# How many atoms of aluminum are there in a piece of foil that has a volume of "2.00 cm"^3? The density of aluminum is "2.702 g/cm"^3

Aug 16, 2017

$1.21 \cdot {10}^{23}$

#### Explanation:

The first thing that you need to do here is to figure out the mass of the sample.

To do that, you can use its volume and the fact that aluminium is said to have a density of ${\text{2.702 g cm}}^{- 3}$, which implies that every ${\text{1 cm}}^{3}$ of aluminium has a mass of $\text{2.702 g}$.

In your case, the mass of the sample will be equal to

2.00 color(red)(cancel(color(black)("cm"^3))) * overbrace("2.702 g"/(1color(red)(cancel(color(black)("cm"^3)))))^(color(blue)("the density of Al")) = "5.404 g"

Next, you must convert the mass of the sample to moles by using the molar mass of aluminium, which is equal to ${\text{26.982 g mol}}^{- 1}$.

5.404 color(red)(cancel(color(black)("g"))) * overbrace("1 mole Al"/(26.982 color(red)(cancel(color(black)("g")))))^(color(blue)("the molar mass of Al")) = "0.2003 moles Al"

Finally, to convert the number of moles to atoms, use the fact that $1$ mole of aluminium must contain $6.022 \cdot {10}^{23}$ atoms of aluminium $\to$ this is known as Avogadro's constant.

You should end up with

$0.2003 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Al"))) * overbrace((6.022 * 10^(23)color(white)(.)"atoms Al")/(1color(red)(cancel(color(black)("mole Al")))))^(color(blue)("Avogadro's constant")) = color(darkgreen)(ul(color(black)(1.21 * 10^(23)color(white)(.)"atoms Al}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the volume of the foil.