Question

How many electrons are in 0.990 oz of a pure gold coin?

Answer 1

Gold (Au) atomic number 79 hence it has 79 electrons per atom. Mass of 1 mole of Au is 196.967 gm .

And 0.990 oz Au = 28.066 gm of Au

Thus,

0.990 oz of Au = ((28.066 gm) / (196.967 gm) ) 1 mole = 0.1425 moles

1 mole of Au will contain = 6.022 10²³ atoms of Au

0.1425 moles of Au = 0.1425 6.022 10²³ = 0.8581 10²³ atoms of Au

If 1 atom of Au contains 79 electrons then,

0.8581 10²³ atoms of Au willl contain = 79 0.8581 10²³ = 67.7899 10²³ electrons.
Answer is: 0.990 oz of a pure gold coin will contain 67.7899 10²³ electrons.