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Hi all,

I am curious how to determine the calculation by external cal and standard addition method. I get quite different result.

For external cal I prep on 1ml dmf and get calibration curve as 20,50, 100 and 200ug/ml hexane STD. Then I add 2g sample (or lesser weight) plus 1ml dmf. Mean the concentration obtain need to multiply 1ml and decide sample weight.

For STD addition method, I prep series of 20,50,100 and 200ug/ml hexane STD. Then each vial I put 2g sample then add 1ml of series hexane concentration. For STD addition, I will plot y=my+c. Then assuming y equal zero to calculate the concentration of the sample.

Question is this two method can not get tally result.
Then I now abit confuse on STD addition do I still need devide my sample weight? Theoritically no need because each vials have 2g sample, but I not able to cancel of the unit to obtain ug/g. Another way is I convert become ug (mean let say stock is 5000ug/ml, I take 0.1ml and top up to 5ml, equal 100ug/ml. From this 100ug/ml, I take 1ml to dissolve either STD addition or external cal, mean my cal point will be 100ug. At this stage if I decide my sample weight will be ug/g. However this three method non of it tally.

Can anyone help to advise on this?

Thanks

For the external standard calibration you would use 20ug/ml for the target analyte and 2g/ml for the sample, when you solve the equation you end up with 20ug/2g and the ml cancel so you don't have to worry about the volume of solvent at all, just the mass of analyte added to the vial and the weight of the sample added to the vial.

Standard addition should work the same, you have unknown amount / 2g then you have unknown amount ug + known added amount ug/2g so final answer will be ug/2g.

It is much easier to ignore the volume of DMF and only look at mass of analyte added to vial. That way the response on the instrument is for x ug of analyte, then once you have the answer for x, you then divide by whatever weight of sample you used. So if instead of 2g you weigh out 3g the response equal to 20ug on the instrument would then result in 20ug/3g. So just calibrate the instrument as ug of analyte in vial, then just divide the result by whatever weight of sample you use.
The past is there to guide us into the future, not to dwell in.

Hi James,

Just to clarify my understanding is correct.
I prep cal as below for STD addition
Spike A: 1ml dmf + 2g sample
Spike B: withdraw 0.5ml of 10000ug/ml hexane top up to 5ml. Then take 1ml to spike on 2g sample ( hence hexane contain inside is absolute weight of hexane is 1000ug = 1000ug+ unknown conc of sample/2g sample

So when plot y = mx+c. Y equal zero. What ever x I obtain is the result? Or the result I still need to divide 2g sample weight. If let said spike B I use 1000ug/ml as cal curve, so I need divide by 2g ( due to 1000ug/ml contain 2g/ml). However if I use absolute weight mean it only contain 1000ug also need to divide by 2 also, right? But isn't it STD addition ur sample in each vial prepare is 2g, hence can cancel of for the sample weight, unless different sample weight apply?

Thanks

seet88 wrote:
Hi James,

Just to clarify my understanding is correct.
I prep cal as below for STD addition
Spike A: 1ml dmf + 2g sample
Spike B: withdraw 0.5ml of 10000ug/ml hexane top up to 5ml. Then take 1ml to spike on 2g sample ( hence hexane contain inside is absolute weight of hexane is 1000ug = 1000ug+ unknown conc of sample/2g sample

So when plot y = mx+c. Y equal zero. What ever x I obtain is the result? Or the result I still need to divide 2g sample weight. If let said spike B I use 1000ug/ml as cal curve, so I need divide by 2g ( due to 1000ug/ml contain 2g/ml). However if I use absolute weight mean it only contain 1000ug also need to divide by 2 also, right? But isn't it STD addition ur sample in each vial prepare is 2g, hence can cancel of for the sample weight, unless different sample weight apply?

Thanks

If you calibrate the instrument as ug on the column, then do all calculations without the sample weight to achieve and answer as ug. Then divide by the sample weight and the final units will be ug/g.

If you incorporate the sample weight through the whole calculation then you have more places to make sure you have it correct. If you calculate the calibration at 2g but you weigh 2.05g, then you have to figure in that ratio through the whole calculation, but if you calculate the calibration as simply ug (or amount on column in theory) then any difference in sample weight is simply corrected for in the final step of ug(instrument reading)/g (sample weight).

If you are plotting the results as ug, then your result will be in ug(sample) + 1000 ug(added to sample)

I have never worked much with standard addition, normally I just make a 5 point calibration curve. If the sample result exceeds the upper point of the calibration curve then use less sample to bring it within the curve.
The past is there to guide us into the future, not to dwell in.

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