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Calculate moles

Hi everyone,

We have just done a tritation calculation for an equilibrium reaction. This involved an acid catalyst, mixed with ethanol and ethanoic acid.

I know the number of moles of the catalyst, and the number of moles of NaOH which reacted with the catalyst and any remaining ethanoic acid, but how I calculate tye number of moles of ethanoic acid which reacted with the alcohol?

Mol. Of HCl = 0.015
Mol. Of NaOH = 0.003725

(edited 8 years ago)

Reply 1

thefatone

Original post by Electrogeek

Hi everyone,

We have just done a tritation calculation for an equilibrium reaction. This involved an acid catalyst, mixed with ethanol and ethanoic acid.

I know the number of moles of the catalyst, and the number of moles of NaOH which reacted with the catalyst and any remaining ethanoic acid, but how I calculate the number of moles of ethanoic acid which reacted with the alcohol?

Mol. Of HCl = 0.015
Mol. Of NaOH = 0.0003725

write out a balanced equation which will tell you how many moles of what reacted with what

Reply 2

Electrogeek

Original post by thefatone

write out a balanced equation which will tell you how many moles of what reacted with what

Is the equation for the HCl or for the ethanoic acid?

Reply 3

username986184

Original post by Electrogeek

Is the equation for the HCl or for the ethanoic acid?

this looks like a classic acid + alcohol / ester + water equilibrium calculation

If you know how much acid catalyst you had (I assume from what you have written above that you used concentrated HCl) and you know how much NaOH you needed to titrate ALL acid present then

moles of ethanoic acid (that didn't react with the alcohol) = moles of NaOH - moles of HCl

EDIT: Your given experimental values don't make any sense. Even if ALL of the ethanoic acid had reacted you would need the same number of moles of NaOH as HCl.

(edited 8 years ago)

Reply 4

Electrogeek

Original post by TeachChemistry

I thought that was what I had to do - and thats why I became confused because it would give a negative value. Thanks. :-)

Reply 5

user382

http://a-z-animals.com/media/animals/images/original/mole2.jpg

Reply 6

username986184

Original post by Electrogeek

I thought that was what I had to do - and thats why I became confused because it would give a negative value. Thanks. :-)

What was the concentration and titre for your NaOH solution?

Reply 7

Electrogeek

Original post by TeachChemistry

What was the concentration and titre for your NaOH solution?

0.5 mol NaOH, with a titre of 7.45 cm^3

Reply 8

username986184

Original post by Electrogeek

0.5 mol NaOH, with a titre of 7.45 cm^3

And looking at your figures you used 1.25 cm³ of conc. HCl?

Reply 9

Electrogeek

Original post by TeachChemistry

And looking at your figures you used 1.25 cm³ of conc. HCl?

We found the problem - the molarity of NaOH wasn't for the volume if our solution - we had to times the value by 5, and then it all works out. Turns out when our teacher produced the sheet they accidently mixed up the values with another question (which we compared our results to)