SME
Mineralogy
Mineral chemistry
Use this for “An olivine is 40% SiO₂, 49% MgO and 11% FeO by weight. Find its formula and forsterite content.”
Table
Oxide
wt
(%)
M
(g/mol)
mol
O
per 4 O
SiO₂
40
60.08
0.6658
1.332
0.9861
MgO
49
40.3
1.216
1.216
1.801
FeO
11
71.84
0.1531
0.1531
0.2268
Sum
2.701
This is an example. Type your own values to start a new problem.
SiO₂
Silica
example · %
MgO
Magnesium oxide
example · %
FeO
Iron oxide
example · %
O
Oxygen per 100 g
worked out (not typed) · mol
f
Normalizing factor
worked out (not typed)
Si
Silicon per 4 O
worked out (not typed)
Mg
Magnesium per 4 O
worked out (not typed)
Fe
Iron per 4 O
worked out (not typed)
Fo
Forsterite
calculated · %
Formulas
O
=
2
×
SiO₂
60.08
+
MgO
40.30
+
FeO
71.84
2.7006
=
2
×
40
60.08
+
49
40.30
+
11
71.84
f
=
4
O
1.4812
=
4
2.7006
Si
=
SiO₂
60.08
×
f
0.9861
=
40
60.08
×
1.4812
Mg
=
MgO
40.30
×
f
1.8009
=
49
40.30
×
1.4812
Fe
=
FeO
71.84
×
f
0.2268
=
11
71.84
×
1.4812
Fo
=
100
×
Mg
Mg + Fe
88.8153
=
100
×
1.8009
1.8009 + 0.2268
Assumptions
•
Olivine has 4 oxygens per formula unit, (Mg,Fe)₂SiO₄; all iron is Fe²⁺.
•
Molar masses: SiO₂ 60.08, MgO 40.30 and FeO 71.84 g/mol; each SiO₂ brings 2 oxygens.
•
Mg and Fe share one site, so Mg + Fe should come to about 2 and Si to about 1.
Step-by-step
WE KNOW
SiO₂ = 40%, MgO = 49%, FeO = 11%
FIND
oxygen per 100 g (O), normalizing factor (f), silicon per 4 O (Si), magnesium per 4 O (Mg), iron per 4 O (Fe), forsterite (Fo)
Step 1 · Find oxygen per 100 g (O)
Use O = 2 × SiO₂ ÷ 60.08 + MgO ÷ 40.30 + FeO ÷ 71.84
Take 100 g: each oxide’s weight over its molar mass gives its moles, and SiO₂ carries two oxygens.
O
=
2
×
40
60.08
+
49
40.30
+
11
71.84
40
60.08
=
0.6658
mol
of
SiO₂
49
40.30
=
1.216
mol
of
MgO
11
71.84
=
0.1531
mol
of
FeO
O
=
2.700557
mol
Step 2 · Find normalizing factor (f)
Use f = 4 ÷ O
Scale the analysis so the oxygens come to 4, one formula unit of olivine.
f
=
4
2.700557
f
=
1.481176
Step 3 · Find silicon per 4 O (Si)
Use Si = SiO₂ ÷ 60.08 × f
Each SiO₂ brings one Si, so its moles scaled by the factor give the Si per 4 oxygens.
Si
=
40
60.08
×
1.481176
40
60.08
=
0.6658
mol
of
SiO₂
Si = 0.9861
Step 4 · Find magnesium per 4 O (Mg)
Use Mg = MgO ÷ 40.30 × f
Each MgO brings one Mg, so its moles scaled by the factor give the Mg per 4 oxygens.
Mg
=
49
40.30
×
1.481176
49
40.30
=
1.216
mol
of
MgO
Mg = 1.800934
Step 5 · Find iron per 4 O (Fe)
Use Fe = FeO ÷ 71.84 × f
Each FeO brings one Fe, so its moles scaled by the factor give the Fe per 4 oxygens.
Fe
=
11
71.84
×
1.481176
11
71.84
=
0.1531
mol
of
FeO
Fe = 0.226795
Step 6 · Find forsterite (Fo)
Use Fo = 100 × Mg ÷ (Mg + Fe)
Forsterite is the magnesium share of the shared site; the rest is fayalite.
Fo
=
100
×
1.800934
1.800934 + 0.226795
Fo
=
100
×
1.8009
2.028
Fo
=
180.0934
2.028
Fo = 88.8153% → Mg + Fe = 2.03, about 2: the formula is (Mg₁.₈₀Fe₀.₂₃)Si₀.₉₉O₄.
Check
Put the numbers back in. Both sides should match.
2.700557
=
2
×
40
60.08
+
49
40.30
+
11
71.84
✓
1.481176
=
4
2.700557
✓
0.9861
=
40
60.08
×
1.481176
✓
1.800934
=
49
40.30
×
1.481176
✓
0.226795
=
11
71.84
×
1.481176
✓
88.8153
=
100
×
1.800934
1.800934 + 0.226795
✓
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