To determine the molecular formula of benitoite, we proceed with the following steps:
1. Determine the molar masses of each element:
- Barium (Ba): 137.33 g/mol
- Titanium (Ti): 47.87 g/mol
- Silicon (Si): 28.085 g/mol
- Oxygen (O): 16.00 g/mol
2. Calculate moles of each element in 100 grams of the compound:
Since the percentage composition of the elements is given, we assume we have 100 grams of the compound. This scenario simplifies calculations as the percentage of each element corresponds directly to grams.
- Moles of Ba = [tex]$\frac{33.21 \, g}{137.33 \, \text{g/mol}}$[/tex]
- Moles of Ti = [tex]$\frac{11.58 \, g}{47.87 \, \text{g/mol}}$[/tex]
- Moles of Si = [tex]$\frac{20.38 \, g}{28.085 \, \text{g/mol}}$[/tex]
- Moles of O = [tex]$\frac{34.83 \, g}{16.00 \, \text{g/mol}}$[/tex]
3. Calculate the exact moles:
- Moles of Ba = [tex]$\frac{33.21}{137.33} \approx 0.2418$[/tex]
- Moles of Ti = [tex]$\frac{11.58}{47.87} \approx 0.2420$[/tex]
- Moles of Si = [tex]$\frac{20.38}{28.085} \approx 0.7261$[/tex]
- Moles of O = [tex]$\frac{34.83}{16.00} \approx 2.1769$[/tex]
4. Determine the mole ratio by dividing each by the smallest number of moles:
The smallest number of moles is approximately 0.2418.
- Ratio of Ba = [tex]$\frac{0.2418}{0.2418} \approx 1$[/tex]
- Ratio of Ti = [tex]$\frac{0.2420}{0.2418} \approx 1$[/tex]
- Ratio of Si = [tex]$\frac{0.7261}{0.2418} \approx 3$[/tex]
- Ratio of O = [tex]$\frac{2.1769}{0.2418} \approx 9$[/tex]
5. Simplify the ratios to the nearest whole numbers:
- Ba: 1
- Ti: 1
- Si: 3
- O: 9
Hence, the molecular formula of benitoite can be written as [tex]\( \text{Ba}_1\text{Ti}_1\text{Si}_3\text{O}_9 \)[/tex].
By matching this with the options provided:
A. [tex]\( \text{Ba}_1\text{Ti}_1\text{Si}_3\text{O}_9 \)[/tex]
Thus, the correct answer is:
A. [tex]\( \text{BaTiSi}_3\text{O}_9 \)[/tex]
