Answer
Hematite: $69.94\%$
Magnetite: $72.34\%$
Siderite: $48.20\%$
The ore with highest iron content is magnetite, with $72.34\%$ iron.
Work Step by Step
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:
$\frac{Total-mass (Fe)}{Molar-Mass} \times 100\% = \frac{ 55.85}{ 115.86} \times 100\% = 48.20\%$
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Therefore, the ore with highest iron content is magnetite, with $72.34\%$ iron.
