The mass percent composition of iron in hematite is approximately 69.9%, in magnetite it is 72.36%, and in siderite, it's 48.20%.To calculate the mass percent composition of iron in various iron ores, we use the formula for mass percent composition: percent composition = (mass of element in 1 mole of compound / molar mass of compound) x 100%.For hematite (Fe₂O₃):The molar mass of Fe2O3 is (2 x 55.85) + (3 x 16.00) = 159.7 g/mol. The mass of iron in one mole of Fe₂O₃ is 2 x 55.85 = 111.7 g. Therefore, the mass percent of iron in hematite is (111.7 / 159.7) x 100% = 69.9%.For magnetite (Fe₃O₄):The molar mass of Fe₃O₄ is (3 x 55.85) + (4 x 16.00) = 231.55 g/mol. The mass of iron in one mole of Fe₃O₄ is 3 x 55.85 = 167.55 g. Thus, the mass percent of iron in magnetite is (167.55 / 231.55) x 100% = 72.36%.For siderite (FeCO₃):The molar mass of FeCO₃ is 55.85 + 12.01 + (3 x 16.00) = 115.86 g/mol. The mass of iron in one mole of FeCO₃ is 55.85 g. Hence, the mass percent of iron in siderite is (55.85 / 115.86) x 100% = 48.20%....

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