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19

The analytical technique is called inductively coupled plasma optical emission spectroscopy. It was invented by a chemist V. A. Fassel, although it is pure physics. The specimen is digested in an acid, and a fine spray is created. That "spray" is heated to a temperature, which is higher than the surface of the Sun. Most of the elements of the periodic table ...


12

Inductively Coupled Plasma (ICP) based techniques are very common for elemental analysis. The goal of such analyses is to quantify the elements present in a sample. The neat thing about this, in my opinion, is that the integrity of the molecular compounds does not need to be maintained throughout the analysis, since we only care about the elements that ...


11

There are many ways to detect mercury, and you can find a more detailed analysis of these methods, along with examples of these detectors and their limitations, here. Gold film sensors Gold Film Sensors were the first reliable forms of mercury detectors due to gold’s affinity for elemental mercury [...] When a mercury rich air sample passes over a thin ...


11

Most commonly they work by pulling air through a UV absorption cell and simply measuring the absorption at 254 nm. Often there is a pre-concentration step involved. This usually involves a gold-plated region prior to the UV cell. At room temperature mercury adsorbs to the gold, then after some period of time the gold-plated region is flash heated to ...


11

There is no chemical symbol for paper since it is not an element but rather a mixture of several different compounds. Remember that paper is made out of trees. Thus, paper is made mostly out of organic compounds: that is carbon, hydrogen and oxygen (C, H and O). Paper also contains non organic materials to improve its properties. These may be chalk (CaCO3) ...


8

TL;DR f-f transitions in lanthanide ions give rise to the colours of the ions. The electronic states can be derived from quantum mechanics, and their energies vary from ion to ion, which give rise to different colours.. The colours are pale because these transitions are Laporte-forbidden and therefore weak. These absorption bands do indeed arise from ...


7

$$\ce{C_{$a$}H_{$b$} + $\left(a+\frac b4\right)$O2 -> $a$CO2 + $\frac{b}{2}$ H2O}$$ Suppose you had $n$ moles of hydrocarbon, then we have $a\cdot n$ moles of $\ce{CO2}$ and $\frac{b}{2}\cdot n$ moles of $\ce{H2O}$ dividing their moles we'll get $2\frac{a}{b}$: $$2\frac ab=\frac{330/44}{135/18}=\frac{7.5}{7.5}=1\implies \frac ab=\frac12$$ So the ...


7

Neutron activation analysis is a technique that can be used for trace mineral quantification in a sample. You start by irradiating your sample in a small experimental nuclear reactor. These were more common in the 1960's and '70's than they are today. Slow neutrons are absorbed by many different isotopes, converting them to radioactive species. The sample ...


6

Crystal solids, however small, have defects in them. These defects are basically irregularities in the arrangement of constituent particles. They are of two types: Point defects Line defects. Point defects themselves are divided into other kinds of defects. Here, I'll explain only the relevant one. One type of point defect is a vacancy defect in which ...


6

Disclaimer: Here I take the molar masses of carbon, hydrogen and oxygen to be $\pu{12g, 1g, 16g}$ respectively, which is what would be expeccted of you at school. For the sake of simplicity, let's assume you have $\pu{100g}$ of the said compound. Which would mean it contains $\pu{21.62g}$ of oxygen. Employing a bit of basic math in the analysis of that ...


5

If you have $0.2\ \mathrm{mol}$ of a compound with $M_\mathrm{r}=82$, then you have $16.4\ \mathrm{g}$ of material. If there is only carbon, hydrogen and nitrogen present, and the carbon and hydrogen total up to $10.8\ \mathrm{g}$, then the amount of nitrogen must be $16.4\ \mathrm{g}-10.8\ \mathrm{g}=5.6\ \mathrm{g}$.


4

You seem to be using a very simplified version of the index of hydrogen deficiency. A more complete expression is: $$IHD = \frac{2C + 2 - H - X + N}{2}$$ Where $C$, $H$, $X$, and $N$ represent the number of carbon, hydrogen, halogen, and nitrogen atoms, respectively. The logic of the IHD is that any simple acyclic alkane has a formula of $\ce{C_{n}H_{2n + ...


4

First step is finding the molar masses of each compound. Since cellulose, hemicellulose and lignin are collections of similar molecules, I picked the following as a crude estimate: cellulose: $\ce{C6H10O5}$ with $M_c = 162~\mathrm{g/mol}$ hemicellulose (L-Arabinose): $\ce{C5H10O5}$ with $M_h = 150~\mathrm{g/mol}$ lignin: $\ce{C31H34O11}$ with $M_l = 582~\...


4

Lehrbuch der analytischen und präparativen anorganischen Chemie by Jander et al. [1] suggests to precipitate borate with $\ce{Ba^2+}$ in neutral solution beforehand as barium(II) thiocyanate is one of the few barium(II) salts actually soluble in water, or use an excess of $\ce{Fe^3+}$ to make sure all the interfering anions are precipitated. From [1, p. 358]...


4

When you ask how to remove borax, I assume it is Sodium tetraborate decahydrate (Wikipedia). Thus, I agree with @andselisk of using $\ce{Ba^2+}$ solution to precipitate it. Even though it is seemingly an excellent answer, the reference given and the text body are in German, and I didn't understand it much. :-) Thus, I want to give some clues to make it sense....


3

It will depend on the size of the particles which are in suspension. If the particles are very large compared with the capillary feeding the solution into the spray chamber then the capillary will block up and you will detect nothing and end up very unhappy with a broken machine. If the particles are much smaller than the capillary then they will be able to ...


3

You can check if the calculated molecular weight corresponds to lactic acid if not try next multiple like 2, then 3 etc. You get $\pu{30g/mol}$ for $\ce{C1H2O1}$ and $(\pu{90g/mol})/(\pu{30g/mol})=3$ So the multiple to be applied is 3: $\ce{C3H6O3}$


3

I suppose that you are referring to a Wiley–McLaren time‐of‐flight mass spectrometer [Rev. Sci. Instr. 65, 3344 (1994)] which is basically a tube containing an electrode stack for accelerating the ions toward a detector. As mentioned in the comments, one can discriminate different masses if one uses a pulsed experiment, for instance using pulsed lasers or ...


3

A sad Ph.D. student had been working on the synthesis of a natural product, with empirical formula $\ce{CH2N}$, for several years. He finally succeeded in synthesising it - or so he thought - because his NMR data seemed to fit the target structure. Excited about his breakthrough (in fact, he was so excited that he didn't even bother taking a mass spectrum!), ...


3

For the determination of calcium (at least) two methods are conceivable: gravimetric analysis Sulfate is a rather bad choice here. As compared to the carbonate, the sulfate has a much higher solubility. A better choice would be the oxalate with a solubility around $6\,\mathrm{mg\,l^{-1}}$. photometry Calcium forms a purple complex with the Arsenazo III dye ...


3

To best understand, let us consider the reactions occurring. First, we have $\ce{Cl-}$ ions in solution. Acidifying the solution serves to remove other anions that might give a false positive for the chloride test. Upon addition of $\ce{AgNO3}$, the following reaction occurs: $$\ce{Ag+(aq) +Cl-(aq) -> AgCl (s)}$$ When enough ammonia is added to ...


3

I did this a little differently. If there were $\pu{4.5897 \times 10^{-3}}$ moles of $\ce{CO2}$ produced, that must mean that there were $1/7$ as many moles of $\ce{C7H16O4}$ present in the original sample, or $$\frac{4.5897\times 10^{-3}}{7}\times 164.1995=\pu{0.10766 g}$$ in the original sample. The rest of the original sample ($\pu{0.01469 g}$) must ...


2

It depends what you mean by "exactly". A single gram of salt contains about 1022 atoms of sodium and chlorine. The important question is how much of an imbalance would be physically noticeable. If you can't detect the difference it isn't significant. A single atom discrepancy would be impossible to detect with the best equipment available to science. So it ...


2

Your budget should allow you to measure $\ce{Mg^{2+}, Ca^{2+}, Li+ and Cl-}$ in-house using ion specific electrode analysis. There will certainly be some training and maintenance involved, but anyone with a bit of a chemistry background will be able to learn this technique. Adding to the simplicity, and thus the ease of training, is the fact that your ...


2

What do you want? If you're just looking for a source of potassium ions you could just buy $\ce{KCl}$ or some other compound, they're relatively cheap if not pro analysis. If you actually want to test for $\ce{K+}$: What else do you know about the product? Tests are completely different for different amounts of $\ce{K+}$. Does the vendor specifies the ...


2

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


2

This is a strangely written text. In a practical setting, the way to get oxygen concentration is by sending an organic sample to a lab for an ultimate analysis. The ultimate analysis chemistry is explained elsewhere. The highlighted text seems to describe anoxic pyrolysis of organic matter well above 450°C (for coals this temperature may be higher). ...


2

In all the (physical) experimental literature I've read, at least for biomolecules (usually enzymes), the technique of choice is Mossbauer spectroscopy. Here is a good starting point.


2

If the oxides are separate grains or compounds (a mixture, not a solid solution), you can find this with SEM microprobe (a geology technique). For tricky new mixtures, may be better than powder x-ray diffraction (find small amounts of new compound, than resolve stoichiometry and do XRD on phase pure samples. It will not give you the oxidation level though. ...


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