Question

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

Answer 1

`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.