Answer to Question #132763 in General Chemistry for grace

Question

if 4.695 g of pure element contains 4.449x10^22 atoms, what is the element?

Answer 1

The molecular mass of the element is unknown hence we let it be $(xg/mol)$

The number of moles of the element $x$ is also unknown hence $(ymol)$

Converting the elements from Mass$\to$ Moles$\to$ Atoms gave us $4.449\times 10^{22}$ atoms of $x$

In finding the molecular weight of the unknown element, we convert from

Atoms$\to$ Moles$\to$ Mass

Atoms of element $x$ $=4.449\times10^{22}$

Moles of the element $x$ (using the Avogadro Constant)

$6.022\times 10^{23}x\over 1mol$ $\times ymol x$ $=4.449\times 10^22 atoms x$

$y=0.0739mol x$

The number of moles of this element is thus $0.0739 mol es$

Converting Moles $\to$ Mass

$4.695gx\times$ $1mol x\over xg$ $=0.0739mol x$

$x=63.53g$

The molecular weight of the element $=63.53g$

From the periodic table the element is Copper(Cu)