# What is the mass of 4.70xx10^22 individual "sulfur atoms"?

Feb 10, 2017

Approx. $2.5 \cdot g$

#### Explanation:

We know that $6.022 \times {10}^{23}$ individual sulfur atoms have a mass of $32.06 \cdot g$. How do we know this?

And thus we work out the product:

$32.06 \cdot g \cdot m o {l}^{-} 1 \times \left(4.70 \times {10}^{22} \cancel{\text{ sulfur atoms ")/(6.022xx10^23cancel" sulfur atoms }} m o {l}^{-} 1\right)$

=??*mol

Note that as far as anyone knows, sulfur, as the yellow powder, is the ${S}_{8}$ molecule. The size of the molecule probably explains its extreme insolubility in most solvents.