The gas sulfur dioxide, SO2, oxidizes to form another gas, sulfur trioxide, SO3. How many oxygen atoms are present in 3.20 g of sulfir trioxide?
How many mols SO3 do you have? mols SO3 = grams/molar mass
O atoms in SO3 = 3 O atoms/1 molecule; therefore, 3x that value = mols O atoms.
Then there are 6.02E23 O atoms in 1 mol O atoms; therefore, there are ..... O atoms.
To determine the number of oxygen atoms present in 3.20 g of sulfur trioxide (SO3), we need to use the concept of moles and the molar mass of SO3.
Here's the step-by-step process:
1. Start by finding the molar mass of sulfur trioxide (SO3).
- Sulfur (S) has a molar mass of 32.1 g/mol
- Oxygen (O) has a molar mass of 16.0 g/mol
- Therefore, the molar mass of SO3 is 32.1 g/mol + (3 × 16.0 g/mol) = 80.1 g/mol.
2. Convert the given mass of sulfur trioxide (3.20 g) to moles.
- Divide the given mass by the molar mass:
Moles of SO3 = (3.20 g) / (80.1 g/mol) ≈ 0.04 mol.
3. Use the mole ratio to find the number of moles of oxygen atoms.
- From the balanced chemical equation: SO2 + O2 → SO3
- The mole ratio between sulfur trioxide (SO3) and oxygen (O2) is 1:1.
- Therefore, there are also approximately 0.04 moles of oxygen atoms.
4. Convert the number of moles of oxygen atoms to the number of atoms.
- Since 1 mole of any substance contains Avogadro's number (6.022 × 10^23) of particles, we can use this value to convert moles to atoms.
- Number of oxygen atoms = (0.04 mol) × (6.022 × 10^23 atoms/mol) ≈ 2.41 × 10^22 atoms.
So, there are approximately 2.41 × 10^22 oxygen atoms present in 3.20 g of sulfur trioxide (SO3).