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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: Snorlax on August 07, 2013, 08:06:23 pm

Title: Vinegar + NaOH prac Q
Post by: Snorlax on August 07, 2013, 08:06:23 pm
So i did a prac late last week and I got confused for one of the questions.

Find the mass (g) of CH3COOH that reacted in the experiment.
Mole of NaOH used was: 22.4x10-3 moles.

And to find the number of moles of CH3COOH: (1:1 ratio)
2.4x10-3 / 0.0299 Litres (of vinegar)
=0.075 mols of CH3COOH

now to find the mass of CH3COOH that reacted in the experiment;
0.075 x Mr(CH3COOH) = 4.50 grams.

But how do i use MORE than i began with? As i only used 0.0299L of it?

I think i'm missing something :S

Help please! :D

PS.
THE CALCULATIONS ARE WRONG. I DID THIS QUICK, AND THEY ARE NOT CONCISE, BUT YOU GET THE IDEA :)
Title: Re: Vinegar + NaOH prac Q
Post by: jgoudie on August 07, 2013, 08:50:13 pm
Quote from: nonsense on August 07, 2013, 08:06:23 pm
So i did a prac late last week and I got confused for one of the questions.

Find the mass (g) of CH3COOH that reacted in the experiment.
Mole of NaOH used was: 22.4x10-3 moles.

And to find the number of moles of CH3COOH: (1:1 ratio)
2.4x10-3 / 0.0299 Litres (of vinegar)
=0.075 mols of CH3COOH

now to find the mass of CH3COOH that reacted in the experiment;
0.075 x Mr(CH3COOH) = 4.50 grams.

But how do i use MORE than i began with? As i only used 0.0299L of it?

I think i'm missing something :S

Help please! :D

PS.
THE CALCULATIONS ARE WRONG. I DID THIS QUICK, AND THEY ARE NOT CONCISE, BUT YOU GET THE IDEA :)

Yeah, you're a little confused about number of mols and concentration.

This part is correct:  ratio 1:1
n(NaOH) = n(CH3COOH) = 2.9*10^-3.

but by dividing by the volume you are finding concentration, you don't need that, (even so, you are you are out by a factor of 10, 0.75M, when multiplying this by the molar mass here you are calculating the mass per litre of solution, which is 45grams.)

What you should have done is go straight from moles of CH3COOH to mass, not divide by the volume.
e.g.

m(CH3COOH) = n.Mr
                     = 2.9*10^-3 * 60
                     =1.344grams

Hope this helps.
Title: Re: Vinegar + NaOH prac Q
Post by: Snorlax on August 07, 2013, 09:02:51 pm
Where did 2.9x10^-3 come from?
Title: Re: Vinegar + NaOH prac Q
Post by: jgoudie on August 07, 2013, 09:30:28 pm
Quote from: nonsense on August 07, 2013, 09:02:51 pm
Where did 2.9x10^-3 come from?

Sorry wrote it wrong,  should have been 2.24*10^-3.  This was your number of mole of NaOH hence the number of mole of CH3COOH. fixed it up now.