# Homework Help: Absorbance and Concentration Question

1. Sep 21, 2007

### fuzzy

1. The problem statement, all variables and given/known data
The protein oxyhemoglobin has the following molar absorbance coefficients:
$$\epsilon$$276 nm = 3.44 x10$$^{4}$$M$$^{-1}$$cm$$^{-1}$$
$$\epsilon$$415 nm = 1.25 x 10$$^{5}$$M$$^{-1}$$cm$$^{-1}$$
If a solutioin of oxyhemoglobin has an absorbance at 415nm of 0.234,
a) What is the concentration at 415nm?
b) What is the absorbance at 276 nm?

2. Relevant equations
A=$$\epsilon$$bc, where A is absorbance, $$\epsilon$$ is the molar absorbance coefficient in M$$^{-1}$$cm$$^{-1}$$, b is the length of light path through the cell in cm, and c is concentration in mol/L.

(I don't know if this one is needed...)
$$\frac{As}{Au}$$=$$\frac{Cs}{Cu}$$
where As and Au are absorbance of the standard and unknown and Cs and Cu are concentration of the standard and unknown.

3. The attempt at a solution
Can someone give me a hint at where to start here? I can't see how to do the problem without at least 1 of the concentration values or a light path length. Do I assume that the concentrations remain the same at 276 and 415?

2. Sep 21, 2007

### chemisttree

Assume that the path length, b, is the same for both samples and is 1 cm. You are given the absorbance, the extinction coefficient and you are asked to solve for 'c'. Seems pretty straightforward.

The second part asks you to substitute the concentration and molar extinction coefficient at a different wavelength and determine the absorbance at that wavelength... again, fairly straightforward.

Will the concentration change when you shine a different color light on it?

3. Sep 21, 2007

### symbolipoint

Check your textbook about how to handle quantitative absorbance measurements. You probably need A=e*C, where A is absorbance, e is your epsilon, C is concentration; but I could be mistaken. Each wavelength has it own A=e*C for the oxyhemoglobin.

... Yes, you DO assume the concentration remains the same at both wavelengths.

For part (a), subsituting the values given for the 415 nm,
0.234 = 1.25*10^5 * C,
Find C.

As I say, this stuff is a bit murky now, so recheck your textbook.

4. Sep 21, 2007

### fuzzy

Thank you for the replies.

So this is always the case if it's not otherwise stated? Because in other problems it was given as 0.5cm, so I was confused a bit.

Other than that I think I get it now :D
So I get
a)
c=$$\frac{A}{\epsilon}$$

c=$$\frac{0.234}{1.25x10^{5}}$$

c=1.87x10$$^{-6}$$M
This isn't too low for a concentration?

b)
A=$$\epsilon$$c
A=(3.44x10$$^{4}$$)(1.87x10$$^{-6}$$)
A=0.064

5. Sep 24, 2007

### chemisttree

No, it is not always the case. If the path length information is absent (an oversight in my opinion), you can assume anything. Assuming 1 cm. (a standard photometric cell dimension) must be part of your answer.

It is fairly low but you can express it as 1.87 ppm as well. ppm levels are determined routinely.

Looks good.