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

1 Answer
Jan 23, 2017

Answer:

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?