February 27, 2021, 01:31:58 AM
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Topic: Analysis of trying to calculate the Mr of the FeSO4.xH2O and thus the X value.  (Read 34696 times)

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Offline sssssaaaallmaan

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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.

Offline sssssaaaallmaan

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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?

Offline Borek

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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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Offline AWK

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Quote
Using ( it is a given expression )  3MnO4- = 5FeC2O2

??? should be 2
AWK

Offline Borek

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Quote
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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Offline sssssaaaallmaan

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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.

Offline Borek

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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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Offline sssssaaaallmaan

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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.

Offline Borek

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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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Offline sssssaaaallmaan

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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.

Offline sssssaaaallmaan

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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.

Offline Borek

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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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Offline sssssaaaallmaan

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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?                                     

Offline Borek

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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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Offline sssssaaaallmaan

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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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