Internal standards
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- Опубліковано 3 жов 2024
- Corrections: At 2 minutes 35 seconds, I accidentally moved the wrong math over. It should be F = (20000/20)/(8000/10) = 1000/800 = 1.25. Thank you viewers for the correction!
Also, the last little bit was cut off. ... you should undo the dilution by (14.61mM)(10.0mL)=(?)(9.0mL).... ?=16.23mM.
I'm afraid this was one of my earliest videos...so a few mistakes. I do apologize.
Well done. Been an analytical chemist for 40 years. Sent this to a colleague to clarify a point. Thanks for posting
Wow, you are a lifesaver. I'm trying to teach myself how to do LC/MS/MS for the first time (first semester PhD with little guidance in my lab) and I was not understanding how to prepare my internal standards. Thanks!!
thank you so much for uploading this. i came over a number of videos and articles but your methodology is simple and easily understandable.
Thank you a lot, so clear, in just 5 minutes you explain everything I needed
Thanks again!
Thanks Robyn for the video. It's really useful when it comes to practical calculations. My understanding from the video is that we use molarity in the equation. But my concern is if we take a 1 ml or 10 ml or any portion from 100mM standard solution, we have the same molarity if it's not diluted. So when you multiplied 10 ml (volume) * 100mM (mMoles/volume) then you get "millimoles" as volumes cancel out each other. Regardless of the volume being sampled, the concentration would be the same. Thanks
I don't get this part at all! What isn't 100mM a concentration? Doesn't it mean 100mM in 1 ml? Can you help me?
I think you may have gotten the proportionality constant wrong, I think its supposed to be 1.25
Hi Sam, you are right. There are a couple of corrections in the video description that address that issue. :-)
hi, i am a bit slower in maths. should it be F is 1.25 instead of 0.8? please correct me if i am wrong
liza wewe Hey - you're right! I accidentally moved the wrong math over. It should be F = (20000/20)/(8000/10) = 1000/800 = 1.25. Thank you for the correction!
@@robyngoacher8521 do You have a same video on the other methods?
@@robyngoacher8521 thanks
I actually got 1.25 when I was calculating myself. Wow. I feel like a genius, bearing in mind I am crap at maths 😯
Is this is how we calculate GC yield?
THANK YOU
Thanks Robyn. I have been trying to understand internal standard since I was an undergrad in 1987, LOL. Now I suddenly get it.
holy shit.
Great video!
Thank you so much
You saved me. Thank you so much.
thank you. it is very clear
why are my thinking i dont need internal standard and 'f' to calculate for the new concentration of A. cos if 20mM of A corresponds to an area of 20000, then an area of of 14581 will correspond to: (14581*20)/20000 = 14.58 mM of A.
The math you describe corresponds to the method of single-point calibration with external standards. Internal standards are used to correct for problems with reproducibility. If you want to learn more about the differences between the calibration methods, please see this video: ua-cam.com/video/oSGM91uMSjg/v-deo.html
great effort...
how to get 0.1% impurity from known sample con mg/ml
Perfect explanation. Thank you.
just what i needed, thanks
Hold on my cat is annoying me classic. Thanks for a great video
I think F value is wrong 😅 it's 1.25
Hi, can I replace the concentration with the weight fraction?
thank you
you could have demonstrated the answer at the end of the video.... otherwise really great.
Thanks, but there are misconceptions about upper calculations: is A a internal standard and s external one. what is the unit of concentration you have been using
Hello Mysara, In this video, A stands for "analyte" (the compound you are trying to quantify) and S stands for "standard", which in this case is the internal standard. An internal standard is a compound this is similar to but different from the analyte and is added into all samples at a known concentration. The units in this video are milli-Molar, abbreviated mM, which is one one-thousandth of molarity. If you would like to learn more about how internal standards compare to external standards, you may be interested in this video: ua-cam.com/video/oSGM91uMSjg/v-deo.html
@@robyngoacher8521 thanks so much I learned a lot from your video. Thanks
Thank you!!
Annoyed by cat. This is very real
Thanks
I don't get the dilution part at all! isn't 100mM a concentration? Doesn't it mean 100mM in 1 ml? Where did the concentration of 10millimole come from? Why did we decreases 100 to 10? I don't get it, Can you help me?
If you mean at about 3 minutes in, we take 1 mL of 100 mM internal standard stock and add it to 9 mL of sample. This is how it would be done in lab and the liquids dilute each other to make 10 mL. So the 1mL×100mM = 10mL ×Y and Y is the diluted standard concentration of 10mM.
@@robyngoacher8521 they diluted each other ok, do you mean the 100mM per ml is thus diluted to 10 per ml? If so, then isn't the sample concentration lesser too?
Also does this happen to any experience? Not just Istd? Like when you add components to eachother the dilute each other? Bc we never did this befor in all calculations and labs I took befor and iam so confused does it mean it was wrong?
Thank you very much by the way, you helped me a lot
@@minabish1783 yes the sample also has been diluted from 9 to 10 mL.
@@robyngoacher8521 thank you ❤️
hahaha the cat, classic
There is a mistake in the end of the video, i think. You use 7982 but the correct value is 79 825
Sjoh! This video has so many mathematical errors that just confused me!
I dont like ur fastes expreswsion on determinding in chemistry be slowly on explainig
useless instructional, in my opinion.