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Can someone help me with this calculation please. I am desperate!

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  • Can someone help me with this calculation please. I am desperate!

    Q) The BP requires Phenobarbital sodium tablets to contin between 92.5 to 107.5 % of the stated amount of C12H11N2NaNO3 (Mr = 254.22).

    20 tablets (labelled 30mg phenobarbital sodium per tablet) with a mass of 2.6235g were powdered and 1.3057g of powder dissolved in sodium hydroxide. A fter acidifying with HCl the solution was extracted a number of times with ether. Evaporation of the other ether extracts to dryness gave a residue of the free acid (C12H12N2O3, Mr = 232.24) with a mass of 0.2873g. Calculate the mass of phenobarbital sodium (mg) in an average tablet. Do the tablets comply with the BP requirement?

    I actuall;y don't even know where to start :s Can someone please help! Thanks.

  • #2
    Re: Can someone help me with this calculation please. I am desperate!

    Originally posted by shenzys View Post
    Q) The BP requires Phenobarbital sodium tablets to contin between 92.5 to 107.5 % of the stated amount of C12H11N2NaNO3 (Mr = 254.22).

    20 tablets (labelled 30mg phenobarbital sodium per tablet) with a mass of 2.6235g were powdered and 1.3057g of powder dissolved in sodium hydroxide. A fter acidifying with HCl the solution was extracted a number of times with ether. Evaporation of the other ether extracts to dryness gave a residue of the free acid (C12H12N2O3, Mr = 232.24) with a mass of 0.2873g. Calculate the mass of phenobarbital sodium (mg) in an average tablet. Do the tablets comply with the BP requirement?

    I actuall;y don't even know where to start :s Can someone please help! Thanks.
    Well, I'm not sure I know where to start either, but I've got one son doing maths at cambrigde and the other training as an actuary after getting his first, so perhaps I should give it a shot! - it may be wrong though!!
    step i: convert weight of acid to equivalent weight of sodium salt

    0.2873*254.22/232.24g = 0.314491g

    Step 2: convert this weight to weight present in entire lot of 20 tablets


    0.314491*2.6235/1.3057g = 0.6318963g

    Step 3: divide by 20!

    = 0.0315948g, or 31.59mg per tablet.

    This looks to me like each tab contains 105% of the stated amount of phenobarb sodium, and is therefore compliant with the BP.


    Pleeeeeeeeeeeeese tell me its correct before one of my kids sees it!
    ....just my opinion

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    • #3
      Re: Can someone help me with this calculation please. I am desperate!

      The response of an Atomic Abspsorption Spectrophotometer is found to be linear over the range 0 - 5 mg/L when using the 214 nm line for the determination of zinc.

      A solution of zinc chloride containing 2.0 mg/L Zn produces an absorption reading of 0.471 on the instrument.
      A solution prepared by dissolving 1.898 g of acetylcysteine in 50 ml 0.1 M HCl gave an absorption reading of 0.272 on the instrument under the same conditions.

      Calculate the level (ppm) of zinc in the acetylcysteine sample giving your answer to 3 sig. fig.

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      • #4
        Re: Can someone help me with this calculation please. I am desperate!

        Originally posted by Msocpharm View Post
        The response of an Atomic Abspsorption Spectrophotometer is found to be linear over the range 0 - 5 mg/L when using the 214 nm line for the determination of zinc.

        A solution of zinc chloride containing 2.0 mg/L Zn produces an absorption reading of 0.471 on the instrument.
        A solution prepared by dissolving 1.898 g of acetylcysteine in 50 ml 0.1 M HCl gave an absorption reading of 0.272 on the instrument under the same conditions.

        Calculate the level (ppm) of zinc in the acetylcysteine sample giving your answer to 3 sig. fig.
        Start by using Beer Lamberts law toget an equation. Using the information given calculate the constants. Use this information to calculate the amount of zinc anfd then convert your answers to ppm . Just one simple method.
        Real stupidity beats artificial intelligence every time.
        (T. Pratchett)

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        • #5
          Re: Can someone help me with this calculation please. I am desperate!

          well, it would be nice to have some feedback. So, here is the outline which I think you need.

          0.471 units is equivalent to 2mg/l Zn
          0.272 2 * 0.272/0.471 mg/L Zn (got to be less, see)

          This was dissolved in 50ml, so the total amount of Zn in the sample will be the amount in mg/L divided by 1000 and multiplied by 50.

          When you have this total amount, you can express it in parts per million of the entire sample of 1.898g.

          I make the answer around 30. Do you? I've no idea what Beer Lambert's Law is, btw, so if you need that I may have done it wrongly.
          ....just my opinion

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          • #6
            Re: Can someone help me with this calculation please. I am desperate!

            Originally posted by DavidS View Post
            well, it would be nice to have some feedback. So, here is the outline which I think you need.

            0.471 units is equivalent to 2mg/l Zn
            0.272 2 * 0.272/0.471 mg/L Zn (got to be less, see)

            This was dissolved in 50ml, so the total amount of Zn in the sample will be the amount in mg/L divided by 1000 and multiplied by 50.

            When you have this total amount, you can express it in parts per million of the entire sample of 1.898g.

            I make the answer around 30. Do you? I've no idea what Beer Lambert's Law is, btw, so if you need that I may have done it wrongly.
            You need to understand Beer Lamberts law, you should have learnt it in A level chemistry, if not use google. Its been a long time so I had to too. Even though you may have come up with the right answer you need to understand why it works and what the exceptions are. I get 30.200ppm. On the back of an evelope.
            Last edited by paul2008; 9, January 2009, 05:58 PM. Reason: arithmatic error
            Real stupidity beats artificial intelligence every time.
            (T. Pratchett)

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            • #7
              Re: Can someone help me with this calculation please. I am desperate!

              Cannot remember Beer lamberts Law either, perhaps it was formulated after the 50s.
              johnep

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              • #8
                Re: Can someone help me with this calculation please. I am desperate!

                Originally posted by johnep View Post
                Cannot remember Beer lamberts Law either, perhaps it was formulated after the 50s.
                johnep
                Beer lambert law is more than 100 years old - i forget it often though even though it haunts us every year...
                http://i620.photobucket.com/albums/t...snroses2-1.jpg

                ”We are real. We are not glam sh*t or anything else. We are Guns N’ Roses.”

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                • #9
                  Re: Can someone help me with this calculation please. I am desperate!

                  Originally posted by paul2008 View Post
                  You need to understand Beer Lamberts law, you should have learnt it in A level chemistry, if not use google. Its been a long time so I had to too. Even though you may have come up with the right answer you need to understand why it works and what the exceptions are. I get 30.200ppm. On the back of an evelope.
                  Ok, so I've looked it up, and I don't think you need it: you are given that there is a linear relationship, so working out the constant is not required.

                  I didn't do it for A level: spectroscopic adsorbance wasn't practical at school level, and I don't remember doing it for the BSc either, although I had discovered table football by then.

                  And, Paul, if I can be a bit picky myself, 3 sig figs would make it 30.2ppm, although I didn't want to give, and don't remember, the exact answer as I think that an up-and-coming pharmacist needs to be able to follow the process.

                  Still, I've had enough calculations for now, and will be delighted to leave questions of this nature to my younger colleagues, with greatful thanks!
                  ....just my opinion

                  Comment


                  • #10
                    Re: Can someone help me with this calculation please. I am desperate!

                    Um err, long time since I did any calculus. Looked up in Wiki and will leave to those whose brains not befuddled by old age.
                    johnep

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                    • #11
                      Re: Can someone help me with this calculation please. I am desperate!

                      Originally posted by DavidS View Post
                      Ok, so I've looked it up, and I don't think you need it: you are given that there is a linear relationship, so working out the constant is not required.

                      I didn't do it for A level: spectroscopic adsorbance wasn't practical at school level, and I don't remember doing it for the BSc either, although I had discovered table football by then.

                      And, Paul, if I can be a bit picky myself, 3 sig figs would make it 30.2ppm, although I didn't want to give, and don't remember, the exact answer as I think that an up-and-coming pharmacist needs to be able to follow the process.

                      Still, I've had enough calculations for now, and will be delighted to leave questions of this nature to my younger colleagues, with greatful thanks!
                      Please accept my apologies, I posted before checking and then corrected imperfectly. I'm not perfect, never pretended to be, just occasionally helpful.
                      Real stupidity beats artificial intelligence every time.
                      (T. Pratchett)

                      Comment


                      • #12
                        Re: Can someone help me with this calculation please. I am desperate!

                        Originally posted by paul2008 View Post
                        Please accept my apologies
                        You are more than occasionally helpful, Paul, and I am more than occasionally snappy, and I think actually that its me who owes you an apology, actually.

                        Sorry!

                        d

                        PS I also get a lot of laughs with your Terry Pratchet quote! Particularly since I recently bought a Webbook!
                        ....just my opinion

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