Amount of bromine.....

That's actually a bit vague, but I have a feeling you mean the mass of bromine in the sample. Just use the unit-factor method.

4.50x10^22 atoms Br x (79.90g Br / 6.022x10^23 atoms Br) = 9.35x10^-4 g Br

The conversion factor of (79.90g Br / 6.022x10^23 atoms Br) comes from the fact that each quantity is equivalent to a mole. Therefore, any two things which are each equal to a mole are equal to each other.

1 mol = 1 molar mass

1 mol = 6.022x10^23 particles

1 mol = 22.414 L of any ideal gas at STP

One mole is 6.02 * 10^23 atoms.

Number of moles = 4.50 * 10^22 ÷ 6.02 * 10^23

This is approximately 0.075 mole. Bromine is a diatomic molecule. This means one molecule contains two atoms. The mass of one mole of Br2 is approximately 159.8 grams. To determine the mass of this many atoms, multiply the number of moles by the mass of one mole.

Mass = 159.8 * 4.50 * 10^22 ÷ 6.02 * 10^23

This is approximately 12 grams. I hope this is helpful for you.

4.50×1022 = 4599 atoms

6.022e23 molecules/mole Avogadro constant

4599 atoms / 6.022e23 molecules/mole = 7.64e-21 mole

35 g/mol x 7.64e-21 mole = 2.67e-19 g