# Calculate the mass of carbon dioxide that contains the same number of atoms of oxygen as present in "1.8 g" of water?

Mar 29, 2018

#### Answer:

${\text{2.2 g CO}}_{2}$

#### Explanation:

Start by calculating the number of moles of oxygen present in $\text{1.8 g}$ of water.

As you know, one molecule of water contains two atoms of hydrogen and one atom of oxygen. This means that $1$ mole of water contains $2$ moles of hydrogen and $1$ mole of oxygen.

Use the molar mass of water to convert the mass of water to moles.

1.8 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.0color(red)(cancel(color(black)("g")))) = "0.10 moles H"_2"O"

You can thus say that this sample contains

0.10 color(red)(cancel(color(black)("moles H"_2"O"))) * "1 mole O"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "0.10 moles O"

Now, for every $1$ molecule of carbon dioxide, ${\text{CO}}_{2}$, you get $2$ atoms of oxygen, so $1$ mole of carbon dioxide will contain $2$ moles of atoms of oxygen.

This means that the number of moles of oxygen present in your sample of water is enough to get you

0.10 color(red)(cancel(color(black)("moles O"))) * "1 mole CO"_2/(2color(red)(cancel(color(black)("moless O")))) = "0.050 moles CO"_2" " " "color(darkorange)("(*)")

Finally, to convert the number of moles of carbon dioxide to moles, use the molar mass of carbon dioxide.

$0.050 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles CO"_2))) * "44.0 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = color(darkgreen)(ul(color(black)("2.2 g}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the mass of water.

Notice that you didn't even have to convert the mass of water to atoms of oxygen because a mole of oxygen is simply a very large collection of atoms of oxygen, $6.022 \cdot {10}^{23}$, or ${N}_{\text{A}}$, to be precise $\to$ this is known as Avogadro's constant.

So if you want to double-check your answer, you can say that your sample of water contains

0.10 color(red)(cancel(color(black)("moles O"))) * (N_"A" quad "atoms of O")/(1 color(red)(cancel(color(black)("mole O")))) = (0.10 * N_"A") quad "atoms of O"

The same logic applies--you need $2$ atoms of oxygen to get $1$ molecule of carbon dioxide, so the number of atoms of oxygen present in your sample will get you

(0.10 * N_"A") color(red)(cancel(color(black)("atoms O"))) * "1 molecule CO"_2/(2color(red)(cancel(color(black)("atoms O")))) = (0.050 * N_"A") quad "molecules CO"_2

To convert the number of molecules of carbon dioxide to moles, use Avogadro's constant again.

(0.050 * color(blue)(cancel(color(black)(N_ "A")))) color(red)(cancel(color(black)("molecules CO"_ 2))) * "1 mole CO"_ 2/(color(blue)(cancel(color(black)(N_ "A"))) color(red)(cancel(color(black)("molecules CO"_ 2)))) = "0.050 moles CO"_2

This brings you right back to $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$, so once again, the answer will be $\text{2.2 g}$ of carbon dioxide.