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?

1 Answer
Nov 8, 2016

Answer:

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.