# 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 $Fe(NH_4)_2.xH_2O$

The value of $x$ in $Fe(NH_4)_2.xH_2O$ can be determined by its reaction with acidified mannganate(VII) ions.

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

Stage 2 - The sample has $25cm^3$ of $1mol dm^-1$ sulphuric acid added to it.

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

In stage 2 the hydrated crystals dissolve

$Fe(NH_4)_2.xH_2O(s) + (aq) ---> Fe^{2+}(aq) + 2SO_4^{2-}(aq) + xH_2O(l)$

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

$5Fe^{2+}(aq) + 8H^+(aq) + MnO_4^-(aq) ---> Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)$

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

In stage 3 the titre was 23.15cm^3

Calculate the relative formula mass of $Fe(NH_4)_2.xH_2O$. Hence determine the value of x.
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
Kevlar
Need Help Q3b on trends and patterns jan09 if you have the paper question.

Question
________

Hydrated iron(II) ammonium sulphate has the formula $Fe(NH_4)_2.xH_2O$

The value of $x$ in $Fe(NH_4)_2.xH_2O$ can be determined by its reaction with acidified mannganate(VII) ions.

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

Stage 2 - The sample has $25cm^3$ of $1mol dm^-1$ sulphuric acid added to it.

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

In stage 2 the hydrated crystals dissolve

$Fe(NH_4)_2.xH_2O(s) + (aq) ---> Fe^{2+}(aq) + 2SO_4^{2-}(aq) + xH_2O(l)$

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

$5Fe^{2+}(aq) + 8H^+(aq) + MnO_4^-(aq) ---> Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)$

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

In stage 3 the titre was 23.15cm^3

Calculate the relative formula mass of $Fe(NH_4)_2.xH_2O$. 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)
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 $Fe(NH_4)_2.xH_2O$

The value of $x$ in $Fe(NH_4)_2.xH_2O$ can be determined by its reaction with acidified mannganate(VII) ions.

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

Stage 2 - The sample has $25cm^3$ of $1mol dm^-1$ sulphuric acid added to it.

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

In stage 2 the hydrated crystals dissolve

$Fe(NH_4)_2.xH_2O(s) + (aq) ---> Fe^{2+}(aq) + 2SO_4^{2-}(aq) + xH_2O(l)$

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

$5Fe^{2+}(aq) + 8H^+(aq) + MnO_4^-(aq) ---> Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)$

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

In stage 3 the titre was 23.15cm^3

Calculate the relative formula mass of $Fe(NH_4)_2.xH_2O$. 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...
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) 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)
Divide by Mr of water to find the moles of h2o= 0.0139. The moles of Fe is smaller so divide 0.0139 by 2.315×10-3. 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) 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