Chemistry Forums for Students > Analytical Chemistry Forum

how to calculate concentration? chromatography

(1/1)

**old_dog_learning_new_trks**:

Hi All,

I am having a hard time figuring out how to calculate a concentration based on a measurement on a liquid chromatograph.

If I start with 1.487 g of a sample and dillute it in 20 ml of ethanol. I take that and pipette 5 um into 995 um of ethanol, so a 200 dilution. I measure this dilute material on an LC and get 196 ppm.

196 ppm is also 0.196 mg/g. and 0.196 mg/g * 1.487 g means there is 0.29 g in the sample.

How do I calculate the concentration in the original material? Is it 0.29 * 20 * 200?

**mjc123**:

What are you actually trying to calculate? You say you want to calculate a concentration (of what in what?), but it looks like you're trying to calculate the amount (in g) of analyte in the original sample. If you're not clear what you want to do, you've not much hope of doing it correctly.

--- Quote ---I take that and pipette 5 um into 995 um of ethanol

--- End quote ---

I assume you mean 5 μL into 995 μL.

--- Quote ---196 ppm is also 0.196 mg/g.

--- End quote ---

0.196 mg of WHAT per g of WHAT? It's really important to understand this, or you will talk nonsense like your next sentence.

**old_dog_learning_new_trks**:

You are absolutely right mjc123. As I am not chemist, this is not straight in my mind, and it is unfamiliar territory for me - so I am trying to understand. Let me know if this makes more sense:

I would like to know what the mg/g concentration of an analyte is, undiluted in its original form. I was given a sample of a plant extract, totaling 1.487 g. 1.487 g was dissolved in 20 ml of ethanol, then diluted 5 ul into 995 ul. This was measured on an LC, and resulted in 196 ppm based off of peak integration.

If I would like to know the total mg of the analyte in a 1 g sample...how would I calculate this based on this information? If I am thinking about this correctly, is it:

1. Convert ppm to mg/g -> 300 ppm is 0.30 mg/g

2. 20 ml ethanol is a 20 dilution

3. 5 ul and 995 ul is a 200 dilution

0.30 mg/g * 20 * 200 = 1200 mg/g in an undiluted sample

if I them divide by the mass, 1.487 g

is that the mg in 1 g of sample?

806.99 mg

**Corribus**:

Just to be clear - your original 1.487 g of extract is the unknown mixture? I.e., it's 1.487 g of a mixture, and you are trying to find out what fraction of that mixture is your analyte? Expressed in terms of xx g per gram of extract?

As a minor detail, you should know that ppm is an ambiguous unit. If the concentration was determined by LC to be xx ppm, that is probably a m/v basis not a m/m basis. You may convert this from m/v to m/m basis using the assumption that 1 g of (aqueous) solution ~ 1 mL of solution, but you should be aware this is not strictly correct except in idealized conditions (pure water, standard temperature/pressure, etc.). Assuming 1 g = 1 mL will therefore introduce some error into you determined concentration value.

**mjc123**:

The answer would be right (approximately) if your LC concentration was 300 ppm, which it is not. However, I don't like how you've stated step 2, which is close to being meaningless. Dissolving a neat sample in solvent is not generally referred to as "dilution", and it would be impossible to quantify the "dilution" factor unless you know the volume of the neat sample, which I assume you don't. Work with volume, not "dilution".

The concentration of your LC sample is 196 ppm. Let's assume, as Corribus suggests, that this is ppm m/v, so 196 ppm = 0.196 mg/mL. So your 1 mL sample contains 0.196 mg analyte. This was all originally in the 5μL aliquot, so the concentration of this was 0.196 * 1000/5 = 39.2 mg/mL.

So the amount in your original solution was 39.2*V, where V is the volume of the solution. But we don't know this. If you add 1.487 g solute to 20 mL solvent, the solution volume will be significantly greater than 20 mL, but you don't know a priori exactly what it is. You could assume that the solute has a density of 1 g/mL, and that the volumes are additive. Or you could assume that what you actually did (and what you should have done) was to dissolve the sample in less than 20 mL ethanol, and make up the solution volume to 20 mL with ethanol. Or you could actually measure the solution volume, if you have accurate enough equipment, and this was a real experiment, not a homework question.

Let's assume, though, that your solution volume was 20 mL. The mass of analyte in this solution would be 39.2*20 = 784 mg. So the mass fraction (we wouldn't usually say "concentration") of analyte in your original sample was 784/1.487 = 527 mg/g.

This is what we would get following your method, but without the conceptually dodgy statement that "20 mL is 20 dilution". It also involves an assumption, which may be quite inaccurate, about the volume of the solution. I hope you appreciate why I've done it this way, and the nature of the assumptions involved.

Navigation

[0] Message Index

Go to full version