Question

The mole fraction of forsterite in olivine with MgO = 29.17 weight %, FeO = 34.65 weight % and $SiO_2$ = 36.18 weight % is _______ . (Round off to two decimal places) (Use molecular weight, in g/mol, of MgO = 40.31, FeO = 71.85 and $SiO_2$ = 60.00)

Answer 1

Correct Answer: 0.58

Calculating the Mole Fraction of Forsterite

Step 1: Calculate the Moles of Each Component

First, we calculate the moles of each component using the given weight percentages and molar masses.

• Moles of MgO:

The formula to calculate moles of MgO is:

\[ \text{moles of MgO} = \frac{\text{weight\% of MgO}}{\text{molar mass of MgO}} = \frac{29.17}{40.31} = 0.723 \]

• Moles of FeO:

The formula to calculate moles of FeO is:

\[ \text{moles of FeO} = \frac{\text{weight\% of FeO}}{\text{molar mass of FeO}} = \frac{71.85}{40.31} = 0.482 \]

• Moles of SiO₂:

The formula to calculate moles of SiO₂ is:

\[ \text{moles of SiO₂} = \frac{\text{weight\% of SiO₂}}{\text{molar mass of SiO₂}} = \frac{36.18}{60.00} = 0.603 \]

Step 2: Calculate the Mole Fraction of Forsterite

The mole fraction of Forsterite (\(X_{\text{forsterite}}\)) is calculated as follows:

\[ X_{\text{forsterite}} = \frac{\text{moles of MgO}}{\text{moles of MgO} + \text{moles of FeO} + \text{moles of SiO₂}} = \frac{0.723}{0.723 + 0.482 + 0.603} = \frac{0.723}{1.808} \approx 0.58 \]

Final Answer:

The mole fraction of Forsterite is: 0.58 (rounded to two decimal places).