) ammonia
K2CrO4 H2SO3 NH3
- Determine the correct chemical names of the following species:
a) LiH b) FeO c) MgSO4
Lithium Hydride Iron (II) Oxide magnesium sulfate
- A lab student analyzed her nickel-en complex for ethylenediamine, en, content. A 0.499 g sample of the synthesized Ni-en complex was dissolved in about 50 mL of water and titrated with 0.175 M HCl (aq). Endpoint required 37.49 mL of the HCl (aq). What is the mass (in grams) of en in the Ni-en complex analyzed?
37.49mL HCl/1000= 0.03749L HCl
0.03749L 0.175M HCl= 0.00656mol HCl
0.00656mol HCl 1mol en/2mol HCl= 0.00328 mol en
0.00328 mol en 60.10g en/ 1mol en= 0.197g en
- Continuing on with the analysis of the Ni-en complex as found in question #3, 40 mL of dimethylglyoxime, DMG, solution was added to the titrated nickel-en complex solution. The resulting beautiful red complex, Ni(DMG)2 was collected by filtration and found to weigh 0.469 g. What is the mass (in grams) of Ni in the Ni-en complex analyzed?
0.469g Ni(DMG)2 1mol Ni(DMG)2/ 288.94g Ni(DMG)2 1mol Ni2+/ 1mol Ni(DMG)2= 0.00162mol Ni2+
0.00162mol Ni2+ 58.69g Ni2+/ 1mol Ni2+=0.0951g Ni2+
- Determine the empirical formula of the complex synthesized in questions #3 and #4 above.
0.00162mol Ni2+= 0.00162mol SO4
0.00162mol SO42- 96.06g SO42-/1mol SO42-= 0.156g SO42-
0.0951g Ni2+ + 0.197g en + 0.156g SO42-=0.448g Ni2+,en, SO42-
-0.448g Ni2+, en, SO42-
0.051g H2O 1mol H2O/ 18.015g H2O=0.0028mol H2O
0.0028mol H2O/0.00162mol Ni2=1.7 H2O
0.00162mol Ni2+/0.00162mol Ni2+=1Ni2+
0.00162mol SO4/0.00162mol Ni2+=1SO4
0.00328 mol en/0.00162mol Ni2+=2.02 en
NiSO4 6H2O +2en [Ni(en)2(H2O)2]SO4 + 4H2O