# If I have 9.71xx10^22 "platimum atoms", what molar quantity, and what mass of metal are present?

Sep 1, 2016

You have approx $0.16 \cdot m o l$ of platinum atoms, with a mass of approx. $30 \cdot g$.

#### Explanation:

$\text{Moles of platinum}$ $=$ $\left(9.71 \times {10}^{22} \cdot \text{platinum atoms")/(6.022xx10^23*"platinum atoms} \cdot m o {l}^{-} 1\right)$ $=$ ??"moles"?

Now $1$ $m o l$ of $P t$ atoms has a mass of $195.08 \cdot g$.

So you should have a mass of approx. $30 \cdot g$.

I appreciate that it is hard to work with such unfeasibly large numbers. But remember that the mole is just another number, like a score, or a dozen, or 1 gross - it's just that a mole is *&&??! LARGE. $\text{Avogadro's number}$ of platinum atoms specifies $1$ $m o l$ and has a mass of $195.08 \cdot g$. Likewise $\text{Avogadro's number}$ of ""^1H atoms has a mass of $1 \cdot g$ precisely.

This idea of molar and mass equivalence is something that is fundamental to chemistry, and you should take some effort to appreciate it.