.7 x 10^-3Find Molar Mass of Weak Acid

[EvilBunny]

(purpose is to find molar mass)

At the start we dissolved 0.945 grams of unknown weak acid into water and made it up to
100 ml

Then we used 25 ml of that solution and titrated it until the end point.
Now , this is a monoprotic acid so it's going to take 1 mol of acid to neutralise 1 mol of base

22.1 ml of base was used
the conc of the base is 0.20 mol/l

So with this I can find the moles of acid and I got 4.42 x 10^-3

This is the part am unsure of , am I suppose to multiply that number of moles by 4 ?
because that was the number of moles there was in that 25 ml sample and we started it with 100 ml

If I do multiply by four I get 52

 

[Borek]

this is a monoprotic acid

I assume that information was given beforehand?

am I suppose to multiply that number of moles by 4 ?
because that was the number of moles there was in that 25 ml sample and we started it with 100 ml

Correct thinking.

If I do multiply by four I get 52

Strange, that's not the number I got, although my is very close. Perhaps you have rounded down some intermediate result?

Strange thing is I have no idea what acid could it possibly be, solid and with so low molar mass.

 

[GCT]

(purpose is to find molar mass)

At the start we dissolved 0.945 grams of unknown weak acid into water and made it up to
100 ml

Then we used 25 ml of that solution and titrated it until the end point.
Now , this is a monoprotic acid so it's going to take 1 mol of acid to neutralise 1 mol of base

22.1 ml of base was used
the conc of the base is 0.20 mol/l

So with this I can find the moles of acid and I got 4.42 x 10^-3

This is the part am unsure of , am I suppose to multiply that number of moles by 4 ?
because that was the number of moles there was in that 25 ml sample and we started it with 100 ml

If I do multiply by four I get 52

Exactly how are you getting 52 , are you dividing the original grams by the mole result? If so you should be getting a much higher number.

EDIT- Never mind just realized that you sampled the original solution.

 

[EvilBunny]

Okay thanks for your replies the teacher said my answer is reasonable. I had 52 because of signaficative figures and a bunch of roundings.

As for the low molar mass the teacher said its an acid you wouldn't know . So something obscure and uncommon or something of the sort

 

[Borek]

As for the low molar mass the teacher said its an acid you wouldn't know . So something obscure and uncommon or something of the sort

My bet is that in reality it was not monoprotic, so the real molar mass was something like 104. "Acid you wouldn't know" may work for beginning chemistry students, but some of us have seen many acids in their lifes

 

[GCT]

Okay thanks for your replies the teacher said my answer is reasonable. I had 52 because of signaficative figures and a bunch of roundings.

As for the low molar mass the teacher said its an acid you wouldn't know . So something obscure and uncommon or something of the sort

My guess is that it could be some kind of an amine - ammonium chloride - except that the molecular weight would need to be referenced to the original compound that you weighed out , the acid itself is the ammonium. My guess is that this compound is relatively cheap making it a good candidate. It is also solid at room temperature.

It is a wild guess ... since this lab is for high school right?

 

[Borek]

Good idea, NH₄Cl fits molar mass. A little bit twisted