This may be slightly off the chromatography track, but it's something I would like to know how to do.

I've detected an impurity at 0.5 ppm (ie. 0.5 mg/L) using HPLC analysis with a single point calibration curve. My standard concentration is 10mg/L and my sample concentration is 200 mg/L. How do I convert this 0.5 ppm to %(w/w)impurity?

You have units of weight per volume. You want units in weight per weight. You need to know the density of solution (weight per volume). If this is water, the density is essentially 1 g/mL, so

0.5 mg/L = (0.5 mg/1000 mL) / (1 g/mL)
= 5*10^-4 g/10^3 g
= 5*10^-7 (w/w)
= 5*10^-5 %w/w
= 0.00005%w/w

If the density is other than 1, the calculation is the same, but substitute the density for the "(1 g/mL)"

ppm tends to assume like units - either v/v or w/w. In the w/w case, 1 ppm = 0.00001% w/w.

By the way, one point does not constitute a standard curve unless the method has been validated down to the sub ppm range in your case. If you are expressing this unknown as a % w/w of another analyte, you are making assumptions about both the molecular weight of the impurity adn its extinction coefficient. As long as both are close to ythat of your principle analyte, you are probably okay. If you aren't certain, I suggest expressing the impurity in terms of area % (compared to the principle analyte).

DR

Assuming the density = 1

0.5 ppm
0.5 x 100 / 1000000 = 0.00005 %w/w

1.0 ppm
1.0 x 100 / 1000000 = 0.0001 % w/w and not 0.00001%

Your impurity is 0.5mg/L.
Sample conc. is 200mg/L.
Thus, %w/w of impurity w.r.t. the sample=

0.5/200*100=0.25%w/w

Thanks for all your help,

I think my initial question was ambiguous and so the first 3 answers wern't really what I was after. I'd agree with the last response though, and the calculation now seems painfully obvious.

My sample conc is 200 mg/L and I detect the impurity at 0.5 mg/L. Hence the units cancel nicely( I already have my w/w with the mg/mg) and I can convert the content to a percent.

Matthew