What is the mass, in grams, of #3.40*10^22# atoms #He#?

1 Answer
Jun 23, 2016

Answer:

0.226 gram.

Explanation:

You can get to the answer in two steps:

  1. calculate: #3.40*10^22# atoms = #color(red)x# mol ?
    using Avogadro's constant.
  2. calculate: #color(red)x# mol = #color(red)?# grams,
    using the atomic mass of Helium.

Step 1
Avogadro's constant says that #1# mole of any atom contains #6.022*10^23# atoms. In this case you have #3.40*10^22# atoms:

#(3.40*10^22 color(red)cancel(color(black)("atoms")))/ (6.022*10^23 color(red)cancel(color(black)("atoms"))/"mol"# = #5.65*10^-2 mol#

Step 2
The atomic mass of helium (He) will give you the weight of one mole of this molecule: #1# mol = #4.00# gram:

#4.00 g/(color(red)cancel color(black)"mol") * 5.65*10^-2 color(red)cancel color(black)"mol"# = #0.226 g#

So the #3.40*10^22# helium atoms weigh #color(red)(0.226) color(red) " gram"#.