# WE have 1*g masses of gold, lithium, sodium, and chlorine... Which sample contains the greatest number of atoms?

Jul 3, 2017

We know that $1 \cdot m o l$ of stuff contains $\text{Avogadro's Number,}$ ${N}_{A}$, individual items of stuff........and clearly there are more $\text{lithium ATOMS}$ than atoms of the other substances........

#### Explanation:

And ${N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$. I agree that this is an absurdly large number, but it is the link between the micro world of atoms and molecules, with the macro world of grams, and kilograms, and litres, and cubic metres. And in fact ${N}_{A}$ ""^12C $\text{ATOMS}$ have a mass of $12.00 \cdot g$ precisely.

And so we JUST have to work out the molar quantities of each reagent.

$i .$ $\text{Moles of gold}$ $=$ (1*gxxN_A)/(196.97*g*mol^-1)=??*mol.

$i i .$ $\text{Moles of lithium}$ $=$ (1*gxxN_A)/(6.94*g*mol^-1)=??*mol.

$i i i .$ $\text{Moles of sodium}$ $=$ (1*gxxN_A)/(22.99*g*mol^-1)=??*mol.

$i v .$ $\text{Moles of chlorine gas}$ $=$ (1*gxxN_A)/(70.90*g*mol^-1)=??*mol.

Note that with regard to $i v$, I am perfectly justified in the assumption that we deal with DIATOMIC chlorine, i.e. $C {l}_{2}$; and there is thus a 2:1 ratio between the number of chlorine ATOMS and chlorine molecules...In fact all of the elemental GASES (save the Noble Gases) are DIATOMIC.

Anyway, given the denominator in each quotient, clearly lithium metal, as the lightest element of those listed, contains the greatest number of atoms. This is a good question, and it would pay you to go thru the individual calculations and report the numbers of atoms of stuff that you have in each scenario.