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Topic: Beer-lambert law!  (Read 17199 times)

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Offline frustrated

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Beer-lambert law!
« on: August 17, 2009, 10:09:07 AM »
I need to answer this question but I have no idea where to start.

A 1ml stock solution of the amino acid tyrosin was diluted with 9ml of water. A portion of this solution was placed in a quartz cuvette with a light-path of 1cm. The absorbance of the solution was measured using a spectrophotometer and gave an absorbance of .45 at 294.5 nm. Given that the molar absorbance coefficient of tyrosine at 294.5 nm is 2375 M-1 .cm-1, calculate the concentration of the original, undiluted solution of tyrosine.

Thank you for any help you can give, I'm a biology student, maths scares me!

Offline Dan

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Re: Beer-lambert law!
« Reply #1 on: August 17, 2009, 10:36:01 AM »
Start by looking up the Beer-Lambert law and find out what the different terms mean.

1. What the Beer-Lambert law?
2. What do the different terms mean?

Now rearrange the equation to find an expression for concentration in terms of molar absorption coefficient, path length and transmittance. If you have difficulty, read about logarithms.

3. c = ? <--- work this out
4. Plug in your numbers for molar absorption coefficient, path length and transmittance to find your answer.

Have a go at these steps and see how you get on, post your work and we can help you if there are any problems.
My research: Google Scholar and Researchgate

Offline frustrated

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Re: Beer-lambert law!
« Reply #2 on: August 17, 2009, 10:45:06 AM »
I know both 1 and 2, I've read up about logarithms, but I still don't understand. I have never done any chemistry in my life and no maths since I was 14, genuinely retarded...

Thank you for your help


Offline DrCMS

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Re: Beer-lambert law!
« Reply #3 on: August 17, 2009, 11:15:25 AM »
As Dan said just rearrange the Beer-Lambert equation to calculate concentration c.  (Don't worry about logs you don't need them for this.)

Now relate the concentration you've calculated back to the concentration of the original solution prior to the dilution.

Offline frustrated

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Re: Beer-lambert law!
« Reply #4 on: August 17, 2009, 11:21:55 AM »
I don't know how the equation should be arranged or whether or not they should remain as multiply.

Offline Dan

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Re: Beer-lambert law!
« Reply #5 on: August 17, 2009, 11:28:34 AM »
Go back and read up on your basic algebra. Rearranging (transposing) equations is a fundamental skill that is absolutely required by all scientists.

Here is a good pdf (read the entire document): http://www.andy-roberts.net/teaching/ma11/transposing.pdf

Now show us an attempt at rearranging the Beer-Lambert law and we can see where you're going wrong.

You should be putting in at least as much effort as us here - show some attempts and working. You won't be spoon-fed the answer, but you will be guided to it provided you want to be.
My research: Google Scholar and Researchgate

Offline frustrated

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Re: Beer-lambert law!
« Reply #6 on: August 17, 2009, 11:49:29 AM »
I am putting effort in, I have been trying to do this since 11 am, its 16:39 now.

So, would the used concentration (the whole 10ml) be worked out by

Concentration = absorbance x molar absorptivity x light path

That would make it 1.068 for the 10ml solution.

Offline frustrated

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Re: Beer-lambert law!
« Reply #7 on: August 17, 2009, 11:57:52 AM »
wait no it should be

concentration = absorbance/ (molar absorptivity x light path)

should it?

Offline Johnny010

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Re: Beer-lambert law!
« Reply #8 on: August 17, 2009, 12:15:11 PM »
Yes, c=A/el

Offline Dan

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Re: Beer-lambert law!
« Reply #9 on: August 17, 2009, 12:58:31 PM »
concentration = absorbance/ (molar absorptivity x light path)

Good, that's the hardest part done. Now:

1. Plug in your numbers and calculate the concentration of the diluted sample
2. How does the concentration of the diluted sample compare to the concentration of the original sample?
3. So what is the concentration of the original sample?
My research: Google Scholar and Researchgate

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