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nightwalker
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smile.gif posted on 20-10-2008 at 01:40
understanding the math


Many times I see a process that when carried out such as in distillation that results in recv'ng flask containing a known quantity of liquid. For this example say 125ml. In part 2 of that process, it will call for say 35g of the that same chemical extracted in part one. Is it necessary to dry said chemical, or is their a formula to figure out how many grams you would have in your original extracted liquid. I guess though it would be the same idea, but different view. Quite commonly I have seen someone use the term 30 grams from a certain percent HCL. I would think at first glance, if you knew that HCL liquid is 37 percent pure, that would equal 37g per 100 ml? Then go do the ratios to achieve 30g since it would not be advisably to evap HCL to dryness?
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[*] posted on 20-10-2008 at 06:44


What?



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[*] posted on 20-10-2008 at 08:15


Could you rephrase the question/comment in a way people understand?
It would make it easier for us to help you....
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[*] posted on 20-10-2008 at 08:29


It is necessary to make your chemical as pure as you can get it. Then use the amount they specify in the paper/book/sloppy napkin/whatever source you're using, unless they specify how to purify the product themselves, in that case do what they did.



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jarynth
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[*] posted on 20-10-2008 at 08:35


Is the answer 'Hungary'?

EDIT
Synthesis writeups are often very synthetic indeed. They like to give the least necessary information, no more.

Suppose you're meant to add an aqueous HCl solution. They might tell you to use a certain concentration (or molarity); the quantity (indicated by weight or volume) will refer to the solution; you don't need to figure out the actual amount of HCl inside the solution you're pouring.

In other cases, they may not care how concentrated your HCl is, so they'll just tell you how many moles you need and let you work it out.

The most obfuscating practice is to mention how many 'parts' of each reagent you need. This usually refers to volume or weight, and throughout the synthesis you must stick to either one.

All in all, you should be able to infer which situation applies. If this is not the case, then your recipe is unreliable.

Quote:
Originally posted by nightwalker
Many times I see a process that when carried out such as in distillation that results in recv'ng flask containing a known quantity of liquid. For this example say 125ml. In part 2 of that process, it will call for say 35g of the that same chemical extracted in part one. Is it necessary to dry said chemical, or is their a formula to figure out how many grams you would have in your original extracted liquid.


If they say 35g of the (solid) chemical (which lies in solution as it has just been extracted), then you'll have to either figure out the concentration of your solution (e.g. titration, pH measurement, etc) or evaporate it and measure it as dry as possible.

Quote:
I guess though it would be the same idea, but different view.


Exactly, so go the route that suits you best.

Quote:
Quite commonly I have seen someone use the term 30 grams from a certain percent HCL. I would think at first glance, if you knew that HCL liquid is 37 percent pure, that would equal 37g per 100 ml? Then go do the ratios to achieve 30g since it would not be advisably to evap HCL to dryness?


Not very advisable, I agree. In this case you have no choice.

There are various ways to measure concentration. If a solid is said to be 37% pure, that's 37g of pure chemical out of 100g (not 100ml). Otherwise you'd easily get over 100%, as most solids are denser than water (density is weight/volume). What about liquids? Concentrated sulfuric acid is over 90%, i.e. 90g H2SO4 out of 100g of liquid, NOT 90g out of 100ml (since a 100ml sample weighs over 180g...if this were the case then there would be other 90g of water in the sample!). Now what about HCl?

If this is really what your question boils down to, then I'd also like an answer.

[Edited on 20-10-2008 by jarynth]
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nightwalker
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[*] posted on 20-10-2008 at 11:48


Thanks... Your reply helps a lot. After the first few replies I went back and read my original post. I have to admit, it looks as if my frustration had come over into how I worded my question. I thought it could be answered just as general question without having to get too detailed, but I can see that by not having more specific details, confusion evolves, and the much needed help can be hindered, so to expand further here it goes.

I came across a synth. for c2h5no2. This is already something of a tricky procedure (according to the talk on the board) to begin with. The synth. is straight forward and doesnt seem to be out of my range. Problem is that (as already mentioned) after carrying out the various actions and collecting product in receiving flask, the synth. ends with "and store the c2h5no2 liquid in amber jar until it needed" (no instruction on how to dry, recrystallize, etc...). Step 2 calls for 30g of said reagent. What! Now how am I to figure that one out. I am sure they were not asking me to weigh out 30g of solution and hope for the best :)

The reason I followed up with the hcl question was simple really. Another liquid question. I recently came across a thread where the synth. called for adding say 40g of said acid to a sol. over a set period of time. Agh. Yes, it would be simpler if the author had said "well if you acquired that special variety otc in most shops at 31 percent sol. then use 250ml". But not only is that not done but probably very unscientific. Again, instead of asking the obvious of how does that translate to what I am using, I am trying to figure out the math. Is there a basic mathematical or algebraic formula that I am missing to convert liquid into dry weight, ie grams. It just can't be a guessing game.

Someone told me years ago that when possible always use dry reagents. But I am finding out with my experimentation (especially in org. chem), that is not always the case. Someone will always throw out there something that messes me up, like the 20g of a liquid. I am somewhat new to a lot of this. Not looking for a free pass, just wanting to learn. I love figuring out the "how's" and if all goes well the "yeah, I did that". But am no way near university trained. So sometimes even the simpler stuff gets me stuck until someone is cool enough to explain. Thanks, and sorry for the lengthy explanation. New kid on the block and didn't want the bad first impression of what an idiot.

[Edited on 20-10-2008 by nightwalker]

[Edited on 20-10-2008 by nightwalker]

[Edited on 20-10-2008 by nightwalker]
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chemrox
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[*] posted on 20-10-2008 at 13:26


"The most obfuscating practice is to mention how many 'parts' of each reagent you need. This usually refers to volume or weight, and throughout the synthesis you must stick to either one."

And sometimes they mean equivalents- I hate it when they use "parts." Patents are the worst.




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[*] posted on 20-10-2008 at 17:39


If the balanced equation can be inferred you can estimate what you need by
figuring the molar weights. The stipulated percent yield is a good indicator of
side reactions and wether or not this will be stoichiometric. Some proceedures
require a greater excess of one reagent or other, as much as ten times what
is theoretically consumed, in such case the concentrations matter little, just
add a lot of it.

.
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[*] posted on 20-10-2008 at 19:06


The most important piece of equipment a chemist needs is a good electric gram scale or triple beam. If liquids are asked to be used and is refered to in grams you will need to know its concentration, usually the liquid needed for a synth will be mentioned by its molarity. In a well written synth report they will have notes on the bottom and tell you where they got their chemicals so that the same reagents and procedures can be followed so they can have the same results, and quirk around the wth synth if one permits to.

If the synth does not tell you, you will have to do a balanced equation and find out how many moles of that substance you need and convert to grams. Then follow the same procedure but aquire or make the same reagents (in molarity, mass percent, etc...) as in the lab report.

They probably think you have the means to purify your chemicals since it is such a common practice in organic chemistry. As in C2H5NO2 a simple distillation would work, then weight out 30 grams of that for the next step. A general rule in organic chemistry is purify and dry your reagents so less by-products will be formed.

One more thing is do not distill nitroethane to dryness it may explode. It is always good to research all your chemicals before playing with them, I have wounds of my own for being lazy and to cocky to followup with procedures, as even something as mediocre as distillation. There's a good ebook on purifying lab reagents somewhere around here.




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nightwalker
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[*] posted on 22-10-2008 at 19:45


Sorry it's been a couple of days. Had to earn a living type excuse has kept me gone.

Thanks for the input. It has made things a little bit clearer. I was reviewing this synth again and it seems that it has all the bad elements that some of you guys talked about. It starts out in parts, which though it can be a pain still when it tells, 113 parts of a dry reagent to 20 parts of a liquid, that seems easy enough I will start with 113 g of one to 20 g of the other. And if successful, scale up or down accordingly in the future.

Problem is the chemical in question again... is 95 percent wood grain alky. Yeah, not to tripped out with the actual name of chem, since I think of it is as simple as etoh? But the synth calls for 20g of etoh? Hence the whole sol. problem again. The synth. already has done the easy part of telling me in ratio form how much etoh I need. But I am still stuck on how to weigh out 20 g of etoh. Grams have always meant dry weight to me. It seems this type of wording is going to be common problem in trying to figure things out.
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[*] posted on 24-10-2008 at 04:42


Eureka I think I got it.... Well, maybe :(

After much reading, I followed examples set forth by many people concerning other chemicals. After following said examples backwards (don't ask me why, just happened that way) to the point of origin, I decided to try it out for myself with my own wood grain alcohol question. Please let me know if I nailed it or did I screw it up.

Trying not to take short cuts, since I really am wanting to learn mols. I started with the obvious "what is wood grain alcohol". For me that was the easy part wood grain alcohol = etoh = ethanol = C2H5OH (excuse the shouting with capitals but decided to use in order to show my way of thinking, and to keep things accurate).

According to your input the next thing to do is isolate each element and figure out the moles of each:

C=12.011
H=1.0079
O=15.999

with this in hand its a matter of addition (and some mult. if an element contains 2 atoms)

(12.011 x 2) + (1.0079 x 5) + 15.999 + 1.0079 = 46.0684g per mol. Right so far?

However before I went any further I decided that since my sol. is only 95 percent pure, I needed to adjust the numbers to reflect how many grams at 95 percent I would need to equal 1 mol of etoh. I figured that out as follows.

46.0684g (per mol if it was a 100 percent pure sol.) / .95 (since that is the concentration I am working with) or: 46.0684/.95 = 48.493052...... In my head the slightly higher result of this equation is justified that in order to get the equivalent of 1 mol out of a 95 percent solution it is logical you will need more weight than if it it was figured out based on a 100 percent pure reagent.

So now I have a problem that reads something like: I will need 48.493052....g's = 1 mol of etoh (if using a 95 percent sol). Could I still be right? Hope so.

However this is only the first part of what I was trying to acheive, which is converting my solution to grams. I thought it would get a bit dicey from here, but it actually seems like it was a breeze (maybe too easy) once figure out.

To figure out how many grams are in a body of solution I needed to know both the mol. and density of liquid. That is all good except I knew from the start that adjustments needed to be made since I was dealing with a 95 percent sol. and not 100 percent. That is why I took those steps earlier to make it a sort of "level playing field", something I remember from alg. (I think)

To find out what I had from the mls of 95 percent, I need to follow the following equation: mol weight/density. So:

For this example 48.493052g (mol weight) / .789 gcm3 (density) = 61.4614093.... After a little investigating into the term cm3, I found that this term can simple be thought of as 1cm3 = 1ml. Yes?

Meaning after this lengthy explanation, finally in the end with rounded off numbers and the such. 61.46 mls (of 95 percent etoh) should contain just under 49g. After this much is known. It should be a simple proportion equation to get the exact amount of liquid needed for the known quantity of grams desired: 49g/61.46ml to 20g/x = solving for "x", I ended up with 25.085ml = 20g. Or if I was being graded by my crappy teacher on a word problem, I would have to say: It takes roughly 25.085 ml of 95 percent etoh to equal 20 g.

I am not sure if I did this right? Please any input is much valued. The reason for laying it out like this is if I missed the correct answer by doing one step wrong, I with anyone else that has problems with this can figure it out from there. Thanks



[Edited on 24-10-2008 by nightwalker]
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HexJam
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[*] posted on 27-10-2008 at 07:56


Correction time: "understanding the math" should be: "understanding the MATHS" as what you originally said when expanded would be "understanding the mathematic" which doesn’t make sense at all.

Sorry to sound totally pedantic but this is one of my (but I'm not the only one!) pet peeves and it makes people sound really dumb when they say it. I’m not saying that you are dumb, just that it makes you sound dumb when you say that, no offence intended.

Spread the word! The word is "maths"! ;-)
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[*] posted on 27-10-2008 at 08:15


HexJam, do you live in the UK?

I'm guessing that this is merely a colloquial difference in terminology between UK speak and USA speak. No offence if I am wrong. A US resident never says "maths."

I see many differences in terminology and spelling between UK English and US English. One of my favorites is aluminium vs aluminum. Another is sulphate vs sulfate. Or holiday vs vacation. The list is long. ;)
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[*] posted on 27-10-2008 at 09:44


Quote:
Originally posted by Magpie

I see many differences in terminology and spelling between UK English and US English. One of my favorites is aluminium vs aluminum. Another is sulphate vs sulfate. Or holiday vs vacation. The list is long. ;)


In India we mostly follow English English. Even though American English is used by many (especially younger generation) in day-to-day work, our school text books, reputed print media etc are in proper English. So when I see words with dropped "u" such as neighbor / humor etc, it just doesn't feel right.

Here is a webpage compiled by one young software techie after his firsthand encounter with American English:

http://www.onesmartclick.com/usa/american-english.html

:) gsd
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[*] posted on 2-11-2008 at 00:58


I am troubled by your using the term "Wood Grain Alcohol".

Wood Alcohol is Methanol. Grain Alcohol is Ethanol. Which is it?

Since you seem a little raw. I suggest you use 100% grain alcohol. Make it yourself.

Distill it from whatever, and dry it. It will be a good warm up for your future experiments. And, you will be starting your Nitroethane synthesis with the "Right" reagent. There are a number of threads on producing ~!00% Grain Alcohol.


Next, give us exact experimental details. Making Nitroethane without a permit, figures to be a very marginally legal project in the UK. But, it's no big deal to most of us. We just assume you have any required permits, and you need some technical assistance.

Of greatest concern, is your completing your experiment safely. If you provide more experimental details, someone here will recognize the synthesis you are attempting, and alert you to possible hazards, or possibly guide you to a better procedure.
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nightwalker
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[*] posted on 2-11-2008 at 08:24


Thanks, and apologies. I can actually see the confusion in using the term Wood Grain Alcohol. You are 100 percent right, Grain Alchohol is the actual chemical that would be used for this synth.

Yes, I do believe that warming up with a distillation of a chosen alky over to etoh could be a lot of fun, and give a little bit of experience (though purity might be a problem). This also may serve 2 fold since the synth that I am looking at is the one that was posted by Desseigne & Giral, with a introduction to otc sodium ethyl sulfate by evil lurker. I believe that due to the nature of nitroethane (and what some are saying the volatility of the by-products produced), it may be in my best interest to do this under vacuum. I like the idea of keeping the temperature needed for proper distillation, as low as possible. I am still trying to work out the kinks, such as figuring out a rate of change in boiling point over the use of an applied vacuum. Also using a vacuum is very new to me. Thought about trying it out on something as simple as h2o, maybe the etoh distillation, and see how that goes first.

Am a firm believer of seeing it all on paper, and then in my head first before going any further. That is why being as new as I am, presented the grams to liquid problem. I wanted to figure that out before anything else since it seemed like a simple thing that I should know. I have heard a few good things about this nitroethane synth. The best point is the possibility of doing it from scratch. Also not absolutely terrible yields, which seems to go hand in hand with most of the synths out there for c2h5no2. The setup seems pretty straightforward. Basic distillation set-up with use of a vigreaux column and vacuum.

I have seen the synth posted quite a few times, and much of the threads following has seemed to embrace it and not tore it down as worthless. So that alone intrigued me to give it a second look. Also, I have seen a few other posted synths. Most of which either have low yields, hard to come by precursors, complicated procedures, prone to scorching, or worse a chemical time bomb.

Not sure if this is the type of informatin you were asking for. Any help is welcome help. Still a very long way (if at all) from putting the information to good use. But I thought that figuring out the basic stuff along the way was worth dealing with since I had the time. Not sure how often this synth has found its way into the light, or how difficult it really is compared to the literature. I am pretty aware of some of the considerations that one should take into account before attempting such a synth. I dont' believe in cutting corners, that is why I have a million questions.


[Edited on 2-11-2008 by nightwalker]
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