After calculating the concentration in molarity (as he demonstrated in the video) I added the "unknown" (now known concentration and molarity) to the two columns on the left, I created a new chart. I had to rewrite the axes and title but no big deal. I'm sure there is an easier way, but it worked for me.
Hey man, nice video! But I wonder how much important is the R-Squared value (Coefficient of Determination probably). Should I divide the final unkown concentration with the R-Squared or multiply? I cant find it anywhere on the internet and I have exam in few days. Thank you in advance!
You should be able to, yes. Ideally c would be zero or at least a very small number because the absorbance should be minimal when the concentration is zero.
This is great! If my calibration curve is emission% (e.g. sodium) with concentration (ppm) and i want to find the sodium mg/mL, do i follow the same steps?
I have a couple thoughts, but first, I would recommend asking your teacher what they recommend. (1) You can take the average of the two absorbance readings. I assume it is the same solution, so you may have run it twice to get the average. (2) You can plot two lines. This could potentially give you another slope, and another answer in the end. (1) is my advice, but I would recommend asking your teacher.
Can I use same method for GC-MS sample concentration calculation by placing area instead of absorbance i.e. by plotting concentration vs area graph. Then is it ok to calculate sample concentration using y value of trend line as discussed here.
@@mikedavis6274 I did the ratio of know concentration over zero to get absorbance to create my calibration curve but it's not linear at all. I don't know what exactly I did wrong😨
@@mikedavis6274 for each concentration it gave me a number for each wavelength, like at wavelength 380.5 it had absorbance of 1.206, this continues until wavelength 899
Hi friends, I am using HG-C 1100 sensor and this sensor measures the range from -35mm to 35mm. I have doubt in using the formula float val = (float)(map(samplse,0,1023,-3500,3500)/100.0);. I have to take the readings from the sensor and should display it in the serial monitor. The values also should be accurate as in the sensor. Can anyone explain me how to take the formula and calibrate the same? Thanks in advance
@@mikedavis6274 thanks for the answer. If I measured the blank, it shows little absorbance. In that case, can we subtract the blank from the analyte absorbance? If blank has absorbance, then how to write 0 absorbance at 0 concentration?
can you explain why you multiplied the unknown by 10 at the end? is it because it was originally in a 100ml flask and then you pipetted 10 ml from that into a new 100mL flask?
I'm doing a homework assignment where we are given a graph and the concentrations. I plotted concentration vs absorbance and did all the calculations and got 1.4884E-05? Do you need to multiple it by 10 or was that just for your example?
My students had to dilute their sample by a factor of ten, so in the end they had to multiply by ten. If you didn't dilute, you don't need to multiply.
That number (140.33) is the slope of the best fit curve through those points. Excel calculated it. You can also get it by picking an empty cell and typing in =slope(. It will prompt you to highlight your known y's, and then your known x's. So it would look like this for the example we just did. =slope(B2:B5,A2:A5). That will return the slope for you.
@@mikedavis6274 So, first when I prepared the unknown I diluted it by a 1:10 factor. Then, I diluted it again by a 10:25 factor. So, I have to multiply by 2.5?
In 3 and a half minutes you explained what 15 pages of instructions couldn't. Thank you!
Taking intermediate environmental chem and still didn't know what I was doing with this one. Absolute chad, you're a hero Mike.
You don’t even know how much of a saint you are to me right now!! Thankyou!!
lol bro same here, fucking assignment
7 years later this video is still highly appreciated.
What a life saviour, even four years later during a pandemic hahahaha TYSM!!!!
THANK YOU!!!! You simplified this so much to understand!!! Life saver!!!!
2 weeks before the big lab test and you saved me! Thank you man
Thank you so much, explained in 3 minutes what I couldn't get in hours.
Absolute life savior man! Thank you loads!!
You are an angel for ones that cannot make an equation curve. Thank you
Heros get remembered; but, Legends never die. Fantastic content sir 👏👏👏
Wow. Thats what i was looking for. So simply explained. Thank you.
Dude...you are my friggin' hero.
This video helped so much with my lab. Thank you!!!
sir video helps a lot.
our teacher explains it in a very bad manner.
person like is boon to students all over the world.
Super frumos si practic ai explicat. Multumesc frumos!
Straight to the point and well explained. Thanks!
SAVED MY LIFE
This video rocks, thank you
Thank you! This helped a lot!
Thank you for that very straight forward explanation. How do you plot that unknown concentration on the calibration graph?
After calculating the concentration in molarity (as he demonstrated in the video) I added the "unknown" (now known concentration and molarity) to the two columns on the left, I created a new chart. I had to rewrite the axes and title but no big deal. I'm sure there is an easier way, but it worked for me.
thanks so much! it helped me with my report essay
you are a king
Hey man, nice video! But I wonder how much important is the R-Squared value (Coefficient of Determination probably). Should I divide the final unkown concentration with the R-Squared or multiply? I cant find it anywhere on the internet and I have exam in few days. Thank you in advance!
Thank you so much
I appreciate this video thank you
Absolute lifesaver mate!
Simple explanation, so useful. Thanks so much!
In case of y=mx-c..we can use this type of standard curve for calculation or not? ...please reply me sir ..its urgent
You should be able to, yes. Ideally c would be zero or at least a very small number because the absorbance should be minimal when the concentration is zero.
thank you for your awesome tutorial
THANK YOU SO MUCH
This is great! If my calibration curve is emission% (e.g. sodium) with concentration (ppm) and i want to find the sodium mg/mL, do i follow the same steps?
What if we measured absorbance for each concentration twice, how would we be able to plot that?
I have a couple thoughts, but first, I would recommend asking your teacher what they recommend. (1) You can take the average of the two absorbance readings. I assume it is the same solution, so you may have run it twice to get the average. (2) You can plot two lines. This could potentially give you another slope, and another answer in the end. (1) is my advice, but I would recommend asking your teacher.
Thank you very much. Nice explanation
Really good video!
Thank you ! It's very helpful.
Thank you! Your video helped me, to do my job. :)
Can I use same method for GC-MS sample concentration calculation by placing area instead of absorbance i.e. by plotting concentration vs area graph. Then is it ok to calculate sample concentration using y value of trend line as discussed here.
Thank you! very helpful
Thank you! This is very helpful
factor calibration .how many valor it can have ?
Please I'm unable to label my graph on my laptop. Please help me out
How do you get the absorbance, I have the raw data from my concentrations from the UV/Vis spectrometer.
Usually the numbers you get from the spectrometer are unitless absorbance or transmittance numbers
@@mikedavis6274 I did the ratio of know concentration over zero to get absorbance to create my calibration curve but it's not linear at all. I don't know what exactly I did wrong😨
So you had a known concentration, diluted it and ran it through the instrument? What numbers did the instrument give you?
@@mikedavis6274 for each concentration it gave me a number for each wavelength, like at wavelength 380.5 it had absorbance of 1.206, this continues until wavelength 899
I honestly think it's the machine I used. Mabey because it's cheaper.
nice explanation
I don't understand why the absorbance 0.125?
That was a number I made up for my unknown's absorbance. You should use whatever your unknown's absorbance was.
Thank you this is so helpful
how did you get 0.125 as the absorbance?
That is a reading that came off the instrument.
thank you man it was realy helpfull
Hi friends, I am using HG-C 1100 sensor and this sensor measures the range from -35mm to 35mm. I have doubt in using the formula float val = (float)(map(samplse,0,1023,-3500,3500)/100.0);. I have to take the readings from the sensor and should display it in the serial monitor. The values also should be accurate as in the sensor. Can anyone explain me how to take the formula and calibrate the same?
Thanks in advance
do not we need to measure blank sample?
No. Your blank is essential zero. There is zero concentration so there is zero absorbance.
@@mikedavis6274 thanks for the answer. If I measured the blank, it shows little absorbance. In that case, can we subtract the blank from the analyte absorbance? If blank has absorbance, then how to write 0 absorbance at 0 concentration?
Thank you so much.
can you explain why you multiplied the unknown by 10 at the end? is it because it was originally in a 100ml flask and then you pipetted 10 ml from that into a new 100mL flask?
That is right. In this lab, my students had to dilute something by a factor of 10, so we end up multiplying by 10 at the end.
from where u get y value 0.125?
That was my absorbance of my unknown solution. I got it in lab. It is a solution I could get the absorbance of, but didn't know the concentration of.
Thank you for this!!
THANK YOU OMG
How did you get the 0.125?
It was an absorbance for an unknown solution that was similar to what my students were using in lab.
I got it! Thanks!
If I diluted 10ml of my unknown to 100ml would I have to multiply it by 10?
yes
appreciate you, king
Thank you so much!!!
Can u plz tell what should I do to find the concentration at the end if I have used 25 ml sample for the test? the last portion I felt confused?
Do you mean you diluted a 25mL sample to 100mL? You would multiply the answer by 100 and divide by 25 or basically multiply by 4.
No for ammonia test I applied phenol method for the test. No dilution is operated over here? how can I extract its concentration?
Can you use the email address in the video to send me your procedure? Its hard to know without knowing a little more about what you did.
its ok I got the soln thanx .. the video helped a lot
I'm doing a homework assignment where we are given a graph and the concentrations. I plotted concentration vs absorbance and did all the calculations and got 1.4884E-05? Do you need to multiple it by 10 or was that just for your example?
My students had to dilute their sample by a factor of ten, so in the end they had to multiply by ten. If you didn't dilute, you don't need to multiply.
I need to do this too lol
Thank you!!!
Where did you get 140.33 ?
That number (140.33) is the slope of the best fit curve through those points. Excel calculated it. You can also get it by picking an empty cell and typing in =slope(. It will prompt you to highlight your known y's, and then your known x's. So it would look like this for the example we just did. =slope(B2:B5,A2:A5). That will return the slope for you.
Mike Davis 😊
is this the same as a absorbance curve?
Branden Williamson
Yes, it is. You use known concentrations to find an unknown.
Mike Davis thank you!
It is still giving me wierd graph even after following each step.
It could be the data. Send me your file or data.
Thanks a lot....
Thanks!
thank u ❤️❤️❤️
Thank you
Appreciated!
Clutch bro
MIke you are a life saver I want to kiss you. Thank you for helping me solve my reports (University of Nicosia)
I love you
when your prof on analytical chemistry teach also statistics....
My line goes the opposite way
That's odd. What is your data?
u should pass by zero to do this !!!!!!!
King
Thank you so much
If I diluted 10ml of my unknown to 25ml would I have to multiply it by 10?
No. If you diluted your sample 10:25, you would multiply it by 25/10 or 2.5. You diluted it because it was too concentrated.
@@mikedavis6274 So, first when I prepared the unknown I diluted it by a 1:10 factor. Then, I diluted it again by a 10:25 factor. So, I have to multiply by 2.5?
Thanks a lot