# A 2.72*g mass of aluminum constitutes WHAT molar quantity? And how many aluminum atoms does this molar quantity represent?

Feb 10, 2016

Avogadro's number of $A l$ atoms has a mass of $26.98$ $g$. So you tell us.
If there are $6.022$ $\times$ ${10}^{23}$ individual aluminum atoms, there is a mass of $26.98$ $g$ of aluminum. $6.022$ $\times$ ${10}^{23}$ individual aluminum atoms represents $1$ $m o l$ of aluminum.
So, $\frac{2.72 \cdot \cancel{g}}{26.98 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ ?? $m o l$?
Whatever this molar quantity is, we multiply this by Avogadro's number, $6.022$ $\times$ ${10}^{23}$ $m o {l}^{-} 1$, to get the number of aluminum atoms.