How many oxygen atoms are in 6.70 g of Al2(SO4)3? Express your answer using scientific notation with two decimal places.
See your post above.
9.52E23
To determine the number of oxygen atoms in 6.70 g of Al2(SO4)3, we need to consider the molar mass of the compound and the number of oxygen atoms per formula unit.
1. Determine the molar mass of Al2(SO4)3:
The molar mass is calculated by adding up the atomic masses of each element in the compound.
Aluminum (Al) has an atomic mass of 26.98 g/mol, sulfur (S) has an atomic mass of 32.07 g/mol, and oxygen (O) has an atomic mass of 16.00 g/mol.
So, Al2(SO4)3 contains:
2 Al atoms: 2 x 26.98 = 53.96 g/mol
3 S atoms: 3 x 32.07 = 96.21 g/mol
12 O atoms: 12 x 16.00 = 192.00 g/mol
Adding up these masses gives us the molar mass of Al2(SO4)3:
Molar mass of Al2(SO4)3 = 53.96 + 96.21 + 192.00 = 342.17 g/mol
2. Calculate the number of moles of Al2(SO4)3:
To convert grams to moles, divide the given mass by the molar mass.
Number of moles = (mass of Al2(SO4)3) / (molar mass of Al2(SO4)3)
Number of moles = 6.70 g / 342.17 g/mol
3. Determine the number of oxygen atoms in the given mass of Al2(SO4)3:
Now we need to consider the stoichiometry of the compound. From the formula Al2(SO4)3, we can see that each formula unit contains 12 oxygen atoms.
Number of moles of Al2(SO4)3 x (12 oxygen atoms / 1 mole Al2(SO4)3) = Number of oxygen atoms
Substituting the values:
Number of oxygen atoms = (6.70 g / 342.17 g/mol) x (12 atoms / 1 mol)
Calculating this expression will yield the number of oxygen atoms in 6.70 g of Al2(SO4)3.