* Molar mass of KMnO4 = 158.034 g/mol
* Relative atomic mass of manganese (Mn) = 54.938 g/mol
1. Determine the number of moles of manganese in one mole of potassium permanganate:
1 mole of KMnO4 contains 1 mole of Mn.
2. Calculate the mass of manganese in one mole of potassium permanganate:
Mass of Mn = 1 mole of Mn × Relative atomic mass of Mn
= 1 mole × 54.938 g/mol
= 54.938 g
3. Calculate the amount percentage of manganese in potassium permanganate:
Amount percentage of Mn = (Mass of Mn / Molar mass of KMnO4) × 100
= (54.938 g / 158.034 g/mol) × 100
= 34.71%
Therefore, the amount percentage of manganese in potassium permanganate (KMnO4) is approximately 34.71%.
