# How many metal atoms in a 3.4*g mass of copper?

Jul 25, 2017

We know that a $63.55 \cdot g$ mass of copper contains ${N}_{A}$ $\text{copper atoms......}$

#### Explanation:

How do we know this? Because the Periodic Table gives us THE MASS of a molar quantity, i.e. ${N}_{A}$ or $\text{Avogadro's number}$ of ALL the elements. You got a Periodic Table beside you? You should have one because you are doing your chemistry homework.

And thus to get the number of copper atoms, we just find the molar quantity........

$\frac{3.4 \cdot g}{63.55 \cdot g \cdot m o {l}^{-} 1} \times {N}_{A}$, where ${N}_{A} = \text{Avocado number}$,

$= 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

And thus $\text{number of copper atoms.....}$

$\frac{3.4 \cdot g}{63.55 \cdot g \cdot m o {l}^{-} 1} \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1 \equiv 3.22 \times {10}^{22} \cdot \text{copper atoms}$.

Now $\text{Avogadro's number}$ is admittedly a large number. We use it because ${N}_{A}$ ""^1H atoms have a mass of $1 \cdot g$; alternatively, ${N}_{A}$ ""^12C atoms have a mass of $12.00 \cdot g$. It is thus the link between atoms and molecules, what chemists theorize about and of which they conceive, with the macro world of grams, kilograms, and litres, what chemists measure out in the laboratory.

It will be worth your while getting your head around this $\text{principle of equivalent mass}$, of how a given mass of elements or molecules constitutes a SPECIFIC NUMBER of elements or molecules. So please get it right!

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