What is the total mass of 3.01 x 10^23 atoms of helium gas?

Jan 23, 2017

Well $\text{Avogadro's number}$ of helium atoms has a mass of $4.0 \cdot g$.

Explanation:

And $\text{Avogadro's number}$ $\equiv$ $6.022 \times {10}^{23} \cdot m o {l}^{-} 1$

Clearly, we have half a mole of helium atoms, so........

$\left(6.022 \times {10}^{23} \cdot \text{helium atoms")/(6.022xx10^23*"helium atoms} \cdot m o {l}^{-} 1\right) = \frac{1}{2} \cdot m o l$

And $\frac{1}{2} \cdot \cancel{m o l} \times 4.00 \cdot g \cdot \cancel{m o {l}^{-} 1} = 2.00 \cdot g$

The units cancel appropriately to give me an answer in $\text{grams}$ as required.

This principle of equivalent mass, of the use of a number ($\text{the mole}$) to represent a given number of atoms or molecules, is absolutely fundamental to the study of chemistry. The problem is worth reviewing, because if you study chemistry, you will be doing a hell of a lot of problems like this.

If I have $18 \times {10}^{23}$ helium atoms what is the mass?