How many oxygen atoms are present in 40.0 grams of oxygen gas?

Answer 1

Ar of O = 16g/mol

Mr of O2 = 2(16) = 32g/mol

Using the formula : Number of moles = mass / Mr

Number of moles = 40g / 32g/mol = 1.25mols

One mole of substance contains the same number of particles as the Avogadro constant, which is 6.02 x 10^23

Number of Oxygen molecules = 1.25 x 6.02 x 10^23 = 7.525 x 10^23

Each Oxygen molecules contain two Oxygen atoms

Number of Oxygen atoms = 7.525 x 10^23 = 1.505 x 10^24 atoms

Answer 2

There are approximately 0.625 moles of oxygen gas in 40.0 grams. Since the formula for oxygen gas is O2, there are 0.625 moles x 2 = 1.25 moles of oxygen atoms. Therefore, there are approximately 1.25 x 6.022 x 10^23 = 7.53 x 10^23 oxygen atoms in 40.0 grams of oxygen gas.