Is it possible to get yield of a chemical product more than 100%...? If so, how one can defend this situation? I am working on Condensation reactions using organic solvent free synthesis. For some reactants, I am getting more than 100% yield. Kindly let me know if any researcher or professor faced this type of observations during their experience.

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Even with the law of conservation of mass, yield can still be greater than 100%. Consider this reaction:
If you start with 70 g of A and 30 g of B. It is not impossible you get a mass of C > 30 only that the law of conservation of mass must hold, i.e., Mass of C + Mass of D = Mass of A + Mass of B = 70 + 30 = 100.

It is impossible to get more than 100% product. It doesn't mean that there is no impurities, byproducts, salts that aren't seen in NMR. Some compounds absorb water really easily so if you keep them out of exicator for any length of tim, they can soak up a lot. And there's classic "not the product you're looking for" scenario.

Impossible to get a yield more than 100% . If this happen this mean you may have some impurities in your product and you should purify it

If you use an excess amount of the reactant, may be your product still contains the unreacted reactant. Some products could adsorb water or other used solvent during separation and purification process.
I would say that it is impossible. If you calculated and got the yield more than 100%, please check the NMR results, or simply from TLC (HPLC, GC..).
2). Byproducts (self-disintegrated products, oxidation products while conducted the reaction in air...)
Basically, I think maybe some actions to rule out any possibility need to be taken in this case, my friend. :)
Theoretically impossible, while experimental value might be higher than 100% for several reasons. Typical experimental errors might be up to 10%, and you may have the experimentally determined yield up to 110%. Other reasons are already listed in the previous answers. It's also very important how you define "the yield" and how do you measure it.
The maximum or theoretical yield is based on the limiting reactant and cannot exceed 100%. In many cases, when a reaction is worked up, inefficiencies in removing unreacted starting materials, unanticipated side products, inorganics, solvents and moisture might lead to the appearance of "yields" in excess of 100%. Gas chromatography can disclose organic solvents, unreacted starting materials and side products. NMR spectral analyses can reveal the presence of water and solvents.
In fact, it depends on how we define a physical quantity. The value of biomass to substate yield is less than one but the substrate to biomass yield is greater than one. Both quantities are used and the second one is the reciprocal of the first. Regards,
In the meat industry, when producing meat products, the yield counts in relation to the mass of the used meat. The yield can be 100% or even 1000%. There are such products and there are many such situations in the food industry.Returning to the question, if the yield refers to all consumed raw materials, it can not be more than 100%. Regards,
Theoretical yield can never be more than 100%, unless the molecular weight of the formed product is higher than that of the desired product. Further, all information provided above is applicable.
In chemical syntheses, including polymerization reactions, an yield of more than 100% can primarily mean the involvement in the reaction of uncontrolled components: solvent, oxygen, carbon dioxide, water... Regards, VZ
No it's not possible to get more than 100% yield. The yield must be calculated on the basis of reacting components and the molecular formula/ moles of product formed.
Even in the case of the use of reactant excces the yield should be calculated on the relatively to the limiting reactant after elimination of the excess
not feasible , you check some technique , first you will check the TLC, your compound pure are not you clearly you will see. your compound liquid means you can use extra time rota evaparator and use the high vacum pump ,because solvent remove this kinds of method good.
Above all, the law of mass conservation and definitions, but as I wrote above are industry exceptions. This discussion does not lead to anything. Regards,
It is impossible theoretically. You can check again the purity of the product and how you calculate the yield.
While performing recovery test with an extraction method of protein precipitation, the recovery results for all QC levels are always above 100% , the area's of Bio-analytical samples are higher than the aqueous samples . what can be the explanation ?
I prepared a solution containing three estrogens (beta-estradiol, 17alpha-ethynylestradiol and estrone) each with a concentration of 2ug/L. After passing this sample through the cartridge, I evaporated the solvent I used to elute my compounds to get 1000x fold concentration. I then ran the sample through HPLC. Every time I did this, I keep getting a percent recovery much greater than 100%. I wanted to know what were the possibilities of getting this? Why would I get this contamination?
Please tell me how to calculate limit of detection, limit of quantification and signal to noise ratio. Please also explain what is the relation of these parameters with each other. Usually in papers it is mentioned that LOD and LOQ were measured based on signal to noise ratio at about 3 and 10, respectively?
Spiked sample: The same solid sample of 0.5 g was added to 10 mL of HNO3 and 2mL of 1000 ppm Pb standard.

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A calibration curve of Pb was calculated to have the equaiton of y=0.01 x +0.003. y=absorbance and x=concentration of Pb
The concentrations in raw and spiked sample were found using the formula as 5.6 ppm and 6.1 ppm respectively. (This is before considering the DF)
I know it should be (spike result - raw result) / spike added x 100% but I am not sure what their units should be. Thanks!
Could anyone please help me in calculating the Km and Vmax values of an enzyme (I am working on dihydrofolate reductase DHFR) when I have substrate/product inhibition?
I see the inhibition on (specific activity nmol/min/mg to substrate concentration µM) curve, with substrate concentrations near the theoretical Vmax. (I mean the calculated Vmax by ignoring the inhibition).