Is it possible to make metal fulminates? Is it possible to make fulminates such as: magnesium fulminate, copper fulminate, etc.
by the metal + $\ce{HNO3}$ + $\ce{EtOH}$ reaction?

In The Preparatory Manual of Explosives  (paperback at amazon), the author, Jared Ledgard, provides a procedure for the preparation of copper fulminate that can be summarized as follows:

  To 100 mL of 70% $\ce{HNO3}$, 30 g of anhydrous $\ce{Cu(NO3)2}$ is added under stirring. If the copper nitrate doesn't dissolve properly, add water dropwise until it does. (NOTE: Adding water to concentrated acids usually isn't the best idea!) The solution is stirred for 24 hrs at room temperature and then heated to roughly 80 °C. Under stirring, 120 g of 95% ethanol are added. After addition of the ethanol, the heat source is removed and the mixture is alllowed to reach room temperature. The precipitate is rapidly filtered off, washed with several hundred mL od cold water and dried in a desiccator over magnesium sulfate. The product, copper fulminate, should be stored in a desiccator over sodium sulfate in a refrigerator.

NOTE

To my opinion, the procedure leaves out significant information, such as the exact temperature of the aqueous solution while adding the ethanol. It is not clear whether and how much ethanol is supposed to evaporate. While the author suggests to carry out the reaction in a beaker, a setup with a two-neck flask, reflux condenser and a dropping funnel might be a better idea.
Handling fairly concentrated nitric acid requires safety precautions: A fume hood, eye protection, gloves and a lab coat are mandatory!
The reaction product is a primary explosive! While less shock sensitive than mercury fulminate, it still will go off on impact (or upon heating above ~200 °C).

In the case of these or similar sensitive materials, there is a good rule of thumb: If you have to ask on a Q&A site - Don't touch it!


Two reactions which speed each other up Recently I have started looking for two reactions that would speed each other up. That means that the products of the first reaction accelerate the second one, and in turn, the products formed in the second reaction accelerate the first one. Do you know any examples of such reactions?

Have a look at some non-linear reaction schemes such as gas phase radical chain reactions that may lead to explosion. The propagation steps in the hydrogen/bromine gas phase reaction do what you suggest:

$$\begin{align}
\ce{H^. + Br2 &-> HBr + Br^.} \\
\ce{Br^. +H2 &-> HBr + H^.}
\end{align}$$


Buffer Capacity Calculation I know buffer capacity is the following:
$$β=\frac{Δ(\ce{H+})}{Δ(\mathrm{pH})}$$ specifically the amount of acid/base that needs to be added to change pH by 1 unit.

If I have data about how pH of a protein has changed upon adding specific amounts of acid, how do I calculate buffer capacity?
Is there any reason why the change in pH needs to be 1? If I calculate how much acid needs to be added for the pH to change by 1.1, can this be scaled to determine amount of acid needed to change pH by 1?

The buffer capacity of a weak acid-conjugate base buffer is defined as the number of moles of strong acid needed to change the $\ce{pH}$ by 1 unit.
$$\beta = \frac{\mathrm{d}[A]}{\mathrm{dpH}} $$
and the acid is present as $$[A]= \frac{K_\mathrm{w}}{[\ce{H+}]}-[\ce{H+}] +\frac{C_\mathrm{B}K_\mathrm{a}}{[\ce{H+}]+K_\mathrm{a}}$$ where $K_\mathrm{w}$ is the water ionization equilibrium constant, $10^{-14} $, $K_\mathrm{a}$ is the acid dissociation constant, and $C_\mathrm{B}$ is the total concentration of buffer.
You assume that your protein is the weak acid. Change $\ce{pH}$ to $\ce{pH}=-\log_{10}([\ce{H+}])$ to differentiate or use the product rule. You should then find that $$\beta = 2.303\left[ \frac{k_\mathrm{w}}{[\ce{H+}]}+[\ce{H+}] +\frac{C_\mathrm{B}K_\mathrm{a}[\ce{H+}]}{([\ce{H+}]+\ce{K_\mathrm{a}})^2}\right]$$
you can then plot $\beta$ vs $\ce{pH}$ and from this you should be able to find what you want.


Competing resonance and inductive effects in a substituted benzene A molecule of phenol is much more inclined to under go an electrophilic substitution reaction than a molecule of benzene because the $\ce{-OH}$ group is highly reaction favoring. From what I understand, this is because when the $\ce{-OH}$ group is attached, it contributes a resonance structure involving a double bond between the oxygen and the ring, which stabilizes the intermediate arenium ion.

However, from what I also understand, oxygen is also a highly electronegative atom and therefore inductively draws electron density away from the benzene ring. Despite these two competing effects (electron donating via resonance and electron withdrawing via induction), the overall effect of the $\ce{-OH}$ group is electron donating. This also applies to $\ce{-NH2}$ group.

Am I right in concluding that resonance effects are stronger than inductive effects, if not in general, then at least for electrophilic substitution of substituted aryls?

Often, but not always, mesomeric displacement leads to a shift in prevailing over the inductive effect.

Pi-electrons are at the peripheral orbitals. Association with the nucleus is less strong than at the sigma-electrons. The ionization potential of pi-electrons is smaller and chemical bond is more polarizable. Therefore, the dipole moment associated with the mesomeric effect can prevail over the dipole moment associated with the inductive effect.

However, there are groups in which the inductive effect prevails, such as halogens.

Also, you can read about hyperconjugation. Hyperconjugation can have influence to the mesomeric and inductive effects too.

Source: Reutov  O.A., Kurts A.L., Organic Chemistry, vol. 1 of 4, ch. 2.2.2, MSU


Why does the rate of reaction increase less than proportionally to the increase in mass of catalyst? I conducted an experiment where I wanted to speed up the decomposition of hydrogen peroxide with the use of a catalyst (potassium iodide).

I used 5 different test tubes with 5ml of hydrogen peroxide in each, then added different amounts of the potassium iodide: 0.2g, 0.4g, 0.6g, 0.8g, 1.0g. Then I measured the time it took for the reaction to reach a temperature of 30°C. The results I got was that the more of the catalyst I added, the faster the rate of the reaction became. However, the increase in the reaction rate decreased when more of the catalyst I added, and I don't understand why.

I would think that since temperature and amount of hydrogen peroxide are constant variables and the only variable affecting the rate is the amount of potassium iodide, if I doubled the amount the speed would also be doubled. But why is that not the case? Or does it have something to do with errors in the method?

There is a lot going on here. Part of the problem is poor experimental design. I'll just throw this out without evaluating all the energies. (Too lazy...)

(1) The potassium iodide has to be dissolved into the solution since you added it as a solid. This means there is an energy change and this could be endothermic or exothermic. Adding more salt increases the effect obviously.

(2) It takes some time to dissolve the potassium iodide. So before all the potassium iodide is dissolve some is already reacting with the hydrogen peroxide.

Greater surface area (smaller particles of potassium iodide) would allow the crystals to dissolve faster.

(3) The reactions are:
$\ce{H2O2(aq) + I-(aq) = H2O(l) + OI-(aq)}$
$\ce{H2O2(aq) + OI-(aq)= H2O(l) + O2(g) + I-(aq)}$

Overall the reaction is:
$\ce{2H2O2(aq) =  2H2O(l) + O2(g) + \text{heat}}$

But the heat is the combination of the two reactions.

If $\ce{[H2O2](aq) >> [I-](aq)}$ then both reactions will happen.

If $\ce{[H2O2](aq) << [I-](aq)}$ then only the first reaction will happen. The iodide isn't acting as a catalyst but as a reactant.

3.3 % Hydrogen peroxide is

$0.033 \cdot 15 \approx 0.5$ g $\ce{H2O2}$

Moles $\ce{H2O2}$ = $\dfrac{0.5}{34} = 0.015 \text{moles} = 15 \text{millimoles (mM)}$

0.2 g KI = $\dfrac{0.2}{166}$ = 1.2 mM

1.0 g KI = 6.0 mM

So you're adding so much KI that some might get left as $\ce{OI-}$.

(4) so to do the experiment better:

cut the KI concentration
dissolve the KI in 5 ml of water & equilibrate  temp to that of 3.3% H2O2
mix solutions with stirring


Empirical and Molecular Formulae I'm not a chemist, and nor do I study chemistry so please try to be gentle.

Glucose has the molecular formula $\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6$ and the empirical formula $\mathrm{C}\mathrm{H}_{2}\mathrm{O}$.

Starting from the empirical formula and working the other way, am I guaranteed that $\mathrm{C}_n\mathrm{H}_{2n}\mathrm{O}_n$, where $n$ is a positive whole number, are all going to be well-define molecular formulae?

Some examples are Formaldehyde ($n=1$), Acetic Acid ($n=2$) and Ribose ($n=5$). As a starting point, does $n=3$ and $n=4$ make sense, i.e. $\mathrm{C}_3\mathrm{H}_{6}\mathrm{O}_3$ and $\mathrm{C}_4\mathrm{H}_{8}\mathrm{O}_4$ respectively?

Yes they make sense but the same molecular formula can represent different compounds, for the first case $\ce{C3H6O3}$ may rappresent:
Glyceraldehyde:
or lactic acid (and its optical isomers):
$\ce{C4H8O4}$ leads to even more possibilities see the three different tetrose

Generally these compounds may be created when the sum of the formal charges of each atom in the molecular formula is equal to zero. Oxygen, hydrogen normally have formal charge of -2 and -1 carbon +4 so:
$$-2\times n + (-1 \times n \times 2) + (+4 \times n)=-4\times n +4\times n=0$$


Precipitation with NaI I dissolved $\ce{[Pt(NH3)4](NO3)2}$ in water then I added small amount of $\ce{NaI}$ ($1~\mathrm{M}$) to it. Precipitation occurred. I want to know what happened?

The original solution contains the square planar $\ce{Pt(II)}$ complex ion $\ce{[Pt(NH3)4]^2+}$. When sodium iodide is added to the solution, one or more $\ce{NH3}$ molecules are replaced by $\ce{I-}$ ions in a ligand exchange reaction. In particular, replacing two $\ce{NH3}$ molecules with $\ce{I-}$ ions yields the neutral complex $\ce{[PtI2(NH3)2]}$, which is poorly soluble in water, and therefore a precipitate is formed.

Remarkably, two isomers of square planar $\ce{[PtI2(NH3)2]}$ exist: $\ce{$cis${-}[PtI2(NH3)2]}$ and $\ce{$trans${-}[PtI2(NH3)2]}$.
When starting from $\ce{[Pt(NH3)4]^2+}$, the first $\ce{I-}$ can replace any of the four equivalent $\ce{NH3}$ ligands. Because of the larger trans effect of $\ce{I-}$ compared to $\ce{NH3}$, the second $\ce{NH3}$ is preferably replaced trans to the first $\ce{I-}$ ligand and therefore the trans product $\ce{$trans${-}[PtI2(NH3)2]}$ is obtained.
(Note that, if one starts from $\ce{[PtI4]^2- }$ and replaces two $\ce{I-}$ ligands with $\ce{NH3}$, the cis product $\ce{$cis${-}[PtI2(NH3)2]}$ is obtained instead.)


Could one synthesise unsubstituted "cyclohexa-1,3,5-triene"? Is the cyclohexa-1,3,5-triene (non-aromatic benzene) possible to make or will it destabilize (or should I say stabilize?) into benzene?

Cyclohexa-1,3,5-triene is alternative name of benzene better describing its hypothetical, non-aromatic version. However they are similar and I was wondering if you made cyclohexa-1,3,5-triene would the electrons delocalise to form benzene or not?

You cannot make (= synthesise) isolated, nonaromatic cyclohexa-1,3,5-triene[1] much like you cannot make (= synthesise) pseudoaromatic, biradicallic cyclobuta-1,3-diene. Hypothetical unsubstituted cyclohexa-1,3,5-triene will immediately give benzene, as this answer demonstrates.

The only way to actually generate nonaromatic cyclohexa-1,3,5-triene is to apply an intramolecular constraint that disallows the aromatization. Since cyclobuta-1,3-diene is the most obvious antiaromatic, planar system, that will attempt to always adopt a localised double bond structure, structure 1 below immediately suggests itself as a model system for the localised cyclotriene.

Scheme 1: Diercks’ and Vollhardt’s synthesis of a cyclohexa-1,3,5-triene system 1.

Indeed, that was the approach that Diercks and Vollhardt used to synthesise a model of cyclohexa-1,3,5-triene.[2] They report a crystal structure of 4 ($\ce{R}= \ce{TMS}$) wherein the alternating single and double bonds are clearly seen. The cyclobutyl sub-species adopts a cyclobutene-structure, meaning that the central cyclohexatriene ring’s double bonds are localised between the cyclobutane cores as indicated by the structure drawn. The outer benzyl rings are relatively aromatic (but not perfect hexagons).

Note and reference:

[1]: IUPAC nomenclature would allow systematically naming benzene as cyclohexa-1,3,5-triene. This is because IUPAC nomenclature assumes no knowledge of the chemistry of a given structure and requires names to be applicable ab initio and a priori. However, henceforth in this answer I shall make a distinction of (nonaromatic, localised, $D_\mathrm{3h}$-symmetric) cyclohexa-1,3,5-triene and (aromatic, $D_\mathrm{6h}$-symmetric) benzene.

[2]: R. Diercks, K. P. C. Vollhardt, J. Am. Chem. Soc. 1986, 108, 3150. DOI: 10.1021/ja00271a080.


Ketones and nucleophilicity Is there anyone who can give me a convincing argument about the ketone, or the carbonyl oxygen in the ketone being nucleophile or not? . I really don't know what to write anymore. Everyone says something different. My instructor said that the Ketone oxygen can be protonated, thus reacting as a nucleophile. Others says that  Protonating the ketone in fact makes it more electrophilic.

Here is the molecule I'm working with (attached) and I have to mention all the nucleophilic sites - I've done that. This is the only one that really is confusing me. Hoping for some clarification.

http://imgur.com/a/20NAw

Everyone says something different

Well, actually they don't. A ketone has a nucleophilic and an electrophilic center. The carbonyl carbon is electrophilic and can be attacked by nucleophiles, the oxygen itself can act as a nucleophile and can, for example, be protonated. And yes, if we protonate the oxygen the "ketone", or better the carbonyl carbon, get's more electrophilic.

An example would be the Luche reduction. In this case the carbonyl oxygen coordinates to cerium chloride, thus acting as nucleophile, and makes the carbonyl carbon even more reactive towards nucleophiles.


Reaction of an electrode Why is the cathodic reaction of $I^-/AgI/Ag$ is $$\ce{AgI + e-  -> Ag + I-}$$ but not,
$$\ce{Ag+ + e- -> Ag }$$

whereas, in $Ag/AgCl(s)$, $KCl$(0.2M); it is
$$\ce{Ag+ + e- -> Ag}$$

The real reaction is $\ce{Ag+ + e- -> Ag}$ for both cases. Equation written in this way, ${E^0}$ will be equal to the reduction potential of Ag(I) in its 1 M solution.

If it is written as $\ce{AgI + e- -> Ag + I-}$, then ${E^0}$ will be equal to the reduction potential of Ag(I) in 1 M $\ce{I-}$ solution -or in other terms in ${K_{sp}(AgI)}$ Molar Ag(I). solution

Hence, technically, $\ce{Ag+ + e- -> Ag}$ is same with $\ce{AgI + e- -> Ag + I-}$. Latter is more descriptive and practically more relevant, however, because the presence of $\ce{I- /Cl-}$ ions in the medium limits the amount of $\ce{Ag+}$ cations via ${K_{sp}}$ and the notation readily implies this.


Mismatch of plane of symmetry vs Gold Book's definition for a pseudochiral carbon Here is pentane-2,3,4-triol:

At first glance, without deciding the configurations at each chiral carbon, we can clearly see that this molecule has a plane of symmetry, perpendicular to the plane of paper and passing through the $\ce{-OH}$ and $\ce{-H}$ on the 3rd carbon atom.

However, if we then sit down to compute the R/S configuration at carbons 2 and 4, we find them to be R and S respectively. Thus, going by the definition of a pseudo-chiral carbon, given in Gold Book as:

  a tetrahedrally coordinated carbon atom bonded to four different entities, two and only two of which have the same constitution but opposite chirality sense.

we would infer that the 3rd carbon is indeed pseudo-chiral. This is confirmed by ChemSketch which generates the name of the above compound as "(2R,3s,4S)-pentane-2,3,4-triol".

Thus, we have observed that using the plane of symmetry to check chirality, we end up declaring the 3rd carbon to be achiral, rendering the molecule optically inactive. However, if we use the Gold Book's definition, we end up declaring the 3rd carbon to be pseudo-chiral, rendering the molecule optically active.

Another fact confirming that the molecule is optically active is that there exist two isomers, of two opposite configurations of the central carbon atom -  (2R,3s,4S)-pentane-2,3,4-triol and (2R,3r,4S)-pentane-2,3,4-triol. Here are their structures:

I have looked at several questions on pseudochirality here, yet I have been unable to resolve this conundrum. I believe I am using a wrong fact or a misguided inference, but whatever is it, I am unable to grasp it, and thus need help resolving the conundrum.

The Gold Book definition for a meso compound is

  the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members.

So, a meso compound has chiral subsets. Here we would have three such subsets:

Leftmost carbon (2R)
The centermost (pseudochiral 3S)
The rightmost (4S) carbon

Even though  the centermost carbon is pseudo chiral, the plane of symmetry still exists. The compound remains meso, achiral.

Nomenclature is just for naming a compound.

The following two molecules are diastereomers, not enantiomers.


Rationale for assigning antibonding MOs How do people decide if a molecular orbital (MO) is bonding or antibonding? From Wikipedia article on antibonding MO:

  [...] if the antibonding interactions outnumber the bonding interactions, the molecular orbital is said to be antibonding

and vice versa.

How do we define what an "antibonding interaction" is for cyclic molecules? Does one nodal plane count as 1 or 2 antibonding interactions? For example, the π2 molecular orbital of benzene may have 1 nodal plane, but how many antibonding interactions would that MO have?

Similarly, how do we define "bonding interactions" for cyclic molecules?

There are a couple ways that people define bonding and antibonding orbitals. One of the simplest is to just look at the energy of the MOs formed compared to the AOs they were formed from: if the MO is lower in energy, it is bonding and if the MO is higher in energy, it is antibonding.

Like I said, this way of doing thing is relatively simplistic, but you will see it pretty commonly in textbooks on the matter. The method of course isn't perfect, in particular because it essentially assumes that all local bonds are affected equally by a particular orbital.

We can see this is not the case by looking at the $\pi_2$ orbital you show. While the orbital is listed as bonding, the interactions of the p-orbitals where a positive lobe is next to a negative love will locally decrease the bonding between those two atoms (ie the MO is antibonding with respect to those local bonds).

So, for a simple first approximation, you can use the energy of the MO compared to the AO, but for a more complete sense, you should look at the symmetry of the MO and see what effect it has on a given local bond.

In terms of the ordering of the MOs themselves, assuming you don't have experimental way to determine the ordering like ionization energies, you can heuristically use an approach like you suggest where you compare the number of nodes and less nodes will mean lower in energy, more nodes will be higher.

In terms of comparing to the original AOs, it can be tricky without experimental data, however there are again general qualitative ways of ordering the MOs with respect to their original AOs. For example, if you have 2 MOs formed from two overlapping AOs, one will have to be higher in energy and the other will have to be lower.

When you get more orbitals overlapping, the arrangements can vary, but the sum of the energies has to remain the same. So, for example, if you have 3 overlapping AOs, you could form 2 bonding and 1 antibonding MOs or vis versa so long as their combined energy is the same as before.

Beyond this, you really just have to read up on MO theory to get a sense of how qualitative energy ordering is done. There is a lot more to it than what I've written (or could write) and it's probably better to build your understanding of the theory from the ground up rather than getting it piecemeal.


What is the electron dot structure of an alkyne? I am trying to determine the electron dot structure of an alkyne like $\ce{C2H2}$.

I know that hydrocarbons with at least one carbon-carbon triple bond are called alkynes, and their general formula is $\ce{C_{n}H_{2n-2}}$, where $n$ is number of carbon atoms.

So what will the electron dot structure of $\ce{C2H2}$ look like?

Carbon needs four electrons to fill its octet.  In a $\ce{C-H}$ covalent bond carbon shares one electron with hydrogen.  A triple consists of one σ covalent bond and two π covalent bonds.  The carbons share two electrons in each of these bonds; in total six electrons are shared between them.  You can draw this as a dot structure:

$$\ce{H:C:::C:H}$$


Throwing water into lava While discussing magma in class with my students, I showed a video of a man throwing water onto molten, running lava.  A large flame ensued.  I realize that the water is being converted into steam and expanding extremely rapidly, but why does this create a small "explosion," if you will?

Surely none of the lava is being "atomized" into the air in small droplets, is it?  But even so, they would not light aflame.

This is a Steam explosion generally the flame is due to the hydrogen produced when water is superheated so thermal decomposed into $\ce{H2}$ and $\ce{O2}$.

Quoting wikipedia:

  Steam explosions are not normally chemical explosions, although a
  number of substances react chemically with steam (for example,
  zirconium and superheated graphite react with steam and air
  respectively to give off hydrogen, which burns violently in air) so
  that chemical explosions and fires may follow.


Why does a bonding molecular orbtial increase stability? When two atomic orbitals come together, they fuse to form a molecular orbital. There is a large probability that the electrons will spend time between the nuclei. This will reduce the nuclei - nuclei repulsion and hence increase stability.

The part that I do not get is how do the electrons reduce nuclei-nuclei repulsion? Is it because the electrons in the middle attract the positive nuclei. This creates an opposing force to the electrostatic repulsion? Is this the case and if not what is actually causing the stability?

The forces involved are electrostatic, nuclei repelling one another and electrons similarly, but also being attracted to the opposite nuclear charges. Also there is kinetic energy of the electrons (for simplicity we consider the nuclei to be stationary) which form the total energy of the molecule.

The simplest molecule to consider is $H_2^+$. The interactions have potential energy arising from Coulomb interactions $$\displaystyle \displaystyle V(r) = -\frac{e^2}{r_1}-\frac{e^2}{r_2}+\frac{e^2}{R}$$

where $r_1,r_2$ are the separation of the electron from the nuclei and $R$ the separation of the nuclei themselves. In addition there is kinetic energy to consider and this has the form $p^2/2m$ where $p$ is the momentum (operator). Combining these terms the Schroedinger equation is formed and in doing so some integrals have to be solved.

Some of these have classical counterparts and are Coulomb type integrals, but one type, the 'resonance' or 'exchange' integral, is purely quantum mechanical. It is difficult to describe in words but it expresses the possibility of the electron moving from the vicinity of one proton to that of the other. All these integrals are positive but add together according to the signs in the potential.  The upshot of this is that the total energy reaches a (negative) minimum as the separation between nuclei is increased from a very small value, and becomes zero when the nuclei are well separated.

The electron has kinetic energy as well as potential energy so besides the potential energy effect of reducing the repulsive interaction between nuclei, by virtue of having increased density between them, the positive kinetic energy can counteract this effect.

To find out which is most important the Virial Theorem (as result from classical mechanics) can be used. In the special case of a diatomic molecule the average kinetic energy is given by $\langle T\rangle=-E-R(dE/dR)$ and the potential energy $\langle V\rangle=2E+R(dE/dR)$ where $E$ is the total energy at $R$ the internuclear separation.

The plot below shows that it is the minimum in the potential energy that caused the chemical bond to be formed not the smaller minimum in the kinetic energy which is positive anyway. Thus in forming a bond the kinetic energy increases (destabilising) but the potential energy decreases by more and this stabilises the bond.  This is shown below where $\rho$ is bond length in reduced units and $E_I$ is the ionisation energy. $E_0$ is the minimum total energy.


Bicarbonate decomposition and pKa I want to understand why the p$K_{\mathrm a}$ of $\ce{HCO3-}$ is approximately 10.2 and not less.

My hypothesis is this :

Bicarbonate could react in two ways in an aqueous solution:

$[1]\qquad\ce{HCO3- + H2O -> CO3^2- + H3O+}$

$[2]\qquad\ce{HCO3- + H2O -> H2CO3 + OH-}$

Since $\ce{H2CO3}$ is more stable than $\ce{CO3^2-}$ (because it has a lower charge and all the octets are completed), the p$K_{\mathrm a}$ must be high (more alkaline than acid).

In [1], the equilibrium will be shifted to the left and in [2] the equilibrium will be shifted to the right.

Is my line of thought correct?

Carbonic acid is not necessarily more stable than the carbonate ion.

Why carbonic acid ( p$\ce{Ka = 6.3}$ )  is not stable?

The reason for this is same as the one given for the instability of aluminium carbonate, iron (III) carbonate etc. That is, the carbonate ion has a highly polarisable electron cloud, which can be distorted by densely positively-charged species, such as the hydrogen ion, in this case. Thus, carbonic acid is not a very stable species and compared to the bicarbonate ion, which only has one densely-charged hydrogen ion distorting its electron cloud, it is definitely more unstable. Thus, the position of the 2nd equilibrium mentioned in the question should be to the left, favouring the more stable species.

Why is the bicarbonate ion ( p$\ce{Ka = 10.3}$ ) more stable than the carbonate ion  ?

This is due to there being a smaller negative charge on the bicarbonate ion and there being still 2 equivalent resonance structures to spread out the negative charge. Compared to the carbonate ion, which has twice the charge density but yet only 1 more resonance structure, the bicarbonate ion is more stable. Therefore, the position of equilibrium in the first reaction mentioned in the question should be to the left, not to the right.

Therefore, in solution, the dominant species among the three, should be the bicarbonate ion. This should be the correct line of thought.


Changes in pH when adding substances other than acids or bases If I start out with 100 mM acetic acid, what substance (to a final concentration of 1 M) could I add to change the pH considerably? I was thinking of NaCl or ethanol or acetone. Substances added should not be acids or bases, i.e. a 1 M solution of them should have a pH close to neutral. I'm asking because I am trying to create a demonstration on how pH calculations are just approximations under non-ideal conditions.

The acetic acid should not undergo an chemical reaction (other than acid/base). I would use a pH meter to monitor the pH. I would have a control where I increase the volume to match the experiment where I add a substance that influences the pH (or dissolve the substance in 100 mM acetic acid to keep the acetic acid concentration constant).

I am looking for changes in pH based on non-ideal behavior. The concentration of acetic acid should remain the same, and there should not be any neutralization reaction. In other words, the calculate pH should stay at 2.87 (assuming ideal conditions) while the measured pH would change.

You could use an alcohol or acetone but they are slightly acidic (though much less so than acetic acid; also alcohols react to form esters). Now effects from acidic nature can't completely avoided as even a molecule as non-acidic as methane still has a pka of 50. I think an aliphatic ether such as tetrahydrofuran (THF) or diethylether would be the best substance to add as the ethers are only very slightly basic and are miscible with water.

More simply:

  what substance could I add to change the pH considerably?

Why does it have to be a substance? The $K_\mathrm w = 10^{14}$ at $\pu{25^\circ C}$. If you add heat the $K_\mathrm w$ will increase and thus the pH will increase.


What salts are/are not birefringent? Is there a quick way to determine if a salt (or ionic compound in general) crystal should be birefringent from its formula without having to know/look up the crystal structure?  When I'm looking for protein crystals, if I know the condition has a lot of salt in it, but it shouldn't be birefringent (e.g. NaCl), it's a hint that birefringent crystals I do see might be protein.

Is the only non-birefringent crystal type (e.g. not glasses) cubic?  Does it matter if it's primitive, face-centered, or body-centered?

Back to looking for a generalization, are most (all?) salts with polyatomic ions birefringent?    All +1/-1 (NaCl, KBr, etc...) aren't?

Yes, the only crystal system that is guaranteed not to be birefringent is cubic, because all of the other crystal systems have at least one axis distinct from the others. What matters is the symmetry not the centring, so this is equally true for face-centred or body-centred as for primitive crystals.

Sadly, however, there is no way to determine the crystal structure, or even the crystal system, simply from consideration of the formula - indeed, crystal structure prediction and prediction of polymorphism are active research fields. Counterexamples to your first generalisation include ammonium chloride, bromide, and iodide; to your second include ammonium fluoride and silver iodide.


Why should the redox reaction happen between the copper ions and iron atoms? I know that when $\ce{Cu^{2+}}$ ions react to $\ce{Fe}$ atoms, it becomes $\ce{Cu}$ + $\ce{Fe^{2+}}$. My question is, why?
Copper atoms normally have 29 electrons. This means that in the K-shell there are 2 electrons, in $L = 8$ electrons, in $M = 18$ electrons and in $N = 1$ electron. (Yes I know that there is a system with s and other letters, but I haven't been taught in that way at my school). Therefore $\ce{Cu^{2+}}$ should have 27 electrons, which means that the shells are assigned like this: $K = 2, L = 8, M = 17$.
So when the $\ce{Cu^{2+}}$ ions react with $\ce{Fe}$, why does it want to become $\ce{Cu}$? Because $\ce{Cu}$ doesn't have a complete electron shell at the end. Why doesn't it want to become $\ce{Cu+}$, because then it would have a complete shell at the end ($K=2, L=8, M=18$).
I know that when you take away electrons from copper, it does it in a special way. I'm 15 years old and not english so I'm really not an expert or anything, I'm just wondering why $\ce{Cu^{2+}}$ would want to become $\ce{Cu}$ when it meets iron ($\ce{Fe}$).

This is a good question, but you may have gotten off on the wrong track by thinking about the electronic shell structure (although what you have written seems correct, if a bit old-fashioned).  The relevant phenomenon here is the relative abilities of the metal ions to oxidize (or remove electrons from) each other.  For your specific example, the copper ion is a better oxidizing agent than the iron atom.  Therefore it is able to take two electrons away from the iron atom, oxidizing it and leaving the iron 2+ ion behind.

As for why this happens, at your current stage of education the best answer is probably "because iron is easier to oxidize than copper".  You can look up the electrochemical activity series to find out how various metals relate to each other in terms of how easy they are to oxidize.  Copper is one of the harder ones to oxidize, but if you were to mix copper metal with silver ions, you would end up with silver metal and copper ions, because silver is even harder to oxidize than copper.

Does that help?


What signal is measured at the detector in atomic absorption spectroscopy? I understand that a hollow-cathode lamp (HCL) with a specific target is used to excite the sample. The atoms in the sample absorbs the photons and re-emits at the same wavelength and the signals are measured by the detector.
What about the signal from the source? Let's say only 20% of the energy from the source are being absorbed, wouldn't the other 80% also travel through the flame to the detector?

You raised an excellent point, which is not as simple a UV-Vis spectrophotometry because most molecules do no fluoresce or emit light, whereas atoms do. This issue makes an atomic absorption experiment slightly more challenging (at least in principle). The critical question is "what" signal reaches the detector. How does one distinguish the contribution of flame emission from the hollow cathode lamp (HCL) beam? Last but not least, is there is a way to distinguish atomic emission due to HCL?
Have a look at a typical AAS instrument. The critical component is the chopper, which is nothing but a rotating mirror with holes. It merely blocks the light from reaching the flame while rotating at an exact frequency.

You can see only two signals are needed in an atomic absorption spectrophotometry experiment: a) Light beam intensity when there are no atoms in the flame and b) light beam intensity when analyte atoms are present in the flame.

You aspirate a blank (no analyte solution) into the flame- this sets the 100% transmission or absorbance = 0 at the chosen wavelength. Ignore the reference beam; it is for ensuring that the lamp is stable. We are only interested in the beam passing through the flame.

Next, you would aspirate your analyte solution into the flame. Let us take Na as the most straightforward example. There is light absorption by Na atoms in the flame, leading to a reduction in the HCL beam's light intensity after passing through the flame. The ratio of light intensity in steps 2 and 1 gives the transmittance, which is easily converted into absorbance.

However, there is an almost continuous emission spectrum from the flame too.
Here is the smart idea. I do not know who came up with it. The detector continuously sees the background emission from the flame. However, the light beam from the hollow cathode lamp reaches the detector alternately at a particular frequency due to the rotating chopper. The electronics are tuned to register only the light, which is oscillating due to the chopper

The critical but no so critical problem: When atoms absorb a specific wavelength, they emit precisely the same wavelength as you noticed in your post in the flame (called atomic fluorescence). the chopper cannot stop that light. Hence it is part of the transmitted light beam.

The good news is that atomic fluorescence is not directional, unlike the light beam from the HCL. Secondly, the intensity of atomic fluorescence light is so small compared to the HCL beam intensity that it is practically negligible. There is no way to prevent atomic fluorescence from reaching the detector in an atomic absorption experiment, to the best of my knowledge.
Can we see stars during daylight? Stars are always there and "part" of the sunlight. However, stars contribute very little to the daylight. In the same way, atomic fluorescence contribution in an atomic absorption experiment is almost null because the hollow cathode lamp beam is quite strong as compared to the atomic fluorescence in random directions.


Why we should concern with A∞? To determine the order of one reactant at pseudo concentration of another reactant the plot of $\ln(A_t - A_∞)$ versus $t$ is used.
$A_t$ is absorbance of the intermediate at time $t$.
$A_∞$ is absorbance of intermediate at the end of the reaction (approximately after 24 hours).

Why $A_∞$ is important here?

Let's suppose a hypothetical irreversible reaction:
$$\ce{B + H -> D + F}$$
where H is a reactant with constant concentration during the reaction and that don't absorb light in the working frequencies. The absorbance of a species $i$ related with the concentration by the Lambert-Beer Law:
$$A_{i}=c_{i}\varepsilon_{i}l$$
where $c_i$ is the concentration of species $i$, $\varepsilon_{i}$ is the molar absorptivity and $l$ is the length of the cuvette.

At an arbitrary time $t$, the total absorbance of the solution is given by:

$$A_{t}=c_{B,t}\varepsilon_{B}l+c_{D,t}\varepsilon_{D}l+c_{F,t}\varepsilon_{F}l
\tag{1}$$

by the stoichiometry of the reaction, we have: $c_{D,t}=c_{F,t}=c_{B,0}-c_{B,t}$; where $c_{B,0}$ is the initial concentration of the reactant B, so replacing this in Eq. 1 and defining a auxiliary constant $\varepsilon_{i}l=k_{i}$, we have:

$$A_{t}=c_{B,t}k_{B}+k_{D}(c_{B,0}-c_{B,t})+k_{F}(c_{B,0}-c_{B,t})$$
$$A_{t}=c_{B,t}(k_{B}-k_{D}-k_{F})+c_{B,0}(k_{D}+k_{F})\tag{2}$$

At a time $t\rightarrow \infty$  (a time long enough that the reaction goes to completion), $c_{B,\infty}=0$ and $c_{D,\infty}=c_{F,\infty}=c_{B,0}$, so the absorbance is given by:

$$A_{\infty}=c_{D,\infty}k_{D}+c_{F,\infty}k_{F}$$
$$\boxed{A_{\infty}=c_{B,0}(k_{D}+k_{F})} \tag{3}$$

so,replacing Eq.3 in Eq.2 its follows that:

$$A_{t}=c_{B,t}(k_{B}-k_{D}-k_{F})+A_{\infty}$$
$$\boxed{A_{t}-A_{\infty}=c_{B,t}(k_{B}-k_{D}-k_{F})}$$

In conclusion: $A_{\infty}$ is important because is directly proportional to the initial concentration of reactant B. When you subtract $A_{\infty}$ from the absorbance at an arbitrary time $t$ $(A_{\infty})$ it allows you to know (indirectly) the concentration of the reactant at the aforementioned time $t$.


A way to turn KCl into KOH is needed I figured out that KCl + NaOH may result in KOH but I think this reaction can not happen in an aqueous medium. So there I have a couple of questions.

Is there a way to make the reaction I stated above to happen?
Is there a way to turn KCl into KOH?

On an industrial scale, the preferred method for the production of potassium hydroxide from potassium chloride is electrolysis (diaphragm, mercury, or membrane process).

$$\begin{align}
\ce{2Cl- &-> Cl2 + 2e-}\\
\ce{2H2O + 2e- &-> H2 + 2OH-}\\
\hline
\ce{2Cl- + 2H2O &-> Cl2 + H2 + 2OH-}
\end{align}$$

or

$$\ce{2KCl + 2H2O -> 2KOH + Cl2 + H2}$$


How to predict the reason for color in a transition metal complex? I know about two reasons for the color of transition metal complexes that are dd transition and charge transfer phenomenon but I am confused when a complex can show both of the above-mentioned phenomena then how to choose which can be the dominating reason for the color of the complex.

How to predict the reason for color in a transition metal complex?

Nobody can do it just by "looking" at the color of any compound. Spectroscopists have to do calculations to figure that out. There are at least 15 reasons for the "color" of chemical compounds.
Now qualitatively one can "predict" whether it is a d-d vs. it is charge transfer transition for transition metal compounds.
Keep in mind that d-d transitions are "forbidden" (Laporte selection rule). Therefore, majority of the d-d transitions correspond to low molar absorptivity. For example, Cu(II) solution is light blue, manganese (II) is almost invisible pink, chromium (III) is light green.
Charge transfer transitions are very intense! Add two three crystals of potassium permanganate to water, and it is intensely purple.


Which is more acidic BF3 or AlCl3? Which of the species $\ce{BF3}$ and $\ce{AlCl3}$ is more acidic, both are electron deficient.

In case of $\ce{BF3}$ one can argue that there is extensive p-π-p-π back bonding and as a result the electron deficiency is less.
That means that the positive charge on the overall molecule is less.

In case of $\ce{AlCl3}$ one can argue there is p-π-d-π back bonding and as a result the electron deficiency is more in case of aluminiuim.
Therefore the overall positive charge on the molecule is more, meaning it is more acidic.

So $\ce{AlCl3}$ is more acidic that $\ce{BF3}$ is conclusion of the above factors. Is this plausible?

I also faced the counter logic that $\ce{BF3}$ being a small compound tends to react more readily and therefore is a strong acid.

Actually, there are three reasons why BF3 is more acidic than AlCl3.

BF3 is smaller in size and can easily attract the incoming pair of electrons.
Fluorine is more electronegative than chlorine. So boron has less ionisation potential as compare to aluminium thus forms anion easily.
Boron is more electronegative than aluminium, therefore boron has more tendency to attract electrons. It means that BF3 behaves more like Lewis acid than AlCl3.

This photo has been taken from this article.


Why don't ionic compounds show stereoisomerism? When I was reading about crystalline nature of ionic compounds, I came across the statement that ionic compound doesn't show stereoisomerism. What does that mean and can anybody explain the reason with an example?

They can, and they do.  The secret is to have a chiral cation and separately a chiral anion.  Letting $D^+$ be the dextrorotatory form of the cation, $L^+$ be the levorotatory form, and anologously for the anion, we then have four isomeric salts:

$D^+D^-$

$D^+L^-$

$L^+D^-$

$L^+L^-$

The first and second are diastereomers because only the anion is mirror-reflected; the cation is not mirror-reflected.  Similarly for three of the other five possible pairs; only two of the six possible pairs are enantiomeric.

Being diastereomers instead of enantiomers, the first and second salts, for instance, may have different physical properties such as solubility in a given solvent.  This may be used for separating different enantiomers of the anion (or, with a different pair such as first and third, separating different enantiomers of the cation).  And it's actually done.  See here for a brief summary and here for an example of diastereomeric salt formation.


How does Fe2+ become Fe in sacrificial protection? I've searched low and high on the internet, but nothing to help me.

I know that zinc is oxidised in preference to iron, and that it "donates" those two electrons to iron.
$$\ce{Zn -> Zn^2+ + 2e-}$$

But how does iron make use of those two electrons to revert back to iron when it turns into iron oxide, like so:

$$\ce{Fe^2+ + 2e- -> Fe}$$

Doesn't the $\ce{Fe^2+}$ in the $\ce{Fe2O3}$ need to be dissolved in water? How can it be affected by $\ce{2e-}$ travelling along? Maybe I am missing something.

I think the point that you're missing is that $\ce{Fe2O3}$ is never formed when there is $\ce{Zn}$ in the system. When the metal gets oxidized, the electrons come from the $\ce{Zn}$, not from the $\ce{Fe}$. This is true even if the physical location of the oxidation is on the iron, since the $\ce{Fe}$ and $\ce{Zn}$ are connected and conductive, so all of the electrons on both are shared.


How do I have to interpret the input to the OCC flag in Molpro? I'm trying to understand the occupied orbital specification in Molpro. They're specified by the OCC flag, while it's stated in the documentation, that

OCC, m1, m2, ..., mn;

where the mᵢ are the number of occupied orbitals (including core/frozen and closed).

Let's say I want to compute energies in $\ce{N2}$, which has $D_\mathrm{2h}$ symmetry point group, i.e. the following character table:
$$D_\mathrm{2h}\\
\begin{array}{ccc} \hline
\text{No.} & \text{Name} & \text{Function}\\
1& \mathrm{A_g}    & s  \\
2& \mathrm{B_{3u}} & x  \\
3& \mathrm{B_{2u}} & y  \\
4& \mathrm{B_{1g}} & xy \\
5& \mathrm{B_{1u}} & z \\
6& \mathrm{B_{2g}} & xz \\
7& \mathrm{B_{3g}} & yz \\
8& \mathrm{A_u}    & xyz \\
\hline
\end{array}
$$

Then, what exactly does the OCC specification in the following code mean?

{hf-scf;
 occ,3,1,1,0,2,0,0,0;
 wf,14,1,0;
}

In my understanding, it specifies:

3 orbitals of $\mathrm{A_g}$ symmetry, i.e. $s$ orbitals
1 orbital of $\mathrm{B_{3u}}$ symmetry, i.e. $p_x$ orbital
1 orbital of $\mathrm{B_{2u}}$ symmetry, i.e. $p_y$ orbital
2 orbitals of $\mathrm{B_{1u}}$ symmetry, i.e. $z$ orbital

What I find strange with this interpretation is the choice of orbitals - why would I choose 3 $s$ orbitals, while only one $p_x$ and 2 $p_z$?

You are correct in your interpretation of the OCC flag (that command corresponds to $3$ electrons in $\mathrm{a_g}$ orbitals, $1$ in $\mathrm{b_{3u}}$, $1$ in $\mathrm{b_{2u}}$, and $2$ in $\mathrm{b_{1u}}$). (Because these are orbitals, not overall wavefuction symmetries we use lowercase letters)

Your confusion seems to be arising from visualizing the orbitals which you are including. Remember that these representations refer to the symmetry of the molecular orbital, not atomic orbitals. $\mathrm{a_g}$ refers to the set of orbitals which have no phase change under all of the symmetry operations, so that corresponds to the $\sigma(\ce{1s})$, $\sigma(\ce{2s})$, and $\sigma(\ce{p_z})$, orbitals. Similarly, $\mathrm{b_{1u}}$ refers to the filled $\sigma^*(\ce{1s})$, $\sigma^*(\ce{2s})$. Finally the $\mathrm{b_{3u}}$, and $\mathrm{b_{2u}}$ orbitals refer to the filled $\pi(\ce{p_x})$ and $\pi(\ce{p_y})$ orbitals.

As we know, the occupation of $\ce{N2}$ is:
$$
  \ce{\sigma(1s)^2 \sigma(2s)^2 \sigma^*(1s)^2 \sigma^*(2s)^2
  \pi(p_x)^2 \pi(p_y)^2 \sigma(p_z)^2},
$$
and using the $D_\mathrm{2h}$ point group we can rewrite this as:
$$\mathrm{1a_g^2\ 2a_g^2\ 1b_{1u}^2 \ 2b_{1u}^2 \ 1b_{3u}^2\ 2b_{3u}^2\ 3a_g^2}.$$
Hopefully, this rewriting of the occupation pattern in the chosen symmetry makes the choice of orbitals clear.

To really understand which orbitals fall under which irreducible representation you need to visualize them (or draw them in the case of a dimer) and then pull out the character table for the point group.

You can find a $D_\mathrm{2h}$ character table on the homepage of Dr. Achim Gelessus (Jacobs University Bremen).


What are derived names? Q)The derived name of (CH3)4C is

A)Tetramethylmethane

B)2,2-Dimethylpropane

C)Neopentane

D)None of these

The answer was A.

What is a derived name?Never heard of it in any book.

Thanks

There are (were) three main nomenclature systems used for naming organic compounds:

Common Name (or trivial name). Under this convention this molecule is neopentane. For alkanes, this naming system is based on the positioning of methyl substituents, and gives rise to the iso-, sec-, neo- namiong prefixes.
Derived Name. Who uses this method these days???? This system identifies the central carbon atom which has the most carbon groups attached, and then names the groups as a prefix, and ends the name with 'methane'. The formula are often written with the central carbon unit in square brackets. You may still see derived names on MSDS/SDS under the 'Other Names' section. Under this system this molecule is tetramethylmethane. I can think of more important things to teach you in class if this is what you are learning.
IUPAC Name
This is the accepted naming convention in use today, which allows for retained names, which usually have their origins in the trivial naming system. Its naming system is based on a systematic framework and is used universally. In older texts (pre 1970s), it was known as the Geneva system or congress. Under this convention, this molecule is 2,2-dimethylpropane.

I can't even find any decent references for you at this point, but will update if I find some.


If the standard state symbol means that the substance is pure (and at 1 bar) how is it possible to have a standard REACTION enthalpy? For the reaction one mole of substance $A$ in equilibrium with one mole of substance $B$ the standard reaction enthalpy is defined:
$$\triangle_rH^{\theta}(T)=H_m^{\theta}(B;T)-H_m^{\theta}(A;T)$$
However, I have two issues with this.

Firstly: the standard state of a substance (denoted by the symbol $\theta$) is the pure material at a pressure of 1 bar. How can the reaction enthalpy be standard - the reactant have to be mixed (thus are not pure) to react.

Secondly: what is $H_m^{\theta}$? Surely nothing can have a molar enthalpy because only changes in enthalpy can be measured not absolute values. Am I right in thinking that is is because the enthalpy scale has no zero value?

The standard reaction enthalpy you have describe is only for a pure substance if you are talking about a standard enthalpy of formation.  Otherwise the equation wouldn't be valid for reactions that have more than one product, right?  In the specific case of a standard enthalpy of formation, you are talking about the enthalpy change associated with the creation of one mole of a pure substance from its component elements in their standard states.  This gets at your second question ... the zero on the enthalpy scale is established by convention (and this is the reason why the definition of standard enthalpy of formation works) ... the standard enthalpy of formation of an element in its standard state is DEFINED to be zero.

Now, you can estimate the standard reaction enthalpy for any chemical reaction by subtracting the summed standard enthalpies of formation of the reactants from the summed standard enthalpies of formation of the products, taking into account the appropriate stoichiometric coefficients from the balanced chemical equation.  That process is similar to what you have written, but not precisely the same, since you omitted the stoichiometry information (and the chemical equation, for that matter).  The standard enthalpy estimate obtained will pertain to reactions carried out at the given conditions .. standard enthalpies are defined for 1 atm pressure but variable temperature, so you may need to correct tabulated standard enthalpies of formation to match the given reaction conditions.

One last point is that enthalpy is absolutely defined as a thermodynamic quantity .. it is the internal energy plus the product of pressure and volume, $H=U + pV$.  The issue is that the internal energy of a substance is hard to define absolutely .. are you going to include the binding energies of the core electrons to their atoms?  How about the strong forces holding the nuclei together?  Thus for chemical processes it almost always makes sense to deal with just changes in enthalpy, since most of the really low level stuff I mentioned doesn't change between reactants and products.  I think that may have been part of what you were getting at with your second question.


Why does water cause a jet flame when poured over a bunch of tealights? A neat party trick (typically for outdoors) is to build a pyramid of tealights, let them burn for a while and then pour a bit of water over it and see a large jet flame shoot out (see e.g. this youtube video).

I was always told that this is caused by the water spreading the liquid wax out allow more of it to burn at the same time, but I am not completely satisfied with that answer because, in my opinion, that cannot explain the large jet flame. Can someone explain why this jet flame occurs? Is it the wax vapour that catches fire, which can form at 370 $^o\ce{C}$, and is this the reason you need to build a close packed pyramid? But what is the role of the water then?

The wax doesn't need to become gaseous to cause the column of flame (if that's what you were thinking). Water, when in contact with the liquid wax at a couple of hundred degrees above its boiling point, will flash-boil in a very chaotic manner as if it were "exploding" ("transition boiling" in this graph). The water closest to the wax very quickly vaporises and tries to escape, but is temporarily blocked by the layer of liquid water above it, so the system ends up being vigorously churned to make way for the vapour.

The problem is that the turmoil will make tiny liquid wax droplets airborne, as if wax were being shot out of an atomizer, creating a wax aerosol. Suddenly, the same amount of wax has its surface area increased tremendously and is exposed to much more oxygen. Even though the wax in the aerosol quickly loses heat and may become solid again in a fraction of a second, if even a small region of the wax aerosol reaches ignition temperature, the flame quickly propagates as a huge amount of heat is dumped at once in the vicinity and triggers the ignition of all the airborne wax.

This trick is effectively the same as what happens if you try to put out a grease fire with water.


Why does hot steam produce MgO instead of Mg(OH)2? Background research

Magnesium hydroxide exists naturally as brucite (source: wiki)
Passing hot steam over magnesium produces magnesium oxide (source: Edexcel's June 2014 Unit 2 exam (CH02_01): question 18)

Question

Why doesn't magnesium hydroxide form when treated with hot steam, instead of magnesium oxide?

There are two steps in the process:

$\ce{Mg + H2O -> MgO + H2}$

and

$\ce{MgO + H2O <-> Mg(OH)2}$

The equilibrium point in the second step is dependent on the temperature. See file Equilibrium Assemblages for more information.


Search by CAS Registry name to obtain trackable accession: ie InChI? I'm updating a database and trying to move from CAS Registry names to InChIKeys. I have about ~800 compounds to track. Is there a way to do this mappings programatically? I'm looking for a web service or API that already has this implemented, does something like it exist?

Chemical Identifier Resolver can convert several identifiers and has an API so you can convert with a script.

A Python example for CAS to InChIKey:

from urllib.request import urlopen

def CIRconvert(cas):
    try:
        url = 'http://cactus.nci.nih.gov/chemical/structure/' + cas + '/stdinchikey'
        ans = urlopen(url).read().decode('utf8')
        return ans
    except:
        return cas

casno = ['50-78-2', '69-72-7', '95-1E-6', '108-24-7', '31710-30-2']

for casid in casno:
    print(CIRconvert(casid))

Output:

InChIKey=BSYNRYMUTXBXSQ-UHFFFAOYSA-N
InChIKey=YGSDEFSMJLZEOE-UHFFFAOYSA-N
95-1E-6
InChIKey=WFDIJRYMOXRFFG-UHFFFAOYSA-N
InChIKey=CIPFDHFTBYJKQB-UHFFFAOYSA-N


What is a good (small) model system for ligand metal charge transfer practises? I am looking for an easy to calculate model system in which I could see a ligand metal charge transfer. Basically I am looking for a model complex, that might have the metal/ligand in different oxidation states: $\ce{ M^{III}(Lig)}$ versus $\ce{M^{II}(Lig^{\bullet +})}$ or $\ce{M^{IV}(Lig^{\bullet -})}$. I am calling it a charge transfer since I am not sure how else to call it.
I am also quite aware, that the description of these systems is necessarily a multireference case, so ordinary DFT methods will fail. (At least broken symmetry is necessary.)

In principle I was thinking about an iron heme complex that might exhibit that feature.

However, I have the feeling, that this kind of system is already pretty demanding to calculate the electronic structure. Since I would like to use it as an exercise and for a tutorial achieving alternative occupations I would prefer a much simpler case. It should eventually treated with broken symmetry DFT and CAS/MR methods.

Does anyone know of such a system or would suggest one. It does not have to be a real one. Purely hypothetical is just as fine.

AFAIK, permanganate ion is colored because of intense line, corresponding to oxygene-manganese charge transfer. This is noted in wikipedia article on charge-transer complex, but, unfortunately, without quote, and I'm unable to come with one right now

Edit:
Aha. Here there is a MO diagram on $\ce{MnO4-}$ ion and extensive discussion of (mis)understanding of its spectra, including statement I made above.


Are coordination numbers of elements fixed? Can we say that the coordination number of a particular element is fixed? If not please give an example where an element exhibits variable coordination numbers.

No. Coordination numbers of no element are fixed. For example, if you add chloride ions to a pink solution of octahedral $\ce{[Co(H2O)6]^3+}$, a blue, tetrahedral complex is generated according to the following equation:

$$\ce{[Co(H2O)6]^3+ + 4 Cl- <=>> [CoCl4]- + 6 H2O}$$

In fact, if you study the mechanisms of different metal-catalysed organic reactions, you will notice that different coordination numbers of the central metal and specific switching between them is the key to understand these mechanisms.

Finally, for some complexes even establishing their coordination is hard. Zinc(II) forms aquacomplexes like any other cation. But while those complexes of transition metals with any number of d-orbitals unequal to zero or ten are often well-defined and established, that is not the case for zinc. Zinc(II) aquacomplexes rapidly equilibrate from tetracoordinated and pentacoordinated to hexacoordinated species.


Why do the amino groups in EDTA deprotonate first, in contrast to amino acids? We know that in amino acids, carboxylic acid groups deprotonate first than the amine groups. For example, in alanine, $\mathrm{p}K_\mathrm{a,\ce{COOH}} = 2.34$, $\mathrm{p}K_\mathrm{a,\ce{NH3+}} = 9.69$ according to this page.

But according to the structures of EDTA as searched from Google, and from this extract from p.325 of Analytical Chemistry, 8e by Christian, et al.:

  At sufficiently low pH, the nitrogens can also be protonated and this diprotonated EDTA can be considered a hexaprotic acid. However, this occurs at a very low pH and EDTA is almost never used under such conditions.

the hydrogens of these amino groups seem to be deprotonated first before those of the four carboxylic acid groups. Why is this so?

The carboxylates deprotonate first followed by the amines.  The pKa values for the carboxylates ar 0.0, 1.5, 2.0, and 2.66.  The amines have pKa values of 6.16 and 10.24.  Ref. Harris.  Quantitative Chemical Analysis.


Relationship between magnitudes of forward and reverse kinetic rate constants Consider the reversible unimolecular reaction:

$$\ce{A <=>[k_1][k_2] B}$$

We know that the forward reaction is often considerably more thermodynamically favourable than the reverse reaction, and therefore the relationship $k_1 \gg k_2$ holds between the rate constants. The rate constants are in the same units, and so it is possible to write this relationship. A similar example would be a reaction that is bimolecular on both sides, $\ce{A + B <=> C + D}$, etc.

However, consider this reversible reaction:

$$\ce{A + B <=>[k_1][k_2] C}$$

Assume the reaction proceeds at rate $k_1 [\ce{A}][\ce{B}]$ in the forward direction and $k_2 [\ce{C}]$ in reverse. If the overall reaction rate has units $M s^{-1}$, then $k_1$ necessarily has units $M^{-1} s^{-1}$ and $k_2$ has units $s^{-1}$.

My question is: For this second case, we can no longer impose that $k_1 \gg k_2$, due to difference in units, but can we say anything about their relationship? Does setting $k_1$ to some value constrain the choice of $k_2$ in any way?

A chemical reaction may be a very complicated thing, involving many different paths and elementary reactions. Basically it always leads to the formation of an equilibrium, where the lowest possible energy will be the most favoured. Under certain conditions more than one state will be populated and hence there will be enough potential energy to interconvert molecules in certain states. We usually refer to that as an dynamic equilibrium.

Changes under equilibrium conditions will most likely be elementary reactions (or an infinitely complicated chain of these.)

The population of the various states seems to be static at equilibrium, since the overall change of enthalpy is zero. In a dynamic picture that only means, that forward and backward reaction cancel each other. In that sense they are the same reaction in different directions. Therefore they can be described as one trajectory on the potential energy surface. (If there is a chemical reaction involving many different elementary reactions and the equilibrium still occurs, then the mathematical description will be much more complicated as all elementary reactions will be correlated. However, the main conclusions should stay the same.)

\begin{aligned}
\Delta G_r &= \Delta G_r^\circ + \mathcal{R}T\cdot \ln K = 0\\
\Delta G_r^\circ &= − \mathcal{R}T \cdot \ln K
\end{aligned}

The equilibrium constant may be derived from the mass-action law and should be defined as a product of activities (for the standard state). For $\ce{A + B <=> C}$ this will lead to
$$K^\circ= \frac{a(\ce{C})}{a(\ce{A})\cdot a(\ce{B})}$$

The activities are proportional to concentrations (in first approximation) and are unit less ($c^\circ$ being the standard concentration $1 \:\mathrm{mol/L}$)

$$a=\gamma\frac{c}{c^{\circ}}$$

For reasonable dilutions ($c\to0\:\mathrm{mol/L}$) one can assume that the activity coefficient becomes one $\gamma\approx1$ and therefore rewrite the equilibrium constant with concentrations.
$$K^\circ= \frac{c(\ce{C})/c^\circ}{c(\ce{A})\cdot c(\ce{B})/(c^\circ)^2}$$

At equilibrium the forward and backward reaction are coupled and therefore have to reflect the equilibrium constant.
$$K^\circ= \frac{k_f}{k_b} = \frac{c_{\text{eq}}(\ce{C})/c^\circ}{c_{\text{eq}}(\ce{A})\cdot c_{\text{eq}}(\ce{B})/(c^\circ)^2}$$

As one can see, the constants also need to have the same unit. As they are at equilibrium, they are the same reaction with different populated states.

If you are moving away from the equilibrium, thing will start to get a little bit more messy. On reaction will be faster than the other and rates will not be comparable any more.


How is Carboxymethyllysine formed from the heating of olive oil? Olive oil has significant levels of Carboxymethyllysine.

But how does it form from the heating of olive oil, chemically speaking? Olive oil barely has any protein in it to begin with (though it still has some), so where would all the lysine come from?

Given that the levels are expressed in arbitrary units (based on the results of an immunoassay), it's difficult to say how much CML there actually is and accordingly, how much lysine is needed to produce it. It may be that only a small quantity of lysine is required) In fact, a different study found no CML in raw olive oil.

CML is produced during cooking from glycation of the side chain amine, starting with the Maillard reaction between the amine and a reducing sugar's carbonyl at high temperature. However, in the paper you cited, there was already significant CML in raw olive oil that was only somewhat increased by cooking. This paper suggests that CML formation is related to the oxidation of fatty acids and found that oils with increased concentrations of unsaturated fatty acids (olive oil is in the middle of the 5 oils they tested) were more prone to forming CML, but they deliberately added lysine to test it. It could be that this oxidation is something that happens in storage and the study you linked was using old oil or something. Also, just as with any other plant derivative, there could be significant variation in lysine content from different olives. This site indicates a lysine content of 3 mg/8.3 g whole olive, so there is some there. How much actually makes it into oil, I don't know.


SN2: Neopentyl halides and methyl halides Two part question for you. Let's preface this with I'm in organic chemistry 1, so we're just learning the basics of these reactions.

Are neopentyl halides completely inert to any reaction through any pathway, $\mathrm{S_N1, S_N2, E1, E2}$? For $\mathrm{S_N1/E1}$, even after solvolysis of the halide, is there too much steric strain there for anything to attack the carbocation? Even a small base like $\ce{MeO-}$?
Can a methyl iodide/bromide undergo $\mathrm{S_N2}$ with any basically any nucleophile since there's practically no steric hindrance? Even a bulky compound that's not very lewis basic like tButanol?

Carey-Sundberg has some answers (vol. A, chapter 4).  The text is a bit advanced for you at the present stage, but you should give it a try.

It gives as reference JACS 1975, 97, 3694.  That paper has rate constants for nucleophilic substitution of compounds of type $\ce{RCH2Br}$, with $\ce{R=H, Me, Et, i-Pr, t-Bu }$ with $\ce{LiCl}$ in acetone at room temperature.

Rate constants in 105 s-1:
$\ce{R=H}$: 600, $\ce{R=Me }$: 9.9, $\ce{R=Et}$: 6.5, $\ce{R=i-Pr}$: 1.5, $\ce{R=t-Bu }$: 2.6*10-4

Nucleophilic substitution does take place in neopentyl halides, but at a rate of five orders of magnitudes less than in less hindered compounds.


How can mass spectrometry be used to determine concentration? Which forms of mass spec will determine accurate concentrations of metal ions in a sample? Preferable not using GC/MS but can MALDI ToF or quadrupole do this somehow. Also, not relative concentration but actual concentration.

As indicated in the Wikipedia page, Inductively Coupled Plasma Mass Spectrometry (ICP-MS) is the technique of choice to quantify metal ions in a sample. In summary, this technique is based on the atomization and ionization of the sample metal atoms in a plasma torch, and the ions are then separated based on their mass, usually with a quadrupole mass filter, and then directly quantified on the detector. External calibration with standard solutions of known metal concentration or isotopic dilution then affords access to absolute concentrations.


Unknown alcohol sample. Chromic acid negative but iodoform positive? I have a unknown sample of an alcohol (ceric nitrate positive) but when I did chromic acid test it was negative since it never turned into blue/green and against all odds iodoform was positive... yellow precipitate and a lot...

I have no clues of how to make sense into this. So I did lucas test as well but noticed later that it didn't help because my sample is not very soluble in water.

Judging by the great amount of precipitate I got from the iodoform test, I'm thinking that maybe my sample is indeed a primary or secondary alcohol but it has something that won't react with the chromic acid. Are there any considerations to account for when doing this test? or some limitations?

Please help!

I would recommend to repeat all the experiments one more time. The reasons for such results may be:

You had dirty test tube for ceric nitrate test, and it was really false positive. You could have a methyl ketone, which gives negative chromic acid test and positive iodoform test.
You added too much chromic acid, and had low amount of your "alcohol". That caused all alcohol to be oxidized, but that blue-green color was too faint, and you didn't notice it because of chromic acid excess.

Moreover, if your "alcohol" is immiscible with water, that means it is one of the higher alcohols. You could also try to identify it (if it's really an alcohol) by smell: those alcohols have pretty distinctive smells. x.x


Synthetical formation of diamonds As everyone knows a diamond won't last forever. It will eventually turn into a graphite over a long period of time, as the free energy of graphite is lower than that of diamond, hence making the reaction energetically favorable.

However, is it possible to do the reverse, by using graphite and the applying energy onto graphite so as to synthesize diamond. If so, how is it done and can it be done in a lab by a chemist? Does that mean that people who managed to do it can become rich? As a diamond costs so much more than a graphite. Shouldn't we synthesize diamonds in this way? Can I become  rich by synthesizing diamonds and selling them?

Natural Diamonds are formed under high pressure and temperature at depths of 140 to 190 kilometers (87 to 118 mi) in the Earth's mantle. Carbon-containing minerals provide the carbon source, and the growth occurs over periods from 1 billion to 3.3 billion years (25% to 75% of the age of the Earth). Diamonds are brought close to the Earth's surface through deep volcanic eruptions by a magma, which cools into igneous rocks known as kimberlites and lamproites.

According to Wikipedia,

  The formation of natural diamond requires very specific conditions—exposure of carbon-bearing materials to high pressure, ranging approximately between 45 and 60 kilobars (4.5 and 6 GPa), but at a comparatively low temperature range between approximately 900 and 1,300 °C (1,650 and 2,370 °F). These conditions are met in two places on Earth; in the lithospheric mantle below relatively stable continental plates, and at the site of a meteorite strike.

So yes, Diamonds can be produced synthetically if you can simulate almost similar conditions as in Earth's Mantle. (See high-pressure high-temperature process; Synthetic Diamond) An alternative, and completely different growth technique is chemical vapor deposition (CVD).


Oxidation number of phosphorus in phosphine Electronegativity of $\ce{H}$ is greater than $\ce{P}$, but according to my book in $\ce{PH3}$, $\ce{P}$ has oxidation number of -3. How is it possible? Is any of the data wrong?

Here is a way to get the answer. If you place a reagent with oxidation number of H = -1 (hydride) it would produce dihydrogen gas. For example: $\ce{KH + H2O-> KOH + H2}$.

If you place a molecule with oxidation number of $\ce{H} = +1$ dihydrogen gas is NOT formed. $\ce{NH3 + H2O -> NH4+ + OH-}$. Phosphine doesn't form dihydrogen gas on contact with water so $\ce{H}$ is in oxidation state +1 and $\ce{P}$ is in oxidation state -3. Reaction with strong acids leads to $\ce{PH4+}$ which has similar structure to the ammonium ion. For these reasons even for $\ce{As}$ in $\ce{AsH3}$ the oxidation state of $\ce{As = -3}$.

The problem with electronegativity is that we are trying to assign a single number, while we actually need a function. We take average of electronegativity of $\ce{H}$ in $\ce{HF, NaH, H2O, CH4, NH4+}$ and come up with a single number. This works well if you need to guesstimate the behavior, but it is not accurate. It doesn't even account for softness/hardness of a partner.

Analysis "will $\ce{PH3 + H2O}$ form $\ce{H2}$?" (no) vs "will $\ce{PH3 + P2O3}$ form $\ce{P + H2O}$?" (yes) is more accurate way to assign formal charges.


Free energy of solvation and Henry's law constant I am trying to calculate the free energy of solvation of $\ce{CO2}$ from its Henry's law constant. As given on Wikipedia, the dimensionless Henry's law constant is $0.83$. If I try to calculate $\Delta G$ from this, I got

$$\begin{align}
\Delta G &= -RT\ln K \\
  &= -\pu{8.3145 J mol-1 K-1}\times\pu{298.15 K}\times\ln 0.83 \\
  &= \pu{0.46 kJ mol-1} \\
  &= \pu{0.11 kcal mol-1}.
\end{align}$$

However, I saw an article saying that the experimental value is $\pu{0.24 kcal mol-1}$. The article cited from the CRC Handbook of Chemistry and Physics, while Wikipedia cited from this compilation, which I think is also a reliable source.

So what did I do wrong in my calculation of $\Delta G$ from the Henry's law constant?

The CRC Handbook of Chemistry and Physics [1] provides data for the solubility of $CO_2$ in $H_2O$ originally from reference [2]. The reported data are mole ratio solubilities $\chi$ of the gas at a partial pressure of one atmosphere (101.325 kPa), but can readily be converted into Henry's law solubility in other units:

$$H_{CP} = 1000 \delta_{H2O} \chi /M_{H2O} = 997 \times 6.15 \times 10^{-4}/18 = 0.034 mol/L atm$$

or in molal and bar units

$$H_{mP} = 1000\chi /1.013 (bar/atm)M_{H2O}=1000\times 6.15 \times 10^{-4}/(18 \times 1.013) = 0.034 mol/kg bar$$

Reference [2] is one of the original sources of the data tabulated in the Wikipedia, compiled by Rolf Sander in reference [3]. Other references in [3] provide similar values of $H_{CP} \approx 0.034$ mol/Latm. The value of $H_{CP}$ in the wikipedia is therefore consistent with the solubility reported in the CRC Handbook.

The article by Prasetyo and Hofer [4], referred to in the OP, cites the CRC Handbook as the source of the following reported experimental value (see Table 4): $$\Delta_{solv} G^o = 0.24 kcal/mol$$

However, from $H_{mP} = 0.034$ mol/kgbar (or  $H_{CP} =  0.034$ mol/Latm) we obtain $$\Delta_{solv} G^o = -RTlog(H_{CP} \frac{P^o}{m^o})= 2.0 kcal/mol$$ which is far larger than the value reported in the article.

To attempt further progress, note that comparing standard free energies of solvation requires awareness of the associated reference states.

Prasetyo and Hofer indicate (pg 6474) that their choice of standard state is P° = 1 bar, m° = 1 mol/kg, and T = 298.15 K. The experimental values Prasetyo and Hofer compare to simulation results (see Table 4) refer presumably to this standard state. But this is approximately the same as the molar standard state associated with the data reported in the Wikipedia (P° = 1 atm, c° = 1 M, 298.15 K). Certainly the energies computed in the two scales don't differ significantly, as shown above. I therefore suspect that a different standard state was used in their calculation, or free energies for different (chemical) processes are being reported.

It is worth adding that the value $\Delta_{solv} G^o = 0.24 kcal/mol$ is encountered in other articles that also reference the 75th Edition of the CRC Handbook.

References

[1] The CRC Handbook of Chemistry and Physics, 85th Edition, section 8-87, SOLUBILITY OF SELECTED GASES IN WATER, L. H. Gevantman.

[2] R. Crovetto, Evaluation of Solubility Data for the System CO2-H2O, J. Phys. Chem. Ref. Data, 20, 575, 1991.

[3] R. Sander: Compilation of Henry's law constants (version 4.0) for water as solvent, Atmos. Chem. Phys., 15, 4399-4981 (2015).

[4] N. Prasetyo and T.S. Hofer J. Chem. Theory Comput. 2018,  14, 12, 6472-6483


Fitting parameters of Lennard-Jones potential to properties of real materials I'm using the Lennard-Jones potential for some (very simple) protein docking simulation* and couldn't find parameters for different materials (like $\ce{C}$, $\ce{N}$, $\ce{O}$, …). Does it even make sense to differentiate between different materials when using a crude approximation like the Lennard-Jones potential?

* the potential function for each configuration of the two molecules is the sum over all pairwise energies, for which I only consider the van der Waals term by using the LJ-potential

All molecular simulation force fields are essentially "fitted" mathematical functions. The parameters are adjusted until the simulation results match some physical properties we care about. As Geoff says, Lennard-Jones force fields are not particularly realistic, in the sense that you wouldn't expect a simulation of atoms interacting solely through Lennard-Jones potentials to behave exactly like a real system. This becomes very important for things like proteins, where conformation and charge-charge interactions are very important.

However, what might not be clear from the other answers and comments is that all force fields are not 100% realistic. The trick is to intelligently choose your level of realism. It could be that a Lennard-Jones force field that ignores  charge-charge interactions works perfectly well for your application. Without knowing the details, I can't tell you how suitable it would be.

I do know that for polymers all sorts of real phenomena can be modeled with very simple Lennard-Jones-like force fields. A famous example is this article by Kurt Kremer and Gary Grest:

Dynamics of entangled linear polymer melts:  A molecular‐dynamics simulation. J. Chem. Phys. 92, 5057 (1990)

Also, as Geoff mentioned, LJ potentials work pretty well for things that can be modeled by the van der Waals equation of state: noble gases, simple diatomics, and other small molecules at high temperature and low pressure.

Some coarse-grained simulations use LJ potentials for each "blob," and depending again on what you want to study, this can be very successful.

And while we are on the subject of unrealistic force fields, the hard-sphere model has been used to successfully model things like glass transitions (I don't have a good representative link for this, but you can search for "hard sphere glass transition" to find many studies).

So, in response to your actual question:

  Does it even make sense to differentiate between different materials when using a crude approximation like the Lennard-Jones potential?

The answer is:

Maybe - it depends on what the goal of your simulation is. What physical behavior are you trying to model? If you just want to see how a string of soft, slightly sticky beads sticks to another string, then you don't need to differentiate between atoms. If you want to see what happens when some of those beads are really sticky (sort of like hydrogen bonding sites on a protein), then it would make sense to use different parameters for very sticky vs. slightly sticky beads. It probably would never make sense to worry about which beads were carbon and which were oxygen, nitrogen, etc., or to try to select parameters for them that were based on published values (unless those studies were specifically looking at the same system and properties that you are). Most LJ parameters are derived from fitting simulated phase behavior to real phase behavior - which is pretty far removed from what you are doing.

If you want to get a better idea of what real proteins would do, then the problem becomes much more complicated. Real protein behavior has a lot to do with interactions with water, which means you would need to simulate that as well. Charge-charge interactions also become much more important (and the simulations much more computationally expensive)

My advice (and I think what the other responses are saying as well) is to choose your system and goals carefully so that you don't need this level of realism. Or, if you do need it, then do it right; don't try to make a simple model work for a system that it doesn't fit.


Fractional distillation of a product differs from the components listed in the SDS. What am I doing wrong? I recently bought a product called Prestone De-Icer for Windows and Wipers which claimed to contain Methanol. I was puzzled when using it to note that it did not smell much like Methanol so I looked up the SDS for the product which stated that it contained 50-100% Methanol, 1-10% Ethylene Glycol & 1-5% Carbon Dioxide.

Out of curiosity I decided to carry out a fractional distillation to see just what the constituents were and was puzzled to see the thermometer shoot up through the Methanol BP at 66C mark without stopping and the first stop was about 76C which would suggest Carbon Tetrachloride. It then stopped at about the 79C mark which would indicate Ethanol and then carried on up to about 98C which would suggest Heptane.

So my question is this - Is there something wrong with my methodology or interpretation of what I am seeing or is it possible that the manufacturer is in error in the SDS. The latter would be very serious if true since the SDS would be the resource used should it be suspected that someone has been poisoned by the product.

My instinct is to suspect that I am doing something wrong but for the life of me I can't see what it could be. The thermometers I have used ( three in all ) are calibrated to 0.1 degree and were one mercury, one alcohol and one digital and all agreed to within 0.4C max deviation.

I should point out that I am not a professional chemist, just an amateur who has taken the hobby up in retirement so I am quite prepared to be told that I have got it wrong. It's all part of learning in my opinion.

Here is a link to the SDS -
http://www.homedepot.com/catalog/pdfImages/2c/2cd5847a-9dd0-4cf0-a5ef-83efbd532f4c.pdf

I am including a photo of the setup I am using to do the separation. The Vigreaux column is actually wrapped in aluminium foil.

Best Answer is this:
Your observations are consistent with the SDS for the reasons below.

A couple of issues see below.

Fractional distillation is much more complicated than you suppose. You have no way to control reflux ratio, (amount of distillate vs reflux) which right away will limit your separation. The column is not insulated or temperature controlled.
Vigreux columns are good for distilling one component from non-volatiles, but are not made for fractional distillations. You will need a much more efficient column, the cheapest choice being a vacuum jacketed Hempel column with some sort of packing, ie Raschig rings.
With the equipment you are using, a range of 10 C between ideal boiling points would be considered normal, you will not see the sharp cut offs you are expecting.
The first component is likely methanol, and the boiling point range you are seeing is due to the above reasons.
Ethylene glycol does not boil to almost 200 C, when you end the distillation, is there any liquid left in the distilling flask?
As far as I could quickly check, methanol and ethylene glycol do not form an azeotrope, but the tables I checked were on Wikipedia and not thorough, you might want to check that. The formation of an azeotrope definitely explains the observations especially if it boils in the range of 98C +/- 20 C
Not sure on the exact conditions or if the MSDS is wrong, but it is conceivable dimethoxyethane (dimethyl glycol) formed or was the original constituent, this has a boiling point of 85 C, it could have even formed in the distilling flask depending on conditions.
Note the MSDS or SDSs are they are now called are not complete, and there are usually other minor constituents that might appear and could possible catalyze the formation of dimethoxyethane, assuming the SDS was correct in the first place.
Distillation is not the appropriate method here, GC-MS would be the way to go.

Hope that helps! I suggest you purchase Handbook of Laboratory Distillation, it is the fundamental text for distillation in theory and practice, the price is well worth it.

In conclusion your observations could be explained by: azeotrope formation, incorrect SDS, formation of dimethoxyethane, and incorrect set up.


Gaussian16 Raman calculation from IR checkpoint file I already did an IR frequency calculation for my molecule. Is there any possibility to just run a Raman frequency calculation with the chk file from my IR calculation?
Or do I have to do the calculation again with the freq=raman keyword? I checked the Gaussian manual and found the raman=polar keyword, but I think this is not exactly what I am looking for.

Raman intensities require polarizabilities to be calculated, IR intensities only dipole moments. Your checkpoint file may therefore not have the properties you require, in which case you'll need to do more calculations. Using the polar=raman keyword would apparently save you a little bit of time, because the program then already knows your force constants and normal modes from the preceding freq calculation, but there will still be computational effort involved in calculating the polarizabilities.
Note that there are some cases where Gaussian also give you the polarizabilities "for free", as the manual states on the polar=raman page:

Note that Polar is done by default when second derivatives are computed analytically.

You can use the chkchk tool to see what's really stored in the checkpoint file.


Guidelines for Solubility in Water of Organic Compounds in Relation to Carbon Chain Length Is there a way to generalize the solubility of any organic compound based on their functional groups and the length of their carbon chains? I know that beyond 3 carbons, alcohols are insoluble, and beyond 4 carbons, carboxylic acids are insoluble. But what if the alcohol had 2 hydroxyl groups? What if the carboxylic acid had 2 carboxyl groups? How many carbons would they each need for the compound to become insoluble in water again? Are there any quantitative guidelines for determining this?
And what about the solubility of esters and haloalkanes? How many carbons does an ester and haloalkane need to have for them to be insoluble? Again, what if there are multiple ester or halide groups?
Lastly, how do you determine solubility when you’ve got a mixture of functional groups? Take aspirin for example, whose structure contains a benzene ring of 6 carbons, a carboxyl group attached to one corner, an ester group attached to another corner, and one methyl group attached to the ester. How would you systematically determine if this compound was soluble in water or not?

(Image credit: http://www.rxlist.com/images/rxlist/aggrenox2.gif)

To determine the solubility of some compound, a chemist would systematically throw it into water and see what happens. You may draw some correlations like "plus six for two hydroxyls, minus five for five carbons, should be soluble"; they are somewhat useful, but mostly qualitatively.

There is a whole hierarchy of computational methods for prediction of solubility, but these are highly complicated.

BTW, haloalkanes are almost as insoluble in water as alkanes themselves.


Spins spin coupling on proton NMR for an aldehyde? On a proton NMR, does the hydrogen in the CHO group cause coupling, or get coupled itself? Specifically, I a considering 2-methylpropanal. I cannot decide whether the H in the aldehyde group would be split into a doublet or remain as a singlet, and I am unsure as to whether the single proton on carbon number two would be a heptet (from the 6 adjacent protons), an octet ( from the 6 protons plus the proton in the aldehyde group), or if it would be split into some combination of a heptet of doublets, or a doublet of heptets...

Any advice on dealing with spin spin coupling in proton NMR would be much appreciated!

I cannot decide whether the H in the aldehyde group would be split into a doublet or remain as a singlet […]

What would be the argument against $^3J$ coupling between these protons? The aldehyde proton will appear as a doublet.

Don't forget that coupling works both ways! $\ce{(CH3)2C\color\red{H}-}$ will (at least) split to a doublet too. Coupling with this proton will also  split the signal for the methyl protons to a doublet.

What does this mean for the proton at C-2?

Formally, a doublet (from coupling with the aldehyde proton) of a septet (from coupling with six magnetically equivalent methyl protons) is to be expected.

In really, you will observe a smaller number of lines! I suggest to try this with a pen on squared paper to figure out what happens when the two coupling constants are identical or one is twice as big as the other.


Comparison of energy release from different types of fuels I've recently been reading about the different amount of energy release for different types of fuels/hydrocarbons.
Anyways, I was wondering why butanol would release more energy (kj/mol) than methanol? I'm guessing the same theory applies for methane vs butane?

This is essentially a question concerning the trend of molecules in the same homologus series that have differing chain lengths. When we increase chain length, the number of C-C bonds increase in proportion to C-H bonds. Looking at the bond energies, the bond energy of a C-C bond is $347 kJ/mol$, while for C-H it is $414 kJ/mol$. This means to break a C-C bond compared to a C-H bond, less energy is required. Also when increasing the number of C-C bonds, while producing the same products ($CO_{2(g)} + H_2O_{(l)}$), more carbon dioxide is produced in proportion to the amount of water. The enthalpy of formation of carbon dioxide, results in it releasing more energy than that of water. Taking this into account with the lower energy requirement to break the reactants, the chain length is generally directly proportional to the energy produced. Thus, the COMPLETE combustion of Butanol would have a higher energy out put than Methanol and likewise Butane > Methane.
It is important to note that, as the compounds get more energy dense, i.e release more energy upon combustion, that it becomes harder for the compound to reach complete combustion; what I mean is that more oxygen is required for the reaction to release the full energy output. If not enough oxygen is supplied to the system soot ($C_{(s)}$) & Carbon Monoxide ($CO_{(g)}$) will be produced and the energy produced will be less than the energy produced in the complete combustion. (The degree to which this works, depends on how many moles of oxygen is supplied).


Finding electron configuration of lanthanide ions When I have a gadolinium ion ($\ce{Gd^3+}$), how can I calculate its electron configuration?

$$\ce{Gd} : \mathrm{[Xe] (4f)^7 (5d)^1 (6s)^2}$$

Do I need to first subtract the 2 electrons in the $\mathrm{s}$-orbital and then the $\mathrm{d}$-/$\mathrm{f}$-orbital?

I agree with bobthechemist's comment.

When you are determining the electronic configuration of anions/cations, you will first need to determine electronic configuration of neutral atom. Like in the case above:

$$\ce{Gd} : \mathrm{[Xe] (6s)^2 (4f)^7 (5d)^1}$$

Now arrange this configuration according to ascending value of $n$:

$$\mathrm{[Xe] (4f)^7 (5d)^1 (6s)^2}$$

Now, according to your need start subtracting electrons from highest value of $n$, like for $\ce{Gd^3+}$, the electronic configuration will be:

$$\ce{Gd^3+} : \mathrm{[Xe] (4f)^7}$$

See this to know more on calculating electronic configuration of cations and anions.


Using IR to detect / quantify a molecule at low concentrations I want to detect / quantify a specific terpenoid molecule from a microbiological growth medium. The molecule will however be present at very low levels approx. 10-100 mg/L.

Is it viable to use the IR spectrum to detect this molecule at such low levels? If not quantitatively at least in a qualitative sense? The idea is to have some sort of method that is amenable to reasonably high throughput screening.

A GC / GCMS which is the typical assay method is too time consuming for screening a lot of samples.

The IR spectrum is shown below. Can I just monitor a specific wavelength with max absorbance, (say) 3000 cm-1? That way I won't have to scan the whole wavelength range.

Any other ideas?

There isn't enough information to answer the question with certainty.  We'd need an IR spectrum of the culture medium (possibly dried/lyophilized medium)

I suspect that the peak at 3000 $\mathrm{cm}^{-1}$ is an alkyl CH stretch.  You're going to see similar absorption peaks in lipids and proteins, so I don't think that particular peak is going to be too promising for you.

You might look into headspace mass spectrometry.  Many terpenes are volatile and can be analyzed directly from headspace air (i.e. the air directly above your cultures).  You don't necessarily need to do gas chromatography, so the method can be quite fast.  It won't be as diagnostic as to the identity of the particular terpenes formed, however.


Does the protective layer of Aluminum oxide form on the surface of aluminium alloys (such as Al-Zr) as well? Aluminum is an ideal element for usage in airports. Unlike Iron, when oxygenating, there is a protective layer formed that does not allow the rest of the Al metal to oxygenate.Does the protective layer $\ce{Al2O3}$ form on the surface of aluminum alloys (such as Al-Zr) as well as when it happens to Aluminum in its pure state?

Short answer: Yes.

Long answer: Any alloy containing 'passivating' elements such as aluminum, which have the tendency to form a protective oxide as you describe, has the potential to also form the same sort of passivating oxide layer as the pure material.  It's not guaranteed, though.

Pure zirconium also forms a passivating oxide $(\ce{ZrO2})$$^{(1)}$, so the AlZr alloy you mention is almost certain to form a similar protective oxide.

More broadly, a major area of research at present is in the 'alumina-forming' alloys ("alumina" is another word for aluminum oxide, $\ce{Al2O3}$), where aluminum is included in, e.g., steels or nickel superalloys, to try to achieve this passive aluminum oxide layer on the surface in high-temperature, corrosive environments$^{(2-5)}$.  With materials like a stainless steel-plus-aluminum, the passive oxide scale only forms if the alloy has the right percent range of aluminum -- otherwise, the aluminum oxide ends up staying spread throughout the metal and not concentrating at the surface.  If the alumina doesn't end up spread evenly over the surface, you don't get nearly as good of a protective effect.

(1) http://www.mtialbany.com/metals/zirconium/
  (2) http://topaz.ethz.ch/function/web-het-secured/pdfs/Prescott-92.pdf
  (3) http://www.metalle.uni-bayreuth.de/de/download/publications_downloads/Bensch__Sato__Warnken__Affeldt_______Reed__Glatzel__Acta_Mat_.pdf
  (4) http://www.netl.doe.gov/File%20Library/Events/2010/ods/Gordon_Tatlock_Corrosion.pdf
  (5) http://www.ornl.gov/science-discovery/advanced-materials/science-to-energy-solutions/alumina-forming-austenitic-alloys-licensed


Which way does this reaction between NaCl and acetic acid naturally proceed? Here is the overall formula:

$$\ce{CH3COONa  + HCl <=> NaCl + CH3COOH}$$
(sodium acetate + hydrochloric acid $\ce{<=>}$sodium chloride + acetic acid)

Which way does the above reaction naturally proceed? I've tried Googling this, but different sources seem to suggest contradictory answers, and I can't really tell what's correct.

Mostly the reaction will proceed from left to right, unless you have ridiculous amount of acetic acid. Hydrochloric acid is much stronger, so it suppresses dissociation of acetic acid.

However, this is only part of the truth. In fact, in water solutions many compounds (partially) dissociates:

$HCl = H^+ + Cl^-$

The process is strongly endothermic in vacuum, but subsequent interaction of ions with water makes it mostly energy neutral or even highly exothermic for some compounds. This means, that in you case the solution would actually contain $H^+$, $OH^-$ (very little), $CH_3COO^-$, $CH_3COOH$, $HCl$ (almost none, may be ignored in most cases), $Na^+$. The actual reaction would be

$CH_3COO^- + H^+ = CH_3COOH$

The actual direction of the reaction would depend on initial concentraion and guarded by equilibrium constant.

The keywords for googling/textbook reading are 'electrolitic dissociation', 'ionic reaction', 'chemical equilibrium', 'equilibrium constant'. Generally, the procedure of determining which concentration of which ions the system will have requires solution of system of equations, including equilibrium constants, electroneutrality equation and material balance equations. The system often may have no analytic solutions, in which case it is usually approximated by simpler system, dropping some elements of sums in the system.


How to identify the asymmetric carbon in [4-(1-bromoethylidene)cyclohexyl]acetaldehyde? How to identify an asymmetric carbon in the molecule below? Is the carbon identified by the asterisk asymmetric or not? I would say no as it has no 4 different groups attached.

Is this correct?
I will greatly appreciate if anyone can confirm or otherwise explain why it can be asymmetric.

There is probably little reason in discussing the hydrogen and the $\ce{CH2CHO}$ group which are definitely different residues. The question remains whether the sides of the ring are identical or not.

If you go along the top side of the ring, you will encounter:

$\ce{CH2}$
$\ce{CH2}$
$\ce{CR=CMeBr}$

Looking into it a little deeper, we see that we are coming from the (Z) side of the double bond.

Going along the bottom side you will encounter:

$\ce{CH2}$
$\ce{CH2}$
$\ce{CR=CBrMe}$

Here, however, we are arriving from the (E) side of the double bond.

Since (E) is not equal to (Z) the two sides of the ring are not equal, therefore the carbon has four different residues, therefore it is chiral.


How to calculate the concentration of the elements after dissolution of iron(III) chloride? I was presented with the following problem:

  In the laboratory you dissolve $\pu{24.7 g}$ of iron(III) chloride in a volumetric flask in water to a total volume of $\pu{375 ml}$.

  What is the molarity of the solution?
  What is the concentration of iron(III) cation in M?
  And what is the concentration of chloride anion in M?

I know how to do the first part of the problem, the answer is $\pu{0.406 M}\ \ce{FeCl3}$. But when I go to balance the equation, the computer system tells me the equation is wrong.

My equation:
$$
2\ce{FeCl3 -> 2 Fe + 3Cl2}
$$

The computer system:

$$
\ce{FeCl3 -> Fe^3+ + 3Cl-}
$$

I don't understand the computer systems' answer. Simply because when by itself, chlorine is a diatomic molecule therefore, should be $\ce{Cl2}$ not just $\ce{Cl}$. Am I wrong to think this? Am I correct or the computer system? If the computer system is correct, why?

When answering such problems, I recommend writing everything know down, then use the appropriate formula to solve for the unknown. For the first part you already did that correctly.

  In the laboratory you dissolve $\pu{24.7 g}$ of iron(III) chloride in a volumetric flask in water to a total volume of $\pu{375 ml}$.

\begin{align}
  m(\ce{FeCl3 (s)}) &= \pu{24.7 g}\\
  M(\ce{FeCl3}) &= \pu{162.3 g//mol}\\
  V(\ce{FeCl3 (aq)}) &= \pu{375 mL}\\
\end{align}

  What is the molarity of the solution?

Molarity is another word for amount concentration, so you are looking for $c(\ce{FeCl3 (aq)})$.
\begin{align}
  c &= \frac{n}{V}\\
  n &= \frac{m}{M}\\
  c(\ce{FeCl3 (aq)}) &=
    \frac{m(\ce{FeCl3 (s)})}{M(\ce{FeCl3})\cdot V(\ce{FeCl3 (aq)})}\\
  c(\ce{FeCl3 (aq)}) &=
    \frac{\pu{24.7 g}}{\pu{162.3 g//mol}\cdot\pu{375E-3L}}\\
  c(\ce{FeCl3 (aq)}) &=
    \pu{0.406 mol//L}
\end{align}

  What is the concentration of iron(III) cation in M?
  And what is the concentration of chloride anion in M?

You are looking for the ion (amount) concentrations in your solution, i.e. $c(\ce{Fe^3+ (aq)})$ and $c(\ce{Cl- (aq)})$.
For these two questions you need to know how the substance dissolves. The wording here is already very explicit, as it says iron(III) cation, i.e. $\ce{Fe^3+ (aq)}$, and chloride anion, i.e. $\ce{Cl-}$. Otherwise, with some more experience you will know that this compound is a salt and it dissociates into ions when dissolved in polar solvents like water. Therefore the reaction equation is
$$\ce{FeCl3 (s) ->[H2O] Fe^3+ (aq) + 3 Cl- (aq)}.$$
From this you can derive the actual relationships between the ion concentrations.
You can clearly see that there is one iron(III) ion and three chloride ions per formula of $\ce{FeCl3}$; therefore
\begin{align}
  c(\ce{Fe^3+ (aq)})&= c(\ce{FeCl3 (aq)})
  &&=\pu{0.406 mol//L},\\
  c(\ce{Cl- (aq)}) &= 3\times c(\ce{FeCl3 (aq)})
  &&=\pu{1.22 mol//L}.
\end{align}

The other equation would probably be appropriate when you (thermally) decompose the substance (in absence of everything) as it is the reverse equation which can be used to prepare the molecule:
$$\ce{2 Fe (s) + 3 Cl2 (g) -> 2 FeCl3 (s)}.$$