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Chapter 13: Problem 1

Express the following absorbances in terms of percent transmittance: (a) \(0.038\) (d) \(0.241\) (b) \(0.958\) (e) \(0.435\) (c) \(0.399\) (f) \(0.692\)

Short Answer

Expert verified
(a) 91.44%, (d) 57.55%, (b) 10.96%, (e) 36.78%, (c) 39.94%, (f) 20.37%

Step by step solution

01

Understanding the Concept

Transmittance (T) is the fraction of incident light that passes through a sample. It is often expressed as a percentage. Absorbance (A) is related to transmittance by the formula: \( A = -\log_{10}(T) \). To find percent transmittance (%T), first find T using T = 10^{-A}, then multiply by 100 to convert it to a percentage.
02

Calculate Percent Transmittance for Absorbance 0.038

For absorbance \( A = 0.038 \), calculate the transmittance:1. Find T: \( T = 10^{-0.038} \).2. Compute %T: \( %T = T \times 100 \approx 91.44\% \).
03

Calculate Percent Transmittance for Absorbance 0.241

For absorbance \( A = 0.241 \), calculate the transmittance:1. Find T: \( T = 10^{-0.241} \).2. Compute %T: \( %T = T \times 100 \approx 57.55\% \).
04

Calculate Percent Transmittance for Absorbance 0.958

For absorbance \( A = 0.958 \), calculate the transmittance:1. Find T: \( T = 10^{-0.958} \).2. Compute %T: \( %T = T \times 100 \approx 10.96\% \).
05

Calculate Percent Transmittance for Absorbance 0.435

For absorbance \( A = 0.435 \), calculate the transmittance:1. Find T: \( T = 10^{-0.435} \).2. Compute %T: \( %T = T \times 100 \approx 36.78\% \).
06

Calculate Percent Transmittance for Absorbance 0.399

For absorbance \( A = 0.399 \), calculate the transmittance:1. Find T: \( T = 10^{-0.399} \).2. Compute %T: \( %T = T \times 100 \approx 39.94\% \).
07

Calculate Percent Transmittance for Absorbance 0.692

For absorbance \( A = 0.692 \), calculate the transmittance:1. Find T: \( T = 10^{-0.692} \).2. Compute %T: \( %T = T \times 100 \approx 20.37\% \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Absorbance
Absorbance is a measurement of the quantity of light absorbed by a sample. This concept is central to understanding how much light a substance prevents from passing through it. The value of absorbance is directly proportional to the concentration of the absorbing species in the sample and the path length of the light through the sample.

Absorbance is crucial in fields like chemistry and biology, particularly in spectrophotometry, where it is used to determine the concentration of molecules in solution. The unit for absorbance is usually given in terms of optical density. Remember, each unit of absorbance corresponds to a tenfold change in the photon's power traversing through the medium.

In practice, lower absorbance indicates that more light passes through the sample. Conversely, higher absorbance means less light is transmitted. To translate absorbance values into a more intuitive form, we often use the concept of transmittance, relying on simple transformations using logarithms.
Logarithms in Chemistry
Logarithms play a vital role in chemistry, especially when dealing with exponential relationships like those found in spectrophotometry. They help compress large ranges of values into manageable numbers, which is why they are used in the Beer-Lambert Law.

In the context of absorbance and transmittance, we use base 10 logarithms to determine the relationship. The formula:
  • \[ A = -\log_{10}(T) \]
denotes that absorbance (A) is the negative logarithm of the transmittance (T). To make conversions easier, the formula is rearranged to calculate transmittance:
  • \[ T = 10^{-A} \]
By applying this simple transformation, we can quickly move between understanding how much light is absorbed versus how much light passes through the sample. This mathematical approach is essential for chemists analyzing the concentration of solutions using a spectrophotometer.
Light Transmission
Light transmission describes how much light manages to travel through a sample compared to how much initially hits it. Expressed as a percentage, it is a direct measure of the sample's clarity and transparency.

The calculation of percent transmittance (\[ \%T \]) starts with the transmittance (T), derived from the absorbance. The equation:
  • \[ T = 10^{-A} \]
helps find T, which can then be converted into a percentage via:
  • \[ \%T = T \times 100 \]
This calculation offers intuitive insight into the characteristics of the sample being analyzed. For example, a high percent transmittance indicates a very transparent medium, while a lower percent transmittance suggests that the substance is absorbing much of the incident light.

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Most popular questions from this chapter

A compound \(X\) is to be determined by UV-visible spectrophotometry. A calibration curve is constructed from standard solutions of \(X\) with the following results: \(0.50 \mathrm{ppm}, A=0.24 ; 1.5 \mathrm{ppm}, A=0.36 ; 2.5 \mathrm{ppm}, A=0.44 ; 3.5 \mathrm{ppm}, A=0.59\); \(4.5 \mathrm{ppm}, A=0.70\). A solution of unknown \(\mathrm{X}\) concentration had an absorbance of \(A=0.50\). Find the slope and intercept of the calibration curve, the standard error in Y, the concentration of the solution of unknown X concentration, and the standard deviation in the concentration of \(X\). Construct a plot of the calibration curve and determine the unknown concentration by hand from the plot. Compare it to that obtained from the regression line.

Why do quantitative and qualitative analyses often require different monochromator slit widths?

A solution containing the complex formed between Bi(III) and thiourea has a molar absorptivity of \(9.32 \times 10^{3} \mathrm{~L}\) mol \(^{1} \mathrm{~cm}^{1}\) at \(470 \mathrm{~nm}\). (a) What is the absorbance of a \(3.79 \times 10^{-5} \mathrm{M}\) solution of the complex at \(470 \mathrm{~nm}\) in a \(1.00\)-cm cell? (b) What is the percent transmittance of the solution described in (a)? (c) What is the molar concentration of the complex in a solution that has the absorbance described in (a) when measured at \(470 \mathrm{~nm}\) in a \(2.50-\mathrm{cm}\) cell?

A solution containing \(6.23 \mathrm{ppm} \mathrm{KMnO}_{4}\) had a transmittance of \(0.195\) in a \(1.00-\mathrm{cm}\) cell at \(520 \mathrm{~nm}\). Calculate the molar absorptivity of \(\mathrm{KMnO}_{4}\) at \(520 \mathrm{~nm}\).

A portable photometer with a linear response to radiation registered \(63.8 \mu \mathrm{A}\) with the solvent in the light path. The photometer was set to zero with no light striking the detector. Replacement of the solvent with an absorbing solution yielded a response of \(41.6 \mu \mathrm{A}\). Calculate (a) the percent transmittance of the sample solution. (b) the absorbance of the sample solution. (c) the transmittance to he expected for a solution in which the concentration of the absorber is one third that of the original sample solution. (d) the transmittance to be expected for a solution that has twice the concentration of the sample solution.
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