Question

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

Answer 1

Well `"Avogadro's number"` of helium atoms has a mass of `4.0*g`.

Explanation:

And `"Avogadro's number"` `-=` `6.022xx10^23*mol^-1`

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

`(6.022xx10^23*"helium atoms")/(6.022xx10^23*"helium atoms"*mol^-1)=1/2*mol`

And `1/2*cancel(mol)xx4.00*g*cancel(mol^-1)=2.00*g`

The units cancel appropriately to give me an answer in `"grams"` as required.

This principle of equivalent mass, of the use of a number (`"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 `18xx10^23` helium atoms what is the mass?