What exactly is the Horwitz equation? During a laboratory exercise in a course on analytical chemistry, we were asked to calculate the $z$-score.

  $$z=\frac{X_{\text{lab}}-X_{\text{cons}}}{S_{\text{target}}}$$
  where

  $X_{\text{lab}}$ is some measured quantity (e.g., concentration),
  $X_{\text{cons}}$ is the consensus value for that $X_{\text{lab}}$,
  $S_{\text{target}}$ is known as the benchmark standard deviation

One way of optaining the $S_{\text{target}}$ was to use the so-called Horwitz equation:

  $$S_{\text{target}}=2^{(1-0.5\log C)}.$$
  $C$ denotes the concentration as a dimensionless mass fraction.$^{[1]}$

On the mathematical relationship,  M. Thompson comments$^{[2]}$

  As well as being useful, the Horwitz Trumpet is a feature of considerable theoretical interest. It is hard to avoid the assumption that a simple mathematical law that describes the behaviour of large numbers of methods over at least six orders of magnitude of analyte concentration must have some inherent meaning and deserves serious consideration. So far,
  though, nobody has managed to explain the strange empirical exponent from basic principles, although several people have made conjectures. Are we seeing the manifestation of a physical law here, or is there a psychological basis, perhaps to do with our perception of fitness for purpose?

  There is a sure-fire paper in Nature waiting for somebody!

In statistical terms, why have we chosen such a form for $z$?

This might include an explanation of the relationship with the Horwitz equation.

What are the several conjectures that have been made?

[1] Carlos Rivera, Rosario Rodríguez. Horwitz Equation as Quality Benchmark in ISO/LEC 17025 Testing Laboratory

[2] M. Thompson. The amazing Horwitz function. Royal Society of Chemistry. (2004)

The Horwitz function 1, also known as the Horwitz “Trumpet”, is an empirical relationship that has, thus far, not been derivable from fundamental principles. As noted by Thompson 2, and as shown below in Fig. 1 (from 2), the relative standard deviation of reproducibility (RSDR) increases as analyte concentration decreases:

Horwitz formulated his expression after noticing a pattern in the relative standard deviations reported in many (thousands at this point) collaborative trial results. The following figure (a screenshot from Hanley [3, p. 12037]) provides the essential equations and background information:

As Hanley notes, the Horwitz function appears to be in the Tweedie family of exponential dispersion fluctuation scaling relationships (https://en.wikipedia.org/wiki/Tweedie_distribution ). Specifically, the Horwitz function may be a compound Poisson-Gamma process, since the exponent, $\alpha $ in equation 2, is approximately 0.85. (More accurately, Thompson 1 notes that it is $1 - (log2)/2$).

Answer to OP’s first question

Assume measurements are Gaussian distributed, with $\mu $ as the population mean and $\sigma $ as the population standard deviation, and assume both parameters are unknown. Let $X_{lab} $ be a single measurement, i.e., a random sample from the Gaussian distribution. Then $z = (X_{lab} - \mu )/\sigma $. So z, which is the so-called ‘z score’, is simply the signed difference parsed in units of $\sigma $. A small z score means the measured result is close to the true value, i.e., $\mu $, assuming no systematic error. A large z score means the measurement has relatively poor accuracy.

But both $\mu $ and $\sigma $ are unknown, so $X_{cons}$ is used as an estimate of $\mu $ and $S_{target} $ is used as an estimate of $\sigma $. The result is a rough, and possibly biased, estimate of z. Frankly, this is too crude for me.

Answer to OP’s second question

Also as noted by Hanley, attempts have been made to derive the Horwitz function from “the binomial distribution, Zipf’s law, and a heuristic approach”: see 4 and 5 below. However, these have not been generally been accepted and it is fair to say that the final word has not been written on the origin of the Horwitz function. As well, the Horwitz function potentially has a problem at zero analyte concentration, as noted by Holland et al. [6]. This is problematic because analytical blanks do not have infinite standard deviations.

References:

W. Horwitz, “Evaluation of analytical methods used for regulation of foods and drugs”, Anal. Chem. 54 (1982) 67-76.
M. Thompson, “The amazing Horwitz function”, AMC Technical Brief No. 17, July, 2004 Royal Society of Chemistry. (2 pages).
Q.S. Hanley, “Chemical Measurement and Fluctuation Scaling”, Anal. Chem., 88 (2016) 12036-12042 (Open Access).
R. Albert, W. Horwitz, “A Heuristic Derivation of the Horwitz Curve”, Anal. Chem., 69 (1997) 789−790.
P. Hall, B. Selinger, “A statistical justification to relating interlaboratory coefficients of variation with concentration levels”, Anal. Chem., 61 (1989) 1465−1466.
R. Holland, R. Rebmann, C.D. Williams, Q.S. Hanley, “Fluctuation Scaling, the Calibration of Dispersion, and the Detection of Differences”, Anal. Chem., 89 (2017) 11568-11575.


Why do square pyramidal molecules have only two vertical mirror planes? I would have expected molecules like IF5 to have 4, with two planes along the bond axes, and two planes bisecting the bonds. This is similar to how square planar molecules have four vertical mirror planes. However, apparently there are only two planes. Why is this? What am I misunderstanding?

Either your book is wrong or you are misunderstanding what it is trying to say.

There are 4 vertical planes of symmetry.

See for example GROUP THEORY AND ITS APPLICATIONS IN CHEMISTRY, SECOND EDITION , section 1.5.2

See especially:

  Vertical planes are further classified into two, namely $\sigma _v$ and $\sigma _d$ The $\sigma _v$ planes fall along the bonds and the $\sigma _d$ planes bisect the bond angles in molecules. The subscripts v and d stand for vertical and diagonal. However, both $\sigma _v$ and $\sigma _d$ are vertical planes


How to determine the concentration of a base through indirect titration? $5.267~\mathrm{g}$ of $\ce{Na2CO3}$ was dissolved in $250.00\ \mathrm{mL}$ of water. $10.00\ \mathrm{mL}$ of this solution was titrated with $\ce{HCl}$. The end-point occurred at $21.30\ \mathrm{mL}$. This acid was titrated with $25.00\ \mathrm{mL}$ of $\ce{Ba(OH)2}$. The end-point occurred at $27.10\ \mathrm{mL}$. Calculate the concentration of the $\ce{Ba(OH)2}$.
  [Answer: $0.1013~\mathrm{M}$]

Using the mass they gave me ($5.267~\mathrm{g}$) I worked out the moles of $\ce{Na2CO3}$ which was $0.0497\ \mathrm{mol}$.
I then worked out the concentration of $\ce{Na2CO3}$ using the moles I just worked out and the volume they gave me ($0.25~\mathrm{L}$) and I got $0.1988~\mathrm{M}$.
The question says that only $0.01~\mathrm{L}$ was used, so I worked out the number of moles this would be which was $0.001988\ \mathrm{mol}$.
Using stoichiometry, I multiplied this value by two to get the moles of $\ce{HCl}$ that would have reacted with this, which was $0.003976\ \mathrm{mol}$.
It says that the end-point occurred at $21.30~\mathrm{mL}$, which if you subtract the volume of $\ce{HCl}$ which was already there, I worked out that $11.3~\mathrm{mL}$ or $0.0113~\mathrm{L}$ of $\ce{HCl}$ must have been used.
From here, I worked out the concentration of $\ce{HCl}$ (using $0.0113~\mathrm{L}$ and $0.003976\ \mathrm{mol}$) and got $0.35186~\mathrm{M}$.
It says that for the next reaction, $25~\mathrm{mL}$ of $\ce{Ba(OH)2}$ was used and the end-point occurred at $27.10~\mathrm{mL}$. From this information, I worked out that $0.0021~\mathrm{L}$ of $\ce{HCl}$ must have been used in this reaction.
By knowing the volume of $\ce{HCl}$ used, I worked out the amount (moles) of $\ce{HCl}$ used in the reaction, using the previously found concentration. I got $0.0007389\ \mathrm{mol}$.
Using stoichiometry, I worked out the amount of $\ce{Ba(OH)2}$ which must have reacted with this many moles of $\ce{HCl}$. I divided the $\ce{HCl}$ moles by $2$ and got $0.00036945\ \mathrm{mol}$.
By using the volume of $\ce{Ba(OH)2}$ used which they give me ($0.025~\mathrm{L}$) and the moles I just found ($0.00036945\ \mathrm{mol}$) I finally worked out the concentration of $\ce{Ba(OH)2}$, which was $0.0148~\mathrm{M}$.

My answer, $0.0148~\mathrm{M}$ is different to the answer the question gives ($0.1013~\mathrm{M}$), and I'm not sure why.

It says that the end-point occurred at $21.30\ \mathrm{mL}$, which if you subtract the volume of $\ce{HCl}$ which was already there, I worked out that $11.3\ \mathrm{mL}$ or $0.0113\ \mathrm{L}$ of $\ce{HCl}$ must have been used.

I think, this is where you went wrong. If a question says something along the lines of

  The end-point occurs at $\dots~\mathrm{ml}$

then you should generally assume that that volume of whatever you are titrating with was used. In this case, that would mean a full $21.3~\mathrm{ml}\ \ce{HCl}$. The same thing of course happened at your point 7, where you write:

  It says that for the next reaction, $25\ \mathrm{mL}$ of $\ce{Ba(OH)2}$ was used and the end-point occurred at $27.10\ \mathrm{mL}$. From this information, I worked out that $0.0021\ \mathrm{L}$ of $\ce{HCl}$ must have been used in this reaction.

Working through this bit by bit, and starting at your point 5:

$$c(\ce{HCl}) = \frac{n(\ce{HCl})}{V(\ce{HCl})} = \frac{0.003976~\mathrm{mol}}{0.0213~\mathrm{l}} = 0.1867~\mathrm{M}$$

For the titration of $\ce{Ba(OH)2}$:

$$n(\ce{HCl}) = c(\ce{HCl}) \times V(\ce{HCl}) = 0.1867~\mathrm{M} \times 27.10~\mathrm{ml} = 5.059~\mathrm{mmol} = 0.005059~\mathrm{mol}\\
n(\ce{Ba(OH)2}) = \frac{n(\ce{HCl})}{2} = \frac{5.059~\mathrm{mmol}}{2} = 2.529~\mathrm{mmol} = 0.002529~\mathrm{mol}\\
c(\ce{Ba(OH)2}) = \frac{n(\ce{Ba(OH)2})}{V(\ce{Ba(OH)2})} = \frac{2.529~\mathrm{mmol}}{25~\mathrm{ml}} = \frac{0.002529~\mathrm{mol}}{0.025~\mathrm{l}} = 0.1012~\mathrm{M}$$

The difference still there can be attributed to a rounding error somewhere along the way.


Is the Self Consistent Field an actual field? I know this is really stupid. But bear with me.

In everything I have read so far, the Self Consistent Field (SCF) is always used to describe a method or a process, namely, that of iteratively solving the Schrodinger's Equation to generate new wavefunctions, using them for a new potential, and solving until convergence.

If SCF is not an actual field, why is it named as such? And then when we describe the SCF space what are we really talking about? Does the iterative process create a series of objects that spans a field?

For SCF: https://en.wikipedia.org/wiki/Hartree%E2%80%93Fock_method or if you want a textbook for reference, there's always Szabo and Ostlund's Modern Quantum Chemistry.

The notion of a self-consistent field comes from the orbital approximation - that a multi-electron wavefunction can be approximated as multiple one-electron wavefunctions.

At that point, the question remains - how does each of those electrons interact with the remaining $N-1$ electrons?

Well, there's an electric field from the electron density, right? Maybe you want to call this an electrostatic potential -- but Hartree considered it a spherically-symmetric electric field.

The problem, of course, is that since you don't know the final charge density, you have to make a guess and iterate until it's "self consistent."

Thus, SCF has been associated first with HF theory, and other related methods, including Kohn-Sham DFT.


Synthesis of 2-Phenyl-2-Butanol from Grignard Reagent The question is this:

2-Phenyl-2-Butanol can be synthesized by three different combinations of a grignard reagent and a ketone. Show each combination.

I can only think of one.

Butanone with Br-Mg-Ph

I can't think of anything else.

I'm not too sure if this will work:

phenylbutanone withe Br-Mg-Me

now i'm stuck. i can't think of the last one.

Hint: The carbon atom bearing the OH group in the product has three substituents--phenyl, methyl, and ethyl. Any one of these groups could be introduced as a Grignard reagent, with the appropriate ketone. You have already correctly deduced one of the three possibilities.


Exothermic reactions at high temperature and the chemical equilibrium Why are many exothermic reactions, like Haber process, conducted at high temperature, thus  their unfavourable equilibrium position?

There are two competing problems in the Haber Bosch process: the equilibrium position and the reaction rate. The reaction: $\ce{N_2 + 3H_2 -> 2NH3}\ (\Delta H = -92.4 \mathrm{\ kJ/mol})$ is exothermic and the equilibrium is actually favourable at room temperature, but the rate is very slow due to the large energy barrier to breaking the triple bond in $\ce{N2}$. To increase the reaction rate, a catalyst is used at high temperature. As you say, the high temperature pushes the equilibrium towards the reactants (by the Van 't Hoff equation) so there is a competition between yield and rate.

In practice, high pressure is also used to increase the yield (since the products side of the equation has fewer molecules), and like many industrial processes, it's run as a recirculating process. At manageable pressures and temperatures, they may only get 10 or 15% yield after one pass, but the ammonia is removed and the unreacted gases passed back into the reactor, giving quite high overall yield.


Is pentamine or pentylamine the correct name? For $\ce{CH3CH2CH2CH2CH2NH2}$, is the correct name pentylamine or pentamine?

From http://antoine.frostburg.edu/chem/senese/101/organic/faq/yl-ending.shtml

  "The -yl ending means "a fragment of an alkane formed by removing a hydrogen".

How does the above definition apply to this problem?

Does it mean, if the N wasn't there it would be a carbon and 3 hydrogens would attach to the carbon; but with the nitrogen there there is only 2 hydrogens attached to a the nitrogen?

Thank you

The following IUPAC guidelines suggest the correct systematic nomenclature for simple primary amines.

P-62.2.1.2
Primary amines are systematically named in the following ways:
(1) by adding the suffix ‘amine’ to the name of the parent hydride;
(2) by adding the name of the substituent group R - to the parent hydride ‘azane;
(3) by adding the name of the substituent group R - to the term ‘amine’ used as a preselected parent hydride name for NH3; this method is used only with monoamines.

Method (1) gives the preferred IUPAC name (PIN). So in your example this would give pentan-1-amine.
Method (3) gives what is often the more commonly used name, in your example this would give pentylamine.
Both names are 'fine' and in reality method (3) is the one that you will most often see used/written on bottles of reagent etc. But for the correct IUPAC identifier method (1) should be used.
You should also note that pentamine is a wholly different compound:

Other example:

By method (1). Methanamine
By method (2). Methylamine


Determination of Urea and Hydrogen Peroxide in Cow Milk by ATR-FTIR(Attenuated total reflection-Fourier Transform Infrared Spectroscopy) I am doing a research on cow milk adulteration. I am checking Urea and Hydrogen Peroxide content in milk by ATR-FTIR(Attenuated total reflection-Fourier Transform Infrared Spectroscopy). I am using $\ce{1M}$ Urea solution and $\ce{1M}$ Hydrogen peroxide solution. I am adding this to cow milk and testing through FTIR. But i am not getting good IR peaks of $\ce{H2O2}$.

Should i dilute the solutions? As we know, UV spectrophotometer gives better peaks at dilution. But in case of IR what should i do to get better peaks? what concentration of $\ce{H2O2}$ should i use?

In contrast to a typical sample provided by a synthetic lab your sample consists of multiple components; milk -- already consisting of many chemicals for its own, plus urea added, plus hydrogen peroxide added (as aqueous solution).  Using ATR-FT-IR is indeed a good choice for characterizing such samples, both for qualitative, as well as quantitative analysis (even to the point to provide kinetic data) as non-distructive, and potentially on-line method for continous process control.

This being said, you may improve the situation in the way how the ATR data is recorded.  Most often ATR-spectrometers are designed in a single-reflection geometry, wich is fine for not too complex sample compositions.  Potentially better though here is to use a design with a crystal allowing multiple reflections.  (Do not forget to keep the surface of the crystal clean, as (bio)films and absorbed material in general may affect the analysis, esp. with ATR flow cells.)

(source)

As a second hint, after the accumulation of your spectra, you may facilitate recognizing the barycentre of the absorption peaks by looking at their first derivative.  This technique sometimes seen in the field of UV-Vis spectroscopy (ref) is equally helpful if absorption bands are broad, or -- for example because a multi-component system is scrutinized -- partially overlap (e.g., IR-imaging).

At third, data analysis.  Beer-Lambert's Law, likely known from UV-Vis spectroscopy, equally is applicable in IR spectroscopy and you may choose picking an absorption band belonging only to your component of interest.  Because of the complexity of your sample, however, I would suggest that you use Partial Least Squares regression analysis (PLS) as a multi-variate analysis instead.  Very much simplified, instead of performing a regression analysis at one wavelength of interest, you do so for all data points, across the complete spectrum recorded, which equally allows to account for multiple components, too.  Do not worry, there are programs around to assist in this process, either provided with the IR spectrometer (e.g., TQ analyst by ThermoScientific), or independented solutions (e.g., Unscrambler).  Often, PLS allows to lower the detection limit compared to the approach based on a single absorption band quite a lot.

An example of determining hydrogen peroxide in water with PLS by ATR-FT-IR may be found here, of urea in milk here, or milk adultered with Hydrogen peroxide (0.15 g/L) here:

(original source of the article.)


Best way to prepare reference crystal structure with a given force field I was wondering if somebody could give me some insight into best ways to produce crystal structures, with a given force field, at 0 K. I am interested specifically in glycine, and have obtained the experimental structure from the CSD. I have seen in the literature people refer to annealing, where I equilibrate my crystal at a large temperature, and then anneal. However, I am concerned about the effects of such things as annealing rate here. I have also seen people mention performing energy minimization. However, my concern is typical energy minimization routines, at least to my knowledge, do not adjust the simulation volume according to a specified pressure. Thus, if my virial is positive for my experimental atomic coordinates using my force field, I don't think the minimizer would expand the volume to bring the pressure back to 1 atm. Could anyone give me some guidance. I am trying to prepare glycine crystal structures, at 0 K, using the GAFF force field with the CNDO charges.

I'm not entirely sure what you're asking. Are you attempting to match the geometry of a molecule from a crystal structure or the entire crystal packing?

Computational prediction of crystal structures is very much an unsolved problem. There have been multiple "blind tests" from the Cambridge Crystal Database, with varying levels of success. In general, the most accurate results come from dispersion-corrected DFT, not force field methods.

In some cases, to speed results and screening, teams will prepare "tailored force fields" by deriving optimized parameters for a particular compound on-the-fly from the DFT calculations. In no way are these intended for general purpose force field methods.

Also, most methods typically have to optimize both the unit cell parameters and the geometry of the compounds in them. Your note about expanding the volume is important - if you are working in this field, you may wish to use such heuristics to optimize the cell parameters, but typical minimizers will focus on either atomic coordinates or unit cells but not both.


How many samples would be worth when following a desorption? How to relate with limit of detection? I am following the desorption of a compound from a polymer. Let's imagine that it occurs at a certain rate and most of the compound is desorbed after 5 days (as in the figure). Then, I could use many analytical techniques to follow the compound desorption by measuring at, e.g., t = 1 day and t = 4 d. The concentration difference between these two points is large enough to be detected.

However, if I the desorption is much slower (as in the figure below) and I can only take samples for 10 days, the only way to characterise the desorption rate would be with a much more sensitive analytical technique, as I would need to detect changes from, e.g. 0.98 to 0.96

My question is, supposing that I have a good guess of the concentration differences that I will encounter, how can I choose an analytical technique beforehand based on reported limits of detection and/or quantification? How can I relate limits of quantification with the ability to differentiate a sample at 0.98 and another one at 0.96?

I think you are confusing two concepts: the limit of quantification (LOQ) and the sensitivity of the technique.

LOQ is the lowest concentration of the analyte that can be quantified at an acceptable level of uncertainty and accuracy, and is determined (by most methods) as the calculated concentration of the blank + 10*sd of the blank measurements.

Sensitivity, on the other hand, is the method's ability to distinguish between different amounts of analyte.

There are some guidelines that do not make a differentiation between the two (such as in the FDA's Bioanalytical Method Validation guidance document), but this is generally not the case in other situations.

In your case, you will need to find an analytical method that is sensitive enough to differentiate between 0.96 (a.u.) and 0.98(a.u), that is, 0.02 units.

As for the limit of quantification, you need to ask yourself what is the lowest concentration of analyte that you want to quantify. Based on the decay plot you have shown, it seems like the lowest concentration you would encounter is above 0.8 a.u. In that case, a method with an LOQ of 0.8 a.u should be sufficient. If you want to be even more rigorous, your LOQ should be at a level that is low enough that even if it is extended by the expanded standard uncertainty of your method, it should still be low enough to quantify the lowest level of analyte you expect to encounter.


Laboratory preparation of dinitrogen In my book the laboratory preparation of dinitrogen is as follows:

$$\ce{NH4Cl(aq) + NaNO2(aq) -> N2(g) + 2H2O(l) + NaCl(aq)}$$
Small amounts of $\ce{NO}$, $\ce{HNO3}$ are also formed in this reaction: these impurities can be removed by passing the gas through aqueous sulfuric acid containing potassium dichromate. $\ce{N2}$ is collected by downward displacement of water.

So my question is how does it actually happen that impurities can be removed by passing the gas through aqueous sulfuric acid containing potassium dichromate? What are the reactions involved? Any video in which the following process is being illustrated will be helpful to understand.
I have understood what does it mean when it is said that $\ce{N2}$ is collected by downward displacement of water.This video  https://www.youtube.com/watch?v=Ol-2JN01Cfo illustrates it nicely.

The collection of nitrogen $(\ce{N2})$ by downward displacement of water works since nitrogen is poorly soluble in water (at room temperature and standard pressure) and does not chemically react with water.

For the same reason, some impurities (aerosols, water-soluble gases and vapours – for example $\ce{HNO3}$) can be removed by passing the stream of nitrogen through a washing bottle that is filled with water.

Nitric oxide $(\ce{NO})$ is only poorly soluble in water and therefore cannot be removed by this method. However, $\ce{NO}$ can be oxidized. Strong oxidizing agents can convert $\ce{NO}$ into $\ce{NO3-}$:

$$\ce{NO + 2H2O <=> NO3- + 4H+ + 3e-}\quad E^\circ = +0.959\ \mathrm V$$

A suitable strong oxidizing agent is acidic dichromate $(\ce{Cr2O7^2-})$:

$$\ce{2Cr^3+ + 7H2O <=> Cr2O7^2- +14H+ + 6e-}\quad E^\circ = +1.38\ \mathrm V$$

The balanced overall reaction is:

$$\ce{2NO + Cr2O7^2- +6H+ <=> 2NO3- + 2Cr^3+ + 3H2O}$$


Completing inorganic reactions and writing spectator ions Complete the following reactions and write the spectators for each if any
A. cobalt(II) chloride (aq) and cesium phosphate (aq)

I am not quite sure if I have the correct answer to this equation, but I have $\ce{ CoCl2(aq) + Cs(PO4)(aq)}$ (for the equation setup) and for spectator ions $\ce{Co^2+ (aq) + 2 Cl- (aq) + Cs^+ (aq) + (PO4) (aq) -> CsCl(aq) + Co(PO4)2(s)}$ and so for the spectators I have $\ce{Cs^+(aq)\ and\ 2Cl- (aq)}$, Is this correct?

B. sodium phosphate (aq) and calcium chloride (aq)

For this equation I have $\ce{6Na^+\ and\ 6Cl-}$ as spectator ions, $\ce{Ca3(PO4)2(s)}$ (as a solid)

For B, you have correctly identified calcium phosphate as the precipitate, as it is the least soluble of all possible compounds. In the case of A, cobalt phosphate precipitates from the solution containing cobalt(II) and phosphate salts.

$$\ce{3Ca^{2+} + 2PO4^{3-} ->~ Ca3(PO4)2\downarrow}$$
$$\ce{3Co^{2+} + 2PO4^{3-} ->~ Co3(PO4)2\downarrow}$$


Rietveld (FullProf) general background function I was reading through Eric Mittemeijer's book 'Modern Diffraction Methods' when I found the following statement in the Rietveld chapter by Robert Dinnebier.

  It is common practice to use orthogonal Chebyshev polynomials of higher order
  (typically 5–10) in combination with an 1/(2θi) term to describe a steep increase in the background at low scattering angle [...]. If higher-order polynomials are used, different types of correlations can occur, which require careful checking of the correlation matrix.

When I am usually describing my background in FullProf I use the '6-Coefficients polynomial function'. But this seems to cause more problems the more parameters I refine. So I switched to the 'Chebyshev polynomial' and refinded all 24 parameters (just to make sure). This caused a large jump in the X² from about 10 to 6.5 while the background was much better described.

My question is, although it is mentioned in the book, that Chebyshev is more common and I should be careful with using higher order polynomials, is this something I can use in general for my refinements? Can I just automatically assume for most x-ray determined backgrounds that using a Chebyshev polynomial with as many parameters refined as possible will give me useful results? By useful I mean good and especially 'allowed' results. Because sometimes FullProf refines parameters into regions where they are not defined like a negative η for the Pseudo-Voigt function.

If I recall correctly, Chebyshev polynomials are preferred over simple polynomials because of their termination behavior (i.e. they tend not to diverge wildly), although I don't have a reference for this handy.

Dinnebier's admonition to use higher-order polynomials with care is effectively a warning not to over parameterize your fit. If you inspect the two fits with 6- and 24-term backgrounds, my guess is that you will find the over parameterized version nudged your Bragg peaks up or down as necessary to better match your observed data. Hence, your residual has diminished but your fit is not more meaningful.

An overfitted background precludes accurate extraction of the observed Bragg intensities, leading to correlated errors in related factors (e.g. occupancy, position, etc.).


How to reduce the turbidity of a water sample? A task requires us to experiment water quality.
How would you reduce turbidity of a clay water sample with an effective and cost-effective method?

Apart from the article linked in a comment above, possible options are

wait and let settle, then decant
use an established flocculant, such as potassium alum $\ce{KAl(SO4)2*12H2O}$


What reaction will ensue when sulfur trioxide is added to excess barium hydroxide solution? Question:

If $\ce{SO3}$ was added to an excess solution of $\ce{Ba(OH)2}$, what would be the resulting reaction?

Thoughts:
Initially, I suppose it would form barium hydrogen sulfate, which slowly precipitates barium sulfate in water as the hydroxide is present in excess.
$$\ce{2SO3 + Ba(OH)2 <=> Ba(HSO4)2 \\ Ba(HSO4)2 + Ba(OH)2 <=> 2BaSO4 + 2H2O}$$
Is this correct?

Basically yes.
Sulfur trioxide reacts with water to form sulfuric acid (this is extremely dangerous!  do not try it at home) in a strongly exothermic reaction which, unless done under very precise control, produces a fine mist of concentrated sulfuric acid vapor (not nice).
$$\ce{SO3 + H2O -> H2SO4}$$
If you are doing this with an excess of barium hydroxide then the sulfuric acid will be deprotonated in two steps as you suggest.
$$\ce{H2SO4 + OH- <=> HSO4- + H2O}$$ $$\ce{HSO4- + OH- <=> SO4^2- + H2O}$$
Barium sulfate will precipitate from the solution, driving the reaction to completion.
$$\ce{Ba^2+ (aq) + SO4^2- (aq) -> BaSO4 (s)}$$
I couldn't find any mention of barium hydrogen sulfate as a compound anywhere so I take this to mean that it is not stable and therefore will not precipitate.


Why is the ortho isomer a major product in the nitration of toluene? Usually when an ortho-para directing substituent is present on the benzene ring for an electrophilic aromatic substitution reaction, the para product is the major product (exceptions can be there when hydrogen bonding or ortho effect of COOH group makes the ortho product a major one.) But I don't understand why the ortho product in nitration of toluene is the major product (and the para product isn't). The ortho position is supposed to have more steric crowding, right?

A methyl-group is not t-butyl, so it isn't all that crowded.
There are two ortho-positions.
There is also the possibility of an ipso-attack with the nitronium donor attacking the position of the methyl group, then shifting to the ortho-position.

Given all that without any orientation effect, the ratio between ortho and para should be around 3/1 - 2/1. It is commonly observed around 3/2, so the crowding does have an effect.


How does mCPBA convert an amine to an N-oxide? Is this how the mechanism works?
Since it is a very stereoselective reagent I thought maybe it proceed via one step only, but how would that then work?

Firstly, any stereoselectivity in this reaction is not affected by the second step you drew (which is a simple proton transfer). This is because your second step neither creates nor destroys any stereochemistry present in the reactants.

However, mechanistically it is probably easier to deprotonate in the same step using the carbonyl oxygen of mCPBA.


Am I Using the Correct Number of Significant Figures (Heat Problem)? The question is:

  A patient arrives in the emergency room with a burn caused by steam. Calculate the heat, in kilocalories, that is released when $\pu{18.0 g}$ of steam at $\pu{100 °C}$ hits the skin, condenses, and cools to body temperature of $\pu{37.0 °C}$.

1) I solve for the steam as it condenses:

Heat of vaporization: $$\Delta H_\text{vap}(\ce{H2O}) = \pu{2260 J/g}$$

Heat: $$\begin{align}Q_1 &= \pu{18.0 g} \times (\pu{2260 J/g}) \times (\pu{1 cal/4.184 J}) \times (\pu{1 kcal/ 10^3 cal})\\ &= \pu{9.722753346 kcal} \to \text{rounded to 3 SF} = \pu{9.72 kcal}\end{align}$$

2) I solve for the cooling of the $\ce{H2O}$:

Specific heat capacity: $$c(\ce{H2O}) = 4.184\ \mathrm{J/(g\ ^\circ C)}$$

Temperature change: $$\Delta T = \pu{100 °C} - \pu{37.0 °C} = \pu{63 °C}$$ (2 SF)

Heat: $$\begin{align}Q_2&=\pu{18.0 g} \times \pu{63 °C} \times 4.184\ \mathrm{J/(g\ ^\circ C)} \times (\pu{1 cal/4.184 J}) \times (\pu{1 kcal/10^3 cal})\\&= \pu{1.134 kcal} \to \text{rounded to 2 SF} = \pu{1.1 kcal}\end{align}$$

I then add the two results together and get:

$$\begin{align}Q&=Q_1+Q_2\\&=\pu{9.72 kcal} + \pu{1.1  kcal} = \pu{10.82 kcal}\end{align}$$

$\to \text{rounded to 3 SF} = \pu{10.8 kcal}$ (using the addition/subtraction rule for SFs)

The textbook's answer is $\pu{11 kcal}$.

Is my answer right, according to significant figure rules, or is the textbook's provided answer correct?

This is sort of tricky, but I'd say that neither 10.8 nor 11 is right!

You got
$$Q_1 = 9.7\bar22753346\ \mathrm{kcal}$$

which you rounded to $9.72\ \mathrm{ kcal}$. The bar over the first two indicates that is the significant figure and that the excess numbers can be carried in intermediate calculations but would be trimmed for the answer.

Now for the second part you got

$$Q_2 = 1.\bar134\ \mathrm{ kcal}$$

which you rounded to  $1.1\ \mathrm{ kcal}$.

Using the intermediate calculations you are calculating the sum.

$$\begin{align}
9&.7\bar22753346\\
1&.\bar134\\
Q_1+Q_2 =10&.\bar856753346 \end{align}$$

Since 9.72+ answer is good to hundredths, but 1.1134 answer is only good to the tenths, thus the answer must be rounded to the tenths. So the "best" answer using significant figures is:
$$Q_1+Q_2 = 10.9\ \mathrm{ kcal}$$

You should not round intermediate results to their "correct" number of significant figures since that leads to rounding errors in the intermediate calculations.

$$\begin{align}
9&.7\bar2\\
1&.\bar1\\
Q_1+Q_2 =10&.\bar82 \end{align}$$

Obviously carrying the ten digits in $9.7\bar22753346$ is somewhat ridiculous, so a good rule of thumb with a calculator would be to carry two extra digits in intermediate calculations. So round the intermediate number to $9.7\bar228$. The value $1.\bar134$ has two extra digits so its fine as shown.

Notice that this is exactly the same rule you used in calculating the temperature difference.

$$\begin{align}
10\bar0&.\\
-37&.\bar0\\
=6\bar3&.\end{align}$$

Had it been 37.5 degrees for body temperature, then you should have used $6\bar2.5$ in the intermediate calculations.

$$\begin{align}
10\bar0&.\\
-37&.\bar5\\
=6\bar2&.5\end{align}$$

As a matter of curiosity I used the rnd() function in Excel to simulate uniform distribution between all the values. So:

17.95 grams to 18.04999 grams of water
99.5 to 100.49999 degrees C
36.95 to 37.04999 degrees C

$\text{=(2.26)*(18+0.1*(RAND()-0.5))/4.184+(18+0.1*(RAND()-0.5))*(100+(RAND()-0.5)-(37+0.1*(RAND()-0.5)))/1000}$

The 4.184 and 2260 values were assume to be exact. Averaging over a thousand such values I got 10.85703268
with a standard deviation of 0.016370732.

So the gist is that using significant figures doesn't give as good a result as possible. The problem is in the use of 2 significant figures in the calculation of $Q_2$. For the difference between the uniform differences of 100 and 37.0 the relative standard deviation is 0.46%. However the uniform distribution of 1.05 to 1.15 has a relative standard deviation of 2.6%, much worse than what is really happening.


Is the rate of desorption dependent on pressure? While I was learning about Langmuir's adsorption isotherm in my chemistry class, my teacher talked about the situation when the rate of adsorption and desorption will become equal. However, he mentioned that the rate of adsorption is dependent on pressure but the rate of desorption is not and when he gave an explanation I wasn't satisfied with it.
I would imagine that greater the external pressure on the adsorbent lesser will be the tendency of the adsorbate particles to leave the surface. But that isn't what my teacher said. Please explain.

Without knowing exactly what you were told its hard to be exact, but here is a description of the situation.

The Langmuir model makes assumptions (a) adsorption is complete when the surface is filled with one gas molecule  per site, (b) all adsorption sites are equivalent (i.e. the same) and (c) adsorption and desorption are separate processes, i.e. one does not depend on the other, i.e if one site is occupied it does not affect adjacent sites an any way. Note that this is a model and this determines how we look at the problem.
The model is then $$M(g) + S(surface) = GM(surface) $$ where G is gas and S surface and rate constant for adsorption is $k_a$ and for desorption $k_d$.

We let $\theta$ be the extent of adsorption which varies from 0 to 1, i.e. how full the surface is, then the rate of change of $\theta$ depends on the rate constant for adsorption , pressure and fraction of vacant sites or $N(1-\theta)$ for N vacant sites. Thus for adsorption $$ \frac{d\theta}{dt} = k_aPN(1-\theta) $$

and for desorption $$ \frac{d\theta}{dt} = -k_dN\theta $$

At equilibrium we make the rate of change equal to zero, i.e. rate of adsorption equals rate of desorption and with a bit of algebra find that $$\theta=\frac{KP}{KP+1}$$  where K is the equilibrium constant or $K=k_a/k_d$ This equation has the fancy name as the Langmuir Isotherm.

You can see that the rate going onto the surface must be equal to the pressure (no gas, no adsorption) and the area free to adsorb onto $(1-\theta)$. (We assume in this model than no molecule double adsorbs; this is dealt with in more advanced treatments). The rate leaving has to be proportional to the amount already there , i.e. proportional to N the total number of surface sites multiplied by the fraction covered.

Ask more questions in the comments if this is not clear.


Magical reaction tanks I would be interested to know people's opinion on the following matter.

Background

Often, when the far-from equilibrium behaviour of a chemical reaction system is analysed mathematically (in papers or text books), the system is placed in a gradient between a fixed non-zero concentration of "resource" species, and a fixed concentration of "waste" species (which is often zero). The other species in the reaction system are designated as "intermediates", and it is their behaviour which is of prime interest.

For example, for the oscillating Lotka-Volterra reaction system:

$R + X \rightarrow 2X$

$X + Y \rightarrow 2Y$

$Y \rightarrow W$

Species $R$ and $W$ are the fixed concentration resource and waste species respectively, and species $X$ and $Y$ are the (oscillating) intermediates.

Question

It appears as if these reaction systems are being studied in a physical scenario which could not be realised in the lab, a magical reaction tank if you like.

The two reasons most often stated as to why resource and waste are held at constant concentrations, whilst the intermediates can vary, are that:

1) The reaction takes place in a large reservoir, where resource is abundant and where any waste produced by the reaction quickly diffuses away into the vastness

or

2) The reaction has resource species quickly "pumped in" by some external agent, and likewise waste species quickly "pumped away".

My problems with these explanations are as follows:

If in explanation 1, a very large reservoir is keeping resource and waste species at constant concentration, then if follows that the intermediate species can never reach any appreciable concentration, for they too exist in the same large volume. In other words, if the waste diffuses away, then why don't the intermediates?

Conversely, in explanation 2, if a pumping process is artificially maintaining resource and waste concentrations, it is hard to see how this pumping process is being "micro selective" in completely removing the waste compounds, but leaving the intermediate compounds totally unaffected.

More physically realistic models of chemical reactions can be made by also explicitly including the reactor in the model - for example by considering the reaction as taking place in semi-permeable compartments or in bulk in a CSTR. However, additionally modelling the reactor often changes the mathematical properties for the "bare" reaction reported in the literature (for example, in a CSTR, the intermediates as well as the waste are removed).

I would say it's worth investigating. My suspicion is that the answer won't change much, but the way to find out is to try it both ways and see how the assumptions affect the answer under various conditions.

Instead of a CSTR, for this particular reaction a batch reactor might be more appropriate, since that is the easiest setup for a lab. In that case, to model the reaction all we have to do is not assume that some of the rates of change are zero, and see how that changes the reaction.

Fortunately for us, this site has already done the heavy lifting and even has a solution in a mathcad file.

I haven't looked at the solution closely enough to determine if they are making the approximations you mentioned (I don't have mathcad installed), or if they are just directly integrating all four equations. They only show graphs for the concentrations of X and Y. However, it probably wouldn't be too hard to modify in either direction to compare the results under different sets of assumptions.


Deriving the irreps of the d-orbitals under C3 and S6 operations in octahedral symmetry Contextual background: I am trying to reproduce the symmetries in the graph on this page (MO diagram for $\ce{ML6}$ complex). I have reproduced the metal s and p orbital representations as well as the linear combination of representations for the $\ce{L6}$ fragment. All that remain are the metal d orbitals.

So I know the five d-orbitals collectively reduce to the $\mathrm{e_g}$ + $\mathrm{t_{2g}}$ representation. I put my five d-orbitals at the origin and went through the operations for the reducible representation. Everything else adds up, except I can't find that $-1$ in the $C_3$ and $S_6$ operations.

Character table

$$\small\begin{array}{c|cccccccccc|cc}\hline
O_\mathrm{h} & E & 8C_3 & 6C_2 & 6C_4 & \begin{aligned}3C_2 \\ \scriptsize=C_4^2\end{aligned} & i & 6S_4 & 8S_6 & 3\sigma_\mathrm{h} & 6\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_{1g}} & 1  & 1  & 1  & 1  & 1  & 1  & 1  & 1  & 1  & 1  & & x^2+y^2+z^2 \\
\mathrm{A_{2g}} & 1  & 1  & -1 & -1 & 1  & 1  & -1 & 1  & 1  & -1 & & \\
\mathrm{E_g}    & 2  & -1 & 0  & 0  & 2  & 2  & 0  & -1 & 2  & 0  & & \begin{aligned}(2z^2-x^2-y^2,\\ x^2-y^2)\,\,\,\,\,\, \end{aligned} \\
\mathrm{T_{1g}} & 3  & 0  & -1 & 1  & -1 & 3  & 1  & 0  & -1 & -1 & (R_x,R_y,R_z) & \\
\mathrm{T_{2g}} & 3  & 0  & 1  & -1 & -1 & 3  & -1 & 0  & -1 & 1  & & (xy,xz,yz) \\
\mathrm{A_{1u}} & 1  & 1  & 1  & 1  & 1  & -1 & -1 & -1 & -1 & -1  & & \\
\mathrm{A_{2u}} & 1  & 1  & -1 & -1 & 1  & -1 & 1  & -1 & -1 & 1  & & \\
\mathrm{E_u}    & 2  & -1 & 0  & 0  & 2  & -2 & 0  & 1  & -2 & 0  & & \\
\mathrm{T_{1u}} & 3  & 0  & -1 & 1  & -1 & -3 & -1 & 0  & 1  & 1  & (x,y,z) & \\
\mathrm{T_{2u}} & 3  & 0  & 1  & -1 & -1 & -3 & 1  & 0  & 1  & -1 & & \\ \hline
\end{array}$$

Work

$$\begin{array}{c|cccccccccc} \hline
O_\mathrm{h} & E & 8C_3 & 6C_2 & 6C_4 & 3C_2 & i & 6S_4 & 8S_6 & 3\sigma_\mathrm{h} & 6\sigma_\mathrm{d} & & \\ \hline
\mathrm{E_g}    & 2  & -1 & 0  & 0  & 2  & 2  & 0  & -1 & 2  & 0 \\
\mathrm{T_{2g}} & 3  & 0  & 1  & -1 & -1 & 3  & -1 & 0  & -1 & 1  \\ \hline
\Gamma_{\text{d-orbitals}} & 5 & ? & 1 & -1 & 1 & 5 & -1 & ? & 1 & 1 \\ \hline
\end{array}$$

As can be seen, everything matches up but the $C_3$ and $S_6$ operations. I tried using Molecule Viewer and rotating the d-orbitals, but I still don't see the $-1$. They all look like zero to me.

I would expect the $\mathrm d_{x^2-y^2}$ orbital to correspond to $E_g$ from the character table, but I only see it rotate into a "$\mathrm d_{y^2-z^2}$" orbital, with the orbitals along the y and z axes instead of x and y axes. The $\mathrm d_{z^2}$ orbital rotates into a "$\mathrm d_{x^2}$" orbital, lying along the x axis instead of the z axis.

Anyone know which d-orbital rotates into $-1$ for a $C_3$ operation? I assume a $-1$ for $S_6$ is the same orbital, but if not then does anyone which that one is as well?

Thank you

The $C_3$ operation rotates $x \rightarrow y$, $y \rightarrow z$ and $z \rightarrow x$.
Hence,
$$\phi_1=2z^2-(x^2+y^2)$$
rotates into
$$-(1/2)\phi_1 +(3/2)\phi_2$$ and
$$\phi_2=x^2-y^2$$
rotates into
$$-(1/2)\phi_1 -(1/2)\phi_2$$
so, in the basis
 $( \phi_1, \phi_2 )$ the matrix that represents $C_3$ is

$$\left [\begin{matrix} -1/2 & -1/2  \\ 3/2 & -1/2  \end{matrix}\right]$$

Denoting this 2 by 2 matrix by M what we have is
$C_3(\phi_1  \phi_2) = (\phi_1 \phi_2)M$.
The trace of this matrix is $-1$.

This is what you find in the character table.

Since $C_3^3=I$ (identity) its eigenvalues are the cubic roots of unity:
$1, \lambda$ and $ \lambda^*$
where $\lambda = \exp(2\pi i/3)=-1/2+i\sqrt 3/2$,
and $\lambda^*$ is the complex conjugate of $\lambda $.

It is easy to find eigenvectors with the eigenvalue 1 [e.g., an s orbital or x+y+z].

You can find a linear combination of $\phi_1$ and $\phi_2$ that will be an eigenvector of $C_3$ with the eigenvalue $\lambda$ (transform into itself multiplied by $\lambda$)
and another linear combination that will have eigenvalue $\lambda^*$. Now, the matrix representing $C_3$ will be diagonal with $ \lambda$ and $\lambda^*$ along the diagonal. The trace remains $-1$.
Something similar should work for $S_6$.


Hydrolysis of vinyl silanes Alkynyl silanes can be easily hydrolyzed with TBAF like in the last step in the synthesis:

How feasible is the hydrolysis of vinyl silanes with TBAF? Does it actually happen or some other routine must be used?

According to This article, vinylsilanes are stable in TBAF solutions, allowing other silyl groups to be removed selectively. They can be activated in the presence of TBAF using palladium similarly to a halide, though.

This link shows C2-trialkylsilylalkenes being cleaved with potassium t-butoxide, 16-crown-8 and TBAF. I'll look more thoroughly when I get to a computer, though.

I also found this example, but it had no reference, so I'll try to find one:


What are the spectroscopic signatures that differentiate triple state water from water at 1 atm? According to the Wikipedia article Triple point of water, the triple point of water occurs at

  The single combination of pressure and temperature at which liquid water, solid ice, and water vapor can coexist in a stable equilibrium occurs at exactly 273.16 K (0.01 °C; 32.02 °F) and a partial vapor pressure of 611.657 pascals (6.11657 mbar; 0.00603659 atm).

Shown in the illustration below (from the Wikipedia article):

Crucially, to put this into a context, the 611.73 pascals quoted in the article corresponds to the

  minimum pressure at which liquid water can exist. At pressures below the triple point (as in outer space), solid ice when heated at constant pressure is converted directly into water vapor in a process known as sublimation.

Potentially useful information to determine if an exoplanet can possibly have water (if all other conditions are suitable).

What are the spectroscopic signatures that differentiate triple state water from water at 1 atm?

The triple point of water is point of dynamic equilibrium between three phases of water. Any small change in Pressure or Temperature will lead to one phase regions or two phase lines on the phase diagram.

The triple point is not a state of matter rather a combination/mixture of three phases . So i very much doubt we can get any unique spectroscopic signatures . We would get a distorted result as this point has three different phases.
Also Emission and Absorption spectra depends on  temperature and pressure.


Work done in reversible isothermal expansion A cylinder fitted with a frictionless piston is filled with $10~\mathrm{mol}$ of gaseous carbon tetrachloride and immersed in a large bath of water maintained at $300~\mathrm{K}$. The volume of the gas is initially at $V_1~\mathrm{L}$, and the external pressure on the piston is slowly decreased until the volume reaches $V_2~\mathrm{L}$. Calculate the work for this process, in $\mathrm{kJ}$, assuming the gas can be treated as ideal.

I think I have the general idea of how to solve this problem but I'm not sure how to find the external pressure. So far, I'm using the equation

$$\int \mathrm{d}w = \int_{V_2}^{V_1}-p_{\mathrm{ext}}\,\mathrm{d}V$$

since it is pressure-volume work and have tried calculating $p_{\mathrm{ext}}$ using $pV=nRT$ and the final volume of the system, but this doesn't seem to give the correct answer. How does it have to be solved?

I agree with getafix, if you would like an answer that is more tailored to you, you should show us exactly what you've done.

However, I am going to make a (hopefully educated) guess that what you did was to pull $p_\mathrm{ext}$ out of the integral. That is incorrect, because $p_\mathrm{ext}$ is not a constant here.

This process is known as an isothermal expansion - isothermal because the temperature remains constant throughout - and expansion because volume is increased. In thermodynamics it is very important to note which variables are held constant, because then that lets you decide which formula is appropriate to use, or how to derive such formulae).

Since the process is reversible, the external pressure must always be equal to the pressure exerted by the gas, which can be calculated via the ideal gas law $pV = nRT$. Therefore, you have (where 1 and 2 denote the initial and final state respectively)

$$\begin{align}
w &= -\int_1^2 p\,\mathrm{d}V \\
&= -\int_1^2 \frac{nRT}{V}\,\mathrm{d}V
\end{align}$$

and now since $T$ is a constant, you can take it out of the integral (along with $n$ and $R$ which are also constants)

$$\begin{align}
w = &= -nRT\int_1^2 \frac{1}{V}\,\mathrm{d}V \\
&= -nRT\ln\left(\frac{V_2}{V_1}\right)
\end{align}$$

Since you weren't given numerical values for $V_2$ and $V_1$, you can't evaluate a numerical value for $w$. The closest you can get is therefore

$$w = (-24.9~\mathrm{kJ})\ln\left(\frac{V_2}{V_1}\right)$$


Determining the charge on an ion through mole calculation in electrolyis During electrolysis $\pu{0.02 mol}$ of metal $\ce{X}$ is deposited on cathode when $\pu{0.12 mol}$ of electrons are passed through molten chloride of $\ce{X}.$ What is the formula of the chloride?

I have tried solving it and my answer is $\ce{XCl6}$ which is incorrect. The correct answer given is $\ce{XCl3}.$

In the explanation the book says that $\pu{0.06 mol}$ of electrons have flown to cathode and goes on to divide the amount of electrons at cathode by the amount of metal $\ce{X}$ $(0.06/0.02),$ so it's $3$ which means $1$ atom of $\ce{X}$ gains $3$ electrons, giving it $+3$ charge.

What I do not get is what about the other $\pu{0.06 mol}$ of electrons. Isn't metal $\ce{X}$ supposed to gain all of the $\pu{0.12 mole}$ of electrons? Shouldn't cathode have the total $\pu{0.12 mol}$ of electrons?

The suppose total and half reaction are:

$$\begin{align}
\ce{2 MCl_n &-> 2 M(s) + $n$ Cl2} \\
\ce{M^n+ + $n$ e- &-> M(s)} \\
\ce{2 Cl- &-> Cl2 + 2 e-}\\
\end{align}$$

The question is rather tricky.

Note that I am addressing the supposed question in hand, not the electrochemical evaluation.

In some sense, if we consider large scale motion, no electrons passed through the electrolyte, they just got on one side in and on the other side out.

In usual sense, there is 3:1 electron:halogenide molar ratio for $\ce{MCl3},$ therefore result would we $\pu{0.06 mol}.$

But in yet other sense, if we consider nanoscale motion, the $\pu{0.06 mol}$ of electrons passed through the electrolyte at the anode, and the same amount through the electrolyte at the cathode, therefore in total $\pu{0.12 mol}.$

By other words, $\pu{0.12 mol}$ is total molar amount of electrons involved in net electrochemical reaction, passing through the electrolyte between electroactive substances and electrodes.

I would say, the question is either intentionally tricky either badly formulated.


Why is copper (II) oxide used in quantitative analysis of carbon? What is the significance of using cupric oxide in the quantitative analysis of carbon?

As far as I can tell from my research, only $\ce{CuO}$ is used to oxidize carbon to $\ce{CO2}$. Why can't we use other oxidants like $\ce{HNO3}$ or other metal oxides such as hematite? Is the use of $\ce{CuO}$ prevalent because of low cost or does it has some other special property?

You got the crux of it: low cost.  And it works, regardless of what the starting compound was.  And it's not just low cost to purchase, it is easy to regenerate the used cupric oxide multiple times by heating it under oxygen.


Alcohol irregularities in Hydrogen NMR We've started learning about NMR in my ochem class and my instructor told us that in Hydrogen NMR an alcohol can have a fairly wide range of possible values - usually from about 2 to 5.  He said that because of this the actual ppm value is not very diagnostic - but the fact that alcohols tend to have a thicker peak is.

He told us not to really worry about why/how there is such a range or why the reading tends to be thicker, but now I'm curious.

I speculate that there can be such a wide range of possible values because Oxygen is fairly electronegative, and depending on its environment it will exhibit different levels of shielding/deshielding.  I have no idea about why it tends to be a thicker peak.

So I guess I have two questions:

Is my speculation as to the range of possible values correct?
Why is the peak generally thicker with alcohols than with other functional groups?

Alcohols have labile protons which are exchangeable through acid-base equilibria with other similar protons, such as other alcohols, amines, amides, thiols and especially water. This reaction is fast enough, on the NMR timescale, that it can be observed as a reversible exchange process. The most common scenario is that there will be some amount of water in the sample, and so the alcohol and water 1H atoms exchange. The reaction is typically so fast that you don't see the water and alcohol as separate peaks, but as a population-weighted average. Therefore the actual position of the peak depends on a number of variables; temperature, solvent, starting chemical shifts of the exchanging species, populations of the exchanging species, rate of exchange, solvent and so on.

When you say 'thicker', I assume you mean a broader linewidth. Again, this is a consequence of the exchange process, and that you are seeing an average peak for the -OH and (say) H2O. Sometimes this broadening is so large that you can only observe it as a broad hump in the baseline of the spectrum. Again, the linewidth will be a function of the starting chemical shifts of the two components, the rates of exchange, and their relative populations.

It is actually entirely possible to see the alcohol peak with a narrow linewidth, similar to the CH2 and CH3 linewidths. To do this, you need to be able to eliminate the exchange process, either by excluding water from the sample( run ethanol in very dry dmso for instance), or cooling down the sample sufficiently to slow down the exchange (probably not feasible for most common solvents with ethanol and water though). In this case, you will observe your alcohol as a narrow triplet (approx 1.7Hz coupling by memory), with the CH2 appearing as a doublet of quartets. A great example of observing alcohol protons is to run sucrose dissolved in pyridine-d5. All OH protons, and their couplings are observable.

It is really important to understand that the integrals of these peaks are still valid. In the absence of any exchange (where your alcohol peaks are sharp), the integration of an  ethanol 1H spectrum should still give you 3:2:1. Where there is an exchange process occurring with (say) water, the integrals will be 3:2:X, where X=(sum of -OH plus water). This method of analysis can be used to verify alcohol content of spirits and wine, for instance.


Do lone pairs on substituents (e.g. in aniline) count towards Hückel's rule? Why is aniline aromatic? Doesn't it have 8 π electrons including the lone pair on nitrogen, thereby violating Hückel's rule?

The way I see it, there are 6 π electrons from the benzene ring, and an additional 2 from the $\mathrm{sp^2}$hybridized nitrogen.

I am suspecting that I am mis-applying Hückel's rule. I asked an instructor today and the only rationalization he had to offer was that "we just don't count those electrons on the nitrogen" ... which is deeply unsatisfying. He still insisted that aniline is aromatic. Can someone offer an alternate, more through explanation?

It is only the electrons from atoms in the ring that count when applying Huckel's rule. Electrons from substituents on the ring are only cross-conjugated with the aromatic π-system. When you think of it in terms of perturbational molecular orbital theory the substituent electrons take the part of a perturbation for the ring's aromatic system.

The reason for only using the ring electrons traces back to the origin of aromaticity. Take a look at the π-orbital energy levels of benzene (figure taken from p 36 of Fleming, Molecular Orbitals and Organic Chemical Reactions, Reference Edition):

Here, $\alpha$ is the energy of an isolated p-orbital and $\beta$ is the stabilization energy of the π-bond in ethylene. You can see that for benzene you get the best stabilization if you have 6 π electrons ($4n + 2$ for $n=1$) in the system because then you fill the stabilized MOs $\psi_1$, $\psi_2$ and $\psi_3$ which are lower in energy than an isolated p-orbital and gain the maximum stabilization energy possible. If you'd have more electrons in the system you would have to fill the destabilized MOs $\psi^{*}_{4}$, $\psi^{*}_{5}$ and $\psi^{*}_{6}$ which would cost a lot of energy because of the large energy gap between the bonding and antibonding MOs.

Now, what happens if you add a cross-conjugating substituent to the ring. This can be seen in the (Huckel-)MO-diagram of styrene (Fleming, p 71).

The main effects of the substituent are that the degeneracies of some orbitals are lifted and that the orbital energies shift a bit, especially those of HOMO and LUMO are of interest. In the case of styrene the HOMO lies higher in energy and the LUMO lies lower in energy compared to benzene. Now there are 4 stabilized energy levels instead of 3 - any electron that occupies one of those levels will lead to a stabilization of the molecule. But that is nice and well as you get 2 electrons from the substituent which will want to occupy one of the 4 levels and that leaves you with 3 levels that have to be occupied by ring electrons. Thus, your system will be most stable if the π-system of the ring contains 6 electrons, exactly like in benzene. Any more than 6 electrons will lead to destabilization as there is still quite a big HOMO–LUMO gap (even though it is smaller than in benzene).

The argument I made for styrene will remain valid for other molecules with different substituents (like aniline). The MO scheme of aniline will look a little different but the essential features should be the same.


How to wedge d/hash notation a molecule? This is probably trivia for many of you (I already feel the pressure by even posting this question). Yet funny enough, most of the time the materials I have come across show the wedge-d/hash notation. Then due its cumbersome or what not, the material decides to propose Fisher project as a better notation. But frankly that doesn't give the proper picture of the stereochemistry or least understanding of how the atoms would arrange in space.

When revising Rectus-Sinister (R/S) system for stereochemistry, it has almost always indicated about assigning priority to the groups by descending order of the atomic number and that lowest priority substituent should always point away from the plane of the central atom. Speaking of a simple molecule like $\ce{CH4}$, Wedge-Dash is ideally suited to show structure of sp3 hybridised (tetrahedral atoms).

But what about more complex compounds/molecules like amino acids or things like drugs like morphine etc? or even an imaginary molecule given in exam? I dread at the thought of it.

I went through this question for few days now, before posting this question. Perhaps it has more advanced perspective compared to the simplicity of my question I think.

So are there direct guidelines to draw the wedge-h/dash notation for a given complex molecule? Am I really missing something from preliminary general chemistry here?

Obviously there are online applets that allow to get a SMILE of a molecule and generate the 3D structure like JSmol. I also know there are complex methodologies employed in higher level of physical science to determine the structure of molecules. But at this beginning level, is there anyway to determine the 3D structure of a molecule? At least to get a feeling of it, and to predict that what atom/side chain would pop out of plane and remain in the plane?

I am a beginner in Chemistry, so should I really bother about this at all?

If not using a Fischer projection, structural formulae with hashes and wedges allow to be more specific.
In the structure drawn in your current question, you provide information about the connectivity of atoms within the molecule, up to a point to indicate if the bond between two atoms is a single, double, or (not present here) a triple / quadruple.
Using wedges and dashes, equally provides this information; however additionally informs the reader about the sterochemical relationship of the atoms bound around an atom in a centre, addressing a specific (R)/(S) configuration.  Returning to your example, there are three sterogenic centres that I marked by an asterisk:

and for each centre you may find a pair of either configuration; for example for the hydroxyl group in $\alpha$ position to carbonyl alone

In total (for 3 stereogenic centres), there are $2^3 = 8$ distinct stereoisomers possible with different properties, which all fit into the first "line only" general representation.


Can oxidizing agents be used to increase the combustion efficiency of waste oil? Waste oil is atomized and combined with air to produce a very high temperature flame capable of melting iron. I was wondering if it's possible to further increase its combustion efficiency by mixing it with an oxidizing agent such as Sodium Perchlorate, Hydrogen Peroxide or some other oxidizer in mixable form.
Conceivably, it could enhance the combustion since it adds more oxygen to the fuel, especially at higher temperatures. At the same time, it could also inhibit the combustion since the additive may absorb energy from the reaction than it produces, since it's already an oxide.

It is safer to keep the process as simple as possible without the addition of perchlorates or peroxides to the oil.  Perchlorates are salts with low solubility in oil; hydrogen peroxide equally is a polar compound with little affinity to form a solution with oil.  Thus, the salts could clog the fuel delivery, or in case of $\ce{H2O2}$ simply yield a biphasic liquid mixture if stored in the same container as the fuel.  It equally would increase the potential hazard to run such a combustion engine; perchlorates are heat sensitive and could explode when you don't want them to do so. In addition, over time residues of salts and salt residues facilitate corrosion (already a problem for the transportation of crude oil); here, the elevated temperatures could amplify the problem e.g., at the burner's nozzle, or as deposits in the fire box.
Thus, stay with the delivery of liquid oil and air.  If your are lucky in terms of viscosity of the oil, you still may obtain an aerosol but not the fine mist of air and fuel as characteristic for Otto engines, slow motion at 2:56 min onward).  More often, however, similar to oil driven heaters, you will rely on special burners running which bring pressurized air via fans into contact with a fine film of oil:

(credit)


Symmetry and SALC: choosing coordinate basis When we construct symmetry adapted linear combinations of orbitals (aka SOs) to build an MO diagram, we choose the z axis as the axis of highest symmetry and all other axis to make symmetry operations of a group easier to perform.

If we choose z-axis to point in some other direction, we will get a representation that is clearly not irreducible, but yet we would be unable to reduce it with the chosen basis because all orbitals mix (e.g. considering the case of 3d orbitals of a TM octahedral complex and having $d_{z^2}$ at a slant to the z-axis). I wonder if someone could explicitly explain why in this case the approach fails and highlight its limitations in general.

Thank you!

In general, when dealing with characters tables and with group-operations, it is crucial to choose the right orientation of axes in order to get sensible results. I say in general because it can also occur that, for highly symmetric molecules the "wrong" choice (or better say unconventional) can give the right result. Almost the 90% of characters tables follow the "Mulliken" conventions stated in Mulliken's paper "Report on Notation for the Spectra of Polyatomic Molecules" (The Journal of Chemical Physics 23, 1997 (1955)). Whenever you orient your axes not in agreement with the characters table you are referring for performing operations, there is a high risk to fail and face strange results as you did (all orbitals mixing etc..). I hope my answer is general enough and helpful.


Can a part of a compound be a reducing agent? I have the equation:

$$\ce{2 Cu2O + Cu2S -> 6 Cu + SO2}$$

My doubt is that since copper is reduced (it goes from $+1$ to a $0$ O.S.), then which of the two reactants is reduced? Or why would it be fine to say that only copper is reduced and not an entire reactant? Because from my experience, I've seen that an entire reactant is considered to be reduced or oxidized. So is it right to tell only a part of the compound is oxidized?

For accounting purposes, we can assign oxidation numbers to all the atoms. Oxygen could be -II in all species here, then sulfur would be +IV in $\ce{SO4}$ and -II in $\ce{Cu2S}$ (oxygen and sulfur are in the same group and both are more electronegative than copper, so assigning the same oxidation number in $\ce{Cu2S}$ and $\ce{Cu2O}$ makes sense). Oxidation numbers of atoms in elements are zero, and the copper in both compounds would be +I to balance out the -II of the chalcogens.

Once you are in agreement with your discussion partner on these oxidation states, you can make a statement about which atoms undergo oxidation or reduction: Sulfur gets oxidized, and copper in both copper compounds gets reduced.

In cases where only one atom in a compound changes oxidation states, or when all changes are in the same direction, we do talk about the compound being reduced or oxidized. For example, it is common to say that when you oxidize an aldehyde, you get a carboxylic acid (and when you reduce it, you get an alcohol or an alkane).

On the other hand, if you hydrate an alkene, some folks don't consider that a redox reaction.


How is this sugar classified as an aldose? I have made the following image in paint, so forgive me if it isn't the most appealing sugar. I have a hard time visualizing how the following sugar is an aldose.

My speculation for the Fischer projection of the alleged carbohydrate is:

My speculation is as a result of the lack of primary alcohol at the end of traditional carbohydrates like D-glucose for example means that there is not carbon group alpha or beta to carbon 4 in the furanose ring. The fact that there is still a CHO group on Carbon 1 would make this sugar an aldose? Is this logic correct, and is that the Fischer projection for the sugar shown above in the Haworth projection?

Yes its an aldose it is classified as an aldose because of the aldehyde group (cho). The compound is actually an aldotetrose called L-Threose. Try using acd Chem sketch program for the projection it is a free download for the freeware version...


Is storing diluted (~1M) sulfuric acid, HCl, NaOH, etc. in plastic PET bottles a bad idea? Do I really need glass bottles if the acids and NaOH is so diluted? I thought plastics dont really react with acids. And I already have them in plastic bottles (probably a different kind).

Thanks

Here you can find thorough and clear instructions about using plastic bottles for acids and bases:

http://www.corning.com/lifesciences/emea/en/technical_resources/product_guid/bottles/material.aspx

A common rule is to not store strong acids and bases (like sulfuric and hydrochloric acids, and NaOH) in plastic bottles if you are not sure about the type of plastics.

Polystyrene bottles are OK for strong alkali.

Here is a PDF document on this topic:

http://csmedia2.corning.com/lifesciences/media/pdf/PSUG_Storage_Bottles_CLS_GL_025.pdf


Calculate the rate of decomposition of H2O2 at 25 °C when [H2O2] is exactly 10.0 mmol/L I'm really confused?

$$\ce{2H2O2 -> 2H2O + O2}$$

$$\begin{array}{ccc}\hline
\text{Experiment no.} & c(\ce{H2O2}) & \text{rate} \\
\hline
1 & 14.45 & 3.35 \times 10^{-3}\\
2 & 28.90 & 6.65 \times 10^{-3}\\
3 & 57.80 & 13.35 \times 10^{-3}\\
\hline\end{array}$$

So from previous parts of the question, I found the order of reaction (1st), the rate equation for the reaction ($r = k[\ce{A}]$), the rate constant ($2.3 \times 10^{-4}\ \mathrm{s}^{-1}$, which is correct according to the answers), and now I need to find the rate of decomposition of $\ce{H2O2}$ when $\ce{[H2O2]}$ is exactly $10.0\ \mathrm{mmol\ L^{-1}}$.

I thought it's just $2.3 \times 10^{-4} \times 10.0$, which is $2.3 \times 10^{-3}$, but the answers provided is double that (so $4.6 \times 10 ^{-3}$), so I must have gotten the rate equation wrong?

It is a matter of definition:

The rate of the reaction is defined as: $$r=-\frac{1}{2}\frac{d[\ce{H2O2}]}{dt}$$
The rate of the decomposition of $\ce{H2O2}$ is defined as:$$r_{\mathrm{decomp}}=-\frac{d[\ce{H2O2}]}{dt}= 2r$$


How to rationalise the drop in pH when excess aluminium nitrate is added into an aqueous ammonium solution? Consider the flowing reaction
  $$\ce{NH3 + H2O <=> NH4+ + OH-}$$
  Explain “when excess powdered aluminium nitrate is added to the reaction mixture and the $\mathrm{pH}$ value of the solution is lowered.”

There are 3 questions that confuse me:

When $\ce{OH-}$ is removed by aluminium(III) ions, $\ce{NH4+}$ is produced, which is acidic. In that case, is the decrease in $\mathrm{pH}$ due to the formation of $\ce{NH4+}$ ions or because of the removal of $\ce{OH-}$ ions?
When $\ce{OH-}$ is removed, according to $K_\mathrm{c} = [\ce{H+}][\ce{OH-}]$, the $\ce{H+}$ ions will increase since $K_\mathrm{c}$ is a constant value. Hence, the pH decreases. Is is a correct explanation to the drop in $\mathrm{pH}$ when $\ce{OH-}$ ions are removed?
Does addition of excess aluminium nitrate mean that the equilibrium reaction will go to complete and there is no any ammonia at the end?

As andselisk has already pointed out in a comment, the hexaquo complex of aluminium is a known "cation acid"
$$\ce{[Al(H2O)6]^{3+} + H2O<=> [Al(H2O)5(OH^-)]^{2+} +H3O+}$$

A practical application is the use of solid $\ce{KAl(SO4)2·12 H2O}$ (alaun) for blood clotting during wet shave.


Why is NaCl insoluble in CCl4? Now I know it's something related to the fact that $\ce{CCl4}$ does not have a net dipole moment but I can't connect the dots.

In order for a substance to dissolve, the free energy change for the dissolution must be negative.  For the enthalpy change to be negative the heat of solvation (the solvent solvating the solute) must be greater in magnitude than the heat needed to break up the forces holding the solid solute together (in the case of NaCl lattice energy).  The entropy change of the solution process is accounted for by the gain in entropy from breaking up the solid and the loss of entropy from organizing solvent around the solute.  In forming most solutions, the entropy change is positive.

In this case the ability of carbon tet to solvate sodium and chloride ions is too low to make up for the lattice energy.  Carbon tet is nonpolar.  A good solvent for NaCl would be polar and the negative end of the solvent would surround sodium ions and the positive end would surround chloride ions.

http://www.biology.arizona.edu/biochemistry/tutorials/chemistry/graphics/nacl2.gif


What do different dissociation constants mean? What does first, second, third etc. Acid dissociation constants mean? They are written as $pK_{a_{1}}, pK_{a_2}$ etc. If any one would explain it that would be good.

The first, second, third, etc... acid dissociation constants refer to the equilibrium constant for loss of the first, second, third, and so on proton. For example, $H _2SO_4$ can lose one proton to make HSO4–,  which can then lose another proton to generate SO42–. Ka1 and Ka2 would be the equilibrium constants for these reactions. pKa1 and pKa2 are, of course, -ln(Ka1) and -ln(Ka2).


Would decomposition take place in an atmosphere without oxygen? Let's say you placed some sort of biological material (like raw chicken meat, an apple, etc.) in a very large glass container (large enough that moisture could evaporate at a similar rate to the normal atmosphere); filled it with a heavier-than-air, inert gas, like xenon; then, sealed the container and placed it outside.

Question: Would the biological material decompose?

My natural assumption is that it would decompose slightly, but mostly it would dry out and essentially become "mummified."

I think your experiment is along the same lines as one performed in 1680 by Antonie van Leeuwenhoek. His experiment is described in this Wikipedia article:

  In his letter of 14 June 1680 to The Royal Society, Antonie van
  Leeuwenhoek described an experiment he carried out by filling two
  identical glass tubes about halfway with crushed pepper powder, to
  which some clean rain water was added. Van Leeuwenhoek sealed one of
  the glass tubes by using a flame and left the other glass tube open.
  Several days later, he discovered in the open glass tube ‘a great many
  very little animalcules, of divers sort having its own particular
  motion.’ Not expecting to see any life in the sealed glass tube, Van
  Leeuwenhoek saw to his surprise ‘a kind of living animalcules that
  were round and bigger than the biggest sort that I have said were in
  the other water.’ The conditions in the sealed tube had become quite
  anaerobic owing to consumption of oxygen by aerobic microorganisms.

Obviously his experiment was not as tightly controlled as what you propose.  Also, I think the actual size of your "very large glass container" would be the key to microbial activity in your experiment, as anaerobes do require water to thrive. It sounds like van Leeuwenhoek gave his bugs plenty water for the scale of his experiment.

My guess is that there would be some anaerobic decomposition before it pretty much dried out and "mummified".


Why should duplicates compounds be removed from compound libraries before screening/activity prediction? This is one of the first steps to be done while analysing chemical data or applying training models to predict chemical activity. However, I am unclear as to why is this done. Doe sit have a major effect on the result if duplicates are not removed? What is the main reason why we remove duplicate compounds?
Thanks

Let's consider a data set with three attributes for each compound: a chemical name, a CAS number (unique identifier for a compound) and a single empirically measured numerical interest, such as efficacy in lowering blood pressure in rats. If you had duplicate entries for a single experimental result, that could potentially bias the analysis. If you had two results for the same compound from two different experiments or measurements, you would surely want to include both in the analysis. If you had two results for the same compound (same CAS number) but named differently, you would want to keep them both in your data set, but use only the CAS number as the nominal variable for structure.


Role of 10-camphor sulfonic acid as a catalyst in protecting group chemistry I recently did a reaction in the lab where 10-camphor sulfonic acid was used as the catalyst instead of something more conventional like hydrochloric acid or sulfuric acid. Dry acetontrile was used as a solvent. The scheme is shown below:

I'm not too sure why it is used. My initial thoughts are it's due to the solubility of 10-CSA in water/organic solvents. Alongside this using a conventional acid such as hydrochloric acid, the water content of the acid may affect the equilibrium. Although 10-CSA still has water in it, it would not be comparable to hydrochloric or sulfuric acid. 10-CSA also is quite bulky, which could provide some selectivity on which positions are deprotonated, although I don't see this as being likely due to its conjugate base having such a high pKa. Aside from this what else could be plausible explanation for the usage of 10-CSA?

For a given protecting group, there is no single best way of introducing it, just as there is no single best way of oxidising an alcohol – the conditions you use depends upon the substrate you have and the tolerance of it towards things such as acid, base, and heat.

In your question you ask specifically about conditions to introduce a benzylidine acetal, namely why aqueous acids aren't used – the tl;dr answer is that they are!

Below is an excerpt from the section in Greene's Protecting Groups in Organic Synthesis that deals with formation of such a group:

  PhCHO, ZnCl2, 28°C, 4 h.
  PhCHO, DMSO, concd. H2SO4, 25°C, 4 h.
  PhCHO, TsOH, reflux,  H2O, 72% yield.
  Sulfuric acid on silica gel, PhCH(OMe)2, 75–93% yield.
  PhCH(OCH3)2, I2, CH3CN, 60–89% yield.

  Taken from: Greene's Protecting Groups in Organic Synthesis

What all of these conditions have in common is that some kind of acid catalyst is required. Mineral acids (sulfuric), Lewis acids (zinc chloride or iodine) and organic acids (tosic acid) have all been shown to work in the literature, and indeed most people involved in complex molecule synthesis will have tried most of them.

Which condition is most appropriate is a matter of context (and often a lot of screening). Solid organic acids (tosic acid, camphor sulfonic acid, pyridine para-toluene sufonate) are often used in late-stage synthesis as they're able to be dried, this allows the reactions to be conducted under (almost) anhydrous conditions. Use of sulfuric acid dictates that water (quite a lot of it) is present, which can knock off other delicate protecting groups/functionality in a molecule.


How come I get different Gibbs free energy for the same reduced equation? I want to calculate the Gibbs free energy of this reaction which can be:

$\ce{H2(g) + 1/2O2(g) -> H2O(g)}$

or

$\ce{2H2(g) + O2(g) -> 2H2O(g)}$

I am using the entropy values (J/molK): H2(g) - 130.684, O2(g) - 205.138, H2O(g) - 188.7

And enthalpy values (kJ/mol): H2(g) - 0, O2(g) - 0, H2O(g) - 241.8

With the equation ΔG=ΔH−TΔS and temperature 298.15K. How come I get ΔG = -457kJ/mol for the full equation and -228kJ/mol for the reduced equation, shouldn't  I get the same Gibbs free energy since they're the same reaction (except one is reduced)?

Shouldn't both be considered "correct" or is one "more correct" than the other?

The change in Gibbs free energy is written per mole. For reactions, consider it "per mole of reaction." Since the stoichiometry of the second reaction is exactly double the first, "one mole of the first reaction" is half "one mole of the second reaction." Therefore, the value of the change in free energy is doubled as expected.

For an arbitrary reaction, there is no standard.

For the standard free energy of formation of a substance, the canonical reaction should generate one mole of the substance from its constituent elements (the first reaction here).


Water with Sodium VS Alcohols with Sodium Why is the reaction between water and sodium more vigorous than that between ethanol and sodium?
Also, why does the reaction between sodium and an alcohol becomes less vigorous the longer the hydrocarbon chain?

The dominant issue explaining the less vigorous reaction of long-chain (i.e. greater than about 3 carbons) alcohols with metallic sodium is simply a function of the molecular-mass of the alcohol. I will address this aspect last.

At the lower end of the molecular-mass for this series (i.e. water, methanol, ethanol, isopropanol, etc.) the destabilization of the hydroxyl intermediates by the electron-rich hydrocarbon groups (of course absent in water) also slows the progress of the reaction to some degree.

The reaction of sodium metal with water/alcohols is:

1)$$\ce{2Na(s) + 2ROH(l) -> 2NaOR v + H2 ^}$$
where R=H for water, and is the hydrocarbon chain for alcohols.

The inductive destabilization effect mentioned above also largely controls the relative acidities of the alcohols, which in turn are directly related to the reactivity of the alcohol with sodium.  The relative acidities of pure solutions of water and some selected $\ce{C1}$ to $\ce{C3}$ alcohols goes as:

$$\text{water > methanol > ethanol > isopropanol}$$

In the discussion above, we have considered the simple case in which the reactant of interest is also the solvent.  Any comparison of relative concentrations of the different reactants of course requires that we prepare some concentration of the water or alcohol in an appropriate solvent.  It is also desirable to dilute the water or alcohol because of the vigorous, highly exothermic nature of their reactions with sodium.

Because water is the most reactive of the compounds of interest, any sodium added to an aqueous solution of an alcohol would just preferentially react with the water solvent.  And of course you could not measure different concentrations of water in itself!  The solvent of choice in this case is DMSO.  DMSO is not reactive toward metallic sodium, and is miscible with  water, low molecular weight and even very high molecular weight alcohols.  In this system, the relative acidities of neat solutions of the $\ce{C1}$ to $\ce{C3}$ alcohols shown above holds, but with one exception, water:

$$\text{methanol > ethanol > isopropanol > water}$$

Although the acidity of water is low in DMSO due to the poor solvation of the hydroxyl anion (the alkoxy anions are solvated by DMSO in order of increasing hydrocarbon chain length), its reactivity toward sodium remains high.  This is because in the presence of sodium, the hydroxyl anion is the reactive species.  It reacts quickly and does not need to remain solvated for the reaction to proceed.

Of course the effects discussed thus far are also applicable toward higher molecular weight alcohols.  However, once you have a carbon chain of more than 3 or so, these  effects diminish with increasing molecular weight of the alcohol.  Here, the larger the molecular weight of the alcohol, the fewer molecules you have to react for a given weight or volume concentration of the alcohol.  Even if you have a pure alcohol solution, think about the "concentration" of the reactive hydroxyl functional group.  In the case of water, hydroxyl groups are everywhere.  In the case of say a $\ce{C20}$ alcohol, the sodium is mainly surrounded by hydrocarbons with a few reactive hydroxyl groups mixed in and available for reaction.


Shouldn't mineral NPK fertilizer ingredients add up to 100%? I use a mineral based  fertilizer in powder form which - according to the label - should not contain any fiber and is fully soluble in water (I dissolved 2 times the recommended dosage and found no solids remaining in the solution):

The NPK + Mg + trace elements add up to about 60%.

What constitutes the rest?

There is a specific convention how the nitrogen, potassium and phosphate concentration is given, and it might be surprising because it mentions substances that are not in the fertilizer, namely nitrogen atoms, $\ce{P2O5}$, $\ce{K2O}$ and $\ce{MgO}$. You have to imagine that all of the nitrogen, phosphorous, potassium and magnesium in your sample reacts to form these species, and then you express the masses of those fictitious products as percentage of the original mass.

$$\text{fertilizer}\ce{-> x N + y P2O5 + z K2O + q MgO}$$

For an example, look at this answer: https://chemistry.stackexchange.com/a/64282 which links to this reference https://en.m.wikipedia.org/wiki/Labeling_of_fertilizer


Simple QM9 properties for prediction I was looking into an article (doi: 10.1063/1.5058063) which presented a histogram of radius of gyration for samples of different configurations of alanine dipeptide. This property is fairly simple to calculate using
$$
a_{\mathrm{Rg}}(x)=\sqrt{\frac{\sum_{p} m_{p}|| x_{p}-x_{\mathrm{COM}}||^{2}}{\sum_{p} m_{p}}}
$$
The sum considers all atoms $p = 1, \dots, P$ of the peptide, where $m_{p}$ and $x_{p}$ denote the mass and the coordinates of each atom, respectively. $x_{\mathrm{COM}}$ denotes the center of mass of the peptide.

In a different problem, I have a number of molecular graphs generated by a network trained using QM9 dataset and am looking for properties that are readily calculable for the newly generated molecular graphs using such packages as RDKit and OpenBable.

The properties in the QM9 dataset itself, like LUMO energy and energy of atomization, need DFT calculation, which is not something that I look forward to do. An example of a property of interest is QED, which can be easily calculated using RDKit package. But QED is more of interest in ZINC dataset.

I'd appreciate comments on any easy to compute property of interest to the community for QM9 dataset.

I found 3 properties:

The molecular weight (MolWt)
Wildman-Crippen partition coefficient (LogP)
Topological Polar Surface Area (TPSA)


Distinguishing metals from nonmetals? Please help me. We just did an experiment wherein we used Fe, Al, Zn, Sn, Cu and HCl to see their reaction (e.g. Fe + HCl, Al + HCl...). The result of the experiment:

1.) $\ce{Fe + HCl}$ = formed bubbles (56 min)

2.) $\ce{Al + HCl}$ = formed bubbles (more than 1 hour)

3.) $\ce{Zn + HCl}$ = formed bubbles (42 min)

4.) $\ce{Sn + HCl}$ = No reaction

5.) $\ce{Cu + HCl}$ = No reaction

Now with the result, how can I distinguish metals from non-metals?

Well as santiago has pointed out, they are all in fact metals.

The way you can determine if it is a metal or not is that when any metal reacts with acid, hydrogen gas is released. This is shown in this general equation (it is not balanced): $$\ce{M^{m+} + HCl -> MCl + H2}$$

Meanwhile, generally non-metals don't react with acid. This because when a substance reacts with acid, it electrons to the H+ ions created by the acid. Non-metals are generally electron-accepters, rather than electron donors so there is no way for them to donate electrons to the hydrogen ions in the acid.

In your experiment, the bubbles that formed when reacted are indication that hydrogen gas is being formed, therefore the substance is a metal. Therefore, you should be able to tell that Fe, Al and Zn are metals.

Therefore, you might assume that Sn and Cu have to be non-metals. However they are actually metals. The reason why no bubbling is observed is because these metals are unreactive and don't get oxidised by the acid.


Problem with synthesis of calcium formate I just tried preparing calcium formate by adding $\ce{Ca(OH)2}$ to ammonium formate, but I observed no bubble formation and no smell of ammonia. Why doesn't calcium hydroxide, being a strong base, displace ammonia?

You probably did get your anticipated reaction, but unlike carbon dioxide or hydrogen ammonia is highly soluble in water.  Unless your reactants are more highly concentrated than is possible with the limited solubility of calcium hydroxide, the ammonia remains in solution instead of giving off bubbles or a smell.
There are various ways to render this reaction visible:

Heat the solution, driving out any dissolved ammonia.  If you generate a smell of ammonia, or better if the solution emits a gas that turns moist red litmus paper blue, you found the ammonia. You should compare with a blank where you just have the unreacted ammonium formate.

If you used solid calcium hydroxide, double-check how much you
dissolved.  Calcium hydroxide has only a solubility of 1-2 grams
per liter, so if
you dissolved more than that it must have reacted.  If you were
using magnesium hydroxide the solubility difference is even more
evident, for magnesium hydroxide is also strongly basic enough to
displace ammonia but its solubility without the reaction is very
low.

Check the pH.  Ammonia is a weaker base than the calcium hydroxide that displaces it, so you get a lower pH than you would have gotten just adding the same amount of calcium hydroxide to plain water (which in this case you would do for comparison).

Chemistry is an experimental science.  Therefore experimental design is essential for chemistry.


How can the standard deviation for a set of experimental value be interpreted in terms of precision? I have read from many websites that the sample standard deviation can be used to measure the precision of experimental values. I understand that the lower the standard deviation the better, as it would mean that the values are close together (precise). But what standard deviation is small enough to be considered precise?

As an example, in a redox titration experiment I obtained titrant values of $16.3$, $16.4$, $16.5$ and $\pu{17.1 mL}$.

From these four measurements, I thus calculate an average of $\pu{16.58 mL}$ and a standard deviation of $\pu{0.35 mL}$.  The uncertainty of the burette is $\pm\,\pu{0.05 mL}$.

From this standard deviation, would this experiment's results be considered precise?

If not, why not, and how would I interpret it?

Is there a better method for quantifying precision?

But what standard deviation is small enough to be considered precise?

"Sufficiently precise," ironically enough, is something that can't really be defined precisely in general terms.
In other words, a level of precision that is perfectly fine in one context might be unacceptably low in others.
For example, if I'm trying to estimate how much table salt I need to buy to supply my catering business for the next six months, I might be perfectly happy with a not-very-precise calculation at $\pm\,30\%$, since salt is cheap and will keep just fine if I buy too much.
On the other hand, if I'm a pharmacist compounding a prescription containing a drug where the level in the body leading to nasty side effects is not all that much higher than the level that provides the desired pharmacological activity, I probably want to use (and am probably required by law to use!) methods and apparatus that'll give me, say, $\pm\,3\%$ precision on my measurements.

Is there a better method for quantifying precision?

Not really.  Pretty much every metric I've seen for quantifying precision involves the standard deviation in some way.  Depending on the situation, one might use:

$s$, the standard deviation by itself

$s/\bar x$, the standard deviation divided by the mean, also called the relative standard deviation or coefficient of variation (CV)

$s/\!\left(\bar x \sqrt n\right)$, the relative standard deviation divided by the square root of the number of measurements

This is equal to the standard error $\left(s/\!\sqrt n\right)$ divided by the mean

The "Comparison to standard deviation" subsection of the Wikipedia article on the coefficient of variation lists some situations where these different quantities would be useful relative to others.

From this standard deviation, would this experiment's results be considered precise?
If not, why not, and how would I interpret it?

In my opinion: Yes, these results are precise. In this case, since the mean value is far from zero and there's only one dataset being examined, I would use the coefficient of variation (CV) as my metric of precision.  With these values, I get a CV of $1.88\%$.  On the whole, I think this is an entirely reasonable CV for a titration experiment. The data could exhibit more precision, obviously, but that low of a CV implies very good repeatability to me.
As porphyrin noted, that $\pu{17.1 mL}$ measurement you obtained is perfectly reasonable in a statistical sense, as it's $25\%$ of a small dataset, residing $1.7$ standard deviations away from the mean. This puts it well line with the expected $32\%$ occurrence of values $x$ where $|x-\bar x|/s > 1.0$.
In other words, I disagree with Mithoron: I don't think that value is an outlier at all.  The usual metric I'm familiar with for identifying outliers is where $|x-\bar x|/s > 3.0$, which should occur only $\approx 0.34\%$ of the time for normally-distributed data.

The uncertainty of the burette is $\pm\,\pu{0.05 mL}$.

A brief note: I believe the small uncertainty on the burette as compared to the standard deviation in your data $(0.05 / 0.38 \approx 13\%)$ indicates that the majority of the spread in your measurements comes from elements of the experimental apparatus/procedure other than the burette.

Further reading: This article at Inorganic Ventures is a good writeup of the mean and standard deviation, and how they relate (or not) to accuracy and precision.


Why does electrode potential depend upon enthalpy of hydration and sublimation? It's given in text that "Electrode potential among other things depends upon: enthalpy of sublimation,ionization enthalpy and hydration enthalpy." I could understand why it would depend upon ionization enthalpy as less is the IE easily we could remove the electron thereby reducing the other reactant. But I couldn't figure out why the other two enthalpy comes. But after some thinking I thought maybe we are adding the enthalpy of the three with sign and substituting in ∆G°=∆H°-T∆S° and obtaining ∆G° to substitute in ∆G°=-nFE°cell to find E°cell value. I couldn't think further clearly after this. Could you help me? ( I'm trying to understand why lithium because of its high hydration enthalpy overweighs its high ionization enthalpy and becomes the strongest reducing agent in aqueous solution)

In a thought experiment, you can substitute the electrochemical half reaction by three steps:

Removing atoms from the electrode into the gas phase (sublimation)
Removing electrons from the atoms (ionization)
Solvating the resulting ions

If you have quantitative descriptions of these processes, you get a quantitative description of what is actually going on. This is because many thermodynamic functions (such a enthalpy) are functions of state, i.e. they don't depend on the path from A to B, just on the state of A and B.

I would have said these quantities are related rather than one depends on the other. There is really no causality here.


Zinc in a single displacement reaction with hydrochloric acid I understand that on the metal activity series, zinc can displace hydrogen from acids, but not from cold water. But is there a structural reason for this. For example, why can zinc displace hydrogen from hydrochloric acid, but not from water?

Thanks

The zinc replaces the hydrogen in that is reduces (gives electrons to) $\ce{H^+}$, forming $\ce{H2(g)}$ and $\ce{Zn^2+(aq)}$ according to this reaction: $$\ce{Zn(s) + 2H^+(aq) -> Zn^2+ + H2(g)}$$ This reaction relies on the prescence of $\ce{H^+}$.

The hydrogen in acids exists as $\ce{H^+}$. Acids are also ionic compounds that involve hydrogen as the donor of electrons. For example, hydrogen gas and chlorine gas react to make hydrogen chloride: $$\ce{H2 -> 2H^+ + 2e^-}$$ $$\ce{Cl2 + 2e^- -> 2Cl^-}$$ $$\ce{2H^+ + Cl^- -> 2HCl}$$  The reverse of this happens when the hydrogen chloride is dissolved into water.  this forms hydrochloric acid.The compound breaks apart into hydrogen and chlorine ions:$$\ce{H2O + HCl -> H+ + Cl-}$$ Hydrochloric acid completely dissociates, meaning that every $\ce{HCl}$ molecule yields one $\ce{H^+}$ ion.

Essentialy, acids increase the amount of $\ce{H^+}$ in solution. Normally, in pure water, concentrations of $\ce{H^+}$ are very low. The hydrogen in $\ce{H2O}$ is covalently bonded to the oxygen, so it is difficult for the hydrogen to react. In order to liberate a hydrogen ion from $\ce{H2O}$, this must occur: $$\ce{H2O -> H^+ + OH^-}$$
This reaction happens infrequently in water, so concentrations of $\ce{H^+}$ are very low. Therefore, zinc barely replaces hydrogen in water.


How to determine the order of a reaction with not a commensurate rate equation As an example of such reactions let's take decomposition of $\ce{N2O}$ which I got the equation from a textbook :

$\ce{N2O = N2 + 1/2 O2}$

with the rate equation of $$-r_{\ce{N2O}}=\frac{k_1[\ce{N2O}]^2}{1+k_2[\ce{N2O}]}$$

So how do we determine the overall order of reaction or order of reaction with respect to a reactant/product ?

To be honest with you, the term "order of reaction" is not an exact term. If I want to pull out the rigorous kinetician out of the closet, I'd say that only elementary reactions have orders, and we have first or second (or zero or maybe third) order reactions only, and they are obvious to spot.

What you have is not an elemental reaction, but an overall - somewhat empirical - net reaction rate for a complex mechanism. I personally do not like to talk about orders of non-elementary reactions, but some authors do this. In this context, I'd say your reaction is first order in $\ce{N_2O}$, simply because you have concentration squared divided by approximately concentration. Note that if $\ce{[N_2O]}$ is sufficiently large (so the $1$ in the denominator is negligible), the equation becomes $-r =\ce{\frac{k_1 [N_2O]^2}{k_2 [N_2O]} = k' [N_2O]}$, so in high concentration limit the unimolecular approximation becomes exact.


Why is FEP transparent and PTFE not? fluorinated ethylene propylene (FEP) is close relative of poly(tetra-fluoroethylene) (PTFE), both only contain fluorine and carbon atoms,both have very similiar properties yet FEP is relatively transparent while PTFE is opaque, why?

The transparency of a plastic does not depend on its nature but on how it is made.

To get a transparent plastic it must be cooled very fast when it is in molten state.
If the cooling is too slow, the polymer begins to order it self in more or less crystalline domains.

When it is not isotropic anymore, it is also not transparent.