# Question #553eb

##### 1 Answer

#### Answer:

Actually, the answer depends on how much carbon dioxide you have.

#### Explanation:

Every molecule of *carbon dioxide*,

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

This means that for *every* **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

#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 *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

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