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Chemisty Calculation A2 OCR Trends and patterns jan09

Need Help Q3b on trends and patterns jan09 if you have the paper question.

Question
________

Hydrated iron(II) ammonium sulphate has the formula Fe(NH4)2.xH2OFe(NH_4)_2.xH_2O

The value of xx in Fe(NH4)2.xH2OFe(NH_4)_2.xH_2O can be determined by its reaction with acidified mannganate(VII) ions.

β€’


Stage 1 - A sample of hydrated iron(II) ammonium sulphate of known mass is added to a conical flask.


β€’


Stage 2 - The sample has 25cm325cm^3 of 1moldmβˆ’11mol dm^-1 sulphuric acid added to it.


β€’

Stage 3 - The contents of the flask are titrated against 0.2moldmβˆ’1MnO4βˆ’0.2mol dm^{-1} MnO_4^-



In stage 2 the hydrated crystals dissolve

Fe(NH4)2.xH2O(s)+(aq)βˆ’βˆ’βˆ’>Fe2+(aq)+2SO42βˆ’(aq)+xH2O(l)Fe(NH_4)_2.xH_2O(s) + (aq) ---> Fe^{2+}(aq) + 2SO_4^{2-}(aq) + xH_2O(l)

In stage 3, the equation for the reaction between Fe2+Fe^{2+} and acidified MnO4βˆ’MnO_4^- is shown below.

5Fe2+(aq)+8H+(aq)+MnO4βˆ’(aq)βˆ’βˆ’βˆ’>Mn2+(aq)+5Fe3+(aq)+4H2O(l)5Fe^{2+}(aq) + 8H^+(aq) + MnO_4^-(aq) ---> Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)

In stage 1, a student used 0.907g of Fe(NH4)2.xH2OFe(NH_4)_2.xH_2O

In stage 3 the titre was 23.15cm^3

Calculate the relative formula mass of Fe(NH4)2.xH2OFe(NH_4)_2.xH_2O. Hence determine the value of x.
Reply 1
My calculations

Moles of MnO4- = 0.02 x (23.15/1000) = 0.000463mol dm-3

1:5 Ratio so

Moles of Fe = 5x0.000463 = 0.002315

The Mr of Fe(NH4)2(SO4)2 = 56 + 28 + 8 + 64 + 128 = 284g

284 x 0.002315 = 0.65746

0.907-0.65746 = 0.24954mol of h2o i think

now i get stuckkkkk
Reply 2
Kevlar
Need Help Q3b on trends and patterns jan09 if you have the paper question.

Question
________

Hydrated iron(II) ammonium sulphate has the formula Fe(NH4)2.xH2OFe(NH_4)_2.xH_2O

The value of xx in Fe(NH4)2.xH2OFe(NH_4)_2.xH_2O can be determined by its reaction with acidified mannganate(VII) ions.

β€’


Stage 1 - A sample of hydrated iron(II) ammonium sulphate of known mass is added to a conical flask.


β€’


Stage 2 - The sample has 25cm325cm^3 of 1moldmβˆ’11mol dm^-1 sulphuric acid added to it.


β€’

Stage 3 - The contents of the flask are titrated against 0.2moldmβˆ’1MnO4βˆ’0.2mol dm^{-1} MnO_4^-



In stage 2 the hydrated crystals dissolve

Fe(NH4)2.xH2O(s)+(aq)βˆ’βˆ’βˆ’>Fe2+(aq)+2SO42βˆ’(aq)+xH2O(l)Fe(NH_4)_2.xH_2O(s) + (aq) ---> Fe^{2+}(aq) + 2SO_4^{2-}(aq) + xH_2O(l)

In stage 3, the equation for the reaction between Fe2+Fe^{2+} and acidified MnO4βˆ’MnO_4^- is shown below.

5Fe2+(aq)+8H+(aq)+MnO4βˆ’(aq)βˆ’βˆ’βˆ’>Mn2+(aq)+5Fe3+(aq)+4H2O(l)5Fe^{2+}(aq) + 8H^+(aq) + MnO_4^-(aq) ---> Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)

In stage 1, a student used 0.907g of Fe(NH4)2.xH2OFe(NH_4)_2.xH_2O

In stage 3 the titre was 23.15cm^3

Calculate the relative formula mass of Fe(NH4)2.xH2OFe(NH_4)_2.xH_2O. Hence determine the value of x.


mole of Fe2+ = 5 x mol of Mn2+ = 23.15/1000 x 0.2 x 5 mol = 0.407 g/RMM

RMM = 17.581 g/mol

but the formula shows that is impossible, hmm....(you got iron that is definitely heavier than that)
Reply 3
shengoc
mole of Fe2+ = 5 x mol of Mn2+ = 23.15/1000 x 0.2 x 5 mol = 0.407 g/RMM

RMM = 17.581 g/mol

but the formula shows that is impossible, hmm....(you got iron that is definitely heavier than that)

im confused lol D;
Kevlar
Need Help Q3b on trends and patterns jan09 if you have the paper question.

Question
________

Hydrated iron(II) ammonium sulphate has the formula Fe(NH4)2.xH2OFe(NH_4)_2.xH_2O

The value of xx in Fe(NH4)2.xH2OFe(NH_4)_2.xH_2O can be determined by its reaction with acidified mannganate(VII) ions.

β€’


Stage 1 - A sample of hydrated iron(II) ammonium sulphate of known mass is added to a conical flask.


β€’


Stage 2 - The sample has 25cm325cm^3 of 1moldmβˆ’11mol dm^-1 sulphuric acid added to it.


β€’

Stage 3 - The contents of the flask are titrated against 0.2moldmβˆ’1MnO4βˆ’0.2mol dm^{-1} MnO_4^-



In stage 2 the hydrated crystals dissolve

Fe(NH4)2.xH2O(s)+(aq)βˆ’βˆ’βˆ’>Fe2+(aq)+2SO42βˆ’(aq)+xH2O(l)Fe(NH_4)_2.xH_2O(s) + (aq) ---> Fe^{2+}(aq) + 2SO_4^{2-}(aq) + xH_2O(l)

In stage 3, the equation for the reaction between Fe2+Fe^{2+} and acidified MnO4βˆ’MnO_4^- is shown below.

5Fe2+(aq)+8H+(aq)+MnO4βˆ’(aq)βˆ’βˆ’βˆ’>Mn2+(aq)+5Fe3+(aq)+4H2O(l)5Fe^{2+}(aq) + 8H^+(aq) + MnO_4^-(aq) ---> Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)

In stage 1, a student used 0.907g of Fe(NH4)2.xH2OFe(NH_4)_2.xH_2O

In stage 3 the titre was 23.15cm^3

Calculate the relative formula mass of Fe(NH4)2.xH2OFe(NH_4)_2.xH_2O. Hence determine the value of x.


Moles of KMnO4 = 0.2 x 0.02315 = 0.00463
ratio of reaction = 1:5
therefore moles of Fe2+ = 5 x 0.00463 x 5 = 0.02315

the formula is Fe(NH4)2SO4.xH2O
therefore moles of Fe2+ = moles of Fe(NH4)2SO4
relative mass of Fe(NH4)2SO4 = 188
therefore 0.02315 moles of Fe(NH4)2SO4 alone has a mass of 4.3522 g

this is clearly impossible as only 0.907 g were used!!!

If the concentration of the KMnO4 were 0.02 mol dm-3 (much more likely) then the mass would be 0.43522 g
this leaves 0.907 - 0.43522 = 0.47178 g of water = 0.02621 moles

thus the mole ratio of iron to water = 0.02315 to 0.02621

this suggests that the formula with the water of crystallisation is Fe(NH4)2SO4.H2O

This is an unsatisfactory answer leading me to believe that the question has a typo or the OP has made a mistake...
Reply 5
Kevlar
im confused lol D;


Based on my calculations, supposed that the question was correct, I get a ridiculous answer, that means there must be a typo with the question.
I'm not sure if there's a mistake in the question but in the one I was given, the equation for stage 2 is:

Fe(NH4)2(SO4)2.xH2O(s) + (aq) β†’ Fe2+(aq) + 2NH4+(aq) + 2SO42-(aq) + xH2O(l)


From equation for stage 3, ratio of Fe2+ :MnO4- is 5:1

n(MnO4-)= 0.02 x (23.15x10-3) = 4.63x10-4

n(Fe2+ )=4.63x10-4 x 5 = 2.315x10-3

From equation for stage 2, ratio of Fe2 :Fe(NH4)2(SO4)2.xH2O is 1:1

n(Fe(NH4)2(SO4)2.xH2O) = 2.315x10-3



Mr(Fe(NH4)2(SO4)2.xH2O)= 0.907/2.315x10-3 = 391.8

Mr(Fe(NH4)2(SO4)2)= 284

Mr(xH2O)=391.8-284=107.8

x(16 + 2(1))=107.8

x=6 , which is the answer given in the mark scheme.
(edited 5 years ago)
Reply 7
Divide by Mr of water to find the moles of h2o= 0.0139. The moles of Fe is smaller so divide 0.0139 by 2.315Γ—10-3. This gives you the ratio 1:6, from the hydrated and dehydrated salt stuff. So x=6
Original post by freyaneve
I'm not sure if there's a mistake in the question but in the one I was given, the equation for stage 2 is:

Fe(NH4)2(SO4)2.xH2O(s) + (aq) β†’ Fe2+(aq) + 2NH4+(aq) + 2SO42-(aq) + xH2O(l)


From equation for stage 3, ratio of Fe2+ :MnO4- is 5:1

n(MnO4-)= 0.02 x (23.15x10-3) = 4.63x10-4

n(Fe2+ )=4.63x10-4 x 5 = 2.315x10-3

From equation for stage 2, ratio of Fe2 :Fe(NH4)2(SO4)2.xH2O is 1:1

n(Fe(NH4)2(SO4)2.xH2O) = 2.315x10-3



Mr(Fe(NH4)2(SO4)2.xH2O)= 0.907/2.315x10-3 = 391.8

Mr(Fe(NH4)2(SO4)2)= 284

Mr(xH2O)=391.8-284=107.8

x(16 + 2(1))=107.8

x=6 , which is the answer given in the mark scheme.


Thank you

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