How many atoms are in 183 g of calcium?

Mar 24, 2016

There are approx. $4.5 \times {N}_{A} \text{ calcium atoms}$ where ${N}_{A}$ $=$ $\text{Avogadro's number}$, $6.02247 \times {10}^{23}$ $m o {l}^{-} 1$.

Explanation:

It is a fact that $40.078$ $g$ calcium contains $\text{Avogadro's number}$ of calcium atoms. $\text{Avogadro's number}$ is thus the link between the macro world of grams and kilograms, that which we can directly measure, and the micro world of atoms and molecules, about which we can conceive.

So moles of $C a$ $=$ $\frac{183.08 \cdot g}{40.078 \cdot g \cdot m o {l}^{-} 1}$ $= \text{ approx. } 4.5 \cdot m o l$.

$\text{Number of calcium atoms } = 4.5 \cdot m o l \times 6.02247 \times {10}^{23} \cdot m o {l}^{-} 1$ $=$ ?? "how many calcium atoms"

How did I know that $1$ $m o l$ of calcium has a mass of 40.078 g?