Iron(II) sulfate, is typically the source of iron in tablets prescribed to anemic patients, a condition where a person lacks enough healthy red blood cells. The tablets can be dissolved in dilute sulfuric acid to solvate the Fe2+ ions. To determine the percentage by mass of iron within the tablets, the Fe2+ ions can be oxidised to Fe3+ using potassium permanganate, where MnO4− is reduced to Mn2+.Calculate the percentage by mass of iron in a 500 mg tablet that requires 24.53 cm3 of 0.0100 mol dm−3 KMnO4 to titrate the iron sample. State your answer to 3 significant figures. M(Fe) = 55.85 g mol−1
Question
Iron(II) sulfate, is typically the source of iron in tablets prescribed to anemic patients, a condition where a person lacks enough healthy red blood cells. The tablets can be dissolved in dilute sulfuric acid to solvate the Fe2+ ions. To determine the percentage by mass of iron within the tablets, the Fe2+ ions can be oxidised to Fe3+ using potassium permanganate, where MnO4− is reduced to Mn2+.Calculate the percentage by mass of iron in a 500 mg tablet that requires 24.53 cm3 of 0.0100 mol dm−3 KMnO4 to titrate the iron sample. State your answer to 3 significant figures. M(Fe) = 55.85 g mol−1
Solution
First, we need to calculate the number of moles of KMnO4 used in the titration. We can do this using the formula:
Moles = Concentration x Volume
The concentration of KMnO4 is given as 0.0100 mol/dm^3 and the volume is given as 24.53 cm^3. However, we need to convert the volume from cm^3 to dm^3 by dividing by 1000:
Volume in dm^3 = 24.53 cm^3 / 1000 = 0.02453 dm^3
So, the moles of KMnO4 = 0.0100 mol/dm^3 x 0.02453 dm^3 = 0.0002453 mol
In the reaction between KMnO4 and Fe2+, the stoichiometry is 5:1. This means that for every mole of KMnO4, 5 moles of Fe2+ are oxidised. So, the moles of Fe2+ = 0.0002453 mol x 5 = 0.0012265 mol
Next, we need to calculate the mass of iron. We can do this using the formula:
Mass = Moles x Molar mass
The molar mass of iron (Fe) is given as 55.85 g/mol. So, the mass of iron = 0.0012265 mol x 55.85 g/mol = 0.0685 g
Finally, we need to calculate the percentage by mass of iron in the tablet. We can do this using the formula:
Percentage by mass = (Mass of iron / Total mass of tablet) x 100%
The total mass of the tablet is given as 500 mg. However, we need to convert this to grams by dividing by 1000:
Total mass in g = 500 mg / 1000 = 0.5 g
So, the percentage by mass of iron = (0.0685 g / 0.5 g) x 100% = 13.7% (to 3 significant figures)
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