Calculate the mass of oxygen in a molecule of #"CO"_2# by using percentage composition? please help

1 Answer
Feb 1, 2017

Answer:

#5.3 * 10^(-23)"g"#

Explanation:

The first thing you need to do here is to figure out the mass of oxygen in #1# mole of carbon dioxide. To do that, you must use the compound's molar mass.

Now, carbon dioxide has a molar mass of #"44.01 g mol"^(-1)#. This means that #1# mole of carbon dioxide has a mass of #"44.01 g"#.

You know that #1# mole of carbon dioxide contains

  • one mole of carbon, #1 xx "C"#
  • two moles of oxygen, #2 xx "O"#

Oxygen has a molar mass of #"16.0 g mol"^(-1)#, so #1# mole of oxygen atoms has a mass of #"16.0 g"#.

This means that if you take #"44.01 g"# of carbon dioxide, you know for a fact that it will contain

#2 color(red)(cancel(color(black)("moles O"))) * "16.0 g"/(1color(red)(cancel(color(black)("mole O")))) = "32.0 g O"#

This means that #"100 g"# of carbon dioxide will contain

#100color(red)(cancel(color(black)("g CO"_2))) * "32.0 g O"/(44.01color(red)(cancel(color(black)("g CO"_2)))) = "72.7 g O"#

Therefore, carbon dioxide has a percent composition of #"72.7%# oxygen, i.e. for every #"100 g"# of carbon dioxide you get #"72.7 g"# of oxygen.

Now, you must determine the mass of oxygen present in a single molecule of carbon dioxide. Start by figuring out the mass of a single molecule of carbon dioxide.

To do that, use Avogadro's constant, which tells you that

#color(blue)(ul(color(black)("1 mole CO"_2 = 6.022 * 10^(23)"molecules CO"_2)))#

So, you know what

#"1 mole CO"_2 = "44.01 g" = 6.022 * 10^(23)"molecules CO"_2#

which means that #1# molecule of carbon dioxide has a mass of

#1 color(red)(cancel(color(black)("molecule CO"_2))) * "44.01 g"/(6.022 * 10^(23)color(red)(cancel(color(black)("molecules CO"_2))))#

# = 7.31 * 10^(-23)"g"#

Now you can use the percent composition of carbon dioxide to find the mass of oxygen present in #1# molecule

#7.31 * 10^(-23) color(red)(cancel(color(black)("g CO"_2))) * "72.7 g O"/(100color(red)(cancel(color(black)("g CO"_2)))) = color(darkgreen)(ul(color(black)(5.3 * 10^(-23)"g O")))#

I'll leave the answer rounded to two sig figs.