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    If 0.32g was taken from an equimolar mixture of 1.8g aspirin and 1.38g of salicylic acid. I know the amount of moles in 0.32g should be the same, but how many are there?

    maths isn't my thing, sorry

    pleease help!
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    I think we need the molecular formula for aspirin and salicylic acid, no?
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    well 1 mole of aspirin has a mass of 180.2g
    1 mole of saclicylic acid has a mass of 138.12g

    that might help
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    I don't know if I'm approaching this, but I'll give it a shot.

    Total weight of mixture = 1.8 + 1.38 = 3.18g.

    % of aspirin = 1.8/3.18 * 100 = 56.25%

    56.25% of 0.32g = 0.18g of aspirin

    => No. of moles of aspirin = no. of moles of salclicylic acid = 0.18/180 ~= 0.001mol.
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    thanks, that makes sense
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    (Original post by Saffie)
    thanks, that makes sense
    Cheers. :gthumb: :beer:
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    (Original post by Knogle)
    I don't know if I'm approaching this, but I'll give it a shot.

    Total weight of mixture = 1.8 + 1.38 = 3.18g.

    % of aspirin = 1.8/3.18 * 100 = 56.25%

    56.25% of 0.32g = 0.18g of aspirin

    => No. of moles of aspirin = no. of moles of salclicylic acid = 0.18/180 ~= 0.001mol.
    Does this mean that the mixture has a molarity of 0.002 overall? :cool:

    thanks muchly btw, will rep you when i next can.
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    (Original post by Saffie)
    Does this mean that the mixture has a molarity of 0.002 overall? :cool:

    thanks muchly btw, will rep you when i next can.
    Yes (NB you can check that is the right answer ... which it is ... but doing the same calculation on the other component in the mixture)
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    ok cheers :top:
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    (Original post by Saffie)
    Does this mean that the mixture has a molarity of 0.002 overall? :cool:

    thanks muchly btw, will rep you when i next can.
    I don't think it'd be accurate to say that it has a molarity of 0.002.. because that'll imply that it's 0.002 moles of a SINGULAR substance. But you have the right idea.

    And thanks.
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    urm, if I need the Mr of the equi-molar mixture, is it just the average of the Mr of its components?

    helpp
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    (Original post by Saffie)
    urm, if I need the Mr of the equi-molar mixture, is it just the average of the Mr of its components?

    helpp
    You probably want the EMM (effective molar mass I think)

    EMM = the mass of mixture in mg that produces 1mmol of H+ ions upon titration.

    So work out the mass that reacts with however many moles of NaOH and then convert to the mass required to react with 1mmol of NaOH
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    (Original post by oxymoron)
    You probably want the EMM (effective molar mass I think)

    EMM = the mass of mixture in mg that produces 1mmol of H+ ions upon titration.

    So work out the mass that reacts with however many moles of NaOH and then convert to the mass required to react with 1mmol of NaOH
    Somehow I doubt that's what Saffie was referring to.

    The Mr of the mixture will just be 0.32g, no?
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    (Original post by Knogle)
    Somehow I doubt that's what Saffie was referring to.

    The Mr of the mixture will just be 0.32g, no?
    The molecular mass won't be 0.32g because that's the MASS of the mixture.

    If it is an equimolar mixture then the EMM will be halfway between the two molecular masses of the two acids (as they are both monoprotic). But if you are doing something on the purity of aspirin as your coursework (as Fleff is!) then as the EMM gets closer to the Mr of aspirin, the purity is increasing.
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    yeah i think i want the effective molar mass. But the last bit you said- i don't understand :s: , I might get Fleff to spell it out in laymans terms later.

    And yep, we're doing very similar pieces of coursework. Except I I'm using substances of known purity then testing different methods to see how accurate they are. I know she made her own aspirin, I didn't... :dontknow:
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    (Original post by oxymoron)
    The molecular mass won't be 0.32g because that's the MASS of the mixture.

    If it is an equimolar mixture then the EMM will be halfway between the two molecular masses of the two acids (as they are both monoprotic). But if you are doing something on the purity of aspirin as your coursework (as Fleff is!) then as the EMM gets closer to the Mr of aspirin, the purity is increasing.
    Mmm okay, thanks for the information. Never learnt what EMM is, despite having completed my 'A' levels, heh.
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    (Original post by Saffie)
    yeah i think i want the effective molar mass. But the last bit you said- i don't understand :s: , I might get Fleff to spell it out in laymans terms later.

    And yep, we're doing very similar pieces of coursework. Except I I'm using substances of known purity then testing different methods to see how accurate they are. I know she made her own aspirin, I didn't... :dontknow:
    Yeah she did.

    Lol, I didn't know what EMM was until last night when she gave me that definition then I worked out what to do.

    Let me try and explain:

    First, the definition: EMM of a mixture of two acids is the mass of the mixture in mg which produces 1mmole of H+ ions in the titration.

    Let's first of all assume that you have a pure sample of aspirin (molecular mass = 180).
    Well if you dissolve 0.18g of it in 25cm3 of water and then titrate that against alkali, you should find that it reacts with 0.001mol of alkali (because you have dissolved 0.001mol of the aspirin so it will react with 0.001mol of NaOH).
    This tells you that for this sample, to produce 1mmol (0.001mol) of H+ [and therefore react with 0.001mol of NaOH), you needed to use 180mg of your sample: so the EMM of the sample is 180

    This is a trivial example because the sample is pure so of course the EMM is equal to the molecular mass - but it shows you how it works.

    Say now that you have an equimolar mixture:
    This time if you react with 0.001mol of NaOH, then that is 0.0005mol of asprin and 0.0005mol of salicylic acid, so your EMM will work out as the mass of sample containing 0.0005mol of each of the acids - which should be 0.16mg in your case below - so EMM is 160
    note that to work this out, you take the mass of the mixture in your aliquot used in the titration (in mg) and divide by the NaOH used in the titration (in mmoles).

    Note also that for an equimolar mixture, the EMM is half way between the two molecular masses. The EMM can show you the purity by how close it is to either molecular mass - it should always be between the two molecular masses unless you have other impurities present.

    Does that make sense?

    PS to Knogle, I had never heard of EMM either and I'm in my 2nd year of university having done A level chem!
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    Now I feel better. How'd you find out about it, then? She contacted you privately or something? :p:
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    (Original post by Knogle)
    Now I feel better. How'd you find out about it, then? She contacted you privately or something? :p:
    Yeah I know Fleff so I PM'd her and asked her what her definition was - then I just explained on MSN as it's easier!
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    ah, that makes things so much clearer now.

    I don't have to touch Chemistry for the rest of my life anyway.. I think.
 
 
 
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