Calculate the mass of oxygen in a molecule of "CO"_2CO2 by using percentage composition? please help

1 Answer
Feb 1, 2017

5.3 * 10^(-23)"g"5.31023g

Explanation:

The first thing you need to do here is to figure out the mass of oxygen in 11 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)44.01 g mol1. This means that 11 mole of carbon dioxide has a mass of "44.01 g"44.01 g.

You know that 11 mole of carbon dioxide contains

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

Oxygen has a molar mass of "16.0 g mol"^(-1)16.0 g mol1, so 11 mole of oxygen atoms has a mass of "16.0 g"16.0 g.

This means that if you take "44.01 g"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.