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

Feb 1, 2017

$5.3 \cdot {10}^{- 23} \text{g}$

#### 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 ${\text{44.01 g mol}}^{- 1}$. This means that $1$ mole of carbon dioxide has a mass of $\text{44.01 g}$.

You know that $1$ mole of carbon dioxide contains

• one mole of carbon, $1 \times \text{C}$
• two moles of oxygen, $2 \times \text{O}$

Oxygen has a molar mass of ${\text{16.0 g mol}}^{- 1}$, so $1$ mole of oxygen atoms has a mass of $\text{16.0 g}$.

This means that if you take $\text{44.01 g}$ 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 $\text{100 g}$ 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 $\text{100 g}$ of carbon dioxide you get $\text{72.7 g}$ 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

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{\text{1 mole CO"_2 = 6.022 * 10^(23)"molecules CO}}_{2}}}}$

So, you know what

${\text{1 mole CO"_2 = "44.01 g" = 6.022 * 10^(23)"molecules CO}}_{2}$

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))))

$= 7.31 \cdot {10}^{- 23} \text{g}$

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

$7.31 \cdot {10}^{- 23} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g CO"_2))) * "72.7 g O"/(100color(red)(cancel(color(black)("g CO"_2)))) = color(darkgreen)(ul(color(black)(5.3 * 10^(-23)"g O}}}}$

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