[Help]

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.

[Arkcon]

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.

[JGK]

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.

[Help]

Oh no. I got 1.3335 x 10^-8 ppb Compound A. That's much smaller then yours.

[Help]

I've already handed it in, I wouldn't mind seeing how you arrived at the answer.

[JGK]

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

[JGK]

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 

[Arkcon]

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.

[DrCMS]

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.

[DrCMS]

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

[JGK]

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)

[DrCMS]

JGK please do not remove your posts later as it makes the replies people then give look odd.