In AAS, Sample weight = 1g, AAS real concentration =0.767 ug/ml, Total volume =50 ml, Dilution factor =50 , I got 0.767*50*50/1=1917.5mg/kg final concentration in ppm . Is it correct calculations ?
Hello,
Your calculation will be concentration multiplied with dilution. So final concentration will be 0.767*50/1=38.35ug/ml.
To go from ug/ml to mg/kg - Such conversion is impossible unless you provide density of final solution.
Indicating concentration ug/ml or mg/kg does not require total volume for calculation. Total volume you can include in calculation if you made concentration of sample from 50ml to 1 ml.
Hope this info will help you.
Good luck!
Hello,
Your calculation will be concentration multiplied with dilution. So final concentration will be 0.767*50/1=38.35ug/ml.
To go from ug/ml to mg/kg - Such conversion is impossible unless you provide density of final solution.
Indicating concentration ug/ml or mg/kg does not require total volume for calculation. Total volume you can include in calculation if you made concentration of sample from 50ml to 1 ml.
Hope this info will help you.
Good luck!
Opposing to Karen's answer, you can calculate the mass fraction in your sample.
You determined 0.767 ug/ml. For 50 ml total volume of the solution analyzed this means 0.767 * 50 = 38.35 ug total analyte mass. Because that mass is from 1 g of sample your mass fraction is 38.35/1 ug/g (I would avoid ppm notation when reporting mass fractions or concentrations because it is easily misinterpreted).
According to your notes, your calculation is true
I do not agree with Karen you can get the mass in 50 ml and then converted to mass percent in 1 g
Hello Khaled Muftah
Not agree with what??? What is true ??? If he has 1000000 L solution so according to his calculation he will receive 0.767*50*1000000000=38250000000ug/ml= 38250000mg/kg =3.8250kg/kg.
Wisest man so try to dilute 3.8kg something in 1L water to make solution bravo
I agree with Karen. Dilution factor must be taken into account for the final answer.
Yes your calculations are correct except you have to divide over the initial weight in (ug) not in g
To be clear, when you say "Dilution factor = 50", that is simply the volume (50mL) divided by the weight (1g)?
Or did you need to dilute the digested sample an additional 50 times (v/v) to bring it in range of the AA?
I just taken 1 g of tissue and digested it in 10ml HNO3 , after completion of digestion Process I diluted 10 ml digest into 40 ml DW and made final volume 50 ml. Now I used this 50 ml sample for AAS analysis (No further dilution ). AAS gives real concentration0.767 ug/ml . Then what is actual concetration in ug/ml of study metal in 1 gm tissue which i digested?
Multiply the concentration (ug/ml) with volume of analysed sample (ml) and divide by sample mass (g) gives you the concentration in the sample in ug/g (or mg/kg).
Your original post uses the dilution factor twice. Follow Bodo's instructions.
See how the units work? ug/mL x mL = ug of the metal. Then divide by amount of original sample digested: ug of metal/g of sample = ug/g of metal in the sample. Don't need the density.
The final concentration will be 0.767 (ug/ml) *50(ml)/1(g)=38.35 (ug/g).
It's mean 38.35 ppm
- Badges
- Science topic
- Similar topics
- Correction