Spectroscopy

You should visit www.chemguide.co.uk for detailed and concise explanations of some spectroscopic techniques. Click on the "Instrumental analysis link". There is a good review of chromatography here too (stick to paper prior to the exam)! (AP chemistry does not require NMR, so avoid that link until after the exam) Chemguide does not explain photoelectron spectroscopy, so I have summarized some important points below.

You should understand that types of electromagnetic radiation will interact differently with molecules depending on the energy of that radiation:

microwaves are low in energy, therefore molecules that absorb microwaves will only increase rotational or translational energy. Consequently microwave spectra gives us information about molecular rotations.
Infrared radiation is higher in energy than microwaves, but lower than visible light. IR causes bonds in molecules to stretch or bend.
UV and sometimes visible light causes electron excitation in molecules. UV light will cause excitation in pi-bonds, and in some cases can break a pi-bond or even a weak sigma-bond.

Recall for UV-visible spectroscopy is useful because the absorbance of a solution is directly proportional to the concentration of the solution. This means if a compound absorbs visible light, we can

Calculate the concentration of an unknown solution. We did this to determine:

A. the concentration of FD and C Blue food dye in gatorade and

B. the equilibrium constant of for the reaction Fe³⁺ (aq) + SCN^– (aq) ⇄ FeSCN²⁺(aq)

C. the rate order of a chemical reaction with respect to a colored reactant

In both cases you have to follow the same steps:

Take a full spectrum to find the wavelength of maximum absorbance

2. Set the spectrophotometer to that wavelength. In the above case the maximum absorbance is at 626.1 nm.

Measure the absorbance of a number of solutions to make a standard curve of absorbance vs. concentration.

The slope of this curve is A/c which is constant at 0.257. You can then solve for the calculation of any solution containing that compound by measuring the absorbance, and calculating the concentration using A = 0.257*(concentration)
Monitor the concentration of a substance over time during a chemical reaction (useful to determine the rate law for the reaction) We did this to measure the kinetics of crystal violet and NaOH. Another lab that is common is the reaction of food dye with bleach.
In this case, it is necessary to repeat step 1 but not step 2 if you only wish to find the order. You do have to repeat step 2 if you would like to calculate the rate constant. Since the concentration is directly proporational to the absorbance, you can simply make 3 plots, and determine the order by seeing which plot is linear.
If Absorbance vs. time is linear, it is 0th order with respect to the colored reactant
If ln(Absorbance) vs. time is linear, it is 1st order with respect to the colored reactant
If 1/(Absorbance) vs. time is linear, it is 2nd order with respect to the colored reactant

Only one of the following graphs is possible (for crystal violet, ln[A] vs. time is linear.

For AP Chemistry, you need to have a general idea about some spectroscopic techniques, although interpretation of spectrum is generally limited to

Mass spectroscopy of elements (existance of isotopes and calculation of average atomic mass)
UV visible spectra, and using Absorbance vs. Concentration curves
- Note,
Photoelecron spectra

Photoelectron Spectroscopy--This spectroscopy uses x-rays to ionize atoms of an element. You get peaks of with intensity proportional to the number of electron in each subshell of an atom.

In the spectrum above, you should note that:

The peak at 2.08 MJ/mol, corresponds to the highest energy subshell. This is the 2p electrons.
This peak is three times taller than the peak at 4.68 MJ/mol, which corresponds to the 2s.
There is a huge difference in energy between the 1s and the 2s because the 1s electron is much closer to nucleus, and it is less shielded, and therefore experiences a much stronger coulombic attraction than the 2s electron.

Comparing 2 different elements using PES:

Although elements in the same period have similar photoelectron spectra, you can use PES to explain important differences:

Consider the photoelectron spectrum of potassium:

You can see that the 4s subshell is requires the least energy to ionize (0.42 MJ/mol) because the 4s subshell is furthest from the nucleus and it is more shielded from the nucleus than the inner core electrons. Ca would have the same PES as K, but every peak would be shifted to the left. Look at the table to convince yourself this is the case. You can rationalize why each subshell in Ca requires more energy to ionize than those in K due to the fact that Ca has 20 protons vs. 19 protons in K. So the greater nuclear charge causes as stronger Coluombic attraction (and consequently greater energy to ionize) between the nucleus and each subshell in Ca than the same respective attraction in K.

Mass Spectroscopy

Mass Spectroscopy ionizes an atom or molecule using an electron gun. The cation is then accelerated by an electric field and then deflected by a magnetic field. Lighter ions will experience more deflection than heavy ions. The the graph will give the relative abundance of each isotope of an element (if the sample is an element). If the sample is a molecule, it can decompose into other ions while prior to getting detected. These fragments can provide insight into the structure of the compound. See the Adrian Dingle page for speculation on how mass spectroscopy could be tested in the new AP format.

Report abuse