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Because if you create a calibration curve and solve the equation for the line, you don't get a zero concentration.
Off-topic conversations and chit-chat.
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).Because if you create a calibration curve and solve the equation for the line, you don't get a zero concentration.
Several flaws in that statement:As I understand it, e (molar extinction coefficient) would be 0.7755 from the same equation:
tom jupille wrote:Several flaws in that statement:As I understand it, e (molar extinction coefficient) would be 0.7755 from the same equation:
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.
tom jupille wrote:
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.
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.I'm not sure I understand what the 0.7755 in the equation actually is. Is this a mass e.g. 0.7755mg?
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