Question

2) Chemical analyses of silicate minerals are given as weight percent oxides, to ensure that charges of cations and anions are balanced. It’s important to be able to interconvert mineral chemical formula and chemical analyses (see Klein pp. 99-104). Analysis of an olivine yields (all as weight percent): 36.63% SiO2, 30.96% MgO, 32.41% FeO; express this composition in terms of mole % forsterite (i.e., Fox, find x). You will need to calculate the formula weights of FeO and MgO. Please show your work, attached additional sheets if needed.

Answer 1

Let us consider 100 g of the olivine  

The contents of olivine are given below:

contents	mass in 100 g	molar mass	moles
SiO2	36.63	60.08	0.61
MgO	30.96	40.3	0.768
FeO	32.41	71.85	0.451

We Know that forsterite = Mg2SiO4 is a combination of MgO and SiO2

Let it be present as [Mg2SiO4 ]x = (Fo)x

Having a mass balance

mass of MgO + mass of SiO2 = {molar mass of Mg2SiO4 }x

30.96 + 36.63 = 140.68x

x = 67.59 / 140.68

= 0.48

Thus the formula of the compound is (Fo)0.48 (FeO)0.451

Divide by 0.451 we have (Fo)1.064 FeO

Thus the formula of the olivine expressed as forsterite is (Fo)1.064 . FeO

moles of Fo = 1.064

mole of FeO = 1

Total moles = 2.064

Therefore  the mole % of Fo = [1.064 / 2.064] x 100 = 51.55 %

Thus the mole% of Forsterite in olivine = 51.55 %