muriaticacid
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Nitration Calculations
Hey,
I was wondering how one goes about calculating the amount of chemicals needed for a nitration. NC, for instance. Brainfevert uses 30 ml of H2SO4, 25g
KNO3 and 2g of cellulose. how did he calculate that proportion? is there some sort of molar ratio?
thank you for your time
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Sickman
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| Quote: | Originally posted by muriaticacid
Hey,
I was wondering how one goes about calculating the amount of chemicals needed for a nitration. NC, for instance. Brainfevert uses 30 ml of H2SO4, 25g
KNO3 and 2g of cellulose. how did he calculate that proportion? is there some sort of molar ratio?
thank you for your time |
First things first:
A. Determine what are the reactants and the products.
B. Balance the equation.
Balancing the equation means showing an equal number of atoms for each element on both sides of the equation. Remember that the law of conservation of
mass states that the same amount of matter must be present on both sides of the equation. Take the following example:
(1) C5H12O4 + (4) HNO3-------> (1) C5H8N4O12 + (4) H2O
Pentaerythritol + Nitric Acid------->Pentaerythritol tetranitrate + Water
The above is pentaerythritol reacting with nitric acid.
If we want the "tetra"nitrate, we will need four NO3 groups (which is conveinently available in the form of nitric acid) for the four hydroxyl groups
on the pentaerythritol molecule. The NO3 part of HNO3 (nitric acid) replaces the hydroxyl groups and at the same time the H (hydrogen) left over from
the HNO3, reacts with the OH (hydroxyl groups) to form water (H2O). As you can see above, this equation is balanced and all the atoms can be accounted
for.
Note: As you can see, nitrations which involve replacing hydroxyl groups such as in the synthesis of nitric esters, like PETN, ETN, and
nitroglycerine, all produce a fair amount of water as a byproduct of the reaction. This water dilutes the nitric acid as the reaction progresses and
can greatly interfere with the reaction by diluting the reaction zone resulting in fewer NO3 groups being added to the molecule and lower yields
overall. To counteract this dilution of nitric acid during the nitration, sulfuric acid is often employed in many nitrations and acts as a dehydrator
to the nitric acid. In the case of PETN, nitric acid alone can be used, however the nitric acid must be in excess of what is required in the balanced
equation to get a good yield of PETN.
Secondly: Solve the mass-mass problem.
The above equation is simply a shorthand expression which gives information about the reaction of pentaerythritol with nitric acid. Now that we have
looked at the equation and we know it's balanced, we can use it to help us solve the mass-mass problem. The coeffieciants of our balanced equation
give us the relative amounts (in moles) of reactants and products.
Let's look at this for a moment. Let's say we start off with 10 grams of pentaerythritol. How many grams of nitric acid will we need to completely
convert all 10 grams of pentaerythritol to PETN? To answer this question we need to convert our 10 grams of pentaerythritol into moles. (I have a
handy table of atomic masses for each element in front of me and you can easily find such a table of atomic masses on the internet.) By multiplying
the number of carbon atoms in pentaerthritol which happens to be five with the atomic mass of carbon we get 60.055. Now if we do the same with the
hydrogen we get 12.0948 and the oxygen 63.9976 and if we add all the masses together for pentarythritol we get 136.1474 grams per mole. Now lets do
the same thing for nitric acid HNO3. Nitric acid has one hydrogen atom per molecule. The atomic mass of hydrogen is 1.0079. Nitric acid has one
nitrogen atom per molecule. The atomic mass of nitrogen is 14.0067. Finally, nitric acid has three oxygen atoms per molecule. The atomic mass of
oxygen is 15.9994.
When we multiply the atomic mass of oxygen by three for the three oxygen atoms in a nitric acid molecule and add atomic masses for the hydrogen atom
and the nitrogen atom we get 63.0128 grams per mole.
Now we already know that we need 4 moles of nitric acid for every one mole of pentaerythritol (we want the tetranitrate). So we multiply 63.0128 times
four. We get 252.0512 grams. So we would need 252.0512 grams of 100% nitric acid to convert 136.1474 grams of pentaerythritol to PETN. Now if we add
252.0512 grams of nitric acid to 136.1474 grams of pentaerythritol we can expect to have a total yeild of 388.1986 grams of products from the
reaction. An easy way to find out how many grams of PETN you can theoretically expect as a yield from the reaction is to first find out how many grams
are in 4 moles of H2O which we know will be byproduct of the reaction and subtract it from the total sum of reactants. The answer is 4 moles of water
= 72.0608 grams, we now subtract this number from the total sum of reactants and we get 316.1378 grams of PETN from 136.1474 grams of pentaerythritol.
From this we can determine how many grams of 100% nitric acid we need to convert pentaerythritol to PETN. These same methods of finding the solution
can be applied to any nitration.
Thirdly: Be practical.
You will never acheive the 100% theoretical yield, because you have several factors working against you. It's true you can achieve very high yields,
but you will need to "ALWAYS" use an excess of nitric acid in order to insure not only that enough NO3 groups are available , but that they are
"bombarding" the hydroxyl groups in such a mathimatically abundant way that the chance of a hydroxyl group not being converted to a nitro group is
highly unlikely. As for your example, BrainFever's ratio's for NC are obviously in great excess of nitric acid which forms from the H2SO4 and KNO3,
whose reactions ratios can all be determined by the above methods. The excess nitric acid helps to insure a high yield by limiting the mathimatical
probability that a hydroxyl group won't come into reaction with a nitro group.
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microcosmicus
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| Quote: |
(I have a handy table of atomic masses for each element in front of me and you can easily find such a table of atomic masses on the internet.)
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Along these lines I recommend a handy-dandy little application
"Chemical Calculator" by Jean Brefort which which you can install
on your computer desktop, available along with other chemistry
utilities such as a pop-up periodic table at the following URL:
http://gchemutils.nongnu.org/
You simply type in the chemical formula and it works out the molecular
mass as well as the mass ratios for the constituent elements. For
instance, in our case, typing "C5H12O4 " gives
Molecular weight: 136.148
C 44.11%
H 8.88 %
O 47.01 %
and typing "HNO3" gives
Molecular weight: 63.0128
H 1.60 %
N 22.23 %
O 76.17 % .
Suppose you were going to use KNO3 as your source of nitrate ions
instead of HNO3 and wanted to know how much KNO3 to use in
place of 252.05 g HNO3 to obtain the same amount of nitrate ions.
Here the percent compositions would come in handy --- the right
amount is the one which yields the same amount of nitrogen,
Typing KNO3 into the calculator, we get
Molecular weight: 101.1032
K 38.67 %
N 13.85 %
O 47.47 %
Since HNO3 is 22.23 % N, in 252.05 g of HNO3 we find 56.03 g N.
Since KNO3 is 13.85 % N, to obtain 56.03 g N, we would need
404.55 g KNO3.
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woelen
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A similar program is available on my website. It also can be used to balance chemical equations, if the reactants and products are known. If the
solution space of the equation is multi-dimensional (not a single unique solution exists), then that is reported as well.
http://woelen.homescience.net/science/chem/chemeq/index.html
Please read the tutorial, before asking questions about this software.
Edit(woelen): Made link to my webpage work again.
[Edited on 30-7-16 by woelen]
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muriaticacid
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thanks guys, you lay the whole process out really well, very easy to understand.
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StevenRS
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Hmmm.. Something I have always wondered, is there method to finding out how much KNO3+H2SO4 to use instead of HNO3+H2SO4, or do I just need to wade
through the equations?
edit: In various nitration's, like a formula or conversion factor, like for every (x)ml of nitric acid needed, I use (2x)ml of HNO3 and (3x)g of
H2SO4, plus the Sulfuric I would normally use.
[Edited on 15-4-2008 by StevenRS]
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gregxy
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One thing to keep in mind is that if you use KNO3 then more H2SO4 needs to be used since H2SO4 is consumed in generating HNO3 and in absorbing the
water. For PETN something like:
C5O4H12 + 4(KNO3) + 8(H2SO4) -> C5H8(NO3)4 + 4KHSO4 + 4(H2SO4*H2O)
136 Grams PE , 404 g nitrate, 784 g acid give 316 g PETN
Since H2SO4 is 1.84 g/ml use 426 ml of acid.
These are probably the minimum amounts of KNO3 and H2SO4
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StevenRS
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Ok, I'll just be a leech and ask it outright, because I cant seem to get it right.
How many grams of KNO<sub>3</sub> and ml of H<sub>2</sub>SO<sub>4</sub> for 1 ml of HNO<sub>3</sub> to
be produced, along with potassium bisulfate? (HSO4<sup>-</sup> is bisulfate, right?)
Edit: Ha! super and subscript!
[Edited on 15-4-2008 by StevenRS]
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Formatik
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| Quote: | Originally posted by StevenRS
Ok, I'll just be a leech and ask it outright, because I cant seem to get it right.
How many grams of KNO<sub>3</sub> and ml of H<sub>2</sub>SO<sub>4</sub> for 1 ml of HNO<sub>3</sub> to
be produced, along with potassium bisulfate? (HSO4<sup>-</sup> is bisulfate, right?)
Edit: Ha! super and subscript!
[Edited on 15-4-2008 by StevenRS] |
Like Sickman wrote you have to add the atomic masses of the elements for each compounds, this gives the molar mass. For KNO3 it is K: 39.09 + N:
14.0067 + O: 16*3 = 101.09 g/mol. That gets g/mol or in other words, that's how many grams are in a mole of that compound. Do the same for H2SO4, or
just find the molar mass (98.078 g/mol), and all of the other reactants or products.
KNO3 + H2SO4 = HNO3 + KHSO4
So, so far you know you need 101.09 g of KNO3 and 98.078 g of theoretically pure H2SO4 (density should be about 1.83 g/ccm), which gets 63.012 g
theoretical HNO3 (1 mole) and KHSO4.
98.078 g/101.09 g * amt of KNO3 to be used = amt of acid needed
For 35 g KNO3, thus 98.078 g/101.09 g * 35 g KNO3 = 34 g of acid needed. The volume would be v = m/d, 18.6 mL.
If you want 1 ml HNO3 then do the same estimates but simply switch them around mathematically (note the ratios should be about the same).
Because the density of pure HNO3 is 1.522 g/ccm. The mass of this 1 ml is m = d*v, gives 1.52 g as the mass.
Thus, to get 1.52 g HNO3 one would need 101.09 g KNO3 / 63.012 g HNO3 * 1.52 g HNO3 = 2.44 g KNO3.
And then H2SO4 98.078 g/101.09 g * 2.44 g KNO3 = 2.36 g acid. But since 100 g of 96% H2SO4 has 96 g H2SO4, 100 g/96 g*2.36 g of 96% should be: 2.46 g
(volume: 1.34 mL). Pure acid should not be used anyways because it increases the decomposition of HNO3.
Dilute HNO3 or weak HNO3 can be concentrated by boiling itself until a product of 68.18% HNO3 is obtained which then boils at 121.70 deg.C. at 760 mm
Hg (atmospheric pressure). The highest concentration obtainable from regular distillation is at the highest 96 to 99% HNO3. According to Gmelin the
distillation with nitrate gets 95 to 96% of nitric acid with a concentration of 86 to 90%. Distilling nitric acid with H2SO4 (92-96% conc.) gets
higher concentrations like 96 to 99% HNO3. The distillation of 98.7% HNO3 with double the amount its volume of H2SO4 at a pressure of 20 mm Hg gets a
completley colorless acid with a concentration of 99.70%. That's about as close to pure acid as one would normally get.
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Sickman
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| Quote: | Originally posted by StevenRS
Ok, I'll just be a leech and ask it outright, because I cant seem to get it right.
How many grams of KNO<sub>3</sub> and ml of H<sub>2</sub>SO<sub>4</sub> for 1 ml of HNO<sub>3</sub> to
be produced, along with potassium bisulfate? (HSO4<sup>-</sup> is bisulfate, right?)
Edit: Ha! super and subscript!
[Edited on 15-4-2008 by StevenRS] |
Oky-Doky! Here we go again!
This is pretty simple chemistry, so PAY ATTENTION!
There are THREE basic steps to follow to solve this problem.
1. The first step is a two part movement that can be done simultaneously as you write out the balanced equation.
- A. Determine what are the reactants and the products.
In the above problem we know KNO3 and H2SO4 are the reactants and nitric acid and potassium hydrogen sulfate are the products.
- B. Balance the equation.
Make sure you start and end with the same number of atoms on both sides of the equation. With this in mind I balance the equation thus:
1 H2SO4 + 1 KNO3 -----------> 1 HNO3 + 1 KHSO4
1 mole sulfuric acid + 1 mole potassium nitrate = 1 mole nitric acid and 1 mole potassium hydrogen sulfate.
As you look at the above equation you will see that the only atoms that moved were the two hydrogen atoms in sulfuric acid (one went to making
HNO3 the other to making KHSO4) and the one potassium atom in potassium nitrate went to making KHSO4.
Therefore our equation is now balanced. We start with one mole of each reactant and we end up with one mole of each product. Pretty simple. Now we use
our balanced equation to help us solve step two.
2. The second step to solving our problem is to solve the mass-mass problem.
What we really want to know is how many grams are there in one mole of each reactant. This will help us in planning an experiment as well as in
determining what our theoretical yield of products should be for the reaction.
To speed it up:
H2SO4 98.08
KNO3 101.10
HNO3 63.01
KHSO4 136.17
The above are molecular wieghts. For all practical purposes they can simply be taken as grams.
Hence 98.08 grams of sulfuric acid reacts with 101.10 grams of potassium nitrate to form 136.17 grams of potassium hydrogen sulfate and 63.01 grams of
nitric acid.
We have now balanced our equation and solved the mass-mass problem. We know how many moles we need of each reactant and we know how many moles of each
product we can expect. We have converted our moles into grams for each reactant and product so now we may design our experiment and also we know what
kind of yields to expect.
Now we must set aside our theoretical knowledge and be practical in our experiments.
3. The third step is to be practical. If you think that there is nothing working against you, by all means use the theoretical proportions and expect
theoretical yields, but we live in a real world where many factors are often working against us and we must keep this in mind when designing an
experiment. In this case we start with one liquid reactant and end with one liquid product. We start with one solid reactant and end with one solid
product. To begin, sulfuric acid is our solvent which facilitates this reaction and as the reaction progresses nitric acid is formed which also acts
as a solvent facilitating the reaction between a liquid and a solid. So what I'm getting at here is that you will, if there are factors working
against a perfect yield (and there is), want to use more sulfuric acid than is called for by the equation to act as a solvent during the reaction
phase. Which is sure to give a "practical world" yield which will be closer to the theoretical yield than if the theoretical proportions of reactants
had been used.
Special remarks: Potassium hydrogen sulfate and potassium bisulfate are the same chemical. Don't forget safety. Large amounts of poisonous nitrogen
oxides (fumes) can be given off during this reaction. Not only is this undesirable from a safety point of view, but this would also represent a loss
of fixed nitrogen from the reaction which will result in a lower yield of HNO3.
It's important to study, to educate yourself as much as possible before designing and performing an experiment. So that you have all the practical
considerations of safety and reaction kinetics in mind, before you begin.
That said (again) , let me now answer your question StevenRS.
Roughly, you will need 2.42 grams of anyhydrous KNO3 and 1.27 ml of 100% sulfuric acid to make 1 ml of 100 % nitric acid.
But, remember what I said about being practical.
If you are not using anhydrous reactants you should factor this in to the equation.
A word of practical advice: Things often don't turn out the way we expected that they might. This is simply experiancing the unkown factors in an
experiment and is truly the thing that makes chemistry both exciting and frustrating; both dangerous and rewarding.
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StevenRS
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Not only did I get the answer I wanted, I also learned how to get it, not as hard as I thought. Thanks for the two great, thorough responses. I
appreciate it.
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maxidastier
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Me, tooo 
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Rosco Bodine
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When working out the quantities for nitrations, you also have to look at what will be the composition of the spent reaction mixture, the dilution
caused by byproduct H2O in the residual amount of unused acids which must be present in sufficient excess so that the spent mixture is above the
threshold for achieving nitration of the desired product. As the reaction proceeds and the nitration mixture becomes spent of acids and richened in
water content, it generally becomes an oxidizing mixture rather than a nitrating mixture, which is undesirable and can even be dangerous, as you do
not want a mass of freshly nitrated product to start being "oxidized", as that oxidation could accellerate and become ugly, a cascading decomposition
kind of late stage "runaway". So you can't safely be too stingy with the nitration mixture excess. You have to maintain the reaction condition
through to the end of the nitration, and still have at least a marginally nitrating mixture leftover. I hope this makes sense that there is then
more involved than just the balanced equation for the intended reaction, you also must have a well designed "spent reaction mixture" which is
reasonably stable. It can become complicated working out the "acid economy" for a nitration. And then there has to be experimental observation and
tweaking of the quantites, times, temperatures, ect. because the actual reaction usually does not proceed exactly the same way as calculations
predict, so the equations only get you in the ball park. It can be a lot of work and a lot of experiments to get a procedure optimized. There are
graphs which show the allowable ranges for concentration combinations of acids and water which can be used to check your proposed reaction schemes to
see if the entire reaction runs to completion within the envelope shown on the graph. Attached is a nitration graph for PETN.
[Edited on 12-8-2010 by Rosco Bodine]
Attachment: PETN nitration graph.pdf (286kB)
This file has been downloaded 771 times
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