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Topic: General w/w w/v percentage calculation question  (Read 17184 times)

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Offline colormelovibond

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General w/w w/v percentage calculation question
« on: December 17, 2006, 03:40:31 AM »
Okay,  I'm trying to figure out how to make 10% sodium methoxide.  I know 10g solute in 100g solution.  However, it's the form of w/v which I despise.  Grr.  And I'm probably making this a lot harder than it really is.  But anyhow, here's the gest of it.  I want a gallon of methanol as the solution, so how much NaOH do I need to add?  0.1 gallons right?  Now to convert this to metric (as is the standard).  There are roughly 3,785 mL in a gallon and the density of MeOH is 0.7918g/mL, so in terms of weight I have ~2997g per gallon.  How much of this needs to be NaOH to make 10% sodium methoxide, provided that sodium methoxide is a reaction between NaOH and MeOH in a 1:1 ratio---forming water and sodium methoxide.  10% of 2997g is 299.7g, however, I can't help but think that is a lot of NaOH for a 10% solution of sodium methoxide.  If the sodium is reacting with the methanol (with methanol left in excess), shouldn't the NaOH be reduced by half?  Otherwise the 10% NaOH that was added is going to react with 10% of the methanol and form I guess 10% sodium methoxide.  Hmmm.  But what about the water?  Say I take it out, how much NaOH do I need to add then?  Oh I have got to be making this harder than it really is.  But it's late and doing weight/volume percentage equations should be tackled after dusk.

Any help is greatly appreciated.  Thank you!

Offline Borek

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Re: General w/w w/v percentage calculation question
« Reply #1 on: December 17, 2006, 05:54:06 AM »
% w/v concentrations are faulty by definition.

http://www.chembuddy.com/?left=concentration&right=mass-volume-percentage

http://www.chembuddy.com/?left=concentration&right=concentration-follies

Are you sure it has to be w/v? The way you get 300g looks more or less OK, assuming your procedure is right. Are you sure you have to dissolve NaOH? That won't give methoxide solution.
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Offline colormelovibond

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Re: General w/w w/v percentage calculation question
« Reply #2 on: December 17, 2006, 03:49:17 PM »
Well, it doesn't have to w/v, but sodium hydroxide is in solid form while methanol is in the liquid.  I was using the density of methanol to convert it into a w/w %.  If I dissolve NaOH in methanol what's it going to give me if it's not sodium methoxide?  NaOH + CH3OH -> CH3ONa + H2O

Thank you.

Offline Borek

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Re: General w/w w/v percentage calculation question
« Reply #3 on: December 17, 2006, 04:10:05 PM »
Well, it doesn't have to w/v, but sodium hydroxide is in solid form while methanol is in the liquid.

So you better use w/w from the beginning to the end.

Quote
If I dissolve NaOH in methanol what's it going to give me if it's not sodium methoxide?  NaOH + CH3OH -> CH3ONa + H2O

Equilibrium for this reaction is far on the left. So far, that for practical purposes you may probably safely assume there is no methoxide in the CH3OH/NaOH mixture.
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Offline colormelovibond

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Re: General w/w w/v percentage calculation question
« Reply #4 on: December 18, 2006, 01:04:36 AM »
Yes, I'm aware that the equilibrium is on the left, that's why there is so much excess methanol...to drive the reaction to the right.  The methanol is acting as the solution as well as participating in the reaction.  Just saturating NaOH to make it go to sodium methoxide.  Did you mean to use w/w from beginning to end?  Or w/v, which is the initial form?  I know conversion from one unit to another leaves the door open to making additional mistakes.  And I've thought about this.  If NaOH is reacting with CH3OH in a 1:1 reaction (within of course excess methanol), then I have to consider the molarity.  One mole of NaOH reacts with one mole of CH3OH, but not necessarily 1mL reacts with 1mL, 40 to 32 respectively.  It will still come out to ~300g, but less methanol will be reacting than is in the solution.  I just like to type this out so I can make some sense of it.  I appreciate all of your help.  Thank you.

Offline Borek

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Re: General w/w w/v percentage calculation question
« Reply #5 on: December 18, 2006, 03:52:24 AM »
Yes, I'm aware that the equilibrium is on the left, that's why there is so much excess methanol...to drive the reaction to the right.

It won't help without removing water. Even traces of water will decompose methoxide back to NaOH and methanol  Standard way of preparation requires metallic sodium. Note: check MSDS before attempting to really prepare methoxide. It is a pretty nasty stuff.

As for the w/w w/v thing - if you are told 10%, probably anything between 9-11% will do (most likely even 8-12 will do). Thus you don't need to high accuracy when preparing solution. Still, go for w/w and calculate amount of methanol in terms of mass used, convert it to volume just to measure volume instead of weighting it, as it is easier.
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