# How many atoms in an 197*g mass of calcium metal?

Jan 22, 2017

Approx. $5 \times {N}_{A}$, where ${N}_{A}$ is $\text{Avogadro's number}$.

#### Explanation:

You find the molar mass of calcium metal, it is listed as $40.1 \cdot g \cdot m o {l}^{-} 1$. You should check your copy of the Periodic Table to see if I have got it right. This means that $\text{Avogadro's number}$ of calcium atoms, i.e. $6.022 \times {10}^{23}$ individual calcium atoms have a mass of $40.1 \cdot g$. And thus we can find the number of calcium atoms in a lump of metal, simply by measuring the mass of the lump and doing a simple calculation. And of course, we can also find the number of calcium atoms given a mass, and a formula for a calcium-containing material.

So $\text{Moles of calcium}$ $=$ $\frac{197 \cdot \cancel{g}}{40.1 \cdot \cancel{g} \cdot m o {l}^{-} 1}$

=??mol. We get an answer in $\text{moles}$, because dimensionally $\frac{1}{m o {l}^{-} 1} = \frac{1}{\frac{1}{m o l}} = m o l$ as required. Note that an answer that uses ${N}_{A}$ to represent the given number would be quite acceptable; of course you could multiply it out.

If I were you I would study the relevant section of your text that deals with this principle. The idea of equivalent mass, the use of mass to represent a NUMBER of combining particles, is fundamental to the study of chemistry, and should not require too much angst to incorporate.

How many sodium atoms (approx.) in $23 \cdot g$ of sodium metal?