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Topic: Stoichiometry of an acid/base neutralization  (Read 3308 times)

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Offline organika

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Stoichiometry of an acid/base neutralization
« on: May 14, 2015, 02:10:50 PM »
How many milliliters of 36.2% HCL is required to neutralize 46.26g of C15H19NO3?

I know this is a weak base/strong acid reaction but I have no idea how to calculate it.

Thanks!

Offline Arkcon

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Re: Stoichiometry of an acid/base neutralization
« Reply #1 on: May 14, 2015, 03:12:33 PM »
I'd like to welcome you, organika: to the Chemical Forums.  According to the Forum Rules{click}, we want to see your attempt, so please click on that link, read the contents, and comply with our rules, so we can help you. 

You'll need a balanced chemical equation for a start, at least understanding how many acid equivalents are in your poorly formatted chemical molecule, which could be anything, even if written clearly.  So start there.  You will also have to be able to convert volumes and masses into moles.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline organika

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Re: Stoichiometry of an acid/base neutralization
« Reply #2 on: May 14, 2015, 08:51:51 PM »
This is what I have so far, I've balanced the equation but I'm not sure this is right..

C15H19NO3 + HCL = C15H19NO3.HCL + H2O
C15H19NO3 + HCl = C15H18NO2Cl + H2O

Molar weight of C15H19NO3 is 261.3163 g/mol
46.26g = 0.177 mol

0.177 * 36.46094 (molar mass HCL) = 6.45g 100% HCL / 0.178 mol

Is this correct so far? Next I would need to calculate the amount of moles of HCL in 1 litre of 36.2% and proceed from there?
« Last Edit: May 14, 2015, 09:38:19 PM by organika »

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Re: Stoichiometry of an acid/base neutralization
« Reply #3 on: May 15, 2015, 02:53:00 AM »
Neither of these reaction equations is balanced, so while you can be right about molar ratio of the neutralization reaction, reaction doesn't go as written.

Please note it is impossible to tell what is the neutralization reaction without knowing what the compound is. Chemspider reports over 3000 compound with this formula. Single nitrogen is a strong suggestion reaction goes 1:1, but it is not guaranteed (actually it is not guaranteed that the reaction goes at all).

Assuming 1:1 you are right about the mass of HCl required, and yes now you need to convert that mass to the volume of the 36.2% w/w solution. You don't need number of moles for that - you can easily calculate mass of the solution. However, you will need density of the solution to convert mass to the volume. Going through moles is a detour, and is not going to change need for the solution density.
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Offline organika

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Re: Stoichiometry of an acid/base neutralization
« Reply #4 on: May 15, 2015, 07:40:03 AM »
Please tell me how to balance the equation correctly, am I wrong about the products? I checked C15H19NO3 + HCl = C15H18NO2Cl + H2O with this calculator http://www.webqc.org/balance.php and it balanced.


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Re: Stoichiometry of an acid/base neutralization
« Reply #5 on: May 15, 2015, 09:20:29 AM »
Sorry, yes, only the first reaction is not balanced. Technically second is balanced - that is, there is the same number of atoms on both sides. But - you were told that already twice - not knowing what the compound is (and the formula you gave is not unique) you can't predict the product.
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Offline organika

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Re: Stoichiometry of an acid/base neutralization
« Reply #6 on: May 15, 2015, 10:50:25 AM »
Okay thanks, I understand. I know the molecule has secondary amine and hydroxy moieties, can I predict the product based on that? Am I right in thinking with these two functional groups an ammonium salt would be the product of the neutralization?

Also how are the ratios determined? Are they worked out experimentally through titration or can it be calculated?

Sorry for torrent of questions here  ;D


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Re: Stoichiometry of an acid/base neutralization
« Reply #7 on: May 15, 2015, 11:38:12 AM »
"Hydroxy" in the context of organic chemistry doesn't mean the same "hydroxy" means in metal hydroxides. In such case I would expect hydrochloride to be the product ("ammonium salt" is not the same, but we are talking about analogous case).

Once you know the structure ratio is typically obvious, we know which groups have acidic character and which are basic.
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