HPLC Agilent - mg/L to % (w/w)

Discussions about HPLC, CE, TLC, SFC, and other "liquid phase" separation techniques.

12 posts Page 1 of 1
Hello, could someone please help with converting concentration of my sample from mg/L to % (w/w) with the details below;

I weighed about 1.05 g of my sample to 100 ml and then did further 10x dilution afterwards. the concentration output from the Agilent HPLC (using OpenLab software) was 89.315 mg/L. I did not take density reading of my sample.

I would like to know how to calculate the final concentration to % (w/w)

Thank you
X = 100 x [89.315 x 10 x (100 / 1000)] / 1050 = 8.5 % (w/w). However, this is not exactly. More confident answer can be given only if you describe the procedure of the instrument calibration, the purpose of analysis, etc.
Thank you vmu for your explanations :-)

I've weighed and dissolved and made it up to volume (100 ml) with methanol
and created a 6 point calibrations


Is it possible to calculate the final concentration (% w/w) without the
sample's density values as it is a slightly viscous liquid.

Can you please explain what (100/1000) indicate.

Thank you again :-)
You do not need to know the sample density since you calculate %w/w (not %w/v), and you do not know the volume of your sample, but you know its weight (1.05 g).

(100/1000) converts 100 mL to 0.1 L.

The description of the instrument calibration I mentioned includes not only the number of the calibration points, but it also includes the description of the preparation of the standard solutions, the procedure of constructing the calibration curve, etc. Actually, to confidently answer your question, it is necessary to know the whole analytical procedure you use.
I'm very confused by this thread. The original solution was 1.05g in 100mL, which sounds like approximately 1% (sanity check: the answer must be in the vicinity of 1).

Let's work it out both ways:
The original solution was about 1.05g dissolved in 100mL (of methanol?)
This is 10.5g in 1000mL
Methanol has a density of 792g per 1000mL
Therefore our solution is 10.5g/792g
10.5/792 = 0.01326
The solution is 1.326% by weight (assuming it's in methanol, which is lighter than water)
The ten-fold dilution, in methanol, is 0.1326%

I assume you made the solution approximately, but have assayed it to check its true concentration, so we will now work the other way.

The assay of the ten-fold dilution came out at 89.315 mg/L
This is 0.089315g/L
or 0.089315g per 792g (assuming methanol, where 1L weighs 792g)
0.089315/792 = 0.0001128
= 0.01128%
This is 10-fold away from what we expect, which is worrying. What standard was used for calibration of the LC method? Does the LC method already include any dilution factors in its data-processing?

(sanity check: 89mg/L is about 100mg/L, or about 0.1g/L, or about 0.01g/100mL, which is about 0.01%)

[I have assumed methanol because in your follow-up mail you named methanol as the solvent. There are people who take w/w to mean w/v based on solvents weighing 100g/100mL, but I think if you're going to use w/w, it should mean what it says. 1% w/w should mean that if I take 100g of the solution, I will have 1g of the solute. Therefore, to me, the solvent matters.]
You diluted your sample x 1000 (once by 100, then by 10) - you can do the decimals yourself. So the sample was 1000 x 81.395 mg/L + 81.395 g/L.

Without knowing the density of the sample, that is as far as you can go.

Peter
Peter Apps
good point, sorry, I assumed that the 1.05g was pure material that you'd used to make a standard solution you were checking against something else, not 1.05g of a solution, which is why I got thoroughly confused. The word "about" confused me. Ignore my first post.

Actually your life is simple.
Your 10-fold dilution was 89.315mg/L
Therefore the undiluted stock was 893.15mg/L
This is 89.315mg/100mL
you dissolved about 1.05g in 100mL
therefore your sample is about 89.315mg/1.05g
this is about 8931.5mg/105g
or about 8506mg/100g
=8.506g/100g
=8.506% (w/w) (which is what the first reply said).
You don't need to know the density of anything.

However, the "about" doesn't make any sense in this context. If you're preparing a solution that you will quantify independently, then you can be as approximate as you like with the measuring, but if the weighing is part of the quantification, it has to be accurate. (sorry if I'm still misinterpreting)
Peter apps:

Thank you for your help :)
vmu and Imh:

Thank you both for all of your explanations :D

I'm sorry for the confusion - I have used 'about' to give the weight of my sample in 2 d.p. instead of 4 d.p. (actual weight of sample taken) to simplify the calculation as I was stuck on calculation method...

Would my dilution factor be 1000 (as Peter apps suggested above) as I weighed 1.05 g into 100 ml and then did further 10x dilution?

Is the final calculation for the concentration in % (w/w) be;

89.315 mg/l x 1000 = 8931.5 mg/l
= 893.15 mg/ 100ml
= 893.15 mg / 1.05 g
= 85.06 g / 100g

Thank you :-)
chemgc779, first of all you should define what you are trying to calculate as %w/w. What relative to what? What is the numerator and what is the denominator in your target w/w ratio? I suppose that the denominator is the mass of your sample. However, the denominator might be the nominal content of one of the components of your sample if the sample is multicomponent.
Do you quantify the main component (or one of the main components)? Do you quantify an impurity? Again, to confidently answer your question, it is necessary to know your analytical method in much more detail.
chemgc779 wrote:
89.315 mg/l x 1000 = 8931.5 mg/l
= 893.15 mg/ 100ml
= 893.15 mg / 1.05 g
= 85.06 g / 100g

This is incorrect. First, 89.315 x 1000 = 89315 (not 8931.5). Then you will obtain 850 %.
vmu:

I'm sorry for the typo error in my calculation (it should've been 89315 mg/L)
I was confused by Imh's calculation as I believe my total dilution factor was 1000.

I was trying to find a method for calculating the level of rhodamine b dye in water leak detecting product in (g/100g) - x g of rhodamine b in 100g of the product.

Thank you.
We were able to get Chemstation to deliver sequence summary reports in either Percent or ppm. There are some Agilent PDFs which explain this decently well, but we had to figure out the particulars ourselves.

Later on, we added system suitability data to the sequence summary reports too, and I am NOT a computer guy !
12 posts Page 1 of 1

Who is online

In total there is 1 user online :: 0 registered, 0 hidden and 1 guest (based on users active over the past 5 minutes)
Most users ever online was 1117 on Mon Jan 31, 2022 2:50 pm

Users browsing this forum: No registered users and 1 guest

Latest Blog Posts from Separation Science

Separation Science offers free learning from the experts covering methods, applications, webinars, eSeminars, videos, tutorials for users of liquid chromatography, gas chromatography, mass spectrometry, sample preparation and related analytical techniques.

Subscribe to our eNewsletter with daily, weekly or monthly updates: Food & Beverage, Environmental, (Bio)Pharmaceutical, Bioclinical, Liquid Chromatography, Gas Chromatography and Mass Spectrometry.

Liquid Chromatography

Gas Chromatography

Mass Spectrometry