# What mass of gold, Au, contains the same number of atoms 9.0 g of aluminum, Al?

Nov 26, 2015

#### Answer:

$\text{9.0 g Al}$ contains $2.0 \times {10}^{23} \text{atoms Al}$.
$2.0 \times {10}^{23} \text{atoms Au}$ has a mass of $\text{65 g}$.

#### Explanation:

First determine the number of atoms in $\text{9.0 g Al}$.

Divide the given mass of $\text{Al}$ by its molar mass (atomic weight on the periodic table in g/mol), then multiply times $6.022 \times {10}^{23} \text{atoms/mol}$, which is the same for all elements.

$9.0 \cancel{\text{g Al"xx(1cancel"mol Al")/(26.982cancel"g Al")xx(6.022xx10^23"atoms Al")/(1cancel"mol Al")=2.0xx10^23 "atoms Al}}$

Now reverse the process to determine the mass of gold that contains $2.0 \times {10}^{23} \text{atoms Au}$.

Divide the calculated atoms by the number of atoms in one mole, then multiply times the molar mass of $\text{Au}$.

$2.0 \times {10}^{23} \cancel{\text{atoms Au"xx(1cancel"mol Au")/(6.022xx10^23cancel"atoms Au")xx(196.967"g Au")/(1cancel"mol Au")="65 g Au}}$