# Question #bfcbb

May 19, 2017

$3.8 \cdot {10}^{- 21} \text{g}$

#### Explanation:

Copper has a molar mass of ${\text{63.546 g mol}}^{- 1}$, which means that $1$ mole of copper has a mass of $\text{63.546 g}$.

Now, in order to have $1$ mole of copper, you need to have $6.022 \cdot {10}^{23}$ atoms of copper $\to$ this is known as Avogadro's constant.

So, right from the start, you can say that if $1$ mole of copper has a mass of $\text{63.546 g}$ and $1$ mole of copper contains $6.022 \cdot {10}^{23}$ atoms of copper, then

$6.022 \cdot {10}^{23} \textcolor{w h i t e}{.} \text{atoms Cu " -> " 63.546 g}$

All you have to do now is to use this as a conversion factor to calculate the mass of $6$ atoms of copper

$6 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atoms Cu"))) * "63.546 g"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms Cu")))) = color(darkgreen)(ul(color(black)(3.8 * 10^(-21)color(white)(.)"g}}}}$

I'll leave the answer rounded to two sig figs.