Emergency Chemistry Question

[Tilda]

My sisters stuck on her A level chemistry coursework. Shes not allowed teacher help and our entire family (3x A level chemistry) are stumped.
Tris et al. can you give us a hand?

Question is;

Aim, to calculate the relative formula mass of succinic acid and its molecular formula by titration with sodium hydroxide.
In a titration of succinc acid and sodium hydroxide, what calcuation would you use to find "n" in the equation HOOC(CH[little2])[little]n COOH

Her teacher apparently says to make up values for the titration volume etc.
We know that the "n" should equal 2, but we can't work out how to prove this using titration calcuations.

Can anybody give us a shove/hand in the right direction.(or do it for us?!)

 

[Gamah]

titration tehehehhe!

*Dons dunce hat*

 

[old.Tohtori]

I'm not much of a chemist, but the words emergency and chemistry should NEVER be used in the same sentence.

'cause i got real disappointed that there was no danger of explosion or some cool stuff in this thread....

 

[Bugz]

I know titration calculations to a very simple extent but do you not need a concentration of one of the two to get anywhere?

 

[UrganNagru]

Right hope I remember the formula correctly but I assume you know how to calculate the concentration using the titration and C1V1/R1=C2V2/R2. So use made up values for that.
Then using the concentration and a made up mass value stick it into Rfm=mass/concentration (this where I'm having trouble remembering it correctly, i know its those three values pretty sure its in that order but having trouble at double checking by thinking it through, play around or if you have any books try looking it up to check). Now you hopefully have the RFM, take away the relative masses of the carboxyl groups (90 i think) then divide the remaining mass by the the mass of an Alkyl group (CH2- 14 i think) to get the number of Alkyl (CH2) groups.

I hope that its of some use, sorry I can't be certain of the RFM=mass/concentration formula but hopefully it will ring some bells, also not sure what they mean by a succint acid I'm assuming they monobasic in which case it will be a 1:1 reaction for the R values.
Any trouble drop a line, and I'll try to help as best as I can.

Edit hehe, just seen one fault, my chemistry is pretty sketchy but by having 2 carboxyl groups it will have 2 free hydrogens in which case it is a 1:2 reaction with NAOH.

 

[Tilda]

Right hope I remember the formula correctly but I assume you know how to calculate the concentration using the titration and C1V1/R1=C2V2/R2. So use made up values for that.
Then using the concentration and a made up mass value stick it into Rfm=mass/concentration (this where I'm having trouble remembering it correctly, i know its those three values pretty sure its in that order but having trouble at double checking by thinking it through, play around or if you have any books try looking it up to check). Now you hopefully have the RFM, take away the relative masses of the carboxyl groups (90 i think) then divide the remaining mass by the the mass of an Alkyl group (CH2- 14 i think) to get the number of Alkyl (CH2) groups.

I hope that its of some use, sorry I can't be certain of the RFM=mass/concentration formula but hopefully it will ring some bells, also not sure what they mean by a succint acid I'm assuming they monobasic in which case it will be a 1:1 reaction for the R values.
Any trouble drop a line, and I'll try to help as best as I can.

Edit hehe, just seen one fault, my chemistry is pretty sketchy but by having 2 carboxyl groups it will have 2 free hydrogens in which case it is a 1:2 reaction with NAOH.

Thanks, we're going to go through what you've said and her txt book/notes and will get back to you if we get stuck. Are you up for long?

 

[crispy]

Hmm im not quite sure i understand the question Tilda :>

You wanna know the weight per mole of succinic acid by titrating it?

Ah well i read your question again, more carefully

What numbers do you have? What variables do want to include in the calculation?

 

[Tilda]

Hmm im not quite sure i understand the question Tilda :>

You wanna know the weight per mole of succinic acid by titrating it?

we have to work out the molecular mass of n I think, the problem is, we have no figures so end up making them up.

 

[Tilda]

Can someone explain this bit to me?

titration and C1V1/R1=C2V2/R2.

 

[crispy]

Well to calculate n in the first formula you had, you will need to know the molar mass.

To measure the molar mass by titration you also need to know the concentration (you get this from the titration) and the mass of your succinic acid in water.

Correct so far?

 

[UrganNagru]

Hmm just reading your question again little n isn't a mass it is the number of CH2 groups i think.

C1V1/R1=C2V2/R2 is the most basic formula used for calculating cocentrations from a titration C1=concentration of acid V1=volume of acid R1= parts of acid per alkali (in this case i think its 1) C2=concentration of NaOH V2= Volume of NaOH R2= parts of alakli per acid in this case I think its 2.

I'll be checking around on off for next hour or so.

 

[tamtap]

struggling with titration, not the sharpest knife in the drawer seen as its 20% of the sylabus for A level chemo.

I would lend a hand but your a tit so i don't think i'll bother.

Tam

 

[crispy]

But its quite easy to calculate n in your formula once you know the molar mass.

Molar mass = all the different atom in your formula times their unit weight

Hopefully you got the molar mass from the titration (should be around 120)

so...

120 = HOOC-(CH[little2])[little]n-COOH

=>

120 - COOH (this is ~45) - COOH (again ~45) = (CH2)·n

=>

(120-2·45)/(14) = n <- 14 is the weight of CH2

wich gives you an n of about 2 - corresponding to the real formula of succinic acid.

Now this is the step if you got the molar mass.

If you dont know the molar mass and want to calculate this by data from a titration you also need to know the following: Concentration of NaOH, how big a volume you needed to titrate with and the mass of the succinic acid in the solution you titrate.

You want a formula for the molar weight?

 

[psyco]

isn't the molecular mass on the periodic table?

 

[crispy]

No thats the atom weight, but two sides of the same story hehe

 

[DocWolfe]

I had great fun doing titration calcs at A-Level... thank fuck I can't remember anything about them now. Alcohol made sure of that.

 

[UrganNagru]

Aye if it was was to work out the mass of the CH2 groups could just look it up on periodic table 12+1+1=14, have a feeling instead its the number of CH2 groups instead bit more to it that way. I did shit in A level chem, just as a warning, did decent enough in the exam, but got a U or an E in my coursework due to being a lazy bugger and having to invent my own experiment rather than being allowed to use a tried n tested one out of a book a basically spent 2 weeks boiling vinegar stinking the class room out before realise ethanoic acid forms an azeotrope

 

[Vladamir]

struggling with titration, not the sharpest knife in the drawer seen as its 20% of the sylabus for A level chemo.

I would lend a hand but your a tit so i don't think i'll bother.

Tam

Wow, looking back at your old posts too you really are a tard

 

[psyco]

No thats the atom weight, but two sides of the same story hehe

there's 4 or 5 values on the periodic table, 1 is the 'molecular' mass... i think

just want to make a separate point now... tilda, your family has 4 a alevels in chemistry, now why did you pay money, to forget it 2-3 years later?

 

[Tilda]

UrganNagru you are correct, its the number of CH2 groups we're trying to work out.

crispy, to get the molar mass we need data, the problem is it seems there is none! Do we just make it up?

We figured out how to do it once we got the RMM, because then its simply a matter of taking away the end COOH groups and figuring it out from there. The problem is, there is no data, we think we're supposed to make it up, which means we can't work out the RMM.

If you dont know the molar mass and want to calculate this by data from a titration you also need to know the following:Concentration of NaOH, how big a volume you needed to titrate with and the mass of the succinic acid in the solution you titrate.

So do we just make the NaOH conc, and the volumes up?


psyco, you're right, if we knew the formula it would be piss easy, the problem is we have an unknown number of CH2 groups in the middle of our acid. We know its a 1:2 tritation with NaOH it seems the rest you have to make up as you go along unless theres something huge we're missing.

 

[Tilda]

Going on the C1V1/R1 thing and making stuff up we get this:
We figured that if you have an unknown acid the things you could work out would be the conc and vol of your alkali and the vol of the acid.
So we said

Alkali Conc = 0.1mol
Alkali Vol = 0.002dm3

Acid vol = 0.0025dm3

The ratio which we have is 1:2

So we get;

C1 * 0.0025 = 0.1 * 0.002
------------ -------------
1 ``````````````````2

When you work it out;

You get 0.04mol/dm3

Then we have moles =vol * conc
and so moles = 0.0025 * 0.04

Which equals moles = 0.0001

Then we do RMM = Mass /Moles

But we don't have a mass, so the problem is, whatever mass we use changes the RMM.

So we could do a mass of 5, and get RMM = 5 / 0.0001
and get an RMM of 50000 but if we change the 5, then the RMM changes?!

 

[UrganNagru]

Aye looks like your going to have to cheat and work backwards from:
HOOC–CH2–CH2–COOH
Which gives and RFM of 118.
Invent a concentration for it say 1 mol/dm^3
Invent a concentration of NaOH, say 1.5 mol/dm^3
Pick a volume suitable for a pippette eg 25ml.
I think having 2 carboxyl group will make it dibasic so:
C1V1/R1=C2V2/R2
1*0.025/1=1.5*V2/2
So about 53.33 ml of NaOH will be used, now you can just make up a table of results which repeat 3 times, which give you a concentration for the acid of 1mol/dm^3 keep the conc of NaOH constant and the volume of acid constant just alter the volumes of NaOH either way by a couple of decimal places.

Right to get the last value you need the mass of acid used: 0.025*118=2.95g of acid in each 25ml measure of it.

Hopefully the figues from that combine with my first post should give you something that makes sense. Again if you can double check the RFM=mass/concentration is correct as thats the only point I'm not too sure of.

 

[crispy]

So do we just make the NaOH conc, and the volumes up?

Well not exactly make the volumes and masses up, rather call them variables.

To calculate the molar mass you need to know how much succinic acid you got. Call this variable m.

Now you need to know how many moles is this you got. Call this variable n.

To calculate n you need to know the volume V you titrate with, and the concentration M of the NaOH.

Now since you need twice the number of moles of NaOH to titrate succinic acid you would calculate n like this:

(1) n = ½·V·M (remember to use the correct units! V in litres and M in moles pr litre for example)

Then your molar mass is given by this equation:

(2) molar mass = m/n

which combines with (1) gives:

(3) molar mass = m/(½·V·M)

the molar mass is then used to calculate n as i did in my previous post.

edited!

should be ½ instead of 2 in my equations. Also the molar mass is the molar mass of succinic acid just to be clear about that...

 

[psyco]

if we knew the formula it would be piss easy, the problem is we have an unknown number of CH2 groups in the middle of our acid. We know its a 1:2 tritation with NaOH it seems the rest you have to make up as you go along unless theres something huge we're missing.

as i understand it each chemical has a formulae, so maybe its alot less complicated than you think

just to refresh my memory, titration is about neutralizing the solution?

and maybe its just a case of balancing it... but hey, what do i know i got a E in double science for GSCE

 

[UrganNagru]

Hmmm looking back at it cheating might be a bad idea won't show how you got the mass being as I was using the bloody answer to caculate it. Guess just have to cheat again to get that by saying you made up the sample of acid by dilluting the solid form of the acid.
so if it was 2.95g per 25 ml it will be 59g of acid with 0.5 dm^3 of water for example.

Edit: Just out of curiosity did they actually do this experiment or not, if not I'd be worried about their teacher, bit daft expecting them to do a report for coursework with out doing the experiment.

 

[crispy]

They are not interested in hard numbers, the teacher is interested in a formula to calculate the n if you were to determine this by titration. So you will end up with a formula with 3 variables to calculate n. Mass, volume and concentration which is the data you collect during a potential titration.

 

[psyco]

Edit: Just out of curiosity did they actually do this experiment or not, if not I'd be worried about their teacher, bit daft expecting them to do a report for coursework with out doing the experiment.

sister probably fucked around any moron would know that the kid cant do the work if it hasn't been taught

 

[UrganNagru]

You'd be suprised what some teachers try to pull off through blind insistance that it has to be done their way, even if it makes fook all sense, luckily mine never tried that with anything important.

 

[Tilda]

How did you get the 53.33ml ?
We plugged in the 0.025 and got 0.03recurring.

If you messed up units and didn't convert the 25cm^3 to dm^3 then you get 33.3
However, we're struggling to see where the 53 came from?!
Unless we've messed up?

She hasn't done this specific experiment it seems like a conceptual thing to make sure they acctually understand the entire titration topic.

 

[UrganNagru]

They are not interested in hard numbers, the teacher is interested in a formula to calculate the n if you were to determine this by titration. So you will end up with a formula with 3 variables to calculate n. Mass, volume and concentration which is the data you collect during a potential titration.

Fair point. Though now they have both though would stick with the C1V1/R1=C2V2/R2 rather than n=1/2VM, I know yours is the same thing just cancelled down but they like to have it in it basic form before playing round with how its arranged.