Expert: Mike Fulcher - 10/3/2006

Question
Hey,

Everyone in the chemistry area was maxed out and I thought that you might be able to help me out with this since you wouldve have taken Regular Chemistry Course.

My question is this. I am doing an experiment for my sceince fair research where I need to obtain 10mg/L and 20 Mg/L Chromium.

My Question is that I am using Chromium (III) Sulfate Hydrated. I wanted to what procedures/Calculations are necessary in order to obtain 10 Mg/L and 20 Mg/L of Chromium III ion in the compound.

Regards
Manthan  

Answer
This sound pretty straightforward.  First you need to know the molecular (or formula) weight of the chromium sulfate hydrate.  I don't know this off the top of my head, but if you know the chemical formula (should be on the jar) you can add up the atomic weights of each element in the formula to get the formula weight.  Don't forget to include the weight of the hydrate (water).  Now, take the  formula weight of chromium sulfate (FW) and divide it by the molecular weight of chromium (MW).  Multiply this ratio by you desired concentration (i.e. 10 mg/L), and you will get the necessary mgs of chromium sulfate to add to 1 L of water.  If you only want to make 100 mL, then you can just divide that number by ten.

I would be happy to check your numbers for you once you do the calculations, since I did them myself while I was answering your questions.  Good luck.