Can anyone help with A=2-log %T ??

11/13/2010 | Me

[paul544]

Was wondering if there are any chemistry whizzes here who can help me with a problem in a lab I am working on?

We were given a list of %T values and told to use A=2-log %T to get the absorbency (A). I'm struggling with this. Can someone take me step by step through one? Say 25%?

[Vendome]

Did it 30 years ago at my first job Taco Bell. But the toilet cleaned itself while I was outside gasping for air. LOL

[paul544]

Yep. I’ve been around long enough to know there is an expected amount of crap one can expect on these types of threads.. But I also know that FR has very helpful knowledgeable folks.

[cripplecreek]

I saw a word I recognized and bounty was the first absorbant thing I could think of.

[Vendome]

Wish I could help.

If you ever have a computer related question though, I’m good for em.

Good luck.

[allmost]

What are you absorbing?

[CharlesWayneCT]

First: Beers law: Beers Law

The web site shows how you get from a standard transmittal equation T=(P/Po) to the equation you are using.

A = log10 P0 / P
A = log10 1 / T 
A = log10 100 / %T
A = 2 - log10 %T 

As to your question, it seems you should just plug in the percent given, and you'll get an "A" value.

For example, for 25% transmission, the answer is:

A= 2 - log10 25
A= 2 - 1.39794
A= 0.60206

You use the percent number directly (that's the point of changing the 1/T to 100/%T -- 1/.25 is the same as 100/25).

If you had a different question, please ask again.

[CharlesWayneCT]

Yes, you leave it as a whole number (see my prior post)

[devere]

The next step would be to apply the log function, but I suspect the %T should be 25 not .25

2 is log(100) so for 100% A = 2- log(100) =0

Perhaps A is supposed to be a number from 0 to 2.

Look in your book for examples of the definition and values of A. You’ll probably figure it out quickly.

[paul544]

Thanks very much...

[paul544]

Thanks as well.

[Scutter]

I don’t know the chemistry, but I think I can answer your question.

I did a search on this, and the formula for absorbency seems to be -log(T). However, a percentage is often expressed as 100 times the fractional amount. Log base 10 of 100 is 2. And log(A x B) = log(A) + log(B).

So 2 - log(T) = -log(1/100) - log(T) = - log(T/100)

So if you express T as a percentage from 0 to 100, you use the formula (2 - LOG(T)). If instead you express T as a fraction, from 0 to 1.0, you would use the formula (-LOG(T)).

[CharlesWayneCT]

Note that the “trick” is that Log10(A/B) = Log10(A)-Log10(B)

[JRandomFreeper]

I’m not even smart enough to understand the question.

That's where we all start. If we stay there, that's up to us.

Me? I'm just a cook. Want bearnaise sauce on your poached egg on muffin with Canadian bacon?

I can do that.

/johnny

[martin_fierro]

Also check out Wolfram Alpha

[bigbob]

Nicely done!

[Tolerance Sucks Rocks]

Well, if your absorbancy values appear to be way off, compared to any examples offered, then you might indeed need to use the whole number for % (e.g. 25 = 25%). I’m surprised that your lab text apparently did not make that clear.

[paul544]

Thanks. The light bulb finally went on... I have it now.

[JRandomFreeper]

No-one talks about moving the decimal point two places to the right anymore...

/johnny

[GunsAndBibles]

Just thinkin’ issues like this probably don’t come up to often over at DU.

[paul544]

I was being thrown by the basic layout of the formula. As simple as it is I was just looking at it wrong. Thx