# What is the mass of aluminum that would have the same number of atoms as 6.35 g of cadmium?

Mar 14, 2016

Approx. $1.5$ $g$. We work out the molar quantities, in that we know the mass of Avogadro's number of metal atoms. (How do we know this?)

#### Explanation:

We want the same number of each metal atom; therefore, we require an equality in the number of moles of cadmium and aluminum.

$\text{Moles of cadmium}$ $=$ $\frac{6.35 \cdot g}{112.41 \cdot g \cdot m o {l}^{-} 1}$ $=$ ??mol

$\text{Mass of aluminum}$ $=$ $\frac{6.35 \cdot g}{112.41 \cdot g \cdot m o {l}^{-} 1}$ $\times$ $26.98 \cdot g \cdot m o {l}^{-} 1$ $=$ ??g

In each case I have used the concept of the mole. Avogadro's number (1 mole) of each metal atom has a specific mass; and these masses are tabulated on the Periodic Table.