Carry out a quantitative analysis of Aluminum tetraoxosulphate(vi)-(So4)3
To carry out a quantitative analysis of Aluminum tetraoxosulphate(VI), also known as Al2(SO4)3, we need to determine the percentage composition of sulfur in the compound.
Here's the step-by-step process of how to perform the quantitative analysis:
1. Determine the molar mass of Al2(SO4)3:
- Aluminum (Al) has a molar mass of 26.98 g/mol.
- Sulfur (S) has a molar mass of 32.07 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
The compound Al2(SO4)3 consists of two aluminum atoms, three sulfate (SO4) groups, and four oxygen atoms. Therefore, its molar mass is:
Molar mass of Al2(SO4)3 = (2 × 26.98 g/mol) + (3 × [(32.07 g/mol) + (4 × 16.00 g/mol)])
= 213.00 g/mol.
2. Calculate the molar mass of sulfur in Al2(SO4)3:
The molar mass of sulfur can be determined by dividing the molar mass of sulfur by the molar mass of Al2(SO4)3 and multiplying by 100%.
Molar mass of sulfur = [(3 × (32.07 g/mol)) / 213.00 g/mol] × 100%
Simplifying: Molar mass of sulfur = (96.21 g/mol / 213.00 g/mol) × 100%
Molar mass of sulfur = 0.452 × 100%
Molar mass of sulfur = 45.2%.
Therefore, the quantitative analysis shows that sulfur constitutes approximately 45.2% of the total mass of Aluminum tetraoxosulphate(VI) (Al2(SO4)3).