Question 9fbfb

Jun 21, 2017

$6.0 \cdot {10}^{22}$ $\text{atoms S}$

Explanation:

The trick here is to realize that sulfur exists as an octatomic molecule under standard conditions.

In other words, sulfur exists as molecules that contain $8$ atoms of sulfur. The chemical formula for octatomic sulfur is ${\text{S}}_{8}$.

This implies that $1$ mole of sulfur molecules will be equivalent to $8$ moles of sulfur atoms. Consequently, the molar mass of octatomic sulfur will be equal to

8 * overbrace("32.065 g mol"^(-1))^(color(blue)("the molar mass of elemental sulfur")) = "256.52 g mol"^(-1)

You can thus say that $\text{3.2 g}$ of sulfur will contain

3.2 color(red)(cancel(color(black)("g"))) * "1 mole S"_8/(256.52color(red)(cancel(color(black)("g")))) = "0.012475 moles S"_8

The number of octatomic sulfur molecules present in this sample will be equal to

0.012475 color(red)(cancel(color(black)("moles S"_8))) * overbrace((6.022 * 10^(23)color(white)(.)"molecules S"_8)/(1color(red)(cancel(color(black)("mole S"_8)))))^(color(blue)("Avogadro's constant")) = 7.51 * 10^(21)# ${\text{molecules S}}_{8}$

Since each molecule of octatomic sulfur contains $8$ atoms of sulfur, you can say that your sample contains

$7.51 \cdot {10}^{21} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{molecules S"_8))) * "8 atoms S"/(1color(red)(cancel(color(black)("molecule S"_8)))) = color(darkgreen)(ul(color(black)(6.0 * 10^(22)color(white)(.)"atoms S}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the mass of the sample.