For the lab of Determining the % of iron in an iron ore sample
The instructions were to:
1) Weigh out btw 0.3-0.4g of iron ore
2) disolve in concent HCl (10ml) and 3ml of 0.25M SnCl2
3) add enough tin so the yellow color disapeared
4) heat until sample decomposes and disolves
5) add 3-5ml SnCl2 to eliminate yellow color
6) after ore dissolves remove excess SnCl2 by adding 0.02M KMnO4 dropwise until the sol turns yellow
7) add 15ml dis water
8) heat each solution to boiling
9) add SnCl2 dropwise until the yellow color disapears
10) cool sample to room temp
11) treat each sample individually, add 10ml 5% Mercury chloride to get cloudy white precipitate. Absence of white precipitate indicates insufficent recduction of Fe.
12) After adding Mercury chloride, wait 2 minutes then add 25ml Zimmerman reagent and 300ml dis water
13) Titrate with 0.02M KMnO4
untill solution stays pink for 30 sec.
Thing is that I was curious as to what is happening at each step, like why does the solution turn yellow after adding KMnO4, clear after adding tin chloride, and have precip after adding the Mercury chloride.
Would adding too much of SnCl2 would cause anything to happen?
Well besides what happens in each part I did steps 1-10, and then left the solutions in the storage cabinet. The solution was clear when I left it in there but upon taking it out today I observed that the solution was yellow again. I don't understand why that is but, I was told to heat it and add more SnCl2, thus leading to the question of can you add too much SnCl2.
Another reason for me asking this is b/c when I added the Mercury Chloride to the solution each solution looked different from each other. The first one had very little precipitate and the precipitate was not shiny at all just the clear solution was at best misty. Basically you could see through it. The second beaker, upon adding the Mercury Chloride became milky and had swirly, shiny precipitate. It had so much precipitate that I couldn't see through it compared with the first beaker. The third beaker after adding the Mercury Chloride had shiny, swirly precipitate but the bottom of the beaker could still be seen. In general somewhere in btw the 1st and second beaker's precipitate concentration from the looks of it anyways.
Well anyway, I wasn't sure of how to do the calculations so I'm not sure my they're correct. and I'm not exactly sure what constitutes iron ore exactly, but this is what I did.
concent KMnO4: 0.02002M
Mwt Fe: 55.85g/mol
Ratio KMnO4 to Fe: 1:5
beaker# mass(g) iron ore unknown
1------0.3945g
2------0.3291g
3------0.3501g
amount used (ml) KMNO4
1------46.6ml
2------38.90ml
3------41.28ml
%Fe
1) [(46.60mlKMNO4)(1L/1000ml)(0.02002MKMNO4)(5molFe/1molKMnO4)(55.85g/molFe)(100)]/ 0.3945g iron ore=
65.95%Fe
2)[(38.90mlKMNO4)(1L/1000ml)(0.02002MKMNO4)(5molFe/1molKMnO4)(55.85g/molFe)(100)]/ 0.3291g iron ore=
65.99%Fe
3)[(41.28mlKMNO4)(1L/1000ml)(0.02002MKMNO4)(5molFe/1molKMnO4)(55.85g/molFe)(100)]/ 0.3501g iron ore=
65.86%Fe
RSD = 0.0667
Well I guess they're pretty close to each other but I still don't know what a favorable RSD should be...
Thanks =)
The purpose of the SnCl2 is to reduce the iron(III) to iron(II). It's the iron(II) you want to titrate with permanganate. You MUST add enough SnCl2 to change all of the iron(III) to iron(II); otherwise, low results will result because not all of the iron is being titrated with permanganate. Therefore, you always add a little excess of SnCl2. (Yes, you CAN add too much SnCl2 and you will know if you did for instead of getting a white milky ppt when adding HgCl2, the ppt turns grey or black.) You don't want to titrate the SnCl2, also; therefore, adding the HgCl2 reaagent oxidizes the excess Sn(II) to Sn(IV) and reduces the Hg(II) to Hg(I). HgCl2 is soluble, Hg2Cl2 is not and Hg2Cl2 is the shiny white ppt you see. (A grey or black ppt instead of the shiny white one indicates that the Hg(I) was reduced to Hg(o)(the metal) and that means either (1) too much SnCl2 was added OR (2) the HgCl2 reagent was added too slowly.) The white ppt means everything is ok. The addition of the Zimmerman-Reinhardt reagent is to reduce the tendency of Cl^- to be oxidized during the titration part with permanganate. As for the iron ore solutions being white or colorless upon putting them away, then yellow when you came back: air oxidized some of the iron(II) to iron(III); therefore, your lab instructor just had you do the SnCl2 bit over again and that was the proper thing to do. The difference in the amount of milky ppt you saw in the three samples reflects the amount of excess SnCl2 you added in reducing the iron(III) to iron(II). It really doesn't matter how much Hg2Cl2 ppt you have; BUT you must have some and it must be white or milky. No ppt means you didn't have an excess of SnCl2 which means all of the iron wasn't reduced and low results would be the order of the day. As for your precision, it looks pretty good to me.
There are a number of iron ores used; i.e., hematite, magnetite, limonite. Real iron ores are difficult to dissolve and many student labs now use synthetic iron ores made of FeO and some diluent that dissolves easily. I still remember determining iron in hematite AND limonite when I took quant umpteen years ago and when I say some iron ores are difficult to dissolve I know because I've been there. I spent most of one day just getting my samples into solution. And you know how difficult that was because it left a LASTING impression.
Thanks
Now I get why I had to add tin chloride and the other chemicals to the solution..
Just today I saw someone doing this lab and she had grey and black precipitate and the lab tech was trying to figure out why this was. Now I know why their solution looked like that...interesting
I know that my iron ore was definately the synthetic type since it dissolved readily
But, basically what iron ore is composed of is iron + something else??
That's right. Hematite is largely Fe2O3 and magnetite is largely Fe3O4, but there are other things there too, such as silica, aluminum (and other iron like materials with a valence of 3+ usually simply called R2O3.) Also there is Ca, Mn, Mg, alkalies, P, S, Ti, V and Cr to name just some of them. Other ores are Fe2O3.H2O (goethite), Fe2O3.H2O.xH2O (limonite), and FeCO3 (siderite). Siderite is a minor ore. All of this information comes from Textbook of Quantitative Inorganic Analysis, Kolthoff and Sandell, 3rd edition, 1952. (I use these OLD books often for some of the new ones don't have all that good information about the kinds of ores and their names.)
Just as a side note, if the black or grey ppt forms, the sample should be discarded for there is no way to "go back" and redo. Titrating the grey or black sample will lead to high results for permanganate oxidizes the metallic mercury. Another by the way, the black color is caused by tiny tiny droplets of mercury metal. The grey is caused by some black and some white mixing to form grey. So if you get no ppt when adding HgCl2 reagent the sample is lost. And if you get a black or grey color the sample is lost. Yours were white but you got some (even though one was a small amount) for all of them. All three of yours looked good to me.
that's good to hear that my results looked fine
Another thing I noticed was that in the beakers with the gray solutions there was a rather thick layer of precipitate on the bottom (400ml beakers) about 1/2 an inch of chalky like precipitate that was settled on the bottom . I definately didn't get this much precipitate and not of that nature either.
I was wondering what could that precipitate have been?
Wow, I didn't know that there was so many types of iron ores...
The ppt was more than likely Hg2Cl2 but the appearance was changed due to the mixing of the metallic Hg with it.
Okay, I'm just glad mine didn't look like that...=/
Thanks Dr.Bob =)