Moles question! Molecules vs atoms?

Well, the number of atoms of hydrogen and the number of formula units can be determined by using the fact that each formula unit of aluminium hydroxide -- you're dealing with an ionic compound, so there are no molecules of aluminium hydroxide -- contains

• one atom of aluminium, `1 xx "Al"`
• three atoms of oxygen, `3 xx "O"`
• three atoms of hydrogen, `3 xx "H"`

This basically means that for every `7` atoms present in your sample, you get `3` atoms of hydrogen. In this case, the number of atoms of hydrogen will be equal to

`7.90 * 10^(24)color(red)(cancel(color(black)("atoms"))) * "3 atoms H"/(7color(red)(cancel(color(black)("atoms")))) = color(green)(bar(ul(|color(white)(a/a)color(black)(3.39 * 10^(24)"atoms H")color(white)(a/a)|)))`

Now, in order to find the number of moles of aluminium hydroxide present in the sample, calculate the number of formula units first

`7.90 * 10^(24) color(red)(cancel(color(black)("atoms"))) * ("1 f. unit Al"("OH")_3)/(7color(red)(cancel(color(black)("atoms"))))`

`= 1.1286 * 10^(24)"f. units Al"("OH")_3`

To answer your question, you have to divide by `7` because that's how many atoms are present in `1` formula unit.

SIDE NOTE

For example, if I said that you have `48` months, how would you calculate how many years you have? You would use the fact that `1` year has `12` months, which basically means that you would divide the number of months by the number of months present in `1` year.

That's exactly what you are doing here.

END OF SIDE NOTE

Now simply use Avogadro's constant to convert the number of formula units to moles

`1.1286 * 10^(24)color(red)(cancel(color(black)("f. units Al"("OH")_3))) * ("1 mole AL"("OH")_3)/(6.022 * 10^(23)color(red)(cancel(color(black)("f. units Al"("OH")_3))))`

`=color(green)(bar(ul(|color(white)(a/a)color(black)("1.87 moles Al"("OH")_3)color(white)(a/a)|)))`

The answers are rounded to three sig figs.

Now, let's double-check the result by calculating the number of moles of hydrogen atoms present in the sample.

`overbrace(3.3857 * 10^(24))^(color(blue)("unrounded to three sig figs")) color(red)(cancel(color(black)("atoms H"))) * "1 mole H"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms H"))))`

`= " 5.622 moles H"`

Since you know from the chemical formula that every mole of aluminium hydroxide contains `3` moles of hydrogen, you can say that the sample will contain

`5.622 color(red)(cancel(color(black)("moles H"))) * ("1 mole AL"("OH")_3)/(3color(red)(cancel(color(black)("moles H"))))`

`=color(green)(bar(ul(|color(white)(a/a)color(black)("1.87 moles Al"("OH")_3)color(white)(a/a)|)))`

As expected, the result checks out.