# Moles question! Molecules vs atoms?

## There are $7.90 \cdot {10}^{24}$ atoms in a sample of $A l {\left(O H\right)}_{3}$. How many H atoms are there? How many moles of $A l {\left(O H\right)}_{3}$ 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?

Nov 8, 2016

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, $1 \times \text{Al}$
• three atoms of oxygen, $3 \times \text{O}$
• three atoms of hydrogen, $3 \times \text{H}$

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"))))

$= \text{ 5.622 moles H}$

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.