Iron is found in Earth's crust as several iron compounds

Question

Iron is found in Earth's crust as several iron compounds. Calculate the mass (in kg) of each compound that contains 6.9*10^3 kg of iron.

Express your answer in kilograms to two significant figures.

Part A- Fe2O3 (hematite)

Part B- Fe3O4 (Magnetite)

Part C- FeCO3 (siderite

Answer 1

When attempting to follow this logic yourself, pay attention to the sig figs, your units, and your math.

1). Fe₂O₃ --> oxygen is -2, -2*3 = -6 . Therefore Fe is Fe(III), being +2, +2*3=6. Compound is not negative or positive therefore this checks out, -6 + 6 = 0.

Molar Mass Fe(III) => 55.845 g/mol , Molar Mass Oxygen = 16. 55.845 * 2 + 3 * 16= Molar Mass hematite=159.69 g/mol. Weight Fraction of Fe(III) per mole hematite = (55.845*@2)/(159.69) = approx 0.699.

Weight fraction O per mole Hematite = 16*3 / 159.69 = approx .300.

69.9/100 = (6.9* 10^3)/x, where x= weight of total hematite.

2). Magnetite consists of 1 Fe(II) and 2 Fe(III). ( Molar Mass Fe(II) + 2(Molar Mass Fe(III) + 4(Molar Mass O)) = Molar mass magnetite --> approx 231.533 g/mol. (Molar Mass Fe(II) + 2*Molar Mass Fe(III))/ 231.533 = weight fraction of iron in magnetite. (weight fraction of iron * 100)/100 = (6.9*10^3)/x, where x = weight of total magnetite.

3). Molar Mass of Fe(II) / Molar mass siderite = weight fraction of iron in siderite. (Weight fraction of Iron * 100)/100=(6.9*10^3)/x where x is total weight of siderite