Determination of the formula of Hydrated Iron (II) Sulphate crystals

Authors Avatar

Determination of the formula of Hydrated Iron (II) Sulphate crystals

Analysis

Method one.
In order to find out the ratio of H2O to FeSO4, the mass of water needs to be calculated, which is done by weighing the iron sulphate before and after the heating. This will tell us the weight that was lost and thus the weight of the water.
 
Grams (g)

Empty crucible 12.01
Crucible and Iron (II) sulphate crystals 13.48
Mass of Iron (II) sulphate crystals (2-1) (13.48-12.01) 1.47
Crucible and Anhydrous Iron (II) sulphate 12.82
Mass of Anhydrous Iron (II) sulphate (5-1) (12.82-12.01) 0.81
Mass of water (13.48-12.82) 0.66

Repeat
7. Crucible and Anhydrous Iron (II) sulphate (after re-heating) 12.80
8. Mass of Anhydrous Iron (II) sulphate (after re-heating) (7-1) (12.80-12.01) 0.79
9. Mass of water lost during repeat (4-7) (12.82-12.80) 0.02
10. Mass of water lost altogether (2-7) (13.48-12.80) 0.68
Finding the molar ratio between FeSO
4 and H2O requires finding the empirical formula using their masses. To find the number of moles of each of the compounds, the equation

moles= mass/molar mass is used.

Finding the Atomic masses of FeSO4 and H2O:
Atomic Masses Fe= 55.8 S=32.1 O4= 16x4
H2=1x2 O=16
Molar mass of FeSO
4= 55.8 + 32.1 + (16x4)
= 151.9
Molar mass of H
2O = (1x2) + 16
=18


FeSO
4 H2O
Mass
 0.79g 0.68
 
Moles 0.79/151.9= 0.005 0.68/18=0.038
Finding the Ratio: 0.005/0.005 0.038/0.005
=1 =7.4
Ratio of FeSO
4 to H2O is: 1:7
 
 
Therefore, the formula for Iron(II) Sulphate in this experiment is;

FeSO4.7H2O

Thus, the overall equation for the reaction was:
FeSO4.7H2O(s) -> FeSO4(s) + 7H2O(l)
Observations and comments

As heated, the pale green crystals became whiter and joined together to form one block, which became grey around the sides of the crucible. This was because this part of the FeSO4 closest to the heat and so was beginning to be decomposed.
When the ‘anhydrous’ iron sulphate taken out of the crucible after procedure completed;
red on the underneath of the iron sulphate. This suggests that some of the Fe2+ had been decomposed to iron oxide and oxidised to fe3+. When the substance turns red, this is what has taken place:

Fe(II)SO4.7H20(s) -> FeSO4(s) + H2O(l)

2Fe(II)SO4(s) -> Fe2(III)O3(s) + SO2(g) + SO3(g)

The Fe2O3 is a red solid.

Method two
Weight of FeSO4.xH2O: 2.92g
Results of Titre: Rough Titre 1 Titre 2 Titre 3
First reading (
cm³) 0.00 0.00 0.00 0.00
Second reading (cm³) 28.90 28.45 28.35 28.40
Titre (cm³) 21.10 21.55 21.65 21.60

Chosen titres to find average titre from are
: 21.55cm³, 21.65 cm³ and 21.60cm³
To find the average titre of KMnO4 used to oxidise the Fe2+ ions, add the three titres together that are within 0.10cm³ difference of each other and divide by three:
Average titre:= (21.55 + 21.65 + 21.60)/ 3
= 21.60cm³
Therefore, it requires an average volume of 21.60cm³ of 0.01 mol dm-³ KMnO4 to oxidise all the Fe2+ ions in 25cm³ of FeSO4.xH2O (dissolved in 1mol dm-³ concentration of H2SO4).
To find the ratio of FeSO4 to H
2O requires firstly to calculate the number of moles of KMnO4 required to oxidise the Fe2+ ions. In order to calculate this, the equation

no. of moles= (volume in cm3/1000) x concentration

is used.
No. of moles of KMnO
4 = (21.60/1000) x 0.01
=
0.0002 moles
Redox half equations that occur during the reaction:
MnO4-+ 5e- + 8H+ -> Mn 2+ + 4H2O
Fe2+ -> Fe3+ + e-
Full balanced ionic half equation:
MnO4-(aq) + 5Fe2+(s) + 8H+ -> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
There are 5 Fe2+ ions to every one MnO4- ion in order to balance the number of electrons required by for each Mn2+ to form. The full ionic equation shows that to every 1 mole of MnO4- ions, there were 5 moles of Fe2+ ions. Therefore, to calculate the number of moles of Fe2+ ions; multiply the number of moles of KMnO4 by 5:
No. of moles of Fe2+ ions = 0.0002 x 5
=
0.001moles
However, that is only the number of moles of Fe2+ ions in 25cm³ but that is only a 10th of the number of ions in the total 250cm³ (250/25=10). To find the number of moles of Fe2+ ions in the whole of the solution multiply the number of moles of Fe2+ in 25cm³ by 10.
No. of moles of Fe2+ in 250cm³= 0.0010825 x 10
=
0.010 moles
0.01 moles of Fe2+ means the same as saying there is 0.01 moles of FeSO4. Therefore, in order to find the mass of the FeSO4, you can use the equation;

Join now!

mass= no. of moles x molar mass.

Mass of FeSO4= 0.01 x 151.9
=
1.52g
As the total mass of the FeSO4.xH2O was 2.92g in order to find the mass of water, you can subtract the weight of FeSO4:
Weight of water = 2.92-1.52
=
1.40g
Now that I have the mass of both the FeSO4 and H2O, I can find the ratio of the two by finding the number of moles of both.
FeSO4 H2O
Mass (g)
1.52 1.40
Moles= mass/molar mass 1.52/151.9 1.40/18
= 0.01 = 0.08
Ratio= both molar values divided by the smaller molar value of the two
0.01/0.01 0.08/0.01
=1 = 8
1:8

Therefore, the formula for Iron(II) sulphate in ...

This is a preview of the whole essay