HELP confused...

[destain]

I'm not sure if you'll get what I mean so I'll just write out the question and then ask.

To determine the amount of MnO2 in an ore of manganese, 5.25g of the ore was allowed to react with 100mL of 0.250M oxalic acid (H2C204).
Mn02 + H2C2O4 +2H(+) ------>Mn(2+) + 2CO2 + 2H20

The remaining amount of oxalic acid is determined by titration with tri-iodide ions.
I3(-) + H2C2O4 -----> 2CO2 + 2H(+) + 3I(-)

20mL of the resulting solution is then titrated with 0.1M solution of the triiodide ion. The average titre found was 12.54mL. Determine the percentage by mass of MnO2 present in the ore.


So...when you find the number of moles you'll have to multiply it by 5 becasue only 20mL was taken...and that's because it was allowed to react with 100mL which is 20x5. BUT this is only because the starting sample was in grams like...5.25g...What if the question was...40ml of blabalh was allowed to react with 100ml of balblahb...So then when you get the number of moles in 20mL of the sample taken...Do you multiply it by 7 times then?

[Mr. Study]

If we had a 20 ml of alcohol and we pipetted it into a 250 ml standard flask and made the solution up to the mark. Which we then take a 20ml aliquot of the diluted alcohol, we can find the original sample's mole by doing the mole of the diluted alcohol times 250/20.

It's 250 over 20 as we had 250 ml and we took 20 ml of it, which is the dilution factor. If you think about it, the original sample must have a higher mole than the diluted sample, so it would make sense if the diluted samples mole is multiplied by something to increase it back to the undiluted, original mole.

If anything I said is wrong, Someone please correct me.

tl:dr: If we have an original sample, place it in a known volume, fill it up to the mark and we took an aliquot, a known volume of the diluted sample. We can find the original samples mole by multiplying the diluted samples mole by the known volume divided by the aliquot.

Does that answer your question? If not, I'll scan a question I did on dilution factors and annotate it.

[destain]

nono, i get WHY its diluted and the multiplication thing.

BUT im asking that because the it was added on to 5.25G <--- GRAMS we could just multiply by 5 to to go from 20ml to 100ml

BUT IF it was 100ml added onto 40ml...then im asking from the 20ml would you multiply by 7 to get back to original?

Basically do you consider the 'added onto' part as part of the volume.

[Shenz0r]

I'm not sure if you'll get what I mean so I'll just write out the question and then ask.

To determine the amount of MnO2 in an ore of manganese, 5.25g of the ore was allowed to react with 100mL of 0.250M oxalic acid (H2C204).
Mn02 + H2C2O4 +2H(+) ------>Mn(2+) + 2CO2 + 2H20

The remaining amount of oxalic acid is determined by titration with tri-iodide ions.
I3(-) + H2C2O4 -----> 2CO2 + 2H(+) + 3I(-)

20mL of the resulting solution is then titrated with 0.1M solution of the triiodide ion. The average titre found was 12.54mL. Determine the percentage by mass of MnO2 present in the ore.


So...when you find the number of moles you'll have to multiply it by 5 becasue only 20mL was taken...and that's because it was allowed to react with 100mL which is 20x5. BUT this is only because the starting sample was in grams like...5.25g...What if the question was...40ml of blabalh was allowed to react with 100ml of balblahb...So then when you get the number of moles in 20mL of the sample taken...Do you multiply it by 7 times then?

This question looks like a Back Titration to me, I'm a bit confused on what you're trying to say in the end though. If you're confused on what a Back Titration is, I explained it at Back titration Confused on  the information given

In order words, this Back Titration does not take into account dilution or whatever, but it takes into account n(initial) = n(reacted) + n(excess).

Step 1: Work out the initial amount of mol of oxalic acid from the first equation.
n(oxalic acid)initial = 0.1 x 0.250 = 0.025 mol
Step 2: Work out the number of mol of Tri-iodide ions in order to work out how much oxalic acid was in excess.
n(I3) = 0.01254 x 0.1 = 0.001254 mol
n(oxalic acid)excess = 1/1 x 0.001254 = 0.001254 mol
Step 3: Work out the number of mol of oxalic acid that reacted with MnO2
n(oxalic acid)reacted = n(initial) - n(excess) 0.025 - 0.001254 = 0.023746 mol
n(MnO2) = 1/1 x 0.023746 = 0.023746 mol
Step 4: Work out the mass of MnO2
m(MnO2) = (54.9 + 32) x 0.023746 = 2.06353 g
Step 5: Work out % of MnO2 in the iron sample
% of MnO2 = 2.06353 / 5.25 x 100 = 39.3% (correct to 3 sf)

[edwardo]

Ok so what you're asking is if the question told you the volume of the sample taken instead of grams(40ml) was added to 100ml solution, how would you workout the number of moles in the original sample? Well I doubt it would ask you that because normally in an analysis you would measure it in grams.

Even if the starting sample was 5.26g, you wouldn't need to be using that to calculate the number of moles in the MnO2, you can find the number of moles by multiplying the number of moles found in the 20ml solution by 5(100ml/20ml). the 5.26g is the weight accurately measured when rocks or something was grinded up, not the weight of the solution, in case you got that mixed up.

but if it was 40ml diluted to 100ml where 20ml aliquots were taken from that and you want to find the number of moles in the 100ml, you would multiply the number of moles in the 20ml by 5. Then if you want to find how much moles were in 40ml solution i think you would then need to multiply the number of moles in the 100ml by 0.4(40ml/100ml). -That might be a possibility, but what you're asking is if it started with giving you the sample taken measured as 40ml, well like I said before, a question will never ask that as that's not even logical since that is not how making a standard solution works. How would you be able to find the mole in 140ml if 40ml of something was added on to 100ml of something completely different, or are you saying that the 40ml and 100ml are the exact same solution? Because that would be weird if you were given 40ml of it in the first place.

[destain]

oh okay thanks, now that makes more sense.
So the question won't exist hopefully. THANKS
and I know my question was so confusing sorry haha but I didn't really know how else to explain it...
And yes it was a back titration