# How many oxygen atoms in a 300*g mass of calcium carbonate?

Jun 8, 2016

We work out (i) the molar quantity of $C a C {O}_{3}$, and (ii) the constituent quantity of oxygen.

#### Explanation:

$\text{Moles of calcium carbonate, } \frac{300 \cdot g}{100.09 \cdot g \cdot m o {l}^{-} 1}$ $\cong$ $3 \cdot m o l$.

Given the elemental makeup, in such a quantity there are $3$ moles of calcium, $3$ moles of carbon, and $9$ moles of oxygen.

But what is a mole? It is simply a LARGE number that relates to a given mass of stuff. In $1$ $m o l$ of stuff there are $6.022 \times {10}^{23}$ individual particles of that stuff.

So given 3 moles of calcium carbonate, there are 3 moles of calcium, 3 moles ofcarbon, and 9 moles of oxygen atoms.

So $\text{number of oxygen atoms}$ $=$ $9 \cdot m o l \times {N}_{A}$

$=$ $6.022 \times {10}^{23} \cdot m o {l}^{-} 1 \times 9 \cdot m o l \text{ oxygen atoms}$ $=$ "oxygen atoms"??

What is the mass of this quantity of oxygen atoms, and how do you know?

See here for another examples of this process.