Question

When a dunite comprising pure forsterite undergoes melting, the weight % of MgO in the melt is ........ (answer in one decimal place). Given molecular weights of \( \text{SiO}_2 = 60.08 \), \( \text{MgO} = 40.30 \).

Hint:

To calculate weight %, use the molar masses of the components and determine their proportion in the total mass.

Answer 1

Correct Answer: 57 - 57.5

Step 1: Understanding the composition of forsterite. 
Forsterite has the formula \( \text{Mg}_2\text{SiO}_4 \). When it melts, it will produce a melt primarily consisting of magnesium oxide (\( \text{MgO} \)) and silicon dioxide (\( \text{SiO}_2 \)).

Step 2: Calculate the molecular weights. 
- Molecular weight of forsterite \( \text{Mg}_2\text{SiO}_4 \): \[ \text{Molar mass of } \text{Mg}_2\text{SiO}_4 = 2 \times 40.30 \, \text{g/mol (Mg)} + 60.08 \, \text{g/mol (SiO}_2) = 140.68 \, \text{g/mol} \]

Step 3: Calculate the weight % of MgO in the melt. 
The weight of MgO in the melt will be contributed by the magnesium part of forsterite, which is 2 atoms of magnesium (80.60 g/mol from \( \text{Mg}_2 \)). The weight % of MgO in the melt is: \[ \text{Weight \% of MgO} = \frac{80.60 \, \text{g/mol}}{140.68 \, \text{g/mol}} \times 100 = 57.3\% \]

Step 4: Conclusion. 
The weight % of MgO in the melt is 57.3%.