Question 15857

Jun 14, 2017

72.71%

Explanation:

The idea here is that you need to figure out how much oxygen is present in $\text{100 g}$ of carbon dioxide $\to$ this number will give you the percent composition of oxygen in carbon dioxide.

Since you know that every mole of carbon dioxide contains

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

you can say that a mass of carbon dioxide that is equivalent to $1$ mole of this compound will contain the mass of $2$ mole of oxygen.

Now, carbon dioxide has a molar mass of ${\text{44.0095 g mol}}^{- 1}$, which means that $1$ mole of this compound has a mass of $\text{44.0095 g}$. Oxygen has a molar mass of ${\text{15.9994 g mol}}^{- 1}$, and so $1$ mole of oxygen has a mass of $\text{15.9994 g}$.

Therefore, you can say that $2$ moles of oxygen will have a mass of

2 color(red)(cancel(color(black)("moles O"_2))) * "15.9994 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = "31.9988 g"

Since this is how much oxygen you get in $\text{44.0095 g}$ of carbon dioxide, you can say that $\text{100 g}$ of carbon dioxide will get you

100color(red)(cancel(color(black)("g CO"_2))) * "31.9988 g O"/(44.0095color(red)(cancel(color(black)("g CO"_2)))) = "72.71 g O"#

This means that the percent composition of oxygen in carbon dioxide is

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{% composition CO"_2 = "72.71% O}}}}$

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