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determining wt% of Iron Sample titrated with dichromate

I need to determine the wt% of this redox reaction. Here is the procedure. Please help.


1. 1.3009g K2Cr2O7 (MW- 255.0863g) diluted in 250mL mark with dionized water.
(concentration is 0.0203M K2Cr2O7)

2. 0.6295g of unknown Ore
added in a 500mL flask with 20mL HCl
3. Add 134 drops of SnCl2 (added in dropwise until color disappears)
4. add 10mL DI water, 10mL HgCl2, 10mL 3M H2SO4, 20mL 6M H3PO4, and 5 drops of Sodium diphenylamine Sulfonate. (this is ffor titration, dont think its necessary for calculations, but not sure)
5. Titrate with the K2Cr2O7 from step 1
total added was 17.76mL to achieve end point

6. a Blank titration was used and 0.09mL was added (1 drop added)
(blank titration, 20ml HCl, 2 drops SnCl2, 100mL DI water, 10mL HgCl2, 10mL 3 M H2SO4, 20mL 6M H3PO4, diluted with 17.76 mL)

Note: the blank value needs to be subtracted from the volumes of dichromate standard used in the preceding titrations.

QUESTION:
DETERMINE THE %Fe OF TITRATED SAMPLE

spock Sun, 03/04/2012 - 16:36

A quick Google search provided me with the equation for the titration:

Cr2O72- + 6 Fe2+ + 14H+ → 2Cr3+ + 6 Fe3+ + 7H2O

This indicates that the Fe concentration will be 6 times the dichromate concentration.

In Step 5, did you titrate the entire sample that you prepared?  Usually smaller samples are taken from the main sample?  If that was the case, it would be important to know what the volume of the sample being titrated was. I will assume here that you titrated the entire sample.

In Step 6, I'm assuming that when you titrated the blank a volume of .09mL) generated the color change.

So, you would find the moles of Cr2O7-2 by multiplying the molarity times the corrected volume (17.76 - .09 mL) in Liters..

    moles of dichromate =     .0203M  x    .01767 L

 

   moles of iron =    6  x   moles of dichromate

   mass of iron =   moles of iron x molar mass of iron

   % iron in sample =  mass of iron / mass of iron ore sample*      x  100

 

* If the entire sample was titrated this would be .6295 g.