Question

Moles question! Molecules vs atoms?

There are `7.90 * 10^(24)` atoms in a sample of `Al(OH)_3`. How many H atoms are there? How many moles of `Al(OH)_3` are there?

When doing the problem, why do you have to divide the number of atoms by 7 and use the number of molecules to calculate the number of moles?

Answer 1

Here's what I got.

Explanation:

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.