Question

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

Answer 1

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"`.