To find the molecular formula of the mineral benitoite based on the given data, we need to go through the following steps:
### Step 1: Convert Percent Composition to Grams
Given the total molecular weight of benitoite as [tex]\( 413.44 \, \text{grams/mole} \)[/tex], we first convert the percent composition of each element into grams.
- Ba (Barium):
[tex]\[ \frac{33.21}{100} \times 413.44 = 137.303 \, \text{grams} \][/tex]
- Ti (Titanium):
[tex]\[ \frac{11.58}{100} \times 413.44 = 47.876 \, \text{grams} \][/tex]
- Si (Silicon):
[tex]\[ \frac{20.38}{100} \times 413.44 = 84.259 \, \text{grams} \][/tex]
- O (Oxygen):
[tex]\[ \frac{34.83}{100} \times 413.44 = 144.001 \, \text{grams} \][/tex]
### Step 2: Calculate Moles of Each Element
Next, we calculate the moles of each element using their atomic weights.
- Ba:
[tex]\[ \frac{137.303 \, \text{grams}}{137.327 \, \text{grams/mole}} = 0.9998 \, \text{moles} \][/tex]
- Ti:
[tex]\[ \frac{47.876 \, \text{grams}}{47.867 \, \text{grams/mole}} = 1.0002 \, \text{moles} \][/tex]
- Si:
[tex]\[ \frac{84.259 \, \text{grams}}{28.085 \, \text{grams/mole}} = 3.0001 \, \text{moles} \][/tex]
- O:
[tex]\[ \frac{144.001 \, \text{grams}}{15.999 \, \text{grams/mole}} = 9.0006 \, \text{moles} \][/tex]
### Step 3: Normalize the Moles
We normalize the number of moles by dividing each by the smallest number of moles calculated. In this case, the smallest value is approximately 0.9998 moles for Ba.
- Ba:
[tex]\[ \frac{0.9998}{0.9998} = 1.0 \][/tex]
- Ti:
[tex]\[ \frac{1.0002}{0.9998} = 1.0004 \][/tex]
- Si:
[tex]\[ \frac{3.0001}{0.9998} = 3.0007 \][/tex]
- O:
[tex]\[ \frac{9.0006}{0.9998} = 9.0018 \][/tex]
### Step 4: Determine the Empirical Formula
We round the normalized ratios to the nearest whole numbers to determine the empirical formula.
- Ba: 1
- Ti: 1
- Si: 3
- O: 9
Therefore, the empirical formula for benitoite is [tex]\( \text{BaTiSi}_3\text{O}_9 \)[/tex].
### Conclusion
Given the calculations, the molecular formula of benitoite is:
A. [tex]\( \mathbf{BaTiSi_3O_9} \)[/tex]
