Spectrophotometry Question

[walker8476]

If you get a zero absorbance for a sample, is the concentration therefore, zero?

Because if you create a calibration curve and solve the equation for the line, you don't get a zero concentration.

[Johnny Rod]

Depends if you have subtracted your solvent blank and you are in the linear range. Are you talking about UV-vis?

[rb6banjo]

It doesn't really matter. All you can ever say about zero response (or response that's in the noise of the detector) is that it's undetected or less than the limit of detection (LOD).

[tom jupille]

Because if you create a calibration curve and solve the equation for the line, you don't get a zero concentration.

The y-intercept will almost never be *exactly* equal to zero. You need to look at the standard error as well. If the standard error interval does not include zero, then your calibration curve has issues (e.g., a contaminated blank, or outside the linear range, or significant heteroscedasticity in your data).

[walker8476]

OK thanks. I have another query about the equation for a calibration curve. I have run some standards through a UV/VIS spectrophotometer. I have a calibration curve with concentration in mg/L.

I get this equation from the calibration curve:

y = 0.0.7755x - 0.0780

So to calculate the concentration of my sample I have substituted in the absorbance of the standard into y which is 0.0225 and solved for x.

0.0325 = 0.7755x - 0.0780

0.0325 + 0.0780 / 0.7755 = x

I get a concentration of 0.142mg/L


I would also like to calculate concentration by using the formula: c = A/el

As I understand it, e (molar extinction coefficient) would be 0.7755 from the same equation:

0.0325 = 0.7755x - 0.0780

c = 0.0325/0.7755 x 1 = 0.045

I get a concentration of 0.045mg/L

Why do I get a different concentration value? There's something I'm missing.

[tom jupille]

As I understand it, e (molar extinction coefficient) would be 0.7755 from the same equation:

Several flaws in that statement:

1. That would be true only if the y-intercept were zero.
2. What you're looking at is not *molar* extinction coefficient, but *mass* extinction coefficient (you need to know the molecular weight to convert one to the other). This is a nitpick, because the difference in your value comes from the y-intercept.

[walker8476]

As I understand it, e (molar extinction coefficient) would be 0.7755 from the same equation:

Several flaws in that statement:

1. That would be true only if the y-intercept were zero.
2. What you're looking at is not *molar* extinction coefficient, but *mass* extinction coefficient (you need to know the molecular weight to convert one to the other). This is a nitpick, because the difference in your value comes from the y-intercept.

So how do I convert mass extinction coefficient to molar extinction coefficient?

[tom jupille]

By converting mass to moles (i.e., divide mass my molecular weight).

It actually doesn't matter whether you do your calibration in mass or moles, and it's more convenient to use mass. Just be consistent.

[walker8476]

By converting mass to moles (i.e., divide mass my molecular weight).

It actually doesn't matter whether you do your calibration in mass or moles, and it's more convenient to use mass. Just be consistent.

I'm not sure I understand what the 0.7755 in the equation actually is. Is this a mass e.g. 0.7755mg?

[tom jupille]

I'm not sure I understand what the 0.7755 in the equation actually is. Is this a mass e.g. 0.7755mg?

No, it's the slope of your calibtation line. If you remember from your first-year algebra class, the slope is "the rise over the run" (i.e., y/x), so the units in this case would be (mg/mL)/AU.

[Fernando]

Hi walker8476

Why your absorbance is so small I think 0.0325 is very low. You must use a value of absorbance in the middle of your calibration curve, say 0.3 or 0.4, is this an assay?

Fernando.