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Chapter 24: Problem 15

Convert the accompanying transmittance data to absorbances. "(a) \(27.290\) (b) \(0.579\) *(c) \(30.6 \%\) (d) \(3.98 \%\) "(e) \(0.093\) (f) \(63.7 \%\)

Short Answer

Expert verified
Absorbances: (a) 0.564, (b) 0.238, (c) 0.514, (d) 1.400, (e) 1.032, (f) 0.196.

Step by step solution

01

Understanding Transmittance and Absorbance

Transmittance \, \(T\), is a measure of how much light passes through a sample compared to the incident light. It can be expressed as a percentage or a decimal fraction. Absorbance \, \(A\), is a measure of how much light is absorbed by the sample, calculated using the formula: \[A = -\log_{10}(T)\] where \(T\) should be in decimal form rather than a percentage.
02

Convert Percentage to Decimal

Before using the absorbance formula, convert percentage transmittance to decimal by dividing by 100. For example, if transmittance is \(30.6\%\), convert it to \(0.306\).
03

Calculate Absorbance (a)

With a transmittance of \(27.290\), use the formula: \[A = -\log_{10}(0.27290)\] Calculate to find \(A \approx 0.564\).
04

Calculate Absorbance (b)

For a transmittance value of \(0.579\), calculate the absorbance: \[A = -\log_{10}(0.579)\] Resulting in \(A \approx 0.238\).
05

Calculate Absorbance (c)

Convert \(30.6\%\) transmittance to decimal \(0.306\) first. Then, \[A = -\log_{10}(0.306)\] Compute to get \(A \approx 0.514\).
06

Calculate Absorbance (d)

Convert \(3.98\%\) transmittance to decimal \(0.0398\). Use the formula: \[A = -\log_{10}(0.0398)\] Which gives \(A \approx 1.400\).
07

Calculate Absorbance (e)

For transmittance \(0.093\), compute: \[A = -\log_{10}(0.093)\] Resulting in \(A \approx 1.032\).
08

Calculate Absorbance (f)

Convert \(63.7\%\) transmittance to decimal \(0.637\). Then calculate: \[A = -\log_{10}(0.637)\] Giving \(A \approx 0.196\).

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Key Concepts

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

Transmittance
Transmittance describes how much light manages to pass through a sample compared to the amount of light that first hits the sample. If you have a very clear sample, a lot of light will pass through, resulting in high transmittance. Transmittance can be recorded in two forms:
  • Percentage
  • Decimal fraction
For practical calculations, we often convert percentage transmittance to a decimal. For example, a transmittance of 30.6% would be changed to 0.306 by dividing by 100. This standard conversion helps in later calculations when finding absorbance, making our math simpler.
Absorbance
Absorbance is used to quantify the amount of light absorbed by a sample as light passes through it. It's a critical calculation in analytical chemistry to understand the properties of solutions. The formula used to determine absorbance from transmittance is:\[A = -\log_{10}(T)\]where:
  • \(A\) is the absorbance and
  • \(T\) is the transmittance in decimal form.
This formula inverts the transmittance value on a logarithmic scale. A higher absorbance value indicates a greater light absorption by the sample. This data can help uncover concentrations and characteristics of solutions, as different substances absorb light uniquely at varying wavelengths.By converting the percentage transmittance to decimal and plugging it into the formula, the absorbance can be accurately calculated.
Logarithmic Calculations
Logarithmic calculations transform large multiplicative challenges into manageable additive ones, which simplifies many mathematical problems in chemistry. When working with absorbance, logs turn varying scales of transmittance into a linear range of absorbance values. This is significant because:
  • It allows for easier interpretation and comparison of data.
  • It translates exponential data into linear data.
When using the formula \(A = -\log_{10}(T)\), we're dealing with a base 10 logarithm, which is a common choice in scientific work as it nicely represents orders of magnitude.To perform a logarithmic calculation:- Identify the transmittance in decimal form.- Apply the logarithmic function to find the log of that number.- Negate the result to get the absorbance.This method transforms how data is understood in analytical chemistry, linking light properties to substance properties.

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Most popular questions from this chapter

. One common way to determine phosphorus in urine is to treat the sample after removing the protein with molybdenum (VI) and then reducing the resulting 12-molybdophosphate complex with ascorbic acid to give an intense blue-colored species called molybdenum blue. The absorbance of molybdenum blue can be measured at \(650 \mathrm{~nm}\). A 24-hour urine sample was collected, and the patient produced \(1122 \mathrm{~mL}\). in 24 hours. A \(1.00 \mathrm{~mL}\) aliquot of the sample was treated with Mo(VI) and ascorbic acid and diluted to a volume of \(50.00 \mathrm{~mL}\). A calibration curve was prepared by treating \(1.00 \mathrm{~mL}\) aliquots of phosphate standard solutions in the same manner as the urine sample. The absorbances of the standards and the urine sample were obtained at \(650 \mathrm{~nm}\) and the following results obtained: \begin{tabular}{lc} Solution & Abserbance at \(650 \mathrm{~nm}\) \\ \hline \(1.00 \mathrm{ppm} \mathrm{P}\) & \(0.230\) \\ \(2.00 \mathrm{ppm} \mathrm{P}\) & \(0.436\) \\ \(3.00 \mathrm{ppm} \mathrm{P}\) & \(0.638\) \\ \(4.00 \mathrm{ppm} \mathrm{P}\) & \(0.848\) \\ Urine sample & \(0.518\) \\ \hline \end{tabular} (a) Find the slope, intercept, and standard error in \(y\) of the calibration curve. Construct a calibration curve. Determine the concentration number of phosphorus in ppm in the urine sample and its standard deviation from the least-squares equation of the line. Compare the unknown concentration to that obtained manually from a calibration curve. (b) What mass in grams of phosphorus was eliminated per day by the patient? (c) What is the phosphate concentration in urine in \(\mathrm{mM}\) ?

A solution with a "true" absorbance \(\left[A=-\log \left(P_{0} / P\right)\right]\) of \(2.10\) was placed in a spectrophotometer with a stray light percentage \(\left(P_{J} / P_{0}\right)\) of \(0.75\). What absorbance \(A^{\prime}\) would be measured? What percentage error would result?

A sophisticated ultraviolet/visible/near-IR instrument has a wavelength range of 185 to \(3000 \mathrm{~nm}\). What are its wavenumber and frequency ranges?

. Calculate the frequency in hertz and the energy in joules of an \(X\)-ray photon with a wavelength of \(2.70 \AA\).

The complex formed between \(\mathrm{Cu}(\mathrm{I})\) and 1,10phenanthroline has a molar absorptivity of \(7000 \mathrm{~L}\). \(\mathrm{cm}^{-1} \mathrm{~mol}^{-1}\) at \(435 \mathrm{~nm}\), the wavelength of maximum absorption. Calculate (a) the absorbance of a \(6.17 \times 10^{-5} \mathrm{M}\) solution of the complex when measured in a \(1.00-\mathrm{cm}\) cell at \(435 \mathrm{~nm}\). (b) the percent transmittance of the solution in (a). (c) the concentration of a solution that in a \(5.00-\mathrm{cm}\) cell has the same absorbance as the solution in (a). (d) the path length through a \(3.13 \times 10^{-5} \mathrm{M}\) solution of the complex that is needed for an absorbance that is the same as the solution in (a).
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