Mercury(II) nitrate synthesis with subsequent Al/Hg producing. Video manual.

[ASheSChem]

@HIGGS BOSSON , @Marvin "Popcorn" Sutton ,
if in the thermometer there is only mercury... (1,5 gr or 2 gr? do yo know?)
I suppose we can use a vial of pure mercury?

 

[ACAB]

if in the thermometer there is only mercury... (1,5 gr or 2 gr? do yo know?)
I suppose we can use a vial of pure mercury?

ASheSChemIn any case, I would determine the amount of mercury with a balance and then calculate it according to the formula
Hg + 4HNO3 ----> Hg(NO3)2 + 2NO2 + 2H2O,
calculate the amount of nitric acid plus a small excess. For this, however, it might be necessary to filter the mercury beforehand because of the glass. Larger amounts of mercury can be separated well from impurities with a large syringe and absorbent cotton, but a thermometer provides too little mercury for this method.

 

[malayboy]

hi sir, how much of gram mercury is needed on 20ml nitric acid? what is the ratio?

sorry i just saw pennywise reply, i would try calculate myself

 

[Toomuchname]

Hey, can I use 60% nitric acid?

 

[ASheSChem]

@Saymynamehsb
yes 2gr mercury + 20ml nitric acid 60% it's what i do (the result is ok for 4x12gr aluminium)

 

[malayboy]

@Saymynamehsb
yes 2gr mercury + 20ml nitric acid 60% it's what i do (the result is ok for 4x12gr aluminium)

ASheSChem
May I know how do you calculate to get the 2gr mercury + 20ml nitric acid?

as for @Pennywise provided formula Hg + 4HNO3

 

[Toomuchname]

@Saymynamehsb
yes 2gr mercury + 20ml nitric acid 60% it's what i do (the result is ok for 4x12gr aluminium)

ASheSChemCool, thank you!

 

[ASheSChem]

May I know how do you calculate to get the 2gr mercury + 20ml nitric acid?

malayboyI just relied on the info from the video. After research a thermometer has on average 2gr of mercury.

So I tested and it works. liquid mercury nitrate can be stored in the fridge in an opaque glass vial. (I tested one week without problem, not yet more)

 

[malayboy]

hmmm, but I need to know why its 2gr mercury and 20ml nitric acid according to the equation Hg + 4HNO3
Maybe some pros can explain here? Because this will help us understanding and can apply on other synthesis understanding
@G.Patton

 

[ACAB]

May I know how do you calculate to get the 2gr mercury + 20ml nitric acid?

as for @Pennywise provided formula Hg + 4HNO3

malayboyThe theoretical way:

Molar mass Hg: 200.59g

Molar mass HNO3: 63.01g *4 = 252.04g / 1.51g/ml = 166.91ml HNO3 = 278.18ml 60%

20ml 60% = 12ml HNO3 * 1.51g/ml =18.12g HNO3

Needed Hg 14.42g / 13.55g/ml =1,06ml Hg

But normally you weigh your mercury and then you calculate with an excess of acid.

 

[ACAB]

And you have to consider what concentration has my solution, because the amount of aluminum and mercury salts should be in a ratio, which is quite imprecise. But it is enough as a guideline.

 

[malayboy]

The theoretical way:

Molar mass Hg: 200.59g

Molar mass HNO3: 63.01g *4 = 252.04g / 1.51g/ml = 166.91ml HNO3 = 278.18ml 60%

20ml 60% = 12ml HNO3 * 1.51g/ml =18.12g HNO3

Needed Hg 14.42g / 13.55g/ml =1,06ml Hg

But normally you weigh your mercury and then you calculate with an excess of acid.

Pennywisehello sir, many thanks to your reply, from you calculation i come out on my own case, please help me out:

I have 68% Nitric Acid 500ML, Mercury 100 g, the density i got from

Mercury (element) - Wikipedia

en.wikipedia.org

Nitric acid - Wikipedia

en.wikipedia.org

with Hg + 4HNO3, I want to scale down to around 50g Mercury, so can I have (Hg/4) + HNO3 ?

if yes,

solution of HNO3(63.01g / 1.51 g/ml) = (41.73ml / 68%) = 61.37ml
mass of mercury needed, 200.59g / 4 = 50.1475g, because I wanted to have gram instead of ml (possible?)

this seems really really wrong, and I am so embarrassed, please help me out

and your last two lines of the calculation,

20ml 60% = 12ml HNO3 * 1.51g/ml =18.12g HNO3

Needed Hg 14.42g / 13.55g/ml =1,06ml Hg

where are those calculation needed? its HNO3 in ml needed and gram of Hg right?

 

[G.Patton]

solution of HNO3(63.01g / 1.51 g/ml) = (41.73ml / 68%) = 61.37ml
mass of mercury needed, 200.59g / 4 = 50.1475g, because I wanted to have gram instead of ml (possible?)

malayboyFor 50g of Hg you have to take of 92.39g HNO3 68% or 65.77 mL
v(Hg)=50/200.59=0,249 mole
v(HNO3)=0.249*4=0,997 moles
m(HNO3)= 0.997*63.01g/mole=62.82 g pure but you have 68% HNO3. m(HNO3)=62.82*100/68=92.39g of 68% HNO3.
V(HNO3 68%)= 92.39g/1.4048g/mL=65.77mL

 

[MadHatter]

But ... it's always nice with perfect stochiometry and all, but is it really necessary? i just put the amount of mercury I want to convert in a beaker and add some nitric acid. If the reaction stops, I add a little more until all mercury is gone. In the end, there will only be acidic water and mercury nitrate anyway.

 

[malayboy]

For 50g of Hg you have to take of 92.39g HNO3 68% or 65.77 mL
v(Hg)=50/200.59=0,249 mole
v(HNO3)=0.249*4=0,997 moles
m(HNO3)= 0.997*63.01g/mole=62.82 g pure but you have 68% HNO3. m(HNO3)=62.82*100/68=92.39g of 68% HNO3.
V(HNO3 68%)= 92.39g/1.4048g/mL=65.77mL

G.Pattonhi sir, thanks for your reply, I got it now

one last thing, the density of HNO3 68% is given 1.14048g/mL (or 1.141g/mL in wikipedia),
what if I have 75%? sure thats not 1.151g/mL(100%) in wiki, thus how to get the density?

 

[malayboy]

But ... it's always nice with perfect stochiometry and all, but is it really necessary? i just put the amount of mercury I want to convert in a beaker and add some nitric acid. If the reaction stops, I add a little more until all mercury is gone. In the end, there will only be acidic water and mercury nitrate anyway.

DocX
yes, credits to @Pennywise

But normally you weigh your mercury and then you calculate with an excess of acid.

I understand always prepare excess of solvent is necessary, i can follow the video anyway, as long as the mercury is fully dissolved

but come to think of the reaction rate and quality of product, I will get the product that definitely close of 100% concentration and I don't waste any more acid right?
where I also think of the reaction rate will be faster than what the video shows due to the concentration

I just wanted to know what I am doing, that my study in chemistry pays off, do a little math before experiment so I don't need to prepare multi kilos every time just to drops drops drops until all solute is gone when comes to large scale

 

[MadHatter]

In any case, I would determine the amount of mercury with a balance and then calculate it according to the formula
Hg + 4HNO3 ----> Hg(NO3)2 + 2NO2 + 2H2O,
calculate the amount of nitric acid plus a small excess. For this, however, it might be necessary to filter the mercury beforehand because of the glass. Larger amounts of mercury can be separated well from impurities with a large syringe and absorbent cotton, but a thermometer provides too little mercury for this method.

PennywiseSorry, but this equation doesn't add up. You need to put H2O in the left side also..

 

[MadHatter]

Or use this one instead, which is balanced:

3Hg + 8HNO3 → 3Hg(NO3)2 + 2NO + 4H2O​

 

[malayboy]

hi sir, thanks for your reply, I got it now

one last thing, the density of HNO3 68% is given 1.14048g/mL (or 1.141g/mL in wikipedia),
what if I have 75%? sure thats not 1.151g/mL(100%) in wiki, thus how to get the density?

malayboyfound it here

The Complete Aqueous Nitric Acid Solutions Density-Concentration Calculator

 

[ACAB]

Sorry, but this equation doesn't add up. You need to put H2O in the left side also.

DocX

Or use this one instead, which is balanced:

3Hg + 8HNO3 → 3Hg(NO3)2 + 2NO + 4H2O​

No, sir, the formula is right, because we don't have nitrogen oxide in the reaction, which is colorless and odorless. Our gas is brown and smells like car exhaust and is called nitrogen dioxide.