12

Yes. For some chiral compounds, each enantiomer in a racemic mixture absorbs certain frequencies of light differently depending on the circular polarization type of the light. That means that if you can pump in sufficiently intense and energetic (e.g. ultraviolet) circular polarized light pulses into the racemic mix of such compounds, you may in some cases ...


11

Any molecule can absorb UV light. What the question is probably going for is why those three molecules absorb at longer wavelengths than other amino acids. This has to do with the conjugated pi bonds from aromaticity. Looking at a list of amino acids, we can see that most of them don't have conjugated pi ystems. amino acids (source) Conjugated pi bonds ...


10

The broad absorption bands of the electronic transitions observed in UV/Vis spectroscopy/spectrophotometry, as well as the myriad types of electronic transitions that might lead to absorption at a given wavelength, indeed make the technique much less powerful for identification of unknown compounds as compared to, e.g., IR, NMR, microwave, and mass$^*$ ...


8

It seems that the results of the calculations are more or less fine and the OP just misinterpreted the NIST data. As I said in my comment above, NIST does not claim that $\lambda_{\mathrm{max}}=276 \, \mathrm{nm}$. Clearly only a small region of wavelength is shown on the graph and in the paper[1] referenced on the NIST page it is said (emphasis mine): ...


8

Internet consensus seems to define "high concentration" for Beer's Law as >0.01M Keep in mind that Beer's law is an approximation. Look at the more advanced version. Beer's law-advanced version. The statement that a concentration of 0.01 M is the upper limit of Beer's law is incorrect due to several reasons. It is the absorbance value which we should ...


7

In an absorption experiment a collimated beam of light is passed through the sample. The intensity of this is measured before and after the sample. The absorption is measured as optical density which is $D=[c]\epsilon _{\lambda} l$, where $[c]$ is the concentration, $\epsilon_\lambda$ the extinction coefficient at wavelength $\lambda$ (a measure of the ...


7

Absorbance is useful because it is additive. That is, it's absorbance which is used in Beer's law: $$A = \epsilon \cdot c \cdot l$$ While you can certainly make a version of this law which uses transmittance instead of absorbance, the math becomes much less straightforward. The explanation of why absorbance is additive falls out of the derivation of Beer'...


6

What the plot shows is the reciprocal path length needed for a given absorbance to apply, for example slightly less than 100 m in the visible to $10^{-8}$ m in the uv, as you point out, but this is for $55.345$ molar water which is very, very concentrated compared to most solutions from which an absorption spectrum is obtained, typically $<< 1 $ molar....


6

The primary reason is that when soda warms up, the $\ce{CO2}$ would escape slowly as bubbles. These bubbles would significantly scatter light affecting the absorption spectra. Usually, the effect of such scattering is quite pronounced and the absorption bands of individual molecules will be (might be) buried below the broadened feature in the lower ...


6

No it doesn’t. Measurements can in principle be made at any wavelength. If there are two colored specimens in the same solution, it is actually better to measure at the wavelength where the difference between the spectra is as high as possible. Measuring at a peak has some advantages: the absorbance is high, giving a good signal/noise ratio, and if the ...


5

When you can, use TDDFT, in part because of the oscillator strengths I like to point out to my group that optical excitations consist of two things consistently (e.g., in a UV/Vis spectra). There are the excitation energies (the x-axis) and the intensity or extinction coefficient (the y-axis). To some degree, the peak widths give a bit more information (e.g....


5

Here's how I approached the problem. Plot the Data Below is the plot I quickly hacked together, and a fit of the form $g(x) = a\cdot x$. (Note that I do want the line to go through (0,0) because the absorption at zero concentration should be zero. And it's a pity you didn't check that with a measurement... One never knows whether the calibration was ...


5

I found it quite hard to follow both the question and the answer, so I am going to state a few relevant things and paraphrase the answers of @LDC3 to add some clarity. Chemical shift (ppm) and peak area (integral) These are independent quantities. One tells you about the environment (shift) the protons are in, and the other tells you how many protons are ...


5

Classical approach Absorption occurs when there are "oscillators" that can resonate with the frequency of the electromagnetic wave. In classical weave physics, these frequencies are called natural frequencies. These oscillators can involve electrons (UV, VIS, XR), bonds (MIR, FIR), nuclei (gamma rays). They are able to absorb electromagnetic waves changing ...


5

The optical rotation is certainly used to follow the separation of optical isomers (enantiomers if they are mirror images, or diastereomers if you have more than one stero-center and the two isomers are not mirror images). You can see how the optical rotation changes as you use the normal separation techniques (chromatography, distillation, crystallization, ...


5

To know the chemical name, molecular structure, or best: the unique registry number provided by the American Chemical Society (CAS-number) is beneficial in your search as trade names may vary depending on the supplier of the material you are interested in. The index number of the Colour Index (like C.I. 73000 for indigo) frequently used in the business of ...


5

Since this is pretty well covered in countless number of books and other sources I will try to just emphasise the key points. The emission spectrum of atomic hydrogen was well known at that time, the only problem was that no one was able to explain it. :D Where does all this lines come from, i.e. what physical process gives rise to such discrete spectrum? ...


5

Beer's law relates transmitted intensity to concentration $[c]$ of the solution at each wavelength $\lambda$ as $$ I_{trans_\lambda}=I_0e^{-\epsilon_\lambda [c]L}$$ where L is the sample path-length and $\epsilon$ the extinction coefficient of the molecule at wavelength $\lambda$ for an incident intensity $I_0$. (The extinction coefficient measures how ...


5

The problem is a bit strange since the details of the absorbance measurement aren't detailed very well. I'll assume the book defines the relationship as: $$A = \epsilon bc$$ where: A=absorbance $\epsilon$ = molar attenuation coefficient $b$ = path length $c$ = concentration So the concentration, c, of the $\ce{[Fe(SCN)]^2+}$ is: $$ c = \dfrac{A}{\epsilon ...


5

To start with, its often helpful to make an ICE table when dealing with problems of chemical equilibrium. $$\begin{array}{cccc} \begin{array}{c|ccc} \hline & \ce{Fe^3+} & \ce{SCN-} & \ce{[Fe(SCN)]^2+} \\ \hline \mathrm{initial} & 1.32\times10^{-5} & 8.40\times10^{-6} & 0\\ \mathrm{change} & -x & -x &+x \\ \mathrm{equil} &...


4

Working wavelength is chosen by analysing the spectrogram $A(\lambda)$. Such diagram is obtained by measuring absorbance $A$ in function of wavelength $\lambda$. Practically, you test a moderate concentration of your analyte and sweep the complete wavelength bandwidth of your analyser. Then you select wavelength where a maximum of absorbance occurs. If ...


4

A compound S follows the Lambert-Beer law if the absorbance $E_\lambda$ at a particular wavelength $\lambda$ is proportional to the concentration $c$ of S. \[E_\lambda = \left(\frac{I_0}{I} \right) = \epsilon\cdot c \cdot d \] A "deviation" is observed when the real concentration $c$ is different from the expected/assumed. This may happen if S is an acid ...


4

An isosbestic point is found by maintaining the sum of the concentrations of the two (or more, though it's very unlikely) species the same and varying the individual contributions of each, and looking for the point at which the absorbance spectra intersect. This is easy for something like a pH indicator where a pH change just converts one species into the ...


4

Actually that's not the correct definition of the isosbestic point. From IUPAC goldbook: Isosbestic point - Wavelength, wavenumber or frequency at which the total absorbance of a sample does not change during a chemical reaction or a physical change of the sample. A simple example occurs when one molecular entity is converted into another that has ...


4

The usual approach is: Find the equilibrium geometry (within the Born-Oppenheimer app.). Expand the energy in Taylor series till second order term. The first order will be 0 because of the minimum energy condition. Include this expression in the Hamiltonian. Generate a coordinate transformation to get normal coordinates. Relate second order derivative of ...


4

You can build molecules and calculate vibrational frequencies with Molcalc. Molcalc does have a molecular size limit, so for bigger molecules you must download some programs. I have written about computing vibrations with the GAMESS program.


4

Generally speaking, the typical auxochromic groups possess (at least) one pair of non-bonded $n$-electrons and -- if taken alone -- do not absorb in the UV. If attached, for example to a benzene ring, their pair(s) of $n$-electron interact(s) with the $\pi$-electrons, and the delocalization of the $\pi$-system is extended. Now the catch: Bathchromic ...


4

This is absolutely not true. Many ligands can strongly dictate the color of transition metal solutions. Your example picks three simple salts and then questions whether the color of the solutions (dictated largely by $\ce{[Cu(H2O)6]^{2+}}$) is very different. No, because the resulting majority complex in aqueous solution is likely identical. Even taking a ...


4

Both your conclusions are absolutely correct. I'll point out that you'll never get "all" the $\ce{Cu^{2+}}$ to the amine complex. You're trying to get greater than 99% of the $\ce{Cu^{2+}}$ to the amine complex. Because 99% and 99.9% and 99.99% would have virtually the same absorption reading. To give even more detail, in mildly acid solution the copper ...


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