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Topic: Analysis of trying to calculate the Mr of the FeSO4.xH2O and thus the X value.  (Read 34697 times)

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Borek

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Re: Analysis of trying to calculate the Mr of the FeSO4.xH2O and thus the X value.
« Reply #15 on: April 21, 2009, 10:08:19 AM »
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.

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sssssaaaallmaan

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Re: Analysis of trying to calculate the Mr of the FeSO4.xH2O and thus the X value.
« Reply #16 on: April 21, 2009, 01:32:55 PM »
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.

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

Borek

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Re: Analysis of trying to calculate the Mr of the FeSO4.xH2O and thus the X value.
« Reply #17 on: April 21, 2009, 03:01:14 PM »
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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sssssaaaallmaan

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Re: Analysis of trying to calculate the Mr of the FeSO4.xH2O and thus the X value.
« Reply #18 on: April 21, 2009, 05:26:21 PM »
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

Borek

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Re: Analysis of trying to calculate the Mr of the FeSO4.xH2O and thus the X value.
« Reply #19 on: April 21, 2009, 05:52:45 PM »
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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sssssaaaallmaan

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Re: Analysis of trying to calculate the Mr of the FeSO4.xH2O and thus the X value.
« Reply #20 on: April 21, 2009, 06:02:10 PM »
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?

sssssaaaallmaan

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Re: Analysis of trying to calculate the Mr of the FeSO4.xH2O and thus the X value.
« Reply #21 on: April 21, 2009, 06:39:52 PM »
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
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
« Last Edit: April 21, 2009, 07:08:28 PM by sssssaaaallmaan »

UG

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Re: Analysis of trying to calculate the Mr of the FeSO4.xH2O and thus the X value.
« Reply #22 on: April 21, 2009, 08:46:22 PM »
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...

sssssaaaallmaan

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Re: Analysis of trying to calculate the Mr of the FeSO4.xH2O and thus the X value.
« Reply #23 on: April 21, 2009, 09:04:34 PM »
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.

sssssaaaallmaan

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Re: Analysis of trying to calculate the Mr of the FeSO4.xH2O and thus the X value.
« Reply #24 on: April 21, 2009, 10:08:24 PM »
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 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.

UG

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Re: Analysis of trying to calculate the Mr of the FeSO4.xH2O and thus the X value.
« Reply #25 on: April 21, 2009, 10:40:28 PM »
I am just getting used to some of your notation (we use different ones in this part of the world you see  )
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?

Quote
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.

sssssaaaallmaan

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Re: Analysis of trying to calculate the Mr of the FeSO4.xH2O and thus the X value.
« Reply #26 on: April 21, 2009, 11:16:06 PM »
Quote
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

Quote
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

Sal.

UG

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Re: Analysis of trying to calculate the Mr of the FeSO4.xH2O and thus the X value.
« Reply #27 on: April 21, 2009, 11:17:45 PM »
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

--------
EDIT: Are you sure the mass used wasn't 2 grams? In which case x will be 2.

sssssaaaallmaan

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Re: Analysis of trying to calculate the Mr of the FeSO4.xH2O and thus the X value.
« Reply #28 on: April 21, 2009, 11:34:25 PM »
Quote
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 Volume of KMnO4 solution used       32.82 cm³

UG

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Re: Analysis of trying to calculate the Mr of the FeSO4.xH2O and thus the X value.
« Reply #29 on: April 22, 2009, 12:09:11 AM »
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.