Analytical chemistry practice — Why is absorbance (A) generally preferred over percent transmittance (%T) as the measure of absorption in quantitative spectrophotometry?

Difficulty: Easy

Correct Answer: Because absorbance is proportional to analyte concentration whereas %T is not

Explanation:


Introduction / Context:
In UV-Vis spectroscopy, one can report data as percent transmittance (%T) or as absorbance (A). For quantitative analysis via Beer–Lambert law, absorbance is the preferred metric because it relates linearly to concentration.



Given Data / Assumptions:

  • Beer–Lambert law: A = ε b c.
  • %T = 100 × (I / I0), and A = −log10(T) where T = I / I0.
  • Calibration methods require linear relationships for straightforward regression.


Concept / Approach:
Absorbance is directly proportional to concentration at constant path length and wavelength because ε is constant for a given analyte. %T is logarithmically related to concentration; equal concentration changes do not produce equal %T changes, complicating calibration and error analysis. Therefore, using A simplifies quantitation and statistical treatment.



Step-by-Step Solution:

Start with T = I / I0 and A = −log10 T.Apply Beer–Lambert: A = ε b c → linear in c.Recognize %T varies nonlinearly with c (exponential dependence).Conclude that absorbance is preferred for linear calibration.


Verification / Alternative check:
Typical calibration plots of A vs. c yield straight lines with slopes ε b; plots of %T vs. c are curved.



Why Other Options Are Wrong:

  • %T can be measured accurately; accuracy is not the core issue.
  • Dependence on incident power cancels when T = I / I0 is used properly; modern instruments compensate.
  • “None” is incorrect because there is a clear quantitative reason to prefer A.


Common Pitfalls:
Confusing precision with linearity; the critical advantage of absorbance is proportionality to concentration.



Final Answer:
Because absorbance is proportional to analyte concentration whereas %T is not

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