Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: Help on April 12, 2008, 11:17:33 AM
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I just can't seem to wrap my head around this for some reason.
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In order to solve this problem, you have to build a series of equations, made up of conversion factors, that are detailed in the various steps of the problem.
Useful hint: Don't scan-n-post a complex question like this. Attempt to write it out, on scrap paper, and see what bit of info is needed, and when. See if you can type out a question on this forum, that's built up of important parts of this question. We'll let you know if you're getting close. Or you can ask us for one specific problem that you're having with the question.
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Well, where to start would be helpful. I just get overwhelmed with the information given, and I'm unsure of what values to take from it. I don't know what I need and what I don't need.
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I know you're overwhelmed, but you've let the complexity of the problem dazzle you, and you've just shut yourself down. If I was at work, and someone said, "Here you go, do it now," I'd be a little ticked off too, this is the sort of thing you have to work at, over time, to solve. Look at it this way:
"Ground grain is extracted into acetonitrile + water by shaking in a centrifuge bottle on a mechanical shaker"
"A crushed ore sample is digested with hot nitric acid by vortexing in a quartz extraction tube in a Vortexicon-2000"
Those two statements above, mean, essentially nothing, to a problem like this. They are exactly the same, yet totally different. Many statements in your scan add no useful information. It is your job to start this problem, by writing on paper or typing in a clear question.
Listen, I have a bias, and I'm not shy about expressing it -- I don't care much for people who scan their textbook and say, where do I begin. You haven't done any work, at all.
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I understand that. And I'm not asking you to solve it for me. I'm simply asking for some direction.
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What is the question, that the problem asks. How much raw material do you start with. How much analyate do you find. How much is the raw material diluted/concentrated at each step. You've let it confuse you, but it is a lot more simple than you give the problem credit for. Can you write anything along those lines for us?
Look at my made-up example above. Digest the text into simple dilutions and concentration steps. The question starts and ends with numbers -- turn the descriptive text into numbers.
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Well, it wants the concentration of compound a in ug/L.
I'm starting with 100 g of raw material, or, 1.0 x 10 ^8 ug.
I find 0.02 ng from 2 uL of sample injected.
0.2 x 10^-7 g from from 0.002 ml of sample
0.00001 g/ml compound A
Diluted: 100g diluted in 600 ml= 0.167 g/ml
Then 10 mls of that is taken and diluted into 40 mls of buffer. Now, is this where I do a C1V1=C2V2 equation, or, do I not care about that?
If so, then, (0.167)(10)= C2 (40)
C2= 0.04175 g/ml extract
I don't quite know where to go from here. I don't even know if I'm doing it right.
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Sorry, scratch that, the C1V1 equation should have had 50 as a final volume.
giving me 0.0334 g/ml
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You can do it that way, but for me, C1V1=C2V2 is best reserved for simple, one step dilutions, but go ahead if it works for you.
I like to think of a problem like this one in terms of :
100g/600 ml*10ml/50 ml*15ml ...
Do you see how I got those numbers? Try it out, on some scrap paper.
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Yes, although I have to admit, I didn't think the 15 ml part was needed.
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Yes, although I have to admit, I didn't think the 15 ml part was needed.
Well, it's hardly complete, there are many more steps for you to do.
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This is one of those problems that adds more information than necessary for the actual solution, but I still think it's important to understand the procedure by translating what they said in technical jargon into practical sense.
It's a good exercise. For me, before I begin a problem like this, I don't pay any attention to numbers. Rather, I try to translate the basic procedure they're doing. "Add this"; "dilute that"...etc.
Then I refer back to the numbers given and insert those into the necessary calculations given the question. Translating the procedure can really help you dispose of excess information when determining the actual solution.
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Exactly, but the O.P. should have noticed, unless the class is actually recently talking about multiple extractions and HPLC sample prep and fluorescence quantification, that the whole question had to all be a trick, and it was to chip away at the monster question, a little at a time, until it was broken down into the simplest elements -- something similar to something the class had done recently.
That's not to say the question isn't pretty cool, taken as a whole. Such procedures are done, in quantitative analysis. Someone on these boards will often as for an extraction or analysis procedure, so this is a great example of the bridge between sample prep and real numbers that answer a question. Sometimes in industry we're called on to work out the math and program the HPLC's integrator, so that a meaningful value related back to the original sample and outputted.
But I do stand by my point of view, that the ease of the scanning and posting of text book questions, has ruined the current batch of chemistry students.
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If you must know, this is for a written online test that I have to do for a job, before I can even get an interview. It's for the Canadian Grain Commision as a lab tech. I've answered their other technical questions, such as what would you do if the retention time of the HPLC was decreasing, and how to make solutions and whatnot. But then I cam across this and I'm panicking because I'm not sure how to work it all out, and that really makes me feel like maybe I shouldn't work there then. I've always needed a bit of direction when it comes complex math questions, and apparently, that hasn't changed. It's very discouraging. And I just can't work it out.
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Well, don't let it bring you down, quite a few people would be stymied by this question, similar questions are quite common on this board.. I told you, if they plopped this on my desk at work, I'd be overwhelmed too -- for a little while. Usually, once the calculation has been worked out, it's been worked into the integration program of the HPLC software, and no one has to look at it again. It's nice, from the employer's point of view, to see that you're versatile enough to handle the problem, in case something goes wrong.
You'll notice, that I didn't work out the whole thing for you -- if I were to, and double and triple check it to be sure it worked right -- I'd expect to be paid, and of course, so do you, eventually. Give them this freebie, and let them know, you'll be looking forward to more problems like this, so you get the job. Once you've done that, you can brush up on these skills, and be in high demand for jobs with these sorts of tasks.
And if you get it wrong -- own the error, make it yours, make it work for you. Admit you're not perfect, show them you want to learn, and let them know they can't spook you with a big block of text.
Oh, and uh, when you get a chance, can you point me to the online test? Don't worry, I'm not a Canadian citizen, so I'm not gunning for your job, I just want to take the test myself, and see if I'm good enough. It could come in handy with my own job search.
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Haha. I could send it to you if you'd like. They sent it to my e-mail, with a form to fill out answers for.
Thankyou for your advice. I bothered one of my old professers for some direction. I managed to figure it out by working backwards, which is good, because it was due at 11:00 am this morning.
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Sure, drop me a PM on this forum, and I'll send you my email address. I went and browsed their web site, and didn't see an obvious form for submitting yourself for a job, so I couldn't find it on my own. I wouldn't worry too much about this question, they state, right at the top, that this question won't make or break your application. IMHO, working something like this out this isn't a technician task, but more an associate chemist or level 1 chemist, if not a more senior level chemist.
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I have arrived at some answers: 10,000 ppb and 13,333.3 ppb (adjusted for efficiency)
I can post the work-through if needed.
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Oh no. I got 1.3335 x 10^-8 ppb Compound A. That's much smaller then yours.
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I've already handed it in, I wouldn't mind seeing how you arrived at the answer.
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I have arrived at some answers: 10,000 ppb and 13,333.3 ppb (adjusted for efficiency)
I can post the work-through if needed.
Here it is:
No correction for extraction efficiency
2 µL injection of extract (ACN:0.5% HAc) yields 0.02 ng = Concentration 10 µg/mL
Extract Volume = 0.5mL which was obtained from 15 mL extract added to affinity column. Extract Concentration = 0.333 µg/mL
The 15 mL was a sub-sample of a 50 mL volume (Amount in 50 mL volume 16.6667 µg) which in turn was made from a 10 mL portion of a 600 mL extract (total in extract 1000 µg, equivalent to 10 µg/g or 10,000 ppb)
If recovery is 75%
Then original concentration is 13,333.3 ppb
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Oh no. I got 1.3335 x 10^-8 ppb Compound A. That's much smaller then yours.
1 ppb = 10 ng/g
Must be one hell of a LC-fluorescence method if it's capable of measuring in the Femtogram range
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That's the great thing about fluorescence, either with HPLC or a cuvette. I've used it from time to time my whole career, and doing a serial dilution, trying to find the limit of detection, is always astounding to people on the sensitivity. Of course, you have to find a excitation and emission maximum, but there usually is one, for large molecules, at least in the UV. Yet, may people, when confronted by a new substance that they can't detect with an absorbence detector, just won't buy a fluorescence detector, preferring an RI or a ESD, despite their shortcomings.
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I get 10ppb
2 µL injection of extract (ACN:0.5% HAc) yields 0.02 ng = Concentration 10 µg/mL
I think 0.02ng in 2µL is 10µg/L not per ml.
0.02ng = 2x10-11g
2µL = 2x10-6L
So concentration is 1x10-5g/L = 10µg/L or 10ng/ml
Besides an answer of 10000ppb is best written 10ppm.
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No you've got it wrong again!
Go back to maths class before answering these questions again.
1 ng = 1x10-9g i.e. 1 billionth of a gram
so 1ng/g = 1ppb
10ng/g = 10ppb
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No correction for extraction efficiency
2 µL injection of extract (ACN:0.5% HAc) yields 0.02 ng = Concentration 10 ng/mL
Extract Volume = 0.5mL which was obtained from 15 mL extract added to affinity column. Extract Concentration = 0.333 ng/mL
The 15 mL was a sub-sample of a 50 mL volume (Amount in 50 mL volume 16.6667 ng) which in turn was made from a 10 mL portion of a 600 mL extract (total in extract 1000 ng, equivalent to 10 ng/g or 10 ppb)
If recovery is 75%
Then original concentration is 13.33 ppb
Sorry for the earlier errors, my confusion with ppb (US definition)
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JGK please do not remove your posts later as it makes the replies people then give look odd.