Determine the empirical and molecular formula for chrysotile asbestos.
Chrysotile has the following percent composition: 28.03% Mg, 21.60% Si, 1.16%
H, and 49.21% O. The molar mass for chrysotile is 520.8 g/mol.
To determine the empirical formula for chrysotile asbestos, we need to find the ratio of the elements in the compound.
First, we need to find the number of moles for each element present in 100g of chrysotile:
- Mg: (28.03g Mg) / (24.31g/mol Mg) = 1.15 mol Mg
- Si: (21.60g Si) / (28.09g/mol Si) = 0.77 mol Si
- H: (1.16g H) / (1.01g/mol H) = 1.15 mol H
- O: (49.21g O) / (16.00g/mol O) = 3.08 mol O
Next, we need to divide the number of moles of each element by the smallest number of moles to find the simplest whole-number ratio:
- Mg: 1.15 mol / 0.77 mol ≈ 1.49
- Si: 0.77 mol / 0.77 mol = 1
- H: 1.15 mol / 0.77 mol ≈ 1.49
- O: 3.08 mol / 0.77 mol = 4
Next, we need to multiply these ratios by whole numbers to get the simplest whole-number ratio:
- Mg: 1.49 ≈ 1.5
- Si: 1
- H: 1.49 ≈ 1.5
- O: 4
Therefore, the empirical formula for chrysotile asbestos is approximately Mg1.5SiH1.5O4.
To find the molecular formula, we need to know the molar mass of the empirical formula. The molar mass of the empirical formula can be calculated as follows:
- (1.5 mol of Mg) * (24.31 g/mol) = 36.47 g/mol of Mg
- (1 mol of Si) * (28.09 g/mol) = 28.09 g/mol of Si
- (1.5 mol of H) * (1.01 g/mol) = 1.52 g/mol of H
- (4 mol of O) * (16.00 g/mol) = 64.00 g/mol of O
The total molar mass of the empirical formula is:
36.47 g/mol + 28.09 g/mol + 1.52 g/mol + 64.00 g/mol = 130.08 g/mol
To find the molecular formula, we divide the molar mass of the compound (520.8 g/mol) by the molar mass of the empirical formula (130.08 g/mol):
520.8 g/mol / 130.08 g/mol = 4
Therefore, the molecular formula for chrysotile asbestos is 4 times the empirical formula:
Mg6Si4H6O16