Moles of FeSO4 = 0.004736842
This value quoted for moles of FeSO4 is quoted to higher degree of accuracy than I could calculate so Instead I will quote this value to 3 significant figures. Therefore:
Moles of FeSO4 = 0.00474 correct to 3 significant figures
I now need to calculate the moles of H2O using the Mr of H2O and the mass
Moles of H2O = Mass
Mr
Moles of H2O = 0.652
(2 x 1) +16
Moles of H2O = 0.0362 correct to 3 significant figures
To calculate the formula of FeSO4. xH2O the smallest number of the two values for the moles of FeSO4 and H2O needs to be made to 1 and the ratio then calculated. To do this both numbers are divided by the smallest number.
FeSO4 H2O
0.00474 0.0362
0.00474 0.00474
= 1 = 7.64
The ratio of FeSO4 to H2O must be in whole numbers as reacting ratios need to be integers. From this experiment I can therefore say the ratio is 1:8 so in FeSO4. xH2O x = 8 and the formula is;
FeSO4. 8H2O.
Method 2
Results
My average titration was taken from the 2nd and 3rd titrations and not the 1st as the 1st was only a test to see where roughly the end point was.
Analysis
To calculate the formula of the FeSO4.xH2O I need to first calculate the moles of Fe2+. To do this I need to first calculate the moles of MnO4-. I can do this using the following equation.
Moles of MnO4- = volume x concentration
1000
I know the volume of the MnO4- as this is what I had to find out with the titration. The concentration of this solution was given as 0.01 mol dm-3. The equation is therefore:
Moles of MnO4- = 22.25 x 0.01
1000
Moles of MnO4- =0.0002225
This is the amount of moles of MnO4- However I need to find out the moles of Fe2+. I can now do this using the ratio of Fe2+ to MnO4- .
5Fe2+ : MnO4-
So moles of Fe2+ in 25cm3 of solution = 5 x 0.0002225
= 0.00113
So moles of Fe2+ in 25cm3 of solution = 0.00113
This is the number of moles of Fe2+ in 25cm3. I used 250 cm3 so the number of moles in 25cm3 needs to be multiplied by 10 to give the number of moles in 250cm3 of the Fe2+ solution.
0.00113 x 10 = 0.0113
So in the Fe2+ solution there were 0.0113 moles.
I need to convert the value for the number of moles of Fe2+ into the mass of FeSO4 used. This can be done using the following formula:
Moles = mass
Mr
This needs to be rearranged to give;
Mass = moles x Mr
I already know the moles of FeSO4 . The Mr is the total of the atomic masses for all of the atoms in FeSO4 . This means my equation is this
Mass = 0.0113 x (55.8 + 32.1 + (16 x 4))
Mass = 1.71647g correct to 3 significant figures.
The value for the mass is quoted more accurate than I could measure to. I will therefore make the value for the mass correct to 3 significant figures
Mass of FeSO4 = 1.72g correct to 3 significant figures.
I now have the mass of FeSO4 used.
For the first step of the experiment I had to weigh the mass of FeSO4. xH2O. This weighed 3.00g. Therefore I can work out the mass of the H2O by subtracting the mass of the FeSO4 from the mass of FeSO4. xH2O.
Mass of H2O = mass of FeSO4. xH2O – mass of FeSO4
Mass of H2O = 3.00 – 1.72
Mass of H2O = 1.28g
Now that I have the mass of the H2O I can calculate the moles of H2O using the equation:
Moles = Mass
Mr
Moles = 1.28
((1x2) +16)
Moles = 0.0713 correct to 3 significant figures
Number of moles of H2O = 0.0713
I also know the moles of FeSO4 from previous equations earlier.
Number of moles of FeSO4 = 0.0113
To calculate the formula of FeSO4. xH2O the smallest number of the two values for the moles of FeSO4 and H2O needs to be made to 1 and the ratio then calculated. To do this both numbers are divided by the smallest number.
FeSO4 H2O
0.0113 0.0713
0.0113 0.0113
= 1 = 6.31
The ratio of FeSO4 to H2O must be in whole numbers as reacting ratios need to be integers. From this experiment I can therefore say the ratio is 1:6 so in FeSO4. xH2O x = 6 and the formula is;
FeSO4. 6H2O.
