You are Here: Home

# is this reaction equation correct? watch

1. (Original post by charco)
The compound itself is usually abbreviated as dmgH2

In this complexation the base removes one hydrogen ions from each dimethylglyoxime to give dmgH-

It is this that forms a planar bis(dmgH) complex with nickel(II)

The negative charge on the deprotonated oxygen then hydrogen bonds to the hydrogen of the other ligand, This happens in two places in the complex, giving a pseudo tetradentate complex which is very stable and very insoluble.

On your structure of the final complex are there 2 hydrogens missing/not drawn in? (bonded to the the two nitrogens without formal charges). The structure on pubchem has 2 hydrogens there thats why?
2. ughh i messed up again

when i calculated the yield i got 300+ % which is obviously wrong.... i did the moles of the product divided by the moles of the deficit reactant (starting nickel complex).

Can anyone help? Im so sorry for being annoying and asking for help all the time :/
3. (Original post by kiiten)
ughh i messed up again

when i calculated the yield i got 300+ % which is obviously wrong.... i did the moles of the product divided by the moles of the deficit reactant (starting nickel complex).

Can anyone help? Im so sorry for being annoying and asking for help all the time :/
calculate moles of starting nickel complex by dividing the mass by the Mr

Find theoretical yield by multiplying mass of nickel product by moles above

Percentage yield = 100 x actual mass product/theoretical mass product
4. (Original post by charco)
calculate moles of starting nickel complex by dividing the mass by the Mr

Find theoretical yield by multiplying mass of nickel product by moles above

Percentage yield = 100 x actual mass product/theoretical mass product
Im confused by your second line. Im gonna state values to make it easier to explain.

Mass starting nickel complex = 0.25g, mmol = 0.75

Actual mass of end nickel complex = 0.84g, mmol = 2.89

So youre saying the theoretical mass in this case = 0.84 x (0.75 / 1000)
= 6.3 x 10^-4 which doesnt make sense?

The way i did it:
0.75 x 290.93 = 218.197... mg theoretical
= 0.22 g

yield = (0.84 / 0.22) x 100
= 385% ????

EDIT: Apparently that yield is correct? but i didnt dry the product for long enough so its higher than expected?
5. (Original post by kiiten)
Im confused by your second line. Im gonna state values to make it easier to explain.

Mass starting nickel complex = 0.25g, mmol = 0.75

Actual mass of end nickel complex = 0.84g, mmol = 2.89

So youre saying the theoretical mass in this case = 0.84 x (0.75 / 1000)
= 6.3 x 10^-4 which doesnt make sense?

The way i did it:
0.75 x 290.93 = 218.197... mg theoretical
= 0.22 g

yield = (0.84 / 0.22) x 100
= 385% ????

EDIT: Apparently that yield is correct? but i didnt dry the product for long enough so its higher than expected?
so that's good news then ...
6. (Original post by charco)
so that's good news then ...
I guess .... but now i will get marked down for the yield xD
7. (Original post by kiiten)
I guess .... but now i will get marked down for the yield xD
Marks are not the be all and end all.

The most important thing is to understand.
8. (Original post by charco)
Marks are not the be all and end all.

The most important thing is to understand.
Hey i just wanna ask you one more thing - if you can still remember this experiment haha.

So i did some research but im unsure about how the complex (final nickel complex w DMG ligands) is insoluble in aqueous solution.

Am i right to say its neutral and very stable so it doesnt form hydrogen bonds with water. This makes it insoluble.

Thanks
9. (Original post by kiiten)
Hey i just wanna ask you one more thing - if you can still remember this experiment haha.

So i did some research but im unsure about how the complex (final nickel complex w DMG ligands) is insoluble in aqueous solution.

Am i right to say its neutral and very stable so it doesnt form hydrogen bonds with water. This makes it insoluble.

Thanks
Yes. Neutral & no H bonds.
10. Me again

Just a quick ques

[Ni(H2O)6]Cl2 + 6NH3 + 2NH4BF4 --> [Ni(NH3)6](BF4)6 + 6H2O + 2NH4Cl

So with this equation how do i know which reactant is in deficit. I calculated the mmoles of reactants as 12.6, 620, 23.9 respectively. Ammonia is in excess but how do i know ammonium tetrafluoroborate is the one in deficit when there are more moles than the nickel complex?!
11. (Original post by kiiten)
Me again

Just a quick ques

[Ni(H2O)6]Cl2 + 6NH3 + 2NH4BF4 --> [Ni(NH3)6](BF4)6 + 6H2O + 2NH4Cl

So with this equation how do i know which reactant is in deficit. I calculated the mmoles of reactants as 12.6, 620, 23.9 respectively. Ammonia is in excess but how do i know ammonium tetrafluoroborate is the one in deficit when there are more moles than the nickel complex?!
The limiting reagent is the one with the fewest moles after the equation stoichiometry is taken into consideration.
12. (Original post by charco)
The limiting reagent is the one with the fewest moles after the equation stoichiometry is taken into consideration.
So why isnt it the nickel complex because it has 1 mole compared to 2 of NH4BF4 in the reaction eq. And also a lower amount of calculated moles
13. (Original post by kiiten)
So why isnt it the nickel complex because it has 1 mole compared to 2 of NH4BF4 in the reaction eq. And also a lower amount of calculated moles
You can hardly expect anyone to comment when you have not provided the data.
14. (Original post by charco)
You can hardly expect anyone to comment when you have not provided the data.
Oh! My apologies - i just assumed i already added it earlier in the thread xD Sorry!
15. Bump

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: April 17, 2018
Today on TSR

### He broke up with me because of long distance

Now I'm moving to his city

### University open days

1. Norwich University of the Arts
Thu, 19 Jul '18
2. University of Sunderland
Thu, 19 Jul '18
3. Plymouth College of Art