[welsh_Ajo]

I am analysing  biomass material using HPLC and I have totally confused my self over dilution. I have an answer to the following but would love someone else to confirm it. And maybe show working out.

I had a 0.01g sample of pure material (compound X) which I diluted in 1 ml of water from this solution I took 0.1 ML of liquid and diluted it in 1 ML of water. This gave me a 1 mg in 1 ML standard (well almost) which I used to calibrate the Machine; the machine gave an integration of 52500 for 1 mg in 1 ML.

For my unknown sample I macerated 60g in 90 ML water, from this solution I took 0.1ML and diluted this with 1 ML of water. The machine gave this a value of 2656.

So my question what % of my unknown sample is compound X?

[Borek]

So my question what % of my unknown sample is compound X?

Show your calculations.

[chiralic]

If enough one point to calibrate your Liquid Chromatograph?

[profmsg]

just one point is not enought for calibration however, a simple logical calculation can be made
calculation :
 1mg = 1.5ml
1/1.5 = 1ml
0.066mg = 0.1ml


now 1 mg = 52500
0.0666mg = 52500*0.0666 = 3496.5
theoritical value = 3496.5
actula value obtained = 2656
so percentage of your compound = 2656*100/3496.5 = 75.961/*%

this simple calculation should work but im not 100% sure whether it will work or not
thanks