What is the difference between a stoichiometry problem that has moles or one that has molecules?

I know how to solve stoichiometry problems, but I've always been confused about the difference between a mole and molecule in the problem. Maybe I'm just not doing the problem correctly since my answer doesn't match the ones provided in the key bank, but I'd like for someone to explain this before my semester exam. Thank you!

1 Answer
Dec 6, 2017

Answer:

Here's how you can think about this.

Explanation:

For starters, you know that you can think of a balanced chemical equation as using both moles and molecules.

For example, let's say that you have #2# molecules of hydrogen gas and #1# molecule of oxygen gas in the right conditions to produce a reaction.

#2"H"_ (2(g)) + "O"_ (2(g)) -> 2"H"_ 2"O"_ ((l))#

So when the reaction takes place, #2# molecules of hydrogen gas react with #1# molecule of oxygen gas to produce #2# molecules of water.

This is true because a balanced chemical equation must make sense at a molecular level.

Now, you always need #6.022 * 10^(23)# particles of a substance in order to have #1# mole of that substance, as given by Avogadro's constant.

So if we use #N_"A"# to designate Avogadro's constant, or #6.022 * 10^(23)# particles per mole, you can say that

  • #"1 mole H"_ 2 = N_ "A" color(white)(.)"molecules H"_2#
  • #"1 mole O"_ 2 = N_ "A" color(white)(.)"molecules O"_2#
  • #"1 mole H"_ 2"O" = N_ "A" color(white)(.)"molecules H"_2"O"#

This means that you can express the molecules that take part in the reaction in moles by going

#2 color(red)(cancel(color(black)("molecules H"_ 2))) * "1 mole H"_ 2/(N_ "A"color(red)(cancel(color(black)("molecules H"_ 2)))) = (2/N_ "A")color(white)(.)"moles H"_2#

#1 color(red)(cancel(color(black)("molecule O"_ 2))) * "1 mole O"_ 2/(N_ "A"color(red)(cancel(color(black)("molecules O"_ 2)))) = (1/N_"A") color(white)(.)"moles O"_2#

#2 color(red)(cancel(color(black)("molecules H"_ 2"O"))) * ("1 mole H"_ 2"O")/(N_ "A" color(red)(cancel(color(black)("molecules H"_ 2"O")))) = (2/N_"A") color(white)(.)"moles H"_ 2"O"#

If you rewrite the balanced chemical equation using moles instead of molecules, you will end up with

#(2/N_"A") "H"_ (2(g)) + (1/N_ "A") "O"_ (2(g)) -> (2/N_ "A") "H"_ 2"O"_ ((l))#

But since #N_ "A"# is present in the denominator on all species that take part in the reaction, you can simplify it

#(2/color(red)(cancel(color(black)(N_"A")))) "H"_ (2(g)) + (1/color(red)(cancel(color(black)(N_"A")))) "O"_ (2(g)) -> (2/color(red)(cancel(color(black)(N_"A")))) "H"_ 2"O"_ ((l))#

and get

#2"H"_ (2(g)) + "O"_ (2(g)) -> 2"H"_ 2"O"_ ((l))#

As you can see, the balanced chemical equation can be understood to refer to both molecules and moles, i.e. the mole ratio is equivalent to the molecule ratio.

So, for example, let's say that you start with #10# moles of hydrogen and excess oxygen gas and you want to find out how many molecules of water are produced.

The balanced chemical equation tells you that you get #2# moles of water for every #2# moles of hydrogen gas, so you will get #10# moles of water.

Since #1# mole of water contains #N_"A"# molecules of water, you can say that you will get #10 xx N_ "A"# molecules of water.

Similarly, if you start with #10# molecules of hydrogen gas and excess oxygen, you will get #10# molecules of water, which is equivalent to #10/N_"A"# moles of water.

To sum this up, you have

#"moles of reactant " stackrel(color(white)(acolor(blue)("use the mole ratio")color(white)(aaa)))(->) " moles of product " stackrel(color(white)(acolor(blue)("multiply by N"_"A")color(white)(aaa)))(->) "molecules of product"#

and

#"molecules of reactant " stackrel(color(white)(acolor(blue)("use the molecule ratio")color(white)(aaa)))(->) " molecules of product " stackrel(color(white)(acolor(blue)("divide by N"_"A")color(white)(aaa)))(->) "moles of product"#