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## ppm determination of stock solution

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### ppm determination of stock solution

I am making a 10,000 ppm stock of 2-CHLOROBIPHENYL
The 2-CHLOROBIPHENYL is a crystalline powder and is >99% pure. I am treating the powder as if it is 100% 2-CHLOROBIPHENYL

If I want a 10,000 ppm solution I was told to weigh out 0.5grams of the 2-CHLOROBIPHENYL powder into a volumetric flask and fill the flask to 50 ml with methanol. a weight/volume measurement. This doesn't seem to account for the density of methanol being less than water though.

My question is would it more accurate to weigh out 0.5g of the 2-CHLOROBIPHENYL into 49.5 grams of methanol? Total weight would be 50.0 grams, and the 2-CHLOROBIPHENYL would be at a concentration of 10,000 ppm.

### Re: ppm determination of stock solution

I think you mean 49.5 g of methanol though. Otherwise, yes, this seems more accurate given the density of methanol.

### Re: ppm determination of stock solution

That's the problem with "ppm"; it's ambiguous. Do they mean wt/vol or wt/wt? you need to check the details of the method (or whoever gave you the procedure) to verify which one was intended.

For solutions, much better to specify wt/volume (e.g., in this case 10 g/L).
-- Tom Jupille
LC Resources / Separation Science Associates
tjupille@lcresources.com
+ 1 (925) 297-5374

### Re: ppm determination of stock solution

Blazer wrote:
I think you mean 49.5 g of methanol though. Otherwise, yes, this seems more accurate given the density of methanol.

Sorry that was a typo. You are correct it should have been 49.5g and not 49.95g

### Re: ppm determination of stock solution

tom jupille wrote:
That's the problem with "ppm"; it's ambiguous. Do they mean wt/vol or wt/wt? you need to check the details of the method (or whoever gave you the procedure) to verify which one was intended.

For solutions, much better to specify wt/volume (e.g., in this case 10 g/L).

The method only specifies to spike each sample with 5ul of 10,000ppm TFB.

Since the stock of TFB used in the curve was made using wt/volume I will do the same for this stock, I just wasn't sure if this was standard practice when making stock solutions to calculate ppm using the density of water rather than whatever the actual solvent density is. I would think this would be a problem for many applications.

### Re: ppm determination of stock solution

"ppm" is not a proper unit - in this case it could mean micrograms per gram, micrograms per ml, (implausibly) microlitres per ml, or micomoles per mole, each of which will give you a different composition.

All the confusion would be eliminated by specifying the composition of solutions in SI units.

Peter
Peter Apps

### Re: ppm determination of stock solution

It seems to me that since you're adding 5 µL of this standard at the end, you'd really want to know how many µg if the chlorobiphenyl you're really adding to the sample. Wt./vol makes more sense to me. That's how I do it as well.

The density of methanol doesn't vary that much at room temperature. I found data online that indicate:

10 °C = 0.801 g/mL
20 °C = 0.792 g/mL
30 °C = 0.782 g/mL

I don't know about you but my lab temperature never varies this much. Mostly, I'm around 20-23 °C. Your 49.5 g of methanol translates to:

10 °C (50 °F) = 61.8 mL
20 °C (68 °F) = 62.5 mL
30 °C (86 °F) = 63.3 mL

Even for this wide swing in temperature, you're only off by about 1.2% (61.8/62.5 = 0.989, 63.3/62.5 = 1.013). I doubt seriously that you'll ever experience this type of swing in temperature unless you're operating outside.

0.5 g/50 g x 10^6 = 10,000 µg/g
0.5 g/62.5 mL x 10^6 = 8,000 µg/mL

But if I'm adding 5 µL of this solution to sample, and I want to know that I'm adding 50 µg of the chlorobiphenyl to that sample, I'd go with 0.5 g chlorobiphenyl to 50 mL of solution with methanol as the dilution. The actual mass of methanol that you're adding at 20-23 °C doesn't vary a whole lot (<< 1.2%).

Just my \$0.02.

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