Chemistry Help

[Lioness]

It's prac report writing time again, and once again we haven't learnt stuff we need to.

First thing: What is the molecular formula for Tartaric Acid? Google gives it as C4H6O6. However, my chemistry workbook gives it as C4H8O6. I need to know in order to work out the relative molecular mass.

Second thing: What on earth does it make when reacted with NaOH. Salt + water. Great. H2O and some complicated salt. What is the complicated salt? I tried googling it, and came up with Sodium tartrate. Is this correct, and if so, what is the formula?

I think that's all. Any help much appreciated.

[Adumbration]

The correct formula, as far as I can tell, is C4H6O6. If you draw out the formula - or look it up from say, Wikipedia, you can count the possible places for hydrogen, and it is 6.

I think Sodium Tartrate is the logical outcome of the reaction. The NaOH is a powerful base - IIRC - and so it should react completely with the acid until either one runs out. Na2C4H4O6 ought to be the correct formula.

Wikipedia is an excellent aid with these things, really. (Although, depending on what education level you are currently, it might be more beneficial to reason these things out by drawing the structure of the molecule yourself. It's the best way to learn, really.)

[Slayn82]

I second Adumbration. Also, here are a couple links:

http://en.wikipedia.org/wiki/Tartaric_acid

http://www.byui.edu/chemistry/Lab_Manuals/Chem_105/Chem_105_Exp_8_The_Titration_of_an_Acid_and_a_Base .pdf

Remember that the reaction happens at the group -OH that is part of the carboxyl group. In the wikipedia picture, its the outer -OH groups, with the exchange of the proton for a sodium cation. For a quick review:

http://en.wikipedia.org/wiki/Carboxylic_acid

[Strawberries]

The correct formula, as far as I can tell, is C4H6O6. If you draw out the formula - or look it up from say, Wikipedia, you can count the possible places for hydrogen, and it is 6.

That's correct. It's C4H6O6.



I think Sodium Tartrate is the logical outcome of the reaction. The NaOH is a powerful base - IIRC - and so it should react completely with the acid until either one runs out. Na2C4H4O6 ought to be the correct formula.

Correct once again. What happens is that you have a base (NaOH) reacting with an acid (your tartaric acid). Now, what a base usually does is to deprotonate the acid: OH- catches a proton and turns into H2O. Your tartaric acid loses two protons, and the total negative charge is balanced by acquiring the two Na+ ions. The reaction is C4H6O6 + 2NaOH -----> Na2C4H4O6 + 2 H2O

Now, what you need to understand is why the molecule loses just two H+ (it has, after all, 6 hydrogen atoms) and which ones are they. Slayn82's links can help you there. The OH- reacts only with the two protons on the -COOH (carboxylic) groups, because they are the more "acid", which is, the more likely to be lost. The explanation of this can be found reading the theory on carboxylic acids.

[Prodan]

Now, what you need to understand is why the molecule loses just two H+ (it has, after all, 6 hydrogen atoms) and which ones are they. Slayn82's links can help you there. The OH- reacts only with the two protons on the -COOH (carboxylic) groups, because they are the more "acid", which is, the more likely to be lost. The explanation of this can be found reading the theory on carboxylic acids.

Carboxylic acids are weak acids who can deprotonate more easily than alcohols or water due to resonance stabilization of the oxygen atoms, wherein the negative charge resulting from the loss of the proton is delocalized over the functional group, thus stabilizing the resulting anion.

[Lioness]

That's correct. It's C4H6O6.



Correct once again. What happens is that you have a base (NaOH) reacting with an acid (your tartaric acid). Now, what a base usually does is to deprotonate the acid: OH- catches a proton and turns into H2O. Your tartaric acid loses two protons, and the total negative charge is balanced by acquiring the two Na+ ions. The reaction is C4H6O6 + 2NaOH -----> Na2C4H4O6 + 2 H2O

Now, what you need to understand is why the molecule loses just two H+ (it has, after all, 6 hydrogen atoms) and which ones are they. Slayn82's links can help you there. The OH- reacts only with the two protons on the -COOH (carboxylic) groups, because they are the more "acid", which is, the more likely to be lost. The explanation of this can be found reading the theory on carboxylic acids.

Thankyou. That confirmed all of my suspicions, and told me what I didn't know. It's been a while since we did organic acids.

[Lioness]

Another quick question: Would a standard solution not being homogeneous be a systematic or random error?

[Don Julio Anejo]

Think about it this way: a systematic error is error due to methodology or instruments or whatever and is repeatable if you do the experiment again under the same conditions. Random error is, well, random and non-repeatable.

Also, a non-homogeneous standard solution would be random error by definition - even if you use the same solution again, the end result would be different each time as you'd get a different mixture of constituents every time.