Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: sssssaaaallmaan on April 17, 2009, 03:10:04 PM
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Greetings.
I was wondering if you can go over my results and analysis and see if i am moving in the right direction.
1. Results and calculations for the Standardization of KMnO4
Results
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)2SO4.6H2O 392.16
Concentration of standard Fe(NH4)2SO4.6H2O solution
N of moles =mass/Mr ∴ 9.794/392.16= 0.249745002
∴ C=1000xn/v =0.0998980008
Titrations
Average KMnO4 solution titre value 24.82
Volume of pipette used 25.00
Calculations:
n(KMnO4)= Cxv/1000 ∴ 0.02 ( concentration is given) x 24.82 /1000
= 0.0004964
n(KMnO4)= n(Fe2+) x 5
∴ 0.0004964 x 5 = 0.002482
c(Fe(2+) = nx1000/ v ∴ 0.002482 x 1000/ 25.00 = 0.09928
Using the following equation( Have to use this 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= 0.09928 X 25.00 / 5x24.82
∴ m2= 0.02
2.Results and Calculations for the Relative molecular mass, Mr, of FeC2O4.xH2O and hence the value of x
Mass of compound used 0.200 g
Final Bruette reading 32.82 cm³
Final Volume of KMnO4 solution used 32.82 cm³
A) Ratio Mass/Titre/ g cm³ 1:164
B) Calculating the Relative molecular mass, Mr of the hydrate and hence the value of x, the number of moles of water of crystallization in one mole of the hydrate.
Please note that the following few sentences and expressions are given to us and have to be used to work out the problem
Using ( it is a given expression ) 3MnO4- = 5FeC2O2
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
End of the given stuff
Calculations
n(MnO4-) 0.02 x 32.80/1000 = 0.000656
Using 3MnO4- = 5FeC2O2
0.000656 x 5/3 = 0.0010933333....
∴ c=0.0010933333.... x1000 /32.80 =0.03
Mr= 600 x 0.200 /0.03 x 25.00
=160
x=(160 -143.87)/18.02 = 0.89511... =1
Thank you so much for your time and consideration.
Sal.
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After discussing my calculations with my Tudor, i found out that i was suppose to get between 2-7 for my X value, but i got a 1 for the answer. Can you please see if/where I am doing anything wrong?
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You are looking for x in FeSO4.xH2O, but you are titrating Fe(NH4)2SO4.6H2O and FeC2O4.xH2O?
Please describe precisely the experimental procedure you have followed, as now trying to understand it is mostly a guesswork.
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Using ( it is a given expression ) 3MnO4- = 5FeC2O2
??? should be 2
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Using ( it is a given expression ) 3MnO4- = 5FeC2O2
??? should be 2
1 electron from Fe2+, 2 electrons from (COO-)2, 3 together.
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You are looking for x in FeSO4.xH2O, but you are titrating Fe(NH4)2SO4.6H2O and FeC2O4.xH2O?
Please describe precisely the experimental procedure you have followed, as now trying to understand it is mostly a guesswork.
Sorry, i should have provided more information.
I hope this is enough.
Outline of the experiment: http://img293.imageshack.us/img293/7590/outline.jpg
The theory behind my calculations http://img179.imageshack.us/img179/6379/12397571.jpg
The procedure: http://img150.imageshack.us/img150/8236/68373584.jpg
Thanks,
Sal.
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You are not standardizing Fe2+ solution with exactly 0.02 M permanganate, you are standardizing about 0.02 permanganate solution to know its exact normality.
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You are not standardizing Fe2+ solution with exactly 0.02 M permanganate, you are standardizing about 0.02 permanganate solution to know its exact normality.
Could you please elaborate more ( if its even possible) on what you have said here. Are you suggesting that the concentration shouldn't be 0.02 M ?
After doing some research i know that my Mr value should be 179.8951 g/mol (Molar mass (molecular weight) of FeC2O4*2H2O ). I am still not sure where i have gone wrong in my calculations.
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Titrations
Average KMnO4 solution titre value 24.82
Volume of pipette used 25.00
Calculations:
n(KMnO4)= Cxv/1000 ∴ 0.02 ( concentration is given) x 24.82 /1000
= 0.0004964
n(KMnO4)= n(Fe2+) x 5
∴ 0.0004964 x 5 = 0.002482
c(Fe(2+) = nx1000/ v ∴ 0.002482 x 1000/ 25.00 = 0.09928
You calculated results of your first titration wrong way. What you did you first assumed concentration of permanganate to be 0.02, you used it to calculate concentration of Fe2+, you used calculated concentration of Fe2+ to calculate back concentration of permanganate. Not surprisingly you got 0.02. You weighted accurately your solid to be able to calculate concentration of Fe2+ from the solid mass, not from the titration.
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Thanks for the reply.
So you are saying to work out n(Fe2+) = 9.794/55.8 and then dividing that answer by 5, due the fact that n(KMnO4)= n(Fe2+) x 5. Then using this value, i calculate the concentration of Fe 2+ which then goes in the formula m2=m1v1/5v2
,where m1 and m2 is concentration of Fe(2+) and MnO4- and v1 and v2 is the corresponding volumes used, which then i get conc. of KMnO4.
Sal.
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I think ( I hope) I am getting what you are saying. Could you see if this is what you were talking about?
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)2SO4.6H2O 392.16
n(Fe2+) = mass / Mr therefore: 9.794/55.8 = 0.1755197133
n(KMnO4)=0.1755197133/5 ( because n(KMnO4)= n(Fe2+) x 5)
therefore: n(KMnO4)=0.0351039426
C(Fe2+) =nx1000/v therefore: 0.1755197133x1000/250 = 0.7020788532
So now: m1 ( conc. of Fe2+) = 0.7020788532
v1 (Volume of Fe2+) = 25.00
v2 ( Volume of MnO4-) = 24.82
m2( Conc. Of KMno4-) = ???
Using the equation m2=m1v1/5v2
m2= 0.141434069
Part Two
Mass of compound used 0.200 g
Final Bruette reading 32.82 cm³
Final Volume of KMnO4 solution used 32.82 cm³
n( FeC2O4) = mass /mr = 0.200/143.8= 0.001390
3 x n(MnO4-)=5 x n( FeC2O4)
Therefore:
n(MnO4-) = 0.002318
c(MnO4-)= nx1000/32.80 = 0.072438
v( FeC2O4) =n x 1000/0.141434069 = 9.833702
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
Mr = 600 x 0.200 / 9.833702 x 0.072438 = 172.6705083
Therefore X = 1.059825
Sal.
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Mr of Fe(NH4)2SO4.6H2O 392.16[/color]
n(Fe2+) = mass / Mr therefore: 9.794/55.8 = 0.1755197133
Think again. Mass of what, molar mass of what?
Even after correcting that, check your math, 0.141434069 doesn't look correct to me (and not just because of blatant abuse of significant digits ;) ).
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Mr of Fe(NH4)2SO4.6H2O 392.16[/color]
n(Fe2+) = mass / Mr therefore: 9.794/55.8 = 0.1755197133
Think again. Mass of what, molar mass of what?
Even after correcting that, check your math, 0.141434069 doesn't look correct to me (and not just because of blatant abuse of significant digits ;) ).
Mr of Fe(NH4)2SO4.6H2O == 392.16
Mr of the Fe part is 152.06 Hoping this is the one. If yes, do/will the rest of my workings work out ? or is there a problem here or there?
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Mr of Fe(NH4)2SO4.6H2O == 392.16
Mr of the Fe part is 152.06 Hoping this is the one. If yes, do/will the rest of my workings work out ? or is there a problem here or there?
What you are dissolving? What have you weighted? 9.794g is mass of what?
Think this way: a car weights 1000kg. A person weights 70kg. Mass of cars with drivers is 3210 kg. How many drivers? Will you divide 3210kg by 70kg, or by 1070kg?
Now imagine Fe(NH4)2SO4.6H2O is a car with driver, and Fe is a driver. You have 9.794g of 'cars with drivers' - how many drivers?
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Mr of Fe(NH4)2SO4.6H2O == 392.16
Mr of the Fe part is 152.06 Hoping this is the one. If yes, do/will the rest of my workings work out ? or is there a problem here or there?
What you are dissolving? What have you weighted? 9.794g is mass of what?
Think this way: a car weights 1000kg. A person weights 70kg. Mass of cars with drivers is 3210 kg. How many drivers? Will you divide 3210kg by 70kg, or by 1070kg?
Now imagine Fe(NH4)2SO4.6H2O is a car with driver, and Fe is a driver. You have 9.794g of 'cars with drivers' - how many drivers?
If i don't get this right i will have embarrassed myself. Thanks for being patient dude.
Is it?
392.16/(9.794x55.8 ) = 0.7175778...
If this is correct, is the rest of my workings alright ? or am i a long way from home.
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Still wrong.
Seems like you have problems with simple calculation of concentration, and instead of trying to do it in a systematical way, you are guessing throwing numbers... sigh.
Let's start with something - hopefully - simpler. What is molar concentration of your Fe(NH4)2SO4.6H2O?
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Still wrong.
Seems like you have problems with simple calculation of concentration, and instead of trying to do it in a systematical way, you are guessing throwing numbers... sigh.
Let's start with something - hopefully - simpler. What is molar concentration of your Fe(NH4)2SO4.6H2O?
Sorry man. I'm very new to the concept of calculations in chemistry. Hope this is right.
N=mass /Mr = 9.794/392.16 = 0.0249745
C=m/vx1000 = 0.0249745 x1000 / 250 = 0.099898 therefore = 0.1
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N=mass /Mr = 9.794/392.16 = 0.0249745
C=m/vx1000 = 0.0249745 x1000 / 250 = 0.099898 therefore = 0.1
Please, be consistent. First, you calculated some N, then, in the next formula, you used some m. It is not easy to followe, you know.
At least your concentration is OK (almost - I would put it as 0.0999).
Now, if concentration of Fe(NH4)2(SO4)2.6H2O is 0.1, what is concentration of Fe2+? Before you will start to multiply/divide anything by anything, just think - how many moles Fe2+ per mole of Fe(NH4)2(SO4)2.6H2O?
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N=mass /Mr = 9.794/392.16 = 0.0249745
C=m/vx1000 = 0.0249745 x1000 / 250 = 0.099898 therefore = 0.1
Please, be consistent. First, you calculated some N, then, in the next formula, you used some m. It is not easy to followe, you know.
At least your concentration is OK (almost - I would put it as 0.0999).
Now, if concentration of Fe(NH4)2(SO4)2.6H2O is 0.1, what is concentration of Fe2+? Before you will start to multiply/divide anything by anything, just think - how many moles Fe2+ per mole of Fe(NH4)2(SO4)2.6H2O?
number of moles of salt = 9.794/392.16 = 2.497 x 10^(-2)
Since each molecule of the salt contains one ferrous ion, the number of moles of that is exactly the same right?
n(Fe2+) = 0.0249745
I am unsure as to what i put for the volume in the following equation: and i hate to guess but i hope the following is alright:
C( Fe2+) =nx1000/v = 0.0249745x1000 / 50.00 =0.4994 which is 0.5
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Since each molecule of the salt contains one ferrous ion, the number of moles of that is exactly the same right?
So, if th enumber of moles is exactly the same, how can the concentration be different?
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Since each molecule of the salt contains one ferrous ion, the number of moles of that is exactly the same right?
So, if th enumber of moles is exactly the same, how can the concentration be different?
The concentration is the same right?
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Hey , i was wondering...Do we even need the concentration of Fe2+? When i find the number of moles, i just divide it by 5 to give me the number of moles of KMnO4- and then find the concentration that way without having to use "the equation m2=m1v1/5v2 ".
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
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I haven't read through the thread but that calculation on the last post looks correct, although I haven't type it out on my calculator...
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I haven't read through the thread but that calculation on the last post looks correct, although I haven't type it out on my calculator...
Thanks. Good to know.
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If part 1 of my calculations are correct then would you please have a look over to the following:
Part Two
Mass of compound used 0.200 g
Final Bruette reading 32.82 cm³
Final Volume of KMnO4 solution used 32.82 cm³
I have to use the following to work out the second part of my calculations.
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
I know the following information.
W=0.200 g
m=0.2012 ( from part 1 of my calculations)
v=???
I am uncertain as to how to get the value for v. I don't know what the role of FeC2O4 is in the calculation, other than the following rule
3 x n(MnO4-)=5 x n( FeC2O4)
If you can point me in the right direction, it would be sound.
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I am just getting used to some of your notation (we use different ones in this part of the world you see ;D)
But why do I get the feeling you are making this a lot more complicated than it really is? Did you write this or your teacher give it to you?
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.
W=0.200 g
m=0.2012 ( from part 1 of my calculations)
v=???
Isn't 0.2012 the concentration of permanganate? You've got the volume and the concentration so you can work out the number of moles.
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But why do I get the feeling you are making this a lot more complicated than it really is? Did you write this or your teacher give it to you?
Yes it was. And i have to use that process to work out the Mr, of FeC2O4.xH2O and hence the value of x.
The theory behind my calculations, which have two parts to it, and i think i already have done the 1st part. The second part is holding me down.
http://img179.imageshack.us/img179/6379/12397571.jpg
Outline of the experiment: http://img293.imageshack.us/img293/7590/outline.jpg
Isn't 0.2012 the concentration of permanganate? You've got the volume and the concentration so you can work out the number of moles.
When i find the number of moles of permanganate using ( 0.2012x 32.82 /1000) , and then i would use the answer to get the number of moles of FeC2O2 using the following expression right?
3MnO4- = 5FeC2O2
If yes where do i go from here?
Sorry for the notations us English people use. But they work well for us :P
Thanks for your time.
Sal.
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Hence W/Mr=5mv/3000
So that Mr=600W/mv
Moreover, since Mr=143.87 + 18.02x
Then x= (Mr-143.87)/18.02
If I followed this, I end up with a negative number. ???
W/Mr=5mv/3000 :
0.2/Mr = (5 x 32.82 x 0.2012)/3000
0.2/Mr = 0.01100564
Mr = 18.1725
x= (Mr-143.87)/18.02:
In which case I'll get
x = (18.1725-143.87)/18.02
x = -6.98 ~ 7
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EDIT: Are you sure the mass used wasn't 2 grams? In which case x will be 2.
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Are you sure the mass used wasn't 2 grams? In which case x will be 2.
Yes, totally sure.
Here is the info for my second titration of FeC2O4 V MnO4-
Mass of compound used 0.200 g
Final Bruette reading 32.82 cm³
Final Volume of KMnO4 solution used 32.82 cm³
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When i find the number of moles of permanganate using ( 0.2012x 32.82 /1000) , and then i would use the answer to get the number of moles of FeC2O4 using the following expression right?
3MnO4- = 5FeC2O4
If yes where do i go from here?
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. ???
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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?
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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?
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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?
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?.
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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?
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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?
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So you diluted up to 250mL and then took a 25mL sample?
Aye. Using a pipette
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Then the rest should be correct! :D
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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.