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

Jun 23, 2016

0.226 gram.

#### Explanation:

You can get to the answer in two steps:

1. calculate: $3.40 \cdot {10}^{22}$ atoms = $\textcolor{red}{x}$ mol ?
2. calculate: $\textcolor{red}{x}$ mol = color(red)? grams,
using the atomic mass of Helium.

Step 1
Avogadro's constant says that $1$ mole of any atom contains $6.022 \cdot {10}^{23}$ atoms. In this case you have $3.40 \cdot {10}^{22}$ atoms:

(3.40*10^22 color(red)cancel(color(black)("atoms")))/ (6.022*10^23 color(red)cancel(color(black)("atoms"))/"mol" = $5.65 \cdot {10}^{-} 2 m o l$

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 $3.40 \cdot {10}^{22}$ helium atoms weigh $\textcolor{red}{0.226} \textcolor{red}{\text{ gram}}$.