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Topic: balancing an equation  (Read 26377 times)

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sundrops

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balancing an equation
« on: October 03, 2005, 10:10:33 PM »
Okay, I know balancing equations is the bacics of chemistry - but I'm having a little bit of trouble.

ok so I have: methyl salicylate + sodium hydroxide -> sodium salicylate + methanol + water

C8H8O3 + 2NaOH _> NaC7H5O3 + CH3OH + H2O

thats the given equation.

my question is with the methanol - can i just write CH4O instead of CH3OH ? because it makes a bit of a difference. lol

I'll keep an update of my progress - but if someone could help me out that'd be fabulous :D


sundrops

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Re:balancing an equation
« Reply #1 on: October 03, 2005, 10:13:10 PM »
well exactly whats given is this:

methyl salicylate + 2NaOH -> sodium salicylate + CH3OH +H2O so I probably do need to keep it in the proper formula...

urgh in that case - does anyone have any tips on where to begin solving?


Offline Mitch

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Re:balancing an equation
« Reply #2 on: October 03, 2005, 10:19:08 PM »


my question is with the methanol - can i just write CH4O instead of CH3OH ? because it makes a bit of a difference. lol



It doesn't matter. They are equivalent formulas.
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Offline mike

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Re:balancing an equation
« Reply #3 on: October 03, 2005, 10:29:04 PM »
I am not sure why there is water (H2O) as a product of the reaction, I haven't seen this before usually the products are the salt of the carboxylic acid and the alcohol (saponification).
There is no science without fancy, and no art without facts.

sundrops

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Re:balancing an equation
« Reply #4 on: October 03, 2005, 10:33:51 PM »
can I just get rid of the water on the products side?

2C8H8O3 + 2NaOH -> 2NaC7H5O3 + 2CH4O

Reactants: 16 C,  18H,  8 O
Products: 16 C, 20H, 8O

sundrops

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Re:balancing an equation
« Reply #5 on: October 03, 2005, 10:35:33 PM »
oops:
products 18H

because without the water the equations are balanced, but then again - can i just remove a water?

can anyone see what I'm doing wrong?

thx mitch - i was a lil confused with the formula aspect...

Offline mike

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Re:balancing an equation
« Reply #6 on: October 03, 2005, 10:35:36 PM »
It looks better to me without the water.

You don't need the 2 in front of everything as this is the same as 1:1:1:1  ;)

I think this is right then...
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sundrops

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Re:balancing an equation
« Reply #7 on: October 03, 2005, 10:39:19 PM »
I am not sure why there is water (H2O) as a product of the reaction, I haven't seen this before usually the products are the salt of the carboxylic acid and the alcohol (saponification).

not quite sure what saponification is, but apparently I'm going the hydrolysis of a methyl salicylate.

and I'm just going by what's written in my lab manual - however I can;t for the life of me balance it!  :o

so what you're saying is that my products should be sodium salicylate and methanol only?

sundrops

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Re:balancing an equation
« Reply #8 on: October 03, 2005, 10:40:05 PM »
haha yeah - I guess I don't need the two's up front after all :)

sundrops

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Re:balancing an equation
« Reply #9 on: October 03, 2005, 10:48:45 PM »
if its a hydrolysis - then doesn;t water have to be present?

Offline mike

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Re:balancing an equation
« Reply #10 on: October 03, 2005, 10:49:08 PM »
so what you're saying is that my products should be sodium salicylate and methanol only?

yes.

Saponification is like the opposite of esterification.

It is the base hydrolysis of an ester (you can also do acid hydrolysis which is reversible).

It is also how you make soap  :)
« Last Edit: October 03, 2005, 10:49:57 PM by mike »
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Offline mike

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Re:balancing an equation
« Reply #11 on: October 03, 2005, 11:08:32 PM »
if its a hydrolysis - then doesn;t water have to be present?

hydrolysis is usually the breaking of a bond by the addition of water across the bond, I assume because hydro- is water and lyse- is to break.

You won't quite see this in your reaction because it usually involves a second step which is to add acid after the base hydrolysis to get a carboxylic acid (from the salt) and the alcohol.

So yes there is water involved but it doesn't need to be written in the equation, the reaction would be done in water.
There is no science without fancy, and no art without facts.

sundrops

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Re:balancing an equation
« Reply #12 on: October 03, 2005, 11:14:55 PM »
ok - i get what you're saying and it makes sense, however what I'm doing is a two part problem. the products from this reaction are my reactants for the next one.

and for the second equation I have 2NaC7H5O3 +H2SO4 -> 2C7H6O3 + NaSO4
so clearly I need 2 sodium salicylates in the product part of the preious section.
so do I give everything a 2:2:2:2 ratio or is there a way to incorporate that water and end up with 2 sodium salicylates.

phew. i'm not enjoying this problem.  :-\

sundrops

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Re:balancing an equation
« Reply #13 on: October 03, 2005, 11:17:46 PM »
but i do add acid in the second step H2SO4 (sulfuric acid)
so where does this silly water come into play?

Offline mike

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Re:balancing an equation
« Reply #14 on: October 03, 2005, 11:30:33 PM »
No, you don't need 2 in the previous reaction. This second equation is a new one which you can balance separately to the first.

Basically your first reaction is hydrolysing the ester (methyl salicylate) to form the sodium salt (sodium salicylate). The sodium salt of the carboxylic acid (C6H4(OH)COONa or C6H4(OH)COO-) is just protonated (H+) by the acid to make the carboxylic acid (salicylic acid) C6H4(OH)COOH.

If you look carefully at all of the reactants and products you will see that you have taken methylsalicylate and broken the ester bond to make salicylic acid and methanol. Now salicylic acid and methanol are just methyl salicilate with H2O added between them. To make that less confusing, think of the ester reaction between salicylic acid  and methanol:

C6H4(OH)COOH + CH3OH --> C6H4(OH)COOCH3 + H2O

can you see where the water comes into it?

now look at the same reaction in reverse (you are just doing it in two steps is all).

So don't worry about the water.
There is no science without fancy, and no art without facts.

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