Calculate the mass of carbon dioxide that contains the same number of atoms of oxygen as present in #"1.8 g"# of water?
1 Answer
Explanation:
Start by calculating the number of moles of oxygen present in
As you know, one molecule of water contains two atoms of hydrogen and one atom of oxygen. This means that
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
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 color(red)(cancel(color(black)("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,
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
#(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