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  • Need some help on these chem calculations

    What weight of the following soluts is required to produce a 1 molar solution of each.

    CaCL2
    KCL
    MGCL2
    C12H22O11

    Ca= 40, CL= 35.5, K=39, Mg=24, C=12, O=16, H=1

    do you just use moles=mass/mr?
    3rd yr pharmacy student - bath

  • #2
    Re: Need some help on these chem calculations

    A 1 molar solution is equivalent to the molecular weight of the constituent ions.

    I.e. For CaCl2 it would be 40 + 35.5 + 35.5 = 111g

    Thus a 1 molar solution of CaCl2 will have 111g of CaCl2

    It's as simple as that.

    You could use the equation as mentioned:

    Moles = Mass/Mr
    1 = ?/111
    And ? = 111

    Comment


    • #3
      Re: Need some help on these chem calculations

      REmember material should be dried to constant weight.
      johnep

      Comment


      • #4
        Re: Need some help on these chem calculations

        Originally posted by johnep View Post
        REmember material should be dried to constant weight.
        johnep
        If you know it, you can account for the water of crystallisation.

        Apologies to Admin for advertising, but as I don't have a financial interest here goes...

        You need introduction to pharmaceutical calculations

        Amazon.co.uk: Introduction to Pharmaceutical Calculations: Books: Judith A. Rees,Ian Smith,Brian Smith

        There is also a work book.

        If you are still at Uni they should have it in the library.

        If you are a pre-reg you can borrow books by "mail order" from the RPSGB library.
        RPSGB.org: Using the library

        Make sure you send them back though - they check if you have any outstanding or overdue books when you try to register!

        Alternatively, try looking around the internet its got loads of 'science and stuff' on it <g>
        47 BC : Julius Cesar : Veni Vidi Vici : I came, I saw I conquered.
        2018 AD : Modern Man : I shopped, I clicked, I collected.
        How times change.

        If you find you have read something that has upset or offended you an anyway please unread it at once.

        Comment


        • #5
          Re: Need some help on these chem calculations

          Originally posted by Pharmanaut View Post
          Apologies to Admin for advertising, but as I don't have a financial interest here goes...
          There's no need to apologise for this type of reference (or advertising as you've put it) my friend.......

          The forums "policy" on advertising is all about common sense really. If you were blatently advertising your own website (selling drugs, recruitment, whatever else!) then that's unacceptable, and people often message me to complain before I've even seen it. Either that or one of the moderators just deletes it.

          This is a case of you pointing out a useful book to attempt to help a member, which is obviously fine. We do live in a commercial world, and often need to refer to books or companies to forward the discussion.

          I believe we all know when somebody is honestly trying to help somebody else, and when they are trying to help themselves by attempting to sneak in adverts. Once the post is removed, and they get a gentle warning e-mail, they invariably never come back.

          The book you link to looks good. I was actually thinking of starting a section for book reviews. The NPA section is not used any more, and the person who requested it has not made a post since May 2007, so the interest has obviously gone there. I was thinking of changing it to a book review section, where we could have for example a section for pharmacy books, and another for all type of books to keep it interesting. I thought the pharmacy book part would be useful to us all (especially the students) and the "any other" books might kick off some debates and keep it interesting.

          I think we'll give it a go and see if it's popular. If it turns out to be another NPA section we can always change it again.
          Lively debate is encouraged but please respect the opinions and feelings of others.
          Please help keep the forum vibrant by spreading the work to friends and colleagues via word of mouth or social media.
          Thank you for contributing to this site.

          Comment


          • #6
            Re: Need some help on these chem calculations

            Originally posted by Pharmanaut View Post
            If you know it, you can account for the water of crystallisation.

            Apologies to Admin for advertising, but as I don't have a financial interest here goes...

            You need introduction to pharmaceutical calculations

            Amazon.co.uk: Introduction to Pharmaceutical Calculations: Books: Judith A. Rees,Ian Smith,Brian Smith

            There is also a work book.

            If you are still at Uni they should have it in the library.

            If you are a pre-reg you can borrow books by "mail order" from the RPSGB library.
            RPSGB.org: Using the library

            Make sure you send them back though - they check if you have any outstanding or overdue books when you try to register!

            Alternatively, try looking around the internet its got loads of 'science and stuff' on it <g>
            There is a book here in the U.S. that back in the day helped both me and my wife tremendously. Quite a bit devoted to pharmaceutical calculations and may be worth a look as sometimes different perspectives/presentations can make things easier to learn. It's called: "Comprehensive Pharmacy Review" by Leon Shargel, Alan H. Mutnick, Paul F. Souney and Larry N. Swanson. Last edition was the 5th that we have, but may be another out now. I found it to be a good general overview of pharmacy principles and it is the most common text used to date for passing the NAPLEX exam in the U.S.

            Comment


            • #7
              Re: Need some help on these chem calculations

              ignore thanks
              Last edited by Asterix; 30, July 2008, 08:52 PM.
              3rd yr pharmacy student - bath

              Comment


              • #8
                Re: Need some help on these chem calculations

                Originally posted by giggsy View Post
                1. How much solid drug is required to produce 250ml of a solution when diluted 40 fold to produce a 1 in 9000 diluted solution?
                The NPSA should rapidly recommend a standard way of describing dilutions.
                It's been so long since I had to know exactly what it meant that I no longer know with any certainty what "diluted 40 fold " means


                2. Given 2 batches of syrup (batch x= 1.32g/ml and batch y=1.28g/ml) what volumes of each are required to produce 400ml of a syrup of 1.305g/ml.
                I said I'd try and talk through my thought process.
                (and it's been years since I did any of this so be patient)

                So....
                First thought
                An equal mixture of each would give a syrup of 1.30g/ml
                So we need a little bit more of batch x - the more concentrated.

                We need Xml of syrup x and Yml of syrup y
                X + Y = 400

                (X * 1.32 ) + (Y*1.28) = 400 * 1.305

                Then it's just simultaneous equations (see GCSE maths)
                The answer should have X slightly more than 200 and Y slightly less than 200

                Jeff

                Comment


                • #9
                  Re: Need some help on these chem calculations

                  Originally posted by giggsy View Post
                  Thank you to all who have helped me, I will look into those books. I have some other calculations which I would be grateful if someone showed me how to do. I do apologise for posting these questions but we actually get no support at uni for calculations and as I dont have a level maths I am not sure how I am going to cope on this course. Anyway here are the questions.

                  1. How much solid drug is required to produce 250ml of a solution when diluted 40 fold to produce a 1 in 9000 diluted solution?
                  2. Given 2 batches of syrup (batch x= 1.32g/ml and batch y=1.28g/ml) what volumes of each are required to produce 400ml of a syrup of 1.305g/ml.
                  1. (250 x 40) /9000 = 1.111g

                  so take 1.111g and put it in 250mls

                  to work it back: -

                  1.111/250 = 0.00444g/ml or 4.44mg/ml

                  0.00444 x 9000 = 40 = dilution factor

                  2. If we assume we want equal quantities of actual syrup from each batch, split in two initially: -

                  400 / 2 = 200mls

                  Then calculate how much syrup is needed to give 200mls at each strength: -

                  (strength required/strength we have) x final volume required = volume needed of batch

                  (1.305/1.32) x 200 = 197.73mls (less then 200 as is stronger then solution required)

                  (1.305/1.28) x 200 = 203.90mls (more then 200 as is weaker then solution required)

                  adding these two together gives us more then 400mls (1.97.73 + 203.90) = 401.63mls

                  so we have to reduce by a factor of (400/401.63) = 0.99594

                  so just multiply each volume by this factor

                  197.73 x 0.99594 = 196.92

                  203.90 x 0.99594 = 203.07

                  then add together 196.92 + 203.07 = 400 to check it equals the volume we want

                  so we can use 196.92mls of solution X and 203.07mls of solution Y = 400mls in total

                  I hope this is right - I'm tired!

                  No 2 is a bit long but that's how my brain works I'm afraid! I tried to write the steps down. In reality you can do this a lot quicker then it looks. It's also easier if you stick to your calculator as you can use the memory (and thus the same number of decimal places) for each calculation.
                  Lively debate is encouraged but please respect the opinions and feelings of others.
                  Please help keep the forum vibrant by spreading the work to friends and colleagues via word of mouth or social media.
                  Thank you for contributing to this site.

                  Comment


                  • #10
                    Re: Need some help on these chem calculations

                    Originally posted by Jeff View Post
                    The answer should have X slightly more than 200 and Y slightly less than 200Jeff
                    I make it the other way round - X is stronger then we require, so we need less. Y is weaker then we require, so we need more.

                    A simpler way to do it would be: -

                    in 200mls of a 1.305g/ml solution there is (1.305/2) x 200 = 130.5g of syrup

                    for solution x we thus need (130.5/1.32) x 2 = 197.73mls

                    for solution Y we thus need (130.5/1.28) x 2 = 203.90mls

                    These add up to 401.63 so we still need to reduce by 400/401.63 to get a final volume of 400mls

                    The same but maybe a bit shorter........
                    Last edited by admin; 21, January 2008, 11:31 PM.
                    Lively debate is encouraged but please respect the opinions and feelings of others.
                    Please help keep the forum vibrant by spreading the work to friends and colleagues via word of mouth or social media.
                    Thank you for contributing to this site.

                    Comment


                    • #11
                      Re: Need some help on these chem calculations

                      Originally posted by admin View Post
                      I make it the other way round - X is stronger then we require, so we need less. Y is weaker then we require, so we need more.

                      A simpler way to do it would be: -

                      in 200mls of a 1.305g/ml solution there is (1.305/2) x 200 = 130.5g of syrup

                      for solution x we thus need (130.5/1.32) x 2 = 197.73mls

                      for solution Y we thus need (130.5/1.28) x 2 = 203.90mls

                      These add up to 401.63 so we still need to reduce by 400/401.63 to get a final volume of 400mls

                      The same but maybe a bit shorter........
                      They keep telling us to use aligation to solve these, what is that?!
                      3rd yr pharmacy student - bath

                      Comment


                      • #12
                        Re: Need some help on these chem calculations

                        Originally posted by giggsy View Post
                        They keep telling us to use aligation to solve these, what is that?!
                        A rule relating to the solution of questions concerning the compounding or mixing of different ingredients, or ingredients of different qualities or values.
                        47 BC : Julius Cesar : Veni Vidi Vici : I came, I saw I conquered.
                        2018 AD : Modern Man : I shopped, I clicked, I collected.
                        How times change.

                        If you find you have read something that has upset or offended you an anyway please unread it at once.

                        Comment


                        • #13
                          Re: Need some help on these chem calculations

                          Originally posted by giggsy View Post
                          .

                          1. How much solid drug is required to produce 250ml of a solution when diluted 40 fold to produce a 1 in 9000 diluted solution?
                          2. Given 2 batches of syrup (batch x= 1.32g/ml and batch y=1.28g/ml) what volumes of each are required to produce 400ml of a syrup of 1.305g/ml.
                          Firstly you don't need A Level Maths to answer the questions posed. You just need to look at them and work through logically. The way I would approach these calculations is:

                          1) Work backwards, you have been given the final strength as well as information relating to the dilutions and volume.

                          So: (1/9000) x 40 will give us the original strength
                          If we multiply this by 250mL, we will get the amount required.

                          2) Use alligation or simultaneous equations as posted above.

                          As a footnote, alligation simply means working out how many parts of the said solutions will be required to make the final concentration you require.

                          Comment


                          • #14
                            Re: Need some help on these chem calculations

                            I know a few people have put how they would do these calculations, but could someone else please put the answers they get, so I can know if I've done them correctly or not.

                            Would the original poster please put the answers as well.

                            Thanks
                            Lively debate is encouraged but please respect the opinions and feelings of others.
                            Please help keep the forum vibrant by spreading the work to friends and colleagues via word of mouth or social media.
                            Thank you for contributing to this site.

                            Comment


                            • #15
                              Re: Need some help on these chem calculations

                              how would you go about converting the concentrations to moles per litre.

                              2.9g/25ml?

                              @admin: I will try to get the answers.
                              3rd yr pharmacy student - bath

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