Questions and Answers

Question:  Can you tell me what 2,4-Dichlorophenoxyacetic acid is most commonly used for?

Question:  Jared --  I have been looking around the internet and in some books about some different chromatography methods.  One that I have came across that interests me is size exclusion chromatography.  Can you tell me what SEC is all about and what it is usually used to separate?

Jared's Answer:  I have also researched different methods of chromatography.  I'm sure you have noticed that there are many currently out there.  The one in your question, size exclusion chromatography (SEC), actually is a broad branch of chromatography in itself.  SEC is generally used to separate components of a mixture according to their molecular size, based on the flow of the sample through a porous packing.  As I mentioned previously, there are two sub-branches, if you will, that are classified under SEC.  They are gel filtration chromatography (GFC) and gel permeation chromatography (GPC).  GFC is used to separate molecules in aqueous systems.  On the other hand, GPC is used to separate organic polymers in non-aqueous systems.  GFC's favorable characteristics are its gentle non-interaction with the sample allowing high retention of bimolecular enzymatic activity while separating multimers that are not easily distinguished by other chromatographic methods.  The main stipulation on SEC, however, is that the molecular weights of the biomolecules must differ by at least a twofold in order for successful separations to occur.  Interestingly in SEC, molecules larger than the pores in the packing are completely excluded and elute with the solvent front.  Small molecules, however, which freely pass in and out of the pores are retained.  Molecules who fall in the intermediate size permeate the pores of the matrix in relation to their size.

Chromatography is exciting, so if this sounds interesting take a look at some literature that explains the function of some of the separation methods.  Also, a great source of information and chromatographic supplies is TosoHaas.  You can find them on the web at:  http://www.rohmhaas.com/tosohaas/

Question:  Hello, I was just wondering what explination could be found about this topic-->  If iron* is a element (Fe)  than how can it rust.  My basic understanding was that an element was a substance in its purest form, and could not be broken down further by basic chemical means.  This is a puzzler
to me, and if you could explain the technicalities of the process to me I would be very thankful.

           --a puzzled high school chemist.

Jared's Answer:  You e-mail me a commonly asked (and wondered) question.  You are absolutely right that elements cannot be broken down any further...they exist in their purest form.  However, when we talk about iron existing as a solid metal, Fe(s); and then iron turning to rust, we are not forming a new element, we are merely changing the composition of that element (Fe).  In chemistry, we call two possible changes that matter (elements) can undergo.  They are:  physical and chemical changes.  When solid iron forms rust, the iron atoms are undergoing a chemical change.  The complete equation describing this change appears below.  It may help in your understanding.

           4 Fe(s) + 3 O2(g) + 6 H20 (l) --> 2 Fe2O3 * 3 H20

To explain this equation, solid (molecular) iron is coming in contact with with oxygen and water, which OXIDIZE (make iron lose 3 e-).  In any redox chemical reaction, one element is oxidized and one reduced; unless you have a disproportionation rxn, which isn't utilized in analytical methods.  If we assign oxidation numbers, we say that molecular iron is at oxidation state 0, 02 is a molecular state, so it is also 0.  Hydrogen in H20 is +1 and oxygen in water is -2.  When we move to the products, iron (III) oxide trihydrate (the complex consisting of Fe203*3H20...or rust) shows us where the electrons went.  Iron is becoming oxidized (going from 0 to +3; the ferric ion), where oxygen (from the diatomic 02 on the reactants) is becoming reduced (going from 0 to -2 oxidation).  Therefore, this is a simple oxidation-reduction reaction.

     I feel that I may have went into too much depth to completely make things totally clear, but imagine this example. Your body breaks down excess starches into sugar, which is stored as glucose (C6H12O6) in your liver.  This is a chemical change, as the composition is the same as what you started with, it only appears in different arrangement or composition (ions, etc.)

     I hope that I made this clear.  If not, please let me know where you are still vague.  Thanks for your question!  Good luck with your high school chemistry class.  Keep me in mind when more questions arise.

     Jared Anderson
     organic_chemist@hotmail.com

  Question: How does the buckyball compare with the other forms of carbon ( Physical and chemical properties). Thank you for any help you can provide

- A junior chemist

Jared's Answer: Dear Junior Chemist: Although I am only still a junior undergraduate chemistry student, I will try my best to answer your question. The bucky ball is one of the most interesting aspects of "carbon" chemistry, in my mind. Why? Well, it is well known that all carbon atoms tend to catenate, or form bonds to themselves. We see long chains of carbon systems (alkanes, alkenes, alkynes). What is very interesting to me is how does carbon form the soccer ball shape as in the Bucky Ball. Let's take a look at how the Bucky Ball is formed.

The first bucky ball was synthesized by vaporizing a sample of graphite (very similar to pencil lead) with an intense pulse of laser light and using a stream of helium gas to carry the vaporized carbon into a mass spectrometer (similar to analytical techniques using organic solvents in Gas Chromatography/Mass Spectrometry). However, in 1990 scientists found that appreciable amounts of buckyball can be prepared by electrically evaporating graphite in an atmosphere of helium gas. The U.S. and German scientists in the synthesis contend that approximately 14% of the resulting soot consists of C(60) and a related molecule C(70)--C(70) deviates from the soccer ball shape to a more egg-shaped geometry. Because the bucky balls are forms of elemental carbon bound in molecules, they are more easily dissolved then diamond or graphite. Hopefully this sheds a little light on the physical differences. (Note that when the bucky ball is formed, two forms result. Doesn't this remind you of an organic synthesis when different products may form and a purification process may be necessary?)

As far as chemical properties are concerned, we are embarking on a new frontier with the Bucky Ball. Recently I talked to a professor who worked with Bucky Balls and substituting metal atoms inside the ball. The bucky ball results by the carbon sphere surrounding the metal atom. (Pretty cool, huh?!) Work continues on this field of bucky ball chemistry. We have been able to substitue metal atoms (organometallic substitution) onto regular organic molecules, but rest assured that the characteristics differ! I am assuming that you have read about the scanning tunneling microscope being used to submit a potential across the bucky ball. This is only one some of the technology that will be utilized by the Bucky Ball. I know that more literature will be coming out about the Bucky Ball and when it does, I hope to keep you all informed. I hope that I answered your question. If you have any comments, please don't hesitate to let me know. Please direct any future chemistry related topics my way. Also, coming soon, I will be posting laboratory reports that I have written in Physical Chemistry. Thank you for your question!

Question: What is the concentration (mol) of 6ml of H2O2 if it produces 190ml of O2 at 295K and 99KPa?
                -Jr. Chemist
 

Jared's Answer: Dear Jr. Chemist:

       In your question, I am assuming that hydrogen peroxide is being decomposed as the following equilibrium illustrates:

We know the decomposition forms:  Volume of Oxygen (0.190 L) under the following conditions:  Pressure of 99 kPa (or 99 kPa / 101.325 kPa) = 0.977 atmospheres of pressure at 295 K.  At this pressure and temperature, we may assume that oxygen is an ideal gas.  Using the ideal gas law:  PV = nRT  (0.977 atm)(0.190 L O2) = (n moles O2)(0.08206 L-atm/mol-K)(295 K)

Knowing the number of moles of the oxygen, we can then use the molar stoichiometry and say that for every two moles of peroxide that are decomposed, one mole of oxygen is produced.  Or, in equation form:   0.00767 moles O2 x( 2 moles H202/1 mole 02)

=0.01534 moles H202

This means, in order to carry out this reaction and obtain 190 mL of oxygen at the specified conditions, we must have 0.01534 moles of peroxide to decompose.  Therefore, when we look at the molarity of the solution (moles H202/L H202), you said that 6 mL of the peroxide was used to proceed the given decomposition.  That is, the volume of H202 multiplied by the molarity of the peroxide solution MUST equal the number of moles required to produce the 190 mL of the ideal gas O2.  Or, in equality form:

0.01534 moles H202/(0.006 L H202)

This equality above will give us the molarity of the hydrogen peroxide solution needed, which turns out to be:  2.56 M H202Therefore, take 6 mL of a 2.56 M H202 solution of H202 and decompose it at 295 K and 0.977 atm of pressure and you should obtain 190 mL of O2 gas.

PLEASE NOTICE that I made an assumption when completing the calculations of this lab when I treated O2 as an ideal gas in this situation.  However, some gases are known to deviate from ideal gas behavior.  To look at the error associated with this assumption, I calculated the number of moles that would be produced if the peroxide were decomposed with the conditions.  The van der Waals equation was used.  The van der Waals equation corrects for the volume of the molecules and the correction for molecular attractions.  The van der Waals equation follows below:

In this case, for oxygen, a = 1.36 (L2-atm/mol2) and b = 0.0318 L/mol).  Plugging all values into the preceding equation for determination of the number of moles (n), we obtain a nasty equation.  However, solving algebraically, the equation can be reduced to a cubic equation:

Using a TI-85 calculator and the poly function, I came up with a mol (n) value of:  0.007675 mol.  When we compare this to the above value of 0.00767 mol, we achieve an extremely small and perfectly negligible percent difference.

     I realize that this was a long, long process to analyze our answer, but please remember that we made the assumption that oxygen was and ideal gas that those conditions.  When making any assumption, one should always justify that assumption as I did when I solved the van der Waals equation.

     Thank you for your question.  What an excellent review of Physical Chemistry!

    Jared organic_chemist@hotmail.com
 
 

Question:  Jared-    This is actually 5 questions, but here goes.  The three elements X, Y, and Z have successive atomic
                  numbers, each increasing by one in the order given.  Atoms of element X form stable ions with the formula X
                  negative and atoms of element Z form stable ions with the formula Z plus.  Which of the following statements is
                  false concurring elements X, Y and Z?  Why are they false?

                  1.  A neutral atom of element Y has one more electron that neutral atom of element X, but one less electron than a
                       neutral atom of element Z.

                  2.  Element X could be in group 17.

                  3.  Element Z could be in-group 1.

                  4.  Elements X, Y, and Z could all be in the same group of the periodic table.

                  5.  Elements X, Y, and Z could be those elements with atomic numbers 9, 10, and 11, respectively.

                   I was just wondering if you could answer these questions for me, because I couldn't.

                   Thanks-
                   confused chemistry student

Jared's Answer:  Confused chemistry student:

     In the riddle it is stated that atom X forms stable ions with the formula X negative (X-); atoms of Z form stable ions with the formula Z plus (Z+).  Nothing is said about Y.  Nevertheless, a key word is STABLE.  Looking at the periodic table, those elements in group I (Alkali Metals) exist as stable ions of Z+.  Those elements in group 17 (Halogens) exist as stable ions of X-.  Knowing this, lets take a look at each question we are faced against:

     1.  If we are assuming that X is a halogen and Z is a alkali metal, then the question states that element Y would have one more electron than X and one less electron than Z.  This is true if Y is a noble gas.  If we take fluorine (F) as X and sodium (Na) as Z, then one electron added to fluorine would give us the next element, namely Neon (Ne).  One electron less than sodium would also give us neon, so this assumption must be TRUE.

     2.  Yes, this is true.  We are saying that X (the halogens) are in group 17.

     3.  Yes, this is also true.  We are also saying that since Z is a stable positive cation that it exists as a group 1 element.

     4.  This is definitely FALSE.  For X, Y, and Z to be in the same group of the periodic table would be for a given group to possess atoms with positive one cation, neutral atom, and a negative one anion.  No such group on the periodic table possesses these characteristics.

     5.  This would follow our generalization, so we would say this is TRUE.  X, Y, and Z could be fluorine, neon, or sodium; respectively.  With what we know, they could also be chlorine, argon, or potassium as this also fits the consecutive trend and each atom possesses the charge as stated in the riddle.

     Therefore, the only question that is false is question #4.  All other questions are true.

     I hope that this all makes sense.  If it still doesn't make sense, please e-mail me once again and tell me what you don't understand.

     Thanks for the chemistry riddle.  That was fun and it required some knowledge of the periodic table!  Please e-mail me again when another question arises!  Have fun with your future chemistry endeavors.

     Jared Anderson
     organic_chemist@hotmail.com