# If one has 4.50 xx 10^22 atoms of silver, what mass is available?

Aug 2, 2017

$\text{8.06 g Ag}$,

i.e. the mass of a handful of silver.

This is easiest in steps:

$\text{atoms" stackrel(N_A" ")(->) "mols" stackrel(M" ")(->) "mass}$

where ${N}_{A} = 6.0221413 \times {10}^{23} {\text{mol}}^{- 1}$ is Avogadro's number and $M$ is the molar mass in $\text{g/mol}$ of the thing we are looking at.

So, to get to mols, we utilize the number of whatchamacallits in $\text{1 mol}$:

4.50 xx 10^(22) cancel"atoms" xx ("1 mol")/(6.0221413 xx 10^(23) cancel"whatevers")

$=$ $\text{0.0747 mols of any atom}$

To be specific to silver, we then need to use its molar mass $M$ to identify the atom as silver. Look up the molar mass on a periodic table to find $M = \text{107.8682 g/mol}$. Thus:

0.0747 cancel"mols unknown element" xx ("107.8682 g Ag")/(cancel"1 mol Ag")

$=$ $\textcolor{b l u e}{\text{8.06 g Ag}}$