Question

Question #06e09

Answer 1

`"3.80 g"`

Explanation:

The idea here is that you need to work backwards from the number of atoms of nitrogen to find the number of moles of nitrogen, the number of moles of aluminium nitrate, and finally the mass of aluminium nitrate.

So, the first thing to do here is to convert the number of atoms of nitrogen present in your sample to moles of nitrogen.

As you know, the link between number of atoms and number of moles is made by Avogadro's constant, which basically acts as the definition of a mole.

`color(purple)(bar(ul(|color(white)(a/a)color(black)("1 mole" = 6.022 * 10^(23)"atoms")color(white)(a/a)|)))`

You can use Avogadro's constant as a conversion factor to get you from atoms of nitrogen to moles of nitrogen

`3.22 * 10^(22)color(red)(cancel(color(black)("atoms N"))) * "1 mole N"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms N"))))`

`= " 0.05347 moles N"`

Now, examine the chemical formula of aluminium nitrate, `"Al"("NO"_3)_3`. Notice that one formula unit of this compound contains

• one aluminium cation, `1 xx "Al"^(3+)`
• three nitrate anions, `3 xx "NO"_3^(-)`

In turn, each nitrate anion contains

• one nitrogen atom, `1 xx "N"`
• three oxygen atoms, `3 xx "O"`

This means that for every mole of aluminium nitrate you get `3` moles of nitrogen atoms, one for each nitrate anion present.

In your case, the sample will contain

`0.05347 color(red)(cancel(color(black)("moles N"))) * ("1 mole Al"("NO"_3)_3)/(3color(red)(cancel(color(black)("moles N"))))`

`= "0.01782 moles Al"("NO"_3)_3`

Now all you have to do is use the molar mass of aluminium nitrate, which is listed as `"213.0 g mol"^(-1)`, to find the mass of the sample

`0.01782 color(red)(cancel(color(black)("moles Al"("NO"_3)_3))) * "213.0 g"/(1color(red)(cancel(color(black)("mole Al"("NO"_3)_3))))`

`= color(green)(bar(ul(|color(white)(a/a)color(black)("3.80 g")color(white)(a/a)|)))`

The answer is rounded to three sig figs.