Quick and simple explanation of molar mass, molecular mass and atomic mass As a physics student, I hardly deal with such quantities and when I do, I blunder through.
I think it’s a good time to be given a good explanation for these since I’m in a class of Nuclear physics. I have tried googling but it seems that some sources are ‘misleading’ and most have more than 1 terminology for the same thing!
Horrifying but real.

Please tell me what is the difference between molar mass, molecular mass and atomic mass and what are their associated twin terminology.

The quantity molar mass (quantity symbol: $M$) for a pure sample is defined as
$$M = m/n$$
where $m$ is mass and $n$ is amount of substance.
The dimension of the molar mass is
$$\dim M = \mathsf{M}\;\mathsf{N}^{-1}$$
The coherent SI unit for molar mass is ‘kilogram per mole’ (unit symbol: $\mathrm{kg/mol}$).
(The usually used unit is gram per mole ($\mathrm{g/mol}$) rather than kilogram per mole ($\mathrm{kg/mol}$).)

The quantity relative molecular mass (quantity symbol: $M_\mathrm{r}$) is defined as the ratio of the average mass per molecule or specified entity of a substance to 1/12 of the mass of an atom of the nuclide $\ce{^12C}$; i.e. the relative molecular mass is the ratio of the mass of a molecule to the unified atomic mass constant:
$$M_\mathrm{r} = m_\mathrm{f}/m_\mathrm{u}$$
where $m_\mathrm{f}$ is mass of entity (molecule or formula unit) and $m_\mathrm{u}$ is the unified atomic mass constant.
The relative molecular mass is a quantity of dimension one (for historical reasons, a quantity of dimension one is often called dimensionless):
$$\dim M_\mathrm{r} = 1$$
The coherent SI unit for relative molecular mass is the unit one (symbol: $1$).
For historical reasons, the IUPAC accepts the use of the special name ‘molecular weight’ for the quantity relative molecular mass. The use of this traditional name is deprecated.

For atoms, the relative molecular mass $M_\mathrm{r}$ is the relative atomic mass, and the quantity symbol $A_\mathrm{r}$ may be used. The quantity relative atomic mass is defined as the ratio of the average mass per atom of an element to 1/12 of the mass of an atom of the nuclide $\ce{^12C}$:
$$A_\mathrm{r} = m_\mathrm{a}/m_\mathrm{u}$$
where $m_\mathrm{a}$ is the atomic mass (see below) and $m_\mathrm{u}$ is the unified atomic mass constant.
The relative atomic mass is a quantity of dimension one:
$$\dim A_\mathrm{r} = 1$$
The coherent SI unit for relative atomic mass is the unit one (symbol: $1$).
For historical reasons, the IUPAC accepts the use of the special name ‘atomic weight’ for the quantity relative atomic mass. The use of this traditional name is deprecated.

The quantity atomic mass (quantity symbol: $m_\mathrm{a}$) is defined as rest mass of a neutral atom in the ground state.
The dimension of the atomic mass is
$$\dim m_\mathrm{a} = \mathsf{M}$$
The coherent SI unit for atomic mass is ‘kilogram’ (unit symbol: $\mathrm{kg}$).

I've been studing this last week because I'm on the 10th grade. Siply the atomic mass is the mass of the atom which is "n" times bigger than standard mass (1/12 of the mass of carbon-12). For example, oxigen, the atomic mass of oxigen is 16 because is 16 times bigger than 1/12 of the mass of carbon-12, and 1/12 of 12 is 1.
Molecular mass is the sum of the atomic mass of the atoms in the molecule.
Finally, molar mass is the mass of 1 mol of particles which has the same number of relative mass.

I hope I could help someone with this.


Suitable coating material for protection of sensors I'm building a moisture/temperature sensor for outdoor use in soil. It uses capacitive sensing for moisture and an NTC thermistor for the temperature. It's going to be used for vegetables so food safety is a concern.

I don't think the PCB is food safe, or safe from corrosion/degradation due to the soil and moisture would creep into the FR4 material in the PCB and cause my measurements to drift so I would like to coat the entire thing to protect it from the elements, keep my vegetables free from toxins, and keep the measurements from drifting.

So these are the requirements in order of importance for the coating:

Not electrically conductive
Food safe
Doesn't degrade/react in/to soil/sun light
Hydrophobic
Resistant to abrasion when insertion/removing from the soil
Affordable
Low relative permeability (electromagnetic)
Decent thermal conductivity
Not a host for bacteria

I first considered a food safe epoxy but I had difficulties finding a reputable supplier of food safe epoxies in Germany (my German sucks). I did find a supply of food safe silicone which is nice in that it has all the properties above but I could only find it in large quantities and it's quite expensive. I also considered L/HDPE which seems to have all the criteria apart from the bacterial aspect (I'm not sure about that) but it seems like melting it to have a low enough viscosity to be able to form a solid but thin coating seems hard/impossible.

My knowledge on the topic of epoxies, plastics, rubbers and chemistry in general is limited, so I'm turning to chemistry@sx for recommendations to what I can use for a coating and what process to use for coating. And maybe I've misunderstood something of the materials I've considered so far... Also I'm sorry if this is the wrong place for this question, I'm not sure where else to post.

Various DIY capacitive moisture meters use a PCB for the sensor:

While not directly related to your question I thought I'd mention that you can buy these sensors for several dollars:

Purchasing a readymade kit and relying on the discount of mass production (along with a manufacturer's warranty) is probably easier than designing your own, and etching your own PCB.

I think how you plan to make the probe plays a part in how you plan on coating it; for example you could fit the parts in a test tube, that would meet all the requirements in your list. Such a capacitive sensor would operate similar to the touchscreen of a cellphone.


Find the molarity of the filtrate I can't figure out this stoichiometry question!

10.0 mL of 1.15M copper(II) nitrate reacted with 20.0 mL of .85M ammonium hydroxide.  The solution is filtered. Determine the molarity of the filtrate.

I figured out that the net ionic equation is Cu + 2OH -> Cu(OH)2, so the 2NH4NO3 dissolves, and the Cu(OH)2 probably doesn't pass through the filter, but I don't know where to go from there. Thanks in advance!

$$\ce{Cu(NO3)2}+2\ce{NH4OH} =>\ce{Cu(OH)2}(s)+2\ce{NH4NO3} $$
Essentially, we're trying to find the Molarity of $\ce{NH4NO3}$ since the copper filters out(as you stated)

First solve for limiting reagent. $$10mL\ce{Cu(NO3)2}*\frac{1L}{1000mL}*\frac{1.15mol}{1L}=0.0155mol\ce{Cu(NO3)2}$$
$$20mL\ce{NH4OH}*\frac{1L}{1000mL}*\frac{0.85mol}{1L}=0.0170mol\ce{NH4OH}*\frac{1}{2}=0.0085$$

Why do we divide mols of $\ce{NH4OH}$ by 2? Remember that for every molecule of $\ce{Cu(NO3)2}$, there must be 2 molecules of $\ce{NH4OH}$. $$\frac{2\ce{NH4OH}}{\ce{Cu(NO3)2}}$$ for this reason we need to divide the amount of $\ce{NH4OH}$ by 2 to compensate for stoicheometry.
So actually, $\ce{NH4OH}$ our limiting reagent.
Since you didn't give my any solubility constants, I'm going to assume 100% of the Copper became  $\ce{Cu(OH)2}(s)$ and all of the ammonium became $\ce{NH4NO3}$. In the real world, perciptates never form at 100%.

0.0085 mols of $\ce{Cu(OH)2}$ is produced. You want $\ce{NH4NO3}$  Can you use stoichometric ratios to proceed from there? Molarity - (mol/L). Make sure you add up the two volumes to find the Total volume.


Use of Δ (delta) to describe alkenes In a lot of (mainly older) papers discussing organic synthesis,  Δ is used to describe alkenes. As far as I know, this is no longer (if ever) IUPAC recommended nomenclature, so I haven't been able to find any reference to its intended use.

The main area I'd seen it used was in steroid synthesis, where I assumed it had some specific use relating to that class of natural product (i.e. related to the biosynthesis etc.). However it does appear in other places, such as in the synthesis of Rifamycin:

 Scheme from Nicolau Rapamycin synthesis. Taken from Classics in Total Synthesis I

"… the Δ35,36 double bond …" is used in the text to describe the alkene that gets reduced. Given that this could easily be referred to as the (E)-alkene or even just 206, I suspect that there is a special reason why Nicolau chose Δ35,36 however that reason isn't apparent to me.

@Loong's encyclopaedic knowledge of IUPAC nomenclature prompted me to have another look through the (irritatingly unsearchable) 2013 recommendations and the 2004 draft upon which they were built.
The only use of the uppercase Δ listed is when describing a double bond that links two cyclic systems:

P.28.2.2: Ring assemblies with a double bond junction
When two cyclic systems are linked by a double bond [...] the presence of a double bond is indicated by the Greek letter Δ, and the point of attachment of the ring is given by superscript locant numbers. This method is not continued in these recommendations

The example given is 1'1-bi(cyclopentylidene) (PIN) which should be used in preference of Δ1,1'-bicyclopentylidene.

Based on this, the example given in the question, using Δ to describe an unsaturated carbonyl would appear invalid, and used for convenience rather than to convey anything specific.

Yes, he could certainly have said substance 206 but in my opinion he uses Δ35,36 just so it's clear which position it is in the final molecule.


Reason for inability to find the absolute value of internal energy Today in my thermodynamics lecture, my teacher told me that it is not possible to find the absolute value of internal energy and so we have to calculate the change in internal energy.
So, my question is why isn't it possible to find the absolute value of internal energy? Is this statement true?

In order to answer this question, one needs to define what the absolute energy of a system is. Energy can be trapped in a system in ways not yet discovered or fully understood. Think of the energy associated with mass ($E=mc^2$), which is a result that was not known to the early founders of thermodynamics.

We need to define reference points, which we can use to compare our products with. Usually this "zero" point is defined as the energy of the individual atoms in their most energetically favorable state (phosphorus is an exception, since red phosphorus is much more available), such as $\ce{H2}$ and $\ce{He}$.

The statement that it is impossible to calculate the absolute energy at an ambient T seems to be intuitively incorrect, difficult possibly, but not impossible. Since U is a state function the value of U for a given set of conditions is invariable and can be set as an arbitrary zero and changes in U calculated. So all one must do is work in reverse to absolute zero. This is done for entropies using the 3rd Law. The general approach in thermodynamics is that it is entirely accurate to work with the changes in state functions and there is NO NEED to calculate absolute internal energies from absolute zero.


Does acid anhydride give a positive iodoform test? I thought that acetic anhydride $\ce{CH3COOCOCH3}$  would give a positive Iodoform test because it has a $\ce{CH3-(CO)}-$ group.But it is given that this does not happen.

Is this true? If so, can anybody explain the reason? Thanks

According to this source, both acid halides and anhydrides can form enols reasonably readily and can be halogenated. I'd imagine, for anhydrides, this to be related to the potential for intramolecular hydrogen bonding in the enol form (as shown below):

This allows halogation at the alpha position by treatmeant with the halogen. I'd also imagine this process can be catalysed in anhydrous acidic conditions, avoiding hydrolysis.
The same cannot be said for base-facilitated enolate formation. As user55119 implied, the highly reactive $C=O$ is likely to be attacked before the alpha hydrogens (or even if deprotonation occurs first, proceed through an E1cB pathway due to the presence of a good leaving group; implied by Clayden Organic Chemistry), decomposing the anhydride instead of forming the enolate.
Enolate formation in basic conditions is usually required for multiple halogenation, and generally three substitutions are required to form a suitable leaving group for the haloform reaction to proceed. This is why it is a test for a methyl group next to a $C=O$ (as Mathew has said); as there are three alpha hydrogens.
Considering basic conditions are not an option, only mono-halogenation is feasible (perhaps even in acidic conditions) and the haloform test is made virtually impossible.

Iodoform test is used to detect the carbonyl carbon and since acid anhydride is not a carbonyl compound , it does not give this test . The lone pair of oxygen ( the middle one ) and the π bond of carbonyl group are in resonance . When you actually  see the reaction mechanism , you observe that the methyl group attached to the carbonyl should have  acidic hydrogen but this is not so in the case of acid anhydride .


Are there substances that cannot undergo combustion? Are there substances that cannot undergo combustion? If so, what are they?

It depends a little on your definition of combustion.

Most combustion we see involves things reacting with the oxygen in air. Most combustion reactions are a self-sustaining exothermic reaction of some substance with oxygen gas. But atmospheric oxygen is relatively dilute (only about 21% of air is oxygen) and this limits what will burn sustainably in air (also, at slightly lower concentrations many commonly flammable things won't stay lit). The requirement for things to be combustible in air is, crudely, a reaction path with oxygen that releases energy.

Examples include many types of compound. Hydrogen reacts with oxygen to give water; hydrocarbons react to give carbon dioxide and water. Many metals react to give metal oxides (eg magnesium, spectacularly when lit; iron, only when hot and finely divided).

Many things won't burn in air for two reasons: there are activation barriers to a potential oxidation meaning a potential reaction can't be sustained (aluminium has a strong oxide layer on its surface protecting the reactive metal from further oxidation, for example); other compounds are already highly oxidised an no further oxidation can happen (eg silicon dioxide, a major ingredient of sand and glass).

But there are many more things that will combust in pure oxygen as Apollo 1 discovered, tragically. And as many demonstrations show spectacularly (see demo here). But highly oxidised things like silica can't be oxidised further and won't burn not matter how concentrated the oxygen is.

But oxygen isn't the only gas that can support exothermic reactions. Rather more things will burn in fluorine, for example. Even relatively unreactive substances can support combustion if there is an available reaction path that releases enough energy and a reaction can get started. One of the best known examples is that magnesium will burn in carbon dioxide (the reaction producing magnesium oxide releases energy) and even nitrogen (producing magnesium nitride). See the explanation here and the video here.

So, to return to the original question, are there any substances that won't combust? Whatever the definition of combustion is the answer is still yes. Noble gases, for example, though some can be coerced into reacting under extreme circumstances, never combust or support combustion.

If we restrict our definition to combust in air, many substances won't combust though many of them might do so under different circumstances.

If you define combustion as a chemical process that consumes $\ce{O2}$ and produces $\ce{H2O}$, then the answer is: any $\ce{X}$ that does not contain hydrogen is not combustible.

$\ce{X + O2 -> Y + H2O}$


Why is zinc sulfate used in daniell cell Why is zinc sulfate needed in the construction of a daniell cell.  It seems that because the zinc is getting oxidized and not reduced there is no need to have a solution of zinc sulfate as it will produced when the zinc is reduced.  The sulfate ions would simply cross the salt bridge and balance out the charges.

Zinc sulfate is not needed. With any other neutral salt in its place, your cell will work just fine.

Without any salt at all, though, it will not work, since pure water is a poor conductor of electric current.


What is resonance, and are resonance structures real? My teacher told me about resonance and explained it as different structures which are flipping back and forth and that we only observe a sort of average structure. How does this work? Why do the different structures not exist on their own?

This answer is intended to clear up some misconceptions about resonance which have come up many times on this site.

Resonance is a part of valence bond theory which is used to describe delocalised electron systems in terms of contributing structures, each only involving 2-centre-2-electron bonds. It is a concept that is very often taught badly and misinterpreted by students. The usual explanation is that it is as if the molecule is flipping back and forth between different structures very rapidly and that what is observed is an average of these structures. This is wrong!  (There are molecules that do this (e.g bullvalene), but the rapidly interconverting structures are not called resonance forms or resonance structures.)

Individual resonance structures do not exist on their own. They are not in some sort of rapid equilibrium. There is only a single structure for a molecule such as benzene, which can be described by resonance. The difference between an equilibrium situation and a resonance situation can be seen on a potential energy diagram.

This diagram shows two possible structures of the 2-norbornyl cation. Structure (a) shows the single delocalised structure, described by resonance whereas structures (b) show the equilibrium option, with the delocalised structure (a) as a transition state. The key point is that resonance hybrids are a single potential energy minimum, whereas equilibrating structures are two energy minima separated by a barrier. In 2013 an X-ray diffraction structure was finally obtained and the correct structure was shown to be (a).

Resonance describes delocalised bonding in terms of contributing structures that give some of their character to the single overall structure. These structures do not have to be equally weighted in their contribution. For example, amides can be described by the following resonance structures:

The left structure is the major contributor but the right structure also contributes and so the structure of an amide has some double bond character in the C-N bond (ie. the bond order is >1) and less double bond character in the C-O bond (bond order <2).

The alternative to valence bond theory and the resonance description of molecules is molecular orbital theory. This explains delocalised bonding as electrons occupying molecular orbitals which extend over more than two atoms.

What my professor tells me about this is that chemists may have made this just for explaining the following facts:

In benzene, or chloro-benzene for that matter, the single bond length is shorter than the average value, and the double bond length is somewhat larger than the standard value.
A possible explanation for this maybe that a double bond develops a single bond character and vice-versa.
Now, to make it sound more scientific (or more chemistic) the respective chemists tried to explain this using the concept of resonance, and it sounded to be sane, as one can visualize electrons (pi-electrons in the case of benzene) to be moving around in order to stabilize the whole compound, which also made sense from that basic principle of chemistry that everything, EVERYTHING that happens, happens in the direction in which more stability is achieved).


Why do they use graphite electrodes in the Hall–Héroult process? In the Hall–Héroult process of extracting aluminium from aluminium oxide, the electrolyte is a liquid mixture of cryolite $(\ce{NaAlF6})$ and aluminium oxide $(\ce{Al2O3}).$ The cathodes and the anodes are made of graphite.
During the electrolysis process, aluminium is deposited at the cathode and oxygen is liberated at the anode. Some of this oxygen reacts with the carbon in the graphite to form carbon-dioxide, thus slowly burning away the anodes. Thus, the anodes have to replaced periodically.
But if this is the case, why do they use graphite anodes at all? Why don't they use anodes of some other, less reactive material? I'm pretty sure there are many out there that would fit the bill.

According to this source, graphite is used as anode material because carbon is easier to oxidize than $\ce{O^2-}$. The anodic oxidation of carbon

$$\ce{C + 2O^2- \rightarrow CO2 + 4e-}$$

proceeds more efficiently than the oxidation of $\ce{O^2-}$ at an inert anode

$$\ce{2O^2- \rightarrow O2 + 4e-}$$

$\ce{C}$ is a much better reducing agent than $\ce{O^2-}$ whose oxidation state is not changed in the reaction with graphite. As a consequence, anodic current density and the deposition rate of aluminium at the cathode will increase.

Carbon is a better oxidizing agent than oxygen which produces more heat to retain steady temperature


pKa of Tris corrected for ionic strength I have a question regarding the molar fraction calculation of Tris buffer at different temperatures.

Tris dissociates according to the following equation:

$$\ce{Tris + H2O -> Tris+ + OH-}$$

So, in order to calculate the amount of $c(\text{Tris}^+)$ and thus the amount of strong acid needed for titration, one needs to use derivation of Henderson-Hasselbalch eq.:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a}' - \log_{10}\left(\frac{[\text{Tris}^{+}]}{[\text{Tris}]}\right)$$

However, $\mathrm{p}K_\mathrm{a}'$ is $\mathrm{p}K_\mathrm{a}$ corrected for ionic strength ($I_c$) at a certain temperature.

The correction of $\mathrm{p}K_\mathrm{a}$ is done via the Debye-Hückel eqation:

$$\mathrm{p}K_\mathrm{a}' = \mathrm{p}K_\mathrm{a} -0.509\times\left| z-z^{+}\right| \sqrt{I_c}$$

Now, for Ic itself one needs to calculate the molar fraction of $[\ce{Tris+}]$:

$$I_c = 0.5\times \left([\ce{Tris+}] + [\ce{OH-}]\right)$$

which requires the use Henderson-Hasselbalch equation making this problem somewhat circular. or am I missing something?

First decide the Ic you want to work with. Second, you should correct the pka with the Debye Huckel equation and finally you use the pka corrected with the Henderson Hasselbach equation.


Is it safe to lyophilize/Freeze-dry a chelated compound? I have a compound that consists of a metal ion chelated by a linear organic structure with assorted moieties in aqueous medium. I'd like to dehydrate this mixture fully to get solid product, but heating is out of question due to unwanted repercussions on aforementioned "assorted moieties". If I lyophilize the compound, will it remain chelated or is there a risk of the metal ion unbinding?

I lyophilized my sample and redissolved a portion of it and confirmed chelation by verifying the presence of the absorption peak of the Lanthanide I was chelating and using Beer's Law.

Thanks for your help.

Lyophilization is regarded safer as normal drying because it is performed at considerably lower temperature; however in order to achieve this, you have to lower the pressure above the condensed phase, too.  Naturally volatile compounds (including some organic compounds solid at room temperature) may sublime under these conditions, too; which may be hold for hours, even days to remove the water gradually.  This simultaneous process may lower the yield of isolated product.

Other light organic molecules (like methanol, ethanol, THF) equally may leave the vessel, too.  The complexation may be affected if leaving molecules were involved as part of the no-further specified chelate, too.


Why is moist hydrogen made to react with chlorine to give hydrogen chloride? Moist hydrogen reacts with chlorine in diffused sunlight to give out hydrogen chloride. Why is moist hydrogen used? Why not dry hydrogen?

H2O has h ions so it adds more + ions so that the cl- gets reacted faster towards the H+ ions so water is a catalysts or booster
I hope u understnd


Why does the hydrogen atom in HCl (when dissociated in water) bond to the oxygen rather than staying with the chlorine atom? Why does the hydrogen atom in $\ce{HCl}$ (when dissociated in water) bond to the oxygen rather than staying with the chlorine atom. They ($\ce{Cl}$ and $\ce{O}$) have roughly the same electronegativity (3.5).

I saw this question:
Dissociation of HCl in aqueous solution
but it doesn't answer what I'm asking.

Look at it from the point of view of the chloride ion. In $\ce{HCl}$, the chloride is attracted to only one hydrogen. But when dissolved in water, the chloride is attracted to the partial positive charges on the hydrogens of several water molecules. Each $\ce{H2O-Cl}$ attraction is less than the $\ce{H-Cl}$ attraction but many of them act together to attract the $\ce{Cl}$ ion away from $\ce{HCl}$. Once the $\ce{Cl}$ is attracted away, that leaves a hydrogen ion to be attracted to the partially negative oxygen of a water molecule.

These forces between multiple waters and an $\ce{HCl}$ are stronger than forces between adjacent $\ce{HCl}$ molecules. Otherwise, $\ce{HCl}$ would be a liquid at room temperature instead of a gas.

I’m going to answer to the posted question without reading all of the commentary.

First, the answer should NOT be given in terms of electronegativity. I cannot go into detail here, but the failure of electronegativity theory to predict or explain that HI is a strong acid must not be ignored.

Acidity and basicity are properties of the electron pairs of an atom. We know that HF is tetrahedral so we know a proton is attached to an electron pair in HF. We should infer the electron pairs of a fluoride are also tetrahedral. That means we should judge the basicity of fluoride on the basis of each of its valence electrons and not on the basis of its net charge. No change has occurred to the charge of an electron pair.

If dissolving HCl in water gives hydronium ion and chloride ion, this should tell us the electron pairs of oxygen are more basic than an electron pair of a chloride ion. Again, these are properties of the electron pairs and not a property of the ions per se.

The above are consistent with the long bonds of NaCl and the ease at which the ions separate and easily dissolve in water. A theory of ionic attraction suggests they should form strong bonds. Long bonds belie a strong bond. Na(+) and Cl(-) each have completed octets and do not need to be attracted to the electrons of each other. That is different that the much shorter and stronger bonds of CH3Cl in which an electron pair must be shared. It is also consistent with the properties of sodium metal. Metallic sodium has one additional electron beyond its octet. In solution, it likes to donate this electron. Try adding sodium metal to water and see if this reaction is exothermic and spontaneous (everyone knows the answer).

This should tell us things we already know. If we remove a proton from the nucleus of oxygen, will the valence electrons become more or less basic? This should predict nitrogen and ammonia should be more basic than oxygen and water. We should anticipate HCl will form ammonium chloride with ammonia. The attraction of the ions should be low, each have completed octets. The electron pairs of chloride are not very basic which is why HCl is a strong acid. Water is not as acidic, so a mixture of ammonia in water may form some ammonium hydroxide, most does not. It will be hydrogen bonded, but the protons will mainly stay on the oxygen.

On electronegativity theory, I have talked about it at American Chemical Society meetings. Unfortunately, it is not good science although repeated in virtually every chemistry textbook. Acidity is virtually the gold standard to measure ionization. The failure of electronegativity theory to match acidity should have been a clue as to its validity. Electronegativity theory is a quasi scale of reactivity. Fluorine is very reactive and more reactive than chlorine, etc.

I have not updated this in some time, but search for electronegativity and curvedarrowpress and you can see slides of an old talk there. I am writing a manuscript in which I have a fuller explanation of the data Pauling used to advance electronegativity theory. That will explain why HI is a strong acid and why iodide is a good leaving group. The explanations are much like my discussion above.


Ring chain tautomerization of hydroxyacetone My chemistry textbook says that hydroxypropane-2-one and 1,2-epoxy-2-hydroxypropane are ring chain tautomers, but it did not mention the steps for conversion of one into the other. I tried to figure them out but got stuck.

My attempt:

At first, even though this is not related to the question, I want to say that you're confused between double-headed arrow between the two resonance forms and double harpoon in reversible reactions. (The harpoons should be written as below.)

Your incomplete mechanism has two errors:

The hydrogen atom bonded to carbon is far less reactive than the one bonded to oxygen.
The positive carbon, which is bonded to oxygen, a very electronegative element, is very, very unstable, which is not likely to happen.

This is the right mechanism for the reaction:

Step 1: Oxygen is a good nucleophile, so it is most likely to take the most acidic hydrogen.
Step 2: Then, the negative charged oxygen will attack the electrophile carbon atom, forming epoxy ring.

Starting from the aldehyde,

The double bond of the aldehyde breaks, allowing the oxygen to bond with the H in the alcohol. The alcohol subsequently breaks the O-H and forms a ketone.

Note that this molecule is stabilised due to hydrogen bonding; therefore it may be prudent to add a base or acid to encourage tautomerization.


Why is the recipe of Coca Cola still a secret? Why is the recipe for Coca-Cola still a secret?
I think that given the current state of technology, it should be proficient enough to find any of the secret ingredients in Coca Cola.

Any thoughts?
Can we make a 100% identical clone of Coca-Cola next week?

There are different angles this question can be answered:
Chemical point of view:
A full analysis of a totally unknown mixture is painful and extremely costly. It is always helpful to know how many components you are looking for; what types etc. In this case, it is not enough to analyse the elemental composition or some pure elements, but Coke contains a lot of natural products and mixtures like caramel. Imagine how you identify (and correctly describe the production parameters) caramel in a dilute solution...  Also, liquids with extremely high concentrations of sugar and different acids are unpleasant for analysis, as you often have to separate these main component, so you can identify only small components.
That being said, it is not an impossible task; there are many methods to identify e.g. natural products based on DNA traces. But consider another factor:
Business point of view:
The recipe of Coca Cola itself worth nothing.
Most probably any competent soda maker can make a drink that 99% of the consumers cannot distinguish from Coke by the taste (and could do that decades before). However they don't do because no one would buy a drink that tastes like Coca Cola, but made by others.
The value is in the brand. People buy a fake Rolex if they cannot afford a real. But anyone can afford a Coke- why would buy a fake? If you are in beverage business, it is an imperative to make a drink that is somehow DIFFERENT than the other ones!

Universities have HPC's that could probably do it in a week.

Point is, someone has to pay for it and it would be very costly and illegal.  Perhaps the reason it has not been done.


Calculating the potential of an electrochemical cell (Nernst equation) This is the problem:

Calculate the ${E_\mathrm{cell}}$ (not ${E°}$) at 25°C.
I found that the reaction is $$\ce{Cu(s) + Fe^2+(aq) -> Cu^2+(aq) + Fe(s)}$$ $\ce{Cu}$ gets oxidized and $\ce{Fe^2+}$ gets reduced.
I added the ${E°_\mathrm {Cu^{2+}/Cu}}$ and ${E°_\mathrm{Fe^{2+}/Fe}}$ to get ${E°_\mathrm{cell} = -.101}$ . But I'm not even sure if I'm supposed to add them or subtract them because I've seen both done, which is really confusing me. In what situations do you add, and which situations do you subtract half reaction potential values?
I then plugged this calculated value into the Nernst equation.
I said Z = 2 and plugged in the correct constants. However, I am not sure if I calculated Q correctly. I did [anode]/[cathode], but is it supposed to be [products]/[reactants]? It is unclear to me because my instructor did the latter, but I keep seeing the former everywhere else because that is how you would calculate it in an equilibrium problem. I know it doesn't make a difference in this specific problem, but I want to know for future problems when concentration of all components are given (such as if they are all aqueous).
From these steps, I calculated that ${E_\mathrm{cell} = -.164 V}$
How to approach this question?

If you are not sure about the use of the parameter Q, go back to Nernst's law for each electrode, namely $$\ce{E_{M^{2+}/M} = \pu{ E°_{M^{2+}/M}} + \pu{0.0296 V} \times  \log[M^{2+}]}$$ Here the redox potentials of the two half-cells are, according to Nernst's law :$$\ce{E_{Cu} = \pu{+ 0.339 V} + \pu{0.0296 V} \times  \log0.10} = \pu{0.309 V}$$  $$\ce{E_{Fe} = \pu{-0.44 V} + \pu{0.0296 V} \times  \log0.0030} = \pu{- 0.515 V}$$
So the overall performance of the cell is $$\ce{E_{cell} = E_{Cu} - E_{Fe}}=  \pu{0.309 V} - \pu{(-0.515 V)  =  0.824 V}$$


In a substance do atoms actually touch each other? In a substance do atoms actually touch each other?

Quantum mechanics fuzzes out the electrons into clouds of probability distribution (the motion of little electrons orbiting the nucleus is from an old outdated model).  In molecules these electronic clouds do not so much touch, rather they overlap.  See, for instance, https://en.wikipedia.org/wiki/Covalent_bond.

Your question contains a number of conceptual errors that make it clear you have not covered quantum mechanics in any form in your classes yet.

Traditionally, because that was their experience in the real world, scientists would indeed think of atoms as compact spheres (Dalton), hollow spheres with a nucleus and electrons revolving (Rutherford) or the same type of hollow sphere with a nucleus but different orbits on which electrons can be located (Bohr). Using real-world examples, the Dalton model is basically a marble, the Rutherford model a type of cherry and the Bohr model like the solar system.

In all of these models, an atom would have a defined outer boundary which would mark the end and electrons were thought not to go past this boundary except in a reaction. However, these models pose a number of problems that could not be explained using traditional mechanics. For example, an electron revolving around a nucleus would constantly be accelerated towards the nucleus and therefore would emit electromagnetic radiation according to classical physics. This is not observed for atoms. Since there was no way to reconcile the atomic models of Rutherford and Bohr with classical mechanics, an entirely new way to think of atoms needed to be discovered.

This is where quantum mechanics come in. Using the wave-particle dualism and a set of axioms, wave functions for electrons can be calculated. These wave functions give rise to a certain energy (and also to some other observables) that can be measured and that explains the energy levels Bohr had in mind when devising his atomic model very well. Another consequence of this wave function model is that it no longer makes sense to imagine a fixed atom boundary. Theoretically, an electron of an atom can be infinitely far away from the nucleus while still belonging to it. (In mathematical terms: even for very large distances $r$ the absolute value of the wave function $\Phi\ne0$.)

This means that your notion of atoms ‘touching’ does not make sense at microscopic levels, since there is no boundary that could be touching the other side. Instead, the nuclei will come close enough to each other so that the electrons can interact in a favourable way while just far enough apart so that their electrostatic interaction does not dominate.


Charles's Law confusion Charles's Law states that for every 1 degree Celsius increase in temperature of a gas, the volume increases by 1/273 of the original volume (at constant pressure).

What I am wondering is, doesn't gas have an indefinite volume, therefore it is not possible for it to have volume at all? Am I missing something here?

A gas occupies the volume of whatever the container it is in. Take a balloon filled with Helium for example. Now the volume of Helium gas is the volume of the ballon.

If you increase the temperature by 1 °C at constant pressure the volume of the balloon increase by 1/273th of the original volume of the balloon. Hope this might clear things up.

Simply Charle's Law(change in volume is directly proportional to change in temperature) is true with the assumption of constant pressure.


Is photosynthesis endothermic and exothermic at the same time? ... somehow? Photosynthesis is obviously an endothermic reaction, I mean, what would be the point otherwise, right? This is probably just sheer and utter stupidity on my behalf, but why does Wikipedia say the heat of formation of glucose is negative? As you can probably see I'm on a quest of understanding enthalpy as a concept and I'm still quite new, but some sort of ultranoob clarification would be very much welcome.

Consider that when $1$ molecule of glucose forms, this happens at the cost of $6$ molecules of  $\ce{CO2}$ and $6$ molecules of $\ce{H2O}$ :

$
\ce{6CO2(g) + 6H2O(l) -> C6H12O6(s) + 6O2(g)}
$

Considering that at $25\ \mathrm{^\circ C}$ and  $1\ \text{atm}$

$
\Delta H_{\text{f},\ce{CO2}}^\circ=-393.5\ \text{kJ/mol}
$

$
\Delta H_{\text{f},\ce{H2O}}^\circ=-285.8\ \text{kJ/mol}
$

$
\Delta H_{\text{f},\ce{C6H12O6}}^\circ=-1273.3\ \text{kJ/mol}
$

You get, by summing the enthalpies of formation for each of the involved species and considering that for $\ce{O2}$ that enthalpy is zero:

$
\Delta H_\text{r}^\circ=6\times393.5\ \text{kJ/mol}+6\times285.8\ \text{kJ/mol}-1273.3\ \text{kJ/mol}=2802.5\ \text{kJ/mol}
$

Note that, for $\ce{CO2}$ and $\ce{H2O}$, the change of sign is required, since you "lose" them, while you form glucose.

So, finally $\Delta H_\text{r}^\circ$ (heat to be absorbed to form one mole of glucose) is positive, indicating that photosynthesis is indeed endothermic.

Endothermic, exothermic. The difference is that an endothermic molecule is unstable. From the moment you remove the extra energy, it reverts back to its original structure. From original author, Park S. Nobel, 1974, it all depends on the manner you get the four protons in order to get the monomer. (five or six monomers to build a sugar). You get either four protons with two or four water molecules. Up to now, it is utilized two water molecules. So you say that the oxygen of photosynthesis is endothermic. And that can't be. (Park S. Nobel is the father of the expression Electron Chain Transport). He also proposed the possibility that the origin of the four protons is due to four molecules of water but for that one he did not propose a model, so the winning model is Two Water Molecules, only because the other model was not described.


Ethane and ethene: which is easier to burn? Which one burns hotter? Why? Anything that burns "easy", has a low activation energy ($E_\mathrm a$) for the burning process. Anything that burns hotter, will have a lesser enthalpy and thus, will have a more aggressive exothermic reaction.

Ethane is a molecule with two carbon atoms and 6 hydrogen atoms; all bonded with covalent single bonds. Ethene has the same number of carbons, but 4 hydrogen atoms; the bond between carbon atoms, in this one, is a double bond.

If you had to choose a fuel, and heat was your prior factor, which one would you have chosen? I imagined that for this, the activation energy and the enthalpy comparison could lead to an answer. I could easily have looked both $E_\mathrm a$ and enthalpies up, but I didn't.

Why and how do enthalpy and $E_\mathrm a$ change when two single bonds and two H atoms are replaced with a double bond?

Why and how do enthalpy and $E_\mathrm a$ change when two single bonds and two H atoms are replaced with a double bond?

Olefins are less stable than their corresponding alkane analogues.  For example, when ethene is hydrogenated to ethane roughly 33 kcal/mol of heat is given off because ethene has a higher energy content than ethane.

Why is this, why is a molecule with a double bond higher in energy than a molecule without a double bond?

There are two equivalent ways to describe the orbitals in ethene.  One is the "double bond" description where a pi and sigma bond exist together side by side.  We know that a pi bond is not as strong as a sigma bond due to poorer overlap with the pi bond compared to the the sigma bond.  It only takes something on the order of 60 kcal/mol to break a pi bond and produce cis-trans isomerization about the double bond, whereas it takes something like 90 kcal/mol to break a carbon-carbon single bond. This poorer overlap with the pi bond causes ethene to be higher in energy than ethane.  Alternatively one can view ethene as a two-membered ring, no pi bond, just 2 sigma bonds between the 2 carbons forming a two-membered ring.  It is easy to see that such a ring system would contain a significant amount of strain.

Whether you use the pi bond or two-membered ring approach to describe olefinic double bonds, both lead to the conclusion that olefins are destabilized (higher in energy content) due to poor overlap or ring strain (which in itself is really a reflection of poor overlap).  Because of this destabilization, alkenes will generally have lower activation energies and more heat will be given off when they react compared to alkanes.

Example:
Let's consider the bromination of ethene and ethane
$$\begin{align}
\ce{C2H4 + Br2 &-> BrH2C-CH2Br}\\
\ce{C2H6 + Br2 &-> 2 CH3Br}
\end{align}$$

The heats of formation of ethene, bromine and 1,2-dibromoethane are 52.5, 0 and -37.8 kJ/mol respectively; the reaction is exothermic by 90.3 kJ/mol.  On the other hand, in the case of ethane, the heats of formation are -84.7, 0 and (2x) -37.4 kJ/mol; this reaction is endothermic by 10.3 kJ/mol.  The olefinic reaction is significantly favored for the reasons discussed above.

Ethene produce a high temperature. Ethane release more heat per mol. The extra heat released by ethane goes into heating up more water which has a high heat temperature.


Are there atoms with different numbers of protons, but the same number of electrons, with different orbital occupations? Imagine there is an electron-stripped nucleus with $Z$ protons, in high vacuum and isolated. Then you add electrons, one by one, and check which orbitals are filled. Will this depend on $Z$ at all, or does it only depend on the number of electrons that are added?

In other words, will an ion have different orbits filled than the corresponding neutral atom with the same number of electrons?

In other words: are there atoms with numbers of protons $Z_1$ and $Z_2$, both with $n\leq\operatorname{min}(Z_1, Z_2)$ electrons that have different orbital fillings?

Yes, the number of protons in the nucleus does affect how the electrons are filled into orbitals. There are a number of real-world examples that usually stem from transition metals since they do not only have a rather rich oxidation state chemistry but also two energy levels (3d and 4s) which are rather close by each other.

To give you a concrete example, take $\ce{Fe^2+}$ and $\ce{Cr^{\pm 0}}$. They both have the same number of electrons. However, chromium’s ground state electronic configuration is given as $\mathrm{3d^5 4s^1}$, while iron(II) typically has $\mathrm{3d^6 4s^0}$.

The general explanation is that the higher nuclear charge in the iron nucleus attracts the electrons better than the lower nuclear charge in chromium. This difference causes the orbitals to contract. The d-orbitals which are closer to the nucleus, are contracted more and thus experience more stabilisation than the s-orbitals of the next shell.

Since electrons prefer to maintain their lowest energy states, electrons orbiting a stripped nucleus will occupy the lowest available energy levels before filling higher levels (1s before 3p, for example). This is called the Aufbau Principle.

Your question about a striped nucleus is a little confusing. Obviously an electron filling a hydrogen ion will have different energy states compared to an electron rushing toward a helium ion.

Check out https://en.wikipedia.org/wiki/Aufbau_principle as a starting point to your research.

Hope this helps!


What is the platinum doing in a Zippo hand warmer? My Zippo hand warmer requires lighter fluid, platinum catalytic burner, and a heat source to start the flameless catalytic process. What is the platinum doing in this reaction?

Platinum is a catalyst and decreases the activation energy, so the process can start at lower temperatures.


Potassium chlorate from potassium chloride and hydrogen peroxide Why can't you make potassium chlorate using the chloride anion from the potassium chloride and having oxygen bond with it from hydrogen peroxide?

Do not attempt! Peroxide is not a strong enough oxidant to oxidize chloride ion to chlorate. Neither is oxygen. However, peroxide can oxidize chloride to poisonous chlorine gas.

Redox reactions are spontaneous if the two half reactions would produce a positive potential in a galvanic cell. Most redox half reactions have measured standard electrode potentials against a known standard so that they can be compared.

The reaction you want to do is:

$$ \ce{3H2O2 + Cl- -> 3H2O + ClO3-}$$

The two half-reactions are:

Oxidation:   $\ce{Cl- +3H2O -> ClO3- +6H+ +6e-}$
Reduction:   $\ce{H2O2 +2H+ +2e- -> 2H2O} \ \ \ \ E^ \circ =  +1.78 \text{ V}$

The reduction potential of hydrogen peroxide is on the table, but the reduction potential for the other reaction is not. However, the other reaction is the sum of two reactions on the table, so we can use Hess's Law.

$$\ce{ 2ClO3- + 12H+ + 10e- ->Cl2 + 6H2O} \ \ \ \ E^\circ = +1.49 \text{ V}$$
$$\ce{Cl2 +2e- -> 2Cl-} \ \ \ \ E^\circ = +1.36 \text{ V}$$
$$\ce{2ClO3- 12H+ + 12e- -> 2Cl- + 6H2O}  \ \ \ \ E^\circ = +2.85 \text{ V}$$

Since standard potentials do not scale by stoichiometric factor (it's in the Nernst equation instead), the value we just found also works for our oxidation reaction. However, since this reaction is reverse of the "reduction half-reaction", we need a negative sign:

Oxidation:   $\ce{Cl- +3H2O -> ClO3- +6H+ +6e-}\ \ \ \ E^ \circ =  -2.85 \text{ V}$
Reduction:   $\ce{H2O2 +2H+ +2e- -> 2H2O} \ \ \ \ E^ \circ =  +1.78 \text{ V}$
Overall: $\ce{3H2O2 + Cl- -> 3H2O + ClO3-}\ \ \ \ E^ \circ =  -1.07 \text{ V}$

This reaction is not spontaneous. Peroxide cannot oxidize chloride, at least not to the chlorate ion. Note that peroxide can and does oxidize chloride to poisonous chlorine gas in the presence of acid. This redox is spontaneous and will happen.  Do not attempt!

Reduction:   $\ce{H2O2 +2H+ +2e- -> 2H2O} \ \ \ \ E^ \circ =  +1.78
   \text{ V}$
Oxidation:   $\ce{2Cl- + 2e- -> Cl2} \ \ \ \ E^ \circ =
   -1.36 \text{ V}$
Overall: $\ce{ 2Cl- + H2O2 + 2H+ -> Cl2 + 2H2O} \ \ \ \ E^ \circ =  +0.42 \text{ V}$

Peroxide will oxidize chlorine gas to chlorate, but who wants to handle chlorine?

Oxygen atoms want nothing more than to share, borrow, or steal a couple extra electrons.  Chlorine has the same desire for an extra electron.  Therefore, they do not react in solution except in special circumstances I don't know about yet.


The last element's atomic number I was just thinking what can be the last atomic number that can exist within the range of permissible radioactivity limit and considering all other factors in quantum physics and chemical factors.

Nobody really knows. Using the naive Bohr model of the atom, we run into trouble around $Z=137$ as the innermost electrons would have to be moving above the speed of light. This result is because the Bohr model doesn't take into account relativity. Solving the Dirac equation, which comes from relativistic quantum mechanics, and taking into account that the nucleus is not a point particle, then there seems to be no real issue with arbitrarily high atomic numbers, although unusual effects start happening above $Z \approx 173$. These results may be overturned by an even deeper analysis with current quantum electrodynamics theory, or a new theory altogether.

As far as we can tell, however, we will never get anywhere close to such atomic numbers. Very heavy elements are extremely unstable with respect to radioactive decay into lighter elements. Our current method of producing superheavy elements is based on accelerating a certain isotope of an relatively light element and hitting a target made of an isotope of a much heavier element. This process is extremely inefficient, and it takes many months to produce significant amounts of material. For the heaviest elements, it takes years to detect even a handful of atoms. The very short lifetime of the heaviest targets and the very low collision efficiency between projectile and target mean that it will be extremely difficult to go much further than the current 118 elements. It is possible that we may find somewhat more stable superheavy isotopes in the islands of stability around $Z=114$ and $Z=126$, but the predicted most stable isotopes (which even then are not expect to last more than a few minutes) have such a huge amount of neutrons in their nuclei that we have no idea how to produce them; we may be condemned to merely skirt the shores of the islands of stability, while never climbing them.

EDIT: Note that the best calculation presented above is based on quantum electrodynamics alone, i.e. only electromagnetic forces are taken into account. Obviously, to predict how nuclei will behave (and therefore how many protons you can stuff into a nucleus before it's impossible to go any further), one needs detailed knowledge of the strong and weak nuclear forces. Unfortunately, the mathematical description of nuclear forces is still an incredibly tough problem in physics today, so no one can hope to provide a rigorous answer from that angle.

There must be some limit, as the residual nuclear forces are very short-ranged. At some point there will be so many protons and neutrons in the nucleus (and the resulting nucleus will have become so large) that the diametrically opposite parts of the nucleus won't be able to "detect" each other, as they are too far away. Each additional proton or neutron produces a weaker stabilization via the strong nuclear force. Meanwhile, the electrical repulsion between protons has infinite range, so every additional proton will contribute repulsively just the same. This is why heavier elements need higher and higher neutron-to-proton ratios to remain stable.

Thus, at some atomic number, possibly not much higher than our current record of $Z=118$, the electrical repulsion of the protons will always win against the strong nuclear attractions of protons and neutrons, no matter the configuration of the nucleus. Hence, all sufficiently heavy atomic nuclei will suffer spontaneously fission almost immediately after coming into existence, or all the valid reaction pathways to reach an element will require events which are so fantastically unlikely that if even all the nucleons in the entire observable Universe were being collided with each other since the Big Bang in an attempt to synthesize the heaviest element possible, we would statistically expect some sufficiently heavy atom to not have been produced even once.

It might be possible that they can keep on going to find new elements by using cycloterons and fusing atoms together till they reach a blackhole. So  I believe that may be the limit for atomic number can be calculated by the singularity of Kerr Newman metric. But by assuming periodic table pattern if one other element could be found there comes at least 32 others to complete one other row of periodic table to reach may be 150 atomic number.


Configuration of endocyclic double bond in Vitamin A I think the double bond inside the ring of vitamin A should have an (E) configuration. Am I wrong? I don't think I'm assigning the priority correctly.

The endocyclic double bond of vitamin A has a (Z) configuration. Here’s why:

First, let’s determine priorities of the atoms connected on the top side of it:

C

C (ring)
C (methyl 1)
C (methyl 2)

C

C
(C) (ghost atom; due to the double bond)
H

So the ‘ring side’ — the atom on the left in the way it is drawn at the moment — has a higher priority than the side chain (C at (1.3) beats H at (2.3)).

What about the bottom side?

C (ring)

C
H
H

C (methyl group)

H
H
H

Here again the ‘ring side’ has a higher priority than the methyl group: (1.1) carbon beats (2.1) hydrogen.

Both left-hand sides have a higher priority than their corresponding right-hand side. Thus, the double bond is (Z) configured.

There are two groups of compounds with Vitamin A, first is retinoids, also know as vitamin A and provitamin A, known as carotenoids which is a vitamin A precursors( which contain at least one molecule of retinol).

Naturally retinoids exist in the form of retinol, retinal and retionic acid and they are derived from animal sources. There are hundered of Carotenioids which are mainly synthesized by plants and only 50 of them can act as vitamin A precursors.

Technically speaking, large number of geometrical isomers are theoretically possible for retinoids and carotenoids but due to steric hinderence many mono-cis and poli-cis isomer are rarely occurred in nature.
All-trans forms are predominant in nature and more stable, but only those cis isomers occur which offer less steric hinderence and thermodynamically stable.

5-cis-retinol geometrical isomer  is rather unstable and do not occur in that configuration in nature.


How do technicians treat spilt mercury and bromine? As we know, mercury and bromine are in liquid state at room conditions and they are hazardous. The technicians should have ways to treat them if they are spilt.

  How can we safely treat spilt $\ce{Hg}$ and $\ce{Br}$?

I have thought of using dry ice to cool the mercury down into a solid, which can then be easily removed. But I do not think that this is really practical. Maybe some chemical reactions can help to somehow convert the them into a solid in order to remove them.

Any reputable chemical supplier will normally sell a mercury spill kit, and anyone working with mercury these days should have one. (Either a commercially prepared one, or an equivalent one that's assembled from individual supplies.) Generally speaking, they contain some sort of absorbent/adsorbent to pick up mercury, and a chemical which will react with the mercury and keep it from volatilizing. Sulfur is a common choice (recommended by the EPA), as mercury sulfide is pretty darn inert, as well as being very insoluble in water. (Also, "flowers of sulfur" does double duty as an absorbent.)

After the cleanup, the absorbent is collected (by sweeping - vacuuming can spread mercury vapors), packaged securely, and sent to a hazardous waste disposal company.

As elemental bromine is less widely used than elemental mercury, designated spill kits for bromine are less widely available, but the procedure is generally the same: spread a chemical to react with the bromine and render it more inert, then spread an absorbent to contain the liquid. Elemental bromine is rather reactive (an oxidizer), so a number of common reducing agents can be used. A 5% solution of sodium thiosulfate is a common choice (1) (2).

Again, if you're working with elemental bromine, your lab should have a spill response kit already assembled, with sufficient pre-prepared sodium thiosufate and absorbent to handle the amount of bromine you normally deal with.

The previous was for general information purposes only. If you are looking to work with elemental mercury or bromine, consult with your institution's chemical safety officer before purchasing for more information and restrictions which may apply in your area. Your institution likely also has chemical safety and spill response training, which will inform you of current best practices in handling mercury and bromine spills.

As always with accident handling, the most important thing is not to put yourself in greater danger by attempting to fix it. If the size or extent of the spill is such that you're unsure you can handle the spill safely, the best course of action is to evacuate the premises and let the hazardous materials team do the cleanup. (They'll likely use the same techniques of reacting and absorbing the material, but they'll be wearing more extensive personal protective equipment while doing it.)

First, spilled mercury and bromine are emergencies. If this is happening to you - inform colleges and call specialists.

Second, Mg is magnesium, Hg is mercury.

Bromine is very volatile and irritating. It can kill or injure people very fast. If small amount was spilled - leave the open windows, leave the room, close doors, wait. If spilled one someone (this is a real emergency) you should apply 95-100% ethanol to the injured part of the body. This method helps the best.

Hg is chronically toxic (acute toxicity of metallic mercury is low). You can absorb it on iron iron chloride (III) powder to absorb Hg or apply sulfur to the area of the spill.

Mg is relatively safe and can be collected with with a floor mop.


Is there a function to approximate atomic mass from the atomic number? The product of $2Z$ generally approximates the value of $m_\mathrm{a}$ of atoms that comprise relatively few protons. Whereas, the product of $2.5Z$ generally approximates the value of $m_\mathrm{a}$ of atoms that comprise relatively many protons. The product of $((2+2.5)/2)Z$ generally approximates the value of $m_\mathrm{a}$ of atoms that comprise a moderate number of protons.

What function from the $Z$ values produces a value that maximally approximates the $m_\mathrm{a}$ that corresponds to $Z$?

The short answer is that you can find a power-law fit ($1.61Z^{1.1}$) with low average error.

I'd never really thought about it much, but after downloading the IUPAC Atomic Weights, I decided to do some curve fitting.

Here's a linear fit between atomic number and atomic mass:

As you say, the fit isn't very good for small $Z$, but the overall fit isn't bad - the mean absolute error (MAE) is $2.821 \:\mathrm{u}$, and taken as a whole, the data is surprisingly linear. (Well, surprising to me.)

So I thought of a quadratic fit, requiring the intercept to be 0,0 to ensure the best fit for small $Z$:

Looks better, right? Certainly the fit is much better for first and second row elements, but the MAE only reduces to $2.749 \:\mathrm{u}$.

So I went up to a cubic fit, again requiring 0,0 for the intercepts:

Aha, now we're talking! We get the subtle nonlinearities, and the MAE is down to $1.36 \:\mathrm{u}$.

Thanks to the comment by Nicolau below, I performed a power-law fit.

So that gives a power-law fit with MAE of $0.01 \:\mathrm{u}$ and a fairly easy-to remember function:

$m_a \approx 1.61Z^{1.1}$

Atomic mass as a function of atomic number:

(Atomic number)^2/162 + 2*atomic number = atomic mass

It's the same as the answer given above except the final term is 2x

The equation for the first 16 elements is obviously just 2x.

X^2/162 is needed for the rest of the elements.

Thorium has 90 protons so 90^2/162+2*90 = 50+180=230.

The equation breaks down after thorium (140 neutrons)

After thorium 4 halo neutrons are added for every 3 protons.

https://upload.wikimedia.org/wikipedia/commons/d/d1/Uranium_atom.svg


Ester as a side product in the synthesis of bromoalkane from alcohol. Is it correct? About 5 months ago, I answered a very interesting question albeit a very simple one.

  Why do we need sulfuric acid in creating alkyl bromide and not in creating alkyl iodide from alcohol and hydrogen halide?

The question was why sulfuric acid is used to convert any alcohol to bromoalkane but not iodoalkane. Jan's answer quite pointed out the explanation:

  hydrogen iodide is sufficiently acidic to protonate the corresponding    alcohol and drive the reaction. More acids are not
  needed.
  hydrogen bromide is not acidic enough to drive the reaction. Adding    sulphuric acid protonates the alcohol partially and allows the
  reaction to proceed.
  in the case of weakly soluble alcohols and hydrogen iodide, phosphoric acid may be used instead since it does not oxidize iodide (while sulfuric acid does).

Though I am satisfied with Jan's excellent answer, I further pointed out that sulfuric acid may react with hydrobromic acid to form bromine(aq.) and sulfur dioxide. This bromine may oxidize alcohol to form aldehyde and carboxylic acid and they two will form ester. This was my proposed explanation but I couldn't find any experimental result to back my explanation and heavily relied on theory.

Now recently Nurdrage, a renowned chemist famous for his synthesis of pyrimethamine has uploaded a video on Youtube about the synthesis of bromoethane and bromohexane from ethanol and hexanol using sulfuric acid. He also explained the concern of forming bromine as side product from the reaction of hydrobromic acid and sulfuric acid. He showed a purification technique by adding sodium bicarbonate to react with bromine to form sodium hypobromide.

Now my question is:

Is my explanation of forming ester instead of bromoalkane correct enough? Is there any possibility of forming minute amounts of ester? The reaction may be slow but it can happen before any purification is done (reacting bromine with sodium bicarbonate). If so, what can be done to remove any ester or prevent the reaction of ester formation?

Also related: Why is alcohol boiled with HBr for preparing alkyl bromide?

Technically it could form, but it would be hydrolyzed. If it did form, it would be in negligible quantities. Lots of other stuff could form, and they do, but in miniscule amounts ... bromine oxidized the alcohol to the aldehyde. The aldehyde undergoes aldol condensation. Another alcohol undergoes elimination. The compounds  react in a hetero diels-alder. So dihydropyrans can also form ... Yes, but not in quantities that are observable.


Is Ammonia (NH3) base? In my text book, it is written that $\ce{NH3}$ is a base. They mentioned the reason for this is that it produces $\ce{OH-}$ ions in water. They stated this reaction:

$\ce{NH3 + H2O -> NH4OH}$

So That's why, they said that it is a base.
But in a neutralization reaction, mentioned by our teacher,
Acid + Base = Salt + water
So, if $\ce{NH3}$ is a base then,
it should be,
Acid + $\ce{NH3}$ = Salt + Water
But it does not happen. When reacting with acid, it doesn't produce water.
For example,

$\ce{NH3 + HCl -> NH4Cl}$
--  Reference

Again,

$\ce{NH3 + H2SO4 -> (NH4)2SO4}$
--  Reference

So it is not producing water in the neutralization reaction. Hence, I'm confused whether it is base or not.

Your problems are caused by using the Arrhenius theory, which is based on electrolytic dissociation.

According to the Arrhenius definition, acids are compounds that dissociate and release hydrogen ions $(\ce{H+})$ into the solution:
$$\ce{HCl -> H+ + Cl-}$$
Bases are defined as compounds that dissociate and release hydroxide ions $(\ce{OH-})$ into the solution:
$$\ce{NaOH -> Na+ + OH-}$$
The products of a neutralization reaction of an acid with a base are a salt and water:
$$\ce{HCl + NaOH -> NaCl + H2O}$$
This definition works for metal hydroxides such as $\ce{NaOH}$, which can exist as pure compounds. However, it doesn’t work for bases such as $\ce{NH3}$.

In order to solve this problem for $\ce{NH3}$, the fictitious compound “ammonium hydroxide” $\ce{NH4OH}$ was proposed (see also this related answer). Assuming this formula, the Arrhenius definitions of bases and neutralization reactions work again:
$$\ce{NH4OH -> NH4+ + OH-}$$
$$\ce{HCl + NH4OH -> NH4Cl + H2O}$$
However, ammonium hydroxide cannot be isolated; the fictitious solid compound does not exist. Nevertheless, the misleading traditional name “ammonium hydroxide” is still widely used for solutions of ammonia in water.

In order to really solve your problems with $\ce{NH3}$, you need to switch from the definition of bases according to the Arrhenius theory to the definition of bases according to the Brønsted–Lowry theory.

"Acid + base gives salt and water" is only possible when acid and base both can dissociate into two ions. Let's take an example
$\ce{HCl}$ and $\ce{NaOH}$ gives $\ce{NaCl}$ and $\ce{H2O}$
So, in that case, $\ce{HCl}$ and $\ce{NaOH}$ both can dissociate.

But in case of $\ce{NH3}$ and $\ce{H2O}$ only $\ce{H2O}$ dissociate into $\ce{H+}$ and $\ce{OH-}$ but $\ce{NH3}$ remains the same.
And reaction proceeds with the help of lone pair of $\ce{NH3}$.
$$\ce{NH3 + H+ + OH- ->NH4+ +OH-}$$ and in later stages both combine together due to electrostatic force.


Why is water an Arrhenius acid? Is water an Arrhenius acid? We define Arrhenius acidity with respect to water, but what will water itself be? It can donate $\ce{H+}$ as well as $\ce{OH-}$.

We define Arrhenius acidity with respect to water

That definition is new to me, whatever it is. The definition I know is:

An Arrhenius acid is a substance that dissociates in water to form $\ce{H+}$ ions.
An Arrhenius base is a substance that dissociates in water to form $\ce{OH-}$ ions.

The key concept to answer the question is the self-ionisation of water given in equation $(1)$.
$$\ce{H2O <<=> H+ + OH-}\tag{1}$$
This ionisation happens in water, and the ion product as written in equation $(2)$ is $10^{-14}$.
$$K_\mathrm{w} = [\ce{H+}] [\ce{OH-}] = 10^{-14}\tag{2}$$
Thus, water qualifies as a substance that dissociates in water to form $\ce{H+}$ ions. It also qualifies as a substance that dissociates in water to form $\ce{OH-}$ ions. It is both an Arrhenius acid and an Arrhenius base and thus the only Arrhenius amphoteric compound.

water isn't the only arrhenius amphoteric compound. If I am remember correctly Aluminium can also react with water to form both an acid and a base.


Alkyne formation from vicinal dihalide Why must 3 equivalents of sodium amide be used here to form the alkyne? The part I specifically have issue with is in red.

From what I understand, we should just use 2 equivalents of sodium amide to get rid of the bromines and to form the alkyne. Another equivalent of sodium amide will form the acetylide anion. Removing two protons effectively forms the triple bond and eliminates the two halogens. Also, take a look at example 3 here ..

Why then does the book's solution manual want me to use 3 equivalents of sodium amide to get neutral acetylene, and another equivalent to get to the acetylide ion?

As you can see, my book's solution manual contradicts another book's discussion of obtaining alkynes from vicinal dihalides.

I think the comments basically have the answer - if you use only 2 equivalents of sodium amide, you will use 1.33 equivalents creating the alkyne and the remaining 0.67 equivalent will react with the alkyne after it is formed (an acid-base reaction) to form the acetylide salt -- probably a much faster reaction than the dehydrobromination reactions. So about 1/3 of the original dibromide will remain unreacted. You need the third equivalent of sodium amide to drive the reaction to completion.

The third amide is used to deprotonate the alkyne after it is formed. It drives the formation of alkyne to the right, making a lot better yield of alkyne. That is why they also added water as a proton source in the second step to protonate the acetylide ion.


IUPAC name of aldehyde with two functional groups I am confused as to what suffix to use for the following compound:

Should it be carbaldehyde or just -al, like should it be propane-1,2-dicarbaldehyde or should it be propan-1,2-dial?
Or maybe it's 1-methylethane-1,2-dial? I am pretty sure its either of these three or maybe something else.

You have correctly identified the principal characteristic group as aldehyde.
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), simple mono- and dialdehydes are named substitutively using the suffix ‘al’.

P-66.6.1.1.1 Mono- and dialdehydes derived from alkanes are named substitutively using the suffix ‘al’ added to the name of the parent hydride with elision of the final letter ‘e’ of the parent hydride before ‘a’.

For example: pentanal and pentanedial, respectively. Thus, the compound that is given in the question is a butanedial.
The terminal locants of aldehydes, i.e. ‘1,4’ in case of butanedial, are not cited.

P-14.3.4.1 Terminal locants are not cited in names for mono- and dicarboxylic acids derived from acyclic hydrocarbons and their corresponding acyl halides, amides, hydrazides, nitriles, aldehydes, amidines, amidrazones, hydrazidines, and amidoximes, when unsubstituted or substituted on carbon atoms.

Also the locant ‘2’ of 2-methylbutanedial is not cited.

P-14.3.4.3 The locant is omitted in monosubstituted symmetrical parent hydrides or parent compounds where there is only one kind of substitutable hydrogen.

Therefore, the preferred IUPAC name is methylbutanedial.

By way of comparison, the Blue Book contained the similar example ‘2-chlorobutanedioic acid’, which was later corrected to ‘chlorobutanedioic acid’.

The parent hydrocarbon chain is butane. The parent functional group is dialdehyde. The side-chain is methyl. According to the IUPAC nomenclature, the given organic compound is 2-methyl-1,4-butanedial.
According to the IUPAC nomenclature for Aldehydes, the suffix -carbaldehyde is used only when the aldehyde is attached to a ring chain. The suffix -al is used when the aldehyde is attached to a straight/branched chain.


Applying Raoult's Law in volatile solids–liquid solutions My textbook explains applying Raoult's Law for the two cases:

liquid–liquid solutions, in which the vapour pressure of the solution is due to both the components
non-volatile solids–liquid solutions, in which the vapour pressure of the solution is only due to the liquid component of the solution

I am curious as to what will happen (how things change for the application of Raoult's Law) if the solution consisted of some volatile solid like camphor, in some liquid (possibly some liquid hydrocarbon).

Certainly, the solid now also contribute to the total vapour pressure of the solution, but does it still block the escaping of the solvent molecules into vapour phase as much as the non-volatile solid blocks the liquid solvent molecules?

If yes, then how can I apply Raoult's Law for the case to find the total pressure of the solution?

Raoult's Law is only valid when the solute is non-volatile. Therefore, you cannot apply it.


Why doesn't cottage cheese melt? Most kinds of cheese melt but cottage cheese does not at usual cooking temperatures.

During the preparation of various kinds of cheese, I have heard of a coagulant named ‘rennett’ being added along with the acidifying agent. For cottage cheese it is just lemon juice mostly (acidifying agent). Does this have anything to do with the difference in properties?

You essentially make cottage cheese in the process of making cheddar cheese. But cottage cheese is about 80% water. The "cheese" solids are soluble in oil, but not in water. So cheddar cheese will "melt" but cottage cheese won't.

It melts but only if it is in a proper medium. I discovered that shredded cottage cheese melts in my pasta sauce.

I do not know if it is the acidity or the presence of extra ingredients that does this but any macro-physical trace of it is gone, just like mozzarella.


Determining sulfate content in Mohr's salt I am practicing problems for my chemistry test and I am stuck on the following problem:

The value of $x$ in $\ce{Fe(NH4)2(SO4)2 ⋅ xH2O}$ can be found by determining the amount, in moles, of sulfate in the compound.
A $\pu{0.982 g}$ sample was dissolved in water and excess $\ce{BaCl2(aq)}$ was added. The precipitate of $\ce{BaSO4}$ was separated and dried and found to weigh $\pu{1.17 g}.$
Calculate the amount (in moles) of $\ce{BaSO4}$ in the $\pu{1.17 g}$ precipitate. And calculate the amount (in moles) of sulfate in the $\pu{0.982 g}$ sample of $\ce{Fe(NH4)2(SO4)2 ⋅ xH2O}.$

I was able to do the first part of the question:
$$n (\ce{BaSO4}) = \frac{\pu{1.17 g}}{\pu{233.38 g mol-1}} = \pu{0.005 mol}$$
But, I am unsure of the second part. Any help would be appreciated.

If there is $\ce{0.00501 mol}$ of $\ce{BaSO4}$ in the precipitate, this means that there was $\pu{0.00501 mol}$ of sulfate in the Mohr's salt solution and $\pu{0.00501 mol}$ of sulfate in $\pu{0.982 g}$ of Mohr's salt. And there was one half, that is $\pu{0.0025 mol}$ of $\ce{Fe(NH4)2(SO4)2.xH2O}$ in $\pu{0.982 g}$ of Mohr's salt. This allows you to calculate the molar mass of the salt, and to deduce $x$.


How to separate a 5 atom thick layer of Cu from Au I've become very curious about (electrochemical) atomic layer deposition (ALD) via surface limited redox replacement.

Since the technique allows for deposition of metals in atom thick layers it's possible to create films of nearly exact atomic thickness. However, I haven't found any reference that describes how one can remove the film from the (usually Au) substrate that it's deposited on.

Hypothetically speaking, if one used ALD to generate a film of $\ce{Cu}$ that is 5 atoms thick (arbitrary length and width), how would one go about removing the film in-tact? Is it even possible?

In most cases I've seen, ALD is performed on the intended substrate - that is, there's no need to separate it. In other cases, you use a sacrificial substrate that you can remove easily.

If you wanted to separate the film from the substrate, there are various techniques, but nothing comes to mind for separating Au from Cu. I find it incredibly unlikely in the specific case of a 5-atom thick layer of Cu, which would likely not survive typical techniques (vs. bending, mushing, forming a larger packing, etc.)

Some typical techniques for separating two films:

Adhesive tape (as mentioned in a comment) - works for many films, e.g., graphene, other layered materials, where you physically pull apart layers
Swelling (e.g., separating Au from mica) - including various intercalation methods used in graphenes and graphene oxides and other layered materials, so that the layers separate more easily
Dissolving, etching, or eroding the substrate (e.g., back-etching a Si wafer to expose an $\ce{SiO2}$ or thin metal film deposited on the wafer)

In my group, for example, we make "template-stripped gold" where Au is deposited on cleaved mica substrates. Then the thin gold layer is glued to glass substrates and the mica is separated from the Au by swelling / degredation, leaving a very smooth Au surface.

This technique can work with fairly thin layers of gold (~10-30 nm) but there's a large difference between Au and mica, making separation possible. Not true with Au/Cu.

I'd suggest electrostatic repulsion. Charge both layers with electrons and they'd separate easily. Then use a electron microscope to manipulate the layer.


Energy in a Chemical Reaction In a chemical reaction, is energy always either lost or gained by the reactants? As much as I am concerned, changes in energy can occur during the absorption of heat or the emission of light or heat (these are the most often occurring). Is every chemical reaction accompanied by these conditions? Aren't there any chemical reactions without these conditions? Usually when we observe common chemical reactions (sugar that is melting in the water), we don't see any emissions or absorptions. Are these changes so trifling that we can't feel them (while they actually exist), or there are reactions with no emissions or absorptions?

your question is good yet lacks the specifics.
there are different so called "energy fields" in which changes occur (entropy, potential ,gibes and so forth).
the thermodynamic field is endless and can be spanned with any variables (using maxwells equations and Legendre transformation)
there is
a sort of an axiom that states that all matter will try to find a way to a lower state of energy.
but that said its not to scientific.

now for the question:
every reaction creates some amounts of electromagnetic waves (as a price for electron "movement") some can be seen with the naked eye as they emit waves in the Uv-vis spectrum some are just infrared radiations of inner states of electron decay to a lower energetic state via vibrational relaxation.

ideally no reaction will occur in the ideal zero as there is no energy to "pay" for any kind of a reaction.

for further read:
quantitive chemical analysis by harris
chapter 17 fundamentals of spectrophotometry.


How to selectively convert benzene to p-nitrotoluene? I converted benzene into toluene by Friedel-Crafts alkylation reaction.
Then by nitration I got two compounds o-nitrotoluene and p-nitrotoluene.

Is it possible to get only p-nitrotoluene?

I'd doubt that exclusive nitration of toluene in para position is possible in homogenous solution.

But imagine to stuff a toluene molecule into a very narrow "tube". This tube will block any attack at C-2 (and C-3), while a nitrating agent may approach C-4.

Exactly this has been done in the vapour phase nitration of toluene with nitric acid over a zeolite (ZSM-5). See: Kalbasi, R.; Ghiaci, M.; Massah, A. Highly selective vapor phase nitration of toluene to 4-nitro toluene using modified and unmodified H3PO4/ZSM-5. Applied Catalysis A: General 2009, 353 (1), 1–8. DOI: 10.1016/j.apcata.2008.10.013.

I think It can be done. But it's a rather long process.First prepare nitrobenzene from benzene by nitration.Then, convert it to aniline by adding a reducing agent(preferably Sn/Hcl).Then, Convert the NH2 group to NHCOCH3 by adding ethanoic anhydride in acetic acid(Which is a weaker activating group than NH2 but a stronger one as compared to CH3).Then, Add the methyl group by friedel-crafts alkylation. Then, convert NHCOCH3 back to NH2 by hydrolysis.Then, convert aniline to diazonium chloride by adding NaNo2 in Hcl. Finally convert diazonium chloride to Nitro group by Adding NaNo2/Cu. U get p-Nitrotoluene. But, I am not too sure this is acceptable.


Why there is charge on the elements, in the dative bonding? According to the definition of dative bonding electrons are shared which are given by one of the atoms, so if the electrons are shared so how come there is charge on the elements in the dative bonding?

To begin with, the term "dative bond" is not used anymore. It has been replaced by "coordinate bond" :

Let's have an example of this kind of bond.

As you can see, this kind of bond occurs when an atom is giving two electrons from itself to form the bond. The typical coordinate bond is the $\ce {N}$ already bonded with $\ce {3H}$ forming a bond with a proton. Since the proton has no electron, the $\ce {N}$ atom has to fully give one of his electron from his remaining doublet to form the bond. But to respect the charge equality, you had a positive charge at the begining, no electrons were added or removed from the system, so the positive charge should remains on the final molecular edifice.

The question is, where is located this positive charge ? One could start thinking this charge is located on the $\ce {N}$ atom since it is giving an electron. But actually, the $\ce {N}$ atom has a greater electronegativity than the $\ce {H}$ atom, that's say its hability to attract electrons is greater. Therefore the positive charge remaining should be equaly distributed over all the $\ce {4H}$ since they are equivalent.

But, altogether, the net charge of the edifice will be $+1$.

Dative bond/coordinate bond is a type of covalent bond where an atom donate electron and another one does not share any but receive the electron.

Note that i bolded covalent bond. Meaning that this bond happened between 2 non-metals, not between metal and non-metal (that's Ionic bond). In ionic bond there's formal charge because there's electrostatic force occuring between the two atom, but in covalent bond there is no electrostatic attraction at all.

In order to keep track between the transfer of electron we use either formal charge or oxidation number, because covalent bond doesn't have any electrostatic attraction we use oxidation number (also known as imaginary formal charge).

So what you meant might be Oxidation Number and not Formal Charge

Hope that helps.


How does the steric effect affects the electronic density of the central atom? For example, if you have Nitrogen attached to some methyl groups, and other attached to some ethyl groups. Why the electron releasing effect on the ethyl will be lower than for the methyl?

Both methyl and ethyl groups have +I effect and its obvious that the ethyl groups have more electrons than the methyl groups, so in this case the present of more electron density on nitrogen atom with ethyl groups lessen the -I effect of the nitrogen atom toward the ethyl groups compared to that on the methyl groups. I think this should be the reason.


Does liquids that emit light when vibrated exist? I am looking for a liquid that illuminates when vibrations are added.

It does not have to be strictly chemical, it might as well be something else as well.

Maybe a crystal in powder form mixed with some kind of oil, or even algae in a thick:ish liquid.

Does such a liquid exist?

Find compounds that emit light under photolysis.

As that may involve radical production pathways, one can also try to apply sonolysis to the same compounds as:

H2O (with dissolved O2, N2,..) + Sonolysis -> *H + *OH

Reference source: 'Free radical formation from sonolysis of water in the presence of different gases'), to quote:

"In this case, O2 reacts with •H to form •OH, indicating that the presence of dissolved O2 in water sonolysis gives additional amount of •OH. "

Also:

H2O2 + Sonolysis -> *OH + *OH

Apparently, with applications, see, 'Bactericidal effect of hydroxyl radicals generated by the sonolysis and photolysis of hydrogen peroxide for endodontic applications'.

where it is likely that one may be able to replicate to varying degrees the action of light treatment.

Here is a source example citing the action of reactive oxygen species (ROS), which can be created by the UV photolysis of, for example, H2O2 (or sonolysis per reference above):

"In biological systems much attention is focused on the two free radicals, superoxide anion and hydroxyl radical....The generation processes of ROS can be monitored using luminescence. Because of the very weak native luminescence of ROS, both luminol and lucigenin have been used in the past to give measurable signals. However, the luminescence of these two substances is very small compared to that of Pholasin®, which allows precise analysis on very small samples."

So, perhaps try performing a sonolysis of, say, luminol or lucigenin, in H2O2 might be interesting.


Where can I find a downloadable spreadsheet of element properties? Sorry for the FAQ, but I really don't know where to find a full spreadsheet for element properties. I mean a table that has one row for each element, and the columns are its properties, such as name, atomic number, density, melting point, etc. I know that there are many sites or references for that, but I need it in tabular format, because I want to analyze data, make charts, etc.

Well, here's a CSV file that I parsed out of the JSON data provided by Paul Nathan's website, which in turn was produced from gPeriodic data in response to this question. gPeriodic is FOSS, so I can only assume that the element data contained within is good to share, though I have no idea of its provenance.

The parser uses the python json module to read the data, which I then wrote into comma separated values, which should be readable by any halfway competent spreadsheet program.

Here it is: Pastebin Link

save it as elementdata.csv and you're good to go.

Some gotchas:

This reproduces the gPeriodic data, warts and all. I make no guarantees regarding its correctness, recency, etc. I just munged the data.
Some of the data is augmented with tildes, notes about temperature/polymorph/state etc. I've left these as they are but you may need to trim them if you want to plot them as numeric values.
Argon had an atomic radius of '2-', whatever that means. I cut it.
Excel is joke software and habitually interprets numbers wrapped in parentheses as negatives, because apparently some accountants decided that surrounding numbers in parens is a more sensible option than using a minus sign. Needless to say, this is wrong. I've addressed the problem by wrapping the affected numbers in angle brackets.
There are no ionic radii in the source data. At all. Not my fault.
The columns inherit the units of the source data. I've left the units out on purpose because I wanted each column to have no spaces in the name for ease of processing in R or whatever. Furthermore, the units in the source file have some inconvenient characters from an encoding perspective.

Tried plotting Z versus covalent radius in R from this data - looks alright:

Some of the data points are missing, some are not read by R because they are wrapped in <> - pre-process to your heart's content.

Here is a Google Sheets spreadsheet with basic data:
https://docs.google.com/spreadsheets/d/1vEOwbbl-l805_i2oWSVCFxfmFDstwkuRjkZQfBrTh6k/edit?usp=sharing

Hope this is useful!


What does the amount of acetic acid in vinegar depend on? I have different types of vinegar and I'm wondering what factors influence how much acetic acid is present in different vinegars (e.g wine vinegar/white vinegar/apple cider). I was thinking maybe it has something to do with different fermentation processes. Any thoughts?

Vinegar comes from French "vin aigre", aka sour wine. The transformation of wine into vinegar happens when some (possibly unwanted) aerobic bacteria transform the alcohol into acetic acid by oxidation. So any alcoholic mixture which allows growth of those bacteria can be turned into vinegar, like cider, sake...

The amount of acetic acid depends on the alcohol content which in turn depends on the initial concentration of sugar. It is also dependent on the time allowed for fermentation.

In fact it is much more complicated than that. Have a look at Wikipedia for more info.

My 3.785 L, extra strength 6% "Acitity" vinegar (V), means (1), and I just found this out myself. Vinegar produced by the Generator Method, which produces anhydrous acetic acid. The pure Acetic acid (Ac) crystalizes to a solid at less then 16.7 C. (2). You can imagine how easy it is to make any concentration you want, when you have a powder to mix with water = distilled vinegar. That means that in that 3.8 L bottle only about 270g is actually Acetic Acid. However because Acetic Acid is a weak acid in water with a pH of ~ 2.2 (= ) for 6% acidity. Some of that Acetic Acid (moles of Ac - mole of protons = moles of acetic acid left)
dissociates so, there is some (equal to ~moles of protons) Acetate in there as well.

Link 1 seems pretty interesting regarding the making of vinegar
1 - http://www.madehow.com/Volume-7/Vinegar.html
2 - http://antoine.frostburg.edu/chem/senese/101/organic/faq/ethanoic-acid.shtml


Electron configuration and the 2p orbital I looked through the questions already asked about this topic and they were very specific and over my head. I'm having trouble with the basics.

When drawing orbital diagrams I know electrons are placed in the 2p boxes singly first then paired because of they have parallel rotations and the Pauli exclusion principle states that an orbital can only house 2 electrons and those electrons must be of opposite spin. My question is why do electrons behave this way in the 2p orbital but not the 1s or 2s? And why does the p subshell contain three 2p orbitals?

As far as electron configurations and orbital diagrams are concerned I've gotten to Ne and the 2p configuration confused me so I though I'd ask before moving on.

first of all electrons do not rotate but they have spins
next magnetic quantum number decides how many electrons an orbital can have
s can have 2(m=1){consider 1 box}
p can have 6(m=3){consider 3 boxes}
d can have 10(m=5){consider 5 boxes}
f can have 14(m=7){consider 7 boxes}
where each box can have at max 2 electrons and each box represents different orientation of the orbital
and all the orbitals(except s) have degenerate orbitals, i.e they have the same energy so all have to filled with atleast one electron before pairing up. this rule is known as Hund's Rule of Maximum Multiplicity
there are are some more reasons fro it. they are
to maximize exchange energy
to attain symmetry


Which is better: iodine number or bromine number? The degree of unsaturation of fats can be determined with the help of two type of scale namely iodine number and bromine number.

Generally, iodine number is more preferred over bromine number (Is this statement true? Is it because the Wikipedia article of iodine number is more elaborated than that of bromine number?)

Iodine number is typically found out by using iodine monochloride, $\ce{ICl}$ where the iodine atom gets bonded with the carbon atom having double bond. The more iodine  used up to break the $\ce{C=C}$ bonds,  higher is the iodine number.

$$\ce{-CH=CH - + ICl -> -(I)CH-(Cl)CH - }$$

But in the 4th paragraph of the 1st page of this link, it is written that:

  Iodination
  is an endothermic process, resulting in vicinal diiodides that tend
  to revert to alkenes. Consequently, the most common applications of
  alkene halogenation are chlorination and bromination.

So, in this context, bromination is better and thus bromine number is better than iodine number. The statement contradicts the previous statement i.e. using iodide number.

H.P.Kaufmann method of finding iodine number uses bromine. Is this a better method than using $\ce{ICl}$ considering the disadvantage of iodination?

Bromine number (and Bromine Index) are based on theoretically good chemistry; however, that is where their virtues end. If these tests are run according to the current ASTM procedures, you are virtually guaranteed to get an incomplete reaction and low (perhaps as much as 50%) results. When properly applied, off the shelf equipment can result in exceptionally good accuracy and precision, but it does require a deviation from ASTM-accepted technology.

Which is better for measuring unsaturation, Iodine number or Bromine number?  The answer depends on how you define "better". However I can tell you which is commercially useful. If we're dealing with fatty acids or vegetable oils, the industry standard test is Iodine number. For petroleum products, the standard test is Bromine number (or bromine index, slightly different). For the two tests, Reference Bailey's Industrial Oil and Fat Products, sixth ed. vol. 1 pg. 172, and ASTM D-1159.


How many delocalised electrons in gold? How can we determine how many delocalised electrons every atom of gold contributes to the 'sea of delocalised electrons'? More generally, how can we determine the number of electrons any metal contributes? Must we have thermodynamic data to determine this?

Briefly: valence orbitals on adjacent metal atoms interfere to form "bands" of closely spaced molecular orbitals that spread throughout the metal.

If you have N metal atoms with occupied s orbitals, they interfere to form an "s band" with N molecular orbitals. The lowest energy orbital is fully bonding; the orbital on the top of the band is fully antibonding. The spread of the band will be finite, and since N is a huge number, the molecular orbital energies will be very closely spaced, practically continuous.

If you have two valence s electrons in your metal, the s band will be full (each orbital can hold up to 2 electrons, you have 2N electrons, and N orbitals in the band). If you have only one valence s electron, the s band will be half full (the first N/2 orbitals will be filled, and the remaining orbitals will be empty).

Above the s band you'll have a similar p band. In gold, the p band is unoccupied, while the s band is half full (gold has 6s$^1$). However, IF the bottom of the p band overlaps with the top of the s band, the s electrons can easily move into the bonding MO's at the bottom of the p band. They can also move into any vacant orbitals in the s band that are just above the highest occupied orbital.  In either case, the s electron will be delocalized.

To figure out whether the valence electrons will delocalize, then, you'll have to decide what bands are present, where the highest occupied molecular orbital is, and what the width and the spacing of the bands are (does the unoccupied band overlap with the occupied band, or is there a gap between the two?)

Valance electron of s and p orbital take part in delocalisation over the positively charged gold atoms