Question #66091

1 Answer
Jul 12, 2017

Answer:

Here's what I got.

Explanation:

Two things to keep in mind here

  • the definition of a mole is given by Avogadro's constant
  • we use subscripts to show how many atoms of a given element are present in one molecule or formula unit of a compound

So, you should know that in order to have exactly #1# mole of propane molecules, your sample must contain #6.022 * 10^(23)# molecules of propane.

This implies that in order to have #color(blue)(2)# moles of propane, you need to have #color(blue)(2)# times as many molecules of propane as you have in #1# mole of this compound.

You can thus say that your sample contains

#color(blue)(2) color(red)(cancel(color(black)("moles C"_3"H"_8))) * overbrace((6.022 * 10^(23)color(white)(.)"molecules C"_3"H"_8)/(1color(red)(cancel(color(black)("mole C"_3"H"_8)))))^(color(purple)("Avogadro's constant"))#

# = color(darkgeeen)(ul(color(black)(1.2 * 10^(24)color(white)(.)"molecules C"_3"H"_8)))#

Now, a molecule of propane contains

  • three atoms of carbon, #3 xx "C"#
  • eight atoms of hydrogen, #8 xx "H"#

You know that this is the case because of the two subscripts used in the chemical formula of propane.

#"C"_3"H"_8 implies {(3 xx "C"), (8 xx "H") :}#

http://visualizingchemistry-ashley-slinker.blogspot.ro/2012/01/activity-3-visualizing-chemistry.html

You can thus say that your sample contains

#1.2 * 10^(24) color(red)(cancel(color(black)("molecules C"_3"H"_8))) * "3 atoms C"/(1color(red)(cancel(color(black)("molecule C"_3"H"_8))) )#

# = color(darkgreen)(ul(color(black)(3.6 * 10^(24)color(white)(.)"atoms C")))#

and

#1.2 * 10^(24) color(red)(cancel(color(black)("molecules C"_3"H"_8))) * "8 atoms H"/(1color(red)(cancel(color(black)("molecule C"_3"H"_8))) )#

# = color(darkgreen)(ul(color(black)(9.6 * 10^(24)color(white)(.)"atoms H")))#

I'll leave the answers rounded to two sig figs, but don't forget that you only have one significant figure for the number of moles of propane.