Solving the Sugar Content in Potato Flour Puzzle with Chemistry

[rnielsen25]

Homework Statement

Hi everyone, i have a problem. My brain is short-circuited, and it seems like i can't find out the simplest chemistry in the world :D

We made this experiment where we should determine the sugar content in potato flour. We added amylases to break down the starch to glucose. Then we added fehling's solution and isolated the sediments. Then we added some ammoniumiron(III)sulphate. Last we titrated the solution with potassiumpermanganate.

My problem is that i need to find the molar ratio between the sugar/glucose and the permanganate-ion.

Homework Equations

The chemical reactions is given by the following reactions.

1. 5OH^- + RCHO + 2Cu²⁺ --> RCOO^- + Cu₂O + 3H₂O
2. 2H⁺ + Cu₂O + 2Fe³⁺ --> 2Cu²⁺ + 2Fe²⁺ + 2H₂O
3. 8H⁺ + 5Fe²⁺ + MnO₄^- --> 5Fe³⁺ + Mn²⁺ + 4H₂O

The Attempt at a Solution

The molar ratio between RCHO and Cu₂O is 1:1. So is the ratio between Cu₂O in equation 1 and 2.
The molar ration between Cu₂O and Fe²⁺ is 1:2. So the ratio between RCHO and Fe²⁺ is also 1:2. But the ratio between Fe²⁺ in equation 2 and 3 is 2:5. So the ratio between Fe²⁺ and RCHO is 1:(2*2,5) = 1:5? Because there for every Fe(II) in equation 2 there is 2,5 Fe(II) in equation 3.

But the then the ratio between Fe²⁺ and MnO₄^- in equation 3 is 5:1.
So the ratio between MnO₄^- and RCHO is what? 5:5? That cannot be true?

My logic says 1:2,5. But i can't explain why. It maked sense to me at first, but when i tried to explain why, i couldn't. Could you help me figure out what the real molar ratio is, and how to explain why it is that? Sorry, that i suddently can't figure out the simplest chemistry and math in the world :D

 

[rnielsen25]

Well i just tried working it out entering numbers. Which i think makes more sense to me.

If i have 1 mol RCHO, i have 1 mol Cu₂O.
Then i have 2 mols Fe²⁺
In equation 3 i still have 2 mol Fe(²⁺, which means i have 2/5 mol MnO₄^-.

Then the molar ratio between RCHO and MnO₄^- must be 1:0,4. Which means for every mol of MnO₄^- i have 2,5 mol RCHO.