## Chemistry: Molecular Approach (4th Edition)

Hematite: $69.94\%$ Magnetite: $72.34\%$ Siderite: $48.20\%$ The ore with highest iron content is magnetite, with $72.34\%$ iron.
Hematite: 1. Calculate the molar mass of: $Fe_2O_3$ Molar Mass ($Fe_2O_3$): 55.85* 2 + 16.00* 3 = 159.7g/mol 2. Calculate the total mass of $Fe$ in one mole of this compound: Total Mass ($Fe$): $2 \times 55.85g = 111.7g$ 3. Find the mass percent composition: $\frac{Total-mass (Fe)}{Molar-Mass} \times 100\% = \frac{ 111.7}{ 159.7} \times 100\% = 69.94\%$ Magnetite: 1. Calculate the molar mass of: $Fe_3O_4$ Molar Mass ($Fe_3O_4$): 55.85* 3 + 16.00* 4 = 231.55g/mol 2. Calculate the total mass of $Fe$ in one mole of this compound: Total Mass ($Fe$): $3 \times 55.85g = 167.5g$ 3. Find the mass percent composition: $\frac{Total-mass (Fe)}{Molar-Mass} \times 100\% = \frac{ 167.5}{ 231.55} \times 100\% = 72.34\%$ Siderite: 1. Calculate the molar mass of: $FeCO_3$ Molar Mass ($FeCO_3$): 55.85* 1 + 12.01* 1 + 16.00* 3 = 115.86g/mol 2. Calculate the total mass of $Fe$ in one mole of this compound: Total Mass ($Fe$): $1 \times 55.85g = 55.85g$ 3. Find the mass percent composition: $\frac{Total-mass (Fe)}{Molar-Mass} \times 100\% = \frac{ 55.85}{ 115.86} \times 100\% = 48.20\%$ ----------- Therefore, the ore with highest iron content is magnetite, with $72.34\%$ iron.