Question

A gold coin contains 3.47*10^23 gold atoms. what is the mass of the coin in grams

Answer 1

You can use Avogradro's number. Periodic table's name for gold is Au. Mo


`3.47 * 10^(23) atoms Au * (1 mol Au)/(6.022*10^(23) atoms Au) *(196.96657 g Au)/(1 mol Au) = 113.496180322g`

There are 3 sig figs.

113 g Au

Answer 2

Answer: 114.26 grams

Explanation: According to avogadro's law, 1 mole of every substance contains avogadro's number
`6.023* 10^(23)` of particles.

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given atoms}}{\text {Avogadros number}}`

`\text{Number of moles}=(3.47* 10^(23))/(6.023* 10^(23))=0.58moles`

Mass of the gold coin=
`{\text {number of moles}}* {\text {Molar mass of Au}}=0.58* 197g/mol=114.26g`