25g of a sample of ferrous sulphate was dissolved in water contains dilute h2s04 and the volume made up to one litre. 25ml of this solution required 20ml of N/10 kmn04 solution for complete oxidation. Calculte the percentage of f2s04 7 h20 in the sample
My calculation
0.02 x 0.1 = 0.002
0.002 x 1000ml x 293g/mol / 25ml = 23.44
23.44/25g x 100% = 93.76% The right answer 88.96%
To calculate the percentage of FeSO4 · 7H2O in the sample, you need to use the following steps:
1. Calculate the number of moles of KMnO4 used:
- Volume of KMnO4 solution used = 20 mL
- Concentration of KMnO4 solution = N/10
- Number of moles of KMnO4 used = (20 mL) x (N/10)
2. Calculate the number of moles of FeSO4 · 7H2O in the sample:
- Number of moles of FeSO4 · 7H2O = Number of moles of KMnO4 used (by stoichiometry)
3. Calculate the mass of FeSO4 · 7H2O in the sample:
- Mass of FeSO4 · 7H2O = Number of moles of FeSO4 · 7H2O x molar mass of FeSO4 · 7H2O
4. Calculate the percentage of FeSO4 · 7H2O in the sample:
- Percentage of FeSO4 · 7H2O = (Mass of FeSO4 · 7H2O / Mass of sample) x 100%
You performed the calculation steps correctly, except for using a molar mass of 293 g/mol for FeSO4 · 7H2O. The correct molar mass should be 278.01 g/mol. With the correct molar mass, the calculation would be:
1. (20 mL) x (N/10) = 0.002 moles of KMnO4 used
2. 0.002 moles of KMnO4 = 0.002 moles of FeSO4 · 7H2O
3. Mass of FeSO4 · 7H2O = 0.002 moles x 278.01 g/mol = 0.556 g
4. Percentage of FeSO4 · 7H2O = (0.556 g / 25 g) x 100% = 2.224%
Therefore, the correct percentage of FeSO4 · 7H2O in the sample is 2.224%.
To calculate the percentage of Ferrous Sulphate Heptahydrate (FeSO4·7H2O) in the given sample, we can use the following steps:
1. We are given that 25 mL of the solution requires 20 mL of N/10 KMnO4 solution for complete oxidation. This indicates that the molar ratio between FeSO4·7H2O and KMnO4 is 5:1.
2. Next, let's calculate the number of moles of KMnO4 used. We know that N/10 KMnO4 means the solution is 1/10th the strength of a normal KMnO4 solution, which is N/1. So, the concentration of KMnO4 is 10 times lower than normal. Therefore, the number of moles of KMnO4 used can be calculated as follows:
moles_of_KMnO4 = (concentration_KMnO4) x (volume_KMnO4)
= (0.1 mol/L) x (20 ml / 1000 ml/L)
= 0.002 moles
3. Since the molar ratio between FeSO4·7H2O and KMnO4 is 5:1, the number of moles of FeSO4·7H2O present in 25 mL of the solution can be calculated by multiplying the number of moles of KMnO4 by 5:
moles_of_FeSO4·7H2O = 5 x 0.002 moles
= 0.01 moles
4. Now, we can calculate the molar mass of FeSO4·7H2O to convert the number of moles to grams. The molar mass is the sum of the atomic masses of its constituent elements:
molar_mass_FeSO4·7H2O = (molar_mass_Fe) + (molar_mass_S) + (molar_mass_O x 4) + (molar_mass_H x 14)
= (55.845 g/mol) + (32.06 g/mol) + (16 g/mol x 4) + (1.01 g/mol x 14)
= 278.01 g/mol
5. Using the equation moles = mass / molar mass, we can find the mass of FeSO4·7H2O present in 25 g of the sample:
mass_of_FeSO4·7H2O = (moles_of_FeSO4·7H2O) x (molar_mass_FeSO4·7H2O)
= 0.01 moles x 278.01 g/mol
= 2.7801 g
6. Finally, we can calculate the percentage of FeSO4·7H2O in the sample:
percentage_of_FeSO4·7H2O = (mass_of_FeSO4·7H2O / total_mass_of_sample) x 100%
= (2.7801 g / 25 g) x 100%
= 11.12%
Therefore, the correct percentage of FeSO4·7H2O in the sample is 11.12%.