killjoy1990xb9

2022-01-06

Use Avogadro’s number to find the mass of a helium atom.

Annie Gonzalez

Step 1
Avogardo's number is ${N}_{A}=6.02×{10}^{23}$ particles/mol. A mole of anything, by definition, consists of an Avogadro's number of particles.
The mass M of one mole of a pure substance in grams is the same, numerically, as that substance's atomic (or molecular) mass.
The mass of a single molecule is then  $m=\frac{M}{{N}_{A}}$.
Use the table in Appendix B:
For helium, $M=4.00\frac{g}{mol}=4.00×{10}^{-3}\frac{kg}{mol}$.
So, the mass of a helium molecule is
Step 2
$m=\frac{4.00×{10}^{3}kg/mol}{6.02×{10}^{23}molecule/mol}=6.64×{10}^{-27}kg/molecule$.
Since a helium molecule contains a single helium atom,
the mass of a helium atom is
$m=6.64×{10}^{-27}kg$

Debbie Moore

One mole of helium contains Avogadro's number of molecules and has a mass of 4.00g. Let us call ${m}_{0}$ the mass of one atom, and we have
$NA{m}_{0}=4.00\frac{g}{mol}$
or ${m}_{0}=\frac{4.00gmol}{6.02×{10}^{23}molecule/mol}=6.64×{10}^{-24}g/molecule,$
$=6.64×{10}^{-27}kg$

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