Question #9fbfb

1 Answer
Jun 21, 2017

Answer:

#6.0 * 10^(22)# #"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 #"S"_8#.

http://www.oilamerica.com.pa/en/products/sulfur.html

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 #"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)# #"molecules S"_8#

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

#7.51 * 10^(21) color(red)(cancel(color(black)("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.