Question

How many oxygen atoms in a mass of `132*g` with respect to carbon dioxide?

Answer 1

There are `3` `mol` of `CO_2` in such a mass. Thus there are `6xxN_A` individual oxygen atoms.

Explanation:

What is `N_A`? It is simply `"Avogadro's number"`, `6.022xx10^23*mol^-1`.

If I have `6.022xx10^23` individual items of stuff, I have a mole of that stuff. `1` `mol` of `CO_2` is equivalent to a mass `44.0*g`; `12.01` `g` of `C`, and `32.0` `g` of `O`.

Answer 2

`36.132 times 10^23` oxygen atoms

Explanation:

If we work out the relative formula mass of `CO_2`

Relative atomic mass, `A_r` of `C= 12`
Relative atomic mass, `A_r` of `O=16`, so....

`M_r` of `CO_2 =44`

The number of moles of a substance is given by :

`"mass in grams" /(M_r)`

So in this case:

`132/44=3*"moles"`

One mole contains `6.022times 10^23` molecules (Avogadro's number).

So 3 moles contains `18.066 times 10^23` molecules.

Each molecule contains 2 oxygen atoms, so there will be `36.132 times 10^23`oxygen atoms