2.7 Sample Calculations

Atoms, Formulas, Atomic Weights, Formula Weights

These basic chemical relationships do not depend on the particular rock.

Atomic Weight Si:     aw_Si = 28.085 g/mole
Atomic Weight Fe:     aw_Fe = 55 g/mole
Atomic Weight O:     aw_O = 15.999 g/mole

Moles of Si:     mol_Si = mass_Si/aw_Si
Moles of Fe:     mol_Fe = mass_Fe/aw_Fe
Moles of O:     mole_O = mass_O/aw_O

Quartz (Qz) formula:     SiO2
Magnetite (Mt) formula:     Fe3O4

Formula Weight SiO2:     fw_SiO2 = 60.093 g/mole = aw_Si + 2*aw_O
Formula Weight Fe3O4:     fw_Fe3O4 = 231.531 g/mole = 3*aw_Fe + 4*aw_O

Moles of SiO2:     mol_SiO2 = mass_SiO2/fw_SiO2
Moles of Fe3O4:     mol_Fe3O4 = mass_Fe3O4/fw_Fe3O4
Density of Quartz:     den_Qz = 2.65 g/cm3
Density of Magnetite:     den_Mt = 5.175 g/cm3

Constraints

These constraints apply if the rock consists of only the minerals magnetite and quartz.

Wt%_Fe3O4 + Wt%_SiO2 = 100
Mole%_SiO2 = 100*(Wt%_SiO2/fw_SiO2)/((Wt%_SiO2/fw_SiO2)
+ (Wt%_Fe3O4/fw_Fe3O4)))
Mole%_Fe3O4 + Mole%_SiO2 = 100
Atom%_Fe3O4 + Atom%_SiO2 = 100
Oxygen%_Fe3O4 + Oxygen%_SiO2 = 100

Moles of Si = Moles of SiO2
Moles of Fe = 3*(Moles of Fe3O4)
Moles of O = 4*(Moles of Fe3O4) + 2*(Moles of SiO2)

Wt%_Si = ((Wt%_SiO2)*(aw_Si/fw_SiO2))
Wt%_Fe = ((Wt%_Fe3O4)*(3*aw_Fe/fw_Fe3O4))
Wt%_O = 100 - (Wt%_Si + Wt%_Fe)

Mole%_Si = 100*((Mole%_SiO2)/(3*Mole%_SiO2 + 7*Mole%_Fe3O4))
Mole%_Fe = 100*((3*Mole%_Fe3O4)/(3*Mole%_SiO2 + 7*Mole%_Fe3O4))
Mole%_O = 100 - (Mole%_Si + Mole%_Fe)

If the rock has no porosity, there is an additional constraint.

Volume%_Fe3O4 + Volume%_SiO2 = 100
Percentage Calclations

If the Wt%_SiO2 is known, the formulas and constraints listed above can be used to solve for many other percentage values.

Wt%_Fe3O4 = (100 - Wt%_SiO2)
Mole%_SiO2 = 100*(Wt%_SiO2/fw_SiO2)/((Wt%_SiO2/fw_SiO2)
+((100 - Wt%_SiO2)/fw_Fe3O4)))
Mole%_Fe3O4 = (100 - Mole%_SiO2)
Atom%_SiO2 = 100*(3*Mole%_SiO2)/((3*Mole%_SiO2)+(7*(100-Mole%_SiO2))
Atom%_Fe3O4 = 100 - Atom%_SiO2
Oxygen%_SiO2 = 100*(2*Mole%_SiO2)/((2*Mole%_SiO2)+(4*(100-Mole%_SiO2)))
Oxygen%_Fe3O4 = 100 - Oxygen%_SiO2

Wt%_Si = ((Wt%_SiO2)*(aw_Si/fw_SiO2))
Wt%_Fe = ((100 - Wt%_SiO2)*(3*aw_Fe/fw_Fe3O4))
Wt%_O = 100 - (Wt%_Si + Wt%_Fe)

Mole%_Si = 100*((Mole%_SiO2)/(3*Mole%_SiO2 + 7*Mole%_Fe3O4))
Mole%_Fe = 100*((3*Mole%_Fe3O4)/(3*Mole%_SiO2 + 7*Mole%_Fe3O4))
Mole%_O = 100 - (Mole%_Si + Mole%_Fe)

If we know that the rock has a mass of X grams, additional quantities may be calculated. Some of them are:

Mass of SiO2 (g) = X*(Wt%_SiO2)/100
Mass of Fe3O4 (g) = X*(Wt%_Fe3O4)/100 = X*(1-Wt%_SiO2)/100
Moles of SiO2 = X*(Wt%_SiO2)/(100*fw_SiO2) = X*(Wt%_SiO2)/(100*60.083)
Moles of Fe3O4 = X*(Wt%_Fe3O4)/(100*fw_Fe3O4)
= X*(1-Wt%_SiO2)/(100*231.531)
Volume of Quartz (cm3) = (Mass of SiO2)/(den_Qz) = X*(Wt%_SiO2)/(100*2.65)
Volume of Magnetite (cm3) = (Mass of Fe3O4)/(den_Mt)
= X*(1-Wt%_SiO2)/(100*5.175)
Mass of Si (g) = (Mass of SiO2)*(aw_Si/fw_SiO2)
= (X*(Wt%_SiO2)/100)*(28.085/60.083)
Mass of Fe (g) = (Mass of Fe3O4)*(3*aw_Fe/fw_Fe3O4)
= (X*(1-Wt%_SiO2)/100)*(3*55.875/231.531)

If the rock has a mass of 50 grams (X = 50) and a Wt%_SiO2 of 50%, these quantities are:

Mass of SiO2 (g) = 50*(50)/100 = 25 g
Mass of Fe3O4 (g) = 50*(50)/100 = 25 g
Moles of SiO2 = 50*(50)/(100*60.093) = 0.416
Moles of Fe3O4 = 50*(50)/(100*231.531) = 0.108
Volume of Quartz (cm3) = (50*(50)/100)/(2.65) = 9.43 cm3
Volume of Magnetite (cm3) = (50*(50)/100)/(5.175) = 4.83 cm3
Mass of Si (g) = (50*(50)/100)*(28.085/60.083) = 11.7 g
Mass of Fe (g) = (50*(50)/100)*(3*55.845/231.531) = 18.09 g