Question #553eb

Every molecule of carbon dioxide, #"CO"_2#, contains one atom of carbon and two atoms of oxygen.

This means that every mole of carbon dioxide will contain 1 mole of carbon and two moles of oxygen.

Now, carbon dioxide's molar mass, which tells you exatly what the mass of one mole of carbon dioxide is, is equal to #"44.01 g/mol"#.

This means that for every #"44.01 g"# of carbon dioxide, you get one mole of carbon dioxide, which is equivalent to one mole of carbon and two moles of oxygen.

Since you didn't provide a mass of carbon dioxide, I"ll use a 100-g sample. So, how many moles of carbon dioxide will you get in this sample?

#100color(red)(cancel(color(black)("g"))) * ("1 mole CO"""_2)/(44.01color(red)(cancel(color(black)("g")))) = "2.27 moles CO"""_2#

Since you get one mole of carbon for every mole of carbon dioxide, you will also have 2.27 moles of carbon.

To get the number of atoms of carbon, use Avogadro's number, which tells you that one mole of an element contains exactly #6.022 * 10^(23)# atoms of that element.

#2.27color(red)(cancel(color(black)("moles"))) * (6.022 * 10^(23)"atoms of C")/(1color(red)(cancel(color(black)("mole")))) = 1.4 * 10^(24)"atoms of C"#

Now, if you say that the answer is #9 * 10^(23)#, then you can work backward to get the mass of carbon dioxide that will contain this many atoms of carbon.

#9 * 10^(23)color(red)(cancel(color(black)("atoms of C"))) * "1 mole C"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms of C")))) = "1.5 moles C"#

This is of course equivalent to 1.5 moles of carbon dioxide, which means that the mass of #"CO"""_2# that contains #9 * 10^(23)# atoms of carbon will be

#1.5color(red)(cancel(color(black)("moles CO"_2))) * "44.01 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) ~= "66 g CO"""_2#