March 08, 2021, 12:50:00 AM
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Topic: Analysis of trying to calculate the Mr of the FeSO4.xH2O and thus the X value.  (Read 34734 times)

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Offline sssssaaaallmaan

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You got n(MnO4-) as 6.603384 x 10-3 right? Then multiply this number by 5/3 and you'll get the number of moles of FeC2O4.xH2O. But then when you are working out the molar mass you get about 18.2 which will give you x as roughly -7.  ???

Aye. I came to the same conclusion. Is it possible that part 1 of my calculations were incorrect?

Part One

Mass Container plus Fe(NH4)2SO4.6H2O                12.824g
Mass Of container less some Fe(NH4)2SO4.6H2O                 3.030g
Therefore Mass of Fe(NH4)2SO4.6H2O                              9.794g
Volume of Graduated flask used                                        250 cm³
Mr of Fe(NH4)2(SO4)2.6H2O                                               392.16

n(Fe2+) = mass / Mr      therefore: 9.794/392.16  =  0.024974


n(KMnO4)=0.024974/5 ( because n(KMnO4)= n(Fe2+) x 5)

therefore:  n(KMnO4)=0.00499

c(KMnO4) = nx1000/ 24.80 =  0.2012


I used a 0.02 instead of 0.2012 in the calculations, and i got a value of x of about 2.167 which seems about right.

What do you think?

Offline UG

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No, I've done it on my calculator and everything checks out.  ???
I have no idea what is wrong  :'(

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EDIT: What is the: Volume of Graduated flask used                                        250 cm³ for?

Offline sssssaaaallmaan

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No, I've done it on my calculator and everything checks out.  ???
I have no idea what is wrong  :'(

------
EDIT: What is the: Volume of Graduated flask used                                        250 cm³ for?


Mass Container plus Fe(NH4)2(SO4)2.6H2O                12.824g
Mass Of container less some Fe(NH4)2(SO4)2.6H2O                3.030g
Therefore Mass of Fe(NH4)2(SO4)2.6H2O                               9.794g
Volume of Graduated flask used                                        250 cm³
Mr of Fe(NH4)2(SO4)2.6H2O                                               392.16

I used this info to work out the concentration of Fe(NH4)2(SO4)2.6H2O. And i assumed that since each molecule of the salt contains one ferrous ion, the number of moles of that is exactly the same as number of moles of Fe(NH4)2(SO4)2.6H2O and so the concentration should also be the same right?.

Offline sssssaaaallmaan

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OMG  :o i just had an Epiphany.

Part 1 of my calculations were wrong i think.

It should look like this:

Part One

Mass Container plus Fe(NH4)2SO4.6H2O                12.824g
Mass Of container less some Fe(NH4)2SO4.6H2O                 3.030g
Therefore Mass of Fe(NH4)2SO4.6H2O                              9.794g
Volume of Graduated flask used                                        250 cm³
Mr of Fe(NH4)2(SO4)2.6H2O                                               392.16

n(Fe2+) = mass / Mr      therefore: 9.794/392.16  =  0.024974
C(Fe2+) =nx1000/v ( volume is 250 because concentration of Fe(NH4)2(SO4)2.6H2O is the same as Fe2+)  =  0.099896

m1 ( conc. of Fe2+) = 0.099896
v1  (Volume of Fe2+) = 25.00
v2 ( Volume of MnO4-) = 24.82

Using the equation m2=m1v1/5v2  (Using the following equation ...m2=m1v1/5v2
Where m1 and m2 is concentration of Fe(2+) and MnO4-  and v1 and v2 is the corresponding volumes used.)
 

m2= 0.0201240934

Part Two

Mass of compound used                        0.200 g
Final Bruette reading                            32.82 cm³

Final Volume of KMnO4 solution used       32.82 cm³


Thus if a sample of W(g)of the Iron (II) ethanedioate hydrate requires a volume v ( cm³ ) of MnO4- solution of concentration m (mol dm-3), then the number of moles of MnO4-  reacting is mv/1000 and this number of moles will react with 5mv/3000 moles of FeC2O4.xH2O.

Hence  W/Mr=5mv/3000
So that Mr=600W/mv
Moreover, since Mr=143.87 + 18.02x
Then x= (Mr-143.87)/18.02


m=0.0201240934
v=32.80
W=0.200

Therefore using    Mr=600W/mv  =   600 x 0.200/ 32.80 x 0.0201240934
Mr= 181.7988275

And so, using x= (Mr-143.87)/18.02   =     (181.7988275-143.87)/18.02

= 2.104818398 which  is 2.1 

If you spot any mistakes, could you let me know please?

Offline UG

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Mass Container plus Fe(NH4)2SO4.6H2O                12.824g
Mass Of container less some Fe(NH4)2SO4.6H2O                 3.030g
Therefore Mass of Fe(NH4)2SO4.6H2O                              9.794g
Volume of Graduated flask used                                        250 cm³
Mr of Fe(NH4)2(SO4)2.6H2O                                               392.16

n(Fe2+) = mass / Mr      therefore: 9.794/392.16  =  0.024974
C(Fe2+) =nx1000/v ( volume is 250 because concentration of Fe(NH4)2(SO4)2.6H2O is the same as Fe2+)  =  0.099896

m1 ( conc. of Fe2+) = 0.099896
v1  (Volume of Fe2+) = 25.00
v2 ( Volume of MnO4-) = 24.82
So you diluted up to 250mL and then took a 25mL sample?

Offline sssssaaaallmaan

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So you diluted up to 250mL and then took a 25mL sample?

Aye. Using a pipette

Offline UG

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Then the rest should be correct!  :D

Offline sssssaaaallmaan

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Finally!!!

Like to thank Borek for being patient and for his help on this problem and thanks to  UG for putting on the finishing touches.

Really appreciate your help guys.

Sal.
 


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