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Chapter 16: Problem 37

A compound of formula weight 180 has an absorptivity of \(286 \mathrm{~cm}^{-1} \mathrm{~g}^{-1} \mathrm{~L}\). What is its molar absorptivity'?

Short Answer

Expert verified
Molar absorptivity is 51480 cm鈦宦� mol鈦宦� L.

Step by step solution

01

Understand the concept

To find the molar absorptivity, we need to understand that absorptivity refers to how well a compound absorbs light at a particular wavelength. Molar absorptivity is a specific term used when dealing with molarity (moles per liter) instead of grams. It is denoted by \(\varepsilon\).
02

Use the relationship between absorptivity and molar absorptivity

The relationship between absorptivity \(a\) (given in \(\mathrm{cm}^{-1} \mathrm{~g}^{-1} \mathrm{~L}\)) and molar absorptivity \(\varepsilon\) (given in \(\mathrm{cm}^{-1} \mathrm{~mol}^{-1} \mathrm{~L}\)) is given by \(\varepsilon = a \cdot \text{molecular weight}\).
03

Calculate molar absorptivity

Substitute the given values into the formula: \(\varepsilon = 286 \, \mathrm{cm}^{-1} \, \mathrm{g}^{-1} \, \mathrm{L} \times 180 \, \mathrm{g/mol}\). Simplify to find the molar absorptivity.
04

Solve and Simplify

Calculate: \(\varepsilon = 286 \, \mathrm{cm}^{-1} \, \mathrm{g}^{-1} \, \mathrm{L} \times 180 \, \mathrm{g/mol} = 51480 \, \mathrm{cm}^{-1} \, \mathrm{mol}^{-1} \, \mathrm{L}\).

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Key Concepts

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

Absorptivity
Understanding absorptivity is key to interpreting how substances absorb light. It's a measure of how well a compound can absorb light at a specified wavelength. This property helps in understanding the light-absorbing behavior of a substance when present in a solution. Absorptivity values are often given in \(\mathrm{cm}^{-1} \, \mathrm{g}^{-1} \, \mathrm{L}\), which means the absorbance per path length per concentration unit.
  • The higher the absorptivity, the more effectively the substance absorbs light.
  • It is influenced by the compound's nature and structure, as each molecule absorbs light differently.
This concept is pivotal in many analytical techniques, helping scientists and students alike to quantify the concentration of substances in a mixture.
Molarity
Molarity is a vital concept in chemistry, referring to the concentration of a solution. It is defined as the number of moles of solute per liter of solution, often denoted by the capital letter 'M'.
  • Molarity = \(\frac{\text{moles of solute}}{\text{liters of solution}}\).
  • It helps in standardizing solutions and is a key parameter in calculations involving chemical reactions and solution concentrations.

By understanding molarity, students can better comprehend how reactants interact in solution, offering insights into reaction dynamics and solution behaviors.
Molecular Weight
Also known as molecular mass, molecular weight is the sum of the atomic masses of all atoms in a molecule. It鈥檚 expressed in atomic mass units (amu) or grams per mole (g/mol). This is a fundamental property used to convert grams to moles, and vice versa, in chemical calculations.
  • Molecular weight is crucial for calculating molar absorptivity from regular absorptivity.
  • Knowing the molecular weight aids in stoichiometric calculations, enabling precise measurements of reactants and products in chemical equations.

Thus, mastering the concept of molecular weight is essential for effective problem-solving in chemistry.
Spectrophotometry
Spectrophotometry is a technique used to measure how much a chemical substance absorbs light by measuring the intensity of light as a beam of light passes through the sample solution. It's a cornerstone in laboratory analysis and plays a crucial role in quantifying concentrations of substances.
In spectrophotometry:
  • Absorbance is measured directly and related back to concentration using a calibration curve or Beer's Law.
  • Instrument settings and sample preparations significantly affect results, as do wavelength selections.
This method is widely applicable in chemical, biochemical, and environmental analysis, making it an indispensable tool in research and industry for determining compositions of solutions.

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Most popular questions from this chapter

The core of a simple single-cavity bandpass-type interference filter consists of a thin layer of a dielectric material with thin reflective metal layers on each face. Only the wavelength \(\lambda\) that meets the constructive interference criterion $$ n \lambda=2 t n $$ is transmitted, where \(n\) is the order of interference, and \(t\) and \(n\) are, respectively, the thickness and the refractive index of the dielectric layer, and light is incident perpendicularly on the filter. For a core dielectric of 200 -nm thickness and a refractive index of \(1.377,\) what is the transmitted wavelength through this filter if the emission from a white LED (Figure 16.13 ) is focused on it?

The most widely used wavelength region for infrared analysis is about 2 to \(15 \mu \mathrm{m}\). Express this range in angstroms and in wavenumbers.

Most spectrophotometers can display either absorbance or in percent transmittance. What would be the absorbance reading at \(20 \% T ?\) At \(80 \% T ?\) What would the transmittance reading be at 0.25 absorbance? At 1.00 absorbance?

Express the wavelength \(2500 \AA\) in micrometers and nanometers.

Phosphorus in urine can be determined by treating with molybdenum(VI) and then reducing the phosphomolybdate with aminonaphtholsulfonic acid to give the characteristic molybdenum blue color. This absorbs at \(690 \mathrm{nm}\). A patient excreted \(1270 \mathrm{~mL}\) urine in \(24 \mathrm{~h}\), and the \(\mathrm{pH}\) of the urine was \(6.5 .\) A \(1.00-\mathrm{mL}\) aliquot of the urine was treated with molybdate reagent and aminonaphtholsulfonic acid and was diluted to a volume of \(50.0 \mathrm{~mL}\). A series of phosphate standards was similarly treated. The absorbance of the solutions at \(690 \mathrm{nm}\), measured against a blank, were as follows: $$ \begin{array}{lc} \text { Solution } & \text { Absorbance } \\ 1.00 \mathrm{ppm} \mathrm{P} & 0.205 \\ 2.00 \mathrm{ppm} \mathrm{P} & 0.410 \\ 3.00 \mathrm{ppm} \mathrm{P} & 0.615 \\ 4.00 \mathrm{ppm} \mathrm{P} & 0.820 \\ \text { Urine sample } & 0.625 \end{array} $$ (a) Calculate the number of grams of phosphorus excreted per day. (b) Calculate the phosphate concentration in the urine as millimoles per liter. (c) Calculate the ratio of \(\mathrm{HPO}_{4}^{2-}\) to \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) in the sample: $$ K_{1}=1.1 \times 10^{-2} \quad K_{2}=7.5 \times 10^{-8} \quad K_{3}=4.8 \times 10^{-13} $$
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