I have 200mL sample, evaporated it to a small volume, then diluted to 100mL. How do i calculate concentration?

Question

I have a 200mL water sample from a river. I heated it up until it reaches a minimum concentration. Then I transferred the small volume to a 100mL volumetric flask and filled it with deionized water until the total volume reached 100mL. Then I analyzed it using Flame Atomic Absorption Spectrophotometry for it's lead absorbance. I already calculated the concentration based on a calibration graph that I made. But how do I use that concentration to calculate for the actual concentration in the 200mL sample, accounting for the volume changes?

Answer 1

Provided that you don't care for anything volatile it is very simple:

C1V1=C2V2 =>

C1=(V2/V1)C2 = 100/200*C2= 0.5*C2
So the initial concentration is half of what you calculate from the standard curve.

Remember that you only lost some water, the mole of lead are always there.

Or if you prefer:

you concentrated your solution to an unknown voulme Vx and unknown concentration Cx. So your lead content C1V1 =CxVx
then you diluted to 100 mL and found C2 through the standard curve, so CxVx=C2V2
Combine the 2 and C1V1=C2V2