# How many calcium atoms are present in a mass of 169*g of this metal?

Feb 5, 2017

$\text{Number of atoms}$ $= \text{Mass"/"Molar mass"xx"Avogadro's number.}$

#### Explanation:

By definition, $40.1 \cdot g$ of calcium atoms contains $\text{Avogadro's number of molecules}$. We specify this quantity as $1 \cdot m o l$ of $\text{calcium atoms}$.

And so we take the quotient, $\frac{169 \cdot g}{40.1 \cdot g \cdot m o {l}^{-} 1}$, and multiply this by ${N}_{A} , \text{Avogadro's number of molecules} ,$ where ${N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

And thus the product, $\frac{169 \cdot \cancel{g}}{40.1 \cdot \cancel{g} \cdot \cancel{m o {l}^{-} 1}} \times 6.022 \times {10}^{23} \cdot \cancel{m o {l}^{-} 1}$ gives a DIMENSIONLESS number as required, approx. $4 \times {N}_{A}$.

If I asked how many eggs were in 3 dozen eggses, I think you would immediately be able to reply. The question here is much the same sort of question, and can be viewed in the same simple light.