Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to calculate moles

  1. Oct 21, 2003 #1
    Hello everyone. I have a slight problem with my chemistry and was wondering if there is someone that could help me. Chemistry is not my strong point so please go easy. Thank you.
    The problem is that I have gotten mixed up with how to calculate moles. Now, I know that I should use the formula:

    n= Mass (g)/ Mr

    The problem is that this doesn’t seem to apply everywhere and I don’t know why. If I have 1.5g of Na2CO3 in 250cm^3 of solution. Then I know that,

    n= 1.5g/ 106 = 0.014150943mol

    So, there is 0.014mol of Na2CO3 in 250cm^3 of solution.
    Now, I did a question in my textbook where this method of calculating moles did not work.

    “A solution of Na2CO3 contains 12.5g of anhydrous salt in 1000cm^3 of solution. When 25cm^3 of this solution was titrated with a solution of HCl using methyl orange indicator, 23.45cm^3 of the acid was required.”

    Naturally I would then use the formula to work out the moles…
    n= 12.5g/ 106 = 0.117924528mol
    However, the book states that this gives the concentration, and to calculate the moles I will have to multiply it by the volume (i.e. 25cm^3).

    What am I not understanding? Any help would be greatly appreciated, Thank you.:)
    Last edited by a moderator: Mar 7, 2013
  2. jcsd
  3. Oct 22, 2003 #2
    Re: Please help!

    What is the question asking for? You calculated the number of moles correctly. But if the question is asking for the Molarity of the HCl solution than the volume of the HCl solution is essential.
  4. Oct 22, 2003 #3
    Thank you for your response.

    The question is asking me to calculate the molarity of HCl. If I calculated to moles right, I should be able to then calculate the molarity.

    Na2CO3 + 2HCl ==> 2NaCl + H2O + CO3

    Ratio (1:2)
    So, 0.117924528mol * 2 = 0.235849056mol of HCl reacting with Na2CO3


    c= n/ V

    c= 0.235849056mol/ 23.45cm^3
    c= 0.0101mol cm^-3

    This doesn’t seem to be the concentraton, the answer in the book says it is 2.516mol cm^-3

    I’m completely mystified ....
  5. Oct 22, 2003 #4
    I kind of have a standard way of setting these things out. It helps me remember what comes next.

    "A solution of Na2CO3 contains 12.5g of anhydrous salt in 1000cm^3 of solution. When 25cm^3 of this solution was titrated with a solution of HCl using methyl orange indicator, 23.45cm^3 of the acid was required."

    Na2CO3 + 2HCl ==> 2NaCl + H2O + CO3

    Code (Text):

    Number of moles of Na2CO3  = mass/Mr
                               = 12.5/106
                               = 0.118 mol

    Molarity of Na2CO3         = moles/volume
                               = 0.118/1
                               = 0.118 mol dm^-3

    Moles of Na2CO3 in 25 cm^3 = concentration*volume
                               = 0.118 * 25 * 10^-3
                               = 0.00295 mol

    Number of moles of HCl     = 0.00295 * 2
                               = 0.00590 mol

    Molarity of HCl            = moles/volume
                               = 0.00590/(23.45*10^-3)
                               = 0.251 mol dm^-3

  6. Oct 23, 2003 #5
    Thank you very much, that certainly made it much clearer. :smile:
  7. Oct 23, 2003 #6
    Happy to help. :smile:
  8. Oct 23, 2003 #7
    So methyl orange will indicate in the neutral solution but not the sodium bicarbonate solution?
    Last edited: Oct 23, 2003
  9. Oct 24, 2003 #8
    The solution starts off neutral, and the indicator will colour the solution yellow. As the acid is added, the pH lowers and when it reaches pH=4.4 the methyl orange will start to colour the solution orange, as the pH lowers the colour becomes more and more orange until it gets to pH=3.1 when the soloution will become red.
    This means that you should stop the titration at the first hint of orange in the solution. The chances are that it will go straight from yellow to red on one drop, since the acid is a strong acid.

    So to answer you question, it depends what you mean by "indicate".
  10. Oct 24, 2003 #9
    A solution of sodium carbonate is neutral? I didn't know that.
  11. Oct 24, 2003 #10
    There are two ions that affect the pH, they are H+ ions (cause acidity) and OH- ions (cause alkalinity). Since sodium carbonate contains neither of these, the pH of the solution depends on the H+ ions provided by the water. Since water is neutral, the solution will be neutral.

    Even though the solution is neutral, the pH will be slightly higher than 7 (at s.t.p.). This is because the Na2CO3 (s) that is added will actually increase the volume of the solution just very slightly, which causes a drop in the concentration of H+ ions, and so, although the pH will be above 7, it will probably be in the be of the order 7.0001.

    So for all practicality, the solution will have a pH of 7 and overall it will be neutral, unless H+ or OH- ions are added (or removed).
  12. Oct 24, 2003 #11
    So then in order to make a solution basic you need to add something with a hydroxide, like NaOH of K(OH)2? I didn't know that.
  13. Oct 24, 2003 #12
    To make solution less acidic or more alkaline then, yes, you could add something like those two alkalis.
    There are also things called buffer solutions, these resist changes to their pH, so although when you add acid or alkali to them their pH does change, it only changes a little. These may be a little complicated to explain, but if you want I'll try.
    To put it simply, they contain both the un-dissociated acid/alkali molecules and the acidic/alkaline salt ions in roughly equal proportions. They can only occur as a result of weak acids and bases because for strong acids and bases, all of the molecules dissociate.
  14. Oct 24, 2003 #13
    Let's not worry about buffers for now. What I want to know is, if you add something with hydroxide to water (say Ca(OH)2) it will become basic, but if it does not have hydroxide ions, such as KCO3 or butyl lithium, then it will remain neutral?
  15. Oct 24, 2003 #14
    Yep, you've got it twigged.
  16. Oct 24, 2003 #15
    That's interesting. Because I thought butyl lithum was highly basic and would rip the proton off of water if it looked at it funny.
  17. Oct 24, 2003 #16
    Lol, very funny. OK, you laid a very good trap and I walked straight into it. Since you didn't want to know about buffer solutions, I assume that you also didn't want to know about Lowry-Bronsted acids and bases.
    The definition of an LB acid is a proton donar, or a lone pair acceptor such as an H+ ion.
    The definition of an LB alkali is a proton acceptor, or a lone pair donor such as an OH- ion.

    However not all LB acids and alkalis affect the pH of a solution as some do not contain any H+ or OH- ions.

    It seems that you've had a slightly higher chemistry education than you let on.
  18. Oct 24, 2003 #17
    So that butyllithium or Na2CO3 will not effect the pH of water?
  19. Oct 24, 2003 #18
    Not that I know of, I just know that it isn't just H+ and OH- ions that cause acidity and alkalinity.
    I don't think that Na2CO3 will affect the pH of the solution. I suppose that it could cause the water molecules to dissociate and that would affect the pH, (even though the solution would be neutral overall while at the same time being an acid or an alkali). But we haven't done how LB acids and bases affect pH.
  20. Oct 24, 2003 #19
    HCl is a Bronsted-Lowry acid. NaOH is a BL base. Butyllithium is a BL, it accepts a proton from H2O. Now, if you accept a proton from H2O, what have you got?
  21. Oct 24, 2003 #20
    Well water minus a proton leaves OH- ions. So I suppose the pH would change:

    pH = 14 + log10[OH-]

    Although I must say, I'm not familiar wuth butyl lithum, could you provide a structural formula for it please.

    I'd never really connected those two dots about water losing a proton and the solution becomming alkaline before just now.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook